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acquires a charge of 5.0 109 C. How many electrons did the object gain? with a magnitude twice that of the negative charge on an electron. Is the electrostatic force of attraction on an electron in a helium atom equal to the force acting on the nucleus? Justify your answer. 10. Electrical forces are so strong that the combined electrostatic forces of attraction acting on all the negative electrons and positive protons in your body could crush you to a thickness thinner than a piece of paper. Why don’t you compress? e TEST To check your understanding of Coulomb’s law, follow the eTest links at www.pearsoned.ca/school/physicssource. 538 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 539 CHAPTER 10 SUMMARY Key Terms and Concepts law of conservation of charge net charge conduction induction charge migration charge shift grounding charging by induction plasma Coulomb’s law coulomb (C) electrostatics insulator conductor semiconductor superconductor Key Equation F e 1q q 2 k 2 r Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. types of materials conductors insulators Electrostatics forces between charges Coulomb’s law calculating Fe between 2 charges methods of charging objects calculating Fe between more than 2 charges in one dimension friction induction charge separation grounding Chapter 10 Physics laws can explain the behaviour of electric charges. 539 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 540 CHAPTER 10 REVIEW Knowledge 1. (10.1) What is an electrostatic charge? 2. (10.1) On what property of materials does thermal and electrical conductivity depend? 3. (10.1) Under what conditions does selenium become a good conductor or a good insulator? What are materials with this property called? 4. (10.1) What are three methods of charging objects? 5. (10.1) During the process of charging objects by friction, what determines which object becomes negatively or positively charged? 6. (10.1) How are the processes of charging objects by conduction and friction alike? How are they different? 7. (10.1) State the law of charges. 8. (10.1) Who is credited with first naming the two types of charges as negative and positive charges? 9. (10.1) State the law of conservation of charge. 10. (10.1) Selenium and germanium are both semiconductors. Explain why selenium is used in photocopiers rather than germanium. 11. (10.2) Calculate the electrostatic force acting on two charged spheres of 3.00 C and 2.50 C if they are separated by a distance of 0.200 m. Applications 12. What is the distance between two charges of 5.00 C each if the force of electrostatic repulsion acting on them is 5.00 103 N? 13. Charge A has a charge of 2.50 C and is 1.50 m to the left of charge B, which has a charge of 3.20 C. Charge B is 1.70 m to the left of a third charge C, which has a charge of 1.60 C. If all three charges are collinear, what is the net electrostatic force on each of the following? (a) charge B (b) charge C 14. Why is dust attracted to the front of a cathode- ray tube computer monitor? 15. Why is it desirable to develop materials with low electrical resistance? 16. Explain why a charged ebonite rod can be discharged by passing a flame over its surface. 17. Explain why repulsion between two objects is the only evidence that both objects are charged. 18. Why do experiments on electrostatics not work well on humid days? 19. Why does a charged pith ball initially attract a neutral pith ball, then repel it after touching it? 20. Why can you not charge a copper rod while holding one end with one hand and rubbing the other end with a piece of fur? 21. A person standing on an insulated chair touches a charged sphere. Is the person able to discharge the sphere and effectively ground it? Explain. 22. Two charged spheres, separated by a certain distance, attract each other with an electrostatic force of 10 N. What will be the new force in each of the following situations? (a) The charge on both spheres is doubled and the separation distance is halved. (b) The charge on one sphere is doubled while the charge on the other sphere is tripled and the separation distance is tripled. 23. Calculate and compare the electrical and gravitational forces acting on an electron and a proton in the hydrogen atom when the distance between their centres is 5.29 1011 m. 24. An equilateral triangle with sides of 0.200 m has three charges of 2.50 C each, situated on the vertices of the triangle. Calculate the net electrostatic force on each charge. What assumption did you have to make to complete the calculation? 540 Unit VI Forces and Fields 10-PearsonPhys30-Chap10 7/24/08 2:51 PM Page 541 25. In a Coulomb-type experiment, students were investigating the relationship between the force of electrostatic repulsion acting on two charged spheres and their separation distance. The results of their investigation yielded the results shown in the table below. Separation Distance (r) ( 102 m) Magnitude of Force of Repulsion F (N) 1.00 2.00 3.00 4.00 5.00 360.0 89.9 40.0 27.5 14.4 (a) Draw a graph of the results shown in the table. (b) From the shape of the graph, what is the relationship between the electrostatic force and the separation distance between two charges? (c) Make a new table of values to obtain data to straighten the graph. (d) Draw a graph of the data in your new table of values. (e) Determine the slope of the graph. (f) What value does the slope of this graph represent? (g) If the charges of the two spheres are the same, what is the value of the charge on each sphere? Extensions 26. Can a neutral object contain any charges? Explain. 27. Is it possible for a single negative or a single positive charge to exist in nature under normal conditions? Explain your answer. 28. Explain why it is impossible to charge a coin by rubbing it between your fingers. 29. Compare the production of lightning on Earth with the lightning between the rings of Saturn observed by the Voyager spacecraft on its mission to Saturn. 30. You are given two equally sized metal spheres on insulated stands, a piece of wire, a glass rod, and some silk. Devise and describe a method to do the following without touching the rod to the spheres: (a) give the spheres equal and opposite charges (b) give the spheres equal and like charges 31. Using the principles of electrostatics, explain the causes and effects of the following demonstrations: (a) Two strips of clear adhesive tape are stuck together and then carefully separated. When the two strips are brought close to each other, attraction occurs. (b) Two strips of clear adhesive tape are stuck onto a desktop and then carefully removed. When the two strips are held close to each other, repulsion occurs. Consolidate Your Understanding Create your own summary of the behaviour of electric charges and the laws that govern electrical interactions by answering the questions below. If you want to use a graphic organizer, refer to Student Reference 3: Using Graphic Organizers. Use the Key Terms and Concepts listed on page 539 and the Learning Outcomes on page 510. 1. Create a flowchart to describe how to calculate the electrostatic forces between two or more charged objects in one- or two-dimensional situations. 2. Write a paragraph explaining the three methods of charging objects. Share your report with a classmate. Think About It Review your answers to the Think About It questions on page 511. How would you answer each question now? e TEST To check your understanding of the behaviour of electric charges, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 10 Physics laws can explain the behaviour of electric charges. 541 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 542 C H A P T E R 11 Key Concepts In this chapter, you will learn about: vector fields electric fields electric potential difference moving charges in electric fields Learning Outcomes When you have completed this chapter, you will be able to: Knowledge define vector fields compare forces and fields compare, qualitatively, gravitational and electric potential energy define electric potential difference as a change in electric potential energy per unit of charge calculate the electric potential difference between two points in a uniform electric field explain, quantitatively, electric fields in terms of intensity (strength) and direction relative to the source of the field and to the effect on an electric charge describe, quantitatively, the motion of an electric charge in a uniform electric field explain electrical interactions, quantitatively, using the law of conservation of charge Science, Technology, and Society explain that the goal of technology is to provide solutions to practical problems explain that scientific knowledge may lead to the development of new technologies and new technologies may lead to scientific discovery 542 Unit VI Electric field theory describes electrical phenomena. Figure 11.1 The eerie glow of St. Elmo’s fire on the masts of a ship On Christopher Columbus’s second voyage to the Americas, his ships headed into stormy weather, and the tips of the ships’ masts began to glow with a ghostly bluish flame. Sailors of the time believed that this bluish glow was a good sign that the ship was under the protection of St. Elmo, the patron saint of sailors, so they called the blue “flames” St. Elmo’s fire (Figure 11.1). People throughout history have written about this strange glow. Julius Caesar reported that “in the month of February, about the second watch of the night, there suddenly arose a thick cloud followed by a shower of hail, and the same night the points of the spears belonging to the Fifth Legion seemed to take fire.” Astronauts have seen similar glows on spacecraft. What is the cause of this eerie phenomenon? Why does it most often appear during thunderstorms? You will discover the answers to these questio
ns as you continue to study the phenomena associated with electric charges. In this chapter, you will begin by learning how knowledge of the forces related to electric charges led to the idea of fields, and you will compare different types of electric fields. Then you will learn how force is used to define the strength of electric fields. Finally, you will study the motion of charges in electric fields and explain electrical interactions using the law of conservation of energy. 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 543 11-1 QuickLab 11-1 QuickLab Shielding of Cellular Phones Electronic equipment usually contains material that is used as “shielding.” In this activity, you will discover what this shielding material does. 3 Remove the aluminium foil and again dial the number of the cellular phone. 4 Repeat steps 1 to 3 using the sheets of other materials. Problem How does the shielding of electronic equipment, such as a cellular phone, affect its operation? Materials 2 cellular phones sheets (about 20 cm × 20 cm) of various materials, such as aluminium foil, plastic wrap, wax paper, paper, cloth, fur 1 short length of coaxial cable Procedure Part A 1 Wrap the sheet of aluminium foil around one of the cellular phones. 2 With the other cellular phone, dial the number of the wrapped cellular phone and record any response. Part B 5 Carefully remove the outer strip of insulated plastic around one end of the coaxial cable and examine the inner coaxial cable wires. Questions 1. What effect did wrapping a cellular phone with the various materials have on the operation of the cellular phone? 2. Cellular phones receive communication transmissions that are electrical in nature. Speculate why the transmissions are shielded by certain materials. Which materials are most effective for shielding? 3. What material forms the protective wrapping around the inner coaxial transmission wires? Explain the purpose of this protective wrapping. Think About It 1. Desktop computers or computers in vehicles have sensitive electronic components that must be protected from outside electrical interference. Identify a possible source of outside electrical interference. Describe how computer components may be protected from this interference and explain why this protection is necessary. 2. Sometimes, if your debit card fails to scan, the clerk wraps the card with a plastic bag and re-scans it. Explain why a plastic bag wrapped around a card would allow the card to scan properly. Why do clerks not wrap the card with aluminium foil for re-scanning? Discuss and compare your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 11 Electric field theory describes electrical phenomena. 543 11-PearsonPhys30-Chap11 7/24/08 3:04 PM Page 544 11.1 Forces and Fields Figure 11.2 Forces exerted by the horses attached to the chariot cause the “violent” motion of the chariot. The ancient Greek philosophers explained most types of motion as being the result of either “violent” or “natural” forces. They thought that violent forces cause motion as the result of a force exerted by one object in contact with another (Figure 11.2). They thought that natural forces cause the motion of objects toward their “natural element” (Figure 11.3). However, the Greeks found another kind of motion more difficult to explain. You will observe this kind of motion in the following Minds On activity. M I N D S O N Action at a Distance Charge a rubber rod by rubbing it with fur and slowly bring it close to the hairs on your forearm. Do not touch the hairs or your arm. Observe what happens. 1. What evidence is there that the charged rod affects the hairs on your arm without actual contact? Is the force exerted by the rod on the hairs of your arm attractive or repulsive? 2. The rubber rod seems to be able to exert a type of violent force on the hairs of your arm without visible contact. This type of force was classified as “action at a distance,” where one object could exert a force on another object without contact. To explain “action at a distance,” the Greeks proposed the effluvium theory. According to this theory, all objects are surrounded by an effluvium. This invisible substance is made up of minute string-like atoms emitted by the object that pulsate back and forth. As the effluvium extends out to other bodies, the atoms of the different objects become entangled. Their effluvium eventually draws them toward each other. The effluvium theory helped to explain what seemed to be “action at a distance.” Although the effluvium was invisible, there was still a form of contact between the objects. Figure 11.3 To return to its natural element, a rock falls with “natural” motion to Earth’s surface. info BIT A new theory in physics, called string theory, proposes that objects interact through “strings” that transmit the forces between the objects. This new theory has a striking similarity to the effluvium theory proposed 2500 years ago. 544 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 545 Fields In the 17th century, scientists, including Newton, tried to determine why one object can exert a force on another object without touching it. These scientists attempted to explain “action at a distance,” such as the curved path of a thrown ball or the effect of a charged piece of amber on the hair on a person’s arm. Finding that “natural” or “violent” forces and “effluvium” could not explain gravity or electrical forces, scientists developed the concept of fields to describe these forces. A field is defined as a region of influence surrounding an object. The concept of fields helps explain the laws of universal gravitation, which you studied in Chapter 4. Consider a space module on its way to the Moon (Figure 11.4). Nearing its lunar destination, the module begins to experience the increasing influence of the Moon. As a result, the module’s motion begins to follow a curved path, similar to the projectile motion of an object thrown horizontally through the air near Earth’s surface. As Newton’s laws state, the motion of any object can follow a curved path only when acted on by a non-zero force that has a perpendicular component. In space, this happens to the space module when it is near the Moon, so the space near the Moon must be different from the space where no large objects like the Moon are present. From this, we can infer that a field exists around a large object, such as the Moon. When other objects enter this field, they interact with the Moon. Similarly, Earth has a field. Gravitational force acts on other objects that enter this field. Recall from Chapter 4 that this field around objects is called a gravitational field. field: a region of influence surrounding an object Michael Faraday (1791–1867) developed the concept of fields to explain electrostatic phenomena. He determined that the space around a rubber rod must be different when the rubber rod is charged than when it is not. The charges on the rod create an electric field around the rod. An electrostatic force acts on another charged object when it is placed in this field. An electric field exists around every charge or charged object. It can exist in empty space, whether or not another charge or charged object is in the field. Although field theory is a powerful tool for describing phenomena and predicting forces, physicists are still debating how objects can actually exert forces at a distance. Chapter 17 describes how quantum theory provides an extremely accurate model for describing such forces. Figure 11.4 A space module passing near a large planet or the Moon follows a curved path. Concept Check Use field theory to explain the path of a baseball thrown from outfield to home plate. Chapter 11 Electric field theory describes electrical phenomena. 545 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 546 11-2 Inquiry Lab 11-2 Inquiry Lab Electric Field Patterns — Demonstration Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question What is the shape of the electric field around various charged objects? Materials and Equipment plastic platform with 2 electrode holders overhead projector petri dish canola or olive oil lawn seeds single-point electrode two-point (oppositely charged) electrodes parallel copper plates about 4 cm × 4 cm hollow sphere conductor 4–6 cm in diameter 2 Wimshurst generators connecting wires Procedure 1 Pour some of the canola or olive oil into the petri dish so the dish is about three-quarters full. 2 Place the petri dish with the oil on the plastic platform on the overhead projector. Carefully sprinkle the lawn seeds evenly over the surface of the oil. 3 Attach the single-point electrode, with a connecting wire, to one contact of the Wimshurst generator. Immerse the electrode in the oil in the centre of the dish. 4 Crank the Wimshurst generator several times and carefully observe the pattern of the seeds in the oil. 5 Remove the electrode and allow sufficient time for the lawn seeds to redistribute on the surface. (Gentle stirring with a pencil might be required.) 6 Repeat steps 3 to 5 with each of the following: (a) (b) (c) two electrodes connected to similar contacts on two Wimshurst machines two electrodes connected to opposite contacts on one Wimshurst machine two parallel copper plates connected to opposite contacts on one Wimshurst machine (d) one hollow sphere connected to one contact of one Wimshurst machine Analysis 1. Describe and analyze the pattern of the lawn seeds created by each of the charged objects immersed in the oil in step 6 of the procedure by answering the following questions: (a) Where does the density of the lawn seeds appear to be the greatest? the least? (b) Does there appear to be a starting point and an endpoint in the pattern created by the lawn seed
s? 2. Are there any situations where there appears to be no observable effect on the lawn seeds? 3. Based on your observations of the patterns created by the lawn seeds on the surface of the oil, what conclusion can you make about the space around charged objects? Magnitude and Direction of an Electric Field The electric field that surrounds a charged object has both magnitude and direction. Therefore, an electric field is classified as a vector field. At any point around a charge, the field can be represented by a vector arrow. The arrow’s length represents the magnitude of the electric field and the arrowhead indicates direction at that point. By definition, the direction of the electric field around a charge is the direction of the force experienced by a small positive test charge placed in the electric field (Figure 11.5). A test charge is a charge with a magnitude small enough so that it does not disturb the charge on the source charge and thus change its electric field. test charge: charge with a magnitude small enough that it does not disturb the charge on the source charge and thus change its electric field source charge: charge that produces an electric field 546 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 547 (a) (b) test charge F source charge source charge F test charge Figure 11.5 The direction of the electric field at a point is the direction of the electric force exerted on a positive test charge at that point. (a) If the source charge is negative, the field is directed toward the source. (b) If the source charge is positive, the field is directed away from the source. Concept Check Identify the difference in the electric field strength, E II, as represented by the vector arrows in Figure 11.6. , at points I and II E I E Figure 11.6 You can determine the magnitude of the electric field around a point charge from the effect on another charge placed in the field. If a small positive test charge is placed in the field, this charge will experience a greater force when it is near the charge producing the field than when it is farther away from it. By definition, the electric field (E ) at a given point is the ratio of the electric force (F e) exerted on a charge (q) placed at that point to the magnitude of that charge. The electric field can be calculated using the equation F e E q where q is the magnitude of the test charge in coulombs (C); F e is the electric force on the charge in newtons (N); and E is the strength of the electric field at that point in newtons per coulomb (N/C), in the direction as defined previously. info BIT A tremendous range of field strengths occurs in nature. For example, the electric field 30 cm away from a light bulb is roughly 5 N/C, whereas the electron in a hydrogen atom experiences an electric field in the order of 1011 N/C from the atom’s nucleus. Chapter 11 Electric field theory describes electrical phenomena. 547 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 548 Example 11.1 A sphere with a negative charge of 2.10 × 10–6 C experiences an electrostatic force of repulsion of 5.60 × 10–2 N when it is placed in the electric field produced by a source charge (Figure 11.7). Determine the magnitude of the electric field the source charge produces at the sphere. Practice Problems 1. An ion with a charge of 1.60 10–19 C is placed in an electric field produced by another larger charge. If the magnitude of the field at this position is 1.00 103 N/C, calculate the magnitude of the electrostatic force on the ion. 2. The magnitude of the electrostatic force on a small charged sphere is 3.42 10–18 N when the sphere is at a position where the magnitude of the electric field due to another larger charge is 5.34 N/C. What is the magnitude of the charge on the small charged sphere? Answers 1. 1.60 10–16 N 2. 6.40 10–19 C 5.60 102 N F 2.10 106 C Figure 11.7 source charge Given q 2.10 106 C F e 5.60 102 N [repulsion] Required magnitude of the electric field (E ) Analysis and Solution F e, q Since .67 104 N/C Paraphrase The magnitude of the electric field is 2.67 104 N/C at the given point. q1 q2 r Figure 11.8 A test charge (q2) is placed in the electric field of a source charge (q1). The distance between their centres is r. PHYSICS INSIGHT Equations based on Coulomb’s law only work for point charges. 548 Unit VI Forces and Fields The equation for determining the magnitude of the electric field around a point charge, like that shown in Figure 11.8, can be derived mathematically as follows: F e and F e q 2 kq 1 2 r q2 , then If E q2 kq 1 2 r E k E r q 2 where q is the magnitude of the source charge producing the electric field in coulombs (ignore the sign of the charge); r is the distance from the 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 549 centre of the source charge to a specific point in space in metres; k is Coulomb’s constant (8.99 109 Nm2/C2); and E is the magnitude of the electric field in newtons per coulomb. Example 11.2 Determine the electric field at a position P that is 2.20 10 –2 m from the centre of a negative point charge of 1.70 10 –6 C. Given q 1.70 106 C r 2.20 102 m Required electric field E Analysis and Solution The source charge producing the electric field is q. So, q k E 2 r (1.70 106 C) 8.99 109 N m 2 2 C (2.20 102 m)2 3.16 107 N/C Practice Problems 1. The electric field at a position 2.00 cm from a charge is 40.0 N/C directed away from the charge. Determine the charge producing the electric field. 2. An electron has a charge of 1.60 10–19 C. At what distance from the electron would the magnitude of the electric field be 5.14 1011 N/C? Answers 1. 1.78 1012 C 2. 5.29 1011 m Since the source charge is negative and the field direction is defined as the direction of the electrostatic force acting on a positive test charge, the electric field is directed toward the source charge. Paraphrase The electric field at point P is 3.16 107 N/C [toward the source]. Concept Check Compare gravitational fields and electrostatic fields by listing two similarities and two differences between the two types of fields. Often, more than one charge creates an electric field at a particular point in space. In earlier studies, you learned the superposition principle for vectors. According to the superposition principle, fields set up by many sources superpose to form a single net field. The vector specifying the net field at any point is simply the vector sum of the fields of all the individual sources, as shown in the following examples. Example 11.3 shows how to calculate the net electric field at a point in one-dimensional situations. e MATH The nucleus of an atom exhibits both electric and gravitational fields. To study their similarities and differences graphically, visit www.pearsoned.ca/school/ physicssource. Chapter 11 Electric field theory describes electrical phenomena. 549 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 550 Example 11.3 Practice Problems 1. Calculate the net electric field at a point 2.10 10–2 m to the left of the 1.50 10–6 C charge in Figure 11.9. 2. An electron and a proton are 5.29 10–11 m apart in a hydrogen atom. Determine the net electric field at a point midway between the two charges. Answers 1. 3.67 107 N/C [left] 2. 4.11 1012 N/C [toward the electron] Two positively charged spheres, A and B, with charges of 1.50 10 –6 C and 2.00 10 –6 C, respectively, are 3.30 10 –2 m apart. Determine the net electric field at a point P located midway between the centres of the two spheres (Figure 11.9). 1.50 106 C 2.00 106 C A P 3.30 102 m B Figure 11.9 Given qA qB r 3.30 102 m 1.50 106 C 2.00 106 C Required net electric field at point P (E net) Analysis and Solution As shown in Figure 11.10, the electric field created by qA at point P is directed to the right, while the electric field at point P created by qB is directed to the left. Consider right to be positive. The distance between qA and point P is: rqA to P 3.30 102 m 1.65 102 m 2 To calculate the electric field at point P created by qA, use: (1.50 106 C) 8.99 109 m 2 2 C N (1.65 102 m)2 E qA q A k r 2 q to P A 4.953 107 N/C To calculate the electric field at point P created by qB, use: (2.00 106 C) 8.99 109 m 2 2 C N (1.65 102 m)2 E qB q B k r 2 q to P B 6.604 107 N/C Use vector addition to determine the net electric field at point P: E net E qA E qB 4.953 107 N/C [right] 6.604 107 N/C [left] 1.65 107 N/C [left] Paraphrase The net electric field at point P is 1.65 107 N/C [left]. P Eq B Eq A Figure 11.10 550 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 551 Example 11.4 demonstrates how to determine the net electric field at a point due to two charges in a two-dimensional situation. Example 11.4 Calculate the net electric field at a point P that is 4.00 10–2 m from a small metal sphere A with a negative charge of 2.10 10–6 C and 3.00 10–2 m from another similar sphere B with a positive charge of 1.50 10–6 C (Figure 11.11). P Figure 11.11 4.00 102 m 3.00 102 m A 36.9° 2.10 106 C 53.1° B 1.50 106 C Given qA rA to P A 2.10 106 C 4.00 102 m 36.9 to the horizontal qB rB to P B 1.50 106 C 3.00 102 m 53.1 to the horizontal Required net) net electric field at point P (E Analysis and Solution Since qA is a negative charge, the electric field created by qA at point P is directed toward qA from point P. Since qB is a positive charge, the electric field created by qB at point P is directed away from qB toward point P. Determine the electric field created by qA at point P: E A q k A 2 A r to P 2 m (2.10 106 C) 8.99 109 N 2 C (4.00 102 m)2 1.180 107 N/C Determine the electric field created by qB at point P: E B q k B 2 B r to P 2 m (1.50 106 C) 8.99 109 N C (3.00 102 m)2 2 Practice Problems 1. Calculate the net electric field at point P, which is 0.100 m from two similar spheres with positive charges of 2.00 C and separated by a distance of 0.0600 m, as shown in the figure below. P 0.100 m 0.100 m 72.5° 72.5° 2.00 C 0.0600 m 2.00 C 2. Two charges of +4.00 C are placed at
the vertices of an equilateral triangle with sides of 2.00 cm, as shown in the figure below. Determine the net electric field at the third vertex of the triangle. 2.00 cm 2.00 cm 60° 60° 4.00 C 2.00 cm 4.00 C Answers 1. 3.43 1012 N/C [90.0°] 2. 1.56 1014 N/C [90.0°] 1.498 107 N/C Chapter 11 Electric field theory describes electrical phenomena. 551 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 552 B are shown in Figure 11.12. A and E The directions of E y EB 53.1° 36.9° EA x 36.9° y P x EAy EAx EA y EB EBy 53.1° x P EBx Figure 11.12 Figure 11.13 Figure 11.14 Resolve each electric field into x and y components (see Figures 11.13 and 11.14). Use vector addition to determine the resultant electric field. EAx EBx (1.180 107 N/C)(cos 36.9°) EAy 9.436 106 N/C (1.498 107 N/C)(cos 53.1°) EBy 8.994 106 N/C (1.180 107 N/C)(sin 36.9°) 7.085 106 N/C (1.498 107 N/C)(sin 53.1°) 1.198 107 N/C Add the x components: Enetx EBx EAx (9.436 106 N/C) (8.994 106 N/C) 1.843 107 N/C Add the y components: Enety EBy EAy (7.085 106 N/C) (1.198 107 N/C) 4.895 106 N/C Use the Pythagorean theorem to solve for the magnitude of the electric field: E (1.843 107 N/C)2 (4.895 106 N/C)2 net 1.91 107 N/C Use the tangent function to determine the direction of the net electric field at point P (Figure 11.15). tan 14.9° The direction of the net field is 180° 14.9° 165° 4.895 106 N/C Enet θ x 1.843 107 N/C Figure 11.15 Paraphrase The net electric field at point P is 1.91 107 N/C [165°]. 552 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 553 In chapter 10, you learned that there are two types of electric charges that interact and are affected by electrostatic forces. In this section, you have learned that these charges are surrounded by electric fields—regions of electric influence around every charge. Electrostatic forces affect charges placed in these fields. Fields explain how two charges can interact, even though there is no contact between them. Since electric fields are vector fields, you can use vector addition to determine a net electric field at a point in the presence of more than one charge in one-dimensional and two-dimensional situations. 11.1 Check and Reflect 11.1 Check and Reflect Knowledge 1. What is the difference between an electric force and an electric field? 2. Why was it necessary to introduce a “field theory”? 3. How is the direction of an electric field defined? 4. Why is an electric field classified as a vector field? 5. If vector arrows can represent an electric field at a point surrounding a charge, identify the two ways that the vector arrows, shown below, represent differences in the electric fields around the two source charges. (a) the magnitude and direction of the electric field at a point 0.300 m to the right of the charge (b) the magnitude and direction of the electric force acting on a positive charge of 2.00 10–8 C placed at the point in (a) 8. A small test sphere with a negative charge of 2.50 C experiences an electrostatic attractive force of magnitude 5.10 10–2 N when it is placed at a point 0.0400 m from another larger charged sphere. Calculate (a) the magnitude and direction of the electric field at this point (b) the magnitude and the sign of charge on the larger charged sphere P E 9. A negative charge of 3.00 mC is 1.20 m to the right of another negative charge of 2.00 mC. Calculate P E 6. Describe the effect on the electric field at a point (a) if the magnitude of the charge producing the field is halved (b) if the sign of the charge producing the field is changed (c) if the magnitude of the test charge in the field is halved Applications 7. Given a small sphere with a positive charge of 4.50 10–6 C, determine: (a) the net electric field at a point along the same line and midway between the two charges (b) the point along the same line between the two charges where the net electric field will be zero Extension 10. Four similarly charged spheres of 5.00 C are placed at the corners of a square with sides of 1.20 m. Determine the electric field at the point of intersection of the two diagonals of the square. e TEST To check your understanding of forces and fields, follow the eTest links at www.pearsoned.ca/ school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 553 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 554 11.2 Electric Field Lines and Electric Potential In section 11.1, you learned that the electric field from a charge q at a point P can be represented by a vector arrow, as shown in Figure 11.16. The length and direction of the vector arrow represent the magnitude and direction of the electric field (E ) at that point. By measuring the electric force exerted on a test charge at an infinite number of points around a source charge, a vector value of the electric field can be assigned to every point in space around the source charge. This creates a three-dimensional map of the electric field around the source charge (Figure 11.16). Electric Field Lines For many applications, however, a much simpler method is used to represent electric fields. Instead of drawing an infinite number of vector arrows, you can draw lines, called electric field lines, to represent the electric field. Field lines are drawn so that exactly one field line goes through any given point within the field, and the tangent to the field line at the point is in the direction of the electric field vector at that point. You can give the field lines a direction such that the direction of the field line through a given point agrees with the direction of the electric field at that point. Use the following rules when you draw electric field lines around a point charge: • Electric field lines due to a positive source charge start from the charge and extend radially away from the charge to infinity. • Electric field lines due to a negative source charge come from infin- ity radially into and terminate at the negative source charge. • The density of lines represents the magnitude of the electric field. In other words, the more closely spaced and the greater the number of lines, the stronger is the electric field. Figure 11.16 A threedimensional map of the electric field around a source charge info BIT A lightning rod works because of the concentration of charges on the point of a conductor. This concentration of charge creates an electric field that ionizes air molecules around the point. The ionized region either makes contact with an upward streamer to a cloud, thus preventing the formation of a damaging return lightning stroke, or intercepts a downward leader from the clouds and provides a path for the lightning to the ground to prevent damage to the structure. e SIM Explore the electric fields around a point charge and two charges. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Figure 11.17 shows how to draw electric field lines around one and two negative point charges. Figure 11.17 The field lines around these charges were drawn using the rules given above. 554 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 555 M I N D S O N Drawing Electric Field Lines Rarely is the electric field at a point in space influenced by a single charge. Often, you need to determine the electric field for a complicated arrangement of charges. Electric field lines can be used to display these electric fields. In Figure 11.18, lawn seeds have been sprinkled on the surface of a container of cooking oil. In each case, a different charged object has been put into the oil. • On a sheet of paper, sketch the electric field lines in each situation using the rules for drawing electric field lines given on page 554. • Use concise statements to justify the pattern you drew in each of the sketches. (a) (d) (b) (c) (e) Figure 11.18 (a) one negative charge, (b) two negative charges, (c) one negative and one positive charge, (d) two oppositely charged plates, (e) one negatively charged cylindrical ring Conductors and Electric Field Lines In a conductor, electrons move freely until they reach a state of static equilibrium. For static equilibrium to exist, all charges must be at rest and thus must experience no net force. Achieving static equilibrium creates interesting distributions of charge that occur only in conducting objects and not in non-conducting objects. Following are five different situations involving charge distribution on conductors and their corresponding electric field lines. Solid Conducting Sphere When a solid metal sphere is charged, either negatively or positively, does the charge distribute evenly throughout the sphere? To achieve static equilibrium, all excess charges move as far apart as possible because of electrostatic forces of repulsion. A charge on the sphere at position A in Figure 11.19(a), for example, would experience a net force of electrostatic repulsion from the other charges. Consequently, all excess charges on a solid conducting sphere are repelled. These excess charges distribute evenly on the surface of the metal conducting sphere. Chapter 11 Electric field theory describes electrical phenomena. 555 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 556 Figure 11.19(b) shows the electric field lines created by the distribution of charge on the surface of a solid conducting sphere. Because electric field lines cannot have a component tangential to this surface, the lines at the outer surface must always be perpendicular to the surface. (a) F F A F F F F (b) Figure 11.19(a) solid sphere Charges on a Figure 11.19(b) for a charged solid sphere Electric field lines Solid, Flat, Conducting Plate How do excess charges, either positive or negative, distribute on a solid, flat, conducting plate like the one in Figure 11.20(a)? On a flat surface, the forces of repulsion are similarly parallel or tangential to the surface. Thus, electrostatic forces of repulsion acting on charges cause the charges to spread and distribute evenly along the outer surface of a charge
d plate, as shown in Figure 11.20(b). Electric field lines extend perpendicularly toward a negatively charged plate. The electric field lines are uniform and parallel, as shown in Figure 11.20(c). (a) (b) (c) F F F F Figure 11.20 (a) Forces among three charges on the top surface of a flat, conducting plate (b) Uniform distribution of charges on a charged, flat, conducting plate (c) Uniform distribution of charges, shown with electric field lines Irregularly Shaped Solid Conducting Object For an irregularly shaped solid conductor, the charges are still repelled and accumulate on the outer surface. But do the charges distribute evenly on the outer surface? Figure 11.21(a) is an example of a charged, irregularly shaped object. 556 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 557 On a flatter part of the surface, the forces of repulsion are nearly parallel or tangential to the surface, causing the charges to spread out more, as shown in Figure 11.21(b). At a pointed part of a convex surface, the forces are directed at an angle to the surface, so a smaller component of the forces is parallel or tangential to the surface. With less repulsion along the surface, more charge can accumulate closer together. As a rule, the net electrostatic forces on charges cause the charges to accumulate at the points of an irregularly shaped convex conducting object. Conversely, the charges will spread out on an irregularly shaped concave conducting object. On irregularly shaped conductors, the charge density is greatest where the surface curves most sharply (Figure 11.21(c)). The density of electric field lines is also greatest at these points. info BIT The accumulation of charge on a pointed surface is the explanation for St. Elmo’s fire, which you read about at the beginning of this chapter. St. Elmo’s fire is a plasma (a hot, ionized gas) caused by the powerful electric field from the charge that accumulates on the tips of raised, pointed conductors during thunderstorms. St. Elmo’s fire is known as a form of corona discharge or point discharge. (a) (b) (c) F F y x F F x y Figure 11.21(a) A charged, irregularly shaped convex object Figure 11.21(b) Forces affecting charges on the surface of an irregularly shaped convex object Figure 11.21(c) lines around a charged irregularly shaped convex object Electric field Hollow Conducting Object When a hollow conducting object is charged, either negatively or positively, does the charge distribute evenly throughout the inner and outer surfaces of the object? As you saw in Figures 11.19, 11.20, and 11.21, excess charges move to achieve static equilibrium, and they move as far apart as possible because of electrostatic forces of repulsion. In a hollow conducting object, all excess charges are still repelled outward, as shown in Figure 11.22(a). However, they distribute evenly only on the outer surface of the conducting object. There is no excess charge on the inner surface of the hollow object, no matter what the shape of the object is. The corresponding electric field lines created by the distribution of charge on the outer surface of a hollow object are shown in Figure 11.22(b). The electric field lines at the outer surface must always be perpendicular to the outer surface. (a) (b) Figure 11.22(a) A charged hollow conducting object Figure 11.22(b) Electric field lines on a hollow conducting object Chapter 11 Electric field theory describes electrical phenomena. 557 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 558 info BIT Coaxial cable wires are used to transmit electric signals such as cable TV to your home. To prevent electric and magnetic interference from outside, a covering of conducting material surrounds the coaxial wires. Any charge applied to the conducting layer accumulates on the outside of the covering. No electric field is created inside a hollow conductor, so there is no influence on the signals transmitted in the wires. e WEB Research the operation of an ink-jet printer. What is the function of charged plates in these printers? Begin your search at www.pearsoned.ca/school/ physicssource. Most surprisingly, the electric field is zero everywhere inside the conductor, so there are no electric field lines anywhere inside a hollow conductor. As previously described, this effect can be explained using the superposition principle. Fields set up by many sources superpose, forming a single net field. The vector specifying the magnitude of the net field at any point is simply the vector sum of the fields of each individual source. Anywhere within the interior of a hollow conducting object, the vector sum of all the individual electric fields is zero. For this reason, the person inside the Faraday cage, shown in the photograph on page 508, is not affected by the tremendous charges on the outside surface of the cage. Parallel Plates If two parallel metal plates, such as those in Figure 11.23(a), are oppositely charged, how are the charges distributed? Electrostatic forces of repulsion of like charges, within each plate, cause the charges to distribute evenly within each plate, and electrostatic forces of attraction of opposite charges on the two plates cause the charges to accumulate on the inner surfaces. Thus, the charges spread and distribute evenly on the inner surfaces of the charged plates. (a) (b) Figure 11.23(a) The distribution of net charge on oppositely charged parallel plates Figure 11.23(b) Electric field lines between two oppositely charged parallel plates The magnitude of the resulting electric field can be shown to be the vector sum of each individual field, so it can be shown that the electric field anywhere between the plates is uniform. Thus, between two oppositely charged and parallel plates, electric field lines exist only between the charged plates. These lines extend perpendicularly from the plates, starting at the positively charged plate and terminating at the negatively charged plate. The electric field lines are uniform in both direction and density between the two oppositely charged plates, except near the edges of the plates. Such a system is called a parallel-plate capacitor. This type of capacitor is found in many different types of electrical equipment, including printers and televisions (where it is part of the “instant on” feature). It is also used in particle accelerators, such as cathode-ray tubes and mass spectrometers. You will learn about mass spectrometers in Unit VIII. 558 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 559 THEN, NOW, AND FUTURE Defibrillators Save Lives During a heart attack, the upper and lower parts of the heart can begin contracting at different rates. Often these contractions are extremely rapid. This fluttery unsynchronized beating, called fibrillation, pumps little or no blood and can damage the heart. A defibrillator uses a jolt of electricity to momentarily stop the heart so that it can return to a normal beat (Figure 11.24). Figure 11.24 A defibrillator stops the fibrillation of the heart muscle by applying an electric shock. A defibrillator consists of two parallel charged plates (see Figure 11.23(b)), called a parallel-plate capacitor, connected to a power supply and discharging pads. A typical defibrillator stores about 0.4 C on the plates, creating a potential difference of approximately 2 kV between the plates. When discharged through conductive pads placed on the patient’s chest, the capacitor delivers about 0.4 kJ of electrical energy in 0.002 s. Roughly 200 J of this energy passes through the patient’s chest. A defibrillator uses a highvoltage capacitor to help save lives. Such capacitors have many other applications in other electrical and electronic devices, such as the highvoltage power supplies for cathoderay tubes in older televisions and computer monitors. The charge stored in such capacitors can be dangerous. Products con- taining such high-voltage capacitors are designed to protect the users from any dangerous voltages. However, service technicians must be careful when working on these devices. Since the capacitors store charge, they can deliver a nasty shock even after the device is unplugged. Questions 1. How does the magnitude of the power delivered by the plates compare with the actual power delivered to the chest by the jolt? 2. Identify a feature of televisions that demonstrates an important application of parallel-plate capacitors. 3. If a defibrillator can store 0.392 C of charge in 30 s, how many electrons are stored in this time period? M I N D S O N Faraday’s Ice Pail In the early 1800s, Michael Faraday performed an experiment to investigate the electric fields inside a hollow metal container. He used ice pails, so this experiment is often called “Faraday’s ice pail experiment.” charged rod This activity is called a conceptual experiment because you will not perform the experiment. Instead, you will predict and justify the results of an experimental procedure that duplicates Faraday’s investigation. The purpose of the experiment is to determine what type of electric field exists on the inside and the outside of a hollow metal container. A positively charged rod is placed into position inside the metal container, near the centre, as shown in Figure 11.25. The rod is then moved to a position inside the metal container, near one of the inner surfaces. • Which of the electroscopes would show a deflection when the rod is near the centre of the metal container? • Clearly explain your reasoning and the physical principles you used in determining your answers to these questions. insulated plate Figure 11.25 An ice pail is a metal container. It is placed on an insulated surface, and electroscopes are attached to the inside and outside surfaces of the metal container. Chapter 11 Electric field theory describes electrical phenomena. 559 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 560 Figure 11.26 The charged dome of a Van de Graaff generator exposes a perso
n to very large voltages. Electric Potential Energy and Electric Potential A Van de Graaff generator can generate up to 250 kV. Touching the dome not only produces the spectacular results shown in Figure 11.26, it can also cause a mild, harmless shock. On the other hand, touching the terminals of a wall socket, which has a voltage of 120 V, can be fatal. An understanding of this dramatic difference between the magnitude of the voltage and its corresponding effect requires a study of the concepts of electric potential energy and electric potential. These concepts are important in the study of electric fields. Even though the terms seem similar, they are very different. To explain the difference, you will study these concepts in two types of electric fields: non-uniform electric fields around point charges, and uniform electric fields between parallel charged plates. Electric Potential Energy In previous grades, you learned about the relationship between work and potential energy. Work is done when a force moves an object in the direction of the force such that: d W F where W is work, and F displacement of the object. and d are the magnitudes of the force and the In a gravitational system like the one shown in Figure 11.27(a), lifting a mass a vertical distance against Earth’s gravitational field requires work to stretch an imaginary “gravitational spring” connecting the mass and Earth. Further, because the force required to do the work is a conservative force, the work done against the gravitational field increases the gravitational potential energy of the system by an amount equal to the work done. Therefore: gravitational potential energy gain work done (a) Fapp d mass Fg Ep W (b) Fapp d Fe Figure 11.27(a) Work is required to lift a mass to a certain position above Earth’s surface. Figure 11.27(b) Work is required to move a small positive charge away from a larger negative charge. 560 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 561 Similarly, in an electrostatic system like the one shown in Figure 11.27(b), moving a small charge through a certain distance in a non-uniform electric field produced by another point charge requires work to either compress or stretch an imaginary “electrostatic spring” connecting the two charges. Since the force required to do this work is also a conservative force, the work done in the electric field must increase the electric potential energy of the system. Electric potential energy is the energy stored in the system of two charges a certain distance apart (Figure 11.28). Electric potential energy change equals work done to move a small charge: Ep W q1 r P q2 Figure 11.28 Electric potential energy is the energy stored in the system of two charges a certain distance apart. Example 11.5 Moving a small charge from one position in an electric field to another position requires 3.2 10–19 J of work. How much electric potential energy will be gained by the charge? Analysis and Solution The work done against the electrostatic forces is W. The electric potential energy gain is Ep. In a conservative system, Ep W So, Ep W 3.2 1019 J The electric potential energy gain of the charge is 3.2 10–19 J. Practice Problems 1. A small charge gains 1.60 10–19 J of electric potential energy when it is moved to a point in an electric field. Determine the work done on the charge. 2. A charge moves from one position in an electric field, where it had an electric potential energy of 6.40 10–19 J, to another position where it has an electric potential energy of 8.00 10–19 J. Determine the work necessary to move the charge. Answers 1. 1.60 1019 J 2. 1.60 1019 J Choosing a Reference Point In Chapter 7, you learned that commonly used reference points for zero gravitational potential energy are Earth’s surface or infinity. Choosing a zero reference point is necessary so you can analyze the relationship between work and gravitational potential energy. Chapter 11 Electric field theory describes electrical phenomena. 561 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 562 Consider a zero reference point at Earth’s surface. An object at rest on Earth’s surface would have zero gravitational potential energy relative to Earth’s surface. If the object is lifted upward, opposite to the direction of the gravitational force it experiences, then work is being done on the object. The object thus gains gravitational potential energy. If the object falls back to the surface in the same direction as the gravitational force, then the object loses gravitational potential energy. As with gravitational potential energy, the value of electric potential energy at a certain position is meaningless unless it is compared to a reference point where the electric potential energy is zero. The choice of a zero reference point for electric potential energy is arbitrary. For example, suppose an electric field is being produced by a large negative charge. A small positive charge would be attracted and come to rest on the surface of the larger negative charge, where it would have zero electric potential energy. This position could be defined as a zero electric potential energy reference point (Figure 11.29(a)). Then, the test charge has positive electric potential energy at all other locations. Alternatively, the small positive test charge may be moved to a position so far away from the larger negative charge that there is no electrostatic attraction between them. This position would be an infinite distance away. This point, at infinity, is often chosen as the zero electric potential energy reference point. Then, the test charge has negative electric potential energy at all other locations. This text uses infinity as the zero electric potential energy reference point for all calculations (Figure 11.29(b)). at surface at infinity q q Ep 0 (a) Ep 0 (b) Figure 11.29 Two commonly used reference points for electric potential energy: (a) test charge defined as having zero electric potential energy at the surface of the source charge (b) test charge defined as having zero electric potential energy at infinity 562 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 563 Work and Electric Potential Energy Whenever work is done on a charge to move it against the electric force caused by an electric field, the charge gains electric potential energy. The following examples illustrate the relationship between work and electric potential energy. Electric Potential Energy Between Parallel Charged Plates Except at the edges, the electric field between two oppositely charged plates is uniform in magnitude and direction. Suppose a small positive charge in the field between the plates moves from the negative plate to the positive plate with a constant velocity. This motion requires an external force to overcome the electrostatic forces the charged plates exert on the positive charge. The work done on the charge increases the system’s electric potential energy: Ep W F d Example 11.6 When a small positive charge moves from a negative plate to a positive plate, 2.3 × 10–19 J of work is done. How much electric potential energy will the charge gain? Analysis and Solution In a conservative system, EP EP W 2.3 1019 J W. Paraphrase The electric potential energy gain of the charge is 2.3 1019 J. Practice Problem 1. A charge gained 4.00 105 J of electric potential energy when it was moved between two oppositely charged plates. How much work was done on the charge? Answer 1. 4.00 105 J Electric Potential Suppose two positive charges are pushed toward a positive plate. In this case, twice as much work is done, and twice as much electric potential energy is stored in the system. However, just as much electric potential energy is still stored per charge. Storing 20 J of energy in two charges is the same as storing 10 J of energy in each charge. At times, it is necessary to determine the total electric potential energy at a certain location in an electric field. At other times, it is convenient to consider just the electric potential energy per unit charge at a location. The electric potential energy stored per unit charge at a given point is the amount of work required to move a unit charge to that Chapter 11 Electric field theory describes electrical phenomena. 563 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 564 electric potential: the electric potential energy stored per unit charge at a given point in an electric field point from a zero reference point (infinity). This quantity has a special name: electric potential. To determine the electric potential at a location, use this equation: electric potential electric potential energy charge Ep V q where V is in volts, Ep is in joules, and q is in coulombs. Since electric potential energy is measured in joules and charge is measured in coulombs, o e l u j 1 volt b m o l u co 1 1 Thus, if the electric potential at a certain location is 10 V, then a charge of 1 C will possess 10 J of electric potential energy, a charge of 2 C will possess 20 J of electric potential energy, and so on. Even if the total electric potential energy (Ep) at a location changes, depending on the amount of charge placed in the electric field, the electric potential (V ) at that location remains the same. A balloon can be used as an example to help explain the difference between the concepts of electric potential energy and electric potential. Suppose you rub a balloon with fur. The balloon acquires an electric potential of a few thousand volts. In other words, the electric energy stored per coulomb of charge on the balloon is a few thousand volts. Written as an equation, Ep V q Now suppose the balloon were to gain a large charge of 1 C during the rubbing process. In order for the electric potential to stay the same, a few thousand joules of work would be needed to produce the electrical energy that would allow the balloon to maintain that electric potential. However, the amount of charge a balloon acquires dur
ing rubbing is usually only in the order of a few microcoulombs. So, acquiring this potential requires a small amount of work to produce the energy needed. Even though the electric potential is high, the electric potential energy is low because of the extremely small charge. Concept Check Suppose the magnitude of a charge placed in an electric field were doubled. How much would the electric potential energy and the electric potential change? info BIT The SI unit of electric potential is the volt, named in honour of the Italian physicist Count Alessandro Volta (1745–1827), who developed the first electric battery in the early 1800s. 564 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 565 Electric Potential Difference When a charge moves from one location to another in an electric field, it experiences a change in electric potential. This change in electric potential is called the electric potential difference, V, between the two points and V Vfinal Vinitial Ep since V q Ep V q where Ep is the amount of work required to move the charge from one location to the other. The potential difference depends only on the two locations. It does not depend on the charge or the path taken by the charge as it moves from one location to another. Electric potential difference is commonly referred to as just potential difference or voltage. An electron volt (eV) is the quantity of energy an electron gains or loses when passing through a potential difference of exactly 1 V. An electron volt is vastly less than a joule: 1 eV 1.60 1019 J Although not an SI unit, the electron volt is sometimes convenient for expressing tiny quantities of energy, especially in situations involving a single charged particle such as an electron or a proton. The energy difference in Example 11.6 could be given as (2.3 1019 J) 1.4 eV V e 1 019 J 1 1.60 electric potential difference: change in electric potential experienced by a charge moving between two points in an electric field electron volt: the change in energy of an electron when it moves through a potential difference of 1 V PHYSICS INSIGHT The notation VAB is widely used instead of V to represent the potential difference at point A relative to point B. When the points in question are clear from the context, the subscripts are generally omitted. For example, the equation for Ohm’s law is usually written as V IR, where it is understood that V represents the potential difference between the ends of the resistance R. Example 11.7 Moving a small charge of 1.6 × 10–19 C between two parallel plates increases its electric potential energy by 3.2 × 10–16 J. Determine the electric potential difference between the two parallel plates. Analysis and Solution To determine the electric potential difference between the plates, use the equation Ep .0 103 V The electric potential difference between the plates is 2.0 103 V. Practice Problems 1. In moving a charge of 5.0 C from one terminal to the other, a battery raises the electric potential energy of the charge by 60 J. Determine the potential difference between the battery terminals. 2. A charge of 2.00 10–2 C moves from one charged plate to an oppositely charged plate. The potential difference between the plates is 500 V. How much electric potential energy will the charge gain? Answers 1. 12 V 2. 10.0 J Chapter 11 Electric field theory describes electrical phenomena. 565 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 566 Example 11.8 A small charge of 3.2 1019 C is moved between two parallel plates from a position with an electric potential of 2.0 103 V to another position with an electric potential of 4.0 103 V (Figure 11.30). Practice Problems 1. A sphere with a charge of magnitude 2.00 C is moved between two positions between oppositely charged plates. It gains 160 J of electric potential energy. What is the potential difference between the two positions? 2. An electron moves between two positions with a potential difference of 4.00 104 V. Determine the electric potential energy gained by the electron, in joules (J) and electron volts (eV). Answers 1. 80.0 V 2. 6.40 1015 J or 4.00 104 eV A 2.0 103 V B 4.0 103 V Figure 11.30 3.2 1019 C battery Determine: (a) the potential difference between the two positions (b) the electric potential energy gained by moving the charge, in joules (J) and electron volts (eV) Given Vinitial Vfinal 2.0 103 V 4.0 103 V q 3.2 10–19 C Required (a) potential difference between points B and A (V ) (b) electric potential energy gained by the charge (Ep) Analysis and Solution (a) V Vfinal Vinitial (4.0 103 V) (2.0 103 V) 2.0 103 V (b) To calculate the electric potential energy, use the equation Ep. V q Vq (2.0 103 V)(3.2 1019 C) 6.4 10–16 J Ep Ep Since 1 eV 1.60 1019 J, V e 1 (6.4 1016 J) 019 J 1 4.0 103 eV 4.0 keV 1.60 Paraphrase (a) The potential difference between the two positions is 2.0 103 V. (b) The energy gained by moving the charge between the two positions is 6.4 10–16 J or 4.0 103 eV. 566 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 567 The Electric Field Between Charged Plates Earlier in this section, you determined the electric field strength surrounding a point charge using the following equations: k q or E E 2 r F e q You also learned that the electric field around a point charge is a nonuniform electric field. Its magnitude depends on the distance from the charge. Later, you learned that a special type of electric field exists between two charged parallel plates. The magnitude of the electric field between the plates is uniform anywhere between the plates and it can be determined using the general equation for an electric field, F . You cannot use the equation E E e q k q because it is used only 2 r e WEB One of the technological applications of parallelplate capacitors is in disposable cameras. Research the role of capacitors in these cameras. Begin your search at www.pearsoned.ca/ school/physicssource. for point charges. Now, after studying electric potential difference, you can see how another equation for determining the electric field strength between plates arises from an important relationship between the uniform electric field and the electric potential difference between two charged parallel plates (Figure 11.31). If a small positively charged particle (q) is moved through the uni- form electric field (E q. This force is ), a force is required, where F the force exerted on the particle due to the presence of the electric field. If this force moves the charged particle a distance (d) between the plates, then the work done is: E Figure 11.31 Electrically charged parallel plates W F or W E d qd Since this system is conservative, the work done is stored in the charge as electric potential energy: W Ep E qd The electric potential difference between the plates is: Ep V q q d E q d E To calculate the magnitude of the uniform electric field between charged plates, use the equation E V d where V is the electric potential difference between two charged plates in volts; d is the distance in metres between the plates; and E is the magnitude of the electric field in volts per metre. Chapter 11 Electric field theory describes electrical phenomena. 567 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 568 Note that 1 V/m equals 1 N/C because /C J 1 V//C Example 11.9 Practice Problems 1. Two charged parallel plates, separated by 5.0 10–4 m, have an electric field of 2.2 104 V/m between them. What is the potential difference between the plates? 2. Spark plugs in a car have electrodes whose faces can be considered to be parallel plates. These plates are separated by a gap of 5.00 10–3 m. If the electric field between the electrodes is 3.00 106 V/m, calculate the potential difference between the electrode faces. Answers 1. 11 V 2. 1.50 104 V A cathode-ray-tube (CRT) computer monitor accelerates electrons between charged parallel plates (Figure 11.32). These electrons are then directed toward a screen to create an image. If the plates are 1.2 10–2 m apart and have a potential difference of 2.5 104 V between them, determine the magnitude of the electric field between the plates. 1.2 102 m V 2.5 104 V accelerating plates Given V 2.5 104 V d 1.2 10–2 m screen Figure 11.32 Required ) magnitude of the electric field between the plates (E Analysis and Solution To calculate the magnitude of the electric field between the plates, use the equation V E d 4 V 0 1 .5 2 2 1 1 m .2 2.1 106 V/m 0 Paraphrase The magnitude of the electric field between the plates is 2.1 106 V/m. 568 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 569 11.2 Check and Reflect 11.2 Check and Reflect Knowledge 1. Describe the difference between an electric field vector and an electric field line. 2. Sketch electric field lines around the following charges: (a) a positive charge (b) a negative charge (c) two positive charges (d) two negative charges (e) a positive charge and a negative charge 3. Describe the difference between electric potential and electric potential energy. Applications 4. At a point in Earth’s atmosphere, the electric field is 150 N/C downward and the gravitational field is 9.80 N/kg downward. (a) Determine the electric force on a proton (p) placed at this point. (b) Determine the gravitational force on the proton at this point. The proton has a mass of 1.67 10–27 kg. 9. Determine the magnitude and direction of the net electric field at point P shown in the diagram below. 50 μC 10 μC P 0.30 m 0.15 m 10. A uniform electric field exists between two oppositely charged parallel plates connected to a 12.0-V battery. The plates are separated by 6.00 104 m. (a) Determine the magnitude of the electric field between the plates. (b) If a charge of 3.22 106 C moves from one plate to another, calculate the change in electric potential energy of the charge. Extensions 11. A metal car is charged by contact with a charged object. Compare the charge distribution on the outside and the inside of the me
tal car body. Why is this property useful to the occupants of the car if the car is struck by lightning? 5. A metal box is charged by touching it with 12. Explain why only one of the electroscopes a negatively charged object. (a) Compare the distribution of charge at the corners of the box with the faces of the box. (b) Draw the electric field lines inside and surrounding the box. 6. What is the electric field intensity 0.300 m away from a small sphere that has a charge of 1.60 10–8 C? connected to the hollow conductive sphere in the illustration below indicates the presence of a charge. 7. Calculate the electric field intensity 13. Two points at different positions in midway between two negative charges of 3.2 C and 6.4 C separated by 0.40 m. 8. A 2.00-C charge jumps across a spark gap in a spark plug across which the potential difference is 1.00 103 V. How much energy is gained by the charge? an electric field have the same electric potential. Would any work be required to move a test charge from one point to another? Explain your answer. e TEST To check your understanding of electric field lines, follow the eTest links at www.pearsoned.ca/ school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 569 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 570 info BIT Living cells “pump” positive sodium ions (Na) from inside a cell to the outside through a membrane that is 0.10 m thick. The electric potential is 0.70 V higher outside the cell than inside it. To move the sodium ions, work must be done. It is estimated that 20% of the energy consumed by the body in a resting state is used to operate these “pumps.” 11.3 Electrical Interactions and the Law of Conservation of Energy A charge in an electric field experiences an electrostatic force. If the charge is free to move, it will accelerate in the direction of the electrostatic force, as described by Newton’s second law. The acceleration of the charge in the non-uniform electric field around a point charge is different from the acceleration motion of a charge in a uniform electric field between charged plates. Figure 11.33 shows a charge in the non-uniform field of a point charge. The electrostatic force on a charge placed in the field varies inversely as the square of the distance between the charges. A varying force causes non-uniform acceleration. Describing the motion of the charge in this type of situation requires applying calculus to Newton’s laws of motion, which is beyond the scope of this text. However, to determine the particle’s speed at a given point, you can use the law of conservation of energy. source charge F a E Figure 11.33 The electrostatic force on a point charge in a non-uniform electric field causes non-uniform acceleration of the charge. If the forces acting on an object are conservative forces, then the work done on a system changes the potential energy of the system. Electric potential energy, like gravitational potential energy, can be converted to kinetic energy. A charged particle placed in an electric field will accelerate from a region of high potential energy to a region of low potential energy. According to the law of conservation of energy, the moving charge gains kinetic energy at the expense of potential energy. If you assume that no energy is lost to friction and the forces are conservative, the kinetic energy gained equals the potential energy lost, so the sums of the two energies are always equal: Epi Eki Epf Ekf 570 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:38 AM Page 571 Example 11.10 A pith ball of mass 2.4 104 kg with a positive charge of 1.2 108 C is initially at rest at location A in the electric field of a larger charge (Figure 11.34). At this location, the charged pith ball has 3.0 107 J of electric potential energy. When released, the ball accelerates toward the larger charge. At position B, the ball has 1.5 108 J of electric potential energy. Find the speed of the pith ball when it reaches position B. 1.2 108 C 2.4 104 kg B 1.5 108 J A 3.0 107 J Figure 11.34 Given m 2.4 104 kg 3.0 107 J Epi q 1.2 108 C 1.5 108 J Epf Required speed of the ball at position B (v) Analysis and Solution The pith ball is at rest at A, so its initial kinetic energy is zero. Its electric potential energy at B is lower than at A. Since this system is conservative, the loss of electric potential energy when the ball moves from A to B is equal to a gain in kinetic energy, according to the law of conservation of energy: Epf Ekf Eki Epi Substitute the given values and solve for Ekf (3.0 107 J) 0 (1.5 108 J) Ekf Ekf Since the kinetic energy of an object is Ek 2.85 107 J . 1 mv2, 2 Practice Problems 1. A negative charge of 3.00 10–9 C is at rest at a position in the electric field of a larger positive charge and has 3.20 1012 J of electric potential energy at this position. When released, the negative charge accelerates toward the positive charge. Determine the kinetic energy of the negative charge just before it strikes the larger positive charge. 2. A small sphere with a charge of 2.00 C and a mass of 1.70 103 kg accelerates from rest toward a larger positive charge. If the speed of the sphere just before it strikes the positive charge is 5.20 104 m/s, how much electric potential energy did the negative charge lose? Answers 1. 3.20 1012 J 2. 2.30 106 J Ek 2 v 2 m v 2Ek m 2(2.85 107 J) 2.4 104 kg 4.9 102 m/s Paraphrase The speed of the pith ball at position B is 4.9 102 m/s. Chapter 11 Electric field theory describes electrical phenomena. 571 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 572 It is easier to describe the motion of a charge in a uniform electric field between two parallel plates, as shown in Figure 11.35. In this case, the acceleration is constant because of the constant force, so either the work–energy theorem or the laws of dynamics can be used. (Because the electric field is constant (uniform), the force acting on a charge q is also constant because F e qE .) F E E Figure 11.35 In a uniform electric field between two parallel plates, the acceleration of a charge is constant. Concept Check Electrostatic forces and gravitational forces are similar, so the motion of objects due to these forces should be similar. Consider a charge in an electric field between two parallel plates. Sketch the direction of the motion of the charge when its initial motion is: • perpendicular to the plates (the electrostatic force is similar to the gravitational force on falling masses) • parallel to the plates (the electrostatic force is similar to the gravitational force that causes the parabolic projectile motion of a mass close to the surface of a large planet or moon) Example 11.11 Two vertical parallel plates are connected to a DC power supply, as shown in Figure 11.36. The electric potential between the plates is 2.0 103 V. A sphere of mass 3.0 1015 kg with a positive charge of 2.6 1012 C is placed at the positive plate and released. It accelerates toward the negative plate. Determine the speed of the sphere at the instant before it strikes the negative plate. Ignore any gravitational effects. 2.6 1012 C 3.0 1015 kg 2.0 103 V Figure 11.36 572 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 573 Given q 2.6 1012 C V 2.0 103 V m 3.0 1015 kg Required speed of the sphere at the negative plate (v) Analysis and Solution This system is conservative. You can use kinetic energy of the charge to find its speed. , is 0 J. 0 J. The initial electric potential energy of the sphere at the positive plate Vq. Since the sphere was at rest, its initial kinetic energy, is Epi Eki The final electric potential energy of the sphere at the negative plate is Epf According to the law of conservation of energy, Ekf Epi Vq 0 J 0 J Ekf (2.0 103 V)(2.6 1012 C) 0 J 0 J Ekf Epf Eki 5.2 109 J Ekf Since Ek 1 mv 2, 2 v 2Ek m 2(5.2 109 J) 3.0 1015 kg 1.9 103 m/s Paraphrase The speed of the sphere at the negative plate is 1.9 103 m/s. Practice Problems 1. An alpha particle with a charge of 3.20 1019 C and a mass of 6.65 1027 kg is placed between two oppositely charged parallel plates with an electric potential difference of 4.00 104 V between them. The alpha particle is injected at the positive plate with an initial speed of zero, and it accelerates toward the negative plate. Determine the final speed of the alpha particle just before it strikes the negative plate. 2. If a charge of 6.00 106 C gains 3.20 104 J of kinetic energy as it accelerates between two oppositely charged plates, what is the potential difference between the two parallel plates? Answers 1. 1.96 106 m/s 2. 53.3 V Chapter 11 Electric field theory describes electrical phenomena. 573 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 574 Example 11.12 An electron enters the electric field between two charged parallel plates, as shown in Figure 11.37. electric field Figure 11.37 (a) Copy Figure 11.37 into your notebook and sketch the motion of the electron between the plates. (b) If the electron experiences a downward acceleration of 2.00 1017 m/s2 due to the electric field between the plates, determine the time taken for the electron to travel 0.0100 m to the positive plate. Given a 2.00 1017 m/s2 [down] d 0.0100 m Required (a) sketch of the electron’s motion (b) time (t) Analysis and Solution (a) The electron’s acceleration is downward, so the motion of the electron will follow a parabolic path to the positive plate (Figure 11.38), similar to the projectile motion of an object travelling horizontally to the surface of Earth and experiencing downward acceleration due to gravity. electric field Figure 11.38 Practice Problems 1. Two horizontal parallel plates, 1.2 102 m apart, are connected to a DC power supply, as shown in the figure below. The electric field between the plates is 1.7 105 V/m. A sphere of mass 3.0 1015 kg with a positive charge of 2.6 1012 C is injected into the region between the plates, with an initial speed of 3.3 103 m/s, as shown. It acce
lerates toward the negative plate. Copy the diagram into your notebook, sketch the motion of the positive charge through the region between the plates, and determine the distance the positive charge moves toward the negative plate after 6.0 106 s have elapsed. Gravitational effects may be ignored in this case. electric field 1.7 105 V/m 3.3 103 m/s 2. An electron, travelling at 2.3 103 m/s, enters perpendicular to the electric field between two horizontal charged parallel plates. If the electric field strength is 1.5 102 V/m, calculate the time taken for the electron to deflect a distance of 1.0 10–2 m toward the positive plate. Ignore gravitational effects. Answers 1. 2.7 103 m 2. 2.7 108 s 574 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:40 AM Page 575 (b) Use the equation d vi t 1 a(t)2 to determine the 2 time it takes the electron to fall to the positive plate. Since vi d 1 a(t)2 2 t 2d a 2(0.0100 m) m 2.00 1017 s2 0, 3.16 1010 s Paraphrase (a) The path of the electron between the parallel plates is parabolic. (b) The time taken for the electron to fall to the positive plate is 3.16 1010 s. 11.3 Check and Reflect 11.3 Check and Reflect Knowledge Applications 1. In what direction will a positively charged particle accelerate in an electric field? 2. Electric potential energy exists only where a charge is present at a point in an electric field. Must a charge also be present at that point for there to be electric potential? Why or why not? 4. Calculate the speed of an electron and a proton after each has accelerated from rest through an electric potential of 220 V. 5. Electrons in a TV picture tube are accelerated by a potential difference of 25 kV. Find the maximum speed the electrons would reach if relativistic effects are ignored. 3. Two positively charged objects are an 6. A charge gains 1.92 1014 J of electric equal distance from a negatively charged object, as shown in the diagram below. Charge B is greater than charge A. Compare the electric potential and electric potential energy of the positively charged objects. B A potential energy when it moves through a potential difference of 3.20 104 V. What is the magnitude of the charge? 7. How much work must be done to increase the electric potential of a charge of 2.00 106 C by 120 V? 8. A deuterium ion (H1), a heavy isotope of hydrogen, has a charge of 1.60 1019 C and a mass of 3.34 1027 kg. It is placed between two oppositely charged plates with a voltage of 2.00 104 V. Find the final maximum speed of the ion if it is initially placed at rest (a) at the positive plate (b) midway between the two plates Chapter 11 Electric field theory describes electrical phenomena. 575 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 576 9. A small charge of +3.0 108 C with a mass of 3.0 105 kg is slowly pulled through a potential difference of 6.0 102 V. It is then released and allowed to accelerate toward its starting position. Calculate (a) the initial work done to move the charge (b) the maximum kinetic energy of the returning charge (c) the final speed of the returning charge 10. An electron, travelling horizontally at a speed of 5.45 106 m/s, enters a parallelplate capacitor with an electric field of 125 N/C between the plates, as shown in the figure below. Extensions 11. Determine whether an electron or a proton would take less time to reach one of a pair of oppositely charged parallel plates when starting from midway between the plates. Explain your reasoning. 12. How can the electric potential at a point in an electric field be high when the electric potential energy is low? 13. In question 10, explain why the resulting motion of an electron, initially travelling perpendicular to the uniform electric field between the two charged parallel plates, will be parabolic and not circular. e TEST 5.45 106 m/s To check your understanding of electrical interactions and the law of conservation of energy, follow the eTest links at www.pearsoned.ca/ school/physicssource. (a) Copy the diagram into your notebook and sketch (i) the electric field lines between the plates (ii) the motion of the electron through the capacitor (b) Determine the force due to the electric field on the electron. (c) Ignoring gravitational effects, calculate the acceleration of the electron. (d) If the electron falls a vertical distance of 6.20 103 m toward the positive plate, how far will the electron travel horizontally between the plates? 576 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/28/08 9:42 AM Page 577 CHAPTER 11 SUMMARY Key Terms and Concepts field test charge source charge electric field line Key Equations F e E q Ep V q k E r q 2 Ep V q Conceptual Overview electric potential energy electric potential (voltage) electron volt electric potential difference Ep W Ep W F d V Vfinal Vinitial E V d Epi Eki Epf Ekf The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. E Fe q E kq r2 electric potential energy calculate define magnitude direction calculate E between more than two charges one-dimensional situations vector electric potential Electric Fields electric potential difference field lines electric field between plates relationship between electric field and distance drawing electric fields around charged objects define a reference point calculate electric potential energy define calculate define calculate Chapter 11 Electric field theory describes electrical phenomena. 577 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 578 CHAPTER 11 REVIEW Knowledge 1. (11.1) Identify the three theories that attempt to explain “action at a distance.” 13. (11.3) Describe the key differences between the electric field surrounding a point charge and the electric field between charged parallel plates? 2. (11.1) How can it be demonstrated that the space around a charged object is different from the space around an uncharged object? 3. (11.1) How does a vector arrow represent both 14. (11.3) Assuming forces in a system are conservative, explain how (a) work done in the system is related to potential energy of the system the magnitude and direction of a vector quantity? (b) the kinetic and potential energy of the 4. (11.2) What is the difference between an electric field vector and an electric field line? 5. (11.2) Two hollow metal objects, with shapes shown below, are charged with a negatively charged object. In your notebook, sketch the distribution of charge on both objects and the electric field lines surrounding both objects. cross-section of hollow sphere cross-section of hollow egg-shaped object system are related Applications 15. Compare the electric potential energy of a positive test charge at points A and B near a charged sphere, as shown below. A B 16. A large metal coffee can briefly contacts a charged object. Compare the results when uncharged electroscopes are touched to the inside and outside surfaces of the can. 6. (11.2) How do electric field lines represent the magnitude of an electric field? 17. A point charge has a charge of 2.30 C. Calculate (a) the electric field at a position 2.00 m from 7. (11.2) Where do electric field lines originate for (a) a negative point charge? (b) a positive point charge? 8. (11.2) Identify two equations that can be used to calculate the magnitude of an electric field around a point charge. the charge (b) the electric force on a charge of 2.00 C placed at this point 18. A charge of 5.00 C is separated from another charge of 2.00 C by a distance of 1.20 m. Calculate (a) the net electric field midway between the 9. (11.2) When do electric charges achieve static two charges equilibrium in a charged object? 10. (11.2) Why do electric charges accumulate at a point in an irregularly shaped object? 11. (11.2) State a convenient zero reference point for electric potential energy (a) around a point charge (b) between two oppositely charged parallel plates 12. (11.2) What equation would you use to calculate the electric potential energy at a certain position around a point charge? (b) the position where the net electric field is zero 19. Find the net electric field intensity at point C in the diagram below. C 0.040 m 2.0 μC A 0.060 m 2.0 μC B 578 Unit VI Forces and Fields 11-PearsonPhys30-Chap11 7/24/08 3:05 PM Page 579 20. A force of 15.0 N is required to move a charge of 2.0 C through a distance of 0.20 m in a uniform electric field. (a) How much work is done on the charge? (b) How much electric potential energy does the charge gain in joules? 21. How much electric potential energy would an object with a charge of 2.50 C have when it is 1.20 m from a point charge of 3.00 C? (Hint: Consider how much electric potential energy the negatively charged object would have when touching the point charge.) 22. Two parallel plates are separated by a distance of 3.75 cm. Two points, A and B, lie along a perpendicular line between the parallel plates and are 1.10 cm apart. They have a difference in electric potential of 6.00 V. (a) Calculate the magnitude of the electric field between the plates. (b) Determine the electric potential between the parallel plates. 23. How much work is required to move a charge perpendicular to the electric field between two oppositely charged parallel plates? 24. A cell membrane is 1.0 107 m thick and has an electric potential difference between its surfaces of 0.070 V. What is the electric field within the membrane? 25. A lithium nucleus (Li3) that has a charge of 4.80 1019 C is accelerated by a voltage of 6.00 105 V between two oppositely charged plates. Calculate the energy, in joules (J) and electron volts (eV), gained by the nucleus. 26. How much electric potential energy, in joules (J) and electron volts (eV), does an alpha particle gain when it moves between two oppositely charged parallel plates with a voltage of 20 000 V? 27. Consider a sphere with a known charge in the electric field around a larger unknown charge. What would happen t
o the electric field at a point if Extensions 28. Explain why electric field lines can never cross. 29. A bird is inside a metal birdcage that is struck by lightning. Is the bird likely to be harmed? Explain. 30. Explain why charge redistributes evenly on the outside surface of a spherical charged object and accumulates at a point on an irregularly shaped charged object. 31. Why can there never be excess charges inside a charged conductive sphere? 32. Describe a simple experiment to demonstrate that there are no excess charges on the inside of a hollow charged sphere. 33. Identify a technology that uses the principle that electric charges accumulate at the point of an irregularly shaped object. Describe how the technology applies this principle. Consolidate Your Understanding Create your own summary of electric field theory by answering the questions below. If you want to use a graphic organizer, refer to Student Reference 3: Using Graphic Organizers. Use the Key Terms and Concepts listed on page 577 and the Learning Outcomes on page 542. 1. Create a flowchart to describe the differences between electric fields, electric potential energy, and electric potential, using non-uniform and uniform electric fields. 2. Write a paragraph comparing the electric fields around various objects and surfaces. Include diagrams in your comparisons. Share your report with a classmate. Think About It Review your answers to the Think About It questions on page 543. How would you answer each question now? (a) the magnitude of the test charge were doubled? (b) the magnitude of the charge producing the e TEST field were doubled? (c) the sign of the charge producing the field were changed? To check your understanding of concepts presented in Chapter 11, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 11 Electric field theory describes electrical phenomena. 579 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 580 Properties of electric and magnetic fields apply in nature and technology. Figure 12.1 Aurora borealis or northern lights The spectacular aurora borealis paints the night sky with shimmering colours in northern latitudes (Figure 12.1). Frequently seen above 60° north, its scientific name translates from Latin into “dawn of the north.” In southern latitudes, where it is seen mainly above 60° south, it is called the aurora australis — “dawn of the south.” Many ancient civilizations created stories to explain these dancing lights in the sky. Some Inuit peoples of northern Canada believed that the sky was a hard dome that arched over Earth. Spirits could pass through a hole in the dome to the heavens, where they would light torches to guide new arrivals. Other Aboriginal traditions spoke of the creator of Earth travelling to the north when he finished his task of creation. There he remained, building large fires to remind his people that he still thought of them. The northern lights were reflections of these fires. What are the auroras and what causes them? Why can they be observed only in the far northern or southern latitudes? Is there a relationship between the auroras and surface activity on the Sun, called solar flares? Are they related to other physical phenomena observed on Earth? Finally, how can an understanding of the science of the auroras aid in the development of new technologies? Your studies in this chapter will help answer these questions. C H A P T E R 12 Key Concepts In this chapter, you will learn about: magnetic fields moving charges in magnetic and electric fields electromagnetic induction Learning Outcomes When you have completed this chapter, you will be able to: Knowledge define electric current as the amount of charge passing a reference point per unit of time describe magnetic interactions in terms of forces and fields compare gravitational, electric, and magnetic fields describe how the work of Oersted and Faraday led to the theory relating electricity to magnetism describe a moving charge as the source of a magnetic field and predict the field’s orientation explain how uniform magnetic and electric fields affect a moving charge describe and explain the interaction between a magnetic field and a moving charge and a conductor explain, quantitatively, the effect of an external magnetic field on a current-carrying conductor describe the effects of moving a conductor in an external magnetic field in terms of moving charges Science, Technology, and Society explain that concepts, models, and theories are often used in interpreting, explaining, and predicting observations explain that technology provides solutions to practical problems explain that scientific knowledge may lead to the development of new technologies and vice versa 580 Unit VI 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 581 12-1 QuickLab 12-1 QuickLab Magnetic Fields in a Bottle Problem What is the shape of a magnetic field? Materials 50 mL of iron filings 450 mL of light cooking oil 1 clear plastic 591-mL pop bottle string 1 cylindrical cow magnet (must be able to fit in the bottle) tape Procedure 1 Pour 50 mL of iron filings into the bottle. 2 Pour cooking oil into the bottle until it is about three-quarters full. 3 Replace the cap on the bottle securely and shake the bottle several times so that the iron filings disperse throughout the oil. Remove the cap. 4 Attach the string to one end of the cow magnet and insert the magnet in the bottle. Make sure the magnet is suspended vertically in the middle of the bottle. Tape the other end of the string to the top of the bottle. 5 Replace the cap on the bottle and place the bottle on a table to allow the mixture to settle. Observe the pattern produced by the iron filings. Questions 1. In your notebook, draw a diagram of the pattern created by the iron filings. 2. 3. Is the pattern created by the iron filings one-, two-, or three-dimensional? Explain your answer. Identify where the density of the iron filings is the greatest and the least. Explain why the filings are distributed this way. 4. From the pattern of the iron filings, is it possible to determine the strength and the direction of the magnetic influence around the magnet? Explain your answer. Think About It 1. Describe a probable cause of the pattern of the iron filings in 12-1 QuickLab. 2. What types of substances produce this influence? 3. What types of objects are affected by this influence? Discuss and compare your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes to your ideas. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 581 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 582 12.1 Magnetic Forces and Fields An ancient Greek legend from about 800 BCE describes how the shepherd Magnes, while tending his flock, noticed that pieces of a certain type of rock were attracted to the nails on his shoes and to his metal staff (Figure 12.2). This phenomenon was called magnetism and, as time passed, further studies of the behaviour of this rock revealed several curious effects. For example, a piece of this rock could either attract or repel another similar piece (Figure 12.3). This effect seemed to result from two different magnetic effects, so investigators thought that there must be two different types of “magnetic ends,” or poles, on the rock. This observation led to the law of magnetism, which states: Like magnetic poles repel and unlike poles attract each other. (a) (b) Figure 12.3 A piece of magnetic rock, held near one end of a similar piece of magnetic rock, would attract at one end (a) and repel at the other end (b). In 1269, Pierre de Maricourt was mapping the position of a magnetized needle placed at various positions on the surface of a spherical piece of this rock. He observed that the directions of the needle formed a pattern that encircled the rock, like meridian lines, and converged at two points on opposite ends of the rock. When this rock was then suspended by a string, the two converging points tended to align along Earth’s north–south axis. This property of the rock earned it the name “lodestone” or “leading stone.” Maricourt called the end pointing northward the north-seeking or north pole and the end pointing southward the south-seeking or south pole. All magnets have both poles. Lodestone, which contains the mineral magnetite (Fe3O4), was later used in the development of compass technology. Figure 12.2 The magnetic effects of certain materials were observed by ancient Greeks as early as 800 BCE info BIT Magnetic poles always exist in pairs. In the 1930s, Paul Dirac (1902–1984) suggested the existence of a particle called a magnetic monopole. To date, all experiments to discover this onepoled particle have failed, but these particles are still under experimental investigation. 582 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 583 Concept Check Figure 12.4 A U-magnet, a circular magnet, and a bar magnet Copy the picture of each magnet in Figure 12.4 into your notebook. Since each magnet must have two poles, label the possible positions of the north and south poles of each magnet. The next big advance in knowledge about magnetism came from the work of William Gilbert. In his book De Magnete, published in 1600, he not only reviewed and criticized past explanations of magnetism but he also presented many important new hypotheses. He compared the orientation of magnetized needles on the surface of a spherical piece of lodestone with the north–south orientation of a compass needle at various locations on Earth’s surface. From this study, he proposed that Earth itself is a lodestone with north and south magnetic poles. Concept Check The north pole of a magnetic compass needle points toward Earth’s magnetic north. What can you conclude about this point on Earth? Gilbert was also intrigued by the forces that magnets could exert on other
magnetic objects. If you suspend a magnet on a string and bring another magnet close to one of its poles, the suspended magnet will rotate, even though there is no visible contact between the two magnets. Magnets appeared to have the ability to exert forces that seemed to originate from the magnetic poles, and they could affect another magnetic object even without contact. The ancient Greeks called this effect “action at a distance.” Recall from chapter 11 that they used the same terminology to describe the effects of electric charges. In attempting to explain the action at a distance caused by a magnet, Gilbert suggested that an invisible “orb of virtue” surrounds a magnet and extends in all directions around it. Other magnetic substances react to a force created by this orb of virtue and move or rotate in response. His orbs of virtue were the beginnings of the idea of “fields” that would revolutionize physics. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 583 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 584 magnetic field: a threedimensional region of influence surrounding a magnet, in which other magnets are affected by magnetic forces Michael Faraday (1791–1867) further developed this concept. He defined a magnetic field as a three-dimensional region of magnetic influence surrounding a magnet, in which other magnets are affected by magnetic forces. The direction of the magnetic field at a given location is defined as the direction in which the north pole of the compass needle points at that location. Some materials, such as iron, act like magnets while in a magnetic field. 12-2 QuickLab 12-2 QuickLab Observing Magnetic Fields Problem How can the magnitude and direction of magnetic fields be observed and analyzed? Questions 1. Describe the cause of the pattern produced by the iron filings. 2. 3. 4. Is the pattern created by the iron filings one-, two-, or three-dimensional? Explain. Identify where the density of the iron filings is the greatest and the least. Explain why the filings are distributed this way. Is it possible to determine the strength and direction of the magnetic field surrounding the magnet from the pattern of the iron filings alone? Explain your answer. 5. From your investigation of the effect of a magnetic field on a compass, what appears to be the direction of the magnetic field around a magnet? Materials 1 bar magnet 1 sheet of paper (216 mm 279 mm) 25 mL of iron filings 1 compass Procedure 1 Lay the bar magnet on a table and place the paper over the magnet. Trace the shape of the magnet on the paper and label the poles. 2 Carefully sprinkle the iron filings onto the surface of the paper. 3 Tap the paper lightly to reinforce the alignment of the iron filings on the sheet. Draw the pattern of the iron filings around the magnet. 4 Clean the iron filings from the paper and replace the paper over the magnet. 5 Place the compass at several positions around the magnet and trace the direction of the compass needle. 584 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 585 Magnetic Fields The magnetic field surrounding a magnet is represented by the symbol B and is measured in teslas (T). A typical bar magnet in the classroom can have a magnetic field of approximately 1 102 T, whereas Earth’s magnetic field is about 5 105 T. The magnetic field is a vector quantity, so it is represented by a vector arrow. In diagrams, the length of the vector arrow represents the magnitude of the field, and the direction of the arrow represents the direction of the field at a point. You can also use compasses to show the direction of the magnetic field at any position surrounding a magnet, as illustrated in Figure 12.5. Figure 12.5 shows that, in general, this direction is from the north to the south pole of the magnet. info BIT Magnetic field lines run parallel to Earth’s surface only at the equator. As they reach the magnetic poles, they gradually dip toward the surface. At the poles, the magnetic field lines point perpendicular to Earth’s surface. Navigators in the far north or south must be aware that the magnetic compasses may be of limited use in those areas Figure 12.5 The direction of a magnetic field is the direction of the force on the north pole of a compass placed in the field. To represent the entire magnetic field surrounding a magnet, it would be necessary to draw arrows at an infinite number of points around the magnet. This is impractical. Instead, you can draw a few magnetic field lines with a single arrow head indicating the direction of the magnetic field. To find the field direction at a given point, move the arrow head along the field line through that point so that it keeps pointing in the direction of the tangent to the field line. The field lines in Figure 12.6 are a map of the magnetic field with the following features: • Outside a magnet, the magnetic field lines point away from the north pole of a magnet and toward the south pole. • The closeness of the lines represents the magnitude of the magnetic field. e LAB For a probeware activity where you use a magnetic field sensor to determine the relationship between the distance from a magnet and the intensity of the field, go to www.pearsoned.ca/ school/physicssource. S N S N (a) (b) Figure 12.6 (a) The pattern of iron filings surrounding a bar magnet outlines the magnetic field. (b) Magnetic field lines, representing the direction and magnitude of the magnetic field, can replace the iron filings. The number of magnetic field lines that exit a magnetic material is equal to the number of magnetic field lines that enter the magnetic material, forming closed loops. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 585 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 586 PHYSICS INSIGHT Before the adoption of SI units, magnetic fields were sometimes measured in a CGS unit called the gauss (G). You might see this unit in some older books. 1 T 104 G. Table 12.1 shows some examples of magnetic field strengths. Table 12.1 Magnetic Field Strengths Physical system Magnetic field (T) Earth Bar magnet Sunspots High field magnetic resonance imaging device (MRI) Strongest humanmade magnetic field Magnetar (magnetic neutron star) Concept Check 5 105 1 102 1 101 15 40 1 1011 Figure 12.7 shows the patterns produced by iron filings that are influenced by the magnetic fields of one or two magnets. Sketch the magnetic field lines in each case. S N N N N S N S Figure 12.7 Concept Check List at least two similarities and two differences between gravitational, electric, and magnetic fields. 586 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/28/08 10:04 AM Page 587 Cause of Magnetism The force of magnetic repulsion between like poles of magnets is the same force that causes the almost frictionless ride of the Maglev (magnetically levitated) train (Figure 12.8). What is the source of this “magnetic levitation” on the train? info BIT Oersted was among the first to recognize the talent of the writer Hans Christian Andersen and encouraged him when he began writing his now famous fairy tales. (a) (b) Figure 12.8 (a) The force of magnetic repulsion between like poles can cause one magnet to levitate over another. (b) The Maglev train, developed in Japan, floats several centimetres above the guideway, providing a smooth and almost frictionless ride. Experiments by early investigators revealed many facts about the magnetic fields surrounding magnets and their effects on magnetic objects. However, the actual cause of magnetism eluded scientists until 1820. While demonstrating to students that the current passing through a wire produces heat, Danish professor Hans Christian Oersted (1777–1851) noticed that the needle of a nearby compass deflected each time the circuit was switched on. This experiment led Oersted to the important conclusion that there is a relationship between electricity and magnetism, at a time when electricity and magnetism were considered separate phenomena. He proved that electric current was a cause of magnetism. Following his initial observations, it was later shown that if electric current was in a straight line, the magnetic field formed a circular pattern (Figure 12.9(a)), and if the electric current was circular, the magnetic field was straight within the coil (Figure 12.9(b)). (a) battery N (b) electron flow electron flow S Figure 12.9 (a) A current passing through a straight conducting wire produces a magnetic field, represented by concentric red circular lines around the wire. (b) A current passing through a coil produces a magnetic field, represented by red circular lines, with poles similar to those of a bar magnet. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 587 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 588 PHYSICS INSIGHT The observation of a magnetic field produced by a moving charge depends on the frame of reference of the observer. If you are stationary and the charge moves past you, you observe a magnetic field. However, if you are moving along with the charge, the charge is stationary relative to you, so you do not observe a magnetic field. Left-hand Rules for Magnetic Fields A useful left-hand rule to determine the direction of the magnetic field is the wire-grasp rule described in Figure 12.10. To determine the direction of the magnetic field produced by a moving charge, use the left-hand wire-grasp rule if the moving charge is negative. (If the moving charge is positive, then use the right-hand wire-grasp rule.) direction of magnetic field lines (a) e direction of electron flow left hand conductor magnetic field lines (b) core e direction of electron flow magnetic field line e direction of magnetic field Figure 12.10 Left-hand rule for direction of a magnetic field due to moving charges: (a) If the conducting wire is straight, then the thumb indicates the direction of the straight current a
nd the cupped fingers indicate the direction of the circular magnetic field. (b) If the current is in a coil of conducting wire, the cupped fingers indicate the circular current and the straight thumb indicates the direction of the straight magnetic field within the coil Using the Wire-grasp Rule 1. Sketch the following diagrams into your notebook. (a) e Indicate the direction of the magnetic field lines and the direction of current in the wire, as required. (b) N S Electromagnets As shown in Figure 12.9(b), current in a circular loop or coil of wire produces a magnetic field like that of a bar magnet. An electromagnet uses a current-carrying coil of wire to generate a magnetic field that is easy to switch on and off. The strength of an electromagnet can be influenced by: • increasing the current through the wire • increasing the number of loops in the coil • increasing the size of the loops in the coil • changing the core of the coil Powerful electromagnets have many industrial uses, such as lifting steel parts, machinery, or scrap iron. Electromagnets are widely used to remotely operate switches or valves. Often, a valve is activated by a metal rod that is drawn into the core of the electromagnet when current electromagnet: a magnet having its magnetic field produced by electric current flowing through a coil of wire 588 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 589 flows through the coil. Such mechanisms, called solenoids, are common in washing machines, dishwashers, furnaces, and industrial machinery. Figure 12.11 shows two applications of electromagnets. solenoid: an electromagnet that operates a mechanical device (a) (b) Figure 12.11 (a) A lifting magnet (b) An appliance solenoid Domain Theory and Magnetization In some atoms, the configuration of the electrons is such that their movement generates a tiny magnetic field. In ferromagnetic materials, such as iron, nickel, and cobalt, the magnetic fields of adjacent atoms can align to reinforce each other, forming small regions, or domains, with intense magnetic fields. Domains generally range from 0.001 mm to 1 mm across, and may contain billions of atoms. The orientations of the magnetic fields of the various domains are normally random, so their magnetic fields largely balance each other, leaving the material with little or no overall magnetization. However, the size of a domain and the direction of its magnetic field are relatively easy to change. An external magnetic field can cause the domains to align, thus magnetizing the material (Figure 12.12). The small black arrows in Figure 12.12 indicate the orientation of the magnetic field of an individual domain. S N (a) (b) Figure 12.12 (a) When the magnetic fields of atoms in a region line up, they create a magnetic domain in the substance. (b) Aligning the domains produces a magnet. A typical ferromagnetic object has vastly more domains than the diagrams can show. If you hang an iron nail by a string and bring a magnet close to the nail, the nail will rotate toward the magnet, even before they touch. The nail is not a magnet with distinct poles, yet a magnetic attraction exists between it and the magnet. When the magnet is close to the nail, the domains in the nail that are oriented for attraction to the magnet increase in size while the other domains shrink. When the magnet is moved away again, the domains in the nail tend to return to random info BIT Geophysicists theorize that circulating currents of ions in the molten core of Earth produce its magnetic field. ferromagnetic: having magnetic properties like those of iron domain: a region of a material in which the magnetic fields of most of the atoms are aligned e WEB All magnetic substances can be classified as one of the following: • ferromagnetic • antiferromagnetic • ferrimagnetic • paramagnetic • diamagnetic Find out what distinguishes one type of magnetic substance from another. Begin your search at www.pearsoned.ca/ school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 589 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 590 orientations and the nail loses most of its magnetization. This example illustrates induced magnetization. The nail will be much more strongly magnetized if it is stroked with a pole of the magnet. The magnetic fields of many of the domains in the nail will align along the direction of motion of the magnet. This magnetization is strong enough that the nail will remain somewhat magnetized when the magnet is removed. Concept Check A filing cabinet has been in one position for a long time. It is made of ferromagnetic material, so it can become a permanent magnet. If you hold a compass near the top of the filing cabinet, the compass needle points toward the filing cabinet. If you hold the compass near the bottom of the filing cabinet, the opposite end of the compass points toward the cabinet. Has the cabinet been magnetized by Earth’s magnetic field? Or has the cabinet become magnetized by the magnetic compass? Explain your answer. Magnetism in Nature The effects of magnetism have been known since early civilizations, but the causes of magnetic behaviour are only now being revealed. A modern understanding of magnetic phenomena began with the development of field theory to replace “action at a distance.” The symmetry of nature enabled scientists to use the same field theory to describe the gravitational field surrounding any mass, the electric field surrounding any charge, and the magnetic field surrounding any magnet. Oersted’s investigations, which revealed a relationship between electricity and magnetism, ultimately led to the domain theory to explain a cause of magnetism. As scientists probed deeper into the mysteries of magnetism, many more answers were found. However, the tremendous significance of magnetism has only recently been understood in explaining phenomena and producing technological applications. In the field of biology, for example, researchers have found that certain organisms have ferromagnetic crystals consisting of magnetite in their bodies. Some bacteria use these magnetite crystals to help orient themselves within Earth’s magnetic field. Bees and pigeons have magnetite crystals within their brains to help with navigation. The human brain also has these magnetite crystals, but their function is not clear. It is known that an external magnetic field can disrupt the neural activity in the parietal lobe on one side of the human brain. 590 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 591 Understanding magnetism has also led to important technological advancements. These advancements range from simple applications, such as refrigerator magnets, magnetic stripes on cards, and magnetic audiocassette or VCR tapes, to more complicated applications involving magnetic levitation, such as the Maglev train and magnetic resonance imaging (MRI) machines used as a diagnostic tool in health care. Although much has been achieved, there are still many secrets of magnetism to uncover. THEN, NOW, AND FUTURE Earth’s Magnetic Field William Gilbert’s “Terrella” experiment in the 1500s compared the magnetic field of Earth to that of a bar magnet. From that time, Earth has been considered to be a huge magnet, with similar magnetic properties to a much smaller, ordinary magnet. This observation was successful in explaining many phenomena. However, care must be taken in comparing the causes of magnetic behaviour in Earth and in a bar magnet. If the cause of magnetism in substances is the motion of charges, scientists are not quite convinced that the motion of charges within Earth’s molten core is responsible for Earth’s magnetism. They know that Earth’s molten core is simply too hot for atoms to remain aligned and exhibit any magnetic properties. Other probable causes of Earth’s magnetic field could be convection currents rising to the cooler surface of Earth, or the motion of charges in the upper ionosphere. The most acceptable and probable cause, though, is the motion of charges in the molten part of Earth, just beneath the crust (Figure 12.13). Whatever the cause of Earth’s magnetic behaviour, it is known that the magnetic field of Earth is not stable. Molten rock within the interior of Earth has no magnetic properties. However, when molten rock rises to the surface, it cools and solidifies, and its domains orient themselves in line with Earth’s magnetic field at the time. When samples of rock from different strata formed throughout geological times are tested, evidence shows that there are times when not only the magnitude of Earth’s magnetic field changed, but also its direction. In the past five million years, more than 20 reversals have occurred, the last one about 780 000 years ago. Coincidentally, modern humans emerged during this time period. One possible effect of a zero magnetic field, during a reversal, would be an increase in the cosmic ray intensity at Earth’s surface. Normally, the magnetic field shields Earth from harmful radiation from space. Fossil evidence indicates that periods of no protective magnetic field have been effective in changing life forms. Evidence that these types of changes could have occurred also comes from heredity studies of fruit flies when exposed to X rays. We cannot know precisely when the next reversal will occur. However, evidence from recent measurements indicates a decrease in the magnitude of Earth’s magnetic field of about 5% in the last 100 years. Based on this evidence, Figure 12.13 This computer model of Earth shows the molten outer core surrounding the inner core (the small circle). The right side shows the molten currents. The left side shows the magnetic field lines that extend outward through the rest of Earth’s interior. another reversal of Earth’s magnetic field may occur within the next 2000 years. Questions 1. Can the motion of charges in Earth’s core create domains? Expla
in your answer. 2. What is the most probable cause of Earth’s magnetic behaviour? 3. What evidence is there on Earth that its magnetic field is not stable? Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 591 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 592 12.1 Check and Reflect 12.1 Check and Reflect Knowledge 1. What is the law of magnetism? 2. Explain your answers to the following: (a) Does every magnet have a north and a south pole? 10. List at least two differences and two similarities between (a) gravitational and electric fields (b) gravitational and magnetic fields (c) electric and magnetic fields (b) Does every charged object have positive and negative charges? 11. Using the domain theory, explain the following observations: 3. How did William Gilbert determine that Earth was a magnet? 4. What is the most probable cause of magnetism in (a) a bar magnet? (b) Earth? 5. What accidental discovery did Oersted make? 6. What is the shape of the magnetic field (a) around a straight current-carrying conductor? (b) within a coil of conducting wire carrying a current? (a) A magnet attracts an unmagnetized ferromagnetic material. (b) Stroking a nail with a magnet magnetizes the nail. (c) A metal table leg affects a compass. 12. Why does dropping or heating a bar magnet decrease its magnetic properties? 13. Consider a bar magnet and Earth, as shown below. Describe the similarities and the differences of their magnetic fields. S N Earth Applications 7. What would happen to a magnet if you broke it into two pieces? Extensions 8. A negatively charged sphere is approaching you. Describe the magnetic field surrounding the sphere and its direction. What would happen if the sphere were positively charged? 9. A spinning top is charged negatively and is spinning clockwise, as observed from above. Describe the magnetic field created by the spinning top and its direction. 14. Why is it difficult to get an accurate bearing with a magnetic compass near the poles? 15. Do magnetic field lines always run parallel to the surface of Earth? Explain your answer. 16. If a current-carrying wire is bent into a loop, why is the magnetic field stronger inside the loop than outside? e TEST To check your understanding of magnetic forces and fields, follow the eTest links at www.pearsoned.ca/school/physicssource. 592 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 593 12.2 Moving Charges and Magnetic Fields Near the end of the 1800s, researchers were fascinated by a new technology called the cathode-ray tube (CRT), shown in Figure 12.14. It consisted of a glass tube from which air had been evacuated, and it had a positive plate (anode) at one end and a negative plate (cathode) at the other end. These new tubes used electric fields to accelerate a beam called a cathode ray through a large potential difference. The beam would “light up” the fluorescent screen at the end of the tube. Scientists were unsure whether this beam was a type of electromagnetic radiation (similar to light), a neutral particle, or a charged particle. They initially called it a cathode ray because it appeared to originate from the cathode plate. This technology not only enabled the discovery of the electron in 1897 (see section 15.1), but also led to the later development of many other technologies, including television. Until recently, the image in most TVs was produced by an electron beam striking a fluorescent screen in a CRT. The Motor Effect Deflecting charged particles involves an interaction of two magnetic fields. A charged particle in uniform motion produces a circular magnetic field around it (the wire-grasp rule). Now suppose this charged particle enters an external magnetic field, produced between the faces of two opposite magnetic poles. The interaction of the circular magnetic field of the charge and the external magnetic field produces a magnetic force that acts on the particle to deflect it, as shown in Figure 12.15. This magnetic force is also called the motor effect force (F m) because it causes the rotation of a loop of current-carrying wire. This rotation is fundamental in the operation of an electric motor. Figure 12.14 The image on the fluorescent screen at the left end of this cathode-ray tube shows that the rays originate from the negative terminal at the right end. motor effect force: the deflecting force acting on a charged particle moving in a magnetic field Figure 12.15 (a) The cathode ray accelerates in a straight line when it is only influenced by the electric field produced between the cathode and anode plates in a vacuum tube connected to a high-voltage source. (b) A cathode ray will deflect as shown when it is also under the influence of an external magnetic field. (a) (b) In Figure 12.16, the straight, horizontal lines represent the external magnetic field of the magnetic poles, and the dashed lines represent the magnetic field surrounding the moving charge (using the left-hand rule). In the “replacement magnet” method of illustration, tiny magnets are drawn along the field lines to reinforce the idea of the direction and the effects of the interaction of the two magnetic fields. The represents Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 593 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 594 PHYSICS INSIGHT represents a direction into the page, like the fletching of an arrow receding from you. • represents a direction out of the page, like the point of an arrow approaching you. In Figure 12.16 represents negative charge movement into the page. negative charge moving into the page. In Figure 12.16, below the moving charge, the external magnetic field and the magnetic field surrounding the charge are in the same direction. Above the moving charge, the two magnetic fields are in opposite directions. Fm causing deflection Fm Fm S N Figure 12.16 The combined magnetic forces due to the two magnetic fields cause the moving charge to deflect (F perpendicular to the direction of the external magnetic field. m) in a direction perpendicular to its direction of motion and Since the external magnetic field is fixed, the combined effect of m) on the the two magnetic fields produces a net magnetic force (F moving charge. As a result, the moving charge deflects upward (toward the top of the page). The deflecting force is always perpendicular to the direction of both the external magnetic field and the motion of the moving charge, as shown in Figure 12.16. This property distinguishes a magnetic field from electric or gravitational fields. Since the direction of the electric force or gravitational force can be parallel to their respective fields, these fields can be used to change the speed of a charged particle. The magnetic force, on the other hand, is always perpendicular to the velocity of the charged particle. A magnetic force can never do any work on a charged particle, nor can it change the speed or kinetic energy of a charged particle. Since force is not in the direction of the displacement, then there can be no work done on the object. Only the direction of the charged particle’s path may be changed. Left-hand Rule for Deflection Consider a negatively charged particle travelling perpendicular to an external magnetic field. When it enters the region of a uniform magnetic field, it is deflected in a direction perpendicular to both the original direction of charge movement and the direction of the external magnetic field. A useful hand rule to determine the direction of deflection is the left-hand rule shown in Figure 12.17: 594 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 595 • The thumb indicates the direction of the initial charge movement. • The extended fingers indicate the direction of the external magnetic field, from north to south. • The palm faces in the direction of the magnetic force. If the moving charge is positive, use the righthand rule, with the thumb, fingers, and palm indicating the directions of the same quantities as in the left-hand rule. S magnetic field B e N Fm v Figure 12.17 How to use the left-hand rule to determine the deflection of a charged particle Using the Left-hand Rule for Deflection 1. In your notebook, sketch the direction of the unknown variable in each situation. (a) external magnetic field (b) negative charge Fm e motion B Fm 12-3 Inquiry Lab 12-3 Inquiry Lab Using Hand Rules with a Cathode-ray Tube — Demonstration (c) positive charge motion B Fm Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question Do the hand rules predict the deflection of a cathode beam in an external magnetic field? Materials and Equipment 1 cathode-ray tube 1 high-voltage source 1 strong bar magnet S N cathode-ray tube high-voltage source Figure 12.18 CAUTION! High voltage. Be very careful around electrical equipment to avoid shocks. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 595 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 596 Procedure Analysis 1 Connect the cathode-ray tube to the high-voltage 1. What types of particles would be attracted from the source, as shown in Figure 12.18. Identify the cathode and anode of the CRT. 2 Turn on the current supply. Observe the path of the cathode rays that are produced. 3 Carefully hold the north pole of a bar magnet near one side of the centre of the cathode tube, in the horizontal plane. Note the direction in which the cathode rays are deflected. 4 Repeat the procedure in step 3 with the south pole of the bar magnet. 5 Repeat steps 3 and 4 by holding the magnet on the other side of the cathode tube, in the horizontal plane and then in the vertical plane. Observe the direction of the deflection of the beam in each case. cathode to the anode of the CRT? Does the hand rule applicable for these particles predict all of the deflections that you observed? 2. Wh
at types of particles would be attracted from the anode to the cathode of the CRT? Does the hand rule for these particles predict any of the deflections that you observed? 3. Explain how the observed deflections show that cathode rays consist of charged particles. 4. Can you use the hand rules to determine the type of charge carried by cathode rays? Explain why or why not. Concept Check Compare the magnetic force of an external magnetic field on a moving charged particle with: • the gravitational force of Earth on the mass of the charged particle • the electric force due to another nearby charged particle Charged Particle Motion in a Magnetic Field The direction of the initial motion of a charged particle in an external magnetic field determines how the charged particle will deflect. Figure 12.19 shows what can happen to a charged particle as it enters an external magnetic field: (a) If the initial motion of the charged particle is parallel to the external magnetic field, then there is no effect. (b) If the initial motion of the charged particle is perpendicular to the external magnetic field, the charge is deflected in a circular arc. (c) If the initial motion of the charged particle is at an angle to an external magnetic field, the charge deflects in a circular motion that will form a helical path. e SIM Explore the motion of a charged particle in a uniform magnetic field. Follow the eSim links at www.pearsoned.ca/school/ physicssource. 596 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 597 (a) (c) v v v (b) B B (d) B Fm Fm Fm B Figure 12.19 (a) When the charged particle’s velocity is parallel to the external magnetic field (B), the charged particle’s path is a straight line. (b) The charged particle’s motion is perpendicular to the magnetic field, so the particle is deflected in a circular arc. (c) The charged particle’s motion is at an angle to the magnetic field, so the particle follows a helical path. (d) This side view from the left shows the magnetic force acting as the centripetal force that causes the charge to follow a circular path. Oppositely charged particles deflect in opposite directions in a magnetic field (Figure 12.20). If the magnitude of the external magnetic field is large enough, the field can cause circular motion that remains contained in the magnetic field. In this circular motion, the centripetal force is the magnetic force. Magnetic deflection of charged particles is the underlying principle for useful powerful analytical and research tools such as mass spectrometers and particle accelerators. Unit VIII presents these devices and their applications in science, medicine, and industry. Auroras Tremendous expulsions of magnetic energy from the solar atmosphere, called solar flares, expel streams of charged particles at speeds around 10% of the speed of light (Figure 12.21). When some of these particles strike Earth’s magnetic field, they are deflected by the magnetic force and spiral in a helical path along Earth’s magnetic field lines. These particles enter the atmosphere as they approach Earth’s magnetic poles, and collide with air molecules. These collisions excite the atoms of the air molecules, in a process that will be described in Chapter 15, causing them to emit visible light that we see as the aurora. The process repeats because Earth’s nonuniform magnetic field produces a magnetic force component that causes the charged particles to reverse their direction of motion, travelling to Earth’s opposite pole. The same auroral effect is produced at this pole, and the process continues to repeat as the charged particles oscillate back and forth between the poles, trapped in a type of “magnetic bottle” called the Van Allen belt. N S upward deflection downward deflection Figure 12.20 A magnetic field deflects moving oppositely charged particles in opposite directions, as shown. e WEB Research the formation of the Van Allen belts. Consider the shape of the Van Allen belt on the side of Earth facing the Sun and on the side of Earth away from the Sun. What is the cause of this difference in shape? Begin your search at www.pearsoned.ca/ school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 597 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 598 Figure 12.21 This composite image shows the cause of the aurora borealis. Streams of high-energy charged particles erupt from the Sun (far left). They are deflected by Earth’s magnetic field toward the poles, creating the bright ring shown in the satellite image of Earth (centre). There they interact with air molecules in the atmosphere to produce the aurora (far right). Calculating the Magnetic Force By studying the different types of deflections, scientists can also explain the complex deflection of charged particles entering a magnetic field at an angle, such as the particles that cause the auroras. The magnitude of the deflecting force (F m ) depends on all of the following: • the magnitude of the moving charge (q) • the magnitude of the perpendicular velocity component (v) ) • the magnitude of the external magnetic field (B The magnitude of the deflecting force can be calculated using this equation: F m qvB where q is the magnitude of moving charge in coulombs (C); v is the component of the speed perpendicular to the magnetic field in metres per second (m/s); and B is the magnitude of the external magnetic field in teslas (T). Example 12.1 describes how to calculate the magnetic force on a charge moving perpendicular to an external magnetic field. When the velocity of the charge is not perpendicular to the magnetic field, you can use trigonometry to find the perpendicular component: v v sin where is the angle between the charge’s velocity, v, and the magnetic field, B . v v v 598 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 599 Example 12.1 An electron is travelling at 3.20 105 m/s perpendicular to an external magnetic field of magnitude 2.20 101 T (Figure 12.22). Determine the magnetic force acting on the electron. S N S N e Figure 12.22 Given q charge on 1 electron 1.60 1019 C 2.20 101 T B v 3.20 105 m/s Required m) magnetic force (F Analysis and Solution Determine the magnitude of the magnetic deflecting force: F m qvB (1.60 1019 C)(3.20 105 m )(2.20 101 T) s 1.13 1014 N Since the charge is negative, use the left-hand rule for deflection to determine the direction of the magnetic force. • Thumb points in the direction of the charged particle’s movement, into the page. • Extended fingers point in the direction of the external magnetic field, to the right of the page (north to south). • Palm points in the direction of the magnetic deflecting force, toward the top of the page. Practice Problems 1. A proton with a charge of 1.60 1019 C is travelling with a speed of 3.50 104 m/s perpendicularly through an external magnetic field of magnitude 4.20 104 T. Determine the magnitude of the magnetic deflecting force on the proton. 2. An ion with a charge of 3.20 1019 C and a speed of 2.30 105 m/s enters an external magnetic field of 2.20 101 T, at an angle of 30, as shown in the figure below. Calculate the magnitude of the magnetic deflecting force on the ion. 30° S N 3. A negatively charged sphere travels from west to east along Earth’s surface at the equator. What is the direction of the magnetic deflecting force on the sphere? Answers 1. 2.35 1018 N 2. 8.10 1015 N 3. Downward toward Earth’s surface Paraphrase The magnetic force is 1.13 1014 N [upward] (toward the top of the page). Often, a charged particle may be influenced by a combination of two fields, such as a magnetic field and a gravitational field, or a magnetic field and an electric field. “Crossed-field” devices are technologies that use both magnetic and electric fields. An example is the magnetron, which produces microwaves in microwave ovens. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 599 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 600 Example 12.2 A carbon ion, with a mass of 2.01 1026 kg and a positive charge of magnitude 1.60 1019 C, enters the region of an external magnetic field of magnitude 6.32 105 T, as shown in Figure 12.23. Find the perpendicular speed at which the magnetic deflecting force will balance the gravitational force such that the carbon ion will travel in a straight line. Practice Problems 1. An electron, with a charge of magnitude 1.60 1019 C and a mass of 9.11 1031 kg, is travelling west along the surface of Earth at the equator. If the magnitude of the magnetic field at this location is 5.00 105 T, what minimum speed must the electron maintain to remain at the same height above Earth’s surface? 2. Ions, with a charge of 1.60 1019 C and a mass of 8.12 1026 kg, travel perpendicularly through a region with an external magnetic field of 0.150 T. If the perpendicular speed of the ions is 8.00 104 m/s, determine (a) the magnitude of the deflecting force on the ion (b) the radius of curvature of the motion of the deflected ion Hint: The magnetic deflecting force is the centripetal force. F F m c v2 m qvB r Answers 1. 1.12 106 m/s 2. (a) 1.92 1015 N (b) 0.271 m S N carbon ion Figure 12.23 Given m 2.01 1026 kg 6.32 105 T B q 1.60 1019 C g 9.81 N/kg Required speed (v) at which the magnitudes of the magnetic force, F m and the gravitational force, F g , are equal , Analysis and Solution The gravitational force on the carbon ion has a magnitude of mg and is directed downward (toward the bottom of F g the page). The magnetic force on the carbon ion has a magnitude of qvB F m the page). and must be directed upward (toward the top of F net F m F g But the magnetic deflecting force and the gravitational force balance (Figure 12.24), so F net 0. Therefore, F m qvB F g mg g v m B q (2.01 1026 kg)9.81 N k g (6.32 105 T)(1.60 1019 C) 1.95 102 m/s Fm Fg Paraphrase The carbon atom will travel in a straight line if its speed is 1.95 102 m/s. F
igure 12.24 600 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 601 In this section, you have studied the deflection of a moving charged particle in a magnetic field. Applying this science, you learned not only the importance of this phenomenon in technologies, such as a television and a magnetron, but also the significance of this phenomenon in protecting Earth from harmful cosmic radiations. The magnetic field of Earth, in deflecting dangerous charged particles from striking Earth’s surface, also produces one of the most beautiful and spectacular natural light shows—the aurora. 12.2 Check and Reflect 12.2 Check and Reflect Knowledge 1. Why is a cathode ray called a cathode ray? 2. What is the difference between a magnetic field vector arrow and a magnetic field line? 3. An electron and a proton, both with the same perpendicular velocity, enter a region with a uniform external magnetic field. What can you state about the deflections of both particles? 4. Describe the key differences in how magnetic and electric fields affect a moving charged particle. Applications 5. A positively charged lithium ion is travelling horizontally along Earth’s surface. Describe the deflection due to the magnetic force if the ion travels (a) south to north (b) east to west (c) upward into the atmosphere 6. A proton with a speed of 2.00 105 m/s enters an external magnetic field of magnitude 0.200 T. Calculate the magnitude of the deflecting force if the proton enters (a) perpendicular to the magnetic field (b) at an angle of 35.0 to the field 7. A 0.020-g metal ball with a charge of 3.0 C is thrown horizontally along Earth’s equator. How fast must the ball be thrown so that it maintains the same height, during its motion tangential to Earth’s surface, if the magnitude of Earth’s magnetic field is 5.0 105 T? 8. An alpha particle, with a charge of 2 1.60 1019 C, is travelling perpendicularly through a magnetic field of magnitude 2.00 102 T at a speed of 1.02 105 m/s. What minimum gravitational force is required to suspend the alpha particle at the same position above Earth’s surface? 9. Electrons in the picture tube of a television are accelerated to a speed of 1.30 106 m/s. As they travel through the tube, they experience a perpendicular magnetic field of magnitude 0.0700 T. What is the radius of deflection of the electrons in the tube? 10. A cosmic ray proton travelling through space at 4.38 106 m/s deflects in a circular arc with a radius of 5.50 106 m. What is the magnitude of the magnetic field at that point in space? Extensions 11. Why are auroras seen only at higher latitudes? e TEST To check your understanding of moving charges and magnetic fields, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 601 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 602 12.3 Current-carrying Conductors and Magnetic Fields info BIT The first sensitive meter to measure small currents was developed by Luigi Galvani in the 1700s. While dissecting a frog’s leg, he noticed that an electric current caused the frog’s leg to twitch. He realized that he had accidentally discovered a method of detecting small currents and used this discovery to design the galvanometer. current: the quantity of charge that flows through a wire in a given unit of time ampere: the flow of 1 C of charge past a point in a conductor in 1 s 602 Unit VI Forces and Fields Figure 12.25 A galvanometer and an electric motor, like the one in this lawn mower, apply magnetic fields produced by a flow of charge. Two of the most common applications of magnetic fields acting on moving charged particles are meters (such as ammeters, voltmeters, and galvanometers) and electric motors (Figure 12.25). Although these technologies appear to be different from the technology of the television, the basic operating principle of all these technologies is similar. Recall from earlier science studies that a galvanometer is a device for detecting and measuring small electric currents. How does a galvanometer operate? How is its operation similar to the technologies of the electric motor and television? Electric Current Recall from earlier science courses that electric current is the movement of charged particles. It can be defined more precisely as the quantity of charge that flows through a wire in a given unit of time. The unit for current, the ampere (A), is a measure of the rate of current. The ampere is an SI base unit. A current of 1 A is equivalent to the flow of 1 C of charge past a point in a conductor in 1 s. In other words, 1 A 1 C/s. For example, the effective value of the current through a 100-W light bulb is about one ampere (1 A) of current. The ampere is named in honour of the French scientist André-Marie Ampère (1775–1836), who is renowned for his analysis of the relationship between current and magnetic force. This equation shows the relationship between current and charge: q I t where I is the current in amperes, q is the magnitude of charge in coulombs, and t is the time elapsed in seconds. 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 603 Example 12.3 Calculate the current in a wire through which 20.0 C of charge passes in 4.00 s. Given q 20.0 C t 4.00 s Required current (I) Analysis and Solution To calculate the current, use the equation I q t .0 C 2 0 4 s 0 .0 5.00 C s 5.00 A Paraphrase The current in the conducting wire is 5.00 A. Practice Problems 1. A lightning strike transfers 20.0 C of charge to the ground in 1.00 ms. Calculate the current during this lightning strike. 2. If the current in a household appliance is 5.00 A, calculate the amount of charge that passes through the appliance in 10.0 s. Answers 1. 2.00 104 A 2. 50.0 C Magnetic Force on a Current-carrying Conductor In a CRT picture tube, powerful external magnetic fields are used to deflect moving electrons to produce an image on a screen. To analyze the operation of a galvanometer or electric motor, and to reveal the similarity of their operation to that of a television, consider the movement of electrons as a current in a wire conductor. When there is an electric current in a wire that is perpendicular to an external magnetic field, each electron experiences a magnetic force caused by the interactions of its own magnetic field and the external magnetic field (Figure 12.26). You can observe the effect of this force. The magnetic force causes the electrons to deflect upward. However, the electrons cannot escape the wire, so if the magnetic force on the electrons is great enough, the whole wire will rise upward, opposite to the force of gravity. The magnetic force on a conducting wire is the same as the magnetic m) that you studied in section 12.2. deflecting force on a moving charge (F S N S N e Figure 12.26 A current of electrons passes through a conducting wire lying perpendicular to an external magnetic field. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 603 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 604 PHYSICS INSIGHT Remember: If the moving charges are negative, use your left hand; and if the moving charges are positive, use your right hand. Left-hand Rule for Magnetic Force To determine the direction of the magnetic force, you can use the lefthand rule, as shown in Figure 12.27: • Your thumb indicates the direction of electron flow in the conductor. • Your extended fingers point in the direction of the external magnetic field. • Your palm indicates the direction of the magnetic deflecting force on the wire. S N S N e Fm v B Figure 12.27 The left-hand rule for determining the direction of magnetic force To calculate the magnitude of the magnetic force for a length of current-carrying conducting wire, use the equation F m IlB where I is the current measured in amperes; l is the length of the wire perpendicular to the magnetic field in metres; B is the magnitude of the external magnetic field in teslas; and F is m the magnitude of the magnetic force in newtons. The Galvanometer In the operation of the galvanometer, a coil of wire is mounted to allow for movement within the strong magnetic field of the permanent magnet (Figure 12.28). The coil turns against a spring with an attached needle pointing to a calibrated scale. When there is a current in the coil, the magnetic forces cause the coil to rotate. The greater the current, the greater the rotation, as registered on the scale by the needle. The galvanometer, which measures very small currents, can be made to measure larger currents (ammeter) by connecting a small resistance in parallel, and to measure larger potential differences (voltmeter) by connecting a large resistance in series. The magnetic force produced on a current-carrying wire can be demonstrated in the 12-4 QuickLab on page 606. N S current restoring spring Figure 12.28 A schematic diagram of a galvanometer reveals all the essential components in its operation. 604 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 605 Example 12.4 An 8.50-cm length of conducting wire lies perpendicular to an external magnetic field of magnitude 4.20 mT, as shown in Figure 12.29. If there is a negative charge flow of 2.10 A in the conductor, calculate the magnitude and determine the direction of the magnetic force on the wire. S N S N Figure 12.29 Given l 8.50 cm 8.50 102 m 4.20 mT 4.20 103 T B I 2.10 A Practice Problems 1. A 0.500-m length of conducting wire carrying a current of 10.0 A is perpendicular to an external magnetic field of magnitude 0.200 T. Determine the magnitude of the magnetic force on this wire. 2. A thin conducting wire 0.75 m long has a mass of 0.060 kg. What is the minimum current required in the wire to make it “float” in a magnetic field of magnitude 0.15 T? Required magnitude and direction of the magnetic force on the wire (F m) Answers 1. 1.00 N 2. 5.2 A Analysis and Solution Determine the magnitude of the m
agnetic force: F m IlB (2.10 A)(8.50 102 m)(4.20 103 T) 7.50 104 N Use the left-hand rule to determine the direction of the magnetic force, because the moving charges are negative: • Thumb points in the direction of the charge movement or current, into the page. • Extended fingers point in the direction of the external magnetic field, to the right of the page (north to south). • Palm points in the direction of the magnetic force, to the top of the page. Paraphrase The magnetic force is 7.50 104 N [upward] (toward the top of the page). Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 605 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 606 12-4 QuickLab 12-4 QuickLab Demonstration of a Current-carrying Conductor in a Uniform Magnetic Field Problem How does a uniform magnetic field affect a currentcarrying conductor? 2 Carefully increase the current (amperage) from the power supply. 3 Observe any effects on the current-carrying conductor. Materials 1 piece of stiff insulated conducting wire (6–8 cm long) 2 alligator clips 1 U-shaped magnet thread or light string retort stand and clamp variable low-voltage DC power supply with ammeter Questions 1. Describe any effects on the current-carrying conductor that occurred as the current through the conducting wire increased. 2. Does the hand rule verify the direction of the movement of the conducting wire? Explain which hand rule must be used. Procedure 1 Set up the apparatus as shown in Figure 12.30. 3. What is the effect of an external magnetic field on a current-carrying conductor? 4. Based on what you have just observed, design a lab that would demonstrate the effects of a uniform magnetic field on a current-carrying conductor. string or thread ON OFF retort stand power supply insulated wire N S magnet Figure 12.30 606 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 607 Magnetic Forces Between Two Current-carrying Conductors After Oersted demonstrated that a current-carrying conductor creates a magnetic field around a conductor, the French scientist André-Marie Ampère performed extensive studies to determine the magnitude of the magnetic field at any point surrounding a current-carrying conductor. In addition to his mathematical analysis of magnetic fields, he is also noted for determining that two current-carrying conductors exert magnetic forces on each other. The charged particles in one wire are affected by magnetic forces when placed in the magnetic field of another current-carrying wire. Currents in the same direction attract each other (Figure 12.31(a)), and currents in opposite directions repel each other (Figure 12.31(b)). magnetic field wire x e Fm Fm wire x e magnetic field (a) (b) Fm magnetic field wire x e wire e Fm magnetic field Figure 12.31 From the left-hand rule for magnetic fields, the red dashed arrows indicate the orientation of the magnetic field around each wire. Use the left-hand rule for magnetic force to determine how the wires will move relative to each other. (a) When currents are in the same direction, the wires attract each other. (b) When currents are in opposite directions, the wires repel each other. Through careful experimentation and measurement, Ampère was able to determine that the magnetic force between two current-carrying conductors depends on all of the following: • the length of the conducting wire • the distance between the two conducting wires • the amount of current in each wire The SI unit for current is named in honour of Ampère’s work. This unit, the ampere, is now defined as the current required in each of two current-carrying wires, 1 m long and separated by 1 m in air, to produce a force of 2 107 N of magnetic attraction or repulsion. As you learned at the beginning of this section, an ampere is equivalent to the flow of 1 C of charge in 1 s. So, 1 A 1 C/s, and 1 C 1 As. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 607 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 608 Concept Check In intricate electrical circuits, two conducting wires carrying currents in opposite directions are usually crossed. What is the purpose of this crossing procedure? 12-5 Design a Lab 12-5 Design a Lab Using the Current Balance to Measure the Magnetic Force Between Two Current-carrying Conducting Wires The Question How can you use a current balance to investigate the factors that influence the magnetic force acting on two current-carrying conducting wires? Design and Conduct Your Investigation Study the operation of the current balance in your laboratory and design an experimental procedure to investigate the factors that determine the magnetic force acting on two current-carrying conducting wires. In your experimental design: • Identify the factors that determine the magnetic force acting on two currentcarrying conductors. • Write an “if/then” hypothesis statement that predicts how changes in the variable affect the magnetic force. • Clearly outline the procedure you will perform to investigate the relationship of each factor on the magnetic force. • Describe what you will measure and how the data will be recorded and analyzed. • Explain how the data will be used to answer the question. As a group, identify and designate tasks. Prepare a report that describes your experimental design and present it to your teacher. After approval, conduct the investigation and answer the question. How well did your results agree with your hypothesis? info BIT Advancements in technology make it possible to construct extremely small electric motors. Today, 1000 of the smallest electric motors could fit in the period at the end of this sentence. commutator: a mechanism for maintaining a properly polarized connection to the moving coil in a motor or generator The Electric Motor The most important application of the effect of an external magnetic field on current-carrying conductors is the electric motor. Figure 12.32 illustrates a simple electric motor that works with a current-carrying wire loop between two magnetic poles. The current is in one direction. Recall from earlier science studies that current in one direction is called a direct current (DC). A simple DC electric motor consists of three fundamental components: • a stator—a frame with a coil or permanent magnet to provide a magnetic field • an armature or rotor—a rotating loop of conducting wire on a shaft • a commutator—a split metal ring 608 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 609 As electrons in the current pass through the loop of wire in the armature in a clockwise direction (as seen from above in Figure 12.32), they experience a motor effect deflecting force. When you apply the left-hand rule for magnetic force, electrons on the left side of the loop experience a deflecting force upward, and electrons on the right side of the loop experience a deflecting force downward. The combined effect of both forces results in a rotation of the loop in a clockwise direction. Fm stator N armature e S e brush Fm commutator Figure 12.32 In a simple DC electric motor, the brushes provide a sliding contact between the wires from the battery and the armature. The magnetic field exerts an upward force on the left side of the wire loop and a downward force on the right side, causing the armature to rotate clockwise. Concept Check Describe the changes that must be made to the apparatus, shown in Figure 12.32, to cause the armature to rotate counterclockwise. If the rotation of the loop is to continue, the direction of the motion of the electrons in the loop must change every half-rotation. To accomplish this, the armature is connected to a commutator. A commutator is a split metal ring that is fastened to both ends of the loop of wire in the armature. Each half of the metal ring acts as a contact to the terminals of a power supply. Every half-rotation, the leads of each side of the armature contact a different terminal, changing the direction of the electron movement. Once connected to a steady supply of moving electrons, the armature continues to rotate in one direction. This is the principle of a simple electric motor. The Generator Effect (Electromagnetic Induction) In 1996, NASA did an experiment that involved a satellite attached by a conducting tether wire to a NASA space shuttle orbiting in space around Earth (Figure 12.33). Researchers found that the combination generated a current of about 1 A through the wire. The experiment was of particular significance for space scientists because it showed that Figure 12.33 A satellite tethered to a NASA space shuttle Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 609 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 610 this procedure could provide a method of generating the electric energy necessary to power all the electrical components on a space vehicle. This example is a useful and important application of a scientific phenomenon, but this phenomenon can also produce harmful effects in some situations. For example, engineers constructing the 1280-km north–south gas pipeline from Prudhoe Bay to Valdez in Alaska (Figure 12.34(a)) had to take precautions to eliminate the currents of electricity, called telluric currents, in the pipeline. These currents are caused by fluctuations in Earth’s magnetic field. Special magnesium anodes were installed underground along the pipeline to ground it and eliminate the possibility of electrical sparks. (a) (b) Figure 12.34 (a) A pipeline in Alaska; (b) An airplane in flight Similarly, certain grounding conditions must be incorporated in the construction of an airplane to eliminate the current generated by the wings of an airplane in flight through Earth’s magnetic field. These currents could affect the operation of all electrical components on the aircraft (Figure 12.34(b)). How are these examples related? What physical phenomenon is generating the current? The exa
mples described above all involve conductors moving through magnetic fields. The scientific explanation of how they generate electricity began with investigations over 200 years ago. Faraday’s and Henry’s Discoveries Most scientific discoveries are the result of many years of research and investigations. The process is often convoluted and results are often accidental. However, as you have learned, some scientific discoveries are a result of the symmetry of nature. This symmetry led Coulomb and Faraday to conclude that electrical and magnetic forces could be determined using inverse-square relationships similar to Newton’s universal law of gravitation. Similarly, this symmetry in nature, and Oersted’s discovery that electricity could produce magnetism, led scientists to predict that magnetism could produce electricity. Experiments conducted in 1831 by Michael Faraday in England and Joseph Henry (1797–1878) in the United States demonstrated this effect. e LAB For a probeware activity that demonstrates the principle of electromagnetic induction, go to www.pearsoned.ca/ school/physicssource. 610 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 611 In a simplified version of their experiment, shown in Figure 12.35, a magnet is moved toward a coil of conducting wire connected to a sensitive galvanometer. When the magnet approaches the coil, the galvanometer’s needle deflects in one direction, indicating that a current is being produced in the coil of wire. This current is called an induced current, which is produced by a generated voltage. When the magnet is pulled away from the coil, the galvanometer deflects in the opposite direction, indicating that the induced current in the coil is in the opposite direction. When the magnet is stationary, no current is induced. If the magnet were held stationary while the coil of wire was moved back and forth, similar induced currents would be produced. Evidently, it does not matter whether the magnet or the coil of wire moves, as long as there is relative motion between a coil of conducting wire and an external magnetic field. In their conclusions, Faraday and Henry stated that when a piece of conducting wire cuts through magnetic field lines, an induced current is produced. The production of electricity by magnetism is called the generator effect or electromagnetic induction. Figure 12.36(a) shows a piece of conducting wire being moved perpendicularly upward through an external magnetic field. As a result, electrons in the wire also move perpendicularly upward. Use the left-hand rule for magnetic force: If the wire is moving upward (thumb) through the external magnetic field (fingers), then each electron experiences a motor-effect force (palm). Electrons will gather at one end of the wire with stored electric energy from the work done on the system in moving the wire. Thus, one end of the wire has an accumulation of electrons with stored electric energy while the other end has a deficiency of electrons (Figure 12.36(a)). If this wire is part of an external circuit, as in Figure 12.36(b), the induced voltage causes a current to flow through the external wire. A difference in electric potential drives electrons through an external circuit, from a region of high electric potential to a region of lower electric potential. N S galvanometer Figure 12.35 When a magnet is moved toward a loop of wire connected to a galvanometer, the galvanometer needle deflects. This indicates that an induced current is being produced in the coil of wire. generator effect or electromagnetic induction: production of electricity by magnetism Project LINK How would you apply the principle of electromagnetic induction and the operation of commutators to construct a DC generator wire movement Fm B induced current (a) wire movement induced current Fm B A electron flow (b) Figure 12.36 A current can be induced in a wire by moving the wire through a magnetic field. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 611 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 612 12-6 Inquiry Lab 12-6 Inquiry Lab Magnetic Fields and Moving Conductors — Demonstration Question What factors influence the effect produced when there is relative motion between an external magnetic field and a conducting wire? Materials and Equipment 2 bar magnets 3 different sizes of coils of conducting wire (the diameter of the coils should allow a bar magnet to be inserted) galvanometer that can be projected onto a screen using an overhead projector galvanometer N S N S N S Figure 12.37 Procedure 1 Set up the apparatus as shown in Figure 12.37. 2 Slowly push one bar magnet at a uniform speed into the largest coil. Then pull the bar magnet out in the opposite direction at the same speed. Observe the deflection of the galvanometer’s needle in both cases. 3 Repeat step 2 with the other end of the magnet. Observe the deflection of the galvanometer’s needle in both cases. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 4 Repeat step 2, using two bar magnets. Observe the magnitude and direction of the galvanometer’s deflection. 5 Repeat step 2, using two bar magnets at a faster speed through the largest coil. Observe the magnitude and direction of the galvanometer’s deflection. 6 Repeat step 2, using two bar magnets at the same constant speed through the medium and the smaller coils of conducting wire. Observe the magnitude and direction of the galvanometer’s deflection. Analysis 1. How does the direction of the movement of the magnet affect the direction of the deflection of the galvanometer? 2. How does the polarity of the magnet affect the direction of the deflection? 3. Describe how each of the following factors influences the magnitude of the deflection of the galvanometer: (a) speed of the magnets through the conducting wire (b) strength of the external magnetic field (c) number of loops in the coil of conducting wire 4. What is the effect of relative motion between a conducting wire and a magnetic field? 5. Does it make any difference if the magnet or the conducting wire is moved? 6. What are the factors that determine the magnitude and direction of the induced current when there is relative motion between a conducting wire and an external magnetic field? 7. Based on your observations from this activity, design an experiment that demonstrates the effect of a uniform magnetic field on a moving conductor. 612 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 613 9. A battery supplies a current of 5.20 mA to a circuit. Determine the quantity of charge that flows through the circuit in 2.00 s. 10. Two conducting wires parallel to each other carry currents in opposite directions. Using the appropriate hand rule, determine whether the wires will attract or repel each other. 11. A wire 50 cm long and carrying a current of 0.56 A is perpendicular to an external magnetic field of 0.30 T. Determine the magnitude of the magnetic force on the wire. Extension 12. Could a simple electric generator be converted to a simple electric motor? Suggest any alterations that must be made in the design. e TEST To check your understanding of current-carrying conductors and magnetic fields, follow the eTest links at www.pearsoned.ca/school/physicssource. 12.3 Check and Reflect 12.3 Check and Reflect Knowledge 1. What are the factors that affect the magnetic force on a moving charge through an external magnetic field? 2. What are the factors that affect the magnetic force on a charge moving through a conducting wire in an external magnetic field? 3. What is the relationship between amperes and coulombs? 4. In the operation of a simple electric motor and simple electric generator, identify (a) a similarity (b) a difference 5. What symmetry in nature did Faraday and Henry apply in their discovery of electromagnetic induction? 6. What is the function of a split-ring commutator in the operation of a simple DC motor? 7. How do the electrons in a loop of wire in a generator gain energy? Applications 8. A wire lying perpendicular to an external magnetic field carries a current in the direction shown in the diagram below. In what direction will the wire move due to the resulting magnetic force? S N S N I Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 613 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 614 12.4 Magnetic Fields, Moving Charges, and New and Old Technologies info BIT Michael Faraday built the first electric motor in 1821. This motor had a stiff wire hanging from a stand. The lower end of the wire was immersed in a cup of mercury with a bar magnet upright in the middle. When current from a battery flowed through the wire, it rotated around the magnet. From the old technologies of the simple beginnings of electric motors, electric meters (such as galvanometers and ammeters), loudspeakers, and electromagnets to the new technologies of magnetohydrodynamic (MHD) propulsion systems and magnetic resonance imaging (MRI), the science of the production of magnetism by electricity plays a significant role in our everyday lives. Although examples of some of these technologies have been described in previous sections, following are other examples of old and new technological applications of this principle. Table 12.2 describes old and new technologies that use moving charges or current-carrying conductors to produce magnetic fields that can interact with external magnetic fields to produce powerful magnetic forces. Table 12.2 Loudspeakers and MHD Propulsion Old Technology New Technology Magnetohydrodynamic (MHD) Propulsion paper cone water intake oppositely charged plates Loudspeakers electron flow x x x B F S N S voice coil Figure 12.38 A simplified diagram of a loudspeaker The operating principle of most loudspeakers is that currentcarrying wires produce magnetic fields that can exert magne
tic forces. In the design of the loudspeaker shown in Figure 12.38, a coil of wire, called a voice coil, surrounds the north pole of a very powerful external magnet at the back of the speaker. When your sound system sends an electric signal to the coil, a current is produced in the coil, which produces a magnetic field. As a result, the coil experiences a magnetic force due to the interaction of its magnetic field with the external magnetic field. Depending on the direction of the current in the coil, the magnetic force of attraction or repulsion causes the coil to slide to the left or right. The direction of the current is determined by the electric signal produced by the sound system. As the voice coil slides back and forth, it causes the paper cone to vibrate in or out, creating sound waves as it pushes on the air in front of the cone. The electric signal from the sound system is thus converted to a mechanical sound wave in air. 614 Unit VI Forces and Fields superconducting electromagnet jet of water Figure 12.39 MHD uses magnetic fields as a propulsion system for seagoing vessels. The MHD propulsion system is an experimental system for seagoing vessels to replace conventional propeller systems. MHD uses magnetic fields to produce a jet of water for propulsion. Figure 12.39 is a simplified diagram of this type of system. A powerful superconducting magnet surrounds a thruster tube containing seawater. This magnet produces a magnetic field perpendicular to the tube’s length. Inside the tube, electrodes produce a current of ions, perpendicular to the magnetic field, across the tube from the dissolved salts in seawater. As a result of the perpendicular movement of the ions through an external magnetic field, a magnetic force is exerted on the ions, causing them to deflect along the length of the tube. This movement of the water through the tube provides the necessary thrust to propel the vessel. An advantage of MHD propulsion systems is that they have no mechanical moving parts and thus require minimal maintenance. 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 615 Generator Effect Applications The discovery that moving a conducting wire through an external magnetic field generates an induced current in the conductor (generator effect) also led to many important technological applications. From the old technologies of the simple generators, induction coils, and transformers to the new technologies of infant breathing monitors and others, applications of the scientific principle of the production of electricity from magnetism are found everywhere in our lives. Table 12.3 describes two of these applications. Table 12.3 Induction Coils, Transformers, and SIDS Monitors Old Technology New Technology Induction Coils switch SIDS Monitors galvanometer primary coil iron battery secondary coil Figure 12.40 A simplified diagram of Faraday’s induction coil Figure 12.41 Monitors are designed to detect changes in a baby’s breathing. A change in the current in the primary coil produces a changing magnetic field in the iron core. This changing magnetic field produces an induced current in the secondary coil, causing the needle on the galvanometer to deflect. Such coils can induce current in a wire that has no direct connection to the power supply. Figure 12.40 shows a simplified version of Michael Faraday’s original induction coil. In sudden infant death syndrome (SIDS), an infant stops breathing with no apparent cause. One type of SIDS monitor uses induced currents to measure an infant’s breathing (Figure 12.41). A coil of wire attached to one side of the infant’s chest carries an alternating current, which produces a magnetic field. This alternating field cuts another coil taped to the other side of the chest and induces an alternating current in this other coil. As the chest moves up and down, the strength of the induced current varies. These variations are monitored. A Motor Is Really a Generator, Which Is Really a Motor You have analyzed and studied the motor effect and the generator effect as separate phenomena in this chapter. However, the symmetry of nature suggests that related phenomena are really variations of the same effect. Since electricity can produce magnetism and magnetism can produce electricity, then perhaps the technologies that are derived from these phenomena are also similar. Is a motor really that different from a generator? Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 615 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 616 12-7 Inquiry Lab 12-7 Inquiry Lab The Curious Relationship Between Motors and Generators Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question What is the relationship between the motor effect and the generator effect? Analysis 1. Copy Table 12.5 into your notebook. Use the data from Table 12.4 to complete the calculations in Table 12.5. Materials and Equipment Table 12.5 Calculations for 12-7 Inquiry Lab Magnitude of Acceleration of the Magnet Through the Copper Pipe d a 2 2 t (m/s2) Magnitude of Net Force Causing the Downward Acceleration of the Magnet (Fa ma) (N) Magnitude of the Weight of the Magnet (Fg mg) (N) 2. What is the magnitude of the upward force on the falling magnet? 3. Identify and explain where the generator effect is occurring in this experiment. 4. Identify and explain where the motor effect is occurring in this experiment. 5. Do the generator effect and the motor effect complement each other as the magnet falls through the copper pipe? Explain your answer. 1 100-cm length of copper pipe (internal diameter approximately 1.4 cm) 1 cylindrical rare earth magnet (less than 1.4 cm in diameter) 1 metre-stick 1 stopwatch 1 scale Procedure 1 Copy Table 12.4 into your notebook. Table 12.4 Data for 12-7 Inquiry Lab Mass of Magnet Length of Copper (m)(kg) Pipe (d)(m) Average Time Taken for the Magnet to Fall Through the Pipe (t)(s) 2 Measure the mass of the magnet on the scale, and record it in Table 12.4. 3 Measure the length of the copper pipe, and record it in Table 12.4. 4 Holding the pipe in a vertical position, drop the magnet from the exact top of the pipe. Measure the time for the magnet to reappear out the bottom. Record this time in the table. 5 Repeat the procedure in step 4 several times to obtain an average value for the time taken for the magnet to drop the length of the pipe. 616 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 617 An Accidental Discovery Simple DC electric motors and electric generators have three similar components: • an external magnetic field • a loop of conducting wire • a commutator At the 1873 Vienna Exhibition, the Belgian inventor Zénobe-Théophile Gramme (1826–1901) demonstrated a compact and efficient generator that he had designed. A steam engine provided the power to run the generator. A workman mistakenly connected the output of the generator to a second generator in the display. The shaft of the second generator began spinning even though it was not connected to the steam engine. Gramme immediately realized that the second generator was operating as a motor powered by the first generator. Gramme and his colleagues then moved the generators several hundred metres apart and connected them with long wires. The American writer Henry Adams (1838–1918) described the importance of Gramme’s demonstration: “Suddenly it became clear that ELECTRICITY could now do heavy work, transporting power through wires from place to place.” M I N D S O N Perpetual Motion? Suppose that a motor and generator are connected to the same shaft and wired such that the output of the generator powers the motor. If you spin the shaft, the generator supplies energy to the motor, which turns the shaft. The generator then produces more energy to run the motor. Explain why this process cannot continue indefinitely. Lenz’s Law If you did the 12-7 Inquiry Lab, you discovered what happens when you drop a magnet down a metal tube. When a conductor cuts the magnetic field lines of a falling magnet, it generates an induced current in the conducting pipe (the generator effect). However, the induced current moves in a circular motion around the circular pipe, so it creates its own vertical magnetic field, inside the metal tube (the motor effect). The direction of the magnetic field can be directed either upward or downward. The direction of the magnetic field that is produced by the circular induced current in the pipe can have one of the following orientations: • It will attract the magnet and cause it to fall faster, thus generating a greater induced current. • It will repel and oppose the motion of the magnet, causing it to fall much slower. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 617 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 618 The law of conservation of energy requires that you can never get more out of a system than you put into it. So, the direction of the new magnetic field will always oppose the motion of the magnetic field of the original magnet. This is the principle of Lenz’s law, which states: The direction of a magnetically induced current is such as to oppose the cause of the current. For example, if a magnet falls with its north pole directed downward, then the magnetic field produced by the induced current in the conducting pipe will have its north pole pointing upward to repel and oppose this motion. Concept Check Copy Figure 12.42 into your notebook. Apply Lenz’s law by sketching the direction of the induced current in the metal tube and the resulting orientation of the magnetic field in the tube. How would the induced current and magnetic field directions change if the falling magnet’s south pole were directed downward? S N Figure 12.42 Dropping a magnet down a metal tube induces a current in the tube. Figure 12.43 shows a similar situation. As the north pole of a magnet ap
proaches a coil of wire, the induced current generated in the coil produces a north pole to repel and oppose the approaching magnet. S N N S G Figure 12.43 Lenz’s law helps us explain that the direction of current induced in the coil has a magnetic field that exerts a force on the bar magnet that opposes the magnet’s motion. M I N D S O N Lenz’s Law Balance a dime on its edge on a smooth table. Carefully bring a magnet as close as 1 mm to the face of the dime. Quickly pull the magnet away. What happens to the dime? How is this behaviour an application of Lenz’s law? The principle behind Lenz’s law also hinders the operation of electric motors and generators. For an electric motor to operate, an electric current must first be supplied through a conducting loop of wire in a magnetic field, causing the motor effect, so the loop will rotate. 618 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 619 However, as the loop rotates, the conducting wire cuts the magnetic field lines, causing the generator effect. The generator effect induces a current in the loop of wire. The direction of the induced current must be in an opposing direction to the direction of the original current that was supplied. Similarly, to operate a generator, movement of a conducting wire in a magnetic field must be supplied, which will induce a current. However, as soon as the induced current moves through a conductor in a magnetic field, a force on the conducting wire will be produced that opposes the original force and hinders the movement of the conducting wire. e WEB Research the relationship between Lenz’s law and the operation of most vending machines. Write a short report analyzing the operation of vending machines, and describe whether they operate on the principles of Lenz’s law and the generator effect or Lenz’s law and the motor effect. In your description, include the term “eddy currents.” Begin your search at www.pearsoned.ca/ school/physicssource. THEN, NOW, AND FUTURE Nanotechnology Since the start of the Industrial Revolution and with advances in technology, machines have become increasingly smaller. Scientists at University of California’s Berkeley National Laboratory have developed the smallest synthetic electric motor ever made. Essentially, it is an electric rotor spinning on an axle 2000 times smaller than the width of a human hair. Imagine tiny electric motors so small that one motor could ride on the back of a virus, or thousands could fit in the period at the end of this sentence. This is the world of nanotechnology, which is an umbrella word that covers many areas of research and deals with objects in nanometres, or a billionth of a metre. How do nanomachines work? As you have learned, conventional electric motors use electromagnets or strong external magnets to spin rotors made of loops of wire. The spinning of a rotor provides mechanical energy to do work. In the electric motor of nanotechnology, transistors act as switches to move negative and positive charges around a circle of electrodes. The charges jump around the stator electrodes, that are measured Figure 12.44 A rosette nanotube causing an electrically charged rotor to spin around and rotate a nanotube shaft. This spinning rotor provides mechanical energy, similar to a conventional electric motor. The nanomotor can perform only small functions, such as moving a second hand on a watch, but it has many advantages: It spins without gears or bearings; it is unaffected by gravity or inertia; and it can run for a long time with no breakdowns. Another application of nanotechnology is the rosette nanotube, like the one shown in Figure 12.44. This type of tube was developed at the National Institute of Nanotechnology of the National Research Council, located at the University of Alberta. The nanotube is made of molecules that assemble themselves into this distinctive shape. Possible future applications of these tubes include: nanowires in molecular electronics, drug delivery systems within the body, and environmentally friendly oil sands upgrading additives. The potential of nanotechnology in electronics is now beginning to be realized and applied to many different fields. Imagine, for example, nanorobots that can be injected into the body to attack viruses and cancer cells. In the future, it may even be possible to construct molecules of oil and gas, reducing our reliance on fossil fuels. The science of nanotechnology is the science of the future. Questions 1. What are some advantages of nanotechnology? 2. Describe how a nanomotor works. 3. What could nanotubes be used for? Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 619 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 620 12.4 Check and Reflect 12.4 Check and Reflect Knowledge 1. Identify two technological applications that employ (a) the motor effect (b) the generator effect 2. What are the three basic components of an electric motor and generator? 3. What is Lenz’s law? Applications 4. For an electric motor: (a) Describe what you must supply to start the operation of the device. (b) Describe what you get out of the operation of the device (motor effect). (c) Using Lenz’s law, explain how the operation of the motor also produces the generator effect to hinder its own operation. 5. For an electric generator: (a) Describe what you must supply to start the operation of the device. (b) Describe what you get out of the operation of the device (generator effect). (c) Using Lenz’s law, explain how the operation of the generator also produces the motor effect to hinder its own operation. 6. Explain why you will feel a force of repulsion if you attempt to move a magnet into a coil of wire. 7. The north pole of a magnet is pulled away from a copper ring, as shown in the diagram below. What is the direction of the induced current in the ring? N S Extensions 8. An electric motor requires a current in a loop of wire. However, as the loop rotates, it generates a current which, according to Lenz’s law, must be in an opposing direction. Why must the induced current be in an opposing direction? 9. A hair dryer operates on a very small current. If the electric motor in the hair dryer is suddenly prevented from rotating, the dryer overheats. Why? e TEST To check your understanding of magnetic fields, moving charges, and new and old technologies, follow the eTest links at www.pearsoned.ca/ school/physicssource. 620 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 621 CHAPTER 12 SUMMARY Key Terms and Concepts law of magnetism magnetic field electromagnet solenoid ferromagnetic domain motor effect force current Key Equations F qvB m F IlB m Conceptual Overview ampere commutator generator effect electromagnetic induction Lenz’s law The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. definition types Earth domain theory bar magnet U-shaped magnet magnetization by induction law of magnetism generator effect Lenz’s law 2 like poles 2 unlike poles magnetic forces and fields cause moving charges motor effect moving charges in magnetic fields direction of deflection magnitude of deflection Fm qvB charge wire moving charges in 2 current-carrying wires definition of ampere Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 621 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 622 CHAPTER 12 REVIEW Knowledge 1. (12.1) State the major contribution of each of the following scientists to the study of magnetism: (a) William Gilbert (b) Hans Christian Oersted (c) André Ampère (d) Michael Faraday 2. (12.1) State the definition of (a) a magnetic field (b) the direction of a magnetic field 3. (12.1) Compare a magnetic vector arrow at a point near a magnet and a magnetic field line around a magnet. 4. (12.1) How was it determined that there had to be two different types of magnetic poles? 5. (12.1) Sketch the magnetic field lines around each of the following objects and describe the differences in the magnetic fields of (a) a bar magnet and Earth (b) a current-carrying straight piece of conducting wire and a current-carrying coil of conducting wire 6. (12.2) Identify the sources of the two magnetic fields required to produce the motor effect force on a moving charge. 7. (12.2) Describe the deflection of a moving charge, through an external magnetic field, if the direction of the initial motion of the charge is (a) parallel to the external magnetic field lines (b) perpendicular to the external magnetic field lines (c) at an angle to the external magnetic field lines 8. (12.2) Where is a magnetic bottle formed? 9. (12.3) Describe a difference between a galvanometer and an ammeter. 10. (12.3) State two definitions for a current of one ampere. 11. (12.4) Identify two technologies that use the principles of (a) the motor effect (b) the generator effect 12. (12.4) What is the principle of Lenz’s law? 622 Unit VI Forces and Fields Applications 13. Does every charged object necessarily have a positive and a negative charge? Does every magnetized object necessarily have a north and a south pole? Justify your answers. 14. If the direction of the magnetic field outside a magnet is from the north to the south pole, what is the direction of the magnetic field within the magnet? 15. Using domain theory, describe how an iron nail can become magnetic by (a) the process of magnetization by induction (b) the process of magnetization by contact 16. How will the magnetic force on a moving charged particle change if (a) only the charge is doubled? (b) the magnetic field is doubled and the speed is halved? (c) the mass of the charge is doubled? 17. Use domain theory to explain the difference between a permanent and a temporary magnet. 18. You are told that a straight piece of copper wire has a steady current in it. Given only a compass, describe how you can find the direc
tion of the current in the wire. 19. A drinking straw with a green grape at one end is suspended by a string from a hanging support. When either end of a magnet is brought close to the grape, repulsion occurs. Describe a possible reason for this effect. 20. An electron in a TV tube is moving at 7.00 106 m/s perpendicular to a magnetic field of magnitude 0.0880 T in the tube. What is the magnetic deflecting force on the electron? 21. A proton travelling at 35 to an external magnetic field of magnitude 0.0260 T experiences a force of magnitude 5.50 1017 N. (a) Calculate the speed of the proton. (b) Calculate the kinetic energy of the proton in joules (J) and electron volts (eV). 22. What speed must an alpha particle maintain if it is to remain suspended, relative to Earth’s surface, as it travels on a tangent to Earth’s surface and perpendicularly through Earth’s magnetic field of 50.0 T? 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 623 23. An alpha particle travelling with a speed of 4.30 104 m/s enters a uniform magnetic field of 0.0300 T. Determine the magnetic force on the particle if it enters the field at an angle (a) perpendicular to the magnetic field (b) 30.0 to the magnetic field (c) parallel to the magnetic field 24. A magnetic field is used to bend a beam of electrons. What uniform magnetic field is required to bend a beam of electrons moving at 1.2 106 m/s in a circular arc of 0.25 m? 25. Two parallel current-carrying wires are observed to attract each other. What is the source of the force of attraction? How could you demonstrate that the force of attraction is not electrostatic attraction? 26. The magnetic force between two magnets was measured as the distance between the magnets was varied. The following information was obtained: Separation, r ( 102 m) Magnitude of Force, F (N) 5.00 10.00 15.00 20.00 25.00 4.02 1.01 0.45 0.25 0.16 (a) Draw a graph of the magnetic force as a function of separation distance. (b) From the shape of the graph, what is the relationship between force and separation? 27. An airplane is flying east over Earth’s magnetic north pole. As a result of its motion, one wing was detected as having more electrons than the other. Explain why this phenomenon occurs. Identify which wing will have more electrons. 28. A power line carries a current of 500 A. Find the magnetic force on a 100-m length of wire lying perpendicular to Earth’s magnetic field of 50.0 T. Extensions 29. A magnet is dropped through two similar vertical tubes of copper and glass. Which tube will allow the magnet to fall faster? Explain your answer. 30. A disk magnet on a table has two steel balls in contact with it on either side. The steel balls are slowly moved toward each other while still in contact with the disk magnet. As they move, they repel each other. Describe why the steel balls are attracted to the disk magnet, but repel each other. Consolidate Your Understanding Create your own summary of properties of magnetic and electric fields by answering the questions below. If you want to use a graphic organizer, refer to Student Reference 3: Using Graphic Organizers. Use the Key Terms and Concepts listed on page 621 and the Learning Outcomes on page 580. 1. Create a flowchart to identify technological devices that use electric fields, magnetic fields, or a combination of the two fields to control moving charges. 2. Write a paragraph comparing the effects of electric or magnetic fields on moving charges. Share your report with a classmate. Think About It Review your answers to the Think About It questions on page 581. How would you answer each question now? e TEST To check your understanding of the properties of electric and magnetic fields, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 623 12-PearsonPhys30-Chap12 7/28/08 9:46 AM Page 624 UNIT VI PROJECT Building a Model of a Direct Current Generator Scenario The search for better electrical energy production began with the pioneering work of Faraday and Henry in the 1800s and continues today. Over half of the energy consumed in our world is electrical energy and the average consumption per person is increasing every year, so more efficient methods of electrical energy production are being sought all the time. All DC electrical generators consist of three major parts: coils of wire wrapped around a core to make the armature, a commutator, and an external magnetic field. As you learned in this unit, the operating principle behind generators is the movement of a conducting wire through external magnetic field lines so that a voltage is generated in the wire. This in turn induces a current in an external line. All DC generators operate in this fashion. The only difference among generators is the source of the mechanical energy required to turn the turbines that rotate the coil of wires in the magnetic field. In some places, this is the energy of falling water or tides, while in others it is the energy of moving steam from the combustion of fossil fuels or nuclear reactions. Recently, interest has grown in using wind energy to turn the turbines that operate a generator. A single wind generator can produce about 10 MW of electrical power, which is sufficient for a single small farm. The purpose of this project is to research and investigate the operation of a wind-powered electrical generator and to build a model of a wind-powered DC generator capable of generating enough electricity to operate a mini-bulb. Planning Form a team of four or five members, and decide on and plan the required tasks to complete the project. These tasks may include researching the design of a simple generator, obtaining the necessary materials, constructing the model of the generator, preparing a written report, and presenting the project to the entire class. Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies thoroughness of the experimental design effectiveness of the experimental technique effectiveness of the team’s public presentation Materials • insulated copper wire • iron core • split-ring commutator • external magnets • connecting wires • mini-bulb with support base • stiff paper to construct a turbine • balsa wood for the axle • household fan to produce wind Procedure 1 Using the Internet, library, or other resources, research the operation of a simple DC electrical generator and create a design of the model that you will construct. Pay special attention to the commutator required for DC generation. 2 Construct a working model of the generator that can provide the electrical energy to light a mini-bulb. In your model, investigate the factors that determine the magnitude of the generated voltage. 3 Prepare a report explaining the design and the specific functions of all the components. Thinking Further 1. What modifications did you make in the construction of your model of a generator that affected the magnitude of the generated voltage? 2. What other type of commutator could you have used in the design of your generator? What type of current would be induced by this commutator? 3. 4. Identify at least three risks and three benefits of a wind-powered electrical generator. Is wind-powered electrical generation a viable and desirable method of electrical energy generation for the future? Explain your answer. *Note: Your instructor will assess the project using a similar assessment rubric. 624 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 625 UNIT VI SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 10 Modern theory of electrostatics Coulomb’s law Summary Resources and Skill Building Physics laws can explain the behaviour of electric charges. 10.1 Electrical Interactions Substances can be classified as conductors, insulators, semiconductors, and superconductors. Objects may be charged through the processes of friction, conduction, and induction. 10-1 QuickLab 10-2 Inquiry Lab 10.2 Coulomb’s Law Coulomb’s law states that the electrical force acting on charged objects depends on the charges and the distance between the charges. 10-3 Inquiry Lab Examples 10.1, 10.2 Vector analysis Electrostatic forces can be solved in one- and two-dimensional situations using vector analysis. Examples 10.3–10.6 Chapter 11 Electric field theory describes electrical phenomena. Fields Electric field Electric field lines Electric potential energy Electric potential 11.1 Forces and Fields Fields are used to explain action at a distance. An electric field is a three-dimensional region of influence surrounding every charge. Electrostatic force affects another charge placed in the field. The electric field is a vector quantity that has magnitude and direction. 11.2 Electric Field Lines and Electric Potential Electric field lines can depict the electric fields around different types of charged objects. Electric potential energy is the amount of work done on a charged object to move it from infinity to a position in an electric field. Electric potential energy can be calculated. Electric potential is the amount of electric potential energy stored per unit charge and can be calculated. Electric potential difference When a charge moves from a location where it has one electric potential to a location where it has another electric potential, the charge experiences an electric potential difference. Motion of a charge in an electric field 11.3 Electrical Interactions and the Law of Conservation of Energy When a charge is placed in an electric field, it experiences a force that causes it to accelerate in the direction of the field. The acceleration of the charge is different in a non-uniform field surrounding a point charge than the acceleration of the charge in the uniform field between charged plates. Work done by the system on the charge increases the charge’s potential energy, which can be conver
ted to other forms of energy. Chapter 12 Properties of electric and magnetic fields apply in nature and technology. 11-1 QuickLab, 11-2 Inquiry Lab Examples 11.1–11.4 Minds On activities, eSIM Examples 11.5, 11.6 Example 11.7 Example 11.8 Examples 11.10–11.12 Magnetic fields Cause of magnetism 12.1 Magnetic Forces and Fields Magnetic fields are three-dimensional regions of magnetic influence surrounding every magnet in which other magnets or magnetic substances are affected by magnetic forces. Magnetic fields are vector fields and can be depicted by magnetic field lines. The cause of magnetism is motion of charges and can be explained using the domain theory. If the motion of charges is straight, the magnetic field is circular. If the motion of charges is circular, the magnetic field is straight within the loop. The direction of the magnetic field lines can be described using hand rules. 12-1 QuickLab, 12-2 QuickLab Figure 12.10 Magnetizing objects Objects can be magnetized through contact or induction. Concept Check Motor effect on a moving charge 12.2 Moving Charges and Magnetic Fields A charge moving perpendicularly through an external magnetic field experiences a magnetic force due to two magnetic fields, which can be calculated. This motor effect force can explain the operation of electric motors and other technologies. The magnitude of the motor effect force can be calculated and its direction can be determined using hand rules. 12-3 Inquiry Lab Examples 12.1, 12.2 Motor effect on two current-carrying wires 12.3 Current-carrying Conductors and Magnetic Fields A current-carrying conductor that is perpendicular to an external magnetic field experiences a magnetic force that can be calculated. Generator effect A conductor moving perpendicular to an external magnetic field can produce electricity. Applications of the generator effect and the motor effect 12.4 Magnetic Fields, Moving Charges, and New and Old Technologies The generator effect and the motor effect are used in many technologies. Example 12.4 12-4 QuickLab 12-5 Design a Lab 12-6 Inquiry Lab 12-7 Inquiry Lab Lenz’s law Lenz’s law explains how a motor is really a generator and a generator is really a motor. Minds On Unit VI Forces and Fields 625 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 626 UNIT VI REVIEW Vocabulary 1. Use your own words to define the following terms, concepts, principles, or laws. Give examples where appropriate. ampere charge migration charge shift charging by induction commutator conduction conductor coulomb Coulomb’s law current domain electric field line electric potential (voltage) electric potential difference electric potential energy electromagnet electromagnetic induction electron volt electrostatics ferromagnetic field generator effect grounding induction insulator law of conservation of charge law of magnetism Lenz’s law magnetic field motor effect force net charge plasma semiconductor solenoid source charge superconductor test charge 626 Unit VI Forces and Fields Knowledge CHAPTER 10 2. Explain why silver is a better conductor of electricity than rubber. 3. State a technological advantage of developing materials that are superconductors. 4. A negatively charged rubber rod is brought near a small metal ball hanging from an insulated thread. The metal ball is momentarily grounded, and then the ground and the rubber rod are removed. Identify the procedure used to charge the metal ball, and determine the final charge on the metal ball. 5. Describe a similarity and a difference between (a) charging by friction and charging by conduction (b) charge shift and charge migration 6. In each of the following examples, identify the charge on each object and state the method of charging the object. (a) An ebonite rod is rubbed with fur and then is held near a neutral metal sphere. (b) A glass rod is rubbed with silk and then is touched to a neutral metal sphere. 7. During the rubbing process of charging objects, one object gains a net negative charge. What can you conclude about the charge on the other object? Explain why. 8. How do the following factors affect the electrostatic force of attraction between two charged objects? (a) amount of charge on each object (b) distance between the centres of the two objects (c) sign of the charge on each object CHAPTER 11 9. State the superposition principle as it applies to vector fields. 10. Why must a test charge be small when it is used to determine the direction of an electric field around a larger primary charge? 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 627 11. Draw the electric field lines around 22. Justify the following statement: “A charge (a) a negative point charge (b) a positive charge and a negative point charge in the same region (c) a negatively charged cone-shaped object 12. Why is there no net electric field inside a charged hollow sphere? 13. Explain the following statement: “It is impossible to shield against gravitational fields, but it is possible to shield against electric fields.” 14. Draw a diagram showing two small, equally charged, positive spheres a small distance apart. On your diagram, identify a point A where the electric field is zero and a point B where the electric potential is zero. 15. Describe how electric potential energy changes for positive and negative charges as they move, relative to the reference point at infinity. 16. Two oppositely charged parallel plates are connected to a 120-V DC supply. What happens to the magnitude of the electric field between the plates if the distance between the plates decreases? moving perpendicular to a uniform external magnetic field experiences a force but does not change its speed.” 23. Show how a charge moving through an external magnetic field experiences a deflecting force because of the interaction of two magnetic fields. 24. A positively charged disk is spinning in a clockwise direction as seen from above. What is the direction of the magnetic field? Use a diagram to help you describe its shape. 25. Describe the shape and direction of the magnetic field around a negatively charged dart as it travels directly away from you toward a target on a wall. 26. What factors affect the magnetic force on a charge moving through an external magnetic field? 27. State a difference and a similarity between the motor effect and the generator effect. 28. Describe how Lenz’s law affects the operation of (a) a motor (b) a generator 17. Describe the following for a point a small distance from a positively charged sphere: Applications (a) (b) the magnitude and direction of the electric field at this point the electric potential energy of another small positive charge placed at this point (c) the electric potential at this point 18. Describe the relationship between the work done in moving a charge from one region in an electric field to another region, and the energy gained by the charge. What law governs this relationship? 29. An insulator and a conductor are each contacted by a negatively charged rubber rod. Describe the distribution of charge on each object. 30. Describe how Earth’s surface can become positively charged during a thunderstorm. In your description, include the terms “charging by friction” and “charging by conduction.” 31. Assume you have only a negatively charged ebonite rod. Describe a procedure for charging a neutral electroscope CHAPTER 12 (a) positively (b) negatively 19. Draw a diagram of the magnetic field around a bar magnet and Earth. (a) List the similarities between the two fields. (b) List the differences between the two fields. 20. Explain why a single magnetic pole cannot exist by itself. 21. Describe two simple demonstration techniques used to outline the magnetic field around a magnet. 32. Why does a negatively charged ebonite rod initially attract a piece of thread and then eventually push it away? 33. When you touch a Van de Graaff generator, your hair stands on end. Explain your answers to the following questions: (a) Are you being charged by friction, conduction, or induction? (b) Will the same effect occur if you are grounded? Unit VI Forces and Fields 627 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 628 34. A negatively charged ebonite rod is brought 43. A football-shaped hollow conducting object is near a neutral pith ball hanging on an insulating thread. Describe what happens to the charges in the pith ball and the resulting effect (a) before the rod and the pith ball touch (b) after the rod and the pith ball touch 35. A straight length of conducting wire, lying horizontal to the surface of Earth, delivers a current in a direction from south to north. Describe the deflection of a compass needle held directly over the conducting wire. 36. A negatively charged foam plastic ball is hanging from an insulated thread. Another negatively charged ball is brought near, on the same horizontal plane, and the hanging ball swings away so that the supporting thread makes an angle of 30° to the vertical. (a) Draw a free-body diagram depicting the tension force by the string, the force of gravitational attraction, and the electrical force of repulsion on the hanging ball. If the system is in equilibrium, identify the force that balances the gravitational force on the hanging ball and the force that balances the electrostatic force of repulsion on the hanging ball. (b) 37. Explain how the properties of selenium are essential in the operation of a photocopier. 38. Explain the difference between charge shift and charge migration during the process of charging by induction. 39. The cell membrane of a neuron may be thought of as charged parallel plates. The electric potential difference between the outside and the inside of the membrane is about 0.70 V. If the thickness of the membrane is 5.0 109 m, calculate (a) the magnitude of the electric field between the outside and the inside of the membrane (b) the amount of work necessary to move a single sodium ion, Na1, with a charge of 1.6 1019 C, across the me
mbrane 40. In a chart, compare the similarities and differences between Newton’s law of universal gravitation and Coulomb’s law of electrostatics. 41. Determine the distance between two electrons if the mutual force of repulsion acting on them is 3.50 1011 N. 42. A neutral small hollow metal sphere is touched to another metal sphere with a charge of 3.00 102 C. If the two charges are then placed 0.200 m apart, calculate the electrostatic force acting on the two spheres after they touch. 628 Unit VI Forces and Fields charged negatively. (a) Draw the object and then draw the charge distribution on the surface of the object. (b) Draw the electric field lines surrounding the object. (c) Where will the intensity of the electric field appear to be the greatest? (d) Explain how this effect can be used to describe the operation of a lightning rod. 44. A car with a vertical antenna is driven in an easterly direction along the equator of Earth. (a) Describe how a current is induced in the antenna. (b) Determine the direction of the induced current. 45. A small charge of 2.0 C experiences an electric force of 3.0 105 N to the left when it is placed in the electric field of another larger source charge. Determine the strength of the electric field at this point. 46. Calculate the kinetic energy gained by a proton that is allowed to move between two charged parallel plates with a potential difference of 2.0 104 V. What maximum speed could the proton acquire? 47. Two spheres with charges of 4.00 C and 3.00 C are placed 0.500 m apart. At what point between the two charges must a third charge of 2.00 C be placed so that the net electrostatic force acting on this charge is zero? 48. Two oppositely charged parallel plates have a voltage of 2.5 104 V between them. If 1.24 J of work is required to move a small charge from one plate to the other, calculate the magnitude of the charge. 49. A wire that is 0.30 m long, lying perpendicular to an external magnetic field of magnitude 0.50 T, experiences a magnetic force of 0.11 N. Determine the current in the wire. 50. A charge of 3.0 106 C is placed 0.50 m to the right of another charge of 1.5 105 C. Determine (a) (b) the net electric field at a point midway between the two charges the electrostatic force of attraction acting on the two charges 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 629 51. Touching a Van de Graaff generator can result in a painful shock. If a small length of conducting wire is bent into an L-shape and taped to the ball of the generator, so that a shaft of wire projects outward, you will not get a shock when you touch the ball. (a) How does the ball of the generator initially acquire a negative charge? (b) What process of charging objects is involved in the painful shock you receive initially? (c) How does the shaft of wire that is taped to the ball prevent a shock? (d) What is the function of the conducting strips projecting from the airplane wing tip shown in the photo below? 52. A small foam plastic ball with a mass of 0.015 kg and coated with a conducting material is suspended by a string 0.75 m long. If the ball is initially given a charge of magnitude 1.5 108 C, and another charged ball with a charge of magnitude 2.5 108 C is brought near, the hanging ball swings to a position 1.0 cm from its equilibrium position. Calculate the electrostatic force of repulsion acting on the two charges. Draw a free-body diagram. 53. The largest electric field that can exist between two oppositely charged plates with an air gap between them is 3.00 106 N/C. If this electric field limit is exceeded, then a discharge of charge occurs between the plates, resulting in a spark. If the voltage between the plates is 500 V, what is the minimum distance between the plates before a spark occurs? 54. An electron with a mass of 9.11 1031 kg and a charge of magnitude 1.60 1019 C is 5.29 1011 m from a proton with a mass of 1.67 1027 kg and a charge of magnitude 1.60 1019 C. Calculate (a) the gravitational force of attraction between the two masses (b) the electrical force of attraction between the two charges (c) how many times greater the electrical force is than the gravitational force 55. An alpha particle, a proton, and an electron, travelling at the same speed, enter regions with external fields. Compare the motion of these particles as they travel perpendicularly through the same (a) magnetic field (b) electric field (c) gravitational field 56. Given ebonite and glass rods and strips of fur and silk, describe the procedure to charge an electroscope negatively by (a) conduction (b) induction (include the grounding step) 57. A particle with a mass of 2.00 1026 kg and a charge of magnitude 6.40 1019 C is fired horizontally along the surface of Earth in an easterly direction. If the particle passes through a metal detector with a magnetic field of 2.00 104 T in a northerly direction, at what speed must the charged particle be travelling so that the magnetic force counteracts the gravitational force of Earth on the particle at this position? What is the charge of the particle? 58. A helium ion (He2) with a charge of 3.20 1019 C and a mass of 6.65 1027 kg is injected with a speed of 3.00 106 m/s perpendicularly into a region with a uniform magnetic field of 3.30 T. (a) Calculate the magnitude of the deflecting force on the helium ion. (b) If this deflecting force causes the helium ion to travel in a circular arc, what is the radius of the arc? 59. Sphere A, with a charge of 3.50 C, is 4.35 cm to the left of sphere B, with a charge of 2.44 C. Calculate the net electrostatic force on a third sphere C, with a charge of 1.00 C, if this sphere is placed (a) midway on a line joining charges A and B (b) at a point 2.50 cm to the right of charge B (c) at a point 2.50 cm directly down from charge B 60. Calculate the potential difference required to accelerate a deuteron with a mass of 3.3 1027 kg and a charge of magnitude 1.6 1019 C from rest to a speed of 8.0 105 m/s. Unit VI Forces and Fields 629 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 630 61. A uniform electric field of 7.81 106 N/C exists between two oppositely charged parallel plates, separated by a distance of 3.20 mm. An electron, initially at rest and with a mass of 9.11 1031 kg, is injected into the electric field near the negative plate and accelerates toward the positive plate. The electron passes through a “hole” in the positive plate and then travels perpendicularly through an external magnetic field with a magnitude of 1.50 T. (a) What is the voltage between the oppositely charged plates? (b) Calculate the maximum speed acquired by the electron between the plates. Ignore relativistic effects. (c) Determine the magnetic force on the electron as it passes through the magnetic field. (d) Describe the motion of the electron through the electric field and then through the magnetic field. 62. A bar magnet is moved toward a coil of wire that is connected to a sensitive galvanometer, as shown in the figure below. Extensions 63. Explain why computer hard disks are encased in metal. 64. Explain why it is safe to remain inside a vehicle during a lightning storm. 65. Are gravitational, electrical, or magnetic forces responsible for the formation of a black hole in space? Explain your answer. 66. Why must technicians who work on very sensitive electronic equipment be grounded? 67. Describe why Earth’s magnetic field creates a “magnetic bottle.” 68. Charged particles in cosmic rays from the Sun are trapped in Earth’s magnetic field in two major radiation belts, called Van Allen radiation belts. The first belt is about 25 500 km and the other is about 12 500 km above the surface of Earth. Explain why spacecraft must avoid these radiation belts. 69. Compare the motion of a negatively charged particle as it travels through a gravitational, electric, or magnetic field in a direction (a) perpendicular to the field N S (b) parallel to and in the same direction as the field G (a) Explain what happens to the galvanometer readings as the north pole of the bar magnet approaches the coil of wire. (b) Describe how Lenz’s law influences the movement of the magnet toward the coil of wire. (c) Explain what happens to the galvanometer readings as the north pole of the magnet is pulled out of the coil of wire. (d) Describe how Lenz’s law influences the movement of the magnet away from the coil of wire. 70. Why is lightning more likely to strike pointed objects than rounded objects on the ground? 71. Describe the process by which the steel beams in a high-rise building can become magnetized. Why does this effect not happen in homes built with wood beams? 72. High-voltage power lines operate with voltages as high as a million volts. Explain why a bird can perch on a power line with no effect, but must be careful not to touch two nearby power lines. 73. Explain the difference between the dip angle and the angle of declination of Earth’s magnetic field. How do these two angles affect the operation of a directional compass in Alberta? 630 Unit VI Forces and Fields 12-PearsonPhys30-Chap12 7/24/08 3:35 PM Page 631 74. An electron is at rest. Can this electron be set in motion by a magnetic field? Explain your answer. 75. Given only a loop of wire connected to a sensitive galvanometer, describe a simple experiment that could be conducted to prove the existence of a changing magnetic field near a high-voltage power line. 76. The image of the magnetotactic bacterium, as seen under a microscope, reveals a row of magnetite crystals within its cellular structure. Make a hypothesis regarding the purpose of this row of magnetite crystals, and design and describe an experiment to test your hypothesis. (a) Plot a graph to determine the relationship between ⏐B⏐and r. (b) What is the relationship between ⏐B⏐and r? (c) What quantities need to be graphed in order to straighten the graph? (d) Complete a new table of values to straighten the graph. (e) Plot a new graph with the va
riables needed to straighten the graph. (f) Use the graph-slope method and the data from the graph to determine the value of . (g) Use the formula-data substitution method to determine the value of . Skills Practice 77. Draw a Venn diagram to review the similarities and differences between electric and magnetic fields. 78. Design an experiment to determine the direction of gravitational, magnetic, and electric fields. 79. Construct a concept map for solving a two- dimensional electrostatic force problem involving three charges at the corners of a triangle. 80. Design an experiment to show that electrostatic forces vary with the inverse square law. Self-assessment 82. Describe to a classmate which field concepts and laws you found most interesting when studying this unit. Give reasons for your choices. 83. Identify one topic pertaining to fields studied in this unit that you would like to investigate in greater detail. 84. What concept in this unit did you find most difficult? What steps could you take to improve your understanding? 81. A current-carrying wire has a magnetic field 85. Assess how well you are able to explain electric potential energy and electric potential. Explain to a classmate how you determine a reference point. e TEST To check your understanding of forces and fields, follow the eTest links at www.pearsoned.ca/school/physicssource. surrounding it. The strength of this magnetic I field can be found using the formula ⏐B⏐ , r 2 where ⏐B⏐is the magnitude of the magnetic field surrounding the wire, in teslas, is the constant of permeability for space, in Tm/A, I is the current in the wire, in amperes, and r is the distance to the wire, in metres. An experiment is performed to establish the value of the constant, . The current was constant through the wire at 5.00 A, and the strength of the magnetic field was measured at various distances from the wire. The data recorded are given in the table below. Distance from the wire, r (m) Magnitude of the magnetic field, |B| (T) 0.100 0.200 0.300 0.400 0.500 6.28 3.14 2.09 1.57 1.26 Unit VI Forces and Fields 631 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 632 U N I T VII Electromagnetic Electromagnetic Radiation Radiation (a) (b) (c) This series of photos of a supernova remnant shows the various types of electromagnetic radiation that are being emitted from the supernova remnant. The large image is a composite. Images (a) and (b) were taken by the Chandra space telescope (high-energy and low-energy X ray), image (c) was taken by the Hubble space telescope (visible part of the spectrum), and image (d) by the Spitzer space telescope (infrared). 632 Unit VII 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 633 (d) Unit at a Glance C H A P T E R 1 3 The wave model can be used to describe the characteristics of electromagnetic radiation. 13.1 What Is Electromagnetic Radiation? 13.2 The Speed of Electromagnetic Radiation 13.3 Reflection 13.4 Refraction 13.5 Diffraction and Interference C H A P T E R 1 4 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 14.1 The Birth of the Quantum 14.2 The Photoelectric Effect 14.3 The Compton Effect 14.4 Matter Waves and the Power of Symmetric Thinking 14.5 Coming to Terms with Wave-Particle Duality and the Birth of Quantum Mechanics Unit Themes and Emphases • Diversity and Matter • Nature of Science • Scientific Inquiry Focussing Questions The study of electromagnetic radiation and its behaviour requires interpretation of evidence to form theories and models. As you study this unit, consider these questions: • What roles do electricity and magnetism play in electromagnetic radiation? • Does electromagnetic radiation have a wave or a particle nature? • What experimental evidence is required to decide whether electromagnetic radiation has a wave or a particle nature? Unit Project From Particle to Quantum • When you have finished this unit you will be able to use experimental evidence concerning electromagnetic radiation to describe our current understanding of light. You will be able to craft a multi-media presentation of highlights in the development of this understanding. Unit VII Electromagnetic Radiation 633 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 634 C H A P T E R 13 Key Concepts In this chapter, you will learn about speed and propagation of electromagnetic radiation reflection, refraction, diffraction, and interference total internal reflection and Snell’s Law Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe how accelerating charges produce EMR explain the electric and magnetic fields associated with the propagation of EMR explain how the speed of EMR can be investigated and calculated describe reflection, refraction, total internal reflection, and simple optical systems describe diffraction, interference, and polarization and how this supports the wave model of light Science, Technology, and Society explain that scientific knowledge may lead to the development of new technologies and new technologies may lead to scientific discovery explain that scientific knowledge is subject to change as new evidence is found and as laws and theories are tested and subsequently restricted, revised, or reinforced based on new findings and new technologies 634 Unit VII The wave model can be used to describe the characteristics of electromagnetic radiation. Figure 13.1 A Fresnel lens captures light from a lamp and redirects it into a concentrated beam. This 19th-century lens technology used many prisms to produce a light that could be seen for great distances. When you open your eyes in the morning, the first thing you see is light. In fact, it is the only thing you see. In general, what we perceive with our eyes is a combination of many colours, each with varying brightness, but all coming together through the optical system of our eye. It is light or, more broadly speaking, a form of electromagnetic radiation (EMR), which produces an image in the human eye. But what is EMR? How is it produced? How is it transmitted and at what speed? These questions are historically significant, guiding the direction of research and debate over the ages. Finding answers has helped us understand the fundamental principles of our universe and has enabled advances in technology in areas such as lenses (Figure 13.1), fibre optics, and digital devices. Radios, lasers, global positioning systems, and compact discs (CDs) are examples of devices that depend on an understanding of EMR. The nature of light has long been a topic of intrigue and debate. Early Greek scientific thought was based on the work of Aristotle and Euclid, who concerned themselves with the physical and geometric mechanisms of visual perception. A more modern debate ensued as detailed evidence began to be collected about the wave and/or particle nature of light. Isaac Newton (1642–1727) and Christiaan Huygens (1629–1695) defined this debate with evidence from early experiments. Newton put forward the particle or corpuscular theory, while Huygens supported the wave model. Today, both the particle and wave models of light have some validity and contribute to our present understanding, described by the quantum model of light. 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 635 In this chapter, you will see that visible light is only one form of EMR. You will also discover that the concepts of electrical charge and magnetic fields, explored in Chapters 10–12, come together when you consider EMR. You will investigate the generation, speed, and propagation of EMR, and see how light can be reflected, refracted, diffracted, and polarized. In each instance, the wave and particle nature of light is revealed, helping us to understand these complex phenomena and apply them in new technologies. 13-1 QuickLab 13-1 QuickLab Investigating Light as a Form of EMR Problem How can the properties of light be revealed? Materials light source 2 polarizing filters Procedure 1 Place a light source in front of one polarizing filter and observe whether the light is able to penetrate the filter. ? ? ? Figure 13.2 Think About It 2 Turn the filter 90 degrees as shown in Figure 13.2 and observe whether the light is able to penetrate the filter. 3 Place the filters at 90 degrees to one another and observe whether the light is able to penetrate the filters. 4 Slowly rotate one of the filters and observe whether the light is able to penetrate both filters at a variety of angles and positions. Questions 1. Can visible light pass through a single polarizing filter, regardless of how it is oriented? 2. Can visible light pass through two filters when they are aligned at 90 degrees to one another? 3. Describe one characteristic of EMR indicated by your observations of light in this QuickLab. 4. Do the observations made in this investigation explain what EMR is, or do they simply reveal one characteristic of EMR? 1. Do your observations from 13-1 QuickLab give information on whether electromagnetic radiation has wave or particle characteristics? Explain. 2. What experimental evidence would be needed to decide whether electromagnetic radiation has a wave or a particle nature? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 635 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 636 electromagnetic radiation: radiant energy in the form of a wave produced by the acceleration of electrons or other charged particles. EMR does not require a material medium; can travel through a vacuum. frequency: the number of cycles per unit of time wavelength: the distance between adjacent points that vibrate in phase with one another in a wave info BIT Seeing is a photochemical process that is sensitive to certain wavelengths of el
ectromagnetic radiation. When EMR is absorbed by the tissues in the human eye, a compound called retinal changes in physical form from bent to straight. The retinal molecule, in turn, is connected to a membrane-bound protein called opsin forming the complex molecule called rhodopsin. When the retinal molecule changes its form, it separates from the rhodopsin and the opsin triggers a nerve cell to signal the brain that light has been seen. 13.1 What Is Electromagnetic Radiation? Electromagnetic radiation (EMR) is radiant energy, energy that travels outward in all directions from its source. There are different types of EMR, some very familiar, and others rarely mentioned outside scientific discussions. Electromagnetic radiation includes AM/FM radio waves, microwaves, heat, visible light (red to violet), ultraviolet radiation, X rays, and gamma rays. Electromagnetic radiation types are identified based on their frequency, wavelength, and source. The energy sources that produce EMR vary greatly, from nuclear reactions in the Sun, which generate gamma radiation, to chemical reactions in the human body that generate infrared radiation (heat). EMR is produced by the acceleration of charged particles resulting in transverse waves of changing electric and magnetic fields that can travel through space without the need of a material medium. All forms of EMR travel at the same speed, commonly referred to as the speed of light (c, in a vacuum), equal to 3.00 108 m/s. EMR does not always travel directly from the source to the observer. As a result, it can be observed in one of two ways: directly from the source indirectly, as reflected or transmitted radiation Burning magnesium (Figure 13.3) allows both methods of observing radiation. Other objects, such as the text you are reading now, do not produce radiation. You are able to read the text because it is reflecting radiation (light) from another source. If you are wearing glasses, this radiation is transmitted through your lenses. The evidence obtained in 13-1 QuickLab indicates one unique characteristic of electromagnetic radiation. However, visible light, radiant energy that the eye can detect, only makes up a small portion of the spectrum of EMR present in our universe. Concept Check Figure 13.3 When magnesium burns, you can observe radiation directly from the reaction and indirectly from the radiation that is reflected from the smoke. Can infrared radiation reflect off objects and be observed by the human body? Explain and give examples. 636 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 637 electromagnetic spectrum: all types of EMR considered in terms of frequency, wavelength, or energy e WEB To learn more about the different types of EMR and the trends seen across the spectrum, follow the links at www.pearsoned.ca/school/ physicssource. Types of Electromagnetic Radiation The electromagnetic spectrum is the term applied to all the types of EMR considered together in terms of frequency, wavelength, or energy. All parts of the spectrum are found, with varying intensity, in our natural environment. We are most familiar with the visible spectrum since we sense it directly with our eyes. The infrared spectrum is sensed as heat and the ultraviolet spectrum includes radiation that can damage living cells, often causing a physiological response such as sunburn. Other parts of the spectrum may be present as background radiation. Natural background radiation originates from two primary sources: cosmic radiation and terrestrial sources. Cosmic radiation from deep space interacts with atoms in the atmosphere producing X rays and unstable isotopes. Terrestrial sources of gamma radiation include radioactive isotopes of uranium, radon, potassium, and carbon. Also, aboveground nuclear tests of the 1940s–1960s combined with nuclear accidents such as Chernoybl have scattered a substantial amount of radioactive material within our environment. Figure 13.4 shows the overlap of frequencies and wavelengths among the different forms of EMR in the electromagnetic radiation spectrum. 104 108 1012 1016 1020 1024 frequency f (Hz) radio waves microwaves infrared visible light ultraviolet X rays gamma rays 3 x 104 m 3 m 3 x 10–4 m 3 x 10–8 m 3 x 10–12 m wavelength λ (m) a) Electromagnetic spectrum 800 nm 700 nm 600 nm 500 nm 400 nm wavelength λ (m) infrared red yellow blue b) Visible spectrum orange green violet ultraviolet Figure 13.4 The electromagnetic radiation spectrum showing the visible range The relative energy of the different types of EMR varies with frequency across the spectrum. Table 13.1 compares the various sources and characteristics of the radiation found in the EMR spectrum. e WEB To learn more about the medical uses of different types of EMR, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 637 13-PearsonPhys30-Chap13 7/24/08 3:43 PM Page 638 Table 13.1 The Electromagnetic Spectrum: Characteristics Method of Production Characteristics Problems oscillation of electrons in an electric circuit like an antenna oscillation of electrons in special tubes and solid state devices motion of particles, transitions of valence electrons in atoms and molecules higher-energy transitions involving valence electrons in atoms long wavelength allows a large amount of diffraction making it useful for longdistance communication, e.g., PC broadband shorter wavelength reduces diffraction for short-distance communication; frequency matches the natural resonant frequency of water molecules; used in microwave ovens and cell phones causes object absorbing it to become warm; used for remote sensing, night vision scopes, and identification of sources of heat reflects off small objects, making them visible; diffracts around very small objects, making them invisible requires government regulations to control transmission and avoid interference may be linked to some forms of cancer; causes damage to living tissue due to heating of water molecules within tissues significant exposure can burn tissue limits the size of objects that can be seen even higher-energy transitions involving valence electrons in atoms easily absorbed by objects; causes fluorescence of some materials, tanning in humans; kills bacteria may cause sunburn; prolonged exposure can cause mutations and cancer in humans Type of Electromagnetic Radiation Radio and Radar f 104 1010 Hz 104 102 m relative energy: very low Microwaves f 109 1012 Hz 101 104 m relative energy: low Infrared f 1011 4.0 1014 Hz 103 7.5 107 m relative energy: low Visible f 4.0 1014 7.5 1014 Hz 7.5 107 4.0 107 m relative energy: medium Ultraviolet f 7.5 1014 1017 Hz 4.0 107 109 m relative energy: high X ray f 1017 1020 Hz 109 1012 m relative energy: very high penetrates most matter and is absorbed by denser material (like bone or metal); destroys carcinogenic or mutant cells; used for medical imaging in humans and in industry penetrates matter very deeply; destroys carcinogenic or mutant cells on a local scale; used to probe the structure of matter and in industrial imaging penetrates matter very deeply; study of cosmic rays allows investigators to formulate ideas about the universe can cause mutations and cancer in humans can cause radiation sickness and death can cause radiation sickness and death transitions of electrons in an atom or the sudden acceleration of high-energy free electrons Gamma f 1019 1024 Hz 1011 1016 m relative energy: extremely high decomposition of unstable nuclei, either spontaneously or by the sudden negative accelerations from highenergy particle accelerators Cosmic f 1024 Hz and greater 1016 m and less relative energy: extremely high bombardment of Earth’s atmosphere by extremely high-energy particles from space 638 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 639 Competing Models of Electromagnetic Radiation Historically, investigators have tried to explain transmission, reflection, refraction, and absorption, and the other characteristics common to all types of EMR by using models. The historical particle model describes EMR as a stream of tiny particles radiating outward from a source. A particle is a discrete unit of matter having mass, momentum (and thus kinetic energy), and the ability to carry an electric charge. The particle model of EMR is the simplest of all the models. It is supported by the facts that EMR propagates in straight lines, can be reflected, and can be absorbed. The pool ball (particle) in Figure 13.5 exhibits particle characteristics. It can travel in a straight line, obey the law of reflection when it bounces off a side rail, and be absorbed by the table when it drops into a pocket. info BIT The Particle Theory of Light, also known as the Corpuscular Theory of Light, was put forward in Newton’s Opticks, published in 1704. particle model: describes EMR as a stream of tiny particles radiating out from a source wave model: describes EMR as a stream of transverse waves radiating out from a source Figure 13.5 A pool ball on a pool table exhibits the particle characteristics of linear movement, reflection, and absorption. Figure 13.6 A water wave is a transverse wave that transfers energy in the form of a disturbance. waterbed sponge A second model, the wave model, describes EMR as a stream of transverse waves radiating outward from a source. As you learned in Chapter 8, a wave is a transfer of energy in the form of a disturbance. The energy transfer usually occurs in, but is not limited to, a material medium like water. A water wave travels in a straight line, reflects from surfaces, and can be absorbed (Figure 13.6). For example, water waves, which continually reflect inside a waterbed, transfer unwanted energy that can be absorbed by a sponge (Figure 13.7). EMR, however, does not require the presence of a medium. The modern wave model of light has
its origins in the 17th century when the Dutch mathematician and scientist Christiaan Huygens argued that light consisted of waves. He suggested light waves could interfere to produce a wave front, travelling outward in a straight line. At the time, however, Huygens’ wave theory was overshadowed by the immense scientific popularity of Newton’s particle theories. The wave model gained further support from evidence produced by the work of Thomas Young (Figure 13.8) and his now-famous two-slit experiment of 1801. The details and implication of the experiment are set out in section 13.5, but in general, Young was able to show that a beam of light, when split into two beams and then recombined, shows interference effects that can only be explained by assuming that light has wavelike properties. Figure 13.7 Waves in a waterbed exhibit the wave characteristics of linear movement, reflection, and absorption. Figure 13.8 Thomas Young (1773–1829) Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 639 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 640 e WEB To learn more about Young’s experiment and its relationship to the models of EMR, follow the links at www.pearsoned.ca/school/ physicssource. If light behaves as a particle, then one would expect to observe a bright region wherever the light emanating from the slits reaches a distant screen (Figure 13.9(a)). But Young obtained a result similar to that shown in Figure 13.9(b), where a pattern of light of varying intensity is observed after the light passes through the two slits. This pattern is similar to two-source interference patterns in water waves, as seen in section 8.3. Young’s experiment thus presented strong evidence for the wave model of light. schematic representation of particle theory prediction schematic representation of wave theory prediction (a) (b) screen screen Figure 13.9 When light is incident on two small slits, it is diffracted as it passes through each slit and an interference pattern is observed that supports the wave model of light. a) A schematic representation of the particle theory prediction b) A schematic representation of the wave theory prediction Planck and Einstein By the end of the 19th century, both the particle and wave models of EMR were supported by scientific evidence. In 1900, Max Planck (1858–1947) proposed a radically new model to explain the spectrum of radiation emitted from a perfectly black object. In a mathematical derivation Planck assumed that all of the vibrating molecules, “oscillators,” in the black body could vibrate with only specific, discrete amounts of energy. In doing so, he had to ignore the continuous distribution of energy of classical physics and introduce the concept of quanta, or discrete packets of energy. Planck was awarded the Nobel Prize for Physics in 1918, “in recognition of the services he rendered to the advancement of Physics by his discovery of energy quanta.” In 1905 Einstein extended this quantum theory by proposing that light is emitted in quantized, tiny, massless particles, which are now called photons. Planck’s original theory thus evolved into the currently accepted quantum model of light, which is a combination of both the particle and wave models. This model describes light and all other electromagnetic radiation as discrete bundles or “packets” of energy. Each packet, or photon, has both particle and wave characteristics. In the quantum model, EMR has two aspects of behaviour, one being info BIT A quantum is a discrete unit of energy. The term originates from the Latin quantus, which means “how much.” “Quanta” is the plural of “quantum.” The Quantum Theory was so revolutionary that its formulation distinguishes the shift from classical physics to what we now call modern physics. photon: (from the Greek word meaning “light”); a discrete packet of energy associated with an electromagnetic field quantum model: light and all other EMR are discrete bundles of energy, each of which has both particle and wave characteristics 640 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 641 wavelike and the other being particle-like. Quantum theory encompasses these two types of behaviour while, at the same time, challenging both the classical wave and particle models. Planck’s quantum idea challenged the wave model by proposing that EMR does not deliver energy in a continuous form as a wave would, but rather, it delivers energy in small bundles. At the same time, his idea challenged the particle model by limiting the energy of the particles to certain discrete values, a condition not possible for a Newtonian particle. Concept Check Consider the wave model of light and the particle model of light. 1. Which of the following best describes the word “model”? (a) simplified description of a complex entity or process; (b) a representation of something on a smaller scale; (c) both a and b. 2. Explain your answer from part 1 and provide several examples of models that are used in a similar way. 3. The EMR spectrum includes many classes of radiation. Describe the wave characteristics used to classify each type of radiation on the spectrum. How are these characteristics related to the energy of the radiation? Maxwell’s Electromagnetic Theory, 1865 In 1865, James Clerk Maxwell (1831–1879) proposed his Electromagnetic Theory, which synthesized earlier ideas and theories with the results of experiments to provide a theoretical framework for future studies. Maxwell proposed the idea that a changing electric field produces a changing magnetic field and that the interaction between these fields propagates as a wave through space. In his theory Maxwell linked concepts of electricity and magnetism so that we now call them “electromagnetic.” Maxwell based his theory on key phenomena observed by earlier investigators of electricity and magnetism. You have already met these ideas in Unit 6, but we will restate them here to appreciate the importance of Maxwell’s contribution. The Concept of the Electric Field As Faraday had proposed and as you saw in Chapter 11, an electric field surrounds any charged particle and an electrostatic force will act on another charged particle when it is placed in that field. The electric field strength can be calculated and electric field lines drawn to illustrate the region of influence. Electric field lines begin and end at a charge (Figure 13.10) and the number of field lines at any closed surface is determined by the total charge at the surface. This concept allows the idea of interaction between particles at a distance, even though there is no contact between them. e WEB To learn more about Max Planck, begin your search at www.pearsoned.ca/ school/physicssource. Project LINK How will you present the observations and theoretical concepts that culminated in the ideas of Planck and Einstein? electric field: a three-dimensional region of electrostatic influence surrounding a charged object magnetic field: a threedimensional region of magnetic influence surrounding a magnet Figure 13.10 Electric field lines are used to illustrate the region of influence between two oppositely charged particles. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 641 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 642 S N Figure 13.11 The number of magnetic field lines that exit a magnetic material is equal to the number of magnetic field lines that enter a magnetic material — forming a closed loop. capacitor: two conductors, holding equal amounts of opposite charges, placed near one another without touching Maxwell’s Equations: a series of equations that summarized the relationships between electricity and magnetism and predicted the existence of electromagnetic waves and their propagation through space info BIT Maxwell was working with a system of units for electromagnetic theory, referred to as the CGS system. This system was derived from the base units of centimetre, gram, and second. The CGS system has largely been replaced by the SI system, based on the metre, kilogram, and second. The Concept of the Magnetic Field A magnetic field (as discussed in Section 12.1) is a three-dimensional region of magnetic influence surrounding a magnet. Magnetic field lines form a closed loop and represent the direction and magnitude of the magnetic field (Figure 13.11). Relationships Between Electricity and Magnetism Electrical current in a conductor produces a magnetic field perpendicular to the current (Oersted, Unit VI, page 587) and the strength of the magnetic field depends on the magnitude of the current (Ampere, Unit VI, page 607). This relationship is known as Ampere’s Law. Inversely, moving a conductor connected in a circuit through a magnetic field induces an electric current. The magnitude of the electrical current is directly related to the rate of change in the magnetic field (Faraday and Henry, Unit VI, page 611). This relationship is known as Faraday’s Law. Putting the Laws of Electromagnetism Together From this basis and through his work with capacitors, Maxwell developed the laws of electromagnetism. Maxwell’s work with capacitors showed that the electric field produced by a capacitor could have the same effect as a moving charge. In other words, a changing electric field will produce a changing magnetic field in the same manner as a changing electric current can produce a changing magnetic field. This information was extremely important because it showed that a conductor is not necessary for an electromagnetic wave to exist. Maxwell put these ideas together with incredible ingenuity using calculus, which is beyond the scope of this text. His theory is embodied in a series of equations known as Maxwell’s Equations. To visualise the relationship between the electric and magnetic fields, we will proceed stepwise. Begin with a plane transverse wave that shows an electric field of varyi
ng strength (Figure 13.12). Next, consider two of the electric field lines on the transverse wave, an upward one at one location and the corresponding downward one at another. Imagine a path connecting the tips of these two electric field vectors, as shown by the dotted line in Figure 13.13. You can see that the electric field changes direction continuously from one direction to the opposite direction along a "closed path." E E direction of propagation of wave direction of propagation of wave E Figure 13.12 Plane transverse wave showing the electric field lines. Figure 13.13 A path connecting two oppositely directed electric field lines on the transverse wave shows the change in the electric field. 642 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 643 Now, consider the magnetic field that exists perpendicular to the electric field along this closed path. When the electric field is set in motion, the magnetic field will change along with the electric field, and similarly along a closed path. An electromagnetic wave will propagate in a direction perpendicular to both fields (Figure 13.14). The changing electric and magnetic fields will propagate, or radiate, through space in the form of a wave — an electromagnetic wave (Figure 13.15). Maxwell proposed that the electromagnetic wave consists of periodic variations in the electric and magnetic field strengths and that these variations occur at right angles to one another as the wave propagates. B E direction of propagation of wave Figure 13.15 Three-dimensional view of an electromagnetic wave Maxwell’s Predictions Maxwell’s equations not only correctly predicted the existence of electromagnetic waves, but also allowed him to make some predictions about the waves’ properties. 1. Electromagnetic waves are produced whenever an electric charge is accelerating. Therefore, as an electric charge oscillates, electrical energy will be lost, and an equivalent amount of energy will be radiated outward in the form of oscillating electric and magnetic fields. 2. When the electric charge is accelerated in periodic motion, the frequency of oscillation of the charge will correspond exactly to the frequency of the electromagnetic wave that is produced. 3. All electromagnetic waves will travel at a speed of 310 740 000 m/s and obey the universal wave equation (c f) relating speed, frequency, and wavelength. (Note that Maxwell’s theoretical prediction was not far from today’s currently accepted value of 3.00 108 m/s for the speed of light in a vacuum.) 4. The oscillating electric and magnetic fields will always be perpendicular to each other and perpendicular to the direction of propagation of the wave. 5. Electromagnetic waves should show all the phenomena associated with transverse waves: interference, diffraction, refraction, and polarization. It is Maxwell’s last prediction that supports the wave model of EMR and relates his predictions to experimental evidence. Interference, diffraction, polarization, and refraction, as they relate to the wave model of EMR, will be explored in sections 13.4 and 13.5. E B direction of propagation of wave Figure 13.14 The magnetic field lines (in red) allow us to visualize the magnetic field that exists perpendicular to both the electric field and the direction of the wave. electromagnetic wave: periodic variation in perpendicular electric and magnetic fields, propagating at right angles to both fields e WEB To learn more about ways of representing the relation between electric and magnetic fields, follow the links at www.pearsoned.ca/school/ physicssource. Compare and contrast the representations you find. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 643 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 644 Producing Electromagnetic Radiation — The Story of Accelerating Charge In 1887, the German physicist Heinrich Hertz (1857–1894) set up an experiment designed both to produce and to detect EMR. In his experimental apparatus, Hertz used a radiator, consisting of a pair of wires attached to both a high-voltage induction coil and a pair of capacity spheres. The wires were separated by a small gap and, given a sufficient quantity of opposite charge on each wire, a current would oscillate back and forth across the gap at a frequency of 109 Hz. With each oscillation a spark was produced when the moving charge ionized the air molecules on its way from one wire to the other. From Maxwell’s equations Hertz knew theoretically that this rapidly moving electric charge should produce EMR. A short distance away from the radiator a collector plate containing a small loop of wire, the antenna, was observed to detect the effect of EMR. While the radiator was in operation and when the radiator and the antenna were tuned to the same frequency, a spark was observed at the antenna indicating a potential difference and an electric current (Figure 13.16). 3. Electromagnetic waves create electric current in antenna loop; produces small spark in spark gap. 2. Spark produces electromagnetic waves. 1. Induction coil produces high voltage. e WEB To learn more about details of Hertz’s experiment, follow the links at www.pearsoned.ca/school/ physicssource. Figure 13.16 Hertz’s apparatus consisted of a high-voltage induction coil, a radiator that produces sparks, and an antenna loop. Relating Theory and Practice in Hertz’s Experiment The word “changing” appears a number of times in Maxwell’s original proposal. As Maxwell understood it, a “changing” electric field was crucial to creating an electromagnetic wave. This is where the induction coil was important in Hertz’s experimental design. The induction coil rapidly changed the electric field across the spark gap. When this electric field reached a sufficiently high value, the electrons in the wire “jumped” from one electrode to the other. As the charge was rapidly transferred, the electric field underwent a rapid change that caused a changing magnetic field, which then caused a changing electric field, and so on. An electromagnetic wave was produced and it radiated outward in all directions. 644 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 645 In Hertz’s experiment, when the receiver was tuned to the same frequency as the radiator, the induced current flow in the antenna oscillated at a frequency identical to that of the changing electric field in the radiator. This was conclusive evidence that Hertz’s device had indeed produced the EMR that was being observed at his antenna. Furthermore, he was able to measure the velocity of the waves by using a zinc reflector plate to produce a standing wave and moving a ring antenna within the standing wave. He could determine the magnitude and direction of the wave’s components and therefore the wavelength. Given the frequency of the radiator, the velocity of the wave could be calculated using the universal wave equation (v f). Hertz had produced and measured the wavelength and velocity of non-visible EMR for the first time. Two years later, in 1889, radio pioneers used this method to transmit the first radio waves across the English Channel. By 1901, the first radio waves were transmitted across the Atlantic Ocean from Cornwall, England to St. John’s, Newfoundland and a new age of technology had dawned. The production of EMR was one of the greatest scientific achievements of the 19th century, ushering in new possibilities and technologies that are commonplace today. M I N D S O N Going Wireless e WEB To learn more about Marconi and details of the first trans-Atlantic radio transmission, follow the links at www.pearsoned.ca/school/ physicssource. Wireless communication systems that transmit data are commonplace today. 1. Create a list of wireless data transmission technologies. 2. What process must be common to all wireless transmission devices? 3. What variable is used to control the acceleration, and hence the energy, of the accelerating electrons in a wireless device? 4. One radio transmission tower transmits at 5.0 kV. Another transmits at 10.0 kV. Which tower has a greater signal strength and why? 5. Cell phones communicate with cell towers that are dispersed throughout urban cities and transportation corridors. Explain the relationship between the distance that separates a cell phone from the nearest cell tower and the operating power of the cell phone. 6. Many studies have been conducted to investigate and determine if there is a relationship between brain cancer and cell phone usage. Why would one suspect a relationship between cell phone use and brain cancer? Although modern radio signals are generated without a spark, the technology operates in a way similar to Hertz’s original experiment. Radio waves are generated by rapidly changing the electric potential, or voltage, in the radiator tower. The oscillating voltage produces an oscillating electric field in the radiator tower (Figure 13.17). microphone radio signal (radio frequency) RF amp amp sound waves audio signal (audio frequency) AF mixer radiator tower Figure 13.17 Schematic of a simple radio transmitter audio wave (AF) carrier wave (RF) amplitude modulation (AM) Figure 13.18 Amplitude modulation. The audio signal and the carrier signal are mixed by modifying the amplitude of the carrier signal. audio wave (AF) carrier wave (RF) frequency modulation (FM) Figure 13.19 Frequency modulation. The signal is combined with a carrier wave to create a resultant wave with constant amplitude, but varying frequency. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 645 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 646 THEN, NOW, AND FUTURE From Analog Radio to Digital Wireless Analog Radio Technology The first application of Hertz’s discovery was made by Guglielmo Marconi (1874–1937). He recognized the potential for transmitting information using electromagnetic wav
es, by coding the information into dots and dashes, like the Morse code already used in telegraphy. Today, the dots and dashes have been replaced by an analog signal, which uses a continuous spectrum of values. In radio, sound waves in the range of 20–20 000 Hz (audible to humans) are converted into a weak electrical signal in an analog form. This wave is called the audio frequency signal (AF). The weak audio frequency is then amplified and sent to a mixer. In the mixer, a radio frequency (RF) called a carrier signal is added to the audio signal to produce an analog wave that contains two sets of information: the audio signal and the carrier signal. The carrier signal frequency is determined by the government and is unique to that station for that broadcast area. It is the carrier signal that you select when you tune your radio to the station. Finally, the mixed signal is amplified and delivered to the radiator tower, where the movement of electrons through the radiator wire will produce the corresponding form of electromagnetic wave (Figure 13.17). Television signals follow the same process, except that the video and audio portions are transmitted separately. Radio reception is accomplished by tuning the radio to the carrier signal and then removing, or demodulating, the wave. This reveals the audio signal, which is amplified and delivered to the speakers. AM or FM Radio? The mixing of the audio signal and the carrier signal occurs in one of two ways. The first is called amplitude modulation or AM (Figure 13.18). The second method is called frequency modulation or FM. The audio signal and the carrier signal are mixed by modifying the frequency of the carrier signal (Figure 13.19). Figure 13.20 Digital Wireless Cell Phones The cell phone and text messaging technology of today operates on a very different set of principles. When a cell phone is switched on, it registers with the mobile telephone exchange using a unique digital identifier. Once it has been identified, it stays connected to a cellular network of ground-based stations, commonly referred to as cellular tower base stations, which can be found on radio towers, buildings, in church steeples, or any location which is free of physinterference. ical obstacles and Figure 13.20 shows one such cellular tower base station. A cell phone communicates with the cell tower base station using radio waves. In turn the base station relays all information to and from the cell phone to another subscriber of the same cell phone network or through an interconnected public switched telephone network that uses fibre optic and copper land lines to transmit the data. When in use, a cell phone converts the analog voice signal of the person talking into a stream of digital data, which is received by the nearest cell tower base station that may be anywhere from 0.8 –13 km away. Each cell tower site has a lowpower microwave broadcast which can be picked up by cell phones. As the user moves, the cell phone constantly monitors the signals being received from various nearby cell tower stations, switching from one tower to another in such a way that the signal strength remains as high as possible. The stream of digital data is unique to each type of network technology and each operator is assigned a different radio frequency so that several networks can operate simultaneously at the same location and they will not interfere with one another. Wireless Technology for the Future Future uses of cellular phone technology are emerging in the market place as the networks evolve and consume broadcast frequencies formerly occupied by some television channels. New entertainment features will include: streaming video; music downloads; podcasts; smooth speech recognition and language translations; ebook features that use projection; barcode readers that direct users to Internet urls; global positioning systems and mapping combined with accelerometers to measure position and movements of the cell phone user. Questions 1. Suggest why cell phone user data, including geographical position and movement, could be considered an invasion of privacy. 2. How can multiple cell phone networks and traditional radio broadcasts all exist in the same geographical location and not interfere with one another? 3. Compare the carrier frequency of radio broadcasts with the unique digital identifier of a cell phone. What common purpose do they both serve? 4. Explain why a cell phone only works in certain areas and how changes in the signal strength will be observed by the cell phone user as he or she moves. 646 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 647 Concept Check 1. Consider two electric field lines on a transverse electromagnetic wave. Explain how a changing electric field will produce a changing magnetic field. 2. Why can electromagnetic radiation propagate through empty space, in the absence of a conductor? 3. An electromagnetic wave is produced if an electric charge accelerates. Explain how these two phenomena are linked. 13.1 Check and Reflect 13.1 Check and Reflect Knowledge 1. List two ways that electromagnetic radiation is detected. 2. List the different types of EMR in the spectrum and the ways in which they are classified. 3. Define EMR using the (a) particle model (b) wave model (c) quantum model 4. Describe the evidence that Thomas Young obtained from the two-slit experiment and explain how it supported the wave model of light. 10. If the magnetic field lines held within a closed path are constant and unchanging, will there be an electric field along the closed path? Explain. 11. If an electric charge is not moving, will it produce a magnetic field? Explain. 12. Explain why it is necessary for the technician to leave the room or wear protective clothing when a patient is being X rayed. 13. Describe how an antenna is affected by an electromagnetic wave. 14. Describe one method used to produce EMR with a known frequency. 5. Explain how the observations made by Extensions Hertz support the electromagnetic theories of Maxwell. 6. How did Maxwell’s study of capacitors add to his theories? 7. Why is the process of “changing” critical to the production of EMR? Applications 8. A high-voltage spark gap produced sparks with a frequency of 2.5 103 Hz. What is the frequency of the EMR observed at an antenna located nearby? 9. What properties need to be considered when choosing the material to build an antenna for a radio receiver? 15. Explain, with the aid of a transverse standing wave diagram, how the current in a ring antenna (such as the one used by Hertz) will be affected as the antenna is moved along the axis of the standing wave. How could this data be used to determine the wavelength of the standing wave? 16. Draw a schematic that shows, in general terms, the role of a satellite in delivering digital, high-definition, television signals between the signal provider and the customer’s home receiver. e TEST To check your understanding of the nature and production of electromagnetic radiation, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 647 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 648 info BIT Maxwell predicted the speed of EMR before Hertz experimentally verified it. Maxwell knew that electric and magnetic fields are not independent of one another and predicted a speed of propagation by dividing the constants, ke and km , which describe electric and magnetic forces between moving charged particles. Using today’s currently accepted values of ke and km , we can show that the magnetic field and the electrical field are related by a particular speed. 8.99 109 N 2 m 2 C s2 1.00 107 N 2 C 2 m 8.99 1016 s2 k e k m 8.99 1016 2.99 108 m m2 s2 s 13.2 The Speed of Electromagnetic Radiation In addition to producing and detecting electromagnetic radiation in his 1887 experiment, Hertz was able to show that the radiation was travelling at approximately 3.0 108 m/s. This was an empirical verification of a theoretical speed predicted by Maxwell years earlier, as described on page 643: All electromagnetic waves will travel at a speed of 310 740 000 m/s and obey the universal wave equation (c f) relating speed, frequency, and wavelength. Empirical Determination of the Speed of Electromagnetic Radiation Light travels so quickly that there is no apparent time difference between turning on a light and the light reaching your eyes – even when the light source is many kilometres away. Galileo (1564–1642) found this out in his attempt to measure the speed of light. His experiment was very basic: Galileo and an assistant stood, each with a covered lantern, on adjacent hills separated by about a kilometre. Galileo would uncover his lantern and as soon as his assistant observed Galileo’s lantern, he would uncover the second lantern so that Galileo would see it. Galileo used his pulse rate to measure the time difference. The time interval is actually extremely small – Galileo didn’t know it at the time, but light could travel 2 km in about 0.000 066 67 s, a time you couldn’t even measure with a digital stopwatch. Galileo realized that using his pulse as a timepiece was not appropriate. Earth longer time Sun Earth shorter time Io Jupiter Earth’s orbital diameter approximate Figure 13.21 Earth’s orbital diameter causes the eclipse of Io to occur at different times because of the extra distance the light must travel when Earth is farthest from Jupiter. 648 Unit VII Electromagnetic Radiation Olaus Rømer (1644–1710) and Christiaan Huygens used one of the four easily visible moons of Jupiter to calculate the speed of light. The period of revolution for Io (Jupiter’s brightest moon) had been accurately determined using large amounts of astronomical data collected over many years. Roemer knew that the moon Io should be eclipsed and disappear behind Jupiter at regular, periodic i
ntervals (every 42.5 h). However, he discovered that Io appeared to be eclipsed later than scheduled when Earth was farther away from Jupiter and earlier than scheduled when Earth was closer to Jupiter. The time difference between the longest and shortest periods was 22 min (Figure 13.21). 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 649 Huygens reasoned that this discrepant behaviour could be due to the difference in the time taken for the light to travel from Jupiter to Earth. Following this reasoning, he estimated Earth’s orbital diameter and then determined the speed of light based on the following calculations. Orbital diameter 3.0 1011 m v d t Time difference 22 min 1320 s 3.0 1011 m 1320 s 2.3 108 m/s His calculations produced a value of 2.3 108 m/s — astoundingly close to the currently accepted value of 3.0 108 m/s. The value was so large it was rejected by the scientific community of the time and was not accepted until after both Huygens and Roemer were dead. In 1848, Armand Fizeau became the first person to successfully measure the speed of light at the surface of the planet. His apparatus consisted of a rotating toothed wheel, a light source, some lenses, and a mirror. In his experiment, light was allowed to pass through one of the gaps on the toothed wheel and travel toward the mirror (located on an adjacent hilltop 8.63 km away) where it was reflected back toward the spinning wheel (Figure 13.22). The toothed wheel was rotated such that the reflected light was blocked by a tooth in the wheel as it turned, and the observer would not be able to see the source light. Using the rotational frequency of the spinning wheel, Fizeau was able to determine the time it took the light to travel 8.63 km and back (17.26 km round trip) and therefore to determine the speed at which the light was travelling to the distant mirror and back. His experimentally determined value for the speed of light was 3.15 108 m/s — only 5% more than the currently accepted value. toothed wheel mirror glass observer path A path B light source observation point water in water out Figure 13.22 Fizeau’s original apparatus for measuring the time it takes light to travel the distance (8.63 km) and back Figure 13.23 Fizeau’s apparatus for measuring the speed of light in water light source Three years later Fizeau conducted an investigation to determine if the speed of light was affected by a moving medium such as water. In this experiment, water was pumped in opposite directions through two parallel tubes (Figure 13.23). Light from a single source was directed through both tubes, following paths A and B, of identical lengths, to an Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 649 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 650 observation point. Fizeau argued that if the speed of the light waves changed as a result of travelling through the water tubes, the waves arriving from path A and B would produce an interference pattern at the observation point. A device called an interferometer was used to observe this interference. The experiment proved that light would travel at different speeds in different media. The interference pattern produced was also very strong evidence that light was a wave. octagonal set of mirrors observer telescope path of light Mount Wilson 35 km light from source curved mirror Mount St. Antonio Figure 13.24 Michelson’s experimental apparatus to measure the speed of light Practice Problems 1. Students using a 12-sided set of rotating mirrors in an experiment similar to Michelson’s determine the speed of light to be 2.88 108 m/s. The mirrors are located 30.0 km from the fixed mirror. What is the frequency of rotation if 1/12th of a rotation occurs before the light returns? 2. Light reflected from an 8-sided set of rotating mirrors travels 15.0 km to a distant fixed mirror and back. At what frequency is the set rotating if it has turned 1/8th of a rotation before the light returns? 3. An 8-sided set of mirrors, similar to Michelson’s, rotating with a frequency of 500 Hz, is located 36.0 km away from a fixed mirror. If the returning light is observed in the system, at what speed is the light travelling? Answers 1. 400 Hz 2. 1.25 103 Hz 3. 2.88 108 m/s 650 Unit VII Electromagnetic Radiation Michelson’s Experiment Building on Fizeau’s 1848 experimental design and the work of others, Albert Michelson performed an experiment in 1905, using a rotating set of mirrors instead of a toothed wheel. A very intense light source was directed at an 8-sided, rotating set of mirrors, which reflected the light toward a curved mirror located 35 km away. After travelling 35 km, the light was reflected back toward the rotating mirrors. If the rotating mirrors had made 1/8th of a rotation just as the light returned, the returning light could be observed in a telescope (Figure 13.24). For the returning light to be observed in the telescope, the rotating mirrors had to turn at a very precise frequency. Michelson knew the light had to travel a round trip distance of 70 km, and if the rotating mirrors had made 1/8th of a rotation (or a multiple of an 1/8th rotation) between the time the light left and then returned, the light could be observed in the telescope. All Michelson needed to measure was the frequency of rotation and he could determine the time required for 1/8th of a rotation. When his mirror rotated at 32 000 rpm (533 Hz), he could observe the light in the telescope. Example 13.1 The set of rotating mirrors in Michelson’s experiment was rotating at 533 Hz and the curved mirror was located 35.0 km away. Show how Michelson determined the speed of light from these data. Given distance between rotating and curved mirrors 35.0 km 3.50 104 m f 533 Hz Required speed of light (c) Analysis and Solution c d t The light must travel from the set of mirrors to the curved mirror and back again, so the round-trip distance is twice the given distance. d 2 3.50 104 m 7.00 104 m 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 651 For the light to follow the path illustrated in Figure 13.24, the rotating set of mirrors must complete a minimum of 1/8th of a rotation while the light is travelling between the rotating mirrors and the curved mirror. Therefore, the minimum travel time (t) is equal to 1/8th of the period of rotation1.876 103 s) 8 d c t 4 m 0 .0 7 4 s 0 .3 1 1 4 5 0 2 1 Hz 533 1.876 103 s 2.345 104 s 2.98 108 m/s Paraphrase According to Michelson’s observations, the speed of light in air was found to be 2.98 108 m/s. 13-2 QuickLab 13-2 QuickLab Measuring the Speed of EMR Problem Microwaves are a type of electromagnetic radiation and travel at the speed of light. They have a defined wavelength and frequency when they travel in a vacuum or in air. The universal wave equation v frelates the speed of the wave to its frequency and wavelength, where f is the frequency of the radiation, and is the wavelength. Each microwave oven generates microwaves at a given frequency, usually 2450 MHz. If the wavelength of the microwave can be measured, its speed can be determined using the universal wave equation. Materials microwave oven microwave-safe dish marshmallows Procedure 1 Pack a solid layer of marshmallows in a casserole or microwave-safe dish. 4 Remove the dish from the oven and measure the distance between adjacent soft spots. 5 Determine the average separation distance between several soft spots. This distance is equal to half the wavelength of a microwave. 6 Calculate the wavelength by multiplying the average separation distance by two. 7 Record the frequency of the microwave, which will be indicated on the back of the microwave oven. Questions Use the wavelength and frequency of the microwave to answer the following questions. 1. What is the speed of the microwaves? 2. What is the percent error of your calculated value, when compared with the currently accepted value for the speed of light (3.00 108 m/s)? 3. What are several possible sources of error that 2 Remove the turntable from the microwave oven. could affect your calculation? 3 Place the dish in the oven and cook until the marshmallows begin to melt in four or five locations. 4. Compare the wavelength of microwaves with that of radio waves and visible light waves. Why are they called microwaves? Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 651 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 652 13.2 Check and Reflect 13.2 Check and Reflect Knowledge 1. Why was Galileo’s original experiment to determine the speed of light unsuccessful? 2. Explain why, in a Michelson-type experiment, the rotating mirrors have to turn at a very precise frequency in order for light to reach a stationary observer. 3. Explain why the periodic motion of Jupiter’s moon Io appears to be constantly changing when observed from Earth. Applications 4. In measuring the speed of light, the difference in eclipse times for Jupiter’s moon Io, is measured. If an eclipse occurs 24 min later than expected, and Earth’s orbital diameter is 3.0 1011 m, calculate the speed of light. 5. A communications satellite is in orbit around Earth at an altitude of 2.00 104 km. If the satellite is directly above a groundbased station, how long does it take a signal to travel between the satellite and the station? 6. Using a similar approach to Michelson, a student sets up a 64-sided set of rotating mirrors, 8.00 km away from a fixed mirror. What minimum frequency of rotation would be required to successfully measure the speed of light? 7. A 16-sided set of rotating mirrors is used to measure the time it takes light to travel a certain distance. At what frequency does the mirror need to rotate such that it makes 1/16th of a rotation in the time it takes light to travel 3.5 km and back again? 8. An 8-sided set of rotating mirrors rotates at 545 Hz in an experiment similar to that of Michelson. How far away should the fixed mirror be placed
in order to correctly measure the speed of light? 652 Unit VII Electromagnetic Radiation 9. The speed of light was measured to be 2.97 108 m/s using a 16-sided set of rotating mirrors and a fixed mirror separated by 5.00 103 m. At what frequency was the mirror rotating? 10. Students who measure the speed of light using an experimental design similar to that of Michelson with an 8-sided set of rotating mirrors, make the following observations when light passes through the apparatus: (a) rotating mirror frequency 1.00 103 Hz (b) distance between fixed and rotating mirror 17.5 km Determine the speed of light based on the students’ recorded observations. 11. A 64-sided set of rotating mirrors is turning at 340 Hz. If a fixed mirror is located 6.55 km away from the rotating mirror and the light is reflected correctly, what value for the speed of light would be obtained? Extensions 12. Global positioning satellites (GPS) are used to pinpoint individual locations on the surface of Earth. A minimum of three GPS satellites are required, each with synchronized atomic clocks. Each satellite simultaneously sends a time signal to a GPS receiver on the surface. The time signal of each satellite is compared with the time on the receiver’s clock. (a) Explain how the exact distance between each satellite and the receiver can be determined. (b) Explain how the receiver’s location relative to all three satellites indicates an exact coordinate on Earth’s surface. e TEST To check your understanding of speed and propagation of electromagnetic radiation, follow the eTest links at www.pearsoned.ca/school/physicssource. 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 653 13.3 Reflection On a windless day, the smooth surface of a lake will reflect light and produce stunning images of the surrounding landscape and sky. But introduce the slightest disturbance, a gust of wind or a morning rain shower, and the image becomes distorted and blurred. With enough disturbance, the image on the surface of the water completely disappears. Figure 13.25 a) Reflection in a smooth lake b) Diffuse reflection due to windy conditions How do the images in Figure 13.25 form? Light travels in straight lines through a medium that is uniform. This characteristic is termed rectilinear propagation. The light falling on a flat, smooth, reflecting surface, such as a mirror, undergoes specular or regular reflection. In the case of the smooth lake, the crisp, clear images that form are the result of many parallel incident rays that reflect as parallel reflected rays (Figure 13.26). In the case of the rough lake, the blurred images that form are the result of many parallel incident rays that are scattered after reflecting in many different directions. This behaviour is called diffuse or irregular reflection (Figure 13.27). rectilinear propagation: movement of light in straight lines through a uniform medium Figure 13.26 Ray diagram of specular (regular) reflection Figure 13.27 Ray diagram of diffuse (irregular) reflection The Law of Reflection Particular terms are used to describe reflection. These terms are illustrated in the ray diagram shown in Figure 13.28 on the next page. Regular reflection will occur when an incident light ray contacts a polished, reflecting surface. The incident ray makes contact at the point of incidence and then it leaves the reflecting surface and becomes the reflected ray. If, at the point of incidence, an imaginary line, called the ray diagram: a diagram to show a result of a light ray interacting with a surface Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 653 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 654 e SIM Find out more about the law of reflection from the use of simulations. Follow the eSim links at www.pearsoned.ca/school/ physicssource. plane mirror: a smooth, flat, reflecting surface law of reflection: the angle of reflection is equal to the angle of incidence and is in the same plane mirror plane θ i θ r incident ray reflected ray Figure 13.29 The Law of Reflection image I mirror O point object Figure 13.30 How two different observers see the image of an object located in front of a plane mirror virtual image: an image from which light rays appear to come; cannot be formed on a nonreflective surface or screen real image: an image from which light rays come; can be formed on a diffusely reflecting surface or screen normal line (N), is drawn perpendicular to the surface, then the angle formed between the incident ray and the normal line is called the angle of incidence. The angle formed between the reflected ray and the normal line is called the angle of reflection. point of incidence reflecting surface angle of incidence θ i angle of reflection θ r incident ray reflected ray N normal Figure 13.28 Ray diagram and terminology for regular reflection Light rays that are incident upon a plane mirror at 90 will be reflected directly back to the source. In this case, the incident and reflected rays are parallel to the normal line so that the angle of incidence and the angle of reflection are both zero. In a similar way, when a light ray contacts a mirror at an angle of 25 relative to the normal line, it will reflect at an angle of 25 relative to the normal line. In all cases, the angle of incidence will be equal to the angle of reflection. This is called the law of reflection (Figure 13.29). The angle of reflection is equal to the angle of incidence and is in the same plane. Image Formation in a Plane Mirror Look at the image of an object formed in a plane, flat mirror in Figure 13.30. What do the dashed lines represent? Are they real rays coming from a real object? The dashed lines are an extension of the reflected rays, causing your brain to “believe” that they are rays, originating from an object that is located behind the mirror. But this cannot be so. The light rays appear to be coming from an image, but this image is not real. The image formed in a plane mirror in this fashion is therefore called a virtual image. For example, if you were to hold a piece of paper behind the mirror where the image appears to be, you would not see the image on the paper. By contrast, a real image is one that can be formed on a diffusely reflecting surface, such as a piece of paper. A projector is a familiar device that produces real images on a screen. Ray diagrams can be used to show how images form with a reflecting surface, such as a mirror. 654 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 655 13-3 QuickLab 13-3 QuickLab Drawing a Ray Diagram Problem To construct a ray diagram of an image formed in a plane, flat mirror. Materials observer N ray 1 object mirror Figure 13.31 Procedure 1 Choose one point on the object, such as the tip of the candle flame, in Figure 13.31. 2 Draw an incident ray from the selected point on the object (adding a small normal line where the ray is incident on the surface of the mirror) (Figure 13.31). 3 Based on the law of reflection, the reflected ray leaves the mirror surface such that the angle of incidence is equal to the angle of reflection. Draw the reflected ray. 4 Use a dashed line to extend the reflected ray beyond the back surface of the mirror. 5 On your diagram, repeat steps 2 to 4 for a ray originating from the same location on the object, but at a slightly different angle. 6 Note that the viewer must be located where both reflected rays would enter the eye. 7 Repeat steps 2 to 6 and locate the point where the two dashed lines converge (meet). This is the image location for the first selected point on the object. Questions 1. How could you test whether the image formed in the mirror was real or virtual? 2. Is the image magnified or the same size as the object? 3. Will the image appear right side up (erect) or upside down (inverted)? 4. Could you use a ray diagram to determine if the word “HELP” would appear backwards in a mirror? Explain. M I N D S O N Image in a Mirror G F E D C O B A Figure 13.32 Woman in a Mirror Figure 13.32 shows a woman looking in a mirror. 1. How far behind the mirror does her image form, if she is standing 50.0 cm in front of the mirror? Explain how the ray diagram is used to determine this distance. Is the image real or virtual? 2. 3. According to the ray diagram, two sections of the mirror could be removed and her entire face would still be visible. Which two sections are not needed? Explain your reasoning. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 655 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 656 magnification: the relationship of the size of the image to the size of the object image attitude: the orientation characteristic of an image, whether erect or inverted Image Characteristics When an object is placed in front of a mirror, the image can be described by four general characteristics (Table 13.2). The magnification relates the size of the image to the size of the object. The image can be enlarged, diminished, or the same size as the object. The image attitude describes the image as being either erect (upright) or inverted (upside down) relative to the object. This is termed vertical inversion. Horizontal, or left/right, inversion can also occur. The position describes where the image forms relative to the surface of the mirror. The type of image distinguishes between real and virtual images. A real image is one that can be projected on a screen or surface, and a virtual image is one that can only be seen, or photographed. Table 13.2 Image Characteristics Image Characteristic Description magnification same size, enlarged, diminished attitude position type erect or inverted displacement from mirror surface real or virtual Image characteristics are important for the proper application of signage and warning labels. For example, the word “ambulance” is written in reverse on the front of the vehicle so that other drivers will be a
ble to read the writing in their rear-view mirrors when the ambulance is approaching. M I N D S O N Image Characteristics Figure 13.33 shows (a) an image of an ambulance and (b) its image in a rear-view mirror. 1. What image characteristic is addressed by printing the word “ambulance” in reverse? 2. While standing in front of a plane, flat mirror, raise your right hand. Which hand in your image was raised? In a plane, flat mirror, do images appear backwards horizontally or upside down vertically or both? Explain. 3. 4. Can an image in a plane mirror ever appear “upside down”? Figure 13.33 (a) The front of an ambulance (b) The image it makes in a rear-view mirror 656 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 657 Image Formation in a Curved Mirror Curved mirrors come in a variety of shapes, the most common being spherical. Spherical mirrors, like the ones used for store security, have a unique geometrical shape (Figure 13.34). This shape is derived from a sphere. Imagine a hollow sphere, with a polished mirror surface on the inside and out. hollow sphere polished inside and out convex reflecting surface concave reflecting surface Figure 13.34 A spherical store security mirror Figure 13.35 A curved mirror can be formed from a hollow sphere. By removing a section of the sphere, you produce a double-sided spherical mirror with a concave reflecting surface on one side and a convex reflecting surface on the other (Figure 13.35). The concave reflecting surface is curved inwards. This surface is also referred to as a converging mirror since it causes parallel light rays to converge, or come together, after being reflected. The convex reflecting surface is curved outwards. This surface is also referred to as a diverging mirror since it causes parallel light rays to spread out or diverge after being reflected. As with plane-mirror image formation, ray diagrams are useful to determine how images form from curved surfaces. Ray diagrams for curved surfaces, as in Figure 13.36 for example, are more complex and involve an expanded set of terminology. converging mirror: a concave reflecting surface that causes parallel light rays to converge after being reflected diverging mirror: a convex reflecting surface that causes parallel light rays to spread out after being reflected 1. centre of curvature (C) – the point in space that would represent the centre of the sphere from which the curved mirror was cut 2. radius of curvature (r) – the distance from the centre of curvature to the mirror surface 3. vertex (V) – the geometric centre of the PA curved mirror surface ray 2 ray 1 C ray 3 r F V f 4. principal axis (PA) – an imaginary line drawn through the vertex, perpendicular to the surface of the curved mirror at this point 5. principal focal point (F) – the point where light rays that are parallel to and close to the principal axis converge, or appear to diverge from, after being reflected Figure 13.36 A ray diagram for a converging mirror 6. focal length (f) – the distance from the vertex to the focal point, measured along the principal axis. The focal length is related to the radius of curvature by f r2. This means that as the radius of curvature is reduced, so too is the focal length of the reflecting surface. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 657 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 658 Drawing Ray Diagrams for Curved Mirrors The four general image characteristics of magnification, attitude, position, and type, which were applied to plane mirrors, can also be applied to curved mirrors. For curved mirrors, the normal line at the point of incidence is the same as an extension of the radius of the spherical surface drawn at this point. The law of reflection still describes how all the rays will be reflected. To understand and predict how images are produced from curved mirrors, we will need to use three light rays, the law of reflection, and the mirror’s focal point, centre of curvature, and vertex. Two rays from a point on an object locate the corresponding point on the image and the third ray will verify its location (Figure 13.37). Consider each of the three rays: Incident Ray 1 travels from a point on the object parallel to the principal axis. Any ray that is parallel to the principal axis will reflect through the focal point on a converging mirror, or appear to have originated from the focal point on a diverging mirror. Incident Ray 2 travels from a point on the object toward the focal point. Any ray that passes through the focal point on a converging mirror, or is directed at the focal point on a diverging mirror, will be reflected back parallel to the principal axis. PA PA F C Incident Ray 3 travels from a point on the object toward the centre of curvature. Any ray that passes through the centre of curvature on a converging mirror, or is directed at the centre of curvature on a diverging mirror, will be reflected directly back along the incident path. At what incident angle must this light ray hit the mirror surface in order to be reflected straight back along the original path? These three rays alone will allow you to predict and verify the location and characteristics of the image. Notice that some conventions are used when drawing these ray diagrams: 1. objects are often drawn as erect arrows 2. real rays are drawn as solid lines 3. virtual rays (that only appear to exist behind the mirror) are drawn as dashed lines Figure 13.38 (a) shows that the converging mirror produces a real image that is inverted and diminished. Figure 13.38 (b) shows that the diverging mirror produces a virtual image that is erect and diminished. Why is one image real and the other virtual? For curved mirrors, a real image is formed where the reflected light rays converge or meet. At this location, a focussed image would appear on a sheet of paper or a screen if it were located in the exact location where the light rays meet. If the screen were to be moved slightly, the image would appear Figure 13.37 Reflection of three rays from spherical mirrors ray 1 ray 3 ray 2 C F V a) A concave, converging mirror ray 1 ray 2 ray 3 V b) A convex, diverging mirror PHYSICS INSIGHT All three incident rays obey the Law of Reflection. The resulting rules for drawing ray diagrams are just the application of geometry. 658 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 659 blurred because the reflected rays would not be converging perfectly at the screen’s new location. For the diverging mirror, the reflected rays appear to be originating from behind the mirror, but if a screen were located there, the incident light rays would not reach it (the rays would be blocked by the mirror). Figure 13.38 Steps for drawing a ray diagram in converging and diverging spherical mirrors e WEB To learn more about ray diagrams for curved mirrors and lenses, follow the links at www.pearsoned.ca/school/ physicssource. ray 1 ray 3 ray 2 object C F V PA image 3 2 1 a) A concave, converging mirror 1 ray 1 ray 2 3 2 ray 3 object image b) A convex, diverging mirror Characteristics of an Image in a Curved Mirror If you look at the image of your face in a polished spoon and bring the spoon closer and closer to your nose, you will demonstrate that with curved mirrors, the object’s distance from the mirror’s vertex has an effect on the characteristics of the image produced. For example, what happens to an image as the object is brought closer to a converging mirror surface? In Figure 13.39(a), the object is located outside the centre of curvature, and a real, inverted, and diminished image appears. If the object is brought closer to the vertex of the mirror, such that it is inside the focal length, as in Figure 13.39(b), then the image becomes virtual, erect, and enlarged. Figure 13.39 Object location affects image characteristics. 1 2 3 mirror object C F V PA image a) A converging mirror with object located outside C (real, inverted, diminished) 2 mirror 3 C 1 F object V image PA b) A converging mirror with object located inside F (virtual, erect, enlarged) Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 659 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 660 Table 13.3 Image Characteristics for Converging Mirrors Object Position Image Characteristic distant real, inverted, and diminished, close to F outside centre of curvature real, inverted, and diminished, between F and C at the centre of curvature, C real, inverted, and same size, at C between focal point and centre of curvature real, inverted, and enlarged, beyond C at focal point, F undefined (no image forms), at infinity between focal point and vertex virtual, erect, and enlarged, behind mirror Table 13.3 summarizes the image characteristics for a concave, converging mirror. You can verify the characteristics by placing the concave side of a spoon in front of your nose and slowly moving it outward. Note that your image will disappear briefly just as the distance between your nose and the spoon reaches the focal point of the spoon. In a similar way, the image characteristics produced by a convex, diverging mirror can be investigated by using the other surface of the spoon. 13-4 QuickLab 13-4 QuickLab Converging and Diverging Mirrors Problem How do image characteristics in converging and diverging mirrors change as you change the object distance? Materials polished spoon drawing materials Procedure 1 Place a polished metal spoon on your nose and slowly pull it away from your face while watching the image. 4 Complete ray diagrams for the image of your nose inside F, at F, at C, and at a distance, with respect to the spoon as a diverging mirror. (Remember that a diverging mirror has a virtual focal length, so F and C are located on the opposite side of the spoon to your face.) Questions 1. What happens to the image as the object is brought closer
to the surface of a converging mirror? 2. What happens to the image as the object is brought closer to the surface of a diverging mirror? 3. What characteristics do all images formed in a 2 Complete ray diagrams for the image of your nose diverging mirror share? inside F, at F, at C, and at a distance, with respect to the spoon as a converging mirror. (Use a small upright arrow to represent your nose in the ray diagram.) 3 Reverse the spoon so that you are looking at the other side and move it away from your nose again. 4. Your image disappears when your face is at the focal point of the mirror. Sketch a ray diagram for a converging mirror with the object located at the focal point to explain why an image should not appear when the object is located at the focal point. 5. Why is it not possible to put an object at F for a convex mirror? 660 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 661 M I N D S O N Seeing Is Believing—Or Is It? A parabolic mirror is shaped so that incident light rays that travel parallel to the principal axis will be reflected to a central point, or focus. Optical illusions can be produced by arranging two parabolic mirrors facing one another. Use a ray diagram to explain how a three-dimensional, real image, is produced in Figure 13.40. Hint: The object is located at the focal point of the top mirror. Therefore, all rays that originate from the object will be reflected straight down from the top mirror. Assume the bottom mirror has the same focal length as the top mirror. real image appears here mirror object inside bottom mirror Figure 13.40 Two parabolic mirrors facing one another, with a hole cut around the vertex of one mirror Equations for Curved Mirrors The general rules for drawing ray diagrams of curved mirrors, in combination with the law of reflection, allow us to generate several equations for use with curved mirrors. These equations can be used to determine the characteristics of the image. In any ray diagram, two rays can be used to determine the image characteristics. In Figure 13.41, there are two similar triangles (AOV, above, and DIV, below). Both triangles have an identical angle () and right angles at the principal axis. Thus, DI AO DV AV O ho C A D hi I F θ θ V B f do di where hi ho This translates into di do ho is the height of the object, hi is the height of the image, do is the distance between the mirror vertex and the object, and di is the distance between the mirror vertex and the image. Figure 13.41 Triangles AOV and DIV are similar triangles; triangles FVB and FAO are also similar triangles. When we use the ray OFB, the triangles FVB and FAO (shaded light blue) are also similar. Therefore, AO VB AF VF Since • AO ho • VB hi • AF do • VF f f Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 661 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 662 mirror equation: the equation that relates the focal length of a curved mirror to the image and object distances We can write this relationship as: Since hi ho di do the inverse is also true: Combining the equations gives: Dividing both sides by do and rearranging: We obtain this equation: ho hi ho hi do di d o did o 1 di f do f do di f do f 1 do d o d of 1 f 1 do This is the mirror equation, which relates the focal length of a curved mirror to the image and object distances. Sign Conventions for Use with the Mirror Equation As we have seen previously, there are real and virtual images, which can form either in front of or behind the curved mirror. When using the mirror equation, it is therefore necessary to follow a sign convention, which can distinguish the type of image formed. • Real objects and images have positive distances (measured from vertex). • Virtual objects and images have negative distances (measured from vertex). • Erect images and objects have a positive height. • Inverted images and objects have a negative height. • Converging mirrors have a real principal focal point and the focal length is positive. • Diverging mirrors have a virtual principal focal point and the focal length is negative. With this sign convention, a real image formed by a converging mirror will have a negative height (inverted attitude) while the object height is positive (erect attitude). Both the object and image distances will be positive. Magnification (m) is the ratio of the image height to the object height. A negative sign must be added to the equation for magnification to agree with the sign convention above. m hi ho di do An erect image has a positive magnification and an inverted image has a negative magnification. Concept Check 1. If the object distance is extremely large, approaching infinity, 1 becomes zero. Based on the mirror equation, where will the d o image form in relation to the focal length of the mirror? 2. For a plane, flat mirror, 1 approaches zero. Based on the mirror f equation, where will the image form in relation to the object? 662 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 663 13-5 Problem-Solving Lab 13-5 Problem-Solving Lab Using a Converging Mirror to Heat Water (Determining the Focal Length of a Converging Mirror) Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Recognize a Need A converging mirror can be used to focus the radiant energy of the Sun onto a specific point to capture energy, to heat water, or to generate electrical energy using a photovoltaic cell. The Problem Design an efficient apparatus that uses radiation to heat water. Criterion for Success Observing the maximum increase in water temperature in a given time period. Brainstorm Ideas To design an efficient water heater that uses sunlight for energy, you will need to consider the following questions: • Where would a water tank be placed in relation to a converging mirror in order to maximize the amount of radiation that it could absorb? • What colour should the water tank be? • What effect will the size of the mirror have on the water temperature? • What positive and negative effects would result from insulating the water tank? Build a Prototype Construct a miniature version of your water heater using a 50-mL volumetric flask, a thermometer, and a converging mirror mounted on a base of clay. Test and Evaluate Determine the focal length of the converging mirror using a heat lamp, a metre-stick, and a sheet of paper (Figure 13.42). Recall that rays that travel parallel to the principal axis will be reflected back through the focal point. 1. Place the heat lamp a great distance away so that most of the incident rays travel parallel to the principal axis. Adjust the position of the paper until the light is focussed into a small region. Be careful not to block the light from reaching the mirror. You may need to adjust the position of the mirror to avoid this — try tilting it slightly downward and turning the room lights off to see the image clearly. 2. Record the distance between the paper and the vertex of the mirror. This is the approximate focal length of the mirror and it is also the image distance. 3. Measure the distance between the lamp and the mirror. This is the object distance. Using the mirror equation, with the image and object distances, calculate and verify the approximate focal length of the mirror. heat lamp paper do di converging mirror clay Figure 13.42 Locating the focal point 4. Using the heat lamp in place of sunlight, measure the increase in water temperature for a given time period when the water tank is located beyond the focal point, at the focal point, and inside the focal point of the mirror. 5. Determine which position results in the greatest increase in water temperature and explain why. Communicate Using presentation software, create a brochure that outlines the basic operating principles of your water heater and explains the role of radiant energy in its operation. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 663 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 664 Example 13.2 An object is located 20.0 cm from a converging mirror that has a focal length of 15.0 cm. (a) Where will the image form relative to the mirror vertex? (b) If the object is 5.00 cm high, determine the attitude, height, and magnification of the image. 20.0 cm Given do f 15.0 cm (converging mirror has a positive focal length) ho 5.00 cm (assume that the object is erect) Required image distance (di) image height (hi) magnification (m) Analysis and Solution (a) To locate the image, use the relationship between focal length and object distance, the mirror equation: 1 f 1 di 1 do 1 f 1 di 1 do Now, solve for di: 1 1 1 20. 15. 0 cm 0 cm d i di 60.0 cm or di d do of f (20.0 cm)(15.0 cm) (20.0 cm) (15.0 cm) 60.0 cm (b) To determine the attitude, height, and magnification, use the equation for magnification: m hi ho di do Now, solve for m: m 60.0 cm 20.0 cm 3.0 Practice Problems 1. A diverging mirror of focal length 10.0 cm produces an image of an object located 20.0 cm from the mirror. Determine the image distance and the magnification. Is the image real or virtual? 2. Determine the image distance, magnification, and attributes for the following: (a) a converging mirror with a focal length of 12.0 cm with an object 6.0 cm from the mirror (b) a diverging mirror of focal length 5.00 cm with an object 10.0 cm from the mirror (c) a diverging mirror of focal length 10.0 cm with an object 2.0 cm from the mirror 3. A 5.0-cm-high object is placed 2.0 cm in front of a converging mirror and the image is magnified 4. Where does the image form and what is the focal length of the mirror? 4. A 4.0-cm-high object is placed 15.0 cm from a concave mirror of focal length 5.0 cm. Determine the image characteristics using a ray diagram and the mirror equation. 5. Light from a distant planet is incident on a converging mirror. The
image of the planet forms on a screen 45.0 cm from the vertex of the mirror. Find the focal length of the mirror and the image characteristics. Answers 1. di 6.67 cm; m 0.333 , virtual 2. (a) di 12 cm; m 2.0 , virtual, erect, enlarged (b) di 3.33 cm; m 0.333 , virtual, erect, diminished (c) di 1.7 cm; m 0.83 , virtual, erect, diminished 8 cm; f 1.6 cm 7.5 cm; m 0.50 ; 3. di 4. di 2.0 cm; real, inverted, diminished hi 5. f 45.0 cm; real, inverted, diminished 664 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 665 Then solve for hi: hi ih d o d o (60.0 cm)(5.00 cm) (20.0 cm) 15.0 cm Paraphrase (a) The real image forms 60.0 cm in front of the mirror vertex. (b) It is inverted and magnified 3 times to a height of 15.0 cm. 13.3 Check and Reflect 13.3 Check and Reflect Knowledge 1. Explain why the normal lines in Figure 13.27 are drawn in a variety of directions. 2. Create a concept map using ray diagram terminology including: principal axis, focal point, centre of curvature, and vertex. 3. Explain the difference between “specular” and “diffuse” reflection. 4. Compare “virtual” images with “real” images. 5. Compare “converging” and “diverging” mirrors. 6. Describe the path of the three rays that can be used to determine the characteristics of an image formed in a curved mirror. 7. Why does a diverging, convex mirror have a “virtual” focal point and not a “real” focal point? Applications 8. An object 4.0 cm high is located 10.0 cm in front of a concave mirror. If the image produced is erect, virtual, and 5.0 cm high, what is the focal length of the mirror? 9. Draw a ray diagram for a converging mirror and determine the image characteristics when the object is located at: (a) 2.0 f (b) 0.50 f (c) 3.0 f 10. Draw a ray diagram for a diverging mirror and determine the image characteristics when the object is located at: (a) 0.50 f (b) 1.0 f (c) 1.5 f 11. Some flashlights and headlights use concave mirrors to help generate a light beam. If the light source is positioned at the focal point of the mirror, would all the reflected rays travel outward, parallel to the principal axis? Explain your answer. Include a ray diagram. Extension 12. A reflecting telescope uses a curved mirror to produce the image at the eyepiece. Research the telescope’s design and function. Sketch a ray diagram to explain how the converging mirror is used in the telescope design. e TEST To check your understanding of reflection in plane and curved mirrors, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 665 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 666 13.4 Refraction A paddle appears bent when you put it in water (Figure 13.43); the Sun appears to change shape when it sets; a rainbow appears when it rains; and a mirage occurs over the hot desert sand, or even over a hot Alberta highway during the day. These are all visual effects created by refraction, a change in the direction of a light wave due to a change in its speed as it passes from one medium to another. When light rays pass at an angle from air to water, they immediately change direction. However, not all of the light will be refracted. Light rays are partially reflected and partially refracted when they pass from one medium to the next. Figure 13.44 illustrates this phenomenon.. The process of refraction is described by an incident ray and a refracted ray. The angle between the normal line (drawn perpendicular to the medium interface) and the incident ray is called the angle of incidence. The angle between the normal line and the refracted ray is called the angle of refraction. When the light ray passes from air to water, it is bent or refracted toward the normal line. And conversely, if a light ray travels from water into air, it is bent away from the normal line. The bending of the light ray depends on the refractive index of the medium in which the light travels. The refractive index of a medium is related to the effect on the speed of light when it travels within that medium. A vacuum does not impede or slow down light travelling through it and therefore, the speed of light (c) is defined as its speed in a vacuum. Air slows the light down a small amount and thus has a slightly greater refractive index than a vacuum. Water slows the light down to a greater extent and has a greater refractive index than air. This retarding effect was observed in Fizeau’s experiment on the speed of light in water (page 649). The amount of refraction is related to the magnitude of the change in the speed of light as it passes from one medium to another; the greater the change in speed, the greater the amount of refraction. angle of reflection reflected ray normal angle of incidence Mathematically, the refractive index of a medium (n) can be described as a ratio that compares the speed of light in a vacuum (c) to the measured speed of light in the medium (v). n c v Figure 13.43 A paddle as it appears partially submerged in water refraction: a change in the direction of a light wave due to a change in its speed as it passes from one medium to another refractive index: a ratio comparing the speed of light in a vacuum to the measured speed of light in a given material incident ray less refractive more refractive angle of refraction point of incidence refracted ray Figure 13.44 Ray diagram of partial reflection and partial refraction 666 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 667 Table 13.4 lists the refrac- tive indexes for some common substances. Since the speed of light in air is very close to the speed of light in a vacuum, the refractive indexes for a vacuum and for air are considered to be the same. The following general rules are based on refractive indexes: • When light passes from a medium with a high refractive index to one with a low refractive index, it is refracted away from the normal line. • When light passes from a medium with a low refractive index to one with a high refractive index, it is refracted toward the normal line. Table 13.4 Absolute Refractive Indexes (for Sodium Yellow Light, 589 nm) Medium vacuum air ice water ethanol glycerin quartz glass crown glass light flint glass Index of Refraction 1.0000 1.0003 1.31 1.33 1.37 1.47 1.47 1.52 1.58 Lucite (plexiglass) 1.52 ruby zircon diamond 1.54 1.92 2.42 Snell’s Law of Refraction Although the phenomenon of refraction had been observed for centuries, it was not until 1621 that the Dutch mathematician, Willebrord Snell (1580–1626), identified the exact relationship between the angle of incidence and the angle of refraction. Snell’s Law states that: sin i sin r a constant i / sin This relationship indicates that for any angle of incidence greater than zero, the ratio sin r will be constant for any light ray that passes through the boundary between the two media. In the case of an air-water interface, the constant is 1.33, which corresponds to the refractive index of water. In fact, for any ray that passes from air ( air) into a second medium ( r, refractive index n), Snell’s Law may also be written in a simple form as: sin air sin r n As long as the first medium is air, the Snell’s constant and the index of refraction for the second medium are one and the same thing. If the first medium is not air, then the general form of Snell’s Law applies. In the general form, the angle of incidence ( i) is replaced with 1 and the angle of refraction ( 2. The index of refraction for the first medium is denoted as n1 and the index of refraction r) is replaced with PHYSICS INSIGHT On passing through a medium, each wavelength of EMR has a unique refractive index. For this reason, Absolute Refractive Indexes must be quoted for a specific wavelength, as in Table 13.4. Snell’s Law: For any angle of incidence greater than zero, the ratio sin r is a constant for any light ray passing through the boundary between two media. i/sin info BIT A mirage (derived from the Latin term “mirare,” meaning “to wonder at”) is a naturally occurring phenomenon caused by refraction. Often, mirages appear as large bodies of water on hot desert sand, or small puddles on highways. What appears as water is actually an image of the sky being refracted back up from the hot air just above the road surface. Figure 13.45 Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 667 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 668 e SIM Find out more about refraction through the use of simulations. Follow the eSIM links at www.pearsoned.ca/ school/physicssource. for the second medium is denoted as n2. In all cases, the subscript “1” refers to the incident medium and the subscript “2” is reserved for the refracting medium. The general form of Snell’s Law is written as: sin 1 sin 2 n2 n1 n1 sin 1 n2 sin 2 The general form of Snell’s Law applies in all cases, regardless of the indexes of refraction of the media and/or the direction in which the light travels. Compare the general form with the simple form above. If the first medium is air, or a vacuum, so that n1 is 1.00, substituting this value into the general form results in the simple form of Snell’s Law. Example 13.3 Practice Problems 1. Light passes from a diamond into air. The angle of refraction as the light emerges from the diamond is 25. What was the angle of incidence? 2. Light travelling from air into a transparent material is incident at an angle of 20 and refracted at an angle of 17. Determine the index of refraction for the transparent material. 3. A ray of light passes from air into ruby at an incident angle of 15. Calculate the angle of refraction. 4. A ray of light, travelling in air, is incident on an unknown sample at an angle of 20. If the angle of refraction is 15, determine the index of refraction for the unknown sample. Answers 1. 10 2. 1.2 3. 9.7 4. 1.3 Yellow light travels from water into
crown glass. The light rays are incident on the crown glass at an angle of 35. Calculate the angle of refraction as the light enters the crown glass. Given 1 35 Required the angle of refraction ( 2) Analysis and Solution From Table 13.4, 1.33 • Index of refraction of water, the first medium, is n1 • Index of refraction of crown glass, the second medium, is n2 1.52 Therefore, the general form of Snell’s Law applies. n1 sin 1 sin 2 2 n2 sin 2 in n1 s 1 n 2 sin1(1.33 si ( ) 2 5 . 1 30 n35) Paraphrase The angle of refraction for the light travelling from water into crown glass is 30, when the angle of incidence is 35. 668 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 669 Snell’s Law, Refraction, and Wavelength incident wave fronts λ2 2 d ct D refracted wave fronts Figure 13.46 Wave fronts are refracted as they pass from air into water. vacuum speed c less refractive medium n 1.00 more refractive medium n > 1.00 Can we explain the phenomenon of refraction, and by association, Snell’s Law, by considering light as a wave? Recall that the wave model describes light as a stream of transverse waves radiating outward from a source. Let us assume that the light waves are indeed transverse waves of electromagnetic radiation. As each wave front moves from one medium to the next, there is no change in frequency. In other words, the wave fronts do not “pile up” at the boundary between the two media. The number of waves arriving at the boundary is equal to the number of waves leaving the boundary for any given time interval, so the incident and refracted waves have the same frequency. However, the wavelength changes at the boundary because the speed changes. According to the universal wave equation (v f), if the speed changes and the frequency remains constant, then the wavelength must change as well. To visualize this, consider light waves travelling from air into water as illustrated in Figure 13.46: medium speed v The wavelength in air is given by 1 v1/f; The wavelength in water is given by 2 v2/f. The wave fronts travel faster in air than they do in water (Fizeau’s 1 is experiment, page 649). It follows that v1 is greater than v2, and greater than 2. An enlarged view of two wave fronts, as shown in Figure 13.47, reveals another relationship. When we use similar triangles, the angle of incidence 2 are shown within the coloured right triangles and they share the same hypotenuse, x. 1 and the angle of refraction Therefore, the trigonometric ratio for the sine of an angle when applied to each triangle gives: 1 x opposite hypotenuse sin 1 sin 2 opposite hypotenuse 2 x Since x 1 sin 1 therefore and also x 2 sin 2 1 sin 1 2 sin 2 This equation can also be written as: sin 1 sin 2 1 2 incident ray incident wavefront less optically dense medium more optically dense medium refracted wavefront θ 2 refracted ray Figure 13.47 Enlarged view of two wave fronts Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 669 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 670 When the universal wave equation is used, the wavelength can also be written in terms of the wave speed as follows. sin 1 sin 2 sin 1 sin 2 v 1 f v 2 f v1 v2 Therefore, Snell’s Law can be expanded to read: sin 1 sin 2 v1 v2 1 2 n2 n1 By measuring the incident and refracted angles, velocity, and wavelength, this relationship can be tested. Example 13.4 Practice Problems 1. Determine the speed of light in the following materials: (a) water (b) ethanol (c) ruby (d) crown glass 2. Light with a wavelength of 737 nm enters quartz glass at an angle of 25.0. Determine the angle of refraction and the wavelength of the light in the quartz glass. 3. Light enters an unknown crystal from air with a wavelength of 500 nm. If the wavelength of the light in the crystal is found to be 450 nm, what is the refractive index of the crystal? Answers 1. (a) 2.26 108 m/s (b) 2.19 108 m/s (c) 1.95 108 m/s (d) 1.97 108 m/s 2. 16.7, 501 nm 3. 1.11 670 Unit VII Electromagnetic Radiation A ray of yellow light with a wavelength of 570 nm travels from air into diamond at an angle of 30. Determine the following: (a) the speed of light in the diamond (b) the wavelength of the light as it travels in the diamond Given 1 570 nm 5.70 107 m Required the speed of light in diamond (v2) the wavelength of light travelling in diamond ( 2) Analysis and Solution v1, the speed of light in air 3.00 108 m/s From Table 13.4, • Index of refraction of air, the first medium, is n1 • Index of refraction of diamond, the second medium, 1.00 is n2 2.42 n i The general form of Snell’s Law applies Therefore, n1v1 n2 v2 2 n1 1 n2 1.00 (3.00 108 ms) 2.42 1.00 (5.70 107 m) 2.42 1.24 108 ms 2.36 107 m 236 nm Paraphrase (a) The speed of light in the diamond is 1.24 108 m/s. (b) The wavelength of light in the diamond is 236 nm. 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 671 13-6 Inquiry Lab 13-6 Inquiry Lab Determining the Refractive Index of a Variety of Materials Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this experiment, known values of the refractive indexes of water and ethanol are verified. rays from ray box Question What are the refractive indexes of water and alcohol (ethanol)? Variables Manipulated variable: angle of incidence Responding variable: angle of refraction Controlled variables: refracting substance, wavelength of light 360 0 10 10 20 20 30 30 less refractive (air) more refractive (liquid) 320 40 310 50 300 60 30 210 20 200 10 190 0 180 10 170 3 0 15 60 120 50 130 40 140 semicircular container Materials and Equipment polar coordinate paper graphing paper water ethanol single-slit ray box or laser semicircular plastic dish Procedure 1 Design a data table or spreadsheet with the headings: angle of incidence (i), angle of refraction (r), sin angle of incidence (sin i), sin angle of refraction (sin r), and ratio sin i / sin r. 2 Fill the semicircular dish with water and place it on the polar coordinate paper such that the 0–180° line is perpendicular to the centre of the flat side of the dish. This will make the 0–180 line the normal line (Figure 13.48). 3 Direct a single ray of light along the normal line (0). Record the angle of refraction. (This should be zero if the apparatus is set up correctly.) 4 Record the angles of refraction for 10 to 70, increasing the angle by 10º for each step. 5 Complete the data table or spreadsheet calculations and plot a graph of sin angle of incidence vs. sin angle of refraction. Calculate the slope of this graph. 6 Repeat steps 2 to 5 using ethanol instead of water in the semicircular dish. Figure 13.48 Semicircular dish on polar coordinate paper with several incident ray angles shown Analysis 1. (a) What is the value of the slope for the graph when water was used as the refracting substance? (b) What is the value of the slope when ethanol was used as the refracting substance? 2. According to your slope calculations, which material, water or ethanol, has a higher index of refraction? 3. Calculate the percent error for each index of refraction using the absolute index of refraction given for each substance in Table 13.4. 4. Comment on the sources of error that could have occurred in this experiment. 5. Why is it important that the incident light ray contacts the semicircular dish at the centre? 6. Why is it important that a “semicircular” dish is used, instead of a rectangular dish? 7. Which material is more effective in changing the direction of light when light enters it? 8. Predict what would happen if the semicircular dish were replaced with a semicircular block of glass or Lucite®. If you have time, test your prediction. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 671 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 672 27° 42° 59° 20° 40° 30° 50° 50° 0° 0° light source air n 1.00 water n 1.33 Total Internal Reflection If you look down a long straight tube, you should be able to see what is at the other end. If the tube has a bend in it, a set of mirrors could be used to reflect the light through to your eye. This is how the periscope on a submarine works. But what if the tube has multiple bends in it? What could be done to ensure the light is able to travel through the entire length of the tube? Could you line the inside walls with mirrors and angle the light so that it reflects from side to side all along the tube, entering at one end and emerging at the opposite end? Figure 13.49 Increasing the incident angle for light leaving a water-air interface leads to total internal reflection. In theory, this is exactly what happens in the optical fibres that facilitate the data and telephone communication of computer networks. The process is based on total internal reflection. Recall that when light travels from one medium to another, some of the light is partially reflected and some is partially refracted. When light travels from a medium with a high refractive index, like glass, to a medium with a low refractive index, like air, more of the light is reflected than it would be at an interface where its speed decreases. As illustrated by Figure 13.49, when light travels from the water with a high refractive index to the air with a low refractive index, it bends away from the normal line. As the angle of incidence is increased, the angle of refraction also increases, bending farther and farther from the normal line, and eventually reaching the maximum angle of 90. Beyond this angle, refraction ceases and all incident light will be reflected back into the high-index medium (in this case, the water). Therefore, not all the light that approaches a water-air boundary will be refracted. The light approaching the boundary at large angles will be reflected back into the water. This is why underwater pool lights, for example, are not visible at all angles from above. At particular angles, th
ey cannot be seen. When the angle of refraction is 90, the incident angle will have a value unique to the two media that form the interface. This unique angle of incidence is called the critical angle. For light travelling from a medium with a high refractive index to a medium with a low refractive index, the critical angle is determined by assuming that the angle of refraction is 90. Concept Check 1. Is light refracted toward or away from the normal line when passing from a medium with a low refractive index like air, into a medium with a high refractive index like water? 2. Based on your answer above, can total internal reflection occur when light travels from a low-index medium like air, into a high-index medium like water? Explain why or why not. total internal reflection: reflection of all incident light back into a medium of higher refractive index due to the inability to refract light beyond the maximum angle of 90 critical angle: for any two media, the size of the incident angle for which the angle of refraction is 90 info BIT The sparkle and glitter associated with diamonds is caused by total internal reflection. The high index of refraction of the diamond and the skill of the jeweller in cutting and finishing are used to trap and focus light, which we observe as sparkles. Polar Bear Diamonds™, mined, cut, and polished in the Northwest Territories, are recognized for their quality worldwide. Figure 13.50 672 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 673 Example 13.5 What is the critical angle for the quartz glass-air interface? Given The angle of refraction at the critical angle is defined to be r 90.0. Required the critical angle for this interface ( critical) Analysis and Solution From Table 13.4, 1.00 • Index of refraction of air, the first medium, is n1 • Index of refraction of quartz glass, the second medium, is n2 1.47 Use the form of Snell’s law for the critical angle: n2 sin critical sin critical critical n1 sin 90.0 90.0 n n1 si n 2 90.0 sin1 n n1 si n 2 90.0 sin1 in s 1.00 7 .4 42.9 1 Paraphrase Light at an incident angle of 42.9 or greater will be internally reflected at the quartz glass-air interface. Practice Problem 1. Determine the critical angle for the following interfaces: (a) water and air (b) diamond and air (c) diamond and water Answer 1. (a) 48.8 (b) 24.4 (c) 33.3 e MATH To investigate Snell’s Law, refraction, and the concept of the critical angle, using a graphing calculator or a spreadsheet program, visit www.pearson.ca/ school/physicssource. cladding Total internal reflection has many applications. Most notable are optical fibres (Figure 13.51). An optical fibre consists of a central core of glass with a refractive index of approximately 1.5, surrounded by a cladding material of a slightly lower refractive index. Such fibres are used to transmit data on computer and communication networks. The data are transmitted via the modulation of laser light travelling through a glass fibre, with virtually none of the energy loss associated with electrical transmission. Some of the advantages of fibre-optic networks include: • economically less expensive than copper wire of equivalent length • thinner, more flexible, and made of non- flammable materials core >θ c >θ c Figure 13.51 An optical fibre showing cladding, core, axis, and critical angle • able to handle a higher data-carrying capacity based on the fibre bundle diameter • less signal degradation and interference between multiple signals on the same fibre as compared with copper networks • the glass fibres are highly transparent so that repeaters (amplifiers) can be many kilometres apart, as opposed to coaxial cable repeaters that must be less than 1 km apart e WEB To learn more about fibre optics including applications in audiovisual equipment and in nanotechnology, begin your search at www.pearsoned.ca/school/ physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 673 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 674 info BIT Buckyballs, also called C60 or buckminsterfullarenes, are soccer-ball-shaped molecules made of 60 linked carbon atoms. These carbon molecules may be formed into nanotubes to transmit data using light. info BIT Binoculars consist of two identical telescopes mounted side by side and aligned to point in the same direction. When the images from each side of the binoculars are viewed simultaneously, the user is able to sense depth and distance that is not possible with a single telescope view. eyepiece Porro prisms objective Figure 13.52 The light path in binoculars The Porro prisms in the binoculars use total internal reflection to direct light through the compact, short body of the binoculars with minimal absorption, producing a high-quality image. F F F Figure 13.53 A double Porro prism A double Porro-prism system is used in binoculars to reorient an inverted image while at the same time producing a longer, folded pathway for the light to travel between the objective lens and the eyepiece, producing a greater magnification. Fibre-optic systems also have drawbacks. The complex design of the fibres makes the fibre-optic system relatively expensive to set up and the fibres are also subject to wear and breakage. 13-7 Decision-Making Analysis 13-7 Decision-Making Analysis Fibre Optics: Endless Light The Issue The new knowledge society of today has emerged as a result of our ability to communicate and transmit data on a large scale. And to a very large extent, the infrastructure that supports our communication is based on the principles of total internal reflection and its application in fibre-optic systems. New and innovative advances are being made with fibre optics every day – from nanotechnology to entertainment services. Background Information The flexibility of fibre optics has allowed for applications in a variety of industries. For example, telephone, television, and data networks, originally all separate industries, are merging into one large application, which is supported by a fibre-optic backbone. In medicine, applications include the use of a fibrescope to both illuminate internal organs and capture images of them. For example, to see inside the small intestine, an endoscope, a small, flexible bundle of fibres is inserted down the patient’s throat and through the stomach. Once the fibres reach the small intestine, images can be transmitted through individual fibres while others are simultaneously used to illuminate the tissue. This application prevents the need for more-invasive, risky procedures that can leave the patient with a higher risk of infection and perhaps result in a prolonged hospital stay. Other applications include mechanical imaging. For example, fibre-optic bundles can be used to inspect the interior of long pipes, vessels, and hard-to-reach locations, such as bore holes for oil and water wells, pipelines, and hazardous goods containers. With proper instrumentation devices attached, fibre-optic bundles can be used in dangerous conditions to identify gases, pressures, temperatures, and concentrations. Nanotechnology is an emerging application for fibre optics. In this industry, new nanomaterials, such as C60 ( buckyballs), are being investigated to expand the power and application of data transmission using light. Analyze and Evaluate 1. Research applications of fibre optics using the Internet, research journals, and periodicals. 2. Group all the applications by industry and identify what you believe to be the most important applications in each industry. 3. For each of the most important applications, identify the social, political, economic, and environmental impact of the technology. 4. New applications of fibre optics are emerging in nanotechnology. Identify these applications and describe what technological advances could be expected as a result of merging nanotechnology with fibre-optic technology. 5. Prepare a multimedia presentation that demonstrates advances that have been made in a number of industries as a result of fibre-optic applications. Predict what future applications and social issues will evolve as new technologies begin to merge with fibre optics. 674 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 675 Prisms — Dispersion of White Light Where have you seen a rainbow? Everyone has seen one in the sky, but what about in a diamond, or perhaps on the wall near a hanging crystal in a window. There are many items that produce a rainbow of colours, but all the rainbows have something in common. They all originate from white light. The investigation of such effects was undertaken by Newton in 1666. He placed a transparent prism in a beam of sunlight passing through his window shutter at Cambridge University. On the opposite wall, an elongated band of colours appeared and he called this a spectrum (Figure 13.54). In this spectrum, Newton noted the colours red, orange, yellow, green, blue, indigo, and violet, in the order in which they appeared. info BIT It may be that Newton noted seven colours in the spectrum, in part to go along with the views of society at that time: numerology regarded 7 as a “special” number. Six colours are more commonly used, as in Figure 13.55 and Table 13.5. spectrum: the bands of colours making up white light; in order, red, orange, yellow, green, blue, violet dispersion: separation of white light into its components Figure 13.54 The complete spectrum produced by white light passing through a prism Newton set out to determine which of two things was true: either the colours of the spectrum are added to the white light by the prism, or the prism separates all the colours from the white light. To test this, Newton set up two prisms, the first one exposed to white light, producing a spectrum of colours (Figure 13.55), and the second one, only exposed to the red (monochromatic) light coming from the first prism. The second pri
sm did not produce any more colours; only the red light emerged. As a second test, Newton placed a converging lens into the path of the spectrum of light and observed the resulting white light as an image on a sheet of white paper. slit aperture white light prism Based on his observations, Newton concluded that white light is made up of all the colours in the spectrum and the prism was simply separating the colours from one another. The separation of white light into its components is called dispersion. screen red orange yellow green blue violet visible spectrum Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 675 Figure 13.55 Dispersion of white light by a prism 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 676 The recomposition of the spectrum, therefore, should produce white light, as Newton was able to demonstrate in his second experiment. Recomposition can also be demonstrated by painting all the colours of the spectrum in certain proportions on a disc and spinning the disc at a very high speed (Figure 13.56). At sufficient speed, the disc appears white. Is dispersion consistent with the wave model of light? If we consider light as a wave, each of the colours in the spectrum has a unique wavelength, as listed in Table 13.5. As light enters a medium with a high refractive index, the wavelength is reduced but the frequency is unchanged. The speed of a wave is the product of its frequency and wavelength as described by the universal wave equation (v f ). Therefore, the light must slow down as it enters the highindex medium. Because the refractive index is related to the speed of a wave, it is therefore also related to the wavelength of the wave when the frequency is constant. n o o wavelength in vacuum wavelength in medium This relationship between refractive index and wavelength, for a constant frequency, means that the refractive index is different for each wavelength of light that passes through the same medium. Incident light with a smaller, shorter wavelength will slow down and refract to a greater extent than light with a larger, longer wavelength. As each wavelength of light refracts at a slightly different angle, the wavelengths will separate, producing a continuous spectrum. The wave model of light, therefore, is consistent with the phenomenon of dispersion. Figure 13.56 Recomposition of the spectrum using a high-speed disc Table 13.5 Wavelength of each colour in the spectrum e SIM To learn more about the effects of prisms through the use of simulations, follow the eSim links at www.pearsoned.ca/school/ physicssource. Colour violet blue green yellow orange red Wavelength (nm) 400-450 450-500 500-570 570-590 590-610 610-750 676 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 677 13-8 QuickLab 13-8 QuickLab Dispersion Using a Prism Problem Is white light made up of all the colours we can see? Materials 2 identical prisms convex lens intense light source (flashlight or white light ray box) white paper Procedure 1 Using one prism, observe the colour spectrum on a piece of white paper. A dark room and a very intense white light source will work best. The convex lens may be placed between the light source and prism to intensify the beam of white light if the spectrum is not immediately clear and visible. 2 Position the second prism in the spectrum produced by the first one. Rotate the second prism and position the white paper so that you can observe the light that emerges from the second prism. 3 Explore a variety of positions and orientations with both prisms. Questions 1. What is the best way to make a colour spectrum using a single prism? Draw a picture showing the flashlight, prism, convex lens, paper, and the colour spectrum. 2. List the colours of the spectrum from most refracted to least refracted. 3. Compare the wavelengths of each colour to the amount of refraction observed. Describe the relationship between wavelength and amount of refraction. 4. Are the colours of the spectrum added to, or separated from, white light when white light passes through a prism? Support your answer with observations made using two prisms. 5. The spectrum produced by a prism is similar to that observed as a rainbow. Research and explore the similarities between the spectrum produced by a prism and that which is observed as a rainbow. Thin Lenses Projection systems connected with computers and movie players, and optical systems including microscopes and telescopes, use lenses that refract light in order to generate images. A typical thin lens is a circular piece of transparent material with a spherically shaped surface that refracts, or changes the direction of light that passes through it. The uniformly curved surface will refract light rays to varying extents depending on where they contact the lens. Rays that are incident near the edge of the lens will be refracted at larger angles than those that are incident near the centre, where the two faces of the lens are almost parallel. For a converging lens, rays that travel parallel to the principal axis will be refracted inward, intersecting at the principal focus, F. For a diverging lens, rays that travel parallel to the principal axis will be refracted outward, appearing as though they have originated at a virtual principal focus, F. The distance from each lens to F is the focal length, f (Figures 13.57 and 13.58). Each type of lens has a secondary focus, F at a distance of f, on the opposite side of the lens. converging lens: a lens that refracts rays travelling parallel to the principal axis inward to intersect at the principal focus diverging lens: a lens that refracts rays travelling parallel to the principal axis outward to appear as though they have originated at a virtual principal focus Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 677 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 678 PHYSICS INSIGHT A ray that travels through the optical centre is indeed refracted. This causes the refracted ray to emerge parallel to, but laterally shifted from, the incident ray. For a thin lens, this lateral shift is so small that it is assumed that the ray is not refracted. O F principal focus f focal length Figure 13.57 Converging lens F virtual principal focus PA optical centre F O lateral shift Figure 13.58 Diverging lens f focal length Figure 13.59 Lateral shift Drawing Ray Diagrams for Thin Lenses A ray diagram is a useful tool for predicting and understanding how images form as a result of light rays emerging from a curved lens. Ray diagrams for lenses are similar to the ray diagrams used with curved mirrors; only two rays are needed to predict the image location, and a third is used as verification (Figure 13.60). 1 3 optical centre image object F 2 F 2 3 1 object 1 2 3 F image F Figure 13.60 Ray diagrams for (a) converging, and (b) diverging lenses 678 Unit VII Electromagnetic Radiation Ray 1 travels parallel to the principal axis and is refracted such that it emerges and passes through (or appears to have originated from) the principal focus, F. Ray 2 travels through (or appears to be directed toward) the secondary focus, F, and is refracted such that it emerges and travels parallel to the principal axis. Ray 3 travels straight through the optical centre of the lens and is not bent. The ray diagram is not only used to identify the location of an image relative to the lens, but it can also illustrate the other three image attributes of type, attitude, and magnification, that have previously been described for curved mirrors. Relative to the object, the image produced by a thin lens can be real or virtual, inverted or erect, and enlarged or diminished. 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 679 Example 13.6 2F and 2F represent two points each at a distance of 2f (twice the focal length) from a lens on either side of the lens. Draw a diagram to determine the image attributes for an object located at the following positions relative to a converging lens. a) beyond 2F b) at F c) between the lens and F Given Object position relative to the converging lens Required Ray diagram showing image location and attributes Analysis and Solution Construct a ray diagram and visually inspect the image location relative to the converging lens and the image attributes.. object 2F F image F 2F Figure 13.61 (a) object 2F F F 2F Figure 13.61 (b) image Practice Problems 1. Using a ray diagram, determine the image attributes for the following: (a) an object located between 2F and F relative to a converging lens (b) an object located between 2F and F relative to a diverging lens (c) an object located between F and a diverging lens Answers 1. (a) Image is beyond 2F, real, inverted, and enlarged. (b) Image is between the lens and F, virtual, erect, and diminished. (c) Image is between the lens and F, virtual, erect, and diminished. 2F F object F 2F Figure 13.61 (c) Paraphrase (a) The image is located between F and 2F, and is real, inverted, and diminished. (b) No image is formed. (c) The image is located between F and 2F, and is virtual, erect, and enlarged. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 679 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 680 e WEB Equations for Thin Lenses To learn more about thin lenses, follow the links at www.pearsoned.ca/ school/physicssource. An equation relating object distance, image distance, and the focal length of a curved lens can be derived using an analysis nearly identical to that for curved mirrors. In Figure 13.62, the green triangle XOF and the blue triangle ZYF are similar. Therefore, ho object X F O F Y hi image Z do f di Figure 13.62 Thin lens ray diagram OX OF ho f hi ho hi ho di do di do YZ YF hi f di f di f di do f di f di f f f Dividing both sides by di, and simplifying, gives the thin lens equation. thin lens equation: the equation that relates ob
ject distance, image distance, and focal length of a curved lens 1 di d i d ido 1 do d i d i f 1 di 1 f Notice that this equation is identical to the mirror equation. A sign convention will be used here also, to distinguish between real and virtual distances as well as to identify erect and inverted images. Sign Conventions for the Thin Lens Equation • All distance measurements are relative to the optical centre of the lens. • Positive distances are used for real objects and images. • Negative distances are used for virtual images. • Positive image and object heights are upward relative to the principal axis. • Negative image and object heights are downward relative to the principal axis. • Converging lenses have a positive focal length. • Diverging lenses have a negative focal length. Results with a thin lens are similar to those with a curved mirror in that an erect image has a positive magnification and an inverted image has a negative magnification. Therefore, the same magnification equation that is used for curved mirrors is also used for thin lenses. m hi ho di do 680 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 681 Example 13.7 A 2.5-cm-high object is placed 10.0 cm from a diverging lens of focal length 5.0 cm. Determine the image distance, height, and attributes using the thin lens equation. Verify your answer with a ray diagram. Given ho do 2.5 cm 10.0 cm f 5.0 cm (The lens is a diverging lens.) Required image distance (di) height (hi) image characteristics Analysis and Solution The thin lens equation can be used to determine the image distance, image height, and attributes. A ray diagram can verify the answer hi 1 1 1 hi 10. 5.0 0 cm d i cm ih d o d o (3.33 cm)(2.5 cm) 10.0 cm 3 0 cm 10. di cm 10.0 3 3.3 cm 0.83 cm Paraphrase The image distance is 3.3 cm and height is 0.83 cm, indicating the image is virtual, erect, and diminished. The answers are verified by the ray diagram in Figure 13.63. object 2F F image F 2F Figure 13.63 Practice Problems 1. A 3.00-cm-high object is located 15.0 cm from a converging lens with a focal length of 10.0 cm. (a) How far is the image from the lens? (b) How high is the image? (c) Describe the image characteristics and verify them using a ray diagram. 2. A 10.0-cm-high candle is placed 100.0 cm from a diverging lens with a focal length of 25.0 cm. Determine the following using a ray diagram and the thin lens equation: (a) the image location from the lens (b) the image height (c) the type of image formed 3. A projector uses a converging lens to create a focussed image on a screen located 5.00 m away. The image is generated from a slide located 7.50 cm from the lens. (a) What is the focal length of the lens? (b) Determine the magnification of the image. Answers 1. (a) di (b) hi (c) image is real, inverted, enlarged 30.0 cm 6.00 cm 20.0 cm 2.00 cm 2. (a) di (b) hi (c) image is virtual 3. (a) f 7.39 cm (b) m 66.7 X Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 681 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 682 13-9 Inquiry Lab 13-9 Inquiry Lab Comparing Keplerian and Galilean Telescopes Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork In this investigation the functional differences between a Keplerian telescope, with two converging lenses, and a Galilean telescope, with one converging lens and one diverging lens, are explored. Question What are the differences between Keplerian and Galilean telescopes? Variables manipulated: type of lens used responding: magnification of the image controlled: focal length of double concave and double convex lenses Materials and Equipment long focal length converging lens short focal length converging lens short focal length diverging lens lens holders • • • • • optical bench for measuring distances • • paper, ruler, pencil, and tape lamp Procedure for Galilean Telescope 1 Place the long focal length converging lens at one end of the optical bench. This will serve as the objective lens for both telescopes. 2 If the long focal length is unknown, place a lamp in front of the lens and measure the distance between the lamp and the lens. Next, slide a sheet of paper or screen along the optical bench until a bright spot appears (image of the lamp). Record the distance between the paper and the lens. Calculate the focal length of the objective lens using the lens equation, and image and object distances. 3 Place the short focal length diverging lens halfway between the converging lens and its principal focus as illustrated in Figure 13.64. objective lens eyepiece fo light from object Figure 13.64 Galilean telescope 682 Unit VII Electromagnetic Radiation 4 Draw a series of 5 vertical arrows, 1 cm apart, on a sheet of paper. Tape the sheet to a distant wall so that it will be visible in the telescope. 5 Look through the eyepiece with one eye and compare the image with the sheet seen with an unaided eye. 6 Estimate how many vertical arrows (viewed with the unaided eye) would fit in between two of the vertical arrows observed in the telescope. Record this as the estimated magnification. Procedure for Keplerian Telescope 7 Remove the diverging lens from the optical bench. 8 If the focal length of the short focal length converging lens is not known, place a lamp in front of the lens and slide a sheet of paper or screen along the optical bench until a bright spot appears. Record the focal length for the short focal length lens. 9 Position the short focal length converging lens such that the distance between the two lenses is slightly less than the sum of the focal lengths, as shown in Figure 13.65. fo objective lens fe eyepiece light from object Figure 13.65 Keplerian telescope 10 Estimate how many vertical arrows (viewed with the unaided eye) would fit in between two of the vertical arrows observed in the telescope. Record this as the estimated magnification. Analysis 1. Prepare a table that compares the attitude, magnification, and brightness of the image in each telescope. 2. Complete a ray diagram to show how the image is formed in each telescope. 3. Suggest a different application for each telescope. Compare the telescopes’ efficiency in looking at relatively close terrestrial objects, like mountains, or far away extraterrestrial objects, such as the Moon. 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 683 13.4 Check and Reflect 13.4 Check and Reflect Knowledge 1. Is light bent toward or away from the normal line when it passes from a lowindex medium to a high-index medium? 2. How is the index of refraction measured for a particular medium? 3. Construct a concept map using the following terms: incident ray, refracted ray, angle of incidence, angle of refraction, normal line. 4. What is the difference between the simple and general forms of Snell’s Law? 5. Suggest an experimental design to determine the critical angle for a Lucite®-air boundary. 6. List five advantages of a fibre-optic network when compared to traditional electric networks used for data transmission. 7. How does a prism separate all the colours of white light? 8. Explain the effect of wavelength on the index of refraction for a particular medium, like glass. How does this lead to dispersion? 15. Using a ray diagram, determine the image attributes for an object located at 2F relative to a converging lens. 16. A camera with a converging lens (f 4.50 cm) is used to take a picture of a 25.0-m-high tree that is 50.0 m from the camera. How tall is the image? Is it erect or inverted? 17. Red light of 700 nm and blue light of 475 nm are both incident on a Lucite® block. Which colour will be refracted to a greater extent? 18. The focal length of the converging lens in a computer projector is 8.00 cm. An LCD panel positioned inside the projector serves as the object for the lens. (a) If the LCD panel is located 8.10 cm from the lens inside the projector, how far away should the screen be placed so that a clear image is produced? (b) If the LCD panel is 1.75 cm high, how large is the image on the screen? (c) If the screen is moved 3.0 m closer to the projector, how far must the object now be from the lens in order to generate a focussed image? Applications 9. What is the speed of light in water (n 1.33)? Extensions 19. Explain why jewellery crystals, such as 10. A light ray is incident on a block of quartz diamonds, sparkle. glass (n 1.47) at an angle of 35.0. Determine the angle of refraction. 11. When a ray of light passes from water (n 1.33) into a Lucite® block (n 1.52) it is refracted at an angle of 28.0. Determine the angle of incidence. 12. What is the critical angle for a Lucite®-air interface? 13. A light ray travelling in water approaches the water-air boundary at an angle of 50. What happens to the light ray at the boundary? 14. Light with a wavelength of 540 nm enters a ruby crystal (n 1.54) at an angle of 25.0. Determine the angle of refraction and the wavelength of the light in the ruby. 20. Explain why you can start a fire with a magnifying glass. 21. If a diver is underwater and looks up, a circular “hole” appears on the surface of the water directly above the diver. Other than this hole, the surface appears as a mirror. Explain how this happens. e TEST To check your understanding of refraction, fibre optics, dispersion, and lenses, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 683 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 684 Figure 13.66 Christiaan Huygens Huygens’ Principle: a model of wave theory, which predicts the motion of a wave front as being many small point sources propagating outward in a concentric circle at the same speed as the wave itself 13.5 Diffraction and Interference In section 8.3 you learned that there are many forms of interference patterns produced by mechanical waves that trav
el in a medium like water. And more importantly, you learned that the interference pattern contains information about the properties of the waves that created the pattern. In this section, we will investigate interference patterns produced by electromagnetic radiation and analyze the patterns to further our understanding of the wave model of light. Huygens’ Principle Robert Hooke proposed the Wave Model of Light in his Micrographica of 1665. The first major improvement to this model was made by Christiaan Huygens (Figure 13.66), twenty years later. A Dutch physicist, Huygens contributed to the study of astronomy by advancing techniques in lens grinding and by discovering Saturn’s rings and its largest satellite, Titan. He is also credited with producing the pendulum clock, originally proposed by Galileo. With regard to wave theory, Huygens described an elegant, conceptual model that predicted the motion of a wave front. This model is known as Huygens’ Principle. A wave front consists of many small point sources of tiny secondary waves, called wavelets, which propagate outward in a concentric circle at the same speed as the wave itself. The line tangent to all the wavelets constitutes the wave front. To visualize Huygens’ Principle, consider a point source of light that emits electromagnetic waves outward, in a concentric circle. At an instant in time, the wave front will form the line t1. According to Huygens’ Principle, all the points along line t1 become secondary sources, producing wavelets that radiate outward to form the wave front at a future time t2. Line t2 is the tangent to all the wavelets that make up the wave front a short time later (Figure 13.67). The same analysis can be performed on a straight wave front, as shown in Figure 13.68. t1 t2 point source of EMR point source of EMR t2 t1 circular wave front straight wave front Figure 13.67 Circular wave front emitted by a light source Figure 13.68 Straight wave front emitted by a distant light source 684 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 685 Concept Check Use Huygens’ Principle to predict the wave front shapes that occur after the straight waves pass through the openings shown in Figure 13.69 (a) and (b). (a) barrier (b) barrier t t e SIM Explore Huygens’ Principle through simulations. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Figure 13.69 At each opening, one or more secondary point sources exist. Draw the line tangent across the wavelets that would exist a short time after the secondary point sources pass through the opening. Explain what has happened to the shape and direction of the wave front after it passed through the barriers in Figure 13.69 (a) and (b). According to Huygens’ Principle, a periodic straight wave will continue to propagate in a straight line until it encounters a barrier. If the wave front encounters a small opening or aperture in the barrier, the wave front will change shape and direction. This process is called diffraction. Young’s Experiment Huygens’ wave theory can explain many of the properties of electromagnetic radiation, including reflection and refraction, but initially, the scientific reputation of Newton and his belief in the particle model of light dominated the scientific community. However, in 1801, an experiment by Thomas Young provided significant evidence in support of the wave model of light. Problems Involved in the Design of the Experiment In the years leading up to Young’s groundbreaking work, scientists studying light were attempting to observe an interference pattern that was similar in nature to that of two point sources. They believed that if light were indeed a wave, then two side-by-side light sources should produce an interference pattern similar to that observed with water in a ripple tank (section 8.3). In these early light experiments, the light from the two sources was incident on a nearby screen that was observed for evidence of an interference pattern. What was not well understood at the time was that waves of high frequency and short wavelength result in a very short distance between nodal lines, the regions of destructive interference that appear dark. As we know today, light waves have an extremely high frequency and short wavelength, so the distance between nodal lines would be so small it could diffraction: the change in shape and direction of a wave front as a result of encountering a small opening or aperture in a barrier, or a corner interference: an effect resulting from the passage of two like waves through each other e SIM Learn more about Young’s classic experiment through simulations. Follow the eSim links at www.pearsoned.ca/ school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 685 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 686 Figure 13.70 top down view interference pattern sunlight pinhole card not be observed using traditional means. In addition, to produce an interference pattern, the waves must be in phase. When there are two incandescent lights, each source emits light in random bursts not necessarily in phase with the other source. These bursts of light have a variety of wavelengths that make up all the colours in the spectrum. Therefore, the interference pattern would not be constant, but would vary rapidly over time, making any observation of a pattern more difficult. screen (a) Interference fringes (b) Photo of two-slit pattern from Young’s experiment info BIT The double-slit experiment works for water waves too! When a single wave front is incident on a barrier with two holes it will produce an interference pattern similar to the one observed by Young. This effect is explored in Unit IV. Figure 13.71 Interference pattern produced by water waves passing through two holes in a barrier antinode: a point of interaction between waves, at which only constructive interference occurs; in an interference pattern, antinodes occur at path difference intervals of whole wavelengths The Experiment Thomas Young solved both of these problems and successfully observed an interference pattern produced by light. Young conducted his experiment using a pinhole in a window shutter and a card he described as “a slip of card, about one-thirtieth of an inch in breadth (thickness).” The card was positioned edgewise into a horizontal sunbeam directed into the room by using a mirror. The sunbeam had a diameter slightly greater than the thickness of the card. When the card was positioned edgewise in the centre of the sunbeam it split the sunbeam into two coherent beams separated by a very small distance (Figure 13.70(a)). The effect was equivalent to light passing through two slits that were very close together. The small distance of separation between the two beams of light expanded the interference pattern on the screen so that the distance between nodal lines was large enough that it could be observed. Light coming from both sides of the card was in phase and the wave fronts could create a fixed interference pattern of light and dark bands, called interference fringes, on the screen (Figure 13.70 and also Figure 13.9, page 640). Bright fringes or antinodal lines were regions of constructive interference and dark fringes or nodal lines were regions of destructive interference. Young’s double-slit experiment provided the strong evidence needed for acceptance of the wave model of light. The Interference Pattern To understand how the interference pattern is created, consider three different points. The first point is the central antinode (Figure 13.72). This point occurs at the centre of the pattern, along the perpendicular bisector. The perpendicular bisector is an imaginary straight line that runs from the midpoint of a line joining the two slits to the area of constructive interference on the screen. This area of constructive interference may be called the central antinode, central bright fringe, or central maximum. At the central antinode, both waves travel the same distance from the slits and arrive at the screen in phase. Constructive interference will be observed as a bright band. path difference: 3.00 λ 3.00 λ 0 λ screen path length 3.00 λ d } constructive interference at central antinode Figure 13.72 A path difference of zero at the central antinode results in constructive interference and a bright band on the screen. path length 3.00 λ l 686 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 687 The second point is the first node, also called the first dark fringe or the first minimum. At this location, one wave travels a distance of 1 farther than the other so that the waves arrive out of phase, and 2 destructive interference is observed as a dark band (Figure 13.73). The third point is another antinode or bright fringe. At this location, one wave travels a distance of 1 farther than the other, causing the waves to arrive at the screen in phase once again (Figure 13.74). This pattern of bright and dark fringes repeats as we move out in both directions from the central, perpendicular bisector. path difference: 3.75 λ 3.25 λ λ 1 2 path length 3.75 λ path difference: 4.50 λ 3.50 λ 1 λ path length 4.50 λ d d } destructive interference at 1st node path length 3.25 λ l path length 3.50 λ l } constructive interference at n 1 antinode Figure 13.73 A path difference of 1 , at the first node results in destructive 2 interference and a dark band on the screen. Figure 13.74 A path difference of , at the first antinode results in constructive interference and a bright band on the screen. Mathematical Analysis of Young’s Experiment Young’s experiment and interference patterns in general contain mathematical information about the waves that create them. For instance, the interference pattern that Young observed can be used to determine the wavelength of the light that created it. dark nodes (destructive interference) bright central anti
node bright antinodes (constructive interference) 4 3 2 1 bisector 1 P 2 3 4 screen 10 λ 9 λ S1 S2 d Figure 13.75 Two point sources, separated by a short distance, produce antinodal lines of constructive interference and nodal lines of destructive interference. Point sources, S1 and S2, from the same original source, are separated by a short distance, d, and produce identical waves that are in phase. node: a point of interaction between waves, at which only destructive interference occurs; in an interference pattern, nodes occur at path difference intervals of one-half wavelength PHYSICS INSIGHT Compare Figures 13.74 and 13.75. The absolute path lengths (expressed in number of wavelengths) shown in Figure 13.74 and 13.75 are different, yet the waves are still causing constructive interference at the point where they meet. How can this be? If the distance separating the sources and the screen is constant, then the path length (expressed in number of wavelengths) to the antinodes can and will be different for different wavelengths of light. As long as the “difference” in path lengths is a whole number multiple of the wavelength of the light , constructive interference will occur where the waves meet and this will be observed as an antinode. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 687 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 688 path length: the distance between a point source and a chosen point in space difference in path length: the difference between two path lengths, each measured from a different origin and extending to a common point in space An equal number of antinodal lines, or lines of constructive interference, radiate outward on either side of the perpendicular bisector of the line joining S1 and S2. Beginning at the perpendicular bisector, antinodal lines on each side are numbered 1, 2, 3, etc., resulting in a duplicated series (Figure 13.75). For our analysis, we will chose a point, P, on the n 1 antinodal line. The path length from S1 to P and from S2 to P can be measured in multiples of wavelengths. From Figure 13.75, the path S1P is 10 and the path S2P is 9. The difference in path length from point P to the two sources is 10 9 1 Therefore, for any point along the first antinodal line (on either side of the bisector), the following relationship is true: S1P S2P 1 Recall from section 8.3 that waves that are 1 out of phase will constructively interfere. This explains why the antinodal lines are bright. A similar analysis for a point, P2, on the n 2 antinodal line generates a similar relationship 2 S1P2 – S2P2 An analysis of points on the n 3 and n 4, etc., antinodal lines follows a similar pattern, generating the following relationship between path difference and wavelength along the antinodal lines. S1Pn S2Pn n, where n 1, 2, 3, 4, . . . (1) If the wavelength is sufficiently large, the difference in path length can be calculated as in Equation (1). Light waves have extremely small wavelengths; therefore, another method of calculating path length becomes appropriate. Difference in Path Length for Waves of Short Wavelength Pn (close) X S1 S2 S1 d X θn S2 d Pn far away Consider the triangles illustrated in Figure 13.76, showing the path length from two sources to point Pn. For any point along the antinodal line Pn, the difference in path length will be equal to the segment S1X. When Pn is selected to be very far away, the paths are nearly parallel. In such a case, S1-X-S2 forms a right triangle allowing the difference in path length to be expressed in terms of angle n, Figure 13.76 Two point sources, separated by a short distance, d, with path lengths to point Pn sin n S1X d sin n difference in path length d 688 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 689 Constructive interference For constructive interference or antinodal lines, substitute the generalization for the difference in path length (Equation 1 above): sin n n d d sin n n for n 1, 2, 3, 4, . . . (2) Equation (2) can be used to determine the wavelength of the light used to produce the interference pattern, where n is the number of the nth antinodal line and d is the distance between the two sources. For a point Pn located far away from the two point sources, the angle n is equivalent to the angle , located between the perpendicular bisector and the straight line drawn between the midpoint of the two sources and the point Pn. The angle is known as the angle of diffraction (Figure 13.77). angle of diffraction: the angle formed between the perpendicular bisector and the straight line to a nodal or antinodal point on the interference pattern screen Pn bisector Destructive interference The analysis for destructive interference, which occurs along the nodal lines, is done in an almost identical way. Recall that destructive interference occurs when the path difference between waves is a half number of wavelengths. Therefore, the generalization for the difference in path length becomes (n 1 ) 2 d d sin n (n 1 ) 2 for n 1, 2, 3, 4, . . . (3) sin n where n is the number of the nth nodal line relative to the perpendicular bisector, d is the distance between the two sources, and the angle n is equivalent to the angle between the perpendicular bisector and the straight line drawn between the midpoint of the two sources and the point Pn. θα X S1 θ n S2 C (midpoint) d Figure 13.77 When Pn is far away, line S1Pn is approximately parallel to CPn, making n and equal. is the angle of diffraction. Example 13.8 Monochromatic light is incident on two slits separated by 0.30 mm, and the first bright fringe (n 1 antinode) is located at an angle of 0.080 from the central antinode. What is the wavelength of the light? Given d 0.30 mm 3.0 104 m n n 1 0.080 Required Wavelength () Practice Problems 1. Light of an unknown wavelength is incident on two slits separated by 0.20 mm. The second bright fringe is located at an angle of 0.26 from the central antinode. What is the light’s wavelength? 2. Blue light of 460 nm is incident on two slits that are 0.55 mm apart. What is the angle of diffraction for the third antinodal line? Analysis and Solution The angle given is relative to the central antinode, which occurs along the perpendicular bisector; therefore, the angle given is relative to the perpendicular bisector. The first bright fringe is a region of constructive Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 689 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 690 3. The second nodal line of an interference pattern occurs at 0.095 relative to the central antinode. The two slits are separated by 0.40 mm. What is the wavelength and colour of light producing this pattern? Answers 1. 4.5 107 2. 0.14 3. 4.4 107 m, violet interference, where the path difference must be one full wavelength different. Therefore, Equation (2) above for constructive interference applies: d sin n n (3.0 104 m) sin 0.080 1 4.2 107 m Paraphrase Monochromatic light with a wavelength of 4.2 107 m will produce a bright fringe at an angle of 0.080 from the central antinode. info BIT When the angle is small (10º) sin is nearly equal to tan as illustrated in the calculated values for both shown below: sin Angle tan sin / tan 1 2 5 0.017452 0.017455 0.9998 0.03490 0.03492 0.9994 0.08716 0.08749 0.9962 10 0.1736 0.1763 0.9847 distance between central antinode and nth antinodal line Pn xn screen Finding Wavelength and the Angle of Diffraction Under Experimental Conditions In experimental settings, it is often difficult to measure the angle of diffraction because this angle is very small relative to a point that is very far away from the two slits. It is easier to obtain a value for sin by determining the ratio x/l, as shown in Figure 13.78. Here, x is the distance between the central antinode (where the perpendicular bisector intersects the screen) and the antinodal fringe. The length of the perpendicular bisector, l, is the distance from the midpoint between the slits to the screen, where the interference pattern is observed. In Figure 13.78, x/l tan , but when l is much greater than x, the ratio of x/l is very small (generally less than 0.2) making tan nearly equal to sin (see Infobit). Therefore, it is acceptable to assume sin x/l in this case. By replacing sin with the ratio x/l, we arrive at the following equation for antinodal (bright) fringes: x d l n xd nl Applying the same analysis to nodal (dark) fringes gives: x d l 1 n 2 xd 1 l n 2 l (distance to the screen) n λ S1 θ C d θ S2 Figure 13.78 Determining sin using x and l 690 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 691 Example 13.9 A student measuring the wavelength of light emitted by a krypton gas sample directs the light through two slits separated by 0.264 mm. An interference pattern is created on a screen 3.0 m from the slits and the distance between the second bright (n 2 antinode) fringe and the central antinode is measured to be 1.18 cm. What is one of the wavelengths of light emitted by the krypton gas sample? Given d 0.264 mm 2.64 104 m l 3.0 m x 1.18 cm 1.18 102 m n 2 Required wavelength () Analysis and Solution The bright fringe is a region of constructive interference, where the path difference must be a whole number of wavelengths. Solve for the wavelength by using the equation: xd nl (1.18 102 m)(2.64 104 m) (2)(3.0 m) 5.2 107 m Paraphrase The krypton gas sample emits light with a wavelength of 5.2 107 m. Practice Problems 1. Monochromatic light is incident on two slits separated by 0.15 mm. An interference pattern is observed on a screen 5.0 m away. The distance between the 3rd dark fringe and the central antinode is 4.50 102 m. What is the wavelength of the light? 2. Monochromatic light is incident on two slits separated by 3.00 105 m. The distance between antinodes is 3.10 102 m. If the screen is 1.50 m from the slits, what is the light’s col
our and wavelength ? 3. A student used light of wavelength 5.00 107 m and found that the distance between the third node and the central antinode was 1.00 101 m. If the screen was located 1.20 m away from the slits, how far apart are the slits? Answers 1. 5.4 107 m 2. 6.20 107 m, red 3. 1.50 105 m Poisson’s Bright Spot Using an analysis similar to that in Example 13.9, Young calculated the wavelength of various colours of light and announced his results in 1807. He was still overshadowed by Newton’s reputation and support for the particle model of light. Thus, Young was not taken seriously by the scientific community until 1818, when Augustin Fresnel, a French physicist (Figure 13.79), proposed another mathematical wave theory. This new theory formed a critical turning point in the debate between the wave and particle models of the time. A mathematician by the name of Simon Poisson argued that the equations in Fresnel’s theory could be used to predict a unique diffraction pattern that should be produced when light is incident on a small round disc. Poisson showed that Fresnel’s equations should produce a central bright fringe at the centre of a shadow cast by a solid round disc when it is illuminated by a point source of monochromatic light (Figure13.80(a)). This predicted result was similar to the central bright fringe observed in Young’s double-slit experiment. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 691 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 692 Fresnel’s equations showed that the light that is diffracted around the edge of the disc should constructively interfere at the centre of the shadow. Poisson was unable to observe this bright spot experimentally and was of the opinion that he had refuted the wave model of light. solid disc shadow equal path length bright spot of constructive interference Figure 13.79 Augustin Fresnel (1788–1827). Fresnel’s analysis of diffraction provided the theoretical groundwork for the transverse wave model of light. He also designed the lenses that were used in lighthouses of the 19th century and in overhead projectors, solar collectors, and beacon lights of the 21st century. diffraction grating: a sheet of glass or plastic etched with a large number of parallel lines. When light is incident on the grating, each line or slit acts as one individual light source. Figure 13.81 Diffraction grating info BIT The concentric circles cut into a CD when it is made make it behave as a diffraction grating. So, the rainbow pattern you see when you look at a CD is the product of interference caused by diffraction. Figure 13.80 Poisson’s Bright Spot (a) Constructive interference at the bright spot (b) Photograph of Poisson’s Bright Spot Poisson’s prediction of a bright spot was retested by Dominique Arago in 1818, and this time, the bright spot was verified (Figure 13.80 (b)). The bright spot is known as “Poissson’s Bright Spot” because, even though Poisson was a supporter of the particle model of light, he had predicted the spot’s existence if the wave model of light was correct. Mainly as a result of this verification of Poisson’s Bright Spot, by 1850 the wave model of light was generally accepted by the scientific community. The model was then successfully applied to many of the properties of light. Diffraction Gratings Young’s experiment used only two small point sources of light that were in phase. A diffraction grating (Figure 13.81) has a very large number of equally spaced, parallel lines that act as individual light sources. When light is incident on a multi-slit diffraction grating, an interference pattern, similar to that of a double slit, is produced on a distant screen. But there are several key differences. First, the large number of lines in a diffraction grating can deliver more light energy to the distant screen, increasing the brightness of the interference pattern. Second, the antinodal, bright fringes are more defined, being sharper and narrower. And third, when line separation is very small, the separation between the lines (d) is inversely proportional to the distance between the fringes (x) in the interference pattern according to d xd nl nl x Therefore, the extremely small separation distance between lines on a grating will cause an increase in the separation between the fringes in the pattern. For these reasons, a diffraction grating is a very precise apparatus for investigating the wavelength of light. When a diffraction grating is used with monochromatic light, the interference pattern will have the same colour as the wavelength of light used to produce it. When full-spectrum, white light is used, each antinode will appear as a rainbow because each wavelength is diffracted Figure 13.82 692 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 693 at a slightly different angle. The relationship between the wavelength and the angle is: n d sin n as was derived earlier for Young’s double-slit experiment. Diffraction gratings are produced by etching a large number of parallel lines on a sheet of glass or plastic. Each grating is defined by the number of lines per centimetre etched on it. Tens of thousands of lines per centimetre are common. The distance between lines is the inverse of the number of lines per centimetre, so as the number of lines increases, the distance between any two lines decreases accordingly. Diffraction in Nature — Solar and Lunar Coronas When the Sun is rising (Figure 13.83) or an overexposed photo of the Moon is taken (Figure 13.84), a pattern of rings appears. These are examples of diffraction in the atmosphere. When light from the Sun or Moon enters the atmosphere it encounters uniformly sized droplets of water and ice crystals. Diffraction occurs as the light bends around the edges of the particles at varying degrees depending upon the wavelength of the light. A series of coloured rings surrounding the astronomical object results. e LAB For a probeware activity, go to www.pearsoned.ca/ school/physicssource. e SIM Investigate diffraction gratings through the use of simulations. Follow the eSim links at www.pearsoned.ca/ school/physicssource. e WEB To learn more about diffraction effects in nature, follow the links at www.pearsoned.ca/ school/physicssource. Example 13.10 Figure 13.83 A rising Sun Figure 13.84 A ring around the Moon A diffraction grating has 1000 lines/cm. When light passes through the grating, an interference pattern is produced on a screen 4.00 m away. The first-order bright fringe is 19.2 cm away from the central antinode. What is the wavelength and colour of the light? Given d 1 cm / 1000 lines 1 103 cm/line 1 105 m/line x 19.2 cm 1.92 101 m n 1 l 4.00 m Required wavelength of light () Analysis and Solution The separation between lines is the inverse of the lines per centimetre given for the diffraction grating. Line separation is very small, and the first-order bright fringe is n 1 because it is an antinode, or line of constructive interference. Therefore, this equation applies: Practice Problems 1. An unknown wavelength of light is incident on a diffraction grating with 2500 lines/cm. The distance between the central antinode and the 3rd dark node is 20.0 cm when the screen is located 50.0 cm from the grating. Determine the wavelength of the light. 2. How many lines/cm are there in a diffraction grating if the 3rd dark fringe is located 5.00 cm from the central antinode when the screen is located 60 cm from the grating? Assume 500 nm. 3. A diffraction grating has 1000 lines/cm. When red light ( 750 nm) is incident on the grating, what will be the separation between bright antinodes on a screen 3.0 m away? Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 693 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 694 Answers 1. 6.40 107 m 2. 6.7 102 lines/cm 3. 23 cm xd nl (1.92 101 m)(1 105 m) (1)(4.00 m) 4.80 107 m 480 nm Paraphrase Based on the interference pattern, the incident light is 480 nm, which corresponds to the colour blue. 13-10 Inquiry Lab 13-10 Inquiry Lab Determining the Wavelength of Red, Green, and Blue Light This investigation will verify the known wavelengths of different colours of light. According to the visible spectrum, red, green, and blue light have wavelengths in the following ranges: • red: 650–750 nm • green: 500–550 nm • blue: 450–500 nm Question What are the wavelengths of red, green, and blue light? Variables manipulated: the distance between the diffraction grating and the screen will be indirectly manipulated responding: the position of the first-order antinodes for red, blue, and green light on the “apparent” screen controlled: single-filament incandescent light source and diffraction grating Materials and Equipment thin-film diffraction grating two metre-sticks single-filament lamp masking tape and pen Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork them. A screen is not needed as the angles are identical, so we can assume the distance to the hypothetical screen is identical to the distance between the lamp and the grating (1.0 m). While one person observes the antinode, a second person is directed to place an identification tape on the metrestick where the antinode appears to be. This allows the distance between the central antinode and the position of the firstorder antinode to be measured. This procedure can also be used to find the position of the second-order antinode. Procedure 1 Set up the two metre-sticks such that they make a right angle with one another. 2 Place the lamp at the point where the metre-sticks join as shown in Figure 13.85. 3 Place the thin-film diffraction grating vertically upright at the end of one of the metre-sticks. 1 m x2 x1 lamp n 2 n 1 Experimental Design To investigate the position of the first- and second-order antinodes in an interference pattern produced
by white light, a “simulated” screen will be used. Two metre-sticks and a lamp are arranged as shown in Figure 13.85. In this design, observers look through the diffraction grating to see the antinodes as they would appear on a screen behind 1 m θ2 θ1 thin-film diffraction grating θ1 θ2 Figure 13.85 694 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 695 4 Turn on the lamp and look through the grating, moving your head side to side until you can see the first-order antinode, which should appear as a rainbow. Your eye acts as the wavelength detector, in place of a screen. 5 While looking at the antinode, direct your lab partner to put a piece of tape labelled “B” on the metre-stick where the blue antinode band appears to be. 6 Repeat step 5 using tape labelled “G” for the green band of light and “R” for the red band of light. 7 Repeat steps 5 and 6 for the second-order antinode, if it is visible along the metre-stick. 8 Record the distance between the lamp and each piece of labelled tape. Analysis 1. Using the first antinode (n 1), calculate the wavelength of red, blue, and green light. 2. Determine the mean wavelength of each colour of light using the ranges given. 3. Calculate the percent difference between the mean wavelength of each colour and your experimentally determined value. 4. Explain why each antinode appears as a rainbow. Hint: What effect does the wavelength have on the angle of diffraction? 5. How many antinodes should appear in the diffraction grating on either side of the light source? Assume the largest angle of diffraction that could be visible is 89. M I N D S O N Comparing Spectra: Dispersion vs. Diffraction The rainbow produced when white light is refracted through a prism is similar to the rainbow produced at each antinode when white light passes through a thinfilm diffraction grating. In small groups, prepare a presentation to compare and contrast these two phenomena. Your presentation should consider: 1. The wave nature of light 2. The similarities of refraction and diffraction as they relate to Huygens’ Principle of wavelets 3. The different wavelengths of the visible spectrum and how this leads to the separation of the colours in both dispersion and diffraction 4. The key differences in the causes of dispersion in a prism and the production of antinodes in an interference pattern 5. The reversed order of colours Your presentation should also include: • relevant images of both phenomena • schematics of each phenomenon using ray diagrams • where appropriate the use of animations and simulations found at www.pearsoned.ca/school/ physicssource • a list of references Polarization Young’s double-slit experiment, and interference in general, provided strong evidence that light does exhibit wave properties. However, evidence of interference alone could not distinguish whether the waves were transverse or longitudinal. Recall that in section 13.1 light, and electromagnetic radiation in general, was described by Maxwell as consisting of perpendicular electric and magnetic fields, propagating through space at the speed of light. In other words, Maxwell predicted that light was a transverse wave. Is there evidence that this is indeed the case? Using a mechanical model, such as a rope, one can see that a transverse wave can be linearly polarized when vibrations only occur in one plane. The vertically polarized transverse waves shown in Figure 13.86 can pass through the vertical slit, but are blocked, or absorbed, by the horizontal slit. The longitudinal waves, on the other hand, can pass e LAB For a probeware activity, go to www.pearsoned.ca/ school/physicssource. (a) (b) Figure 13.86 (a) A transverse wave passing through a vertical slit and being absorbed by a horizontal slit (b) A longitudinal wave passing through both a vertical and a horizontal slit Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 695 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 696 info BIT Ancient Inuit hunters developed a unique technology to limit the glare of the Sun’s light reflecting off the spring snow. Snow goggles, similar to those seen in the movie Atanarjuat: The Fast Runner, were the first sunglasses. The goggles were carved mainly from caribou antler whalebone, or ivory; driftwood was also used. Narrow slits in the snow goggles reduced the amount of light, thus protecting the hunter’s eyes and preventing the debilitating effects of snow blindness. A secondary positive effect of the goggles was improved visibility. Figure 13.87 Inuit snow goggles polarizing filter: a filter that allows only one plane of the electric field to pass through it; plane polarized EMR emerges polarization: production of a state in which the plane of the electric field for each electromagnetic wave occurs only in one direction plane polarized light: light resulting from polarization, in which only one plane of the electric field is allowed to pass through a filter through both slits unaffected because longitudinal waves are not linearly polarized. By a process similar to the mechanical model, electromagnetic waves can be blocked by two polarizing filters held at right angles to one another. In 13-1 QuickLab at the beginning of the chapter, you discovered that two polarizing filters, held at right angles to one another, can absorb light. Figure 13.88 Two polarizing filters, one held vertically, the other held horizontally, partially overlap, showing the absorption of electromagnetic waves. The photograph in Figure 13.88 can be explained by considering electromagnetic radiation as perpendicular magnetic and electric fields, with the plane of polarization arbitrarily defined by the direction of the electric field. When light is produced by an incandescent light bulb it is not polarized, meaning that the plane of the electric fields for each wave occurs randomly as light propagates outward from the source in all directions. When unpolarized light is incident on a polarizing filter, only one plane of the electric field is allowed to pass through, causing plane polarized light to emerge. If a second polarizing filter is held at right angles to the plane polarized light, then the plane polarized light also is absorbed (Figure 13.89). randomly oriented EMR vertical polarizer horizontal polarizer no waves vertically polarized EMR Figure 13.89 Unpolarized light incident on two polarizing filters at right angles to one another The blue light in sunlight is partially polarized when it is scattered in the atmosphere. Therefore, in sunglasses and camera lenses, polarized filters are used to reduce the blue polarized light from the sky while allowing other non-polarized colours to pass through and appear brighter. To see this effect, tilt your head from side to side while looking at the blue sky with polarized glasses. The polarizing effect supports the wave model of light in general and in particular, the concept that light is composed of perpendicular, oscillating electric and magnetic fields (Figure 13.90). Figure 13.90 Two pairs of polarized sunglasses, at right angles 696 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 697 13.5 Check and Reflect 13.5 Check and Reflect Knowledge 1. According to Huygens’ Principle, what will happen to the shape of a straight wave front after it passes through a small opening in a barrier? 2. Use Huygens’ Principle to describe how interference can occur when a straight wave front is incident on two narrow openings. 3. Two incandescent white lights are placed close to one another. Explain why an interference pattern is not observed on a nearby screen. 4. What parameters did Young need to control to enable him to observe an interference pattern from two point sources of light? 5. Construct a concept map to show the relationship between path length, nodal fringes, antinodal fringes, wave phase, and interference. Applications 6. In an experiment similar to Young’s, how far apart are two slits if the 3rd antinode is measured to be 20 from the central antinode, when light with a wavelength of 650 nm is used? 7. Determine the angle of diffraction to the 2nd node when light with a wavelength of 425 nm is incident on two slits separated by 6.00 106 m. 8. Light with a wavelength of 700 nm is diffracted by a diffraction grating with 5.00 103 lines/cm. If a screen is positioned 1.00 m away from the grating, what is the distance between the 1st and central antinodes? 9. Monochromatic light with a frequency of 5.75 1014 Hz is incident on a diffraction grating with 60 lines/cm. What is the distance between the 2nd and 3rd dark fringes when the screen is located 1.20 m away? 10. An unknown light source is directed at a diffraction grating with 6.00 104 lines/m. If the nodal lines are 5.50 cm apart when the screen is 1.50 m away, what is the wavelength and frequency of the light? 11. Light emitted from an unknown gas sample is incident on a diffraction grating with 5.00 102 lines/cm. The antinodes appear on a screen 1.50 m away and are separated by 3.10 102 m. What is the wavelength and frequency of the light? Extensions 12. Design an experiment to determine the wavelength of an unknown monochromatic light. Include an experimental design, material list, and procedure. 13. Compare the wavelength of X rays to that of visible light and explain what should happen to the diffraction pattern if X rays were used instead of visible light. 14. Investigate how the process of diffraction, using radiation other than visible light, can be useful for determining the shapes of crystal lattices and structures too small to be seen with visible light. e TEST To check your understanding of diffraction, interference, and polarization, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 697 13-PearsonPhys30-Chap13 7/24/08 3:44 PM
Page 698 CHAPTER 13 SUMMARY Key Terms and Concepts electromagnetic radiation frequency wavelength electromagnetic spectrum particle model wave model photon quantum model electric field magnetic field capacitor Maxwell’s Equations electromagnetic wave rectilinear propagation ray diagram plane mirror law of reflection virtual image real image magnification image attitude converging mirror diverging mirror mirror equation refraction refractive index Snell’s Law total internal reflection critical angle spectrum dispersion converging lens diverging lens thin lens equation Huygens’ Principle diffraction interference antinode node path length difference in path length angle of diffraction diffraction grating polarizing filter polarization plane polarized light Key Equations m hi ho di do sin 1 sin 2 v1 v2 1 2 n2 n1 Conceptual Overview 1 do 1 di 1 f sin n n d n c v xd nl n1 sin 1 n2 sin 2 Summarize this chapter by explaining how the properties of electromagnetic radiation support either the wave model of light or the particle model of light, or both. reflection refraction diffraction Particle Model polarization Wave Model interference electric field magnetic field 698 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 699 CHAPTER 13 REVIEW Knowledge 1. (13.1) Create a table to identify all the major categories of electromagnetic radiation, including the wavelengths and frequencies listed in the spectrum shown in Figure 13.4. Brainstorm common uses for each type of radiation. 2. (13.1) Compare and contrast the particle and wave models of electromagnetic radiation. 3. (13.1) What two critical insights were understood by Maxwell when he developed his theory of electromagnetic radiation? 4. (13.1) Consider two electric field lines on a transverse wave. One field line is up, and a moment later, another is down. What is produced as a result of this “changing” electric field? 5. (13.1) Describe how an electromagnetic wave is able to propagate in empty space. 6. (13.1) Describe the five predictions that Maxwell made regarding the properties of electromagnetic radiation. 7. (13.1) How did Hertz prove that the EMR observed at his antenna was, in fact, produced by the nearby spark gap and did not originate from another source? 8. (13.2) The first significant attempt to measure the speed of light was made by Christiaan Huygens, using the eclipse of Jupiter’s moon Io. Describe this method. 9. (13.2) In addition to measuring the speed of light with a rotating toothed wheel, Armand Fizeau demonstrated that light travelled at different speeds in moving water. Explain how the results of his investigation support the wave model of light. 10. (13.3) Draw a ray diagram to demonstrate the law of reflection. 11. (13.3) Construct a ray diagram for a converging mirror and illustrate the following terms. (a) centre of curvature (C) (b) radius of curvature (r) (c) vertex (V) (d) principal axis (PA) (e) principal focal point (F) (f) focal length (f) 12. (13.3) Can a diverging mirror produce a real image? Explain. 13. (13.4) Using a ray diagram, illustrate partial reflection and partial refraction for a ray passing from air into water at an angle of 15. On your ray diagram, label the normal line, the index of refraction, the angle of incidence, the angle of reflection, and the angle of refraction. 14. (13.4) Light passes from a medium with a high refractive index to one with a low refractive index. Is the light bent away from or toward the normal line? 15. (13.4) Dispersion is the separation of white light into all the colours of the spectrum. Explain two different methods that could be used to separate all the colours in white light. 16. (13.5) Illustrate the process of refraction using a straight wave front that travels from air into water. Based on your diagram, does Huygens’ Principle support the wave model or the particle model of light? 17. (13.5) A straight wave front is incident on two small holes in a barrier; illustrate the shape of the wave front a moment after it makes contact with the barrier. Does your drawing indicate that interference will occur? 18. (13.5) Why is an interference pattern not observed when two incandescent lights are located next to one another? 19. (13.5) How does the evidence from polarizing filters support the transverse nature of the wave model of light? 20. (13.5) Each antinode appears as a full spectrum when white light is incident on a diffraction grating. Explain this phenomenon. 21. (13.5) Explain how path length and diffraction are related to the production of Poisson’s Bright Spot. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 699 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 700 Applications 22. (13.1) High-voltage transmission lines that carry alternating current can interfere with radio waves. Explain how this interference can occur. 31. (13.3) A 20.0-cm-high, inverted image is produced when an object is placed 12.0 cm from a converging lens with a focal length of 11.0 cm. Calculate the height of the object. 23. (13.2) An 8-sided mirror is rotating at 5.50 102 Hz. At what distance should the fixed mirror be placed to replicate Michelson’s experiment? 24. (13.2) A fixed mirror and a rotating mirror are separated by 30.0 km. The 8-sided rotating set of mirrors turns at 600 Hz when the light is able to pass through the experimental apparatus. Calculate the speed of light. 25. (13.3) When you look into a plane mirror, an image is formed. Describe the characteristics of the image based on attitude, type, and magnification. 26. (13.3) A student stands 30 cm from a plane mirror. If the student’s face is 25 cm in length, what is the minimum length of mirror needed for the student to see her entire face? 27. (13.3) An object is located 25.0 cm from a converging mirror with a focal length of 15.0 cm. Draw a scale ray diagram to determine the following: (a) the image location and type (b) the image attitude (c) the magnification of the image 28. (13.3) Construct a ray diagram for a diverging mirror and illustrate the following terms: (a) centre of curvature (C) (b) radius of curvature (r) (c) vertex (V) 29. (13.3) Where must an object be placed relative to the focal point for a converging mirror such that the image produced is virtual? 30. (13.3) A 15.0-cm-high object is placed 20.0 cm from a diverging mirror with a virtual focal length of 10.0 cm. How high is the image and where is it located? 32. (13.4) Calculate the speed of yellow light, 589 nm, in the following materials: (a) water (b) ethanol (c) Lucite® (d) quartz glass (e) diamond 33. (13.4) Light with a wavelength of 610 nm is incident on a quartz glass crystal at an angle of 35. Determine the angle of refraction and the wavelength of the light in the quartz glass. 34. (13.4) Can total internal reflection occur when light travels from (a) air into water? (b) water into air? (c) Lucite® into water? (d) water into diamond? 35. (13.4) Light enters an unknown material and slows down to a speed of 2.67 108 m/s. What is the refractive index of the unknown material? Compare the refractive index of this material to that of water — which one has a higher index? 36. (13.4) Calculate the critical angle of the following boundaries: (a) water-air (b) diamond-air (c) diamond-water (d) Lucite®-air 37. (13.4) A 4.00-cm-high object is located 5.00 cm from a diverging lens with a focal length of 10.0 cm. Using the thin lens equation, determine the image attributes and position. Verify your answer with a scale ray diagram. 700 Unit VII Electromagnetic Radiation 13-PearsonPhys30-Chap13 7/24/08 3:44 PM Page 701 38. (13.5) In an experiment similar to Young’s, light with a wavelength of 630 nm is incident on two slits separated by 5.3 105 m. What is the angle to the 1st, 2nd, and 3rd antinodes? 39. (13.5) Monochromatic light is incident on two slits separated by 0.25 mm. The first dark fringe deviated an angle of 0.050 from the central antinode. What is the wavelength and colour of the light? 40. (13.5) Light from an unknown gas sample is incident on two slits separated by 1.4 104 m. On a screen 1.1 m away, the distance between the 7th node and the central antinode is measured to be 0.025 m. What is the wavelength of the light emitted by the unknown gas sample? 41. (13.5) A screen is located 4.5 m from two slits that are illuminated with a 490-nm light source. If the distance between the central antinode and the first-order antinode is 0.037 m, how far apart are the two slits? Extensions 42. An X-ray machine operates by accelerating an electron through a large potential difference, generating a large amount of kinetic energy. The high-speed electron then collides with a metal barrier. Explain why the collision produces a high-frequency X ray. 43. Cable television wires have a metal shield surrounding the copper wire that carries the television signal. The shielding prevents interference from electromagnetic radiation and it must be grounded in order to effectively block interference. Explain how the shielding prevents interference and why it needs to be grounded. 44. When you place the concave side of a spoon on your nose and slowly pull it away from your face, your image disappears at a certain distance. What is the significance of this distance? 45. After light enters Earth’s atmosphere it encounters a temperature gradient as it approaches the surface of Earth, causing a mirage. If the warm air near the surface of Earth has a lower index of refraction than the cooler air above, which way is the light bent? Show this with a ray diagram. 46. Explain why a fibre-optic network is much more efficient and powerful than a copper-wire network. Consolidate Your Understanding 1. Explain how electromagnetic radiation is able to propagate in the absence of a medium, like air. 2. Why is an accelerating charge required to produce electromagnetic radiation and how does this relate to the word “changing” i
n Maxwell’s explanation of EMR? 3. Describe the three-dimensional shape of an electromagnetic wave. Specify the directions of both the electric and magnetic field variations, and the direction of wave propagation. 4. Has Maxwell’s last prediction been verified by experimental evidence? If so, describe the evidence as it relates to reflection, refraction, diffraction, interference, and polarization. 5. Could Hertz have investigated the phenomenon of diffraction by using the same equipment as in his famous experiment? If so, how? Think About It Review your answers to the Think About It questions on page 635. How would you answer each question now? e TEST To check your understanding of the nature and behaviour of electromagnetic radiation, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 13 The wave model can be used to describe the characteristics of electromagnetic radiation. 701 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 702 C H A P T E R 14 Key Concepts In this chapter, you will learn about: the idea of the quantum the wave–particle duality basic concepts of quantum theory Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe light using the photon model explain the ways in which light exhibits both wave and particle properties state and use Planck’s formula give evidence for the wave nature of matter use de Broglie’s relation for matter waves Science, Technology, and Society explain the use of concepts, models, and theories explain the link between scientific knowledge and new technologies Skills observe relationships and plan investigations analyze data and apply models work as members of a team apply the skills and conventions of science 702 Unit VII The wave-particle duality reminds us that sometimes truth really is stranger than fiction! Up to this point in the course, you have studied what is known as classical physics. Classical physics includes most of the ideas about light, energy, heat, forces, and electricity and magnetism up to about 1900. The golden age of classical physics occurred at the very end of the 19th century. By this time, Newton’s ideas of forces and gravitation were over 200 years old, and our knowledge of physics had been added to immensely by the work of James Clerk Maxwell, Michael Faraday, and others. It seemed as though nearly everything in physics had been explained. In the spring of 1900, in a speech to the Royal Institution of Great Britain, the great Irish physicist William Thomson (Figure 14.1) — otherwise known as Lord Kelvin — stated that “… the beauty and clearness of the dynamical theory of light and heat is overshadowed by two clouds….” You could paraphrase Kelvin as saying “the beauty and clearness of physics is overshadowed by two clouds.” One “cloud” was the problem of how to explain the relationship between the temperature of a material and the colour of light the material gives off. The other “cloud” had to do with an unexpected result in an experiment to measure the effect of Earth’s motion on the speed of light. Kelvin was confident that these two clouds would soon disappear. He was wrong! Before the year was out, the first of these clouds “broke” into a storm the effects of which are still being felt today! In this chapter, you will meet one of the strangest ideas in all of science. In many ways, this chapter represents the end of classical physics. You will learn that light is not only a wave, but also a particle. Stranger still, you will learn that things you thought were particles, such as electrons, sometimes act like waves! Hang on! Figure 14.1 William Thomson (1824–1907) was named Lord Kelvin by Queen Victoria in 1892. He was the first British scientist to be honoured in this way. During his long and illustrious career, Lord Kelvin published over 600 books and papers, and filed more than 70 patents for his inventions. He was one of the driving forces behind the first transatlantic telegraph cable. info BIT This chapter is about the “cloud” that became quantum theory. In 1905, the other “cloud” became Einstein’s theory of special relativity. 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 703 14-1 QuickLab 14-1 QuickLab The Relationship Between Temperature and Colour of an Incandescent Object Problem What is the relationship between the temperature of a hot, glowing object and the colour of light emitted by the object? Questions 1. What happens to the temperature of the filament in the light bulb as you increase the voltage output of the transformer? Materials incandescent (filament-style) light bulb variable transformer, 0–120 V transmission-type diffraction grating Procedure 1 Attach a filament-style light bulb to a variable transformer and slowly increase the voltage. 2 Observe the spectrum produced by the light from the light bulb as it passes through the diffraction grating. For best results, darken the room. 2. How does the spectrum you observe through the diffraction grating change as you increase the voltage through the filament? 3. As you increase the temperature of the filament, what happens to the colour at which the spectrum appears brightest? 4. You may have noticed that the colour of a flashlight filament becomes reddish as the battery weakens. Suggest why. Think About It 1. Describe the relationship between the colour of a hot object and its temperature. Note in particular the colour you would first see as the temperature of an object increases, and how the colour changes as the object continues to heat up. 2. What do we mean by the terms “red-hot” and “white-hot”? 3. Which is hotter: “red-hot” or “white-hot”? 4. Is it possible for an object to be “green-hot”? Explain. Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. info BIT Even a great physicist can be wrong! Despite making extremely important contributions to many areas of physics and chemistry, Lord Kelvin has also become famous for less-than-accurate predictions and pronouncements. Here are a few: • “I can state flatly that heavierthan-air flying machines are impossible.” (1895) • “There is nothing new to be discovered in physics now. All that remains is more and more precise measurement.” (1900) • “X rays will prove to be a hoax.” (1899) • “Radio has no future.” (1897) • “[The vector] has never been of the slightest use to any creature.” Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 703 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 704 14.1 The Birth of the Quantum We take for granted the relationship between the colour of a hot, glowing object and its temperature. You know from sitting around a campfire that the end of a metal wiener-roasting stick slowly changes from a dull red to a bright reddish-yellow as it heats up. For centuries, metalworkers have used the colour of molten metal to determine when the temperature is just right for pouring metal into molds in the metal-casting process (Figure 14.2). The association between colour and temperature is so common that you would expect the mathematical relationship between colour and temperature to be simple. That is certainly what classical physicists expected. Despite their best efforts, however, classical physicists were never able to correctly predict the colour produced by an incandescent object. What is the connection between temperature and a glowing object’s colour? Figure 14.3 shows three graphs that relate the colours produced by hot objects to their temperatures. The relationship between colour and temperature may be summarized as follows: 1. Hot, glowing objects emit a continuous range of wavelengths and hence a continuous spectrum of colours. 2. For a given temperature, the light emitted by the object has a range of characteristic wavelengths, which determine the object’s colour when it glows (Figure 14.2). 3. The hotter an object is, the bluer the light it emits. The cooler an object is, the redder its light is. Figure 14.2 The colour of molten bronze depends on its temperature. incandescent: glowing with heat (a 14 12 10 8.0 6.0 4.0 2.0 0.0 (b) T 10 000 K 0.5 1 1.5 2 2.5 3 3.5 Frequency 1015 (Hz.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 5.0 0.0 (c) T 5000 K 0.5 1 1.5 2 2.5 3 3.5 Frequency 1015 (Hz 14 12 10 8.0 6.0 4.0 2.0 0.0 T 2500 K 0.5 1 1.5 2 2.5 3 3.5 Frequency 1015 (Hz) Figure 14.3 Blackbody curves for three different temperatures (Kelvin): 10 000 K, 5000 K, and 2500 K. Frequency is along the horizontal axis, and energy intensity emitted is along the vertical axis. Note that these graphs do not have the same vertical scale. If they did, graph (a) would be 256 times taller than graph (c)! Concept Check Next time you are under a dark, clear sky, look carefully at the stars. Some will appear distinctly bluish-white, while others will be reddish or orange in appearance. What do differences in colour tell you about the stars? 704 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 705 Physicists call the graphs in Figure 14.3 blackbody radiation curves. The term “blackbody,” introduced by the German physicist Gustav Kirchhoff in 1862, refers to an object that completely absorbs any light energy that falls on it, from all parts of the electromagnetic spectrum. When this perfect absorber heats up, it becomes a perfect radiator. The energy it reradiates can be depicted as a blackbody curve, which depends on temperature only (Figure 14.3). Hot objects, such as the filament in an incandescent light bulb used in 14-1 QuickLab, or a glowing wiener-roast stick, are good approximations to a blackbody. Not only did classical physics fail to explain the relationship between temperature and the blackbody radiation curve, it also made a completely absurd prediction: A hot object would emit its energy most effectively at short wavelengths, and that the sh
orter the wavelength, the more energy that would be emitted. This prediction leads to a rather disturbing conclusion: If you strike a match, it will emit a little bit of light energy at long wavelengths (e.g., infrared), a bit more energy in the red part of the spectrum, more yet in the blue, even more in the ultraviolet, a lot more in the X-ray region, and so on. In short, striking a match would incinerate the entire universe! This prediction was called the ultraviolet catastrophe. Fortunately for us, classical physics was incorrect. Figure 14.4 shows a comparison between the prediction made by classical physics and the blackbody radiation curve produced by a hot object. Quantization and Planck’s Hypothesis In December 1900, Max Planck (Figure 14.5) came up with an explanation of why hot objects produce the blackbody radiation curves shown in Figures 14.3 and 14.4. Planck suggested that the problem with the classical model prediction had to do with how matter could absorb light energy. He discovered that, by limiting the minimum amount of energy that any given wavelength of light can exchange with its surroundings, he could reproduce the blackbody radiation curve exactly. The name quantum was given to the smallest amount of energy of a particular wavelength or frequency of light that could be absorbed by a body. Planck’s hypothesis can be expressed in the following formula, known as Planck’s formula: E nhf where E is the energy of the quantum, in joules, n 1, 2, 3 … refers to the number of quanta of a given energy, h is a constant of proportionality, called Planck’s constant, which has the value 6.63 1034 Js, and f is the frequency of the light. If energy is transferred in quanta, then the amount of energy transferred must be quantized, or limited to whole-number multiples of a smallest unit of energy, the quantum. Even though Planck’s hypothesis could reproduce the correct shape of the blackbody curve, there was no explanation in classical physics for his idea. The concept of the quantum marks the end of classical physics and the birth of quantum physics. blackbody radiation curve: a graph of the intensity of light emitted versus wavelength for an object of a given temperature blackbody: an object that completely absorbs any light energy that falls on it classical theory hottest y t i s n e t n I coolest Wavelength of emitted radiation Figure 14.4 According to classical theory, as an object becomes hotter, the intensity of light it emits should increase and its wavelength should decrease. The graph shows a comparison of the classical prediction (dashed line) and what is actually observed for three objects at different temperatures. Figure 14.5 Max Planck (1858–1947) is one of the founders of quantum physics. quantum: the smallest amount or “bundle” of energy that a wavelength of light can possess (pl. quanta) Planck’s formula: light comes in quanta of energy that can be calculated using the equation E nhf quantized: limited to whole multiples of a basic amount (quantum) Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 705 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 706 PHYSICS INSIGHT Planck’s constant can be expressed using two different units: h 6.63 1034 Js or h 4.14 1015 eVs Recall that 1 eV 1.60 1019 J or 1 J 6.25 1018 eV. photon: a quantum of light Concept Check Show that Planck’s formula for one photon can be written as E .hc Einstein, Quanta, and the Photon In 1905, a young and not-yet-famous Albert Einstein made a very bold suggestion. Planck had already introduced the idea of quantization of energy and the equation E hf. He thought that quantization applied only to matter and how matter could absorb or emit energy. Einstein suggested that this equation implied that light itself was quantized. In other words, Einstein reintroduced the idea that light could be considered a particle or a quantum of energy! This idea was troubling because, as you saw in Chapter 13, experiments clearly showed that light is a wave. In 1926, the chemist Gilbert Lewis introduced the term photon to describe a quantum of light. Planck’s formula, E nhf, can therefore be used to calculate the energy of one or more photons. Examples 14.1 and 14.2 allow you to practise using the idea of the photon and Planck’s formula. Example 14.1 How much energy is carried by a photon of red light of wavelength 600 nm? Practice Problems 1. What is the energy of a photon of light of frequency 4.00 1014 Hz? 2. What is the energy of a green photon of light of wavelength 555 nm? 3. What is 15.0 eV expressed in units of joules? Answers 1. 2.65 1019 J 2. 3.58 1019 J 3. 2.40 1018 J Given n 1 600 nm 1 9 m 6.00 × 107 m 1 0 m n 1 Required photon energy (E ) Analysis and Solution Since wavelength is given, first find the frequency using the equation c f , where c is the speed of light, f is frequency, and is the wavelength. c f m 3.00 108 s 6.00 107 m 5.00 1014 Hz Then substitute into Planck’s formula: E nhf (1)(6.63 1034 Js)(5.00 1014 s1) 3.32 1019 J 706 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 707 It is often more convenient to express the energies of photons in units of electron volts. Since 1 eV 1.60 1019 J, the energy of the red photon is e V 1 3.32 1019 J 2.07 eV 019 J 1 1.60 Paraphrase A red photon of light carries 3.32 1019 J of energy, or about 2.07 eV. Example 14.2 Your eye can detect as few as 500 photons of light. The eye is most sensitive to light having a wavelength of 510 nm. What is the minimum amount of light energy that your eye can detect? Given 510 nm 5.10 107 m n 500 photons Required minimum light energy (E) Analysis and Solution Since only wavelength is given, determine frequency using the equation c f: c f m 3.00 108 s 5.10 107 m 5.88 1014 Hz Then apply Planck’s formula: E nhf (500)(6.63 1034 Js)(5.88 1014 s1) 1.95 1016 J Practice Problems 1. What is the frequency of a 10-nm photon? 2. What is the energy of a 10-nm photon? 3. How many photons of green light ( 550 nm) are required to deliver 10 J of energy? Answers 1. 3.0 1016 Hz 2. 2.0 1017 J 3. 2.8 1019 photons Paraphrase Your eye is capable of responding to as little as 1.95 10–16 J of energy. M I N D S O N What’s Wrong with This Analogy? Sometimes the idea of the quantum is compared to the units we use for money. A dollar can be divided into smaller units, where the cent is the smallest possible unit. In what way is this analogy for the quantum accurate and in what way is it inaccurate? Look very carefully at Planck’s formula to find the error in the analogy. Try to come up with a better analogy for explaining quantization. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 707 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 708 The next example involves rearranging Planck’s formula and applying it to find the relationship between the power of a laser pointer and the number of photons it emits. Example 14.3 Practice Problems 1. How much energy is delivered by a beam of 1000 blue-light photons ( 400 nm)? 2. How many 400-nm blue-light photons per second are required to deliver 10 W of power? How many photons are emitted each second by a laser pointer that has a power output of 0.400 mW if the average wavelength produced by the pointer is 600 nm? Given 600 nm 6.00 10–7 m P 0.400 mW 4.00 10–4 W Required number of photons (n) Answers 1. 4.97 1016 J 2. 2.0 1019 photons/s Analysis and Solution Since 1 W 1 J/s, the laser pointer is emitting 4.00 104 J/s. Therefore, in 1 s the laser pointer emits 4.00 104 J of energy. By equating this amount of energy to the energy carried by the 600-nm photons, you can determine how many photons are emitted each second using the equation E nhf. First use the equation c f to determine the frequency of a 600-nm photon: c f Substitute this equation into Planck’s formula. E nhf c nh n E c h (4.00 104 J)(6.00 107 m) (6.63 1034 Js)3.00 108 m s 1.21 1015 Paraphrase A laser that emits 1.21 1015 photons each second has a power output of 0.400 mW. Photons and the Electromagnetic Spectrum Planck’s formula provides a very useful way of relating the energy of a photon to its wavelength or frequency. It shows that a photon’s energy depends on its frequency. An X-ray photon is more energetic than a microwave photon, just as X rays have higher frequencies than microwaves. Consequently, it takes a much more energetic process to create a gamma ray or X ray than it does to create a radio wave. Figure 14.6 gives the various photon energies along the electromagnetic spectrum. 708 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 709 X rays, for example, can only be emitted by a very hot gas or by a veryhigh-energy interaction between particles. Figure 14.7 shows images of the remnants of an exploded star, taken in different parts of the electromagnetic spectrum. Each image shows photons emitted by gases at different temperatures and locations in the remnant. The Electromagnetic Spectrum 103 102 101 1 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 1010 11 10 12 10 longer lower soccer field house baseball this period cell bacteria virus protein shorter water molecule radio waves infrared microwaves v i s i b l e ultraviolet “hard” X rays “soft” X rays gamma rays AM radio FM radio microwave oven radar people light bulb X-ray machines radioactive elements 10 6 107 108 109 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 higher 9 10 8 10 7 10 6 10 5 10 4 10 3 10 2 10 1 10 1 101 102 103 104 105 106 wavelength (in metres) size of a wavelength common name of wave sources frequency (waves per second) energy of one photon (electron volts) Figure 14.6 The energies of photons (in electron volts) along the electromagnetic spectrum Figure 14.7 These images of a supernova remnant were taken by the Chandra X-ray space telescope, the Hubble space telescope (visible part of the spectrum), and the Spitzer space telescop
e (infrared). Each image is produced by gases at different temperatures. X rays are produced by very-hightemperature gases (millions of degrees), whereas infrared light is usually emitted by low-temperature gases (hundreds of degrees). Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 709 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 710 14.1 Check and Reflect 14.1 Check and Reflect Knowledge Extensions 1. What is the energy of a photon with wavelength 450 nm? 2. What is the wavelength of a photon of energy 15.0 eV? 3. Compare the energy of a photon of wavelength 300 nm to the energy of a 600-nm photon. Which photon is more energetic, and by what factor? 4. (a) What is the frequency of a photon that has an energy of 100 keV? (b) From what part of the electromagnetic spectrum is this photon? Applications 5. How many photons of light are emitted by a 100-W light bulb in 10.0 s if the average wavelength emitted is 550 nm? Assume that 100% of the power is emitted as visible light. 6. The Sun provides approximately 1400 W of solar power per square metre. If the average wavelength (visible and infrared) is 700 nm, how many photons are received each second per square metre? 7. Suppose that your eye is receiving 10 000 photons per second from a distant star. If an identical star was 10 times farther away, how many photons per second would you receive from that star in one second? 8. Estimate the distance from which you could see a 100-W light bulb. In your estimate, consider each of the following: • Decide on a representative wavelength for light coming from the light bulb. • Estimate the surface area of a typical light bulb and use this figure to determine the number of photons per square metre being emitted at the surface of the light bulb. • Estimate the diameter of your pupil and hence the collecting area of your eye. • Use the information in Example 14.2 to set a minimum detection limit for light from the light bulb. Remember that your answer is an estimate. It will likely differ from other students’ estimates based on the assumptions you made. (Hint: The surface area of a sphere is 4r 2.) e TEST To check your understanding of Planck’s formula, follow the eTest links at www.pearsoned.ca/ school/physicssource. 710 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 711 14.2 The Photoelectric Effect The secret agent cautiously inches forward and carefully steps over and around the thin, spidery outlines of laser beams focussed on light sensors scattered around Dr. Evil’s secret lair. Is this scenario only the stuff of spy movies? Perhaps, but every time you walk into a shopping mall or have your groceries scanned at the supermarket, you, like the secret agent, are seeing an application of the way in which photons and metals interact. This interaction is called the photoelectric effect. 14-2 QuickLab 14-2 QuickLab Discharging a Zinc Plate Using UV Light Problem Does ultraviolet light cause the emission of electrons from a zinc metal plate? Materials electroscope UV light source zinc plate glass plate electroscope zinc plate UV source Figure 14.8 Procedure 1 Attach the zinc plate so that it is in contact with the electroscope. 2 Apply a negative charge to the zinc plate and electroscope. What happens to the vanes of the electroscope? (If you are uncertain how to apply a negative charge, consult your teacher for assistance.) 3 Turn on the UV light source and shine it directly on the zinc plate (see Figure 14.8). 4 Place the glass plate between the UV light source and the zinc plate. Note any change in the behaviour of the vanes of the electroscope. Remove the plate and once again note any change in the response of the vanes. Questions 1. Why did the vanes of the electroscope deflect when a negative charge was applied? 2. Explain what happened when UV light shone on the zinc plate. Why does this effect suggest that electrons are leaving the zinc plate? 3. Glass is a known absorber of UV light. What happened when the glass plate was placed between the UV source and the electroscope? 4. From your observations, what caused the emission of electrons from the zinc surface? Give reasons for your answer. CAUTION: UV light is harmful to your eyes. Do not look directly into the UV light source. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 711 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 712 In 1887, German physicist Heinrich Hertz conducted a series of experiments designed to test Maxwell’s theory of electromagnetic waves. In one of the experiments, a spark jumping between the two metal electrodes of a spark gap was used to create radio waves that could be detected in a similar spark-gap receiver located several metres away. Hertz noticed that his spark-gap receiver worked much better if the small metal electrodes were highly polished. Eventually, it was recognized that it was not the polishing but the ultraviolet light being produced by the main spark in his transmitter that greatly enhanced the ability of sparks to jump in his receiver’s spark-gap. Hertz had discovered that some metals emit electrons when illuminated by sufficiently short (high-energy) wavelengths of light. This process is called photoemission of electrons, or the photoelectric effect. Electrons emitted by this process are sometimes called photoelectrons. How could light waves cause a metal to emit electrons? Experiments showed that the electrons required energies of a few electron volts in order to be emitted by the metal. Perhaps the atoms on the surface of the metal absorbed the energy of the light waves. The atoms would begin to vibrate and eventually absorb enough energy to eject an electron. There is a problem with this theory. According to classical physics, it should take minutes to hours for a metal to emit electrons. Experiments showed, however, that electron emission was essentially instantaneous: There was no measurable delay between the arrival of light on the metal surface and the emission of electrons. To further add to the puzzle, there was a minimum or threshold frequency, f0, of incident light below which no photoemission would occur. If the light shining on the metal is of a frequency lower than this threshold frequency, no electrons are emitted, regardless of the brightness of the light shining on the metal (Figure 14.9). photon photon The incoming photon has enough energy to knock loose an electron. Nothing happens! The incoming photon lacks enough energy to cause photoemission. Figure 14.9 If an incident photon has a high enough frequency, an electron will be emitted by the metal surface. If the incoming photon frequency is not high enough, an electron will not be emitted. Another puzzle was the lack of clear connection between the energy of the electrons emitted and the brightness of the light shining on the metal surface. For a given frequency of light, provided it was greater than the threshold frequency, the emitted electrons could have a range of possible kinetic energies. Increasing the intensity of the light had no influence on the maximum kinetic energy of the electrons. photoelectric effect: the emission of electrons when a metal is illuminated by short wavelengths of light photoelectron: an electron emitted from a metal because of the photoelectric effect threshold frequency: the minimum frequency that a photon can have to cause photoemission from a metal Table 14.1 Work Functions of Some Common Metals Element Work Function (eV) Aluminium Beryllium Cadmium Calcium Carbon Cesium Copper Magnesium Mercury Potassium Selenium Sodium Zinc 4.08 5.00 4.07 2.90 4.81 2.10 4.70 3.68 4.50 2.30 5.11 2.28 4.33 712 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 713 Einstein’s Contribution The photoelectric effect remained an interesting but completely unexplained phenomenon until 1905. In 1905, Albert Einstein solved the riddle of the photoelectric effect by applying Planck’s quantum hypothesis: Light energy arrives on the metal surface in discrete bundles, which are absorbed by atoms of the metal. This process takes very little time and all the energy needed to expel an electron is provided at once. However, photoemission only occurs if the frequency of the incident photons is greater than or equal to the threshold frequency of the metal. Since the frequency of a photon is directly proportional to its energy, as given by Planck’s formula, E hf, the incident photons must have the minimum energy required to eject electrons. This minimum energy is known as the work function, W. The work function is specific for every metal. Table 14.1 lists the work functions of some common metals. The work function, W, is related to threshold frequency, f0, by the equation W hf0. Photons with a frequency greater than the threshold frequency have energy greater than the work function and electrons will be ejected. M I N D S O N Light a Particle? Heresy! Suggest reasons why a physicist might argue against Einstein’s idea that light is a particle. One such physicist was Robert A. Millikan, whose important experiments on the photoelectric effect were viewed, ironically, as a brilliant confirmation of Einstein’s “crazy” idea. How is skepticism both an advantage and a disadvantage to the progress of science? Millikan’s Measurement of Planck’s Constant When photons are absorbed by a metallic surface, either nothing will happen — the photons lack the minimum energy required to cause photoemission — or an electron will be emitted (Figure 14.10). photon E hf Kinetic energy of the electron equals the difference between the photon energy and the work function. Ek hf W The incident photon has energy E hf. The photon must be able to provide enough energy to equal or exceed the work function, W, in order to cause emission of an electron. work function: the minimum energy that a photon can have to cause photo
emission from a metal; specific for every metal Figure 14.10 The kinetic energy of an electron emitted during photoemission is equal to the difference between the incident photon’s energy and the work needed to overcome the work function for the surface. One of the most successful experiments to investigate the photoelectric effect was conducted by American physicist Robert Millikan (Figure 14.11) and published in 1916. The main result from Millikan’s work is given in Figure 14.12. The graph shows electron kinetic energy as a function of the frequency of the incident light. When the light frequency is Figure 14.11 Robert Andrews Millikan (1868–1953) was awarded the Nobel Prize in physics in 1923 for his work on determining the charge of an electron, and for his work on the photoelectric effect. Despite his important work on the photoelectric effect, Millikan remained deeply skeptical of Einstein’s particle view of light. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 713 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 714 e MATH Millikan graphed the kinetic energy of the photoelectons as a function of the incident frequency. To explore this relationship closer and to plot a graph like the one shown in Figure 14.12, visit www.pearsoned.com/school/ physicssource. PHYSICS INSIGHT Ek hf W and y mx b Therefore, f0 xint and W yint PHYSICS INSIGHT ) Maximum Ek vs. Frequency Cesium Potassium Calcium Magnesium Mercury 6 8 10 12 14 16 18 Frequency (Hz 1014) Different metals have different threshold frequencies, as shown in this graph. Which metal has the highest threshold frequency? below the threshold frequency, no electrons are ejected. When the light frequency equals the threshold frequency, electrons are ejected but with zero kinetic energy. The threshold frequency is therefore the x-intercept on the graph Light below a frequency of 4.39 1014 Hz or wavelength longer than 683 nm would not eject electrons. The fact that this plot was not dependent upon the intensity of the incident light implied that the interaction was like a particle that gave all its energy to the electron and ejected it with that energy minus the energy it took to escape the surface. 0 0 4 6 8 Frequency 1014 (Hz) 10 12 Figure 14.12 A graph based on the 1916 paper in which Millikan presented the data from his investigation of the photoelectric effect Once the frequency of the light exceeds the threshold frequency, photoemission begins. As the light frequency increases, the kinetic energy of the electrons increases proportionally. You can express this relationship in a formula by using the law of conservation of energy. The energy of the electron emitted by the surface is equal to the difference between the original energy of the photon, given by E hf, minus the work needed to free the electron from the surface. The equation that expresses this relationship is Ek hf W where Ek is the maximum kinetic energy of the electrons and W is the work function of the metal. You may recall that this equation is an example of the straight-line relationship y mx b, where m is the slope of the line and b is the y-intercept. The graph in Figure 14.12 shows the linear relationship between the frequency of the incident light falling on a sodium metal surface and the maximum kinetic energy of the electrons emitted by the metal. The slope of this line shows that the energy of the photons is directly proportional to their frequency, and the proportionality constant is none other than Planck’s constant. Millikan’s photoelectric experiment provides an experimental way to measure Planck’s constant. The y-intercept of this graph represents the negative of the work function of the photosensitive surface. The work function can also be determined by measuring the threshold frequency of photons required to produce photoemission of electrons from the metal. Even though classical physics could not explain the photoelectric effect, this phenomenon still obeys the fundamental principle of conservation of energy, where ETotalinitial . The energy of the photon is completely transferred to the electron and can be expressed by the following equation: ETotalfinal hf W Ek 714 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 715 Another way to interpret this equation is that the energy of the photon liberates the electron from the photosensitive surface, and any remaining energy appears as the electron’s kinetic energy. Concept Check Derive a relationship between energy of a photon (hf) and work function for a metal (W) to determine whether or not photoemission will occur. Stopping Potentials and Measuring the Kinetic Energy of Photoelectrons How did Millikan determine the maximum kinetic energy of electrons emitted by a metal surface? Figure 14.13 shows a highly simplified version of his experimental set-up. An evacuated tube contains a photoelectron-emitting metal surface and a metal plate, called the collector. A power supply is connected to the collector and the electron-emitting metal surface. When the power supply gives the collector plate a positive charge, the ammeter registers an electric current as soon as the incoming photons reach the threshold frequency. Any electrons emitted by the metal surface are attracted to the collector and charge begins to move in the apparatus, creating a current. incoming photons collector electron-emitting metal surface evacuated tube Figure 14.13 A simplified diagram depicting an experimental set-up used to investigate the photoelectric effect. When the power supply is connected as shown, the ammeter measures a current whenever the frequency of the incoming light exceeds the threshold frequency for the metal surface. ammeter ON OFF power supply Concept Check Explain the role of the collector plate in the photoelectric experiment apparatus in Figure 14.13. If the collector plate is not given a charge, would an electric current still be measured if the incoming photons exceed the threshold frequency? e SIM Find out more about the photoelectric effect by doing this simulation. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 715 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 716 Now consider what happens if the collector plate is given a negative charge. Instead of being attracted toward the collector, electrons now experience an electric force directed away from the collector. This electric force does work on the photoelectron (Figure 14.14). Photoelectrons will arrive at the collector only if they leave the metal surface with enough kinetic energy to reach the collector. You can express the final kinetic energy of the electrons in the following way: Ekfinal Ekinitial E where Ekfinal is the final kinetic energy of the electron, Ekinitial is its initial kinetic energy, and E is the work done by the electric force. The electric force opposes the motion of the electron. incoming photons Fe collector The electron is repelled by the collector and attracted to the positively charged plate. Figure 14.14 When the charges on the plates are reversed, the photoelectrons are repelled by the negatively charged collector and pulled back toward the positively charged plate. Only the most energetic electrons will reach the negative plate. Ekinitial In Chapter 11, you saw that the work done in an electric field of potential V on a charge q is expressed by the equation E qV. The final kinetic energy of an electron arriving at the collector can now be written as qV, where q represents the charge of an electron. If the Ekfinal negative potential on the collector plate is increased, then eventually a point will be reached at which no electrons will be able to reach the collector. At this point, the current in the ammeter drops to zero and the potential difference is now equal to the stopping potential. In summary, qVstopping. The maximum the current drops to zero when 0 Ekmax kinetic energy of electrons may now be expressed as Ekmax qVstopping where Vstopping is the stopping potential and q is the charge of the electron. Example 14.4 stopping potential: the potential difference for which the kinetic energy of a photoelectron equals the work needed to move through a potential difference, V Practice Problems 1. What stopping potential will stop electrons of energy 5.3 10–19 J? 2. Convert 5.3 10–19 J to electron volts. 3. What is the maximum kinetic energy of electrons stopped by a potential of 3.1 V? 716 Unit VII Electromagnetic Radiation Blue light shines on the metal surface shown in Figure 14.13 and causes photoemission of electrons. If a stopping potential of 2.6 V is required to completely prevent electrons from reaching the collector, determine the maximum kinetic energy of the electrons. Express your answer in units of joules and electron volts. Given Vstopping 2.6 V q 1.60 1019 C Required maximum kinetic energy of electrons (Ekmax) 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 717 Analysis and Solution Use the equation Ekmax Ekmax (1.60 1019 C)(2.6 V) 4.2 1019 J qVstopping. 1 eV 1.60 1019 J Answers 1. 3.3 V 2. 3.3 eV 3. 3.1 eV or 5.0 1019 J 2.6 eV Paraphrase A stopping potential of 2.6 V will stop electrons of kinetic energy 4.2 1019 J or 2.6 eV. Concept Check Show that the idea of stopping potential can lead directly to the qVstopp i expression h f 0 ng W , where h is Planck’s constant,Vstopping is the stopping potential, W is the work function, and f0 is the threshold frequency for emission of electrons from a metal surface. 14-3 Design a Lab 14-3 Design a Lab Using the Photoelectric Effect to Measure Planck’s Constant The Question How can you use the photoelectric effect and the concept of stopping potential to determine Planck’s constant? Design and Conduct Your Investigation You will need to decide on what equipment to assemble to enable you to relate frequency of
incident light to kinetic energy of electrons and stopping potentials. In your design, be sure to address what you will need to measure and what variables will be involved, how to record and analyze your data, and how to use the data collected to answer the question. Prepare a research proposal for your teacher to determine whether your school laboratory has the necessary equipment for this lab, or if alternative approaches may work. Your proposal should include a worked-out sample of how the data you hope to collect will answer the question. Remember to work safely, to clearly identify tasks, and to designate which group members are responsible for each task. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 717 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 718 The following example shows how to relate the concepts of threshold frequency and work function. Example 14.5 Experiments show that the work function for cesium metal is 2.10 eV. Determine the threshold frequency and wavelength for photons capable of producing photoemission from cesium. Practice Problems 1. Light of wavelength 480 nm is just able to produce photoelectrons when striking a metal surface. What is the work function of the metal? 2. Blue light of wavelength 410 nm strikes a metal surface for which the work function is 2.10 eV. What is the energy of the emitted photoelectron? Answers 1. 2.59 eV 2. 0.932 eV 0 hf0 W h f0 Given W 2.10 eV Required threshold frequency (f0) wavelength () Analysis and Solution The work function is the amount of energy needed to just break the photoelectron free from the metal surface, but not give it any additional kinetic energy. Therefore, from 0 J. Ek hf W, for threshold frequency, f0, set Ek First convert the work function to units of joules. W (2.10 eV)1.60 1019 J V e 3.36 1019 J Now solve for the threshold frequency. W 3.36 1019 J 6.63 1034 Js 5.07 1014 Hz From c f, the wavelength of this photon is c f m 3.00 108 s 5.07 1014 s1 5.91 107 m 591 nm Paraphrase The threshold frequency for photons able to cause photoemission from cesium metal is 5.07 1014 Hz. This frequency corresponds to photons of wavelength 591 nm, which is in the yellow-orange part of the visible spectrum. You can also use the law of conservation of energy equation for the photoelectric effect to predict the energy and velocity of the electrons released during photoemission. 718 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 719 Example 14.6 Using Table 14.1, determine the maximum speed of electrons emitted from an aluminium surface if the surface is illuminated with 125-nm ultraviolet (UV) light. Given 125 nm metal aluminium Required maximum speed of electrons (v) Analysis and Solution From Table 14.1, the work function for aluminium is 4.08 eV. Convert this value to joules. W 4.08 eV (4.08 eV )1.60 1019 J eV 6.528 1019 J To determine the energy of the incident photon, use the equation E hf h c. Practice Problems 1. A photoelectron is emitted with a kinetic energy of 2.1 eV. How fast is the electron moving? 2. What is the kinetic energy of a photoelectron emitted from a cesium surface when the surface is illuminated with 400-nm light? 3. What is the maximum speed of the electron described in question 2? Answers 1. 8.6 105 m/s 2. 1.01 eV 3. 5.95 105 m/s Incident photon energy is E h c (6.63 1034 Js) 1.591 1018 J 3.00 108 m s 1.25 107 m To find the kinetic energy of the electrons, use the law of conservation of energy equation for the photoelectric effect, Ek energy of the electron is hf W. Kinetic Ek hf W 15.91 1019 J 6.528 1019 J 9.384 1019 J Finally, use Ek mass of 9.11 1031 kg. 1 mv2 to solve for speed. Recall that an electron has a 2 The electron’s speed is v 2Ek m 2(9.384 1019 J) 9.11 1031 kg 1.44 106 m/s Paraphrase The electrons emitted from the aluminium surface will have a maximum speed of 1.44 106 m/s. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 719 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 720 Millikan’s work on the photoelectric effect provided critical evidence in eventually demonstrating the particle or quantized nature of light. As you will see in the next chapter, Millikan also performed a key experiment that demonstrated the discrete or “quantized” nature of electrical charge: He showed that the electron is the smallest unit of electrical charge. 14.2 Check and Reflect 14.2 Check and Reflect Knowledge 1. What is the energy, in eV, of a 400-nm photon? 2. Explain how the concepts of work function and threshold frequency are related. 3. What is the threshold frequency for cadmium? (Consult Table 14.1 on page 712.) 4. Will a 500-nm photon cause the emission of an electron from a cesium metal surface? Explain why or why not. 5. What stopping voltage is needed to stop an electron of kinetic energy 1.25 eV? 6. Explain how stopping potential is related to the maximum kinetic energy of an electron. 7. True or false? The greater the intensity of the light hitting a metal surface, the greater the stopping potential required to stop photoelectrons. Explain your answer. Applications The following data are taken from an experiment in which the maximum kinetic energy of photoelectrons is related to the wavelength of the photons hitting a metal surface. Use these data to answer the following questions. Wavelength (nm) Kinetic Energy (eV) 500 490 440 390 340 290 240 0.36 0.41 0.70 1.05 1.52 2.14 3.025 720 Unit VII Electromagnetic Radiation 8. Convert the wavelengths given in the data table to frequency units and graph them along with the kinetic energy of the photoelectrons. Be sure to plot frequency on the horizontal axis. 9. Give the value of the slope of the graph that you just drew. What is the significance of this value? 10. What metal do you think was used in the previous example? Justify your answer. Extensions 11. Explain how the photon model of light correctly predicts that the maximum kinetic energy of electrons emitted from a metal surface does not depend on the intensity of light hitting the metal surface. 12. In several paragraphs, identify three common devices that use the photoelectric effect. Be sure to explain in what way these devices use the photoelectric effect. 13. How long would photoemission take from a classical physics point of view? Consider a beam of ultraviolet light with a brightness of 2.0 10–6 W and an area of 1.0 10–4 m2 (about the area of your little fingernail) falling on a zinc metal plate. Use 3.5 eV as the energy that must be absorbed before photoemission can occur. (Hint: Estimate the area of an atom and determine how much of the beam of UV light is being absorbed each second, on average.) e TEST To check your understanding of the photoelectric effect, follow the eTest links at www.pearsoned.ca/ school/physicssource. 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 721 14.3 The Compton Effect Although Einstein’s photon model provided an explanation of the photoelectric effect, many physicists remained skeptical. The wave model of light was so successful at explaining most of the known properties of light, it seemed reasonable to expect that a purely classical explanation of the photoelectric effect would eventually be found. In 1923, however, an experiment by American physicist Arthur Compton (Figure 14.15) provided an even clearer example of the particle nature of light, and finally convinced most physicists that the photon model of light had validity. Compton studied the way in which electrons scattered X rays in a block of graphite. The X rays were observed to scatter in all directions. This effect was not surprising: Both the wave and particle models of light predicted this outcome. What the wave model could neither predict nor explain, however, was the small change in wavelength that Compton observed in the scattered X ray, and the relationship between the change in wavelength and the angle through which the X ray was scattered. The scattering of an X ray by an electron is now referred to as Compton scattering, and the change in wavelength of the scattered X-ray photon is called the Compton effect (Figure 14.16). To understand the Compton effect, you will need to use two of the most central ideas of physics: the law of conservation of momentum and the law of conservation of energy. The interaction between an X-ray photon and an electron must still obey these laws. By using the particle model of light and Einstein’s mass-equivalence equation E mc2, Compton showed that the momentum of the X ray could be expressed as p h Figure 14.15 Arthur Holly Compton (1892–1962) was a pioneer in high-energy physics. He was awarded the Nobel Prize in 1927 for his discovery of the Compton effect, which provided convincing evidence for the photon model of light. Compton scattering: the scattering of an X ray by an electron Compton effect: the change in wavelength of the scattered X-ray photon where p is momentum, h is Planck’s constant, and is the wavelength of the X ray. Compton was also able to show exactly how the change in wavelength of the scattered X ray is related to the angle through which the X-ray photon is scattered. λ recoil electron electron at rest θ scattered X-ray photon λ f i incident X-ray photon Concept Check Which of the following photons has the greater momentum: A 2 nm? Explain your reasoning. 500 nm or B Compton found that the scattered X ray changed its momentum and energy in a way that was exactly what you would expect if it was a small particle undergoing an elastic collision with an electron. Recall from Chapter 9 that energy and momentum are conserved during an elastic collision. Figure 14.16 When an electron scatters an X ray, both momentum and energy are conserved. Compton scattering behaves like an elastic collision between a photon and an electron. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 7
21 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 722 The laws of conservation of energy and of momentum can be applied to the X ray and the electron in the following way: • The total momentum of the incident X-ray photon must equal the total momentum of the scattered X ray and the scattered electron. • The total energy of the incident X-ray photon and the electron must equal the total energy of the scattered X ray and the scattered electron. Concept Check Study Figure 14.16. Define the direction of the incident X-ray photon as the positive x-direction and the upward direction as the positive y-direction. Suppose the incident X-ray photon has a wavelength of 1. Derive an expression for the x and y components of the i and the scattered X-ray photon has a wavelength f. momentum of the scattered photon. 2. Explain how your answer to question 1 gives you the x and y components of the electron’s momentum. 3. How much energy was transferred to the electron in this interaction? Derive a simple expression for the electron’s final energy. CCompton derived the following relationship between the change in the wavelength of the scattered photon and the direction in which the scattered photon travels: f i h (1 cos ) m c where m is the mass of the scattering electron and is the angle through which the X ray scatters. The full derivation of this equation requires applying Einstein’s theory of relativity and a lot of algebra! The central concepts behind this equation, however, are simply the laws of conservation of energy and of momentum. As well, this equation is exactly consistent with Einstein’s idea that the X-ray photon collides with the electron as if it were a particle. M I N D S O N Heisenberg’s Microscope Problem Suggest how Compton scattering shows that it is impossible to “see” an electron. In particular, why is it that we can only see where an electron was and not where it is? (Hint: Think about what photons are doing when you look at something.) This question is sometimes referred to as Heisenberg’s microscope problem. 722 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 723 Example 14.7 What is the maximum change in wavelength that a 0.010-nm X-ray photon can undergo by Compton scattering with an electron? Does initial wavelength (0.010 nm) matter in this example? Given i 0.010 nm Required change in wavelength () Analysis and Solution Maximum change will occur when the X ray is scattered by the greatest possible amount, that is, when the X ray is back-scattered. From the Compton effect equation, i f h (1 cos ), the maximum value for c m occurs when the term (1 cos ) is a maximum. This occurs when 180° and cos 1, so (1 cos ) becomes (1 (1)) 2. Use this relation to determine the largest possible change in wavelength of the scattered X-ray photon. h (1 cos ) c m 2h c m 2(6.63 1034 Js) (9.11 1031 kg)(3.00 108 m/s) 4.85 1012 m Paraphrase The maximum change in wavelength of a photon during Compton scattering is only 4.85 1012 m. This change is independent of the initial wavelength of the photon. Practice Problems 1. What is the energy of an X ray of wavelength 10 nm? 2. What is the momentum of an X ray of wavelength 10 nm? 3. If a 10-nm X ray scattered by an electron becomes an 11-nm X ray, how much energy does the electron gain? Answers 1. 2.0 1017 J 2. 6.6 1026 Ns 3. 1.8 1018 J Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 723 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 724 You can also use the Compton equation to determine the final wavelength of a photon after scattering, as you will see in the next example. Example 14.8 An X-ray photon of wavelength 0.0500 nm scatters at an angle of 30°. Calculate the wavelength of the scattered photon. Given i 30° 0.0500 nm Required final wavelength ( f) Analysis and Solution Rearrange the Compton equation to solve for final wavelength. Recall that the mass of an electron is 9.11 1031 kg. i f h (1 cos ) c m i f i h (1 cos ) c m 0.0500 nm 6.63 1034 Js (9.11 1031 kg)(3.00 108 m/s) (1 cos 30°) 0.0500 nm 0.000 325 nm 0.0503 nm Paraphrase The X-ray photon changes wavelength by 0.0003 nm to become a photon of wavelength 0.0503 nm. Practice Problem 1. An X ray of wavelength 0.010 nm scatters at 90° from an electron. What is the wavelength of the scattered photon? Answer 1. 0.012 nm 724 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 725 For many physicists, the Compton effect provided the final piece of evidence they needed to finally accept Einstein’s idea of the particle nature of light. The Compton effect also describes one of the most fundamental phenomena — the interaction of light with matter. 14.3 Check and Reflect 14.3 Check and Reflect Knowledge Applications 1. What is the momentum of a 500-nm photon? 6. What is the wavelength of a 100-keV X-ray 2. Photon A has a wavelength three times longer than photon B. Which photon has the greatest momentum and by what factor? 3. A photon has a momentum of 6.00 10–21 kgm/s. What is the wavelength and energy of this photon? 4. Identify the part of the electromagnetic spectrum of the photon in question 3. 5. True or false? One of the major differences between classical physics and quantum physics is that the laws of conservation of energy and momentum do not always work for quantum physics. Explain your answer. photon? 7. An X-ray photon of wavelength 0.010 nm strikes a helium nucleus and bounces straight back. If the helium nucleus was originally at rest, calculate its velocity after interacting with the X ray. Extension 8. In order to see an object, it is necessary to illuminate it with light whose wavelength is smaller than the object itself. According to the Compton effect, why is illumination a problem if you wish to see a small particle, such as a proton or an electron? e TEST To check your understanding of the Compton effect, follow the eTest links at www.pearsoned.ca/ school/physicssource. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 725 14-PearsonPhys30-Chap14 7/24/08 3:57 PM Page 726 14.4 Matter Waves and the Power of Symmetric Thinking If waves (light) can sometimes act like particles (photons), then why couldn’t particles, such as electrons, sometimes act like waves? Louis de Broglie (pronounced “de Broy”) (Figure 14.17), a young French Ph.D. student, explored this question in 1924 in a highly imaginative and perplexing thesis. The idea seemed so strange that despite no obvious errors in his argument, the examining committee was reluctant to pass de Broglie. Fortunately, a copy of his thesis was sent to Albert Einstein, who recognized at once the merit in de Broglie’s hypothesis. Not only was de Broglie awarded his Ph.D., but his hypothesis turned out to be correct! De Broglie’s argument is essentially one of symmetry. As both the photoelectric effect and the Compton effect show, light has undeniable particle-like, as well as wave-like, properties. This dichotomy is called the wave-particle duality. In reality, light is neither a wave nor a particle. These ideas are classical physics ideas, but experiments were revealing subtle and strange results. What light is depends on how we interact with it. De Broglie’s hypothesis completes the symmetry by stating that what we naturally assume to be particles (electrons, for example) can have wave-like properties as well. At the atomic level, an electron is neither a wave nor a particle. What an electron is depends on how we interact with it. De Broglie arrived at his idea by tying together the concepts of momentum and wavelength. Using Compton’s discovery relating momentum and wavelength for X-ray photons, de Broglie argued that anything that possessed momentum also had a wavelength. His idea can be expressed in a very simple form: h p where h is Planck’s constant, p is momentum, and is de Broglie’s wavelength. De Broglie’s Wave Equation Works for Both Light and Electrons De Broglie’s hypothesis states that anything that has momentum must obey the following wavelength-momentum equations: For light: Maxwell’s law of electromagnetism shows that the momentum of a light wave can be written as p E , where E is the energy of the c light and c is the speed of light. But Planck’s formula states that E hf. f h . Substituting this equation into de Broglie’s wave Therefore, p c Louis de Figure 14.17 Broglie (1892–1987) was the first physicist to predict the existence of matter waves. wave-particle duality: light has both wave-like and particle-like properties Project LINK How important is de Broglie’s hypothesis to our current understanding of the nature of light and matter? 726 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 727 equation, you obtain h h f c c , which is the wavelength-frequency f relation. It tells you that a photon of light has a wavelength! For electrons: If an electron is moving with a velocity, v, that is much less than the speed of light, then its momentum is p mv and de Broglie’s relationship is h p h mv . For electrons (or any other particles) moving at velocities approaching the speed of light, the expression h p is still applicable. PHYSICS INSIGHT Einstein showed that, as objects’ speeds approach the speed of light, the familiar expression for momentum, p mv, must be replaced by the more complicated equation p mv –––––––– v2 c2 1 , where c is the speed of light. THEN, NOW, AND FUTURE The Electron Microscope Modern TEMs are capable of reaching very high magnification and imaging at the atomic level. The scanning electron microscope (SEM) is similar to the TEM but differs in one important way: Electrons are reflected off the sample being imaged. SEM images have a remarkable three-dimensional appearance (Figure 14.19). The Electron Microscope The idea of matter waves is not simply abstract physics that has no practical application. The wave nature of electrons has been
used to build microscopes capable of amazing magnification. The reason for their amazing magnification lies in the extremely small wavelengths associated with electrons. The usable magnification of a microscope depends inversely on the wavelength used to form the image. In a transmission electron microscope (TEM, Figure 14.18), a series of magnets (magnetic lenses) focusses a beam of electrons and passes the beam through a thin slice of the specimen being imaged. Figure 14.19 An SEM view of an ant’s head Questions 1. Find out more about the varieties of electron microscopes in use. Search the Internet, using key words such as electron microscope, TEM, or SEM, to learn about at least three different kinds of electron microscopes. Summarize your findings in the following way: Figure 14.18 A modern transmission electron microscope • name (type) of microscope • how it differs from other electron microscopes in use and operation • typical applications and magnifications 2. The magnification of a microscope depends inversely on the wavelength used to image a specimen. The very best quality light microscopes typically have maximum magnifications of 1000 to 4000 times. Modern TEMs use electrons accelerated to energies of over 100 keV to observe specimens. Estimate the possible range of magnifications that can be achieved using a TEM by considering the following: • What is a reasonable choice for the wavelength used in a light microscope? • What is the wavelength of a 100-keV electron? • How do the wavelengths of the light and of the electrons compare? (Note: Your answer will likely be an overestimate. The actual magnification of electron microscopes is limited by the ability of the magnetic lenses to focus the electron beam. TEMs are capable of achieving magnifications as high as 500 000 times!) Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 727 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 728 The next two examples apply the de Broglie relationship between momentum and wavelength. Example 14.9 What is the momentum of a 500-nm photon of green light? Practice Problem 1. What is the momentum of a 0.010-nm X ray? Answer 1. 6.6 × 10–23 kgm/s e MATH De Broglie showed how electrons can be thought of as waves and related the speed of an electron to its wavelength. Einstein’s work showed that as the speed of the electron became greater than 10% of the speed of light, relativistic effects has to be taken into account (see Physics Insight p. 727) To explore how the wavelength of an electron is a function of its speed, including relativistic effects, visit www.pearsoned.ca/school/ physicssource. Given 500 nm Required momentum (p) Analysis and Solution To find the photon’s momentum, apply de Broglie’s equation: p h 6.63 1034 Js 500 109 m 1.33 1027 Ns Paraphrase The photon has a momentum of 1.33 1027 Ns. The next example shows how to calculate the wavelength of an electron, thus illustrating that particles have a wave nature. Example 14.10 What is the wavelength of an electron moving at 1.00 104 m/s? Practice Problems 1. What is the wavelength of a proton moving at 1.0 105 m/s? 2. What is the speed of an electron that has a wavelength of 420 nm? Given v 1.00 104 m/s 9.11 1031 kg me Required wavelength () Answers 1. 4.0 1012 m 2. 1.73 103 m/s Analysis and Solution To find the electron’s wavelength, first find its momentum and then rewrite de Broglie’s equation: mv p h h mv 6.63 1034 Js (9.11 1031 kg)(1.00 104 m/s) 7.28 108 m 72.8 nm Paraphrase The electron has a de Broglie wavelength of 7.28 108 m or 72.8 nm. 728 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 729 De Broglie’s idea completes the concept of the wave-particle duality of light. Wave-particle duality combines two opposing ideas and teaches us that, at the atomic level, it is essential to use both ideas to accurately model the world. De Broglie’s Wave Hypothesis: Strange but True! Experimental proof of de Broglie’s hypothesis came very quickly and by accident. Between 1925 and 1927, American physicists C. J. Davisson and L. H. Germer, and British physicist G. P. Thomson (Figure 14.20, son of J. J. Thomson, discoverer of the electron) independently provided evidence that electrons can act like waves. The original Davisson and Germer experiment was an investigation of how electrons scattered after hitting different kinds of metallic surfaces. To prevent an oxide layer from contaminating the surfaces, the scattering was done inside a vacuum tube. In one test on a nickel surface, the vacuum tube cracked and the vacuum was lost, unbeknownst to Davisson and Germer. The nickel surface oxidized into a crystalline pattern. What Davisson and Germer observed was a very puzzling pattern: Scattering occurred in some directions and not in others. It was reminiscent of a pattern of nodes and antinodes (Figure 14.21). A simplified version of a typical Davisson–Germer experiment is shown in Figure 14.22(a). The graph in Figure 14.22(b) shows the kind of data that Davisson and Germer found. M I N D S O N Interpret the Graph (a) Look at the graph in Figure 14.22(b). Explain why it makes sense to interpret the pattern as one of nodes and antinodes. What is happening at each of the nodes? (b) What would you have to do to change the wavelength of the electrons used in an electron diffraction experiment? Figure 14.20 George Paget Thomson (1892–1975) was codiscoverer of matter waves with Davisson and Germer. One of the great ironies of physics is that Thomson played an instrumental role in showing that electrons can act like waves. Thirty years earlier, his father had shown that the electron was a particle! When Davisson and Germer began their experiments, they were unaware of de Broglie’s work. As soon as they learned of de Broglie’s hypothesis, however, they realized that they had observed electron-wave interference! Over the next two years, they and G. P. Thomson in Scotland refined the study of electron-wave interference and provided beautiful experimental confirmation of de Broglie’s hypothesis. In 1937, Davisson and Thomson received a Nobel Prize for the discovery of “matter waves.” incoming electron beam detector w 1 cm 54 V I 10 cm y t i s n e t n I scattered electrons reflected beam crystal surface (not to scale) 0 5 10 15 20 25 Accelerating voltage (a) d 3 Angstroms (b) Figure 14.21 This image was produced by electrons scattered by gold atoms on the surface of a thin gold film. The bright concentric rings are antinodes produced by the constructive interference of electron waves. Figure 14.22 (a) A schematic of a typical Davisson–Germer experiment in which atoms on the surface of a metal scatter a beam of electrons. For specific angles, the electrons scatter constructively and the detector records a large number of electrons, shown in the graph in (b). (b) In this graph, a high intensity means that more electrons are scattered in that direction, creating an antinode, or constructive interference. Similarly, a low intensity can be interpreted as a node, or destructive interference. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 729 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 730 PHYSICS INSIGHT Manipulating the accelerating voltage in a Davisson-Germer experiment changes the speed of the incident electrons and, therefore, their wavelength. Example 14.11 Explain conceptually how the wave properties of electrons could produce the interference pattern shown in Figure 14.21. Given You know that electrons have wavelike properties and that electrons are being scattered from atoms that are separated by distances comparable to the size of the electron wavelength. Analysis Figure 14.23 shows electron waves leaving from two different atomic scatterers. You can see that path 1 is a little longer than path 2, as denoted by the symbol . This difference means that a different number of electron wavelengths can fit along path 1 than along path 2. For example, if the path difference is 1 , 3 , or any odd half-multiple 2 2 of , then, when the electron waves combine at the detector, a complete cancellation of the electron wave occurs, forming a node. On the other hand, if the path difference is a whole-number multiple of , then constructive interference occurs, forming an antinode. The Davisson– Germer experiment provided graphic evidence of the correctness of de Broglie’s hypothesis. incoming electrons δ path 1 path 2 electron detector scattered electron paths δ length path 1 – length path 2 atomic scatterers Figure 14.23 De Broglie’s Hypothesis — A Key Concept of Quantum Physics Despite its simplicity, de Broglie’s wave hypothesis heralded the true beginning of quantum physics. You will now explore two of the consequences that follow from de Broglie’s equation. De Broglie’s Equation “Explains” Quantization of Energy Imagine that you drop a small bead into a matchbox, close the matchbox, and then gently place the matchbox on a level tabletop. You then ask, “What is the kinetic energy of the bead?” The answer may seem obvious and not very interesting: The energy is 0 J because the bead is not moving. If, however, you could shrink the box down to the size of a molecule and replace the bead with a single electron, the situation becomes very different. You can sometimes model molecules as simple boxes. The particle-in-a-box model shows how the wave nature of electrons (and all other particles) predicts the idea of quantization of energy. 730 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 731 M I N D S O N What’s Different Now? From a quantum point of view, explain why it becomes problematic to put a particle in a box. In Chapter 8, section 8.3, you learned about standing waves and resonance. These concepts apply to all waves. Because an electron behaves like a wave as well as like a particle, it has a wavelength, so the ideas of reso
nance and standing waves also apply to the electron. In order to fit a wave into a box, or finite space, the wave must have a node at each end of the box, and its wavelength must be related to the length of the box in the following way: n l 2 n where n is a whole number (n 1, 2, 3, …). Since there is a node at each end of the box, you can think of n as equivalent to the number of half-wavelengths that can fit in the space l, or length of the box (Figure 14.24). The longest possible standing wave that can fit into the box has a wavelength of 2l, where n 1. l n 1, λ 2l n 2, λ l n 4, λ l 2 n 3, λ 2l 3 Figure 14.24 Standing wave patterns for waves trapped inside a box of length l Because the electron is a standing wave, it cannot be at rest. Consequently, it must have a minimum amount of kinetic energy: mv2 m 1 2 m 2 v2 m 2 m Ek 2 p m 2 since p mv where p is the momentum and m is the mass of the electron. Recall that de Broglie’s equation shows that the momentum of an electron is inversely related to its wavelength: h . p Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 731 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 732 PHYSICS INSIGHT The particle-in-a-box model is very useful. It can be used to describe such diverse phenomena as small-chain molecules, tiny nano-scale electronics, and the nucleus of an atom. Depending on the situation, the “box” can have one dimension (for a long-chain molecule), two, or three dimensions. Models and modelling form an essential part of the physicist’s imaginative “toolbox.” Using de Broglie’s equation to relate the momentum of the electron to the length of the box, you can then write: Ek 2 p m 2 2 h 2m h2 2 2m From n l, when n = 1, 2l. Therefore, 2 n Ek 2 h l)2 2m (2 h2 l2 8m This equation represents the minimum kinetic energy of an electron. What if you wanted to give the electron more energy? To have more energy, the electron must have the right momentum-wavelength relation to fit the next standing wave pattern (Figure 14.24). The electron’s wavelength is, therefore, l l 2 2 Substituting into the equation for the kinetic energy of the electron, 2 p h2 2 h 4En1s l2 l)2 m 2 2m ( 2m En2 n2 The energy of a particle in a box is given by the general formula En n2h2 8ml2 , n 1, 2, 3, . . . These equations demonstrate that energy is quantized for the particlein-a-box model. As with photons, quantization means that the electron can have only specific amounts or quanta of energy. (Refer to section 14.1.) Example 14.12 Nanotechnology is one of the hottest areas in physics today. It is now possible to create tiny electric circuits in which electrons behave like particles in a box. Imagine an electron confined to a tiny strip 5.0 nm Practice Problems 1. What is the maximum wavelength for an electron confined to a box of length l 1.0 nm? 2. How much momentum does the electron in question 1 have? long. What are three possible energies that the electron could have? Given l 5.0 nm n 1, 2, 3 Required electron energies (E1, E2, E3) 732 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 733 3. What is the minimum energy that an electron can have when confined to a box of length l 1.0 nm? Answers 1. 2.0 nm 2. 3.3 1025 kgm/s 3. 6.0 1020 J Analysis and Solution n2h2 8ml2 , n 1, 2, 3, ... En Substitute n 1 into the expression for energy: (12)h2 8ml2 E1 (1)(6.63 1034 Js)2 8(9.11 1031 kg)(5.0 109 m)2 2.4 1021 J Calculate any other energy by noting that h2 n2E1 n2 l2 8m 4(2.4 1021 J) 9.7 1021 J (2)2 E1 (3)2 E1 9(2.4 1021 J) 2.2 1020 J n2(2.4 1021 J) En E2 E3 En Paraphrase An electron confined to a space 5.0 nm long can only have energies that are whole-square multiples of 2.4 1021 J. Three possible energies of the electron are, therefore, 2.4 1021 J, 9.7 1021 J, and 2.2 1020 J. Concept Check Refer to Figure 14.24. What happens to the minimum possible energy of a particle in a box when you shrink the box? How would the minimum energy of particles in the nucleus of an atom (about 10–15 m across) compare to the minimum energy of an electron in the atom itself (about 10–10 m across)? M I N D S O N Planck in a Box Argue that the particle-in-a-box model illustrates Planck’s discovery of quantization, and also demonstrates Planck’s radiation law. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 733 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 734 Heisenberg’s Uncertainty Principle Consider what you have just learned about the minimum energy of a particle in a box. The smaller you make the box, the shorter is the wavelength of the particle. Furthermore, because wavelength and momentum are inversely related, the shorter the wavelength, the greater is the momentum of the particle. The greater the momentum, the faster, on average, the particle is moving at any instant. In 1927, the young German physicist Werner Heisenberg (Figure 14.25) realized that the particle-in-a-box model of quantum mechanics has a troubling limitation built into it. Think of the size of the box as indicating the possible uncertainty in the location of the particle. The smaller the box is, the more precisely you know the location of the particle. At the same time, however, the smaller the box is, the greater the momentum and the greater the range of possible momentum values that the particle could have at any instant. Figure 14.26 illustrates this idea by plotting the uncertainty in position of the particle, x (on the vertical axis), and the uncertainty in its momentum, p (on the horizontal axis), as strips that intersect. The shaded areas in Figure 14.26 represent the product of uncertainty in position (x) and uncertainty in momentum (p). x x or p p Figure 14.26 A graphical depiction of the uncertainty in both position and momentum for a particle in a box Heisenberg’s troubling finding was that, due to the wave nature of all particles, it is impossible to know both the position and momentum of a particle with unlimited precision at the same time. The more precisely you know one of these values, the less precisely you can know the other value. To derive the formula for uncertainty in position and momentum of a particle, note that the length of the box is related to the wavelength of the particle: x length of box l (From n l, 2l when n 1.) 2 n 2l x 2 Figure 14.25 Werner Heisenberg (1901–1976) was one of the most influential physicists of the 20th century and a key developer of modern quantum theory. PHYSICS INSIGHT According to the particlein-a-box model, the smaller the space in which a particle is confined, the greater the kinetic energy, and hence momentum, of that particle. If you think of the length of the box as setting the possible range in location for a particle, then this quantity also tells you how precisely you know the position of the particle. This range is x. In the same way, the possible range in momentum is p. 734 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 735 Similarly, from de Broglie’s equation h , the uncertainty in momentum is p p range in momentum h h The product of uncertainty in position and uncertainty in momentum can be expressed as xp 2 xp h 2 This formula represents Heisenberg’s uncertainty principle. Note that the value of the product, h (representing the shaded areas of the graphs in 2 Figure 14.26), is constant. The symbol means that xp is approximately h . A more sophisticated argument produces the following expression: 2 h xp 4 This version is a common form of Heisenberg’s uncertainty principle. It tells you that the uncertainty in your knowledge of both the position and momentum of a particle must always be greater than some small, but non-zero, value. You can never know both of these quantities with certainty at the same time! This result was very troubling to many physicists, including Albert Einstein, because it suggests that, at the level of atoms and particles, the universe is governed by chance and the laws of probability. Heisenberg’s uncertainty principle: It is impossible to know both the position and momentum of a particle with unlimited precision at the same time. De Broglie’s matter-wave hypothesis and its confirmation by Davisson and Germer had an unsettling effect on physicists. Heisenberg’s work represented a logical extension of these ideas and helped set the stage for the birth of modern quantum theory. M I N D S O N Physics and Certainty Two physicists who were deeply troubled by de Broglie’s, and especially Heisenberg’s, work were Max Planck and Albert Einstein. Suggest why their reaction is ironic and why these discoveries were difficult for physicists to accept. To help with your answer, consider the importance of precision in classical physics, and Einstein’s famous quote concerning the uncertainty principle: “God does not play dice with the universe!” Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 735 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 736 14.4 Check and Reflect 14.4 Check and Reflect Knowledge 1. What is the wavelength of an electron that is moving at 20 000 m/s? 2. Calculate the momentum of a 500-nm photon. 6. If an electron and a proton each have the same velocity, how do their wavelengths compare? Express your answer numerically as a ratio. Extensions 3. What is the uncertainty in momentum of a particle if you know its location to an uncertainty of 1.0 nm? 7. According to classical physics, all atomic motion should cease at absolute zero. Is this state possible, according to quantum physics? 4. An electron is trapped within a sphere of diameter 2.5 10–12 m. What is the minimum uncertainty in the electron’s momentum? Applications 5. In your television set, an electron is accelerated through a potential difference of 21 000 V. 8. Derive the expression En n2h2 , 8ml2 n 1, 2, 3, ... for the energy of a particle in a box, where m is the mass of the part
icle, l is the length of the box, and n is one of the possible quantum states. (Hint: Remember that the wavelength of the nth standing wave confined to a box l.) 2 n of length l is n (a) How much energy does the electron e TEST acquire? (b) What is the wavelength of an electron of this energy? Ignore relativistic effects. To check your understanding of matter waves and Heisenberg’s uncertainty principle, follow the eTest links at www.pearsoned.ca/school/physicssource. 736 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 737 14.5 Coming to Terms with Wave-particle Duality and the Birth of Quantum Mechanics These fifty years of conscious brooding have brought me no nearer to the question of “What are light quanta?” Nowadays every clod thinks he knows it, but he is mistaken. Albert Einstein The wave-particle duality represents a deep and troubling mystery. For some physicists, most notably Einstein, the duality was seen as a flaw in quantum theory itself. Others, including Bohr, learned to accept rather than understand the duality. In this section, we will opt to accept and work with the wave-particle duality. 14-4 QuickLab 14-4 QuickLab The Two-slit Interference Experiment with Particles Figure 14.27 for this experiment? Problem To investigate the pattern that a stream of particles produces when passing through a pair of thin slits Materials marble two-slit apparatus (see Figure 14.27) graph paper (or plot on spreadsheet) entrance slit barrier redirects particles slit 1 ? ? slit 2 particle detector Procedure 1 Place the two-slit apparatus on a level table surface and incline it by a small angle to allow the marble to roll down. Repeat this process 100 times. 2 Record your observations by noting how many times the marble lands in each bin. 3 Graph the results of your experiment by plotting the bin number along the horizontal axis and the number of times the marble landed in a given bin on the vertical axis. Questions 1. Why is it important that the table surface be level 2. For 100 trials, how many times would you expect the marble to pass through slit 1? Did you observe this result? Explain. 3. Where did the marble land most of the time? Did you expect this result? Explain. 4. Where would you expect the marble to be found least often? Do your data support your answer? 5. Would the results of your experiment be improved by combining the data from all of the lab groups in the class? Explain. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 737 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 738 The wave-particle duality of light presents us with many puzzles and paradoxes. Consider, for example, the famous two-slit interference experiment that Thomas Young used in 1801 to convince most physicists that light was a wave (see Chapter 13). This time, however, you are going to put a modern, quantum mechanical twist on the experiment. Since you know that light can behave as a particle (photons) and that particles can behave as matter waves (electrons), it does not matter what you choose to “shine” through the slits. Let us choose light, but reduce its intensity by inserting a filter so that only one photon at a time can enter the box (Figure 14.28). Let the light slowly expose a photographic film or enter the detector of your digital camera. slits ? incoming beam Figure 14.28 Young’s double-slit experiment, modified such that the intensity of the beam entering the box is reduced to a level that allows only one photon at a time to enter What will you observe? If you are impatient and let only a few photons through the slits, your result will be a random-looking scatter of dots where photons were absorbed by the film (Figure 14.29(a)). If you wait a little longer, the film will start to fill up (Figure 14.29(b)). Wait longer yet and something remarkable happens: You will see a two-slit interference pattern like the one in Figure 13.9 (Figure 14.29(c)). Why is this result so remarkable? Figure 14.29 Three different results of the double-slit experiment. Image (a) shows the result of only a few photons being recorded. Image (b) shows the result of a few more photons, and image (c) shows the familiar double-slit interference pattern that forms when many photons are recorded. (a) (c) (b) If light was only a wave, then the explanation would be that waves from the top slit in Figure 14.28 were slightly out of phase with waves from the bottom slit in some locations, causing nodes to form. In other places, the waves would combine in phase to produce antinodes. You arranged, however, to have only one photon at a time enter the apparatus. So, the photon would either go through the top slit or the bottom slit. But even a photon cannot be in two different places at once! If the photons can 738 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 739 only go through one slit (either the top or bottom one), why does a two-slit interference pattern, such as the one shown in Figure 14.30, result? Even though the individual photons go through only one slit, they somehow “know” that there is another slit open somewhere else! American physicist Richard Feynman (Figure 14.31) often used this example to emphasize how strange the quantum world is. 0.4 0.2 0.0 0.2 0.4 Figure 14.30 A two-slit interference pattern Figure 14.31 Richard P. Feynman (1918–1988) was one of the founders of modern quantum theory. He once stated: “I think it is safe to say that no one understands quantum mechanics.” So what does it all mean? To try to understand the double-slit experiment as it applies to individual photons or electrons, it is useful to summarize the key points: 1. When the photon or electron is absorbed by the photographic film or the detector of your digital camera, it exhibits its particle nature. 2. The location where any one photon or electron is detected is random but distinct; that is, the photon or electron always arrives and is detected as a distinct particle. 3. Although the location of individual photons or electrons is random, the combined pattern that many photons form is the characteristic pattern of antinodes and nodes, as shown in Figure 14.30. This pattern shows the wave nature of the photons or electrons. By the late 1920s, scientists developed a bold new interpretation of events. The wave-particle duality was showing that, at the level of atoms and molecules, the world was governed by the laws of probability and statistics. Although you cannot say much about what any one electron, for example, would do, you can make very precise predictions about the behaviour of very large numbers of electrons. In 1926, German physicist Max Born suggested that the wave nature of particles is best understood as a measure of the probability that the particle will be found at a particular location. The antinodes in the double-slit interference pattern exist because the particles have a high probability of being found at those locations after they pass through the double-slit apparatus. This measure of probability of a particle’s Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 739 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 740 quantum indeterminacy: the probability of finding a particle at a particular location in a doubleslit interference pattern info BIT American physicist and Nobel laureate Leon Lederman estimates that quantum physics is an essential part of the technologies responsible for over 25% of the North American gross national product. location is called quantum indeterminacy. This concept is the most profound difference between quantum physics and classical physics. According to quantum physics, nature does not always do exactly the same thing for the same set of conditions. Instead, the future develops probabilistically, and quantum physics is the science that allows you to predict the possible range of events that may occur. Although you may think that quantum behaviour is remote and has nothing to do with your life, nothing could be further from the truth. As you will see in the next three chapters, quantum theory has become one of the most powerful scientific theories ever developed. Virtually all of the electronic equipment we use daily that improves our quality of life, and most of our current medical technologies and understanding, are possible because of the deep insights that quantum theory provides. 14.5 Check and Reflect 14.5 Check and Reflect Knowledge Applications 1. Explain which of the following choices 3. Which of the following examples is the best one. (a) The double-slit experiment illustrates the wave nature of a quantum, and which illustrates the particle nature? demonstrates that light is a wave. (a) Electrons hit a phosphor screen and (b) The double-slit experiment shows that light is a particle. (c) The double-slit experiment illustrates that light has both wave and particle characteristics. 2. True or false? Explain. (a) The results of the double-slit experiment described in this section apply only to photons. (b) The results of the double-slit experiment apply to photons as well as to particles such as electrons. create a flash of light. (b) Electrons scatter off a crystal surface and produce a series of nodes and antinodes. (c) Light hits a photocell and causes the emission of electrons. Extension 4. Imagine that, one night as you slept, Planck’s constant changed from 6.63 1034 Js to 6.63 Js. Explain, from a quantum mechanical point of view, why walking through the doorway of your bedroom could be a dangerous thing to do. e TEST To check your understanding of quantum mechanics, follow the eTest links at www.pearsoned.ca/ school/physicssource. 740 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 741 CHAPTER 14 SUMMARY Key Terms and Concepts incandescent blackbody radiation curve blackbody quantum Planck’s formula quantized ph
oton photoelectric effect photoelectron threshold frequency work function stopping potential Compton scattering Compton effect wave-particle duality Heisenberg’s uncertainty principle quantum indeterminacy Key Equations E nhf hf W Ek Ekmax qVstopping p h i f h (1 cos ) m c Conceptual Overview Summarize this chapter by copying and completing the following concept map. discovery of photoelectric effect blackbody spectrum and failure of classical physics Einstein’s explanation of the photoelectric effect Millikan’s determination of Planck’s constant Compton effect Heisenberg and quantum indeterminacy Planck’s radiation law De Broglie and wave-particle duality Davisson–Germer experiment Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 741 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 742 CHAPTER 14 REVIEW Knowledge 1. (14.1) Explain what is meant by the term “ultraviolet catastrophe.” 2. (14.1) Write the equation for Planck’s formula and briefly explain what it means. 3. (14.1) What is the energy of a 450-nm photon? Express the answer in both joules and electron volts. 4. (14.1) If an X-ray photon has a wavelength 100 times smaller than the wavelength of a visible light photon, how do the energies of the two photons compare? Give a numerical answer. 5. (14.2) Who is credited with discovering the photoelectric effect? 6. (14.2) Who provided the correct explanation of the photoelectric effect? In what way(s) was this explanation radical when first proposed? 7. (14.2) If the threshold frequency for photoemission from a metal surface is 6.0 1014 Hz, what is the work function of the metal? 8. (14.3) Explain why the Compton effect provides critical evidence for the particle model of light. 9. (14.3) If a 0.010-nm photon scatters 90 after striking an electron, determine the change in wavelength () for the photon. 10. (14.4) What is the wavelength of an electron that has a momentum of 9.1 1027 Ns? 11. (14.4) What is the momentum of a 100-nm UV photon? 12. (14.4) If a particle is confined to a region in space 10 fm across, could the particle also be at rest? Explain, using Heisenberg’s uncertainty principle. Applications 13. How many photons are emitted each second by a 1.0-W flashlight? Use 600 nm as the average wavelength of the photons. 14. A beam of 300-nm photons is absorbed by a metal surface with work function 1.88 eV. Calculate the maximum kinetic energy of the electrons emitted from the surface. 15. Modern transmission electron microscopes can accelerate electrons through a 100-V potential difference and use these electrons to produce images of specimens. What is the wavelength of a 100-keV electron? Ignore relativistic effects. Why are electron microscopes capable of much higher magnification than light microscopes? 16. A major league baseball pitcher can throw a 40-m/s fastball of mass 0.15 kg. (a) Calculate the wavelength of the ball. (b) Why can you safely ignore quantum effects in this case? 17. Imagine that you are 100 m from a 100-W incandescent light bulb. If the diameter of your pupil is 2 mm, estimate how many photons enter your eye each second. (Note: You will need to make estimates and provide additional information.) 18. How many photons are emitted each second by an FM radio station whose transmitted power is 200 kW and whose frequency is 90.9 MHz? 19. An electron is trapped in a box that is 0.85 nm long. Calculate the three lowest energies that this electron can have. Why can the electron not have energy values between the values you calculated? 20. Calculate the momentum of a 100-keV X-ray photon. 742 Unit VII Electromagnetic Radiation 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 743 Extensions 21. Argue that photons exert pressure. (Hint: Newton’s p second law, F ma, can also be written as F . t In other words, force is a measure of the rate of change in momentum with respect to time. Also, remember that pressure is defined as force acting F.) over an area: P A 22. After you graduate from university, you take a job in a patent office, assessing the feasibility of inventions. Your boss hands you a file and skeptically tells you it is from a physicist who claims that a 1.0-km2 sail made from highly reflecting Mylar film could produce about 10 N of force simply by reflecting sunlight. You are asked to check the physics. Do the following: (a) Estimate how many photons arrive from the Sun per second per square metre at a distance equal to the Earth-Sun separation. You know that the top of Earth’s atmosphere gets 1.4 kW/m2 of energy from the Sun. (b) Calculate the momentum of each photon and remember that the photons are reflected. (c) Multiply the pressure (force per unit area) by the total area of the sail. 24. Einstein thought there was a fundamental flaw in quantum physics because “God does not play dice.” (a) What do you think he meant by this statement? What part of quantum theory was Einstein referring to? (b) Why is it ironic that Einstein made this statement? Consolidate Your Understanding 1. Describe two significant failings of classical physics that challenged physics prior to 1900. 2. Provide evidence for quantization of energy, and explain this concept to a friend. 3. List and describe at least two crucial experimental findings that support Einstein’s claim that light has a particle nature. 4. Explain why it is incorrect to state that light is either a wave or a particle. Comment on how quantum physics tries to resolve this duality. 5. What is meant by the term “quantum indeterminacy”? Provide experimental evidence for this idea. Does the physicist’s claim make sense? Think About It 23. How many photons per second does your radio respond to? Consider receiving a 100-MHz radio signal. The antenna in an average radio receiver must be able to move a current of at least 1.0 A through a 10-mV potential difference in order to be detectable. Review your answers to the Think About It questions on page 703. How would you answer each question now? e TEST To check your understanding of the wave-particle duality of light, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 14 The wave-particle duality reminds us that sometimes truth really is stranger than fiction! 743 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 744 UNIT VII PROJECT From Particle to Quantum — How did we arrive at our present understanding of light? Scenario In the past 500 years, our understanding of the nature of electromagnetic radiation has grown immensely, from the particle vs. wave controversy between Newton and Huygens, to the strange wave-particle duality described by de Broglie’s hypothesis, and to Heisenberg’s uncertainty principle. The key evidence and theories along the way have opened up a bounty of applications, from fibre-optic communication networks, to scanning and tunnelling electron microscopes. With our understanding of electromagnetic radiation has come a vast wealth of information and new technologies, which in turn, have furthered our ability to probe and investigate the nature of our universe. From humble beginnings with simple lenses, the scientific community has followed a long and difficult pathway to our present understanding. In this project, you will retrace this pathway, highlighting the theories, evidence, and experiments that have contributed to our present understanding of light and electromagnetic radiation. Planning Working in small groups or individually, prepare a presentation that summarizes the intellectual journey from the earliest theories of the particle nature of light, to the more modern theory of wave-particle duality. Your presentation should include simulations and illustrations that identify key experimental evidence, and descriptions of each model and theory, including the scientists who proposed them. Your summary can be presented in chronological order, from early discoveries, to later ones, or it can be organized around models (particle, wave, quantum, wave-particle duality). Materials text and Internet resources simulations, illustrations, photos of evidence collected in experiments presentation software (PowerPoint, html editor, etc.) Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies completeness of evidence and effectiveness of presentation Procedure 1 Define each of the following models: particle, wave, quantum, and wave-particle duality. 2 Identify each of the scientists involved with each model. 3 Using either a table or a timeline, place each model and the related scientists in order from earliest, to most recent. 4 On your table or a timeline, identify each key experiment and the evidence that was used to support each model. Include the following experimental evidence and theories: • reflection, refraction, dispersion, diffraction, interference, polarization, blackbody radiation, photoelectric effect, Compton effect, de Broglie’s hypothesis, and Heisenberg’s uncertainty principle 5 Use your table or timeline as the basis for preparing your presentation. Use simulations, illustrations, and photographs where possible to describe experimental evidence. Thinking Further The evolution of an idea or theory can take place over hundreds of years, with one participant handing off evidence to the next participant. A sort of relay develops, because the race is simply too long for one person to complete alone. With this in mind, consider the following questions that could be answered at the end of your presentation. • A relay race has an end. Is there an end in the race to fully understand the nature of electromagnetic radiation and light? If the relay is not over, where do you think we are going from here? • • How have we used the knowledge of our predecessors in determining where to look next? Explain. *Note: Your instructor will assess the project using a similar assessment rubric. 744 Unit VII Electromagnetic Radiation 14-PearsonPhys30-
Chap14 7/24/08 3:58 PM Page 745 UNIT VII SUMMARY Unit Concepts and Skills: Quick Reference Summary Resources and Skill Building Concepts Chapter 13 Types of electromagnetic radiation Models of EMR Maxwell’s electromagnetic theory The wave model can be used to describe the characteristics of electromagnetic radiation. 13.1 What Is Electromagnetic Radiation? Frequency, wavelength, and source are used to identify types of EMR. Different models were used to explain the behaviour of EMR. Maxwell’s theory linked concepts of electricity and magnetism. Electromagnetic radiation is produced by accelerating charges. Speed of electromagnetic radiation 13.2 The Speed of Electromagnetic Radiation Galileo, Roemer and Huygens, and Fizeau measured the speed of EMR, but Michelson’s experiment made the definitive measurement. The Law of Reflection, image formation, and ray diagrams 13.3 Reflection The angle of reflection equals the angle of incidence and is in the same plane. Ray diagrams show a light ray interacting with a surface. Three rays predict the location and characteristics of the image. Image formation/equations The mirror equation relates the focal length of a curved mirror to the image and object distances. Refraction and Snell’s Law Total internal reflection 13.4 Refraction Snell’s Law relates the refraction of a light wave to the speed with which light travels in different media. All light is internally reflected at an interface if the angle of refraction is 90 or greater. Dispersion and recomposition White light can be separated into its component wavelengths. Image formation with thin lenses The lens equation relates the focal length of a curved lens to the image and object distances. Huygens’ Principle Young’s experiment, interference, and diffraction Diffraction gratings 13.5 Diffraction and Interference Huygens predicted the motion of a wave front as many point sources. Young’s experiment showed that two beams of light produce an interference pattern and that light behaves as a wave. Light on a multi-slit diffraction grating produces an interference pattern. Polarization EMR absorption by polarizing filters supports the wave model of light. Chapter 14 The wave–particle duality reminds us that sometimes truth really is stranger than fiction! 14.1 The Birth of the Quantum Classical physics was unable to explain the shape of the blackbody radiation curve. A quantum is the smallest amount of energy of a particular wavelength or frequency that a body can absorb, given by E = hf. 14.2 The Photoelectric Effect The work function is the minimum energy required to cause photoemission of electrons from a metal surface. Millikan’s photoelectric experiment provided a way to measure Planck’s constant. The photoelectric effect obeys the law of conservation of energy. 14.3 The Compton Effect When an electron scatters an X ray, the change in the X ray’s wavelength relates to the angle of the X-ray photon’s scattering. 14.4 Matter Waves and the Power of Symmetric Thinking Something that has momentum also has wavelength: Particles can act like waves. Particles have a wave nature, so it is impossible to precisely know their position at the same time as their momentum. Quantum Photoelectric effect Planck’s constant Compton effect Wave–particle duality Heisenberg’s uncertainty principle Quantum indeterminacy Figure 13.4; Table 13.1 Figures 13.5–13.9 Figures 13.10–13.15 Figures 13.16-13.19; Minds On: Going Wireless Figures 13.21-13.24; Example 13.1; 13-2 QuickLab Figures 13.28-13.30; 13-3 QuickLab; Minds On: Image in a Mirror; Figures 13.31–13.32; Figures 13.36–13.39; 13-4 QuickLab 13-5 Problem-Solving Lab; Example 13.2 Table 13.4; Examples 13.3–13.4; 13-6 Inquiry Lab Figures 13.49-13.53; Example 13.5, 13-7 Decision-Making Analysis Figures 13.54–13.56; Table 13.5; 13-8 QuickLab Figures 13.57–13.59; Example 13.6; Figure 13.61; Example 13.7; 13-9 Inquiry Lab Figures 13.67–13.69 Figures 13.70–13.78; Examples 13.8–13.9 Figure 13.80 Figure 13.81; Example 13.10; 13-10 Inquiry Lab 13-1 QuickLab; Figure 13.86; Figures 13.88–13.90 14-1 QuickLab, Figures 14.3–14.4 Examples 14.1, 14.2, 14.3; Minds On: What’s Wrong with This Analogy? Figure 14.6 14-2 QuickLab, Table 14.1 Figure 14.12, 14-3 Design a Lab; Examples 14.5–14.6 Figure 14.16; Examples 14.7–14.8 Examples 14.9–14.10 Figures 14.21, 14.22, 14.24, 14.26; Example 14.11 14.5 Coming to Terms with Wave–particle Duality and the Birth of Quantum Mechanics Wave-particle duality illustrates the probabilistic nature of atoms and molecules. Quantum indeterminacy is the measure of the probability of a particle’s location. 14-4 QuickLab, Figures 14.28–14.30 Unit VII Electromagnetic Radiation 745 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 746 UNIT VII REVIEW Vocabulary Knowledge 1. Use your own words to define these terms: CHAPTER 13 angle of diffraction blackbody blackbody radiation curve Compton effect Compton scattering converging critical angle diffraction diffraction grating dispersion diverging electromagnetic radiation focal point frequency Heisenberg’s uncertainty principle Huygens’ Principle image attitude incandescent interference law of reflection magnification node, antinode particle model path length period photoelectric effect photoelectrons photon Planck’s formula polarization quantized quantum quantum indeterminacy refraction refractive index Snell’s Law spectrum stopping potential threshold frequency total internal reflection wave model wave-particle duality wavelength work function 746 Unit VII Electromagnetic Radiation 2. How does the quantum model reconcile the wave model and the particle model of light? 3. How did Maxwell’s work with capacitors influence his theories on electromagnetism? 4. Describe the experimental evidence that supports all of Maxwell’s predictions about electromagnetic radiation. 5. Discuss the significance of the word “changing” in Maxwell’s original description of electromagnetic radiation. 6. Why does a spark produce electromagnetic radiation? 7. If a metal conductor, such as a spoon, is placed in an operating microwave oven, a spark is produced. Why? 8. Using a ray diagram, show three rays that are needed to identify and verify the characteristics of an image. 9. What is the relationship between the focal length and the radius of curvature for a curved mirror? 10. What is a virtual focal point and how is it different from a real focal point? 11. Explain, using a ray diagram, how a real image can be formed when using two concave mirrors. 12. When you place the concave side of a spoon on your nose and slowly pull it away from your face, your image disappears at a certain distance. What is the significance of this distance? 13. When an object such as a paddle is partially submerged in water, why does it appear bent? 14. Explain how Snell’s Law supports the wave theory of light. 15. What happens to the wavelength of monochromatic light when it passes from air into water? 16. Several people holding hands run down the beach and enter the water at an angle. Explain what happens to the speed and direction of the people as they enter the water. 17. How was Newton able to show that a prism separates the colours in the spectrum, rather than adding the colours to white light? 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 747 18. What is Huygens’ Principle? 19. A straight wave front is incident on a barrier with a small hole. Using a diagram, describe the shape of the wave front a moment after it makes contact with the barrier. 20. Using a schematic, illustrate Young’s experiment. 38. A proton and a neutron are both moving at the same speed. Which particle has the shorter de Broglie wavelength? 39. Explain, using wave mechanics, why it is impossible for a particle to have zero kinetic energy when it is confined to a fixed region in space. 21. Explain why diffraction supports the wave model of light. Applications 22. What key evidence was observed by Dominique Arago in 1818? Why was this evidence crucial to the acceptance of the wave model of light? 23. How must two plane polarizing filters be aligned in order to fully block electromagnetic radiation? 24. Is an electromagnetic wave one-dimensional, two-dimensional, or three-dimensional? Explain. CHAPTER 14 25. Is a quantum of blue light the same as a quantum of red light? Explain. 26. How much energy is carried by a photon of wavelength 550 nm? 27. Explain how you can estimate the surface temperature of a star by noting its colour. 28. Arrange the following photons from highest to lowest energy: ultraviolet photon, 10-nm photon, microwave photon, gamma-ray photon, 600-nm photon, infrared photon. 29. What is the frequency of blue light of wavelength 500 nm? 30. Ultraviolet light causes sunburn whereas visible light does not. Explain, using Planck’s formula. 31. Explain what is meant by the term “threshold frequency.” 32. How does the energy of photoelectrons emitted by a metal change as the intensity of light hitting the metal surface changes? 33. What is the maximum wavelength of light that will cause photoemission from a metal having a work function of 3.2 eV? 34. Explain the difference between the Compton effect and the photoelectric effect. 35. What is meant by the term “wave-particle duality”? 36. Even though photons have no mass, they still carry momentum. What is the momentum of a 300-nm ultraviolet photon? 37. What is the de Broglie wavelength of an electron moving at 3000 km/s? 40. If visible light is a particle, predict what would be observed if light passed through two small holes in a barrier. Compare this prediction to what is actually observed when light passes through two small holes in a barrier. What does this suggest about the nature of light? 41. How many radio-frequency photons are emitted each second by a radio station that broadcasts at a frequency of 90.9 MHz and has a radiated power of 50 kW? 42. Explain how an antenna is able to “sense” electromagnetic radiation. 43. Detailed measurements of th
e Moon’s orbit could be calculated after the Apollo mission placed large reflecting mirrors on the surface of the Moon. If a laser beam were directed at the mirrors on the Moon and the light was reflected back to Earth in 2.56 s, how far away, in kilometres, is the Moon? 44. When you increase the intensity of a green light, do you change the energy of the green-light photons? Why does the light get brighter? 45. A Michelson apparatus is used to obtain a value of 2.97 108 m/s for the speed of light. The sixteen-sided rotating mirror completes 1.15 104 revolutions in one minute. How far away was the flat reflecting mirror? 46. An eight-sided mirror like Michelson’s is set up. The light reflects from the rotating mirror and travels to a fixed mirror 5.00 km away. If the rotating mirror turns through one-eighth of a rotation before the light returns from the fixed mirror, what is the rate of rotation? 47. A sixteen-sided mirror rotates at 4.50 102 Hz. How long does it take to make one-sixteenth of a rotation? 48. Why do police and search-and-rescue agencies use infrared cameras for night-time surveillance when looking for people? Explain why infrared is used and not some other part of the electromagnetic spectrum. Unit VII Electromagnetic Radiation 747 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 748 49. The speed of light in a material is determined to 60. Calculate the wavelength of electrons used in a be 1.24 108 m/s. What is the material? 50. Light of wavelength 520 nm strikes a metal surface having a work function of 2.3 eV. Will the surface emit photoelectrons? 51. A student replicating Michelson’s experiment uses an eight-sided mirror and a fixed mirror located 35.0 km away. Light is reflected through the system when the rotating mirror turns at 5.20 102 Hz. What is the experimentally determined speed of light and the percentage error in the measurement? 52. An electrically neutral 1-m2 piece of aluminium is put in orbit high above Earth. Explain why, after a period of time, the piece of aluminium will become electrically charged. Predict the sign of the charge. 53. An object is located in front of a diverging mirror with a focal length of 5.0 cm. If the virtual image is formed 3.0 cm from the vertex of the mirror and is 1.0 cm high, determine the object’s characteristics and position. 54. Photon A has four times the energy of photon B. Compare the wavelengths and the momenta of the two photons. 55. A light ray passes from water into ruby at an angle of 10. What is the angle of refraction? 56. An X-ray photon of wavelength 0.025 nm collides elastically with an electron and scatters through an angle of 90. How much energy did the electron acquire in this collision and in what important way did the X ray change? 57. A 3.0-cm-high object is placed 10.0 cm from a converging lens with a focal length of 5.0 cm. Using the thin lens equation, determine the image attributes and position. 58. Imagine that you are asked to review a patent application for a laser-powered deep space probe. The proposal you are reviewing calls for a 1-kW laser producing 500-nm photons. The total mass of the spacecraft, including the laser, is 1000 kg. Determine (a) if laser propulsion is possible, and the underlying principle of this form of propulsion. (b) how fast the spacecraft would be travelling after one year of “laser-drive” if it started from rest. 59. List two ways to recompose the spectrum into white light. 748 Unit VII Electromagnetic Radiation transmission electron microscope if the electrons are accelerated through an electric field of potential 75 kV. Ignore relativistic effects. 61. In an experiment similar to Young’s, two waves arrive at the screen one half-wavelength out of phase. What will be observed at this point on the screen? 62. What is the minimum or rest energy of an electron confined to a one-dimensional box 1 nm long? 63. A mixture of violet light ( 420 nm) and red light ( 650 nm) are incident on a diffraction grating with 1.00 104 lines/cm. For each wavelength, determine the angle of deviation that leads to the first antinode. 64. Light with a wavelength of 700 nm is directed at a diffraction grating with 1.50 102 slits/cm. What is the separation between adjacent antinodes when the screen is located 2.50 m away? 65. Your physics teacher, eager to get to class, was observed from a police spotting-plane to travel a distance of 222 m in 10 s. The speed limit was 60 km/h, and you can quickly determine that he was speeding. The police issued a ticket, but your teacher decided to argue the case, citing Heisenberg’s uncertainty principle as his defence. He argued that the speed of his car was fundamentally uncertain and that he was not speeding. Explain how you would use Heisenberg’s uncertainty principle in this case and comment on whether your teacher’s defence was good. The combined mass of the car and your teacher is 2000 kg. Extensions 66. Traditional radio technology blends a carrier signal and an audio signal with either frequency or amplitude modulation. This generates a signal with two layers of information—one for tuning and one containing the audio information. Describe the two layers of information that a cell phone signal must contain in order to establish and maintain constant communication with a cell phone network. 67. Use Heisenberg’s uncertainty principle to estimate the momentum and kinetic energy of an electron in a hydrogen atom. Express the energy in electron volts. The hydrogen atom can be approximated by a square with 0.2-nm sides. (Hint: Kinetic energy is related to momentum via the equation Ek 2 p .) m 2 14-PearsonPhys30-Chap14 7/24/08 3:58 PM Page 749 68. Global positioning satellites maintain an orbital altitude of 20 000 km. How long does it take for a time signal to travel from the satellite to a receiver located directly below the satellite? 69. Why do some metals have a higher threshold frequency than others? How is this phenomenon related to electric fields? 70. Explain how an optical fibre is able to transmit a light pulse over a long distance without a loss in intensity. 71. The human eye can detect as few as 500 photons of light, but in order to see, this response needs to occur over a prolonged period of time. Seeing requires approximately 10 000 photons per second. If the Sun emits 3.9 1026 W, mostly in the bluegreen part of the spectrum, and if roughly half of the energy is emitted as visible light, estimate how far away a star like our Sun would be visible. 72. A beam of 200-eV electrons is made to pass through two slits in a metal film that are separated by 50 nm. A phosphor screen is placed 1 m behind the slits. Sketch what you would expect to see. Provide calculations to support your answer. Skills Practice 73. An object is located 25.0 cm from a diverging mirror with a focal length of 10.0 cm. Draw a ray diagram to scale to determine the following: (a) (b) (c) the image location and type the image attitude the magnification of the image 74. The following data are taken from an experiment in which the maximum kinetic energy of photoelectrons is related to the wavelength of the photons hitting a metal surface. Use these data to produce a graph that shows the energy of the incident photons on the horizontal axis and the kinetic energy of photoelectrons on the vertical axis. From this graph, determine the work function for the metal. 75. Use a ray diagram to show why a double convex lens is called a converging lens and a double concave lens is called a diverging lens. Label the principal axis, principal focus, secondary focus, and optical centre. 76. Calculate the momentum and wavelength of an electron that has a kinetic energy of 50 keV. Ignore relativistic effects. 77. Explain, with the aid of a ray diagram, why an image does not form when you place an object at the focal point of a converging lens. 78. Determine the momentum of an X ray of wavelength 10 nm. 79. Prepare a table in which you compare the wave and particle models of light. List as many phenomena as you can think of and decide whether light can be explained best using the wave or the particle model. How would you answer the question, “Is light a wave or a particle?” Self-assessment 80. Describe to a classmate which concepts of electromagnetic radiation you found most interesting when studying this unit. Give reasons for your choices. 81. Identify one issue pertaining to the wave-particle duality of light that you would like to investigate in greater detail. 82. What concept in this unit did you find most difficult? What steps could you take to improve your understanding? e TEST To check your understanding of electromagnetic radiation and the dual nature of light, follow the eTest links at www.pearsoned.ca/school/ physicssource. Wavelength (nm) Kinetic Energy (eV) 200 250 300 350 400 450 3.72 2.47 1.64 1.05 0.61 0.26 Unit VII Electromagnetic Radiation 749 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 750 U N I T VIII Atomic Atomic Physics Physics M A cluster of newly formed stars in a spiral arm of the Milky Way galaxy. Physicists are using atomic theories to understand the structure and evolution of the universe. e WEB To learn more about the role of atomic physics in cosmology, follow the links at www.pearsoned.ca/school/physicssource. 750 Unit VIII 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 751 Unit at a Glance C H A P T E R 1 5 Electric force and energy quantization determine atomic structure. 15.1 The Discovery of the Electron 15.2 Quantization of Charge 15.3 The Discovery of the Nucleus 15.4 The Bohr Model of the Atom 15.5 The Quantum Model of the Atom C H A P T E R 1 6 Nuclear reactions are among the most powerful energy sources in nature. 16.1 The Nucleus 16.2 Radioactive Decay 16.3 Radioactive Decay Rates 16.4 Fission and Fusion C H A P T E R 1 7 The development of models of the structure of matter is ongoing. 17.1 Detecting and Measuring Subatomic Particles 17.2 Quantum Theory and the Discovery of New
Particles 17.3 Probing the Structure of Matter 17.4 Quarks and the Standard Model Unit Themes and Emphases • Energy and Matter Focussing Questions The study of atomic structure requires analyzing how matter and energy are related. As you study this unit, consider these questions: • What is the structure of atoms? • How can models of the atom be tested? • How does knowledge of atomic structure lead to the development of technology? Unit Project How Atomic Physics Affects Science and Technology • Unit VIII discusses radical changes in the understanding of matter and energy. In this project, you will investigate how advances in atomic physics influenced the development of technology and other branches of science. Unit VIII Atomic Physics 751 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 752 Electric force and energy quantization determine atomic structure. Max Planck began undergraduate studies at the University of Munich in 1874. During his first term, he took primarily mathematics courses, although he also had considerable musical talent and an interest in physics. Wondering which field to pursue, the teenaged Planck asked his physics professor, Philipp von Jolly, about the prospects for a career in physics. Jolly told Planck that there would be few opportunities since almost everything in physics had already been discovered, leaving only a few minor gaps to fill in. Despite Jolly’s discouraging advice, Planck went on to complete a doctorate in physics, and became a renowned professor at the University of Berlin. As described in Chapter 14, one of the “gaps” in theoretical physics was the puzzling distribution of energy among the wavelengths of radiation emitted by a heated body. Planck concentrated on this problem for months, and by the end of 1900 concluded that this distribution is possible only if energy is quantized. At first, Planck and his contemporaries did not realize the huge significance of his findings. As you will learn in this chapter, Planck’s discovery led to quantum theory, a concept that revolutionized atomic physics. You will also see that Planck was just the first of many researchers who demonstrated that there was still a great deal to be discovered in physics. Figure 15.1 shows how radically our concept of the atom changed during the 20th century. Planck’s idea of quantization of energy led Bohr to propose his model of the atom in 1913. In this chapter, you will study in detail how the model of the atom evolved. H He Li Be … indivisible raisin bun planetary Bohr quantum Figure 15.1 The evolution of theories of atomic structure C H A P T E R 15 Key Concepts In this chapter, you will learn about: charge-to-mass ratio classical model of the atom continuous and line spectra energy levels quantum mechanical model Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe matter as containing discrete positive and negative charges explain how the discovery of cathode rays helped develop atomic models explain the significance of J.J. Thomson’s experiments explain Millikan’s oil-drop experiment and charge quantization explain the significance of Rutherford’s scattering experiment explain how electromagnetic theory invalidates the classical model of the atom describe how each element has a unique line spectrum explain continuous and line spectra explain how stationary states produce line spectra calculate the energy difference between states, and the characteristics of emitted photons Science, Technology, and Society explain how scientific knowledge and theories develop explain the link between scientific knowledge and new technologies 752 Unit VIII 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 753 15-1 QuickLab 15-1 QuickLab Cathode Rays and Magnetic Fields Problem What type of charge do cathode rays have? Materials gas-discharge tube (Figure 15.2) high-voltage power supply bar magnet or electromagnet bar magnet S N high-voltage power supply gasdischarge tube Figure 15.2 A gas-discharge tube Procedure 1 Your teacher will give directions for setting up the particular tube and power supply that you will use. Follow these directions carefully. Note the location of the cathode in the discharge tube. 2 Turn on the power supply and note the beam that appears in the tube. To see the beam clearly, you may need to darken the room. 3 Bring the magnet close to the tube surface. Note the direction of the magnetic field and the direction in which the beam moves. 4 Repeat step 3 at various positions along the tube. Questions 1. How did you determine the direction of the magnetic field? 2. What do you think causes the visible beam in the tube? 3. What evidence suggests that the magnet causes the beam to deflect? 4. How could you verify that the cathode rays originate from the negative terminal of the discharge tube? 5. What can you conclude about the charge of the cathode rays? Explain your reasoning. 6. Describe another way to determine the type of charge on cathode rays. Think About It 1. What are atoms made of? 2. What holds atoms together? 3. How can physicists measure atomic structure when it is too small to be seen by even the most powerful microscope? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 15 Electric force and energy quantization determine atomic structure. 753 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 754 15.1 The Discovery of the Electron info BIT Dalton’s research included important findings about the aurora borealis, atmospheric gases, and the formation of dew. He also made the first scientific study of colour blindness, which he had himself. Around 1803, the English chemist John Dalton (1766–1844) developed an atomic theory to explain the ratios in which elements combine to form compounds. Although later discoveries required modifications to Dalton’s theory, it is a cornerstone for modern atomic theory. This theory could explain the observations made by Dalton and other chemists, but their experiments did not provide any direct evidence that atoms actually exist. At the end of the 19th century, there was still some doubt about whether all matter was made up of atoms. By 1900, experiments were providing more direct evidence. Current technology, such as scanning tunnelling microscopes, can produce images of individual atoms. M I N D S O N Evidence for Atoms How do you know that atoms exist? Work with a partner to list evidence that matter is composed of atoms. Cathode-ray Experiments During the 1800s, scientists discovered that connecting a high voltage across the electrodes at opposite ends of an evacuated glass tube caused mysterious rays to flow from the negative electrode (the cathode) toward the positive electrode. These cathode rays caused the glass to glow when they struck the far side of the tube. The rays could be deflected by a magnetic field. In 1885, after several years of experiments with improved vacuum discharge tubes, William Crookes in England suggested that cathode rays must be streams of negatively charged particles. In 1895, Jean Baptiste Perrin in France showed that cathode rays entering a hollow metal cylinder built up a negative charge on the cylinder. In 1897, Joseph John Thomson (1856–1940) took these experiments a step further. First, he used an improved version of Perrin’s apparatus to show even more clearly that cathode rays carry negative charge (Figure 15.3). Next, he tackled the problem of why no one had been able to deflect cathode rays with an electric field. Thomson’s hypothesis was that the cathode rays ionized some of the air molecules remaining in the vacuum chamber and these ions then shielded the cathode rays from the electric field. By taking great care to get an extremely low pressure in his discharge tube, Thomson was able to demonstrate that cathode rays respond to electric fields just as negatively charged particles would. Thomson had discovered the electron. cathode ray: free electrons emitted by a negative electrode info BIT The Irish physicist G. Johnstone Stoney coined the term electron in 1891. Thomson called the particles in cathode rays “corpuscles.” 754 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 755 Figure 15.3 Reproduction of Thomson’s diagram and part of his description of the apparatus he used for several of his experiments. To determine the charge-to-mass ratio for an electron, Thomson added an electromagnet beside the parallel plates in the middle of the tube. The rays from the cathode C pass through a slit in the anode A, which is a metal plug fitting tightly into the tube and connected with the earth; after passing through a second slit in another earth-connected metal plug B, they travel between two parallel aluminium plates about 5 cm long by 2 broad and at a distance of 1.5 cm apart; they then fall on the end of the tube and produce a narrow well-defined phosphorescent patch. A scale pasted on the outside of the tube serves to measure the deflexion of this patch. M I N D S O N Are Electrons Positively or Negatively Charged? Outline two different methods for testing whether cathode rays consist of negatively or positively charged particles. Charge-to-mass Ratio of the Electron Thomson did not have a method for measuring either the mass of an electron or the charge that it carried. However, he did find a way to determine the ratio of charge to mass for the electron by using both an electric field and a magnetic field. Recall from Chapter 11 that the electric force acting on a charged particle is F e qE e is the electric force, q is the magnitude of the charge on where F the particle, and E is the electric field. Section 12.2 describes how the left-hand rule gives the direction of the magnetic force acting on a negative charge moving through a magnetic field. The magnitude of this force is F m qvB where q is the magnitu
de of the charge on the particle, v is the component of the particle’s velocity perpendicular to the magnetic field, and B is the magnitude of the magnetic field. For a particle moving perpendicular to a magnetic field, v v and F m qvB Consider the perpendicular electric and magnetic fields shown in Figure 15.4. The electric field exerts a downward force on the negative charge while the magnetic field exerts an upward force. The gravitational force acting on the particle is negligible. If the net force on the charged particle is zero, the electric and magnetic forces must be equal in magnitude but opposite in direction: PHYSICS INSIGHT Dots and s are a common way to show vectors directed out of or into the page. A dot represents the tip of a vector arrow coming toward you, and represents the tail of an arrow moving away from you. Chapter 15 Electric force and energy quantization determine atomic structure. 755 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 756 E Fm Fe Fm v v B Figure 15.4 Perpendicular electric and magnetic fields act on a moving negative charge. The red dots represent a magnetic field directed out of the page. Using the left-hand rule for magnetic force, your thumb points in the direction of electron flow (right), fingers point in the direction of the magnetic field (out of the page), and the palm indicates the direction of the magnetic force (toward the top of the page). Example 15.1 F m F net F net F e 0, so Fm Fe q B E B E qv v The speed of the particle is, therefore, v E B Practice Problems 1. A beam of electrons passes undeflected through a 2.50-T magnetic field at right angles to a 60-kN/C electric field. How fast are the electrons travelling? 2. What magnitude of electric field will keep protons from being deflected while they move at a speed of 1.0 105 m/s through a 0.05-T magnetic field? 3. What magnitude of magnetic field will stop ions from being deflected while they move at a speed of 75 km/s through an electric field with a magnitude of 150 N/C? Answers 1. 2.4 104 m/s 2. 5 103 N/C 3. 2.0 103 T A beam of electrons passes undeflected through a 0.50-T magnetic field combined with a 50-kN/C electric field. The electric field, the magnetic field, and the velocity of the electrons are all perpendicular to each other. How fast are the electrons travelling? Given 50 kN/C 5.0 104 N/C E 0.50 T B Required speed (v) Analysis and Solution 0, so the Since the electrons are not deflected, F net m. e equals the magnitude of F magnitude of F E B 5.0 104 N/C 0.50 T v 1.0 105 m/s Paraphrase The electrons are travelling at a speed of 1.0 105 m/s. Concept Check What would happen to the beam of electrons in Example 15.1 if their speed were greatly decreased? Thomson used mutually perpendicular electric and magnetic fields to determine the speed of the cathode rays. He then measured the deflection of the rays when just one of the fields was switched on. These deflections depended on the magnitude of the field, the length of the path in the 756 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 757 field, and the speed, charge, and mass of the cathode-ray particles. Thomson could determine only the first three quantities, but he could use his measurements to calculate the ratio of the two unknowns, the charge and mass of the particles. Thomson made measurements with a series of cathode-ray tubes that each had a different metal for the electrode that emitted the rays. Since he found reasonably consistent values for the charge-to-mass ratio, Thomson concluded that all cathode rays consist of identical particles with exactly the same negative charge. Thomson’s experiments showed that q m for an electron is roughly 1011 C/kg. This ratio is over a thousand times larger than the ratio for a hydrogen ion. Other physicists had shown that cathode rays can pass through thin metal foils and travel much farther in air than atoms do. Therefore, Thomson reasoned that electrons are much smaller than atoms. In his Nobel Prize lecture in 1906, he stated that “we are driven to the conclusion that the mass of the corpuscle is only about 1/1700 of that of the hydrogen atom.” This value is within a few percent of the mass determined by the latest high-precision measurements. Thomson put forward the daring theory that atoms were divisible, and the tiny particles in cathode rays were “the substance from which all the chemical elements are built up.” Although he was incorrect about electrons being the only constituents of atoms, recognizing that electrons are subatomic particles was a major advance in atomic physics. Determining Charge-to-mass Ratios Thomson measured the deflection of cathode rays to determine the charge-to-mass ratio of the electron. You can also determine mass-tocharge ratios by measuring the path of charged particles in a uniform magnetic field. When a charged particle moves perpendicular to a magnetic field, the magnetic force is perpendicular to the particle’s velocity, so the particle’s direction changes, but its speed is constant. In a uniform is constant, so the magnitude of the magnetic force, magnetic field, B qv, is also constant. As described in Chapter 6, a force of constant B magnitude perpendicular to an object’s velocity can cause uniform circular motion. A charged particle moving perpendicular to a uniform magnetic field follows a circular path, with the magnetic force acting as the centripetal force (Figure 15.5). Fm r v F net F c F m F m qv v2 m B r q v r B m Figure 15.5 The path of an electron travelling perpendicular to a uniform magnetic field. The red s represent a magnetic field directed into the page. info BIT Around 1900, the idea that atoms could be subdivided was so radical that some physicists thought Thomson was joking when he presented his findings in a lecture to the Royal Institute in England. Chapter 15 Electric force and energy quantization determine atomic structure. 757 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 758 Concept Check How do you know that the magnetic force on the particle is directed toward the centre of a circular path in Figure 15.5? Example 15.2 When a beam of electrons, accelerated to a speed of 5.93 105 m/s, is directed perpendicular to a uniform 100-T magnetic field, they travel in a circular path with a radius of 3.37 cm (Figure 15.6). Determine the charge-to-mass ratio for an electron. Practice Problems 1. Find the charge-to-mass ratio for an ion that travels in an arc of radius 1.00 cm when moving at 1.0 106 m/s perpendicular to a 1.0-T magnetic field. 2. Find the speed of an electron moving in an arc of radius 0.10 m perpendicular to a magnetic field with a magnitude of 1.0 104 T. 3. A carbon-12 ion has a charge-tomass ratio of 8.04 106 C/kg. Calculate the radius of the ion’s path when the ion travels at 150 km/s perpendicular to a 0.50-T magnetic field. Answers 1. 1.0 108 C/kg 2. 1.8 106 m/s 3. 0.037 m 5.93 105 m/s Given ve 100 T 1.00 104 T B r 3.37 cm 3.37 102 m Required charge-to-mass ratio q m Analysis and Solution Since the magnetic force acts as the centripetal force, F F c m v 2 qv 100 T Fm Fc v 3.37 cm Figure 15.6 Substituting the known values for the beam of electrons gives q m 5.93 105 m/s (1.00 104 T)(3.37 102 m) 1.76 1011 C/kg Paraphrase The charge-to-mass ratio for an electron is about 1.76 1011 C/kg. Thomson’s Raisin-bun Model Most atoms are electrically neutral. If electrons are constituents of atoms, atoms must also contain some form of positive charge. Since no positively charged subatomic particles had yet been discovered, Thomson suggested that atoms might consist of electrons embedded in a blob of massless positive charge, somewhat like the way raisins are embedded in the dough of a raisin bun. Figure 15.7 shows Thomson’s model of the atom. Figure 15.7 Thomson’s model of the atom: negative electrons embedded in a positive body 758 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 759 Concept Check What characteristics should a scientific model have? Does Thomson’s raisin-bun model of the atom have these characteristics? THEN, NOW, AND FUTURE The Mass Spectrometer fields to determine Thomson used electric and magthe netic charge-to-mass ratio of electrons. In later experiments, he made similar measurements for positive ions. These experiments led to the development of the mass spectrometer, an instrument that can detect compounds, measure isotope masses, and determine molecular structures (Figure 15.8). Many mass spectrometers use a four-stage process enclosed in a vacuum chamber: Ionization: If the sample is not already a gas, the ion source vaporizes it, usually by heating. Heating may also break complex compounds into smaller fragments that are easier to identify. Next, the neutral compounds in the sample are ionized so that they will respond to electric and magnetic fields. Usually, the ion source knocks one or two electrons off the compound to produce a positive ion. Acceleration: High-voltage plates then accelerate a beam of these ions into the velocity selector. Velocity Selection: The velocity selector has crossed electric and magnetic fields arranged such that only the ions that have a speed of pass straight through. Ions v E B with slower or faster speeds are deflected away from the entrance to the detection chamber. Thus, all the ions entering the next stage of the mass spectrometer have the same known speed. Detection: A uniform magnetic field in the detection chamber makes the ions travel in circular paths. The radius of each path depends on the charge and mass of the ion. Ions arriving at the detector produce an electrical current that is proportional to the number of ions. The spectrometer can produce a graph of the charge-to-mass ratios for a sample by moving the detector or by varying the electric or magnetic fields. Mass spectrometers can detect compounds in concentrations as small as a few parts per billion. These versatile machines have a huge range of applications in scien
ce, medicine, and industry. Questions 1. Write an expression for the radius of the path of ions in a mass spectrometer. 2. How does the mass of an ion affect the radius of its path in the detection chamber? 3. Describe how you could use a mass spectrometer to detect an athlete’s use of a banned performance-enhancing drug. ionization chamber velocity selector vaporized sample detection chamber heater acceleration plates detector vacuum pump chart recorder amplifier positive ions Figure 15.8 Mass spectrometer Chapter 15 Electric force and energy quantization determine atomic structure. 759 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 760 15.1 Check and Reflect 15.1 Check and Reflect Knowledge 1. An electron moving at 5.0 105 m/s enters a magnetic field of magnitude 100 mT. (a) What is the maximum force that the 8. A mass spectrometer has the electric field in its velocity selector set to 8.00 102 N/C and the magnetic field set to 10.0 mT. Find the speed of the ions that travel straight through these fields. magnetic field can exert on the electron? When will the magnetic field exert this maximum force? 9. This diagram shows a proton moving at 1.0 105 m/s through perpendicular electric and magnetic fields. (b) What is the minimum force that the B 0.50 T [out of page] magnetic field can exert on the electron? When will the magnetic field exert this minimum force? 2. Explain why improved vacuum pumps were a key to the success of Thomson’s experiments. 3. What experimental results led Thomson to conclude that all cathode rays consist of identical particles? Applications 4. A beam of electrons enters a vacuum v 1.0 105 m/s [right] E 100 N/C [down] (a) Calculate the net force acting on the particle. (b) Will the net force change over time? Explain your reasoning. chamber that has a 100-kN/C electric field and a 0.250-T magnetic field. Extensions (a) Sketch an orientation of electric and magnetic fields that will let the electrons pass undeflected through the chamber. (b) At what speed would electrons pass undeflected through the fields in part (a)? 5. Electrons are observed to travel in a circular path of radius 0.040 m when placed in a magnetic field of strength 0.0025 T. How fast are the electrons moving? 6. How large a magnetic field is needed to deflect a beam of protons moving at 1.50 105 m/s in a path of radius 1.00 m? 7. Use the appropriate hand rule to determine the direction of the magnetic field in question 6 if the protons rotate counterclockwise in the same plane as this page. 10. Suppose that a passenger accumulates 5 C of negative charge while walking from left to right across the carpeted floor to the security gate at an airport. (a) If the metal detector at the security gate exerts an upward force on this charge, what is the direction of the magnetic field within the detector? (b) If the metal detector uses a 0.05-T magnetic field, roughly how fast would the passenger have to run through the detector in order to feel weightless? (c) Explain whether it would be practical to use an airport metal detector as a levitation machine. e TEST To check your understanding of Thomson’s experiments, follow the eTEST links at www.pearsoned.ca/school/physicssource. 760 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 761 15.2 Quantization of Charge As you learned in Chapter 14, Planck and Einstein introduced the concept of quantization in physics in the early 20th century. The discovery of the electron raised the intriguing idea that electric charge might also be quantized. 15-2 QuickLab 15-2 QuickLab Determining a Fundamental Quantity Problem Does a set of containers contain only identical items? Materials 5 sealed containers laboratory balance Procedure 1 Measure the mass of each container to a precision of 0.1 g. 2 Discuss with your partner how to record and present your data. 3 Pool your data with the rest of the class. Questions 1. Look for patterns in the pooled data. Discuss as a class the best way to tabulate and graph all of the data. How can you arrange the data to make it easier to analyze? 2. Consider the differences in mass between pairs of containers. Explain whether these differences indicate that the masses vary only by multiples of a basic unit. If so, calculate a value for this basic unit. 3. Can you be sure that the basic unit is not smaller than the value you calculated? Explain why or why not. 4. What further information do you need in order to calculate the number of items in each container? The American physicist Robert Andrews Millikan (1868–1953) and his graduate student Harvey Fletcher made the next breakthrough in the study of the properties of the electron. In 1909, Millikan reported the results from a beautiful experiment that determined the charge on the electron and showed that it was a fundamental unit of electrical charge. Millikan’s Oil-drop Experiment Millikan and Fletcher used an atomizer to spray tiny drops of oil into the top of a closed vessel containing two parallel metal plates (Figure 15.9). Some of the oil drops fell into the lower part of the vessel through a small hole in the upper plate. Friction during the spraying process gave some of the oil drops a small electric charge. Millikan also used X rays to change the charge on the oil drops. Since these oil drops were usually spherical, Millikan could calculate the mass of each drop from its diameter and the density of the oil. He connected a high-voltage battery to the plates, then observed the motion of the oil drops in the uniform electric field between the plates. By analyzing this motion and allowing for air resistance, Millikan calculated the electric force acting on each drop, and thus determined the charge on the drop. info BIT Millikan won the Nobel Prize in physics in 1923. In his Nobel lecture on the oil-drop experiment, he did not mention Fletcher’s work at all. Chapter 15 Electric force and energy quantization determine atomic structure. 761 15-PearsonPhys30-Chap15 7/24/08 3:59 PM Page 762 atomizer high-voltage battery oil drops microscope metal plates insulated from vessel wall Figure 15.9 Millikan’s oil-drop apparatus Millikan made numerous measurements with oil drops of various sizes. He found that the charged oil drops had either 1.6 1019 C of charge or an integer multiple of this value. Millikan reasoned that the smallest possible charge that a drop can have is the charge acquired either by gaining or losing an electron. Hence, the charge on the electron must be 1.6 1019 C. Millikan showed that charge is not a continuous quantity; it exists only in discrete amounts. This finding parallels Planck’s discovery in 1900 that energy is quantized (see section 14.1). Recent measurements have determined that the elementary unit of charge, e, has a value of 1.602 176 462 0.000 000 063 1019 C. A value of 1.60 1019 C is accurate enough for the calculations in this textbook. Note that a proton has a charge of 1e and an electron has a charge of 1e. Since Thomson and others had already determined the charge-tomass ratio for electrons, Millikan could now calculate a reasonably accurate value for the mass of the electron. This calculation showed that the mass of the electron is roughly 1800 times less than the mass of the lightest atom, hydrogen. Millikan and Controversy In the mid 1970s, historians of science made a disturbing discovery: Millikan had, on several occasions, stated that he used all of his data in coming to the conclusion that the charge of the electron was quantized. In fact, his notebooks contain 175 measurements of which he only reported 58! When all of Millikan’s data are used, his evidence for the quantization of charge is far less conclusive! Was Millikan guilty of scientific fraud, or was it a deeper, intuitive insight that led him to select only the data that clearly supported his claim that the charge on the electron is quantized? Historians will debate this question for many years to come, but we still acknowledge Millikan as the first person to measure the charge of the electron and to establish the quantization of charge. e SIM To see a simulation of Millikan’s oil-drop experiment, follow the links at www.pearsoned.ca/ school/physicssource. elementary unit of charge, e: the charge on a proton info BIT Thomson and others had tried to measure the charge on an electron using tiny droplets of water. However, evaporation and other problems made these measurements inaccurate. e MATH For a simple exercise to determine the fundamental unit of charge using a method similar to Millikan’s, visit www.pearsoned.ca/school/ physicssource. 762 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 763 Concept Check Why is it almost impossible to determine the mass of an electron without determining its charge first? Example 15.3 An oil drop of mass 8.2 1015 kg is suspended in an electric field of 1.0 105 N/C [down]. How many electrons has the oil drop gained or lost? Given 1.0 105 N/C [down] E m 8.2 1015 kg Required number of electrons gained or lost (n) Fe Fg , the charge on the oil drop must Analysis and Solution To balance the gravitational force, the electric force must be directed upward (Figure 15.10). Since the electric force is in the opposite direction to the electric field, E be negative. The oil drop must have gained electrons. For the electric and gravitational forces to balance, F their magnitudes must be equal. Since F g net 0, then 0 F and F net e F F g e mg qE F g F e . Figure 15.10 Practice Problems 1. How many electrons are gained or lost by a plastic sphere of mass 2.4 1014 kg that is suspended by an electric field of 5.0 105 N/C [up]? 2. What electric field will suspend an oil drop with a mass of 3.2 1014 kg and a charge of 2e? Answers 1. The sphere has lost three electrons. 2. 9.8 105 N/C [up] where q is the magnitude of the charge on the drop and g is the magnitude of the gravitational field. q Solving for q gives mg E (8.2 1015 kg)(9.81 m/s2) 1.0 105 N/C
8.04 1019 C The net charge on the oil drop equals the number of electrons gained times the charge on each electron. Thus, q q ne and n e Note that n must be a whole number. q n e 8.04 1019 C 1.60 1019 C 5 Paraphrase The oil drop has gained five electrons. Chapter 15 Electric force and energy quantization determine atomic structure. 763 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 764 In Example 15.3, you studied what happens when the electric force on a charged droplet exactly balances the gravitational force on the same droplet. What happens if these two forces do not balance each other? Example 15.4 Practice Problems 1. Calculate the net force on a sphere of charge 5e and mass 2.0 1014 kg when placed in an electric field of strength 1.0 105 N/C [up] in a vacuum chamber. 2. Calculate the acceleration of the sphere if the direction of the electric field in question 1 is reversed. Answers 1. 1.2 1013 N [down] 2. 14 m/s2 [down] A tiny plastic sphere of mass 8.2 1015 kg is placed in an electric field of 1.0 105 N/C [down] within a vacuum chamber. The sphere has 10 excess electrons. Determine whether the sphere will accelerate, and if so, in which direction. Given Choose up to be positive. 1.0 105 N/C [down] 1.0 105 N/C E m 8.2 1015 kg q 10e g 9.81 m/s2 [down] 9.81 m/s2 Fe Fg Required acceleration of the plastic sphere (a) Figure 15.11 (a) Analysis and Solution Express the charge on the sphere in coulombs: q 10e 10 (1.60 1019 C) 1.60 1018 C Draw a free-body diagram of the forces acting on the sphere (Figure 15.11 (a)). Because the charge is negative, the electric force is in the opposite direction to the electric field, E Calculate the magnitude of both the electric and gravitational forces acting on the sphere. . F g F e mg (8.2 1015 kg)(9.81 m/s2) 8.04 1014 N qE (1.60 1018 C)(1.0 105 N/C) 1.60 1013 N Determine the net force on the sphere. From Figure 15.11 (b), F net g e F F 8.04 1014 N (1.60 1013 N) 7.96 1014 N Now find the acceleration of the sphere: F net a ma F net m 7.96 1014 N 8.2 1015 kg Fe Fg Fnet 9.7 m/s2 Figure 15.11 (b) Paraphrase The sphere will accelerate upward at a rate of 9.7 m/s2. 764 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 765 15.2 Check and Reflect 15.2 Check and Reflect Knowledge Extensions 1. Explain the term quantization of charge. 7. A student attempting to duplicate 2. Which two properties of the electron was Millikan able to determine with the results from his oil-drop experiment? 3. Calculate the charge, in coulombs, on an oil drop that has gained four electrons. 4. Determine the electric force acting on an oil drop with a charge of 5e in a uniform electric field of 100 N/C [down]. Applications 5. (a) What is the net force acting on a charged oil drop falling at a constant velocity in the absence of an electric field? Explain your reasoning, using a free-body diagram to show the forces acting on the drop. (b) Describe the motion of the oil drop in an electric field that exerts an upward force greater than the gravitational force on the drop. Draw a diagram to show the forces acting on the drop. 6. An oil droplet with a mass of 6.9 1015 kg is suspended motionless in a uniform electric field of 4.23 104 N/C [down]. (a) Find the charge on this droplet. (b) How many electrons has the droplet either gained or lost? (c) In what direction will the droplet move if the direction of the electric field is suddenly reversed? Millikan’s experiment obtained these results. Explain why you might suspect that there was a systematic error in the student’s measurements. Droplet # Charge ( 1019 C 10 19.0 17.2 10.0 20.8 26.2 24.4 20.8 22.6 15.4 24.4 8. Some critics of Millikan have noted that he used only a third of his measurements when reporting the results of his oil-drop experiment. Use a library or the Internet to learn more about this issue. Explain whether you feel that Millikan was justified in presenting only his “best” data. e TEST To check your understanding of charge quantization, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 15 Electric force and energy quantization determine atomic structure. 765 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 766 15.3 The Discovery of the Nucleus By the beginning of the 20th century, the work of Thomson, Perrin, and others had provided strong evidence that atoms were not the smallest form of matter. Physicists then started developing models to describe the structure of atoms and devising experiments to test these models. 15-3 QuickLab 15-3 QuickLab Using Scattering to Measure a Hidden Shape Problem What can scattering reveal about the shape of an unseen target? Materials cardboard tube 10–15 cm in diameter small marbles or ball bearings carbon paper white letter-size paper marble or ball bearing cardboard tube with hidden target white paper carbon paper support block Figure 15.12 Procedure 1 Your teacher will prepare several “beam tubes” containing a hidden target. A cover blocks the top of each tube, except for a small opening (Figure 15.12). No peeking! 2 Work with a small group. Place a sheet of carbon paper, carbon side up, on the desk, and put a piece of white paper on top of the carbon paper. Then carefully set the cardboard tube on blocks on top of the paper. 3 Drop a marble or ball bearing through the opening at the top of the cardboard tube. Retrieve the marble or bearing, then drop it through the tube again, for a total of 50 times. 4 Remove the piece of paper, and look for a pattern in the marks left on it. Questions 1. Discuss with your group what the scattering pattern reveals about the shape and size of the target hidden in the cardboard tube. Sketch the likely shape of the target. 2. Explain how you can use your scattering results to estimate the dimensions of the target. 3. Compare your predictions to those made by groups using the other prepared tubes. 4. What is the smallest dimension that you could measure with the equipment in this experiment? 766 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 767 Rutherford’s Scattering Experiment Ernest Rutherford (1871–1937), a brilliant experimenter from New Zealand, was fascinated by radioactivity. By 1909, he had shown that some radioactive elements, such as radium and thorium, emitted positively charged helium ions, which are often called alpha () particles. Rutherford had also observed that a beam of these particles spread out somewhat when passing through a thin sheet of mica. So, he had two assistants, Hans Geiger and Ernest Marsden, measure the proportion of alpha particles scattered at different angles from various materials. Figure 15.13 shows the technique Rutherford devised for these measurements. A few milligrams of radium in a lead block with a small opening produced a pencil-shaped beam of alpha particles. The experimenters positioned a sheet of thin gold foil at right angles to the beam of alpha particles and used a screen coated with zinc sulfide to detect particles scattered by the foil. When an alpha particle struck the screen, the zinc sulfide gave off a faint flash of light, just enough to be visible through the microscope. By rotating the screen and microscope around the gold film, Geiger and Marsden measured the rates at which alpha particles appeared at various angles. beam of particles lead scattering angle zinc sulfide screen info BIT Rutherford was a professor at McGill University in Montreal from 1898 to 1907. During that time, he discovered an isotope of radon, showed that radioactive elements can decay into lighter elements, and developed a method for dating minerals and estimating the age of Earth. Although he considered himself a physicist, he cheerfully accepted the Nobel Prize for chemistry in 1908. microscope gold foil radium Figure 15.13 Rutherford’s scattering experiment Most of the alpha particles travelled through the foil with a deflection of a degree or less. The number of alpha particles detected dropped off drastically as the scattering angle increased. However, a few alpha particles were scattered at angles greater than 140, and once in a while an alpha particle would bounce almost straight back. Figure 15.14 shows the relationship between the number of scattered alpha particles and the angle at which they scattered. Rutherford was startled by these results. He knew that the deflections caused by attraction to the electrons in the gold atoms would be tiny because the alpha particles were fast-moving and roughly 8000 times heavier than electrons. He also calculated that deflection caused by repulsion of the alpha particles by the positive charge in the gold atoms would be no more than a degree if this charge were distributed evenly throughout each atom in accordance with Thomson’s model. So, Rutherford did not expect any alpha particles to be scattered at large angles. In a lecture describing the experiment, he said that seeing this scattering “was almost as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you!” Rutherford spent several weeks analyzing the scattering data. He concluded that the positive charge in a gold atom must be concentrated in an incredibly tiny volume, so most of gold foil was actually empty space! 107 106 105 104 103 102 10 ° 20° 40° 60° 80° 100° 120° 140° 160° Scattering Angle Figure 15.14 The scattering pattern observed by Geiger and Marsden Chapter 15 Electric force and energy quantization determine atomic structure. 767 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 768 Rutherford’s team then tried scattering alpha particles from other metals. Using data from scattering experiments with aluminium foil, Rutherford showed that the positive charge and most of the mass of an atom are contained in a radius of less than 1014 m. Rutherford had discovered the nucleus — and disproved the raisin-bun model. Concept Check On average, atoms have a radius of roughly 1010 m. Use Rutherford’s estimate for the size of the nucl
eus to calculate the proportion of the human body that is just empty space. The Planetary Model Rutherford’s discovery of the nucleus quickly led to the planetary model of the atom (Figure 15.15). In this model, the electrons orbit the nucleus much like planets orbiting the Sun. The electrostatic attraction between the positive nucleus and the negative electrons provides the centripetal force that keeps the electrons in their orbits. This model is also known as the solar-system, nuclear, or Rutherford model. To calculate the size of the nucleus, Rutherford applied the law of conservation of energy and an equation for electric potential energy. He knew that the electric potential energy that a charge q1 gains from the field around charge q2 is Ep kq1q2 d where k is Coulomb’s constant and d is the distance between the charges. This equation can be derived from Coulomb’s law using basic calculus. nucleus orbiting electron nucleus alpha particles Figure 15.15 The planetary model of the atom can explain the results of Rutherford’s scattering experiments. The extreme scattering of some alpha particles could only be explained by having most of the mass and positive charge of the atom concentrated in a very small nucleus. planetary model: atomic model that has electrons orbiting a nucleus 768 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 769 Example 15.5 In a scattering experiment, some alpha particles are scattered almost straight back from a sheet of gold foil. Each of these particles had an initial kinetic energy of 1.6 1012 J. The charge on an alpha particle is 2e, and the charge on a gold nucleus is 79e. Estimate the maximum possible size of a gold nucleus, given that the alpha particles do not hit the nucleus. Given Ek 1.6 1012 J q1 q 2e q2 qgold 79e Required radius of the nucleus of a gold atom (r) Analysis and Solution Since energy is conserved, the total kinetic and potential energy of an alpha particle does not change during scattering. As an alpha particle approaches a nucleus, the force of repulsion between the positive charges causes the alpha particle to slow down. This process converts the particle’s kinetic energy to electric potential energy until the kinetic energy is zero. Then the particle starts moving away from the nucleus and regains its kinetic energy. At the point where the alpha particle is closest to the nucleus, all of the particle’s kinetic energy has been converted to electric potential energy. The electric potential energy of the alpha particle is Ep Practice Problems 1. The charge on a tin nucleus is 50e. How close can an alpha particle with an initial kinetic energy of 1.6 1012 J approach the nucleus of a tin atom? 2. An iron nucleus has 56 protons. What is the electric potential energy of a proton located 5.6 1013 m from the centre of an iron nucleus? Answers 1. 1.4 1014 m 2. 2.3 1014 J kq1q2 d , where k is Coulomb’s constant and d is the distance between the alpha particle and the nucleus. The initial distance between the alpha particle and the nucleus is vastly larger than the distance between them when they are closest together. Therefore, the initial electric potential energy of the alpha particle is negligible for this calculation. Initially, kinetic energy electric potential energy 1.6 1012 J 0 When the alpha particle is closest to the nucleus, kinetic energy electric potential energy 0 kq1q2 d Since the total energy is conserved, Solving for d gives kq1q2 d 1.6 1012 J d kq1q2 1.6 1012 J 2 m N (2 1.60 1019 C)(79 1.60 1019 C) 8.99 109 2 C 1.6 1012 J 2.3 1014 m At its closest approach, the alpha particle is about 2.3 1014 m from the centre of the nucleus. Paraphrase The radius of a gold nucleus cannot be larger than 2.3 1014 m. Chapter 15 Electric force and energy quantization determine atomic structure. 769 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 770 15.3 Check and Reflect 15.3 Check and Reflect Knowledge 1. Explain how the results from Rutherford’s gold-foil experiment disproved Thomson’s model of the atom. 2. Briefly describe the planetary model of the atom. 7. In scattering experiments with aluminium foil, Rutherford found that the alpha particles observed at angles close to 180 did not behave like they had been scattered only by electrostatic repulsion. Rutherford thought these alpha particles might have actually hit an aluminium nucleus. 3. Find the potential energy of an alpha (a) Why did Rutherford see only particle that is (a) 1.0 1010 m from the centre of a gold nucleus (b) 1.0 1014 m from the centre of a gold nucleus electrostatic scattering when he used the same source of alpha particles with gold foil? (b) Rutherford used alpha particles with an energy of about 1.2 1012 J. Estimate the radius of an aluminium nucleus. 4. Why does the scattering angle increase as alpha particles pass closer to the nucleus? Extensions Applications 5. Why did Rutherford conclude that it was just the nucleus that must be extremely tiny in an atom and not the entire atom? 6. (a) By 1900, physicists knew that 1 m3 of gold contains approximately 6 1028 atoms. Use this information to estimate the radius of a gold atom. List any assumptions you make. (b) Compare this estimate to the estimate of the size of a gold nucleus in Example 15.5. 8. According to Rutherford’s calculations, the positive charges in an atom are packed tightly together in the nucleus. Why would physicists in 1900 expect such an arrangement to be highly unstable? What did Rutherford’s results suggest about forces in the nucleus? e TEST To check your understanding of atomic models, follow the eTEST links at www.pearsoned.ca/school/physicssource. 770 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 771 15.4 The Bohr Model of the Atom In 1912, Niels Bohr (1885–1962), a Danish physicist, studied for a few months at Rutherford’s laboratory. Both Bohr and Rutherford recognized a critical flaw in the planetary model of the atom. Experiments had shown that an accelerating charge emits electromagnetic waves, as predicted by the mathematical model for electromagnetism developed by the Scottish physicist James Clerk Maxwell (1831–1879). Electrons orbiting a nucleus are constantly accelerating, so they should emit electromagnetic waves. These waves would take energy from the orbiting electrons. As a result, the electrons in an atom should spiral into the nucleus in a few microseconds (Figure 15.16). But empirical evidence indicates that electrons do not spiral into their atomic nuclei. If they did, stable matter would not exist. Figure 15.16 According to Maxwell’s laws of electromagnetism, the orbiting electron should continuously radiate energy and spiral into the nucleus, which it does not do. info BIT Rutherford and Bohr shared an interest in soccer: Rutherford was a fan, and Bohr was an excellent player. Bohr thought Planck’s concept of quantized energy might provide a solution, and puzzled for months over how to fit this concept into a model of the atom. Then, a casual conversation with a colleague, his former classmate Hans Marius Hansen, gave Bohr the key to the answer. Hansen had recently returned from studying in Germany with an expert in spectroscopy. Hansen told Bohr that the wavelengths of the light in the spectrum of hydrogen have a mathematical pattern. No one had yet explained why this pattern occurs. Bohr found the explanation, and provided the first theoretical basis for spectroscopy. spectroscopy: the study of the light emitted and absorbed by different materials Spectroscopy A prism or diffraction grating can spread light out into a spectrum with colours distributed according to their wavelengths. In 1814, Josef von Fraunhofer (1787–1826) noticed a number of gaps or dark lines in the spectrum of the Sun. By 1859, another German physicist, Gustav Kirchhoff (1824–1887), had established that gases of elements or compounds under low pressure each have a unique spectrum. Kirchhoff and others used spectra to identify a number of previously unknown elements. Kirchhoff’s laws for spectra explain how temperature and pressure affect the light produced or absorbed by a material: info BIT The element names cesium and rubidium come from the Latin words for the colours of the spectral lines used to discover these elements. Cesium comes from cesius, which is Latin for sky blue. Rubidium comes from rubidus — Latin for dark red. Chapter 15 Electric force and energy quantization determine atomic structure. 771 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 772 emission line spectrum: a pattern of bright lines produced by a hot gas at low pressure absorption line spectrum: a pattern of dark lines produced when light passes through a gas at low pressure • A hot, dense material emits a continuous spectrum, without any dark or bright lines. • A hot gas at low pressure has an emission line spectrum with bright lines at distinct characteristic wavelengths. • A gas at low pressure absorbs light at the same wavelengths as the light it emits when heated. Shining white light through the gas produces an absorption line spectrum with dark lines that match the bright lines in the emission spectrum for the gas. Figure 15.17 illustrates these three types of spectra. Continuous Spectrum Bright-line Spectra Hydrogen Sodium Helium Neon Mercury Absorption Spectrum for Mercury (against a continuous spectrum) Figure 15.17 Continuous, emission line, and absorption line spectra 350 400 450 500 wavelength (nm) 550 600 650 15-4 Design a Lab 15-4 Design a Lab Emission Spectra of Elements The Question In what ways are emission line spectra characteristic of the elements that produce them? Design and Conduct Your Investigation Investigate the emission spectra produced by various elements. Here are some ways you can heat different elements enough to produce visible light: • Use a Bunsen burner to vaporize a small amount of an element. • Use commercially available discharge tubes containing gaseous elements. • Observe
forms of lighting that use a vaporized element, such as sodium or mercury arc lamps. Often, a diffraction grating is the simplest method for observing a spectrum. Check with your teacher if you need directions for using a diffraction grating. Note the overall colour of the light from each element, and sketch or photograph its emission spectrum. Extending Investigate absorption lines. You could try comparing the spectra of direct sunlight and sunlight that passes through clouds. Alternatively, you could shine white light at coloured solutions and diffract the light that passes through. Why are distinct absorption lines generally more difficult to produce than emission lines? 772 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 773 Improvements in the resolution of spectrometers made them a powerful analytic tool. For example, scientists have painstakingly matched the dark Fraunhofer lines in the solar spectrum (Figure 15.18) to the spectral patterns of dozens of elements, thus proving that these elements are present in the Sun’s atmosphere. However, at the start of the 20th century, the reasons why elements produce spectral lines were still a mystery. spectrometer: a device for measuring the wavelengths of light in a spectrum Fraunhofer line: a dark line in the spectrum of the Sun Figure 15.18 Fraunhofer lines are the dark lines in the visible part of the solar spectrum. The first hint that spectral lines were not just randomly spaced came in 1885. Johann Jacob Balmer, a Swiss mathematics teacher with an interest in numerology, found a formula for the wavelengths of the lines in the hydrogen spectrum. In 1890, Johannes Robert Rydberg generalized Balmer’s formula and applied it with varied success to other elements. Here is the formula for hydrogen: RH 1 1 22 1 n2 400 450 500 550 600 650 700 750 nm n 6 n 5 n 4 n 3 where RH is Rydberg’s constant for hydrogen, 1.097 107 m1, and n has the whole number values 3, 4, 5, . . . . The emission line spectrum of hydrogen is given in Figure 15.19. Hansen told Bohr about this formula. Bohr later remarked, “As soon as I saw Balmer’s formula, the whole thing was immediately clear to me.” Bohr had realized that the spectral lines corresponded to differences between quantized energy levels in the hydrogen atom. This concept was the foundation of a powerful new model of the atom. The Bohr Model of the Atom In 1913, Bohr published a paper suggesting a radical change to the planetary model of the atom. Here are the basic principles of Bohr’s model: • Electrons can orbit the nucleus only at certain specific distances from the nucleus. These distances are particular multiples of the radius of the smallest permitted orbit (Figure 15.20). Thus, the orbits in an atom are quantized. • Both the kinetic energy and the electric potential energy of an electron in orbit around a nucleus depend on the electron’s distance from the nucleus. So, the energy of an electron in an atom is also quantized. Each orbit corresponds to a particular energy level for the electron. • An electron can move from one energy level to another only by either emitting or absorbing energy equal to the difference between the two energy levels. An electron that stays in a particular orbit does not radiate any energy. Since the size and shape of the orbit remain constant along with the energy of the electron, the orbits are often called stationary states. Figure 15.19 Some of the bright lines in the spectrum of hydrogen e WEB To learn more about spectra, follow the links at www.pearsoned.ca/school/ physicssource. energy level: a discrete and quantized amount of energy stationary state: a stable state with a fixed energy level Chapter 15 Electric force and energy quantization determine atomic structure. 773 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 774 Bohr reasoned that Balmer’s formula shows that the energy of an orbiting electron depends inversely on the square of a quantum number n, now known as the principal quantum number. Bohr then used the equations for uniform circular motion, Coulomb’s law, and electric potential energy to derive expressions for the size of the hydrogen atom and the energy of the electron in the atom. principal quantum number: the quantum number that determines the size and energy of an orbit Figure 15.20 In Bohr’s model of the atom, electrons can orbit only at specific distances from the nucleus. r1 4r1 9r1 16r1 Orbit Sizes Bohr’s model of the hydrogen atom states that electrons can orbit the nucleus only at specific locations given by the expression: rn h2 42mke2 n2 where rn is the radius of the nth possible orbit for an electron and n is the principal quantum number, which can have the values 1, 2, 3, . . . . The other symbols in the equation all represent constants: k is Coulomb’s constant, h is Planck’s constant, e is the elementary charge, and m is the mass of the electron. By combining all the constants, this equation can be simplified to rn r1n2, where r1 h2 42mke2 5.29 1011 m Bohr radius: radius of the smallest orbit in a hydrogen atom The quantity r1 is known as the Bohr radius. It is the radius of the lowest possible energy level or ground state of the hydrogen atom. ground state: the lowest possible energy level Concept Check Sketch the first three orbits for a hydrogen atom to scale. Describe how the size of this atom changes as n increases. Energy Levels An orbiting electron has both kinetic energy and electric potential energy. By combining equations for kinetic and electric potential energies, Bohr derived this expression for En, the total energy of the electron in an energy level: 22mk2e4 1 h2 n2 En 774 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 775 As with the expression for the orbit radii, the constants can be com- bined to simplify the equation: En , where E1 E1 n2 22mk2e4 h2 2.18 1018 J or 13.6 eV E1 is the ground-state energy of the hydrogen atom. When n 1, the electron has greater energy and the atom is in an excited state. When n , the electron is no longer bound to the nucleus because E 0. Thus, Bohr’s model predicts that the ionization energy for hydrogen is 2.18 1018 J or 13.6 eV, corresponding to E1. Concept Check Does it make sense that the energy in the equation En 22m k2e4 1 h2 2 n is negative? Consider what you have to do to remove an electron from an atom. excited state: any energy level higher than the ground state ionization energy: energy required to remove an electron from an atom Example 15.6 How much energy must a hydrogen atom absorb in order for its electron to move from the ground state to the n 3 energy level? Given ninitial 1 nfinal 3 Required energy absorbed by atom (E) Analysis and Solution The energy the atom must absorb is equal to the difference between the two energy levels. The energy for each energy level is En E1 n2 2.18 1018 J n2 Practice Problems 1. How much energy does it take to move the electron in a hydrogen atom from the ground state to the n 4 energy level? 2. How much energy does the electron in a hydrogen atom lose when dropping from the n 5 energy level to the n 2 energy level? Answers 1. 2.04 1018 J 2. 4.58 1019 J E E3 E1 2.18 1018 J 32 1 (2.18 1018 J)1 9 1.94 1018 J 2.18 1018 J 12 Paraphrase The electron in a hydrogen atom requires 1.94 1018 J of energy to move from the ground state to the n 3 level. Chapter 15 Electric force and energy quantization determine atomic structure. 775 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 776 Energy Level Transitions and Line Spectra The Bohr model explains why absorption and emission line spectra occur for hydrogen and other elements. To jump to a higher energy level, an electron in an atom must gain energy. The atom can gain this energy by absorbing a photon. For energy to be conserved, the photon’s energy must match the difference between the electron’s initial energy level and the higher one. Recall from section 14.1 that the energy and frequency of a photon are related by the equation E hf. Thus, the atom can absorb only the frequencies that correspond to differences between the atom’s energy levels. Absorption of light at these specific frequencies causes the discrete dark lines in absorption spectra. Similarly, atoms can emit only photons that have energies corresponding to electron transitions from a higher energy level to a lower one. The arrows in Figure 15.21 illustrate all the possible downward transitions from the first six energy levels for hydrogen. Such energy level diagrams provide a way to predict the energies, and hence the wavelengths, of photons emitted by an atom Paschen series (infrared) Balmer series (visible) 0.38 0.54 0.85 1.5 3. 13.6 Lyman series (ultraviolet) n 1 Figure 15.21 The first six energy levels for hydrogen. Which arrow represents the transition that releases the most energy? 776 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 777 15-5 Inquiry Lab 15-5 Inquiry Lab An Emission Spectrum Question What is the energy difference between two energy states of atoms in a particular discharge tube? Hypothesis At a high temperature and low pressure, an atom emits light at the wavelengths predicted by the Bohr model. Variables • diffraction angle • wavelength Materials and Equipment a discharge tube high-voltage power supply diffraction grating 12 straight pins sheet of paper cardboard or foam-core sheet protractor and straightedge masking tape Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Procedure 1 Draw an x-axis and a y-axis that intersect in the middle of the sheet of paper. Then tape the paper to the cardboard or foam-core sheet (Figure 15.22). 2 Connect the high-voltage power supply to the discharge tube. 3 Switch on the power supply and look at the discharge tube through the diffraction grating. Orient the grating so that the spectral lines are vertical. You may want to darken the room to see the spectrum more clear
ly. 4 Align the bottom edge of the diffraction grating with the x-axis on the paper. Hold the grating vertical by pressing a straight pin into the y-axis on either side of the grating. 5 Sight along the y-axis, and turn the cardboard or foamcore base so that the y-axis points directly toward the discharge tube. Now tape the base to the tabletop. 6 For each spectral line, position a sighting pin so that it is on the line between the spectral line and the origin of your axes. Draw a line from the sighting pin to the origin. CAUTION! Keep well clear of the power supply connections when the power is on. Analysis high-voltage supply discharge tube sighting pins diffraction grating θ Figure 15.22 1. Decide how to record your data. You could use a table similar to the one below. Colour of Line Diffraction Angle, Wavelength (nm) 2. Use the diffraction formula n= d sin to calculate the wavelengths of the spectral lines (see section 13.5). Sometimes the spacing of the slits, d, is marked on the grating. If not, ask your teacher for this information. 3. Determine the energy difference between the two energy states. 4. Draw an energy level diagram showing the electron transition. Chapter 15 Electric force and energy quantization determine atomic structure. 777 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 778 When an electron in an atom absorbs a photon, the final energy level of the atom’s electron is greater than its initial energy level. From the law of conservation of energy, the energy of the photon can be expressed as: Ephoton Efinal Einitial level, n, is given by the formula En Since the energy of an electron in a hydrogen atom at a given energy , it is possible Einitial becomes (using algebra) to show that the expression Ephoton 22mk2e4 h2 Efinal 1 n2 1 RH 1 n2 final 1 n2 initial This result is impressive! Bohr’s model not only explains spectral lines, but leads to a generalized form of Balmer’s formula, and predicts the value of Rydberg’s constant for hydrogen. Example 15.7 Find the wavelength of the light emitted by a hydrogen atom when its electron drops from the n 3 to the n 2 energy level. Practice Problems 1. Find the wavelength of the light emitted by a hydrogen atom when its electron drops from the n 5 to the n 2 energy level. 2. Find the wavelength of light that a hydrogen atom will absorb when its electron moves from the n 3 to the n 7 energy level. Answers 1. 434 nm 2. 1005 nm Given ninitial 3 Required wavelength () nfinal 2 Analysis and Solution Simply substitute the known values for n into the wavelength equation for an emitted photon: 1 1 n2 RH RH 1 n2 initial final 1 22 1 32 (1.097 107 m1) 1 9 1.524 106 m1 1 1.524 106 m1 1 4 6.563 107 m Paraphrase The atom emits light with a wavelength of 656.3 nm. 778 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 779 Concept Check What features do the Bohr model of the atom and the planetary model have in common? In what critical ways do these two models differ? The Northern Lights and the Emission Line Spectrum of Oxygen Alberta skies often display one of nature’s most beautiful phenomena — the aurora borealis, or northern lights, which you first studied in Chapter 12. At altitudes between 100 km and 400 km above the surface of Earth, high-energy electrons, trapped by Earth’s magnetic field, interact with oxygen and nitrogen atoms. During these interactions, the electrons in these atoms are excited and move into higher energy levels. Eventually, the excited electrons return to their ground states. In doing so, they emit light that forms the characteristic colours of the aurora borealis. Figure 15.23 shows a display of the aurora borealis above northern Alberta. You can use the quantized energy levels to help explain the characteristic green colour of the aurora (as well as the subtler shades of red and blue). (a) Figure 15.23 The photo in (a) shows a bright, mostly green aurora. The green colour is due to an energy level transition in oxygen atoms in Earth’s atmosphere. The diagram in (b) shows some of the energy levels in the oxygen atom, including the one that produces the green light that is mostly seen in the aurora. In this diagram, the ground state energy equals 0 eV. (b) ) V e ( 4.17 1.96 1.90 λ 557.7 nm Ground state (0 eV) From Figure 15.23(b), the green colour of the aurora occurs when an 4.17 eV to a lower excited electron drops from an energy level Einitial Eintial energy level Efinal and E 1.96 eV. Using the equations E Efinal , you can show that this change in energy level produces hc the colour you see. info BIT The emission spectra of elements act like nature’s fingerprints. By vaporizing a small sample and looking at the emission lines produced, you can determine the chemical composition of a substance. This technique, called atomic emission spectroscopy, is used routinely in forensics labs around the world. Chapter 15 Electric force and energy quantization determine atomic structure. 779 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 780 Concept Check Use the energy level diagram for oxygen, given in Figure 15.23(b), to 1.96 eV will verify that a transition from Einitial produce green light ( 558 nm). Two other possible transitions are also shown in the diagram. What colours would these transitions produce? Determine their wavelengths. 4.17 eV to Efinal 15.4 Check and Reflect 15.4 Check and Reflect Knowledge 1. Explain what the phrase “quantized process” means. 2. Sketch the first five orbits in a hydrogen atom. Indicate on your sketch which transitions cause the blue, green, and red lines shown in Figure 15.21. 3. List three quantities predicted by Bohr’s model of the atom. 4. Why do electrons in hydrogen atoms emit infrared light when they make transitions to the n 3 energy level, and ultraviolet light when they make transitions to the n 1 energy level? Applications 5. The wavelengths of the first four visible lines in the hydrogen spectrum are 410, 434, 486, and 656 nm. (a) Show that Balmer’s formula predicts these wavelengths. (b) Which of the wavelengths corresponds to a transition from the n 4 energy level to the n 2 energy level? (c) Use this wavelength to calculate the energy difference between the n 4 and the n 2 stationary states. 6. A helium-neon laser produces photons of wavelength 633 nm when an electron in a neon atom drops from an excited energy state to a lower state. What is the energy difference between these two states? Express your answer in electron volts. 7. The diagram shown below shows some of the energy levels for the lithium atom. The designations 2s, 2p, etc., are a common notation used in spectroscopy. (a) Without doing any calculations, sort the four transitions shown from shortest wavelength to longest wavelength. Explain your reasoning. (b) Estimate the energy of each of the four energy levels shown. (c) Calculate the wavelengths produced in these transitions. Indicate which ones are in the visible part of the spectrum, along with their colours. 3p 3s 3 2p 0 1 2 V e 3 4 5 6 3d 2 4 2s 1 780 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 781 Hydrogen Sodium Solar (selected lines) 400 450 500 550 600 650 700 750 nm 400 450 500 550 600 650 700 750 nm 400 450 500 550 600 650 700 750 nm 8. What can you conclude about the Determine: composition of the Sun from the spectra given above? Explain your reasoning. 9. (a) Find the difference in energy between the n 2 and n 3 energy levels in hydrogen. (b) Find the energy difference between the n 5 and n 6 energy levels in hydrogen. (c) What happens to the energy difference between successive orbits as the distance from the nucleus increases? 10. The ionization energy for an atom is the energy required to remove an electron completely from an atom. Show why the ionization energy for hydrogen is equal to E1. (Hint: Consider going from the ground state to an energy level with n 1000.) 11. When an atom’s energy levels are closely spaced, the atom “de-excites” by having one of its electrons drop through a series of energy levels. This process is called fluorescence and is often seen when a high-energy photon, such as an X ray or a UV photon, excites an atom, which then de-excites through a series of longerwavelength emissions. A common example of fluorescence occurs in the colours produced when rocks and minerals containing mercury and many other elements are illuminated by UV photons. Below is the energy level diagram for some of the energy levels in mercury (Hg). The ground state has been assigned an energy of 0 eV. (a) the wavelength of the photon needed to excite the mercury atom from its ground state to the n 5 energy level (b) the longest wavelength of photon that will be emitted as the mercury atom de-excites (c) the number of possible downward transitions that can occur (d) If an atom had 100 possible energy transitions that were very close together, what would the spectrum produced by this atom look like Hg Extension 9.23 eV 8.85 eV 7.93 eV 6.70 eV 4.89 eV 0 eV 12. A laser produces light that is monochromatic, coherent, and collimated. (a) Explain each of these three properties. (b) Describe the spectrum of a laser. e TEST To check your understanding of Bohr’s model of the atom, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 15 Electric force and energy quantization determine atomic structure. 781 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 782 15.5 The Quantum Model of the Atom Bohr’s model explains the spectral lines of hydrogen and accurately predicts the size and ionization energy of the hydrogen atom. Despite these remarkable accomplishments, the theory has several serious failings: • It does not really explain why energy is quantized, nor why orbiting electrons do not radiate electromagnetic energy. • It is not accurate for atoms that have two or more electrons. • It does not explain why a magnetic field splits the main spectral lines into multiple closely spaced lines. The Dutch physicist Pieter Zee
man discovered this effect in 1896. It is known as the Zeeman effect. Physicists solved these problems within 15 years, but the solutions were even more radical than Bohr’s theory! The Wave Nature of Electrons In 1924, Louis de Broglie developed his theory that particles have wave properties. As described in section 14.4, diffraction experiments confirmed that electrons behave like waves that have the wavelength predicted by de Broglie. So, the principles of interference and standing waves apply for electrons orbiting a nucleus. For most sizes of orbit, successive cycles of the electron wave will be out of phase, and destructive interference will reduce the amplitude of the wave (Figure 15.24). For constructive interference to occur, the circumference of the orbit must be equal to a whole number of wavelengths: 2rn n where n is a positive integer, is the electron wavelength, and rn is the radius of the nth energy level. By substituting de Broglie’s definition for wavelength, h mv , into the above equation, the condition for constructive interference becomes 2rn nh mv or mvrn nh 2 This condition is fundamentally the same one that Bohr found was necessary for the energy levels in his model of the atom. Thus, the wave nature of matter provides a natural explanation for quantized energy levels. Figure 15.25 shows the standing waves corresponding to the first three energy levels in an atom. Note that the de Broglie wavelength is longer in each successive energy level because the electron’s speed decreases as the radius of the orbit increases. PHYSICS INSIGHT Recall from section 13.5 that for constructive interference to occur, the path difference between waves must be a whole number of wavelengths, or n. constructive interference destructive interference Figure 15.24 A standing wave is possible only if a whole number of electron wavelengths fit exactly along the circumference of an orbit. 782 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 783 n 1 n 2 n 3 Figure 15.25 Standing waves in the first three energy levels, where 2r1 2r2 2, and 2r3 1, 3 In 1926, Erwin Schrödinger (1887–1961) derived an equation for determining how electron waves behave in the electric field surrounding a nucleus. The solutions to Schrödinger’s equation are functions that define the amplitude of the electron wave in the space around a nucleus. Max Born (1882–1970) showed that the square of the amplitude of these wave functions at any point is proportional to the probability of finding an electron at that point. Each wave function defines a different probability distribution or orbital. info BIT Although Born won a Nobel Prize for his work on quantum theory, Schrödinger never accepted Born’s interpretation of electron wave functions. orbital: probability distribution of an electron in an atom Quantum Indeterminacy Unlike the Bohr model, the quantum model does not have electrons orbiting at precisely defined distances from the nucleus. Instead, the electrons behave as waves, which do not have a precise location. The orbitals in the quantum model show the likelihood of an electron being at a given point. They are not paths that the electrons follow. The idea that electrons within an atom behave as waves rather than as orbiting particles explains why these electrons do not radiate electromagnetic energy continuously. Some physicists, including Einstein and Schrödinger, had difficulty accepting a quantum model that could predict only probabilities rather than clearly defined locations for electrons in an atom. As Niels Bohr noted, “Anyone who is not shocked by quantum theory has not understood a single word.” Despite its challenging concepts, quantum theory is the most comprehensive and accurate model of atoms and molecules yet developed. Concept Check Soon after Bohr’s model was published, physicists discovered that the spectral lines in hydrogen and other elements were not distinct, but could themselves be split into numerous, very closely spaced spectral lines. How does the splitting of spectral lines show that Bohr’s concept of energy levels is incomplete? Chapter 15 Electric force and energy quantization determine atomic structure. 783 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 784 15.5 Check and Reflect 15.5 Check and Reflect Knowledge Extensions 1. Describe three failings of the Bohr model. 2. What is the Zeeman effect? 3. What is an orbital? Applications 4. (a) Find the de Broglie wavelength for an electron in the ground state for hydrogen. Recall that the ground-state radius for hydrogen is 5.29 1011 m. (b) Find the momentum of the electron in part (a). (Hint: Use the formula for the de Broglie wavelength.) (c) Find the kinetic energy and speed for the electron in part (a). 5. (a) Express the de Broglie wavelength of the n 2 and n 3 energy levels of an atom in terms of 1, the de Broglie wavelength for the n 1 energy level. (Hint: Apply Bohr’s equation for the orbital radii.) (b) Make a scale diagram or model of the atom showing the electron waves for the first three energy levels. 6. In more advanced treatments of quantum mechanics, the wave solutions to the Schrödinger equation can describe complex-looking orbitals around atoms. These solutions will sometimes have nodes. As you learned in Chapter 13, a node is a location where waves combine destructively and have zero amplitude. According to Born’s interpretation of a wave function, what meaning would you give to a node in a wave function? 7. Radio astronomers use the 21-cm line to study hydrogen gas clouds in our galaxy. Use a library or the Internet to research this spectral line. What unusual transition causes the 21-cm line? Why does this transition occur naturally in hydrogen in deep space, but not in hydrogen on Earth? e TEST To check your understanding of the quantum model, follow the eTEST links at www.pearsoned.ca/school/physicssource. 784 Unit VIII Atomic Physics 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 785 CHAPTER 15 SUMMARY Key Terms and Concepts cathode ray elementary unit of charge planetary model spectroscopy emission line spectrum absorption line spectrum spectrometer Fraunhofer line energy level stationary state principal quantum number Bohr radius ground state excited state ionization energy orbital Key Equations Velocity selection: no deflection if v E B r1n2, where r1 Emission lines: Ephoton Efinal Einitial and RH 1 1 n2 final 1 n2 initial Quantum model: mvrn nh 2 Bohr model for hydrogen: rn 5.29 1011 m , where E1 2.18 1018 J En E1 n2 , where RH 1.097 107 m1 Conceptual Overview Summarize the chapter by copying and completing this concept map. mass and charge measured by discovery of electron showed quantization of charge led to raisin-bun model disproved by led to solar-system model predicts instability led to is basis of reveals energy quantization led to failings led to quantum model Chapter 15 Electric force and energy quantization determine atomic structure. 785 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 786 CHAPTER 15 REVIEW Knowledge 1. (15.1) What is a cathode ray? What type of electric charge does a cathode ray carry? 2. (15.1) (a) Describe the force acting on a cathode ray moving to the right in an electric field directed down (toward the bottom of the page). (b) Describe the force acting on a cathode ray moving to the right in a magnetic field directed down (toward the bottom of the page). (c) How could the electric and magnetic fields be directed so that the net force on the cathode ray is zero? 3. (15.1) The glowing gas shows the path of the electrons in this gas discharge tube. How can you tell that the beam of electrons is travelling through a magnetic field? 4. (15.1) Why was Thomson’s experiment able to determine only the charge-to-mass ratio for an electron? 5. (15.2) Explain how the results of Millikan’s oil-drop experiment also enabled physicists to determine the mass of the electron. 6. (15.2) What is the electrical charge on a dust particle that has lost 23 electrons? 7. (15.2) Calculate the electrical charge carried by 1.00 kg of electrons. 8. (15.3) What is an alpha particle? 9. (15.3) Why is the Rutherford gold-foil experiment sometimes called a “scattering experiment”? 786 Unit VIII Atomic Physics 10. (15.3) Explain how Thomson’s model of the atom was inconsistent with the results of Rutherford’s gold-foil experiment. 11. (15.4) (a) What is an emission line spectrum? (b) How could you produce an emission line spectrum? 12. (15.4) What are Fraunhofer lines? 13. (15.4) Explain how emission line spectra demonstrate Planck’s concept of energy quantization. 14. (15.4) What is the difference between the ground state of an atom and an excited state? 15. (15.4) Here are four energy-level transitions for an electron in a hydrogen atom: 4 → nf 8 → nf ni ni ni ni (a) For which of these transition(s) does the 1 → nf 2 → nf 5 5 3 3 atom gain energy? (b) For which transition does the atom gain the most energy? (c) Which transition emits the photon with the longest wavelength? 16. (15.4) Calculate the radius of a hydrogen atom in the n 3 state. 17. (15.5) Describe the difference between an orbit in Bohr’s model of the atom and an orbital in the quantum model. 18. (15.5) How does the quantum model explain why electrons in an atom do not continuously radiate energy? Applications 19. An electron is moving at a speed of 1.0 km/s perpendicular to a magnetic field with a magnitude of 1.5 T. How much force does the magnetic field exert on the electron? 20. (a) Use the Bohr model to predict the speed of an electron in the n 2 energy level of a hydrogen atom. (b) Explain why the quantum model can predict the energy of this electron, but not its speed. 15-PearsonPhys30-Chap15 7/24/08 4:00 PM Page 787 21. Calculate the electric field that will suspend an oil droplet that has a mass of 2.0 1015 kg and a charge of 3e. 22. Transitions to or from the n 3 energy level produce the Paschen series of lines in the hydrogen spectrum. (a) Find the
change in energy level for the first three Paschen transitions. (b) Find the wavelengths and frequencies of the first three Paschen transitions. (c) Use E hf to calculate the energy of the photon produced by a transition from the n 5 to the n 3 energy level. Is your calculation consistent with your answer to part (a)? Why or why not? (d) In what part of the electromagnetic spectrum would you find the Paschen lines? 23. The helium-neon laser produces a red-coloured light by exciting a gas that contains a mixture of both helium and neon atoms. The energy level diagram below shows three transitions, A, B, and C, that are involved. Two of these transitions are produced by collisions with other atoms or electrons, and the third is the result of photon emission. (a) Explain which process is involved in transitions A, B, and C. (b) For the transition that produces a photon, determine the wavelength of the photon. collision transfers energy 20.61 eV A 20.66 eV B 18.70 eV C ground state helium neon 24. Determine the electric field that will stop this alpha particle from being deflected as it travels at 10 km/s through a 0.25-T magnetic field. B 0.25 T [out of page] v 10 km/s [right] Extensions 25. Classical electromagnetic theory predicts that an electron orbiting a nucleus of charge q will radiate energy at a rate of P 2kq2a2 3c3 , where k is Coulomb’s constant, a is the electron’s acceleration, and c is the speed of light. (a) Determine the kinetic energy of the electron in the ground state of a hydrogen atom. (b) Use the Bohr model to calculate the acceleration of an electron in the ground state of a hydrogen atom. (Hint: Apply the equations for circular motion.) (c) Show that P has the units of energy divided by time. (d) How long will it take the electron to give off all of its kinetic energy as electromagnetic radiation, assuming that the electron’s acceleration remains constant? (e) Explain how your answer to part (d) shows that classical models of the atom are invalid. Consolidate Your Understanding 1. Explain how the Rutherford gold-foil experiment radically changed the understanding of atomic structure. 2. Explain how Bohr linked spectral lines to Planck’s idea of energy quantization. 3. Outline the successes and failures of the Bohr model. 4. Describe three fundamental differences between the quantum model of the atom and the Bohr model. Think About It Review your answers to the Think About It questions on page 753. How would you answer each question now? e TEST To check your understanding of atomic structure, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 15 Electric force and energy quantization determine atomic structure. 787 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 788 Nuclear reactions are among the most powerful energy sources in nature. “I believe a leaf of grass is no less than the journey-work of the stars” — from Leaves of Grass by Walt Whitman When the American poet Walt Whitman wrote this line in 1855, the reactions within stars were unknown, the nucleus had not been discovered, and there was no clear proof that atoms exist. By the late 1950s, however, a new interpretation of Whitman’s words was possible. Astrophysicists had developed a theory that nuclear reactions inside massive stars created the heavy elements essential to life. Some of these stars exploded into supernovae, scattering heavy elements throughout the galaxy. This chapter describes nuclear reactions, the enormous potential energy in some nuclei, and the hazards and benefits of radioactive materials. You will learn about the processes that power the stars and how every leaf of grass may indeed be “the journey-work of the stars.” C H A P T E R 16 Key Concepts In this chapter, you will learn about: half-life nuclear decay nuclear reactions Learning Outcomes When you have completed this chapter, you will be able to: Knowledge describe the nature and properties of nuclear radiation write nuclear equations for alpha, beta-negative, and beta-positive decays perform half-life calculations use conservation laws to predict the particles emitted by a nucleus compare and contrast fission and fusion reactions relate the mass defect of the nucleus to the energy released in nuclear reactions Science, Technology, and Society explain that the goal of science is knowledge about the natural world explain that technology meets given needs but should be assessed for each potential application Figure 16.1 A portion of the Cygnus Loop, an expanding cloud of hot gas formed by a supernova explosion about 15 000 years ago. This composite image was made using photographs from the Hubble Space Telescope. The blue colour is light from oxygen, green is light from hydrogen, and red is light from sulfur. 788 Unit VIII 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 789 16-1 Inquiry Lab 16-1 Inquiry Lab Radiation Intensity Question Does the intensity of radiation depend on the distance from the source of the radiation? Hypothesis There is a mathematical relationship between the intensity of radiation and the distance from the radiation source. Variables • distance between radiation source and detector • reading on radiation detector Materials and Equipment cobalt-60 radiation source radiation detector metre-stick masking tape optional: interface for computer or graphing calculator CAUTION: The radioactive material is enclosed in a durable casing to prevent accidental absorption into the body. Do not damage this casing. Procedure 1 Make sure the radiation source is at least 3 m away from the radiation detector. Switch on the detector and measure the background radiation level for 5 min or more. Record this measurement, including the units. If you are using an interface with a computer or graphing calculator, check with your teacher about recording your data electronically. 2 Centre the cobalt-60 radiation source over the zero mark on the metre-stick, and tape the source in place. If your radiation source is shielded so that it emits radiation only from one side, align the source to direct the radiation along the metre-stick (Figure 16.2). Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork 3 Place the radiation detector on the metre-stick within a few centimetres of the radiation source. Measure the radiation level for at least 1 min. Record the radiation level and the distance between the source and the detector. 4 Increase the separation between the radiation source and the detector in steps of 5 cm. Measure the radiation level for at least 1 min at each distance. Record measurements for at least six distances. radiation source radiation detector metre-stick Figure 16.2 Analyzing and Interpreting 1. Which variable is the manipulated variable in this experiment? 2. Explain why you need to know the background radiation level in order to determine how the intensity of the radiation varies with distance. 3. Graph your data. What type of relationship do you think the graph shows? 4. Discuss with your lab partners how you could use a different graph to determine the exact relationship between the radiation intensity and the distance from the radiation source. Produce a graph using the method that you think will work best. Explain your choice. 5. List any assumptions you made when analyzing your data. Forming Conclusions 6. Do your data support the hypothesis? Explain. Think About It 1. What is radioactivity? 2. Where does the energy released in a nuclear reaction come from? 3. How can stars create elements? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 789 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 790 16.1 The Nucleus Section 15.3 described how scattering experiments directed by Rutherford showed that more than 99.9% of the mass of an atom is concentrated in a nucleus that is typically only a few femtometres (1015 m) in diameter. In 1918, Rutherford began a new series of experiments in which he bombarded nitrogen gas with alpha particles. He found that some of the nitrogen transmuted into oxygen and that the process also produced hydrogen nuclei. Rutherford concluded that the hydrogen nucleus was a fundamental particle that is a constituent of all nuclei. He called these particles protons, from protos, the Greek word for “first.” However, protons could not account for all of the mass of nuclei. For example, the charge-to-mass ratio for protons is twice that of helium nuclei. In 1920, Rutherford suggested that nuclei might also contain neutrons, neutral particles with about the same mass as a proton. Neutral particles are difficult to detect or measure because they do not interact with electric or magnetic fields. A variety of experiments over the next decade failed to find any neutrons. The breakthrough came in 1932 when James Chadwick showed that alpha rays striking a beryllium target produced radiation consisting of neutral particles. In a similar experiment with a boron target, he determined that the mass of a neutron is about 0.1% greater than the mass of a proton. femto: metric prefix meaning 1015 proton: a positively charged particle found in all nuclei neutron: a neutral particle found in nuclei info BIT Chadwick made two earlier attempts to discover the neutron, in 1923 and 1928. In 1935, he received the Nobel Prize in physics for his discovery. Nuclear Terms and Notation nucleon: a proton or neutron Protons and neutrons are called nucleons because they are both components of nuclei. Three numbers describe the composition of a nucleus: Atomic Number, Z: the number of protons in a nucleus Neutron Number, N: the number of neutrons in the nucleus Atomic Mass Number, A: the number of nucleons in the nucleus, Z N Scientists often indicate the
composition of a nucleus with the notation A ZX, where X is the chemical symbol for the element. For example, a carbon nucleus with 6 protons and 6 neutrons has Z 6, N 6, and A 6 6 12. The notation for the carbon nucleus is 12 6C. Apply these terms and concepts in the next example. 790 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 791 Example 16.1 How many neutrons are contained in a gold nucleus 197 79Au? Given Z 79 A 197 Required neutron number (N ) Analysis and Solution Since A Z N, N A Z 197 79 118 Paraphrase There are 118 neutrons in a nucleus of 197 79Au. Concept Check Practice Problems 1. How many neutrons are in a nucleus of 24 12Mg? 2. Find the atomic mass number for a uranium atom that contains 92 protons and 146 neutrons. Answers 1. 12 2. 238 How do the nuclei 12 6C, 13 6C, and 14 6C differ? How are they the same? Isotopes Many elements have two or more isotopes — forms that have the same number of protons (Z) but differing numbers of neutrons (N). For example, ordinary hydrogen (1 1H) are all isotopes of the element hydrogen. Specific isotopes can be indicated by the element name and the atomic mass number. For example, carbon-12 is another way of writing 12 1H), and tritium (3 1H), deuterium (2 6C. isotopes: atoms that have the same number of protons, but different numbers of neutrons All the isotopes of a particular element have the same number of protons and electrons. So, these isotopes have almost identical chemical properties. However, the physical properties can differ dramatically. In particular, one isotope of an element may be highly radioactive, while another is quite stable. Bombarding materials with electrons, neutrons, or other particles can create radioactive isotopes. Atomic Mass Units Atoms and nuclei are much, much smaller than everyday objects. So, even though a kilogram may be a convenient unit for expressing the mass of apples or oranges, it is not particularly useful for measuring the mass of a proton or a carbon nucleus. For calculations involving nuclei and subatomic particles, it is often convenient to use a mass unit that is much smaller than the kilogram. The atomic mass unit (u) is defined as exactly 1 12 of the mass of the carbon-12 atom: 1 u 1.660 539 1027 kg Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 791 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 792 Table 16.1 lists the masses of electrons and nucleons, in both kilograms and atomic mass units. Table 16.1 Some Properties of Subatomic Particles (to Six Decimal Places) Particle Electron Proton Neutron Charge (C) Mass (kg) Mass (u) 1.602 177 1019 1.602 177 1019 0 9.109 383 1031 1.672 622 1027 1.674 927 1027 5.485 799 104 1.007 276 1.008 665 Forces in the Nucleus Aside from hydrogen, all nuclei consist of two or more protons and a number of neutrons (Figure 16.3). Like charges repel each other, so what keeps these nuclei from flying apart? Example 16.2 Can gravitational force bind two protons in a nucleus together? Given Rounding the values listed in Table 16.1 gives proton mass m 1.67 1027 kg and proton charge q 1.60 1019 C. Required Determine if gravitational force can bind two protons in a nucleus together. Figure 16.3 Any nucleus heavier than hydrogen has protons and neutrons packed closely together. Practice Problems 1. Calculate the gravitational force that two protons exert on each other when they are 5 fm apart. 2. Calculate the electrostatic force that two protons exert on each other when they are 5 fm apart. Answers 1. 7 1036 N 2. 9 N Analysis and Solution Compare the gravitational and electrostatic forces between two protons in a nucleus. The magnitude of the gravitational force is F . g The magnitude of the electrostatic force is F e Fg Fe So, m2 Gm 1 2 r q2 kq 1 2 r Gm1m2 kq1q2 . Gm1m2 r 2 kq1q2 r 2 . This ratio shows that the relative strength of the two forces does not depend on the distance between the protons. In order for the gravitational attraction between the protons to overcome the electrostatic repulsion, the ratio have to be greater than 1. Fg Fe would Substituting the known values into the ratio of the forces gives Fg Fe (6.67 1011 Nm2/kg2)(1.67 1027 kg)2 (8.99 109 Nm2/C2)(1.60 1019 C)2 8.08 1037 Paraphrase The gravitational attraction is vastly weaker than the electrostatic repulsion, so gravity cannot be the force that holds a nucleus together. 792 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 793 strong nuclear force: the force that binds together the protons and neutrons in a nucleus PHYSICS INSIGHT Measurements of interactions between subatomic particles suggest that there is a fourth fundamental force, the weak nuclear force. This force acts on electrons. binding energy: the net energy required to liberate all of the protons and neutrons in a nucleus Since gravity is far too weak, there must be some other force that holds the particles in a nucleus together. Physicists call this force the strong nuclear force, and think that it is a fundamental force of nature, like gravity and the electrostatic force. The strong nuclear force has a very short range. Although it is more powerful than the electrostatic force within a nucleus, the strong nuclear force has a negligible effect on particles that are more than a few femtometres apart. The strong nuclear force acts on both neutrons and protons, but does not affect electrons. Chapter 17 describes fundamental forces in more detail. Binding Energy and Mass Defect Removing a nucleon from a stable nucleus requires energy because work has to be done on the nucleon in order to overcome the strong nuclear force. The binding energy, Eb, of a nucleus is the energy required to separate all of its protons and neutrons and move them infinitely far apart. In other words, the binding energy is the difference between the total energy of the separate nucleons and the energy of the nucleus with the nucleons bound together: Eb Enucleons Enucleus where Enucleons is the sum of the energies of the nucleons when they are free of the nucleus and Enucleus is the energy of the nucleus. Mass-energy Equivalence The equivalence of mass and energy is part of the theory of relativity that Albert Einstein developed in 1905. This theory correctly predicted that mass and energy are related by the equation E mc2 where E is energy, m is mass, and c is the speed of light. Earlier in this section, you learned that physicists commonly use the atomic mass unit, u, for calculations involving nuclei and subatomic particles. For nuclear calculations, it is useful to know the energy equivalent for 1 u: E 1 u c2 (1.660 539 1027 kg)(2.997 925 108 m/s)2 1.492 418 1010 J 1.492 418 1010 J 931.494 1 MeV 1 eV 1.602 177 1019 J Thus, 1 u is equivalent to about 149.2 pJ or 931.5 MeV. The binding energy of most nuclei is equivalent to only a small fraction of an atomic mass unit. Nuclear reactions can involve conversions between mass and energy. The law of conservation of energy still applies if the conversions are taken into account. For any closed system, the total of the energy and the energy equivalent of the mass in the system is constant. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 793 16-PearsonPhys30-Chap16 7/28/08 10:14 AM Page 794 Example 16.3 Calculate the energy equivalent for 0.0034 u of mass, in joules and in electron volts. Analysis and Solution Simply multiply 0.0034 u by the appropriate equivalence factors: 0.0034 u 1.492 1010 J 1 u 5.1 1013 J 0.0034 u 931.5 MeV 1 u 3.2 MeV Paraphrase The energy equivalent for 0.0034 u is 5.1 1013 J or 3.2 MeV. Practice Problems 1. Find the energy equivalent, in electron volts, for 0.221 u. 2. Find the mass equivalent to 250 MeV. Answers 1. 206 MeV 2. 0.268 u Mass Defect Rearranging Einstein’s equation for mass-energy equivalence gives m E c2 . Dividing the equation for binding energy by c2 leads to a formula for the mass defect, m, of a nucleus: Enucleons c2 Eb c2 m mnucleons Enucleus c2 mnucleus where mnucleons is the sum of the masses of the separate nucleons and mnucleus is the mass of the nucleus. Thus, the mass of a nucleus is equal to the total mass of its constituents, less the mass corresponding to the binding energy. Physicists have determined the masses of atoms and nucleons with great accuracy. Tables of atomic data generally list the masses of neutral atoms rather than the masses of nuclei alone without any electrons. The following formula uses atomic masses to calculate the mass defect for a nucleus: m Zm1 1H Nmneutron matom where m1 number, and N is the neutron number. 1H is the mass of a neutral hydrogen atom, Z is the atomic Since m1 1H includes the masses of both a proton and an electron, the 1H includes the mass of Z electrons, matching the mass of the term Zm1 electrons included in matom. The differences in the binding energy of the electrons are small enough to ignore in most nuclear calculations. mass defect: difference between the sum of the masses of the separate nucleons and the mass of the nucleus PHYSICS INSIGHT Nuclear calculations often involve very small differences in mass. Such calculations can require data with six or more significant digits. 794 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 795 Concept Check Show that m Zmproton Nmneutron mnucleus. Example 16.4 Find the mass defect, expressed in kilograms, and the binding energy for a carbon-12 nucleus. Given Z 6 A 12 m 12.000 000 u Required mass defect (m) binding energy (Eb) Analysis and Solution The formula N A Z gives the number of neutrons in the nucleus: N 12 6 6 Practice Problems 1. Sodium 23 11Na has an atomic mass of 22.989 769 u. Find the mass defect for this nucleus. 2. Find the binding energy for 23 11Na. Answers 1. 0.200 286 u 2. 186.6 MeV 6C nucleus consists of 6 neutrons and 6 protons. Thus, the 12 Now, use m Zm1 1H Use mass data from Tables 7.5 and 7.6 on page 881. Recall that 1 u 1.660 539 1027 kg (p
age 791). m ZmH matom to find the mass defect. Nmneutron Nmneutron matom 6(1.007 825 u) 6(1.008 665 u) 12.000 000 u 0.098 940 u 1.660 539 1027 kg 1 u 1.6429 1028 kg Use the mass-energy equivalence to calculate the binding energy from the mass defect. 1 u 1.492 1010 J 931.5 MeV 0.098 940 u 1.492 1010 J 1 u 931.5 MeV 1 u or 0.098 940 u Eb 1.476 1011 J 92.16 MeV Paraphrase The mass defect for 12 the carbon-12 nucleus is 1.476 1011 J or 92.16 MeV. 6C is 1.6429 1028 kg. The binding energy of Binding Energy per Nucleon You can compare the stability of different nuclei by dividing the binding energy of each nucleus by the number of nucleons it contains. The , the more stable the nucleus greater the binding energy per nucleon is. Figure 16.4 is a graph of binding energy per nucleon versus atomic mass number for stable nuclei. This graph peaks at about 8.79 MeV per 58Fe, and nucleon. The three most stable isotopes are nickel 28 iron 26 62Ni, iron 26 Eb A 56Fe. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 795 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 796 Figure 16.4 Binding energy per nucleon for stable isotopes ) 10 8 6 4 2 0 16 8 O 56 26 Fe 120 50 Sn 238 92 U 4 2 He 3 2 2 1 He H 50 100 150 200 250 Atomic Mass Number, A The graph also gives a hint about the process that causes the stars to shine. The binding energy per nucleon is much less for hydrogen than for helium. If hydrogen atoms combine to form helium, the nucleons move to a lower energy level and give off the difference in energy. In section 16.4, you will learn more about such nuclear reactions. 16.1 Check and Reflect 16.1 Check and Reflect Knowledge 10. Show that MeV/c2 has the dimensions 1. How many protons and neutrons do each of the following nuclei contain? (a) 90 13C (c) 56 (b) 6 26Fe 38Sr (d) 1 1H 2. Convert 1.6 1010 J to electron volts. 3. Calculate the energy equivalent of 0.25 u. 4. How much mass is converted into energy by a nuclear reaction that produces 5.00 GJ of energy? 5. Define the term isotope. 6. Explain why the mass of a stable nucleus is a bit less than Zmproton Nmneutron. Applications 7. Determine the binding energy for 10 22Ne. The atomic mass of 10 22Ne is 21.991 385 u. 8. The 19 40K isotope of potassium has an atomic mass of 39.963 998 u. (a) Determine the mass defect for 19 (b) Calculate the binding energy per 40K. nucleon for this isotope. of mass. Extensions 11. (a) Contrast the strength and range of the electromagnetic force and the strong nuclear force. (b) Explain how the nature of these forces limits the maximum possible size for nuclei. 12. Suppose that the electrostatic force were much stronger. Describe how this change would affect the stability of nuclei. 1.20 fm and A is 13. Experiments have shown that most nuclei are approximately spherical with a radius 1 of r r0A , where r0 3 the atomic mass number. Use this formula to determine the radius of the nucleus of a 90Sr atom. Then estimate the distance 38 between adjacent nucleons in this nucleus. What can you conclude about the size of protons and neutrons? 9. Use Figure 16.4 to estimate the binding e TEST energy for each of these nuclei: (a) 13 56Fe (c) 238 (b) 26 92U 6C To check your understanding of nuclei, follow the eTEST links at www.pearsoned.ca/school/physicssource. 796 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 797 16.2 Radioactive Decay The French physicist, Antoine Henri Becquerel (1852–1908), discovered radioactive decay in 1896 while conducting an experiment to see if a fluorescent compound of uranium would emit X rays when exposed to sunlight. During a period of cloudy weather, Becquerel put the uranium compound away in a drawer along with a photographic plate wrapped in black paper. When he developed the plate several days later, he was surprised to find that it was fogged even though the fluorescent compound had not been exposed to sunlight. Becquerel realized that the radiation that fogged the plate must be coming from the uranium in the compound. He also found that a magnetic field would deflect some of this radiation. The husband and wife team of Marie Curie (1867–1934) and Pierre Curie (1859–1906) began an extensive study of this radiation. They showed that thorium was also radioactive, and discovered two new elements, radium and polonium, that were both much more radioactive than uranium. Indeed, Marie coined the term radioactive. She also demonstrated that the intensity of radiation from uranium compounds was not affected by the other elements in the compound or by processes such as being heated, powdered, or dissolved. The intensity depended only on the quantity of uranium. Therefore, the radioactivity must result from a process within the uranium nucleus. Rutherford and others identified three forms of nuclear radiation: Alpha (): the emission of a helium nucleus Beta (): the emission of a high-energy electron Gamma (): the emission of a high-energy photon Initially, this classification was based on how much material each type of radiation could penetrate. In radiation from naturally occurring isotopes, the alpha particles typically do not penetrate much more than a thin metal foil or sheet of paper, whereas beta particles can pass through up to 3 mm of aluminium, and gamma rays can penetrate several centimetres of lead. The three types of radiation result from different processes within nuclei. Concept Check Figure 16.5 shows the paths that , , and rays take when passing through a magnetic field. What can you conclude about the electrical properties of these rays? β γ radiation source α Figure 16.5 The paths of , , and rays in a magnetic field info BIT Marie Curie was the first person to win two Nobel Prizes. She died of leukemia, almost certainly the result of years of exposure to radiation in her laboratory. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 797 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 798 16-2 Design a Lab 16-2 Design a Lab Radiation Shielding The Question What common materials provide effective shielding against , , and radiation? Figure 16.6 Radiation meters Design and Conduct Your Investigation Check with your teacher about the radiation sources and radiation meters (Figure 16.6) available for this investigation. Then design your experiment. List the materials you will need and outline the procedure. Try this procedure and modify it if necessary. Keep careful records of your results. Then analyze your data, and explain your conclusions. Conservation Laws and Radioactive Decay In addition to conserving momentum and energy, all radioactive decay processes obey these additional conservation laws: • Charge: The net electrical charge cannot change in a decay process. Any change in the electrical charge of the nucleus must be exactly offset by an opposite change elsewhere in the system. For example, if the charge on a nucleus decreases by 2e, then a particle with a charge of 2e must be emitted. • Atomic Mass Number: The total of the atomic mass numbers for the final products must equal the atomic mass number of the original nucleus. In other words, the total number of nucleons remains constant. 798 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 799 Example 16.5 Determine which of these radioactive decay processes are possible. (a) 214Po → 82 84 208Pb 4 2 226Ra 4 2 28Ni 1 0n (b) 230Th → 88 90 27Co → 60 (c) 60 (1 0n represents a neutron) Analysis and Solution Compare the charge and atomic mass number of the original nucleus to those of the decay products. (a) Charge: 84 82 2 Atomic mass number: 214 208 4 The decay process 214 (b) Charge: 90 88 2 84Po → 208Pb 82 4 is not possible. 2 Atomic mass number: 230 226 4 The decay process 230 (c) Charge: 27 28 0 90Th → 88 226Ra 4 is possible. 2 Atomic mass number: 60 60 1 The decay process 27 60Co → 60 28Ni 1n is not possible. 0 Practice Problems Determine whether these decay processes are possible. 210Rn 4 2 0 233U 1 1H 212Po → 86 84 233Pa → 92 91 14C → 7 6 14N 1 2. 3. 1. Answers 1. Impossible 2. Possible 3. Impossible Concept Check Why are electrons not considered when applying the conservation law for atomic mass number? Alpha Decay In 1908, Rutherford showed that alpha particles are helium nuclei spontaneously emitted by unstable large nuclei. In these nuclei, the electromagnetic force repelling the outer protons is almost as great as the attractive strong nuclear force. Such nuclei can spontaneously emit alpha particles. Because a cluster of two protons and two neutrons forms a highly stable helium nucleus, these unstable large nuclei decay by emitting alpha particles rather than separate protons and neutrons. The emission of an alpha particle decreases the atomic number by 2 and the atomic mass number by 4. For example, alpha decay of uranium-238 produces thorium: 238 92U → 90 234Th 4 2 In this example, uranium is the parent element and thorium is the daughter element. Applying the conservation laws gives this general form for alpha decays: A ZX → A4 Z2Y 4 2 parent element: the original element in a decay process daughter element: the element produced by a decay process where X is the chemical symbol for the parent element and Y is the symbol for the daughter element. Here, A is the atomic mass number of the parent element and Z is its atomic number. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 799 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 800 Example 16.6 Practice Problems Write the -decay process for these elements, and name the parent and daughter elements. 1. 2. 3. 230Th 90 238U 92 214Po 84 Answers 1. 90 2. 92 3. 84 230Th → 226 238U → 90 214Po → 82 88Ra 4 ; thorium, radium 2 234Th 4 ; uranium, thorium 2 210Pb 4 ; polonium, lead 2 Predict the daughter element that results from alpha decay of radium-226. Analysis and Solution From a periodic table, you can see that the atomic number for radium is 88. So, the parent elemen
t is 226 Since the alpha particle carries away four nucleons, including two protons, A decreases by 4 and Z decreases by 2: ZX → A4 A So, the daughter element is 882 The periodic table shows that the element with Z 86 is radon. 2264Y 222 Z2Y 88Ra. 86Y. 4 2 Paraphrase For alpha decay, the daughter element of radium-226 is radon-222. Energy Released During Alpha Decay You can apply the concepts of energy conservation and mass-energy equivalence to alpha decay, using a method similar to the calculation of nuclear binding energy. The mass-energy of the parent nucleus is equal to the sum of the mass-energy and the kinetic energies of both the daughter nucleus and the alpha particle: mparentc2 mdaughterc2 mc2 E The difference in energy, E, appears as the total kinetic energy of the alpha particle and of the daughter nucleus. If the parent nucleus was at rest, the law of conservation of momentum requires the momentum of the alpha particle to be equal in magnitude and opposite in direction to the momentum of the daughter nucleus. Usually, the mass of the daughter nucleus is much greater than the mass of the alpha particle. So, the speed of the alpha particle is correspondingly greater than the speed at which the daughter nucleus recoils: mv mdaughtervdaughter v mdaughtervdaughter m 800 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 801 The kinetic energy of the alpha particle is also correspondingly greater than the kinetic energy of the daughter nucleus: 1 E mv 2 2 1 m 2 mdaughter m mdaughter m mdaughtervdaughter m 2 mdaughterv 2 daughter 1 2 Edaughter Concept Check Explain why E must be positive in order for -decay to occur. Example 16.7 Show that -decay of radium-226 is possible, and estimate the maximum kinetic energy of the emitted alpha particle. Given Parent atom is radium-226. Required maximum kinetic energy of the alpha particle Practice Problems Calculate the energy released during -decay of these nuclei: 1. 2. 3. 230Th 90 238U 92 214Po 84 Answers 1. 7.641 1013 J 2. 6.839 1013 J 3. 1.255 1012 J Analysis and Solution Example 16.6 showed that the daughter element is radon-222. The energy released is equivalent to the difference between the mass of the parent atom and the total mass of the products. m mparent Table 7.5 on page 881 lists the atomic masses for radium-226, radon-222, and helium-4. As in section 16.1, you can use atomic masses instead of nuclear masses because the masses of the electrons will balance out. A radon nucleus has over 50 times the mass of an alpha particle. So, the alpha particle will have over 98% of the total kinetic energy, E. m m226 mproducts 88Ra (m 222 m4 ) 2 226.025 410 u 222.017 578 u 4.002 603 u 0.005 229 u 86Rn E 0.005 229 u 1.492 1010 J 1 u 7.802 1013 J or 0.005 229 u 4.871 MeV 931.5 MeV 1 u Paraphrase Since E 0, alpha decay of radium-226 is possible. The maximum kinetic energy of the alpha particle when emitted is about 7.802 1013 J or 4.871 MeV. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 801 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 802 THEN, NOW, AND FUTURE Ionization Smoke Detectors Most household smoke detectors (Figure 16.7) contain a small amount of americium-241. This isotope emits -particles, which ionize air molecules between two metal plates within the smoke detector. One of the plates has a positive charge, and the other plate has a negative charge. The plates attract the ions, so a small current flows between the plates. If smoke particles enter the smoke detector, they absorb some of the -particles. So, the alpha radiation ionizes fewer air molecules Figure 16.7 This smoke detector uses alpha radiation to sense smoke particles. and the current between the metal plates decreases. This drop in current triggers the alarm circuit in the smoke detector. Questions 1. Why is it safer for a smoke detector to use alpha radiation, instead of beta or gamma radiation? 2. Suggest reasons why most manufacturers of smoke detectors recommend replacing them after 10 years. beta-negative () decay: nuclear decay involving emission of an electron beta () particle: electron emitted by a nucleus Beta Decay Sometimes, a nucleus decays by emitting an electron. This process is termed beta-negative or decay. During decay, a neutron in the nucleus transforms into a proton, electron, and an extremely small neutral particle known as antineutrino, symbol ¯ (Figure 16.8). So, the atomic number of the atom increases by 1, but the atomic mass number does not change. Charge is conserved because the charge on the new proton balances the charge on the electron emitted from the nucleus. This electron is often called a beta () particle, a name that originates from the 0 in equaearly classification of types of radiation. It is often written as 1 tions. For example, decay of thallium-208 produces lead, and the equation is: 208Tl → 81 208Pb 82 0 υ– –1 n e p υ Figure 16.8 During decay, a neutron changes into a proton, electron, and antineutrino. Concept Check Why is the mass of the neutron slightly larger than the sum of the proton and electron masses? 802 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 803 Example 16.8 What element will the decay of thorium produce? Analysis and Solution A periodic table shows that the atomic number for thorium is 90. decay increases the atomic number by 1, so Adaughter The element with an atomic number of 91 is protactinium, the element immediately after thorium in the periodic table. 91. Paraphrase For decay of thorium, the daughter element is protactinium. Practice Problems 1. Find the elements produced by decay of (a) 228 88Ra (b) 212 82Pb Answers 1. (a) 228 (b) 212 89Ac 83Bi As with alpha decays, you can use atomic masses to calculate how much energy a beta decay will release. Example 16.9 How much energy would you expect the decay of a thorium-234 nucleus to release? Given Parent element is 234 90Th. Required Energy released by decay (E) Practice Problems 1. (a) What element does the decay of cobalt-60 produce? (b) How much energy would you expect the decay of a cobalt-60 nucleus to release? Answers 1. (a) 60 28Ni (b) 2.823 MeV Analysis and Solution As shown in Example 16.8, the daughter element is protactinium. However, this daughter atom has only the 90 electrons from the original thorium atom because the electron emitted by the thorium nucleus leaves the atom as beta 91Pa. The energy released radiation. The result is a positive ion, 234 is equivalent to the difference between the mass of the parent atom and the total mass of the decay products. Together, the masses of the protactinium ion and the beta particle equal the mass of a neutral protactinium atom. Table 7.5 on page 881 lists the atomic masses. m mparent m 234 m 234 91 Pa 234.043 601 u 234.043 308 u 0.000 293 u mproducts (m 234 m 234 m 0 1 90 Th 90 Th 91 Pa ) 1 u is equivalent to about 931.5 MeV, so E 0.000 293 u 931.5 MeV 1 u 0.2729 MeV Paraphrase The decay of a 234 90Th nucleus should release 0.2729 MeV. PHYSICS INSIGHT In calculating energy produced in nuclear decay, it is common to use atomic masses because these data are readily available. You see this being done in Examples 16.9 and 16.10. In both of these examples, as an intermediate step an ion notation has been used to account for the change in nuclear charge that happens during beta decay. In reality, in beta decay it is most likely that the atom will not end up ionized. The atom will either lose or gain an electron as appropriate. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 803 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 804 info BIT The name neutrino comes from the Italian word for “little, neutral one.” The word was coined by Enrico Fermi, a renowned physicist who developed a theory to explain beta decay. neutrino: an extremely small neutral subatomic particle e WEB To learn more about the Sudbury Neutrino Observatory, follow the links at www.pearsoned.ca/school/ physicssource. info BIT Each second, more than 100 trillion neutrinos pass through your body! Almost all of these neutrinos were formed by nuclear reactions in the core of the Sun. weak nuclear force: fundamental force that acts on electrons and neutrinos antimatter: form of matter that has a key property, such as charge, opposite to that of ordinary matter ): an antipositron (e or 0 1 electron; a positively charged particle with its other properties the same as those of an electron The Elusive Neutrino Since the daughter nucleus has vastly more mass than an electron, there is practically no recoil of the daughter nucleus during beta decay. Consequently, physicists expected that virtually all of the energy released during decay would appear as the kinetic energy of the electron emitted by the nucleus. However, measurements found that most electrons emitted during decay had somewhat less kinetic energy than expected, and a few had almost no kinetic energy. During decay, small portions of the mass of the parent nuclei seemed to just disappear! In 1930, the Austrian physicist Wolfgang Pauli (1900–1958) suggested that the missing energy in beta decay was carried away by a tiny, as-yetundiscovered neutral particle, now called the neutrino, . Neutrinos are so small that physicists have yet to determine their size and mass. These “ghost-like” particles can pass through Earth with only a slight chance of being absorbed! Indeed, it was 1956 before an experiment using the intense radiation at a nuclear power plant finally proved conclusively that neutrinos actually exist. Eventually physicists discovered that there are actually two kinds of neutrinos given off in beta decay. In decay an antineutrino is released. As you will soon see, in decay a neutrino is released. The neutrino and antineutrino are identical in all respects except for their opposite spins. Many astrophysicists now think that neutrinos play a critical role in the cores of stars and perhaps in the
structure of the cosmos as well. Concept Check How did physicists know that the neutrino must be neutral? Beta Decay, the Weak Nuclear Force, and Antimatter Careful study of beta decays revealed two further important differences from alpha decay. First, the transformation of a neutron into a proton involves a fundamental force called the weak nuclear force. Although it is less powerful than the strong nuclear force, the weak nuclear force acts on electrons and neutrinos, whereas the strong nuclear force does not. The second difference is that beta decay involves antimatter. An antimatter particle has a key property, such as charge, opposite to that of the corresponding particle of ordinary matter. For example, an anti0), has a positive charge but the same mass electron, or positron (e or 1 as an electron. Section 17.2 presents antimatter in more detail. In decay, the transformation of a neutron into a proton produces an antineutrino rather than a neutrino: n → p 1 0 where is the symbol for the antineutrino. Thus, decays have the general form ZX → A A Z1Y 1 0 (Z increases by 1) 804 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 805 A second form of beta decay also produces an antiparticle. In decay, a proton transforms into a neutron, and the parent nucleus emits a positron and a neutrino: beta-positive () decay: nuclear decay involving emission of a positron ZX → A A Z1Y 0 (Z decreases by 1) 1 Sometimes, you will see the electron in these decay processes represented by the symbol e. To distinguish this electron from those orbiting the 0 to represent an electron nucleus, this chapter uses the symbol 1 0 is used to represent an emitted by a nucleus. Similarly, the symbol 1 emitted positron. Example 16.10 Nitrogen-13 (13 by decay. 7N) transmutes into carbon-13 (13 6C) 13N → 7 13C 6 0 1 Calculate the energy released if the atomic masses are 13.005 739 for nitrogen-13 and 13.003 355 for carbon-13. Given Nitrogen-13 transmutes into carbon-13 by decay. Atomic masses: 13.005 739 for nitrogen-13, 13.003 355 for carbon-13 Required Energy released in the decay Analysis and Solution The energy released is equivalent to the difference between the mass of the parent atom and the total mass of the products. Practice Problems 1. (a) What isotope will decay of thallium-202 produce? (b) Write the process for this decay. (c) How much energy will be released by the decay of the thallium-202 nucleus if the mass of the thallium nucleus decreases by 0.001 463 u? Answers 1. (a) mercury-202 81Tl → 80 (b) 202 (c) 0.3400 MeV 202Hg 0 1 m mparent m 13 7N m 13 7N mproducts 6C m 0 (m 13 m 0 (m 13 6C 1 ) m 0 1 Note that again we use an ion notation indicating the presence of a carbon ion. In fact, at the end of the decay process, the carbon ion will lose an electron, and its mass can be written as (m 13 ) as shown above. ) 1 m 0 1 6C Since electrons and positrons have the same mass, m 0 1 Therefore, m m 13 7N 2m 0 1 (m 13 6C ) 13.005 739 u [13.003 355 u 2(0.000 549) u] 0.001 286 u m 0 1 . 1 u is equivalent to 931.5 MeV, so E 0.001 286 u 931.5 MeV 1 u 1.198 MeV Paraphrase The decay of a 7 13N nucleus should release 1.198 MeV of energy. PHYSICS INSIGHT In Example 16.10, the energy released when a nitrogen-13 nucleus decays to form carbon-13 is calculated to be 1.198 MeV. Most nuclear decay data tables, however, will indicate that the total energy released in this decay is 2.221 MeV. Both are correct! When a decay occurs, a positron is emitted. This positron could combine with an electron to release the energy equivalence of 2 electron masses, an additional 1.023 MeV. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 805 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 806 Gamma Decay (-decay) Many nuclei have a series of energy levels that correspond to different configurations of the nucleons. In the excited states, the nucleons are farther apart. As a result, their binding energy is less than when in the ground state, and the total energy of the nucleus is greater. When making a transition to a lower-energy state, a nucleus emits a gamma-ray photon, similar to the photon emitted when an electron in an atom moves to a lower energy level (Figure 16.9). However, the difference in energy is much greater for a nucleus. Gamma () decay does not change the atomic number or the atomic mass number. Gamma decays can be written using this general form: A AX Z ZX* → where * indicates an excited state and represents a gamma ray. Often, alpha or beta decay leaves the daughter nucleus in a highly excited state. The excited nucleus then makes a transition to its ground state, and emits a gamma ray. For example, when decay of boron-12 produces carbon-12, the carbon nucleus is highly excited and quickly emits a gamma ray: 12 5B → 6 6C* → 6 12 12C* 0 1 12C The energy of a gamma ray depends on the energy levels and the degree of excitation of the particular nucleus. Gamma rays can have energies ranging from thousands to millions of electron volts. Stability of Isotopes Figure 16.10 shows that stable isotopes form a relatively narrow band when plotted by their proton and neutron numbers. Other than hydrogen, all stable isotopes have at least as many neutrons as protons. As Z increases, the isotopes require an increasing ratio of neutrons to protons in order to be stable. There are no completely stable isotopes with more than 83 protons. The stable isotopes have greater binding energies than the unstable isotopes. Radioactive decay transmutes unstable nuclei into nuclei with higher binding energies. For example, heavy nuclei above and to the right of the stable band will emit alpha particles (larger red arrows), heavy nuclei below and to the right of the band will emit positrons, or particles (small red arrows), and lighter nuclei to the left of the band will emit electrons, or particles (blue arrows). All of these decay processes produce isotopes that are either in the stable band or closer to it. A nucleus may undergo several successive decays before it reaches the stable band 15 10 5 0 15.1 12.7 9.64 7.65 4.44 ground state Figure 16.9 Nuclear energy levels for carbon-12: How do these energy levels differ from those for electrons in hydrogen? gamma () decay: emission of a high-energy photon by a nucleus e WEB To learn more about nuclear energy levels, follow the links at www.pearsoned.ca/ school/physicssource. transmute: change into a different element α decays β decays N Z β decays 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 10 20 30 40 50 60 70 80 Atomic Number, Z 90 100110 Figure 16.10 The black dots represent the band of stable isotopes. 806 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 807 Radioactive Decay Series Often, a radioactive nucleus will decay into a daughter nucleus that is itself radioactive. The daughter nucleus may then decay into yet another unstable nucleus. This process of successive decays continues until it creates a stable nucleus. Such a process is called a radioactive decay series. β decay α decay 238 234 230 226 222 218 214 210 206 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn 87 Fr 88 Ra 89 Ac 90 Th 91 Pa 92 U Atomic Number, Z Figure 16.11 How many different decay paths are there from uranium-238 to lead-206? Radioactive decay series beginning with 238 92U and ending with 206 82Pb. The dots in Figure 16.11 represent nuclei that are part of the decay series. A decay series can have several branches that lead to the same final product. Figure 16.11 shows that 218 84Po by three different combinations of decays. All of the intermediate isotopes in a decay series are unstable, but the degree of instability is different for each isotope. For example, 218 86Rn usually lasts for only a fraction of a second whereas 222 90Th takes thousands of years. Although not shown in Figure 16.11, many of the intermediate isotopes undergo gamma decay. 86Rn takes several days to decay and 230 84Po can transmute into 214 Concept Check Explain why gamma decays cannot be shown as paths on a decay series graph like the one in Figure 16.11. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 807 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 808 Potential Hazards of Nuclear Radiation Alpha, beta, and gamma radiation are all invisible, and most of their effects on the human body are not immediately apparent. As a result, it was not until years after the discovery of radioactive decay that researchers realized how dangerous radiation can be. Radiation poses two major types of risk: • Radiation Sickness: Radiation can ionize cellular material. This ionization disrupts the intricate biochemistry of the body, resulting in radiation sickness. Large doses of ionizing radiation can kill cells. Blood cells and the lining of the intestine are particularly vulnerable. Symptoms include nausea, vomiting, diarrhea, headache, inflammation, and bleeding. Severe radiation sickness is often fatal. • Genetic Damage: High-energy particles and gamma rays can alter DNA, and lead to the development of cancers or harmful mutations. These effects often appear 10 to 15 years after radiation exposure. Everywhere on Earth, there is some naturally occurring radiation from cosmic rays and from radioisotopes in the ground. This background radiation causes some minor damage, but normally the body can repair such damage without any lasting harm. The effect of radiation on living organisms depends on the energy it carries, its ability to ionize atoms and molecules, and the depth to which it can penetrate living tissue. The charge and energy of the radiation determine how ionizing it is. The energy also affects how far the radiation can penetrate. The energy that a radiation has depends on the process that produces it. Table 16.2 compares the hazards posed by typical radiations from natural sources. Table 16.2 Radiation Hazards from Natural Sources Outside the Body Radiation Typical P
enetration Ionization Hazard alpha beta gamma Travels about 5 cm in air. Cannot penetrate skin. high Travels about 30–50 cm in air. Penetrates about 1 cm into the body. moderate Travels great distances in air. Penetrates right through the body. low low low high Although and particles are much less penetrating than gamma radiation, they can still be extremely harmful if emitted by material absorbed into the body, because the nearby tissue has a continuing exposure to the radiation. For example, health scientists have calculated that breathing in a speck of dust containing just 1 g of plutonium is virtually certain to cause lung cancer within 30 years. The introduction of radioactive isotopes into the food chain is also a serious concern because these materials can accumulate in the body. For example, strontium-90, a by-product of nuclear weapons and power reactors, is absorbed into bones because it is chemically similar to calcium. Radiation from strontium damages bone marrow, reduces the production of blood cells, and can lead to bone cancer and leukemia. info BIT Both Marie and Pierre Curie suffered from radiation sickness. Some of Marie’s laboratory notebooks are still dangerously radioactive. radioisotope: an isotope that is radioactive 808 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 809 Despite its potential hazards, nuclear radiation is not always harmful. As you will see in section 16.3, nuclear radiation has many beneficial industrial and medical applications. Measuring Radiation Exposure The effects of a given dose of radiation depend on the type of radiation. For example, a dose of infrared radiation that delivered 1 J/kg to living tissue would do little more than heat the tissue slightly. The same quantity of energy from X rays would ionize some molecules within the tissue, whereas the same quantity of energy from alpha radiation would be far more ionizing and disruptive. For this reason, SI has two units for measuring radiation exposure: The gray is the unit for absorbed dose and the sievert is the unit for equivalent absorbed dose. Gray (Gy): 1 gray is the dose of ionizing radiation that delivers 1 J of energy to each kilogram of material absorbing the radiation. Sievert (Sv): 1 sievert is the absorbed dose of ionizing radiation that has the same effect on a person as 1 Gy of photon radiation, such as X rays or gamma rays. The absorbed dose in sieverts is equal to the dose in grays multiplied by the relative biological effectiveness (RBE), a measure of how harmful the particular kind of radiation is. For example, the RBE for high-energy alpha particles is about 20, so an absorbed dose of 1 Gy of alpha radiation is equivalent to 20 Sv. An equivalent dose of 6 Sv in a short time is usually fatal. Typical radiation exposure for North Americans is less than 0.5 mSv annually. Table 16.3 summarizes some common sources of radiation exposure. Table 16.3 Common Sources of Radiation Exposure relative biological effectiveness (RBE): a factor indicating how much a particular type of radiation affects the human body Source Natural Artificial Total Radon from ground Cosmic rays Radioactive rocks/minerals, common building materials Ingested from natural sources Medical/dental X rays Nuclear weapons testing Consumer products All other Typical Exposure (Sv/year) 200 44 40 18 73 4 1 2 <400 Figure 16.12 Dosimeters: How do these devices measure exposure to radiation? Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 809 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 810 16.2 Check and Reflect 16.2 Check and Reflect Knowledge 1. What are the three basic radioactive decay processes and how do they differ from each other? 2. What is the ratio of neutrons to protons for the heaviest stable isotopes? 3. (a) Write the alpha-decay process for 234 91Pa. (b) Identify the parent and daughter nuclei in this decay. 4. (a) Which type of beta decay transmutes carbon-14 into nitrogen? 11. Identify each type of decay in this series, and name the parent and daughter elements. (b) (a) 232 (d) 228 90Th → 228 22Na → 10 11 88Ra* → 228 (c) 228 88Ra → 228 89Ac → 228 90Th → 224 228 0n 1p → 0 (e) 228 (g) 0 (f) 88Ra* 4 22Ne 2 0 1 88Ra 0 89Ac 1 90Th 1 0 88Ra 4 2 0 1 (b) Write the process for this decay. Extensions 5. (a) Which type of beta decay transmutes the sodium isotope 11 22Na into 10 (b) Write the process for this decay. 22Ne? 6. Explain why the daughter nucleus in an alpha decay often emits a gamma ray. 7. Which form of radioactive decay has the 12. In a process called electron capture, a nucleus absorbs an electron and emits a neutrino. (a) What effect does electron capture have on the atomic number? (b) Use nuclear notation to write the general form for electron capture. greatest penetrating power? (c) Compare electron capture with beta decay. 13. Devise an experiment to test the hypothesis that gamma rays are emitted by nucleons jumping from higher energy levels to lower ones, similar to the energy-level transitions of electrons in an atom. What would you expect the spectrum of gamma rays emitted by a nucleus to look like? 14. Use library or Internet resources to learn how radon forms in the ground. Explain how radon can accumulate in basements in some areas. Why is this accumulation a health concern? e TEST To check your understanding of radioactive decay, follow the eTEST links at www.pearsoned.ca/school/physicssource. Applications 8. How much energy is released when 22Ne? The mass of 11 22Na decays to 10 11 is 21.994 436 u and the mass of 10 21.991 385 u. 22Na 22Ne is 9. Explain whether the atomic number can increase during nuclear decay. Support your answer with an example. 10. Compare the annual average radiation exposure from natural sources with the dose you would receive from a dental X ray. 810 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 811 16.3 Radioactive Decay Rates How can an archaeologist confidently tell you that a bison head found in southern Alberta provides evidence that First Nations peoples were here more than 5000 years ago? Why do doctors sometimes inject patients with radioactive dyes? In this section, you will be introduced to the concepts of radioactive decay rate and half-life, and begin to see how understanding the behaviour of radioactive elements can provide us with a glimpse into the past or give us powerful techniques to diagnose and combat disease. 16-3 QuickLab 16-3 QuickLab Simulating Radioactive Decay Problem How can decay rates of atoms be predicted? Materials container with 100 pennies graph paper Procedure Work in groups of two or three. 1 (a) Pour the pennies onto a flat surface and spread them out. Put aside any pennies that are tails up. These pennies have “decayed.” (b) Count the remaining pennies and put them back into the container. Record this count in a table. 2 Predict how many pennies will remain if you repeat step 1 two more times. 3 Repeat step 1 a total of eight times. 4 Pool your results with the other groups in the class. 5 Use the pooled data to draw a graph of how the number of pennies remaining varies with time. Questions 1. How many pennies were left after you had done step 1 three times? Does this result match your prediction? 2. If you repeat the experiment, will you get exactly the same results each time? Explain. 3. Suppose that step 1 takes 2 min each time. (a) How long would it take for the number of pennies remaining to decrease by half? How long will it take until only about an eighth of the pennies remain? How are these two time intervals related? (b) Try to find a formula to predict how many pennies will remain at any given time. Activity and Decay Constant The radioactive decay of a specific nucleus is unpredictable. The nucleus could decay in the next minute, or tomorrow, or thousands of years from now. However, you can accurately predict how many nuclei in a sample will decay in a given time. The decay constant () is the probability of any given nucleus decaying in a unit of time. The decay constant is a property of each particular isotope. For example, radium-226 has a decay constant of 1.4 1011 s1, indicating that each individual nucleus in a sample of radium-226 has a probability of 1.4 1011 of decaying in 1 s. The greater the decay constant, the faster an isotope will decay. decay constant: probability of a nucleus decaying in a given time Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 811 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 812 activity or decay rate: the number of nuclei in a sample that decay within a given time becquerel (Bq): unit of activity, equal to 1 decay per second The activity (A) or decay rate is the number of nuclei in a sample that decay within a given time. Activity is usually measured in decays per second, or becquerels (Bq). A highly radioactive sample has many radioactive decays each second. Activity and the decay constant are related by this formula: A N t N where N is the number of radioactive nuclei, t is the time interval, and is the decay constant. Example 16.11 Carbon-14 has a decay constant of 3.8 1012 s1. What is the activity of a sample that contains 2.0 1015 carbon-14 nuclei? Practice Problems 1. Cobalt-60 has a decay constant of 4.1 109 s1. Find the activity of a sample containing 1.01 1022 cobalt-60 atoms. 2. A sample containing 5.00 1020 atoms has an activity of 2.50 1012 Bq. Find the decay constant of this sample. Answers 1. 4.1 1013 Bq 2. 5.00 109 s1 Given 3.8 1012 s1 N 2.0 1015 atoms Required activity (A) Analysis and Solution Substitute the given values into the formula for activity: A N (3.8 1012 s1)(2.0 1015) 7.6 103 Bq The negative sign indicates that the number of carbon-14 nuclei is decreasing. Paraphrase The initial activity of the sample is 7.6 kBq. The activity of a radioactive material decreases over time. The reason is simple: Radioactive decay “uses up” the unstable nuclei in the sample.
Half-life Half-life is the time required for one-half of the radioactive nuclei in a sample to decay. For example, to diagnose thyroid problems, doctors sometimes inject patients with the radioactive isotope iodine-131, which has a half-life of about 192 h. Out of a dose of 20 g of iodine-131, 10 g will decay within 192 h. Only 5 g of iodine-131 will remain after the next 192 h, then 2.5 g after the next 192 h, and so on (see Figure 16.13). A common symbol for half-life is t1/2 . half-life: the time it takes for half of the radioactive nuclei in a sample to decay e SIM To see a simulation of half-life, follow the links at www.pearsoned.ca/ school/physicssource. 812 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 813 The number of nuclei of the original radioisotope left in a sample is given by the equation where t is the time elapsed, N0 is the number of nuclei of the original radioisotope when t 0, is the half-life of the isotope. and t1/2 N N0 t t1 /2 1 2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0. half-life 0 5 10 15 Time (days) Figure 16.13 A graph showing the radioactive decay of iodine-131 Example 16.12 Carbon-14 has a half-life of 5730 years. How long will it take for the quantity of carbon-14 in a sample to drop to one-eighth of the initial quantity? 5730 years Given t1/2 N 1 N0 8 Required time (t) Analysis and Solution N N0 1 N0 8 1 2 t t1 /2 In 3 half-lives, N will decrease to N0 t 3 Therefore, t1 /2 t 3t1/2 3 5730 years 1.719 104 years N0 1 2 1 2 1 8 1 2 Practice Problems 1. Astatine-218 has a half-life of only 1.6 s. About how long will it take for 99% of a sample of astatine-218 to decay? 2. Radium-226 has a half-life of 1600 years. What percentage of a sample of radium-226 will remain after 8000 years? Answers 1. about 11 s 2. 3.125% Paraphrase It will take just over 17 thousand years for the amount of carbon-14 in a sample to drop to one-eighth of its original value. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 813 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 814 Example 16.13 Radon-222 has a half-life of 3.82 days. What percent of a sample of this isotope will remain after 2 weeks? Practice Problems 1. Strontium-90 has a half-life of 29.1 years. What percent of a sample of this isotope will be left after 100 years? 2. Tritium (3 1H) has a half-life of 12.3 years. How much of a 100-mg sample of tritium will be left after 5.0 years? Answers 1. 9.24% 2. 75% Given t1/2 3.82 days t 14 days Required percent remaining after 14 days Analysis and Solution The percent remaining is calculated from the ratio N N0 t t1 /2 1 2 N N0 . N0 4 1 .8 2 3 1 2 1 2 3.66 N0 0.079N0 N N0 0.079 or 7.9% Note that you can use the exponent or ^ key on a scientific or graphing calculator to evaluate powers of enter (1/2)^(14/3.82). 1 2 . On a graphing calculator, you could Paraphrase Only 7.9% of a sample of radon-222 will remain after 2 weeks. Applications of Radiation The Rutherford gold-foil experiment that you learned about in Chapter 15 was one of the first examples of the use of nuclear energy (the release of alpha particles in the decay of radium nuclei) to study the inner working of atoms. Scientists apply radioactive decay in many other fields of scientific research, including archaeology. Radioactive compounds also have numerous industrial applications and are routinely used to diagnose and treat diseases. 814 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 815 THEN, NOW, AND FUTURE Radiotherapy During their first experiments with radium, Pierre and Marie Curie noticed that its radiation could burn the skin, but the wound would heal without forming scar tissue. They realized that radium could therefore be used to treat cancer. To ensure that this treatment was readily available to cancer patients, the Curies refused to patent their discovery. Radiotherapy is particularly useful for treating cancer because cancer cells are more susceptible to the effects of radiation than healthy tissue is. Also, the radiation is concentrated on the cancer, and kept away from the surrounding tissue as much as possible (Figure 16.14). γ rays Figure 16.14 Rotating the radiation source around the patient minimizes damage to normal tissue. There is now a wide variety of radiation treatments. Often, a carefully focussed beam of gamma rays is directed at the tumour. Another common method is to inject the tumour with a short-lived radioisotope that emits alpha-particles. Questions 1. Give two reasons why gamma rays are used for the beam type of radiotherapy. 2. Why does injected radiotherapy use an isotope that undergoes alpha decay rather than one that gives off beta or gamma radiation? Radioactive Dating Nearly 6000 years ago, First Nations people of southwestern Alberta devised an ingenious method for hunting the vast herds of bison on the plains. By setting up barriers along a carefully chosen route, the First Nations people funnelled the bison toward a hidden cliff and then drove them over the edge. There were about 150 buffalo jumps in Alberta. The most famous, Head-Smashed-In Buffalo Jump, is now a United Nations World Heritage Site (see Chapter 2, Figure 2.68). By carefully measuring the ratio of carbon-12 to carbon-14 in bones found at this site, archaeologists have shown that it was used continuously for over 5500 years. e MATH To plot the decay rate of carbon–14 and other radioactive elements, and to learn how to mathematically determine a radioactive sample’s age based on the percentage of the sample remaining, visit www.pearsoned.ca/school/ physicssource. How did this carbon ratio indicate the age of these bones? High-energy neutrons in cosmic rays produce the radioisotope carbon14 by colliding with nitrogen atoms high in the atmosphere: 7N → 1n 14 0 14C 6 1H 1 This carbon-14 diffuses throughout the atmosphere. Some of it is absorbed by plants and enters the food chain. So, a small proportion of all the carbon metabolized by plants and animals is carbon-14. Carbon-14 undergoes decay to form nitrogen-14, whereas carbon-12 is completely stable. When living matter dies, it stops absorbing carbon, and the proportion of carbon-14 gradually decreases as it decays (see Figure 16.15). The half-life of carbon-14 is 5730 years 100 90 80 70 60 50 40 30 20 10 0 t1 2 t1 2 10 000 t1 2 t1 2 20 000 Time (years) 30 000 40 000 Figure 16.15 Carbon-14 content as a function of the age of an artifact Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 815 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 816 e WEB To learn more about radioisotope dating, follow the links at www.pearsoned.ca/ school/physicssource. For archaeologists, bone fragments and other artifacts found at Head-Smashed-In Buffalo Jump are like clocks that show when the living matter stopped absorbing carbon. Suppose, for example, that the proportion of carbon-14 in a bone fragment is about 40% 1.32 40%, the carbon-14 has been of that in living tissue. Since decaying for about 1.3 half-lives, provided that the ratio of carbon-14 to carbon-12 in the atmosphere is the same now as when the buffalo was alive. Thus, the age of the bone fragment is roughly 1.3 5730 7500 years. Accurate estimates require more detailed calculations that take into account factors such as variations in the proportion of carbon-14 in the atmosphere through the ages. 1 2 Geologists estimate the age of rocks and geological formations with calculations based on isotopes with much longer half-lives. Such calculations are one of the methods that scientists use to estimate the age of Earth. Industrial Applications Manufacturers of sheet materials such as paper, plastics, and metal foils often monitor the thickness of the material with a gauge that measures how much of the beta radiation from a calibrated source passes through the material. Unlike mechanical thickness gauges, such gauges need not touch the material they measure, so they do not get worn down and have less risk of marking the material. Gamma rays can pass through thick metal parts to expose a photographic plate. The resulting image can reveal hidden air bubbles or hairline cracks, similar to the way X rays produce images of the inside of a patient’s body. Gamma-ray photographs are a non-destructive way of testing items that X rays cannot penetrate, including structural materials, jet engines, and welded joints in pipelines. Radioactive tracers are also used in pipelines to measure flow and to detect underground leaks. Some uses of radiation are controversial. For example, beta radiation from tritium powers runway lights and emergency exit signs that require no electricity. However, several people have received harmful doses of radiation when tritium lights have been damaged. Critics of these lights argue that other technologies can provide reliable lighting during power failures without any risk of radiation exposure. Perhaps the most controversial application is the irradiation of food to kill bacteria, insects, and parasites. Although this process sterilizes the food and thereby prolongs its shelf life, there are concerns that the radiation might also alter the food in ways that make it harmful or less nutritious. Concept Check Why is beta radiation used for measuring the thickness of sheet materials, whereas gamma radiation is used for testing structural materials? 816 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 817 16.3 Check and Reflect 16.3 Check and Reflect Knowledge 1. What fraction of a radioactive material remains after four half-lives? 2. How many decays per second occur in a radioactive sample containing 6.4 1023 atoms of a material that has a decay constant of 5.8 1012 s1? 3. Which has the greater activity, 1 g of material with a half-life of 1 ms or 1 g of material with a half-life of 1 year? Explain your answer. Applications 4. Analysis of a rock sample shows that only of the original amount of chlorine-36
1 16 remains in the rock. Estimate the age of the rock given that the half-life of chlorine-36 is 3.0 105 years. 5. A radioactive tracer used in a medical test has a half-life of 2.6 h. What proportion of this tracer will remain after 24 h? 6. An archaeologist finds a wooden arrow shaft with a proportion of carbon-14 that is about 25% of that in a living tree branch. Estimate the age of the arrow. 7. A radioactive sample has an activity of 2.5 MBq and a half-life of 12 h. What will be the activity of the sample a week later? 8. Graph the data in this table. Then use your graph to estimate (a) the half-life of the material (b) the activity of the sample at time t 0 Time (h) Activity (decays/min) 1 2 4 6 8 10 3027 2546 1800 1273 900 636 Extensions 9. A dealer in antiquities offers to sell you an “authentic” dinosaur bone for a mere $100. He shows you a certificate indicating that carbon-14 dating determined that the bone is 65 million years old. Why should you be suspicious? 10. Do a Web search on use of irradiation in food production and distribution. Prepare a summary of the arguments for and against this technology. 11. (a) What is depleted uranium? (b) Why is depleted uranium used in armour-piercing shells and in ballast for aircraft? (c) Why are these applications controversial? e TEST To check your understanding of radioactive decay, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 817 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 818 Figure 16.16 The doomsday clock from the Bulletin of the Atomic Scientists 16.4 Fission and Fusion In 1945, a group of the scientists who had designed and built the atomic bomb founded a magazine as part of an ongoing campaign to prevent this weapon from ever being used again. The Bulletin of the Atomic Scientists features a doomsday clock that symbolizes their estimate of the risk of a nuclear war (Figure 16.16). Since 2002, the clock has showed just seven minutes to midnight — a sobering reminder of the dangers posed by the enormous energy that nuclear reactions can release. The graph in Figure 16.4 (page 796) shows that binding energy per nucleon has a maximum value of about 8.7 MeV when the atomic mass number, A, is from 58 to 62 — the values for isotopes of iron and nickel. Up to this maximum, the binding energy per nucleon generally increases as A increases. Then, as A increases further, the binding energy per nucleon gradually decreases. The shape of this graph indicates that two distinct types of reactions can release energy from nuclei. Fission: When a nucleus with A > 120 splits into smaller nuclei, they have greater binding energy per nucleon. This fission reaction gives off energy equal to the difference between the binding energy of the original nucleus and the total binding energy of the products. Fusion: When two low-mass nuclei combine to form a single nucleus with A < 60, the resulting nucleus is more tightly bound. This fusion reaction gives off energy equal to the difference between the total binding energy of the original nuclei and the binding energy of the product. For both nuclear fission and fusion, the energy released, E, is E Ebf (net change in mass defect) c2 Ebi where Ebi is the total binding energy of the original nucleus or nuclei, and Ebf is the total binding energy of the product(s). Since the binding energies correspond to the mass defects for the nuclei, the energy released corresponds to the decrease in the total mass defect. This change in the total mass defect equals the change in the total mass. Thus, the energy released corresponds to the mass that the reaction transforms into energy: mi) c2 E (mf where mi is the total mass of the original nucleus or nuclei, and mf is the total mass of the product(s). Concept Check Why does a nuclear reaction that increases the binding energy per nucleon release energy? Use an analogy to help explain this release of energy. 818 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 819 Nuclear Fission Often, fission results from a free neutron colliding with a large nucleus. The nucleus absorbs the neutron, forming a highly unstable isotope that breaks up almost instantly. Figure 16.17 shows one of the ways that uranium-235 can split into two lighter nuclei. In the next example, you will calculate the energy released during a fission reaction. 1 0 n compound nucleus 141 56 Ba 235 92 U 236 92 Figure 16.17 in a CANDU nuclear reactor. Absorbing a neutron causes uranium-235 to undergo fission 92 36 Kr Example 16.14 Calculate the energy released by the fission reaction 92Kr 3 0 92U 235 1n → 141 0 56Ba 1n. 36 Given Initial mass: 235 Final mass: 141 92U plus one neutron 56Ba, 36 92Kr, and three neutrons Required energy released (E) Analysis and Solution First, use the atomic mass data on page 881 to calculate the net change in mass resulting from the reaction. mi mf mi U mn m 235 92 235.043 930 u 1.008 665 u 236.052 595 u m 141 56 140.914 412 u 91.926 156 u 3(1.008 665 u) 235.866 563 u mf 0.186 032 u 236.052 595 u 235.886 563 u 3mn m 92Kr 36 Ba Practice Problems 1. Calculate the energy released by the reaction 92U 1 235 0n → 94 40Zr 139 52Te 3 1 0n. 35Br), a 2. A uranium-235 nucleus absorbs a neutron and then splits into a bromine nucleus (87 146La), and lanthanum nucleus ( 57 additional neutrons. How many neutrons are released in this fission reaction? Express this reaction as a balanced equation. 3. How much energy is released in the reaction in question 2? Answers 1. 172.9 MeV 92U 1 2. 235 3. 167.8 MeV 0n → 87 35Br 146 57La 3 1 0n Now, use mass-energy equivalence to calculate the energy released. 1 u is equivalent to 931.5 MeV, so 931.5 MeV 1 u E 0.186 032 u 173.3 MeV Paraphrase The fission of an atom of uranium-235 into barium-141 and krypton-92 releases 173.3 MeV of energy. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 819 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 820 Comparing Chemical Energy with Nuclear Energy When you sit by a campfire, the warmth that you feel is due to the chemical energy released by the combustion of wood. All chemical processes, including combustion, involve electrons moving from one energy level to another. In Chapter 15, you learned that such transitions typically release no more than a few tens of electron volts. Example 16.15 shows that a nuclear process can release a vastly greater amount of energy. Example 16.15 Burning 1 kg of gasoline releases about 4.4 107 J. Compare this energy to the energy released by the fission of 1 kg of uranium-235 into barium-141 and krypton-92. Practice Problems 1. A typical family car requires approximately 1600 MJ of energy to travel 500 km. (a) How many kilograms of gasoline does it take to provide this energy? (b) What mass of uranium-235 would provide the same energy? Answers 1. (a) 36 kg (b) 22 mg Given chemical energy content of gasoline 4.4 107 J/kg Required ratio of the energy content of gasoline to that of uranium-235 Analysis and Solution From Example 16.14, you know that uranium-235 has about 173.3 MeV of nuclear potential energy per atom, assuming fission into barium and krypton. Use the atomic mass of uranium-235 to calculate the number of atoms in 1 kg of this isotope. Then calculate the potential energy per kilogram for comparison with gasoline. m235 92 U 235.043 930 u 1.660 539 1027 kg 1 u 3.902 996 1025 kg Number of atoms in 1 kg of 235 92U 1 kg 3.902 996 1025 kg Energy content of 235 92 U (2.562 134 1024 atoms kg )(173.3 MeV atom ) 2.562 134 1024 4.4402 1026 MeV/kg 4.4402 1032 eV kg 019 J 1 1.60 V e 1 7.10 1013 J/kg Energy content of 235 Energy content of gasoline 92U 7.10 1013 J/kg 4.4 107 J/kg 1.6 106 Paraphrase The nuclear potential energy of 1 kg of uranium-235 is about 1.6 million times greater than the chemical potential energy of 1 kg of gasoline. 820 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 821 Concept Check A nucleus is much smaller than an atom. How does this difference in size make nuclear reactions much more energetic than chemical reactions? Fusion What powers the Sun? The discovery of the nucleus and of mass-energy equivalence provided the key to this question, which had puzzled scientists for thousands of years. In the early 1920s, the British-American astrophysicist Cecilia Payne-Gaposchkin (1900–1979) showed that the Sun consists primarily of hydrogen (about 73%) and helium (about 27%). Noting that four protons have 0.7% more mass than a helium nucleus, the British astrophysicist Arthur Stanley Eddington (1882–1944) suggested that a fusion process might power the stars. In the 1930s, the young German physicist Hans Bethe (1906–2005) worked out the details of how hydrogen nuclei could release energy by fusing together to form helium. In the Sun and smaller stars, the process, called the proton-proton chain (Figure 16.18), has four steps. First, two hydrogen nuclei combine to form deuterium (an isotope of hydrogen with one neutron), an antielectron, and a neutrino. Next, another hydrogen nucleus combines with the deuterium nucleus to produce a helium-3 nucleus and a gamma ray. Then, two of the helium-3 nuclei combine to produce a helium-4 nucleus, two hydrogen nuclei, and a gamma ray. In the final step, annihilation of two positron-electron pairs occurs. Each of these annihilations produces a pair of gamma photons. In order for these reactions to occur, the nuclei must have enough kinetic energy to overcome the electrostatic repulsion between them. Step Reaction Energy Released info BIT In the mid-1930s, Hans Bethe won a $500 prize for a paper on fusion in stars. He used the money to get his mother out of Nazi Germany. Bethe won the Nobel Prize for physics in 1967 and helped found the Bulletin of the Atomic Scientists. proton-proton chain: fusion process in which four hydrogen nuclei combine to form a helium nucleus 1H → 1 2H 0 1 (tw
ice) 0.42 MeV (twice) (twice) 5.49 MeV (twice) 1 2 3 4 Total 1 2 1 1H 2 3 2H → 3 1 2He 2He 2 1 2He → 4 0 1 1H → 4 4 1 0 1 → 2 2He 2 0 1 neutrino positron H2 1 H2 1 positron neutrino 1H (twice) 2 7 ray He3 2 He3 2 ray 12.85 MeV 1.02 MeV (twice) 26.71 MeV ray He4 2 Figure 16.18 The protonproton chain Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 821 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 822 Example 16.16 The Sun radiates about 4 1026 W and has a mass of 1.99 1030 kg. Astronomers estimate that the Sun can convert only the innermost 10% of its hydrogen into helium. Estimate how long the Sun can continue to shine at its present intensity. Given Power 4 1026 W Hydrogen available for conversion 10% of total hydrogen 1.99 1030 kg mSun Required Time the Sun will take to convert 10% of its hydrogen into helium (t) Analysis and Solution The fusion of four hydrogen atoms produces 26.71 MeV. To find the rate at which helium nuclei are produced, divide the Sun’s power by the energy released during the formation of each helium nucleus: Rate of helium production power of Sun energy released per helium atom Practice Problems 1. (a) How many helium nuclei does a star with a power of 1.6 1025 W produce every second? (b) Estimate how much helium this star has produced if it is 4 billion years old. Answers 1. (a) 4.1 1036 (b) 3.4 1027 kg 4 1026 W 26.71 MeV/atom J 4 1026 s V1.60 26.71 0 1 e M V e M 1 m to a 13 J 9.36 1037 atoms/s Since 4 hydrogen atoms are needed for each helium produced, multiply by 4 to find the rate at which the Sun converts hydrogen atoms into helium. Then, convert this rate to mass per second by multiplying it by the mass of a hydrogen atom: k g ms1.67 1027 Rate of hydrogen conversion 49.36 1037 ato o at s m 6.25 1011 kg/s Hydrogen makes up 73% of the mass of the Sun, but only 10% of this hydrogen can be converted into helium. The lifespan of the Sun approximately equals the amount of hydrogen that can be converted divided by the conversion rate. t amount of hydrogen available rate of conversion 1.99 1030 kg 73% 10% 6.25 1011 kg/s 2.32 1017 s or about 7 109 years Paraphrase The Sun can continue to produce energy at its present rate for about 7 billion years. 822 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 823 Concept Check Why can fusion reactions occur only at extremely high temperatures? The cores of massive stars can reach temperatures high enough for helium nuclei to combine to form carbon and oxygen. In some stars, these elements can undergo further fusion. The extent of this nucleosynthesis depends on the star’s density, temperature, and the concentration of the various elements. Current theory suggests that synthesis of elements heavier than iron and nickel occurs only during the explosion of supernovae. Such explosions distribute these elements throughout the cosmos. So, the uranium fuel for today’s nuclear power stations may have come from the explosion of a massive star billions of years ago. A hydrogen-fusion reactor might be an almost ideal energy source. Hydrogen is the most abundant of elements, and the end product, helium-4, is harmless. However, controlling and sustaining a fusion reaction for generating power is extremely difficult. To start the fusion process, the hydrogen has to be heated to a temperature between 45 million and 400 million kelvins, depending on which isotopes are used. Then, this extremely hot gas has to be contained so that the fusion reactions can continue. Some researchers are using powerful lasers to generate the necessary temperatures and magnetic fields to contain the fusion reactions. However, the latest experiments have sustained fusion for only a few seconds and produced only slightly more energy than it took to run the reactor (see Figure 16.19). It will take major technological advances to make fusion power practical. nucleosynthesis: formation of elements by the fusion of lighter elements supernova: sudden, extremely powerful explosion of a massive star e WEB To learn more about fusion reactors, follow the links at www.pearsoned.ca/school/ physicssource. Figure 16.19 The Joint European Toroid (JET) fusion reactor Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 823 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 824 Concept Check There are concerns that tritium could leak from a fusion reactor. Why would tritium be a serious environmental hazard? 16.4 Check and Reflect 16.4 Check and Reflect Knowledge 1. (a) Complete this nuclear reaction: 235 92U → 54 140Xe ? 2 0 1n (b) Does this reaction involve fission or fusion? Justify your answer. 2. What happens to the binding energy per nucleon in a nuclear reaction that releases energy? 3. An iron nucleus of binding energy 492 MeV fuses with a silicon nucleus of binding energy 237 MeV to form a nucleus with binding energy 718 MeV. Will this reaction release energy? Explain why or why not. 4. (a) Which elements are most likely to undergo fission? (b) Which elements are most likely to undergo fusion? 5. A neutron is emitted when aluminium-27 absorbs an alpha particle. (a) What isotope does this reaction create? (b) Write the process for the reaction. Applications 6. (a) Write the reaction formula for the fusion of helium-4 with oxygen-16. (b) How much energy does this reaction release? 7. (a) What particle is emitted when 1H) and tritium (3 deuterium (2 fuse to form helium? 1H) (b) How much energy does this reaction release? 8. A CANDU-6 nuclear reactor can generate 700 MW of electrical power. A CANDU power plant transforms about 27% of its nuclear energy into electrical energy, with the rest being lost primarily as heat. 824 Unit VIII Atomic Physics (a) If the plant uses uranium-235 as fuel and the average energy released per uranium nucleus is 200 MeV, how many nuclei undergo fission each second when the reactor is running at full power? (b) Estimate how many kilograms of uranium-235 a CANDU-6 reactor uses in a year. List any assumptions you make. Extensions 9. (a) In stars much more massive than the Sun, iron-56 will eventually be produced in their centres. Suppose that two iron-56 nuclei fuse. Complete the following reaction and identify the element produced: 56 26Fe → (b) The element formed in the reaction 26Fe 56 in (a) has a mass of 111.917 010 u. Show that this reaction absorbs rather than releases energy. (c) Explain why stars like the Sun do not produce elements heavier than iron. 10. (a) Research the radioactive wastes produced by nuclear reactors. List the major isotopes produced and their half-lives. (b) Briefly outline some of the methods for storing and disposing of these wastes. 11. Compare and contrast the risks and benefits of generating electricity with coal and with nuclear reactors. e TEST To check your understanding of fission and fusion, follow the eTest links at www.pearsoned.ca/school/physicssource. 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 825 CHAPTER 16 SUMMARY Key Terms and Concepts femto proton neutron nucleon atomic number neutron number atomic mass number isotope atomic mass unit (u) strong nuclear force binding energy mass defect alpha radiation beta radiation gamma radiation transmute parent element daughter element beta-negative () decay beta () particle neutrino weak nuclear force antimatter ) positron (e or 0 1 beta-positive () decay gamma () decay radioactive decay series radiation sickness genetic damage radioisotope gray (Gy) sievert (Sv) relative biological effectiveness (RBE) decay constant activity (A) or decay rate becquerel (Bq) half-life fission fusion proton-proton chain nucleosynthesis supernova Key Equations Binding energy: Eb Enucleons Enucleus decay: A ZX → A4 Z2Y 4 2 decay: A ZX* → AX Z Activity: A N t N Nuclear energy released: E (mi mf) c2 Mass defect: m mnucleons mnucleus Nmneutron matom Zm1 1H decay: A ZX → A Z1Y 1 0 decay: A ZX → A 1 Half-life: N N0 Z1Y 0 t t1 /2 1 2 Conceptual Overview Summarize this chapter by copying and completing this concept map. nuclear structure nuclear notation mass-energy equivalence types of radiation nuclear decay binding energy conservation laws discovery of neutrino decay constant decay rate activity half-life mass defect fission fusion Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 825 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 826 CHAPTER 16 REVIEW Knowledge 1. (16.1) Do all nuclei contain more neutrons than protons? Justify your answer. 2. (16.1) Is the atomic mass number for an atom always greater than the atomic number? Justify your answer. 3. (16.1) What is the term for elements that have the same atomic number but different neutron numbers? 4. (16.1) Explain how these nuclei are similar and how they differ: 233 92U, 92 235U, 238 92U. 17. (16.2) Compare the radiation dose that North Americans typically receive each year from radon and from diagnostic X rays. Which of these sources poses the greater health hazard? 18. (16.3) What is the activity of a sample that contains 1.5 1020 nuclei of an element with a decay constant of 1.2 1012 s1? 19. (16.3) After 1.5 h, the number of radioactive nuclei in a sample has dropped from 5.0 1020 to 2.5 1020. How many of these nuclei will remain after another 6 h? 20. (16.3) Explain why carbon-14 dating is not 5. (16.1) How many neutrons are in a nucleus useful for determining the age of a rock sample. of 115 55Cs? How many protons? 21. (16.4) Why do all of the elements used as fuel in 6. (16.1) Convert 50 MeV to joules. nuclear power plants have A > 200? 7. (16.1) Calculate the energy equivalent for 22. (16.4) What is the primary energy source for 1 g of matter. most stars? 8. (16.1) Calculate the energy equivalent for 23. (16.4) List the steps in the proton-proton chain. Applications 24. Calculate the binding energy per nucleon for the following nuclei: (a) 4 (b) 28 (c) 58 (d) 235 2He 14Si 26Fe 92U 25. (a) Write the process for the decay of 52 26Fe. (b) Sh
ow that this process conserves charge and atomic mass number. 26. (a) What parent element decays into lead-208 by emitting an alpha particle? (b) Estimate the kinetic energy of the alpha particle. 27. (a) Write a complete decay process for the transmutation of 30 15P into 30 14Si. (b) Calculate the energy released in this decay. 28. In the oldest campsites yet discovered in Alberta, archaeologists have found materials that contain about a quarter of their original carbon-14. Estimate the age of these campsites. Give your answer to two significant digits. 2.3 u of mass. 9. (16.1) Calculate the mass equivalent for 300 MeV. 10. (16.1) Calculate the binding energy for a nucleus that has a mass defect of 0.022 u. 11. (16.2) Which decay processes do not change the atomic number of a nucleus? 12. (16.2) What is the charge on (a) a beta particle? (b) an alpha particle? (c) a gamma ray? 13. (16.2) Explain how each of these decay processes changes nuclear structure: (a) alpha decay (b) beta decay (c) gamma decay 14. (16.2) Describe this decay in words, identifying the parent element, the daughter element, and 19K → 43 the type of decay: 43 15. (16.2) Which of , , and radiation is the most penetrating, and which is the least penetrating? 20Ca 1 0 . 16. (16.2) Explain why physicists think that radioactivity originates from nuclei. 826 Unit VIII Atomic Physics 16-PearsonPhys30-Chap16 7/24/08 4:32 PM Page 827 29. Until the early 1950s, a paint containing radium-226 was used to make the dials on some clocks, watches, and aircraft instruments glow in the dark. Radium-226 has a decay constant of 1.98 1011 s1. (a) If the activity of one of these clocks is 0.10 MBq, how many atoms of radium-226 are on the dial? (b) Calculate the mass of radium on the dial. (c) The half-life of radium is 1600 years. Calculate the activity that the clock will have in 5000 years. 30. Graph the data in this table. Use your graph to estimate (a) the activity of the sample when t 5 h (b) the half-life of the radioactive material in the sample Time (h) 0 2 4 6 8 10 12 Activity (Bq) 1000.0 697.7 486.8 339.6 236.9 165.3 115.3 Time (h) Activity (Bq) 14 16 18 20 22 24 80.5 56.1 39.2 27.3 19.1 13.3 31. Calculate the energy released when three helium-4 nuclei combine to form a carbon-12 nucleus. 32. You are designing a thermoelectric power supply for a space probe. The probe will need 20 W of electricity for 14.5 years. The efficiency of thermal to electrical energy conversion is 15%. You are considering using polonium-208 as the fuel for the power supply. (a) What is the key advantage of polonium over a chemical fuel? (b) How much polonium will you need? Polonium-208 has a decay constant of 7.57 109 s1 and a half-life of 2.9 years. 84Po decays into 204 208 82Pb. Extensions 33. In 1918, Rutherford observed that bombarding nitrogen atoms with alpha particles produced oxygen and hydrogen. Use nuclear notation to write two reactions that could account for these products. Which reaction is more likely to occur? Explain your reasoning. How could you check your conclusion? 34. There have been over 2000 tests of nuclear weapons, including 711 conducted in the atmosphere or in the ocean. What radioactive products did these tests release? What health hazards result from this radioactive fallout? 35. Research nucleosynthesis in stars. List a sequence of fusion reactions that produces iron-56, and explain why smaller stars do not complete this sequence. How are the fusion reactions in the Sun likely to end? Consolidate Your Understanding 1. Explain how atomic number, atomic mass number, and neutron number are related to the structure of the nucleus. 2. Use the concept of binding energy to explain why some nuclei are more stable than others. 3. Describe the differences between the alpha, beta, and gamma decays. 4. Explain how you can use conservation principles to predict the daughter elements created by a radioactive decay. 5. Distinguish between nuclear fission and nuclear fusion, and explain how to calculate the energy yield from either process. Think About It Review your answers to the Think About It questions on page 789. How would you answer each question now? e TEST To check your understanding of nuclear reactions, follow the eTEST links at www.pearsoned.ca/school/physicssource. Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. 827 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 828 C H A P T E R 17 Key Concepts In this chapter, you will learn about: charge-to-mass ratio quantum mechanical model standard model of matter Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain the discovery and identification of subatomic particles explain why high-energy particle accelerators are required describe the modern model of the proton and neutron compare and contrast elementary particles and their antiparticles describe beta decays Science, Technology, and Society explain the use of concepts, models, and theories explain the link between scientific knowledge and new technologies Skills observe relationships and plan investigations analyze data and apply models work as members of a team apply the skills and conventions of science 828 Unit VIII The development of models of the structure of matter is ongoing. Antimatter, quarks, particles appearing out of nowhere! Although these concepts may seem like science fiction, they are crucial for understanding the nature of matter. You are about to enter the world of undetectable particles that blink in and out of existence. You will see that a calculation by a theoretical physicist in the 1920s led to sophisticated new medical technology that uses a previously unknown form of matter (Figure 17.1). You will learn about the peculiar properties of quarks, the elusive building blocks for protons, neutrons, and many other subatomic particles. Quantum effects can make the subatomic world seem very strange indeed. This chapter introduces some of the most unusual and challenging ideas in all of physics. You will learn that experiments are showing that in some profound ways the universe is stranger than anyone could have imagined a century ago. The theories that you will explore next are exhilarating, difficult, weird, and yet elegant. They are a key to the next century of atomic physics. ▲ Figure 17.1 Recent findings in atomic physics may seem strange, but they have led to amazing advances in technology, as well as better models of the structure of matter. 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 829 17-1 QuickLab 17-1 QuickLab Particle Tracking Simulation Problem What can magnetic tracking reveal about the properties and collisions of objects? Materials lid from shirt box or shoe box magnetic metal marbles glass marbles iron filings metre-sticks or wood slats iron filings box top metre-stick supports magnetic or glass marble ▲ Figure 17.2 Procedure 1 Turn the lid upside down and use metre-sticks or wood slats to support it above a smooth surface such as a tabletop. The gap should allow the marbles to roll freely under the lid. 2 Spread iron filings evenly over the lid (Figure 17.2). 3 Roll a glass marble and a magnetic marble under the lid and observe how they affect the filings. 4 Set the lid aside and place a line of five magnetic marbles spaced about 3 cm apart across the middle of the space between the supports. Estimate what percentage of glass marbles rolled between the supports will hit one of the five magnetic marbles. 5 Shake the lid to spread the filings evenly again and put it back on the supports. Then, roll glass marbles under the lid at least 10 times. After each collision, put the magnetic marbles back in line and spread the filings evenly. Note the number and shape of any tracks resulting from collisions between the glass marbles and the magnetic ones. Watch for any pattern in the formation of the tracks. Questions 1. What can you conclude about the magnetic field from the glass marbles? 2. Calculate the percentage of glass marbles that appeared to collide with the magnetic marbles in step 5. How close was your estimate? Account for any difference between your estimate and your observations. 3. Did any factor appear to affect the length of the collision tracks? 4. How would you expect the tracks to change if you repeated step 5 using round plastic beads instead of glass marbles? 5. How could you use the electric field from charged particles to detect these particles? How could you detect uncharged particles? Think About It 1. How can you tell if a particle is fundamental? 2. What did the measurement of beta decays reveal about the structure of matter? 3. How many fundamental particles are there? Discuss your answers in a small group and record them for later reference. As you complete each section of this chapter, review your answers to these questions. Note any changes in your ideas. Chapter 17 The development of models of the structure of matter is ongoing. 829 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 830 17.1 Detecting and Measuring Subatomic Particles A skilled wilderness guide can tell a great deal about an animal from its tracks, not just identifying the animal but also estimating its age and how fast it was moving (Figure 17.3). In a similar way, physicists use tracks left by subatomic particles to identify the particles, study their interactions, and deduce the structure of matter (Figure 17.4). cloud chamber: a device that uses trails of droplets of condensed vapour to show the paths of charged particles ▲ Figure 17.3 Tracks of an adult snowshoe hare. What do these tracks tell you about the hare’s speed? ▲ Figure 17.4 Tracks of subatomic particles. The heavier particles have straighter tracks. Cloud Chambers and Bubble Chambers A cloud chamber contains dust-free air supersaturated with vapour from a liquid such as water or ethanol. The amount of vapour air can hold depends on temperature and pressure. Air is supersaturated when it conta
ins more vapour than it would normally hold at a given temperature and pressure. So, the liquid and vapour in a cloud chamber are not in equilibrium, and a tiny disturbance can trigger condensation of vapour into droplets of liquid. A charged particle speeding through the supersaturated air will ionize some molecules along its path. The ions trigger condensation, forming a miniature cloud along the trajectory of the speeding particle. This cloud track shows the path of the particle the way a vapour trail formed by condensing exhaust gases shows the path of a jetliner through the sky. Figure 17.5 One of Charles ▲ Wilson’s cloud chambers. The glass sphere is an expansion chamber used to lower the pressure in the cylindrical cloud chamber. 830 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 831 Charles Thomson Rees Wilson (1869–1969) made the first observations of particle tracks in a cloud chamber in 1910 (Figure 17.5). For the next 50 years, cloud chambers were the principal tools of atomic physics. They are to atomic physics what telescopes are to astronomy. The bubble chamber (Figure 17.6) was developed in 1952 by the physicist Donald Glaser (b. 1926). It contains a liquefied gas, such as hydrogen, helium, propane, or xenon. Lowering the pressure in the chamber lowers the boiling point of this liquid. When the pressure is reduced so that the boiling point is just below the actual temperature of the liquid, ions formed by a charged particle zipping through the liquid cause it to boil. Thus, the particle forms a trail of tiny bubbles along its path. Bubble chambers reverse the process used in cloud chambers: particle tracks are formed by a liquid turning into vapour instead of a vapour turning into liquid. info BIT Charles Wilson built the first cloud chamber in 1894 to study how clouds form. He shared a Nobel Prize for his contribution to particle physics. Wilson was a renowned meteorologist and an avid mountaineer. bubble chamber: a device that uses trails of bubbles in a superheated liquid to show the paths of charged particles info BIT CERN stands for Conseil Européen pour la Recherche Nucléaire. It is the world’s largest particle physics laboratory. ▲ Figure 17.6 One of the large bubble chambers at the CERN laboratory near Geneva, Switzerland Neutral particles will not create tracks in a cloud or bubble chamber. However, it is possible to calculate some of the properties of neutral particles from the tracks of charged particles that interact with them. Concept Check Outline possible reasons why neutral particles will not show up in a bubble chamber. How could you tell if a neutron were involved in a particle collision in a bubble chamber? e SIM To see an animation of particle tracks, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 831 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 832 17-2 Inquiry Lab 17-2 Inquiry Lab Building a Cloud Chamber Question Can types of radiation be identified by the characteristics of their tracks? Hypothesis Since alpha, beta, and gamma radiations have different properties, the tracks they produce in a cloud chamber will be different. Materials and Equipment clear glass container flat glass or plastic cover black blotting paper to fit the bottom of the container dry ice (frozen carbon dioxide) reagent grade ethanol (ethyl alcohol) foam plastic insulation tape silicone grease lamp with reflector radiation sources CAUTION: The temperature of dry ice is 78 °C. Handle it only with tongs or thick gloves. Be careful not to damage the casing on the radioactive samples. Variables Identify the manipulated, responding, and controlled variables in this experiment. Procedure Work with a partner or a small group of classmates. 1 Cut a piece of black blotting paper to fit the bottom of the glass container. 2 Saturate this blotting paper with alcohol, but avoid having a pool of alcohol in the container. 3 Cover the container using silicone grease to ensure a good seal between the cover and the container. 4 Use a piece of foam plastic insulation as the base for your cloud chamber. Place a piece of dry ice at least 2.5 cm thick in the centre of this base, then put the 832 Unit VIII Atomic Physics Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork glass container on top of the dry ice. Placing more insulation around the sides of the dry ice will make it last longer. 5 Position the lamp so it shines down from the side of the chamber (Figure 17.7). Darken the room and wait several minutes. Note any changes that you observe in the cloud chamber. 6 Now tape an alpha-radiation source onto the inside of the container near the bottom. Write a description of any tracks that appear. If the tracks have a consistent shape or pattern, sketch a typical track. 7 Repeat step 6 with beta- and gamma-radiation sources. transparent cover T18-01 lamp glass container radiation source blotting paper foam plastic insulation dry ice ▲ Figure 17.7 A simple cloud chamber Analyzing and Interpreting 1. Were all of the tracks you observed produced by the three radiation sources? What else could produce tracks in your cloud chamber? Explain your reasoning. 2. Describe any relationship you see between the appearance of the tracks and the type of radiation that produced them. 3. Suggest improvements to the design of this experiment. Forming Conclusions 4. Do your observations support the hypothesis? If so, which properties of the radiation might be responsible for any differences in the tracks? 5. Under what conditions will subatomic particles travelling through the ethanol cloud not produce observable tracks? 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 833 Extending 6. Hold a strong magnet against the side of the cloud chamber and observe the magnetic field’s effect on tracks from the three radiation sources. Explain whether you could use the magnet to help distinguish between different types of radiation. 7. Make a hypothesis about how taping the radiation sources to the outside of the glass container would affect the tracks produced by each source. Test your hypothesis. Could your results help you distinguish between different types of radiation? What other methods could you use? Analyzing Particle Tracks Physicists use cloud and bubble chambers as a key part of a controlled environment for studying subatomic particles. Applying a magnetic field across the chamber causes charged particles to follow curved or spiral paths. Measurements of the resulting tracks can be used to determine the mass and charge of the particles. For example, Figure 17.8 shows the path of a particle moving in a cloud chamber in which a magnetic field is coming out of the page. The particle entered the chamber from the left. Applying the right-hand rule to this track shows that the particle must have a positive charge. Often, a photograph of a cloud or bubble chamber will show tracks from a number of particles entering the chamber. Once in a while, a single track will suddenly branch into several diverging tracks, as shown in Figure 17.9. Such tracks suggest that the original particle has transformed into two or more different particles. B v F ▲ Figure 17.8 The right-hand rule shows that the particle must have a positive charge. Orient your right hand as shown and then rotate your hand so that your fingers point out of the page. Your palm points in the same direction as the force on a positively charged particle. ▲ Figure 17.9 These tracks suggest that a particle interaction can form two or more different particles. The following example demonstrates how a particle’s track can reveal its charge-to-mass ratio. Example 17.1 Assume that the tracks shown in Figure 17.10 were made by particles moving at a speed of 0.10c through a uniform magnetic field of 30 mT [out of the page]. The initial radius of each track is 5.7 mm. Determine the charge-to-mass ratio for the particles. Then, make a hypothesis about what the particles are. What is unusual about this pair of particles? ▲ Figure 17.10 Why are these particle tracks spiral rather than circular? Chapter 17 The development of models of the structure of matter is ongoing. 833 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 834 Given v 0.10c 3.0 107 m/s 30 mT 0.030 T B r 5.7 mm 0.0057 m Required q m identification of each particle Analysis and Solution • Applying the right-hand rule shows that the particle spiralling clockwise has a positive charge. Similarly, the left-hand rule shows that the particle spiralling counterclockwise has a negative charge. • Since there was no track before the two particles appeared, they must have originated from a photon or a neutral particle. For charge to be conserved, the net charge on the two new particles must be zero. Therefore, these particles must have equal but opposite charges. • The charge-to-mass ratio for a particle moving perpendicular to a PHYSICS INSIGHT The tesla is a derived unit that can be expressed in terms of SI base units: k g 1 T 1 s2 A • m magnetic field can be derived from F F c v2 qv • Since the values of q, v, r, and B are the same for both particles, their masses must also be equal. Practice Problems Substituting the known values gives 1. Measurement of a particle track shows a radius of deflection of 8.66 104 m for a proton travelling at a speed of 4.23 105 m/s perpendicular to a 5.10-T magnetic field. Calculate the charge-to-mass ratio for a proton. 2. Determine the radius of the path of an electron moving at a speed of 3.2 105 m/s perpendicular to a 1.2-mT magnetic field. Answers 1. 9.58 107 C/kg 2. 1.5 mm q m 3.0 107 m/s 0.030 T 0.0057 m 3.0 107 m/s g k 0.0057 m 0.030 s2 A • 1.8 1011 A•s/kg 1.8 1011 C/kg The charge-to-mass ratio for an electron is 1.60 1019 C 9.11 1031 kg 1.76 1011 C/kg. The ratios for protons or small ions are about four orders of magnitude smaller. Para
phrase The charge-to-mass ratio of the negative particle is 1.8 1011 C/kg. Since this ratio matches the ratio for an electron, this particle very likely is an electron. However, the other particle has a charge-to-mass ratio of 1.8 1011 C/kg. This particle appears to be a positron, an antimatter particle. 834 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 835 Concept Check Can you tell whether momentum was conserved in the subatomic process in Example 17.1? Explain your reasoning. Example 17.1 illustrates how a conservation law can be a powerful tool for understanding the interactions of subatomic particles. Physicists often apply the conservation laws for charge, momentum, and mass-energy in this way. Experiments and theoretical calculations have shown that several other quantities are also conserved when particles interact. 17.1 Check and Reflect 17.1 Check and Reflect Knowledge 1. Compare the process for forming tracks in a cloud chamber with the process in a bubble chamber. 2. (a) List two subatomic particles that will leave tracks in a bubble chamber. (b) List two subatomic particles that will not leave tracks in a bubble chamber. 3. (a) Why does applying a magnetic field cause the particle tracks in a cloud or bubble chamber to curve? (b) What can the curvature of a particle’s track in a magnetic field reveal about the particle? Applications 4. Will X-ray photons produce tracks in a bubble chamber? Justify your answer. 5. (a) Determine the type of charge on each particle moving through the magnetic field in this diagram. (b) What information would you need to determine which particle is moving faster? 6. Describe and explain the differences in the tracks made in a bubble chamber by the particles in each pair: (a) protons and alpha particles (b) protons and electrons 7. In this bubble-chamber photograph, a particle enters from the bottom and collides with a helium nucleus. (a) Use conservation of momentum to show that the incoming particle was an alpha particle rather than a proton. (b) Describe how you could show that the particles have a positive charge. Extension 8. Bubble chambers have replaced cloud chambers in many research laboratories. What advantages do bubble chambers have over cloud chambers? e TEST To check your understanding of methods for detecting and measuring subatomic particles, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 835 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 836 17.2 Quantum Theory and the Discovery of New Particles fundamental particle: a particle that cannot be divided into smaller particles; an elementary particle Early in the 20th century, many scientists thought that there were just three fundamental particles: the electron, the proton, and the neutron. However, developments in quantum theory in the 1920s and 1930s suggested the possibility of other subatomic particles, some with peculiar properties. The Discovery of Antimatter In 1928, British physicist Paul Adrien Maurice Dirac (1902–1984) predicted the existence of peculiar particles such as the positive electron in Example 17.1 (pages 833–834). Dirac combined Einstein’s theory of relativity with Schrödinger’s wave equation (described in section 15.5). Dirac’s calculations, with the resulting relativistic wave equation, predicted that antimatter could exist. As mentioned in section 16.2, a particle of antimatter has a key property, such as charge, that is opposite to that of the corresponding particle of ordinary matter. In 1932, the American physicist Carl Anderson (1905–1991) provided the first evidence that antimatter really does exist. He photographed a cloud chamber track of a positron, as shown in Figure 17.11. For this achievement, Anderson won the Nobel Prize for physics in 1936. Concept Check How could Anderson tell that the particle track in Figure 17.11 showed a positively charged particle going down rather than a negative particle going up? His ingenious solution was to pass the particle through a thin lead plate. This plate slowed the particle a bit. Explain how Anderson could use this change in speed to confirm that the particle had positive charge. (Hint: The magnetic field for the cloud chamber in the photograph was directed into the page.) Quantum theory predicts that each kind of ordinary particle has a corresponding antiparticle. One of the startling properties of antimatter is that a collision between a particle and its antiparticle can annihilate both particles and create a pair of high-energy gamma-ray photons travelling in opposite directions. For example, an electronpositron collision can be written as e e → 2 Such electron-positron annihilations are part of the nuclear processes in stars. Note that e is the symbol for a positron. In general, charged antiparticles are represented by simply reversing the sign of the charge on the symbol for the corresponding ordinary particles. Antiparticles for neutral particles are indicated by adding a bar over the symbol for the corresponding ordinary matter. Thus, the symbol for an antineutron is n. ▲ Figure 17.11 A picture worth a Nobel Prize: Anderson’s photograph provided evidence for the existence of the positron. Anderson used this path (the white streak in the photo) to show that the particle that made it had a positive charge but a mass equal to that of the electron. annihilate: convert entirely into energy 836 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 837 Concept Check Consider the example of a head-on collision between a positron and an electron travelling at equal speeds. Explain why momentum would not be conserved if all the energy of the two particles transformed into a single photon. Scientific Knowledge Can Lead to New Technologies The discovery of the positron made it possible to develop a powerful new medical diagnostic instrument. Positron emission tomography (PET) is an imaging technique that uses gamma rays from electron-positron annihilations to produce images of cross sections through a patient’s body. A computer can then generate a three-dimensional image by combining successive plane images (Figure 17.12). The patient receives an injection of a radioactive tracer containing an isotope, usually fluorine-18, that gives off positrons as it decays. As these positrons meet electrons within the patient’s body, they create pairs of gamma-ray photons. Several rings of gamma-ray detectors rotate around the patient. As the photon pairs register on diametrically opposite detectors, a computer builds up an image of the location and concentration of the radioactive tracer. These images can show a wide variety of vital information, such as blood flow, brain function, and the location of tumours. Quantum Field Theory By 1930, Dirac, Heisenberg, Born, and others had established the foundations of quantum field theory. In this theory, mediating particles are the mechanism by which the fundamental forces act over the distance between particles. Particles that mediate a force exist for such a brief time that they cannot be observed. For these virtual particles, energy, momentum, and mass are not related as they are for real particles. To help understand this concept, imagine two people tossing a ball back and forth while standing on a very slippery surface, such as a smooth, wet sheet of ice. Throwing and catching the ball pushes the two people farther and farther apart (Figure 17.13(a)). In this analogy, the people correspond to ordinary particles and the ball corresponds to a mediating particle. For an attractive force, picture the same two people handing a somewhat sticky candy apple back and forth. The force that each person exerts to free the candy apple from the other person’s hand pulls the two people toward each other (Figure 17.13(b)). Note, however, that quantum field theory is a complex mathematical model with aspects that cannot be explained by such analogies. (a) (b) ▲ Figure 17.12 A PET scanner e WEB To learn more about PET scanners, follow the links at www.pearsoned.ca/ school/physicssource. quantum field theory: a field theory developed using both quantum mechanics and relativity theory mediating particle: a virtual particle that carries one of the fundamental forces virtual particle: a particle that exists for such a short time that it is not detectable Figure 17.13 (a) Throwing ▲ a ball back and forth while on a slippery surface pushes these people apart. (b) Handing a sticky object back and forth pulls them together. Chapter 17 The development of models of the structure of matter is ongoing. 837 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 838 quantum electrodynamics: quantum field theory dealing with the interactions of electromagnetic fields, charged particles, and photons gluon: the mediating particle for the strong nuclear force graviton: the hypothetical mediating particle for the gravitational force Project LINK For your unit project, you may want to describe the search for gluons and gravitons. The concept of mediating particles was first applied to the electromagnetic force, in a theory called quantum electrodynamics. This theory states that virtual photons exchanged between charged particles are the carriers of the attractive or repulsive force between the particles. For example, consider the electromagnetic repulsion between two electrons. One electron emits a virtual photon in the direction of the other electron. According to Newton’s third law, the first electron will recoil and its momentum will change by an amount opposite to the momentum of the photon. Similarly, when the second electron absorbs the photon, this electron will gain momentum directed away from the first electron. You can think of the photon for an attractive force as acting a bit like the shared electron holding two atoms together in a covalent chemical bond. In the latter part of the 20th century, calculation
s using a refined version of quantum electrodynamics gave results that matched observed values with amazing accuracy — sometimes to 10 significant digits. Mediating Particles By 1970, research with high-energy particle accelerators led physicists to suggest that the strong nuclear force is mediated by zero-mass particles called gluons. So far, there is only indirect evidence for the existence of gluons. Advances in quantum theory also led to the conjecture that the weak nuclear force is mediated by three particles, designated W, W, and Z0. Experiments using extremely powerful accelerators detected these three particles in 1983. Some physicists think that the gravitational force also has a mediating particle, which they call the graviton. As yet, there is no experimental evidence that gravitons exist. Table 17.1 summarizes the current thinking about mediating particles. ▼ Table 17.1 The Fundamental Forces and Their Mediating Particles Force Range Relative Strength for Protons in Nucleus Particles Mediating Particle Observed? Electromagnetic infinite Weak nuclear <0.003 fm Strong nuclear <1 fm Gravitational infinite 102 106 1 1038 photons W, W, Z0 yes yes gluons indirectly gravitons no 838 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 839 17.2 Check and Reflect 17.2 Check and Reflect Knowledge 1. Describe the difference between ordinary matter and antimatter. 2. Outline how Anderson provided evidence for the existence of the positron. 3. (a) Which fundamental force is the strongest over large distances? (b) Which fundamental force is the weakest at nuclear distances? 4. (a) List the mediating particle for each of the fundamental forces. (b) Which of these mediating particles has not been detected at all? 5. Explain why a PET scan is like being X-rayed from the inside out. Applications 6. (a) What event is represented by the equation e e → 2? (b) Why is the event e e → not possible? 7. (a) Under what conditions will two protons attract each other? (b) Under what conditions will they repel each other? 8. The tracks in this diagram show the creation of two particles in a bubble chamber. Initially, the two particles have the same speed. (a) What evidence suggests that a photon created the two particles? (b) Describe the path of this photon. (c) Which of the tracks shows the path of a positively charged particle? (d) Give two reasons why the other track must show the path of a negatively charged particle. (e) How are the mass and charge of the two particles related? (f) Why is it likely that the interaction involves an antiparticle? 9. Explain how the stability of helium nuclei demonstrates that the electromagnetic force is weaker than the nuclear forces. Extension 10. Research the Lamb shift and the Casimir effect at a library or on the Internet. Explain how these phenomena support the quantum field theory. e TEST To check your understanding of quantum theory and antimatter, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 839 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 840 17.3 Probing the Structure of Matter Our understanding of the structure of matter comes from a series of remarkable technological advances over the past century. For example, the first circular particle accelerator (Figure 17.14(a)) was about 12 cm in diameter and generated particles with energies up to 13 keV. Now, the most powerful accelerators are up to 8.5 km in diameter and can reach energies of a teraelectron volt (1012 eV). This huge increase in energy reflects an interesting overall trend: To probe matter at smaller and smaller scales, physicists need bigger and bigger machines! This trend results from the nature of matter: All of the fundamental forces become markedly stronger at distances less than the diameter of a nucleus. ▲ Figure 17.14 (a) The first circular particle accelerator (b) Fermilab near Chicago, Illinois. Its Tevatron accelerator ring is 2 km in diameter. Energy Requirements 13.6 eV is sufficient to ionize a hydrogen atom. With energies of a few hundred electron volts, you can study electron shells of atoms and of molecules (using a spectrograph, for example). To determine the size of a nucleus, you need charged particles with enough energy to get close to it despite strong electrostatic repulsion. For his ground-breaking scattering experiment, Rutherford used alpha particles with energies in the order of 10 MeV. To examine the structure of the nucleus, the energy requirements are much greater because the probe particles have to overcome the strong nuclear force. Within the nucleus, this short-range force is about a hundred times stronger than the electromagnetic force. The fundamental forces within individual subatomic particles are stronger still. So, probing the structure of stable particles such as protons and neutrons requires even more energy. With early, relatively low-energy accelerators, physicists could conduct experiments in which accelerated particles scattered from nuclei or split nuclei into lighter elements (hence the nickname “atom-smasher”). With high-energy particles, physicists can also study interactions that create new types of particles. Producing some of the heavier particles requires a minimum of several gigaelectron volts. 840 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 841 Natural Sources of Energetic Particles Some naturally radioactive isotopes emit particles that are useful for probing the structure of the atom. For example, Rutherford used polonium and radium as particle sources for his experiments. However, the maximum energy of particles from natural radioactive decay is roughly 30 MeV, which is not enough to probe the structure of nuclei. The other major natural particle source is cosmic radiation. Cosmic rays are high-energy particles that stream into Earth’s atmosphere from outer space. Astronomers are not certain about the origin of these particles. Some of them may come from solar flares and from distant supernovae. About 90% of cosmic rays are protons and most of the rest are alpha particles with a few electrons, positrons, antiprotons, and other particles. The energies of these particles range from roughly 102 to 1014 MeV. The particles from space (primary cosmic rays) rarely reach the ground because they interact with atoms in the atmosphere, producing less energetic secondary cosmic rays. Particle Accelerators The first particle accelerators were built around 1930. These accelerators, and the much more powerful ones developed since then, use electric and magnetic fields to accelerate and direct charged particles, usually in a vacuum chamber. Here is a brief description of some of the major types of particle accelerators. • Van de Graaff: A moving belt transfers charge to a hollow, conductive sphere, building up a large potential difference. This potential difference then propels ions through an accelerator chamber. • Drift Tube: An alternating voltage accelerates charged particles through a series of electrodes shaped like open tubes. The applied voltage reverses as the particles pass through each tube, so the particles are always attracted to the next tube in the line. • Cyclotron: A magnetic field perpendicular to the paths of the charged particles makes them follow circular paths within two hollow semicircular electrodes. An alternating voltage accelerates the charged particles each time they cross the gap between the two electrodes. The radius of each particle’s path increases with its speed, so the accelerated particles spiral toward the outer wall of the cyclotron. • Synchrotron: This advanced type of cyclotron increases the strength of the magnetic field as the particles’ energy increases, so that the particles travel in a circle rather than spiralling outward. Some of the largest and most powerful particle accelerators are synchrotron rings. Concept Check Explain the advantages and disadvantages of studying nuclei with protons from a large accelerator as opposed to alpha particles produced by radioactive decay. primary cosmic rays: high-energy particles that flow from space into Earth’s atmosphere secondary cosmic rays: the shower of particles created by collisions between primary cosmic rays and atoms in the atmosphere Chapter 17 The development of models of the structure of matter is ongoing. 841 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 842 e WEB To learn more about particle accelerators, follow the links at www.pearsoned.ca/ school/physicssource. info BIT Marietta Blau published a number of papers on cosmic rays in the 1920s and 1930s. She was nominated for the Nobel Prize several times. muon: an unstable subatomic particle having many of the properties of an electron but a mass 207 times greater pion: an unstable subatomic particle with a mass roughly 270 times that of an electron lepton: a subatomic particle that does not interact via the strong nuclear force hadron: a subatomic particle that does interact via the strong nuclear force meson: a hadron with integer spin baryon: a hadron with halfinteger spin spin: quantum property resembling rotational angular momentum fermion: particle with halfinteger spin boson: particle with integer spin PHYSICS INSIGHT How Small Are Electrons? Experiments have shown that electrons are less than 1018 m across, while protons are roughly 1.6 1015 m in diameter. Leptons might be mathematical points with no physical size at all! 842 Unit VIII Atomic Physics Although particle accelerators were originally developed for pure research, they now have medical and industrial uses as well. Many hospitals use accelerated particles for generating intense beams of X rays that can destroy cancerous tumours. Bombarding elements with particles from cyclotrons produces radioactive isotopes for diagnostic techniques, radiation therapy, testing structural materials, and numerous other applications. Particle ac
celerators can make a variety of specialized industrial materials by, for example, modifying polymers and implanting ions in semiconductors and ceramics. Accelerators are also powerful tools for analyzing the structure and composition of materials. Particle accelerators have even been used to verify the authenticity of works of art. The Subatomic Zoo In 1937, Carl Anderson and Seth Neddermeyer used a cloud chamber to discover muons in cosmic rays. These particles behave much like electrons, but have a mass 207 times greater and decay rapidly. Ten years later, Cecil Frank Powell discovered π-mesons, or pions, by using a photographic technology that Marietta Blau had developed. This method records tracks of particles on a photographic plate coated with a thick emulsion containing grains of silver bromide. Pions are much less stable than muons, and have some properties unlike those of electrons, protons, or neutrons. Improved particle accelerators and detectors led to the discovery of many more subatomic particles. Over 300 have now been identified. Most of these particles are highly unstable and have lifetimes of less than a microsecond. Studies of the interactions and decays of these particles show that there are two separate families of particles: leptons, which do not interact by means of the strong nuclear force, and hadrons, which do. The term lepton comes from leptos, a Greek word for “thin” or “small,” and hadron comes from hadros, a Greek word for “thick.” The diameters of leptons are much smaller than those of hadrons. The hadrons are divided into two subgroups, mesons (from meso, Greek for “middle”) and baryons (from barus, Greek for “heavy”). One of the key quantum properties for classifying particles is their spin. This property is like angular momentum from rotation of the particle. The spin of a particle can be either an integer or half-integer multiple of Planck’s constant divided by 2. Particles that have half-integer spin or 3 (such as 1 ) are called fermions, while those that have integer spin 2 2 (such as 0, 1, or 2) are called bosons. Leptons and baryons are fermions. Mesons and mediating particles are bosons. Spin can affect the interactions and energy levels of particles. ▼ Table 17.2 Classification of Subatomic Particles Leptons Hadrons Meditating Particles Fermions all leptons Bosons baryons mesons all mediating particles 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 843 There are far more types of hadrons than types of leptons. In fact, physicists have found only six leptons plus their corresponding antiparticles. Table 17.3 compares the mass and stability of the leptons and some of the more significant hadrons. You are not required to memorize this table. Its purpose is to show a tiny set of the dozens of particles that physicists had discovered by the 1960s. What they were desperately seeking, and what you will learn about in the next section, was an underlying theory that could help make sense of this “subatomic zoo.” ▼ Table 17.3 An Introduction to the Subatomic Zoo Particle Symbol Mass (MeV/c2) Lifetime (s) Leptons electron electron neutrino muon muon neutrino tauon tauon neutrino Mesons pions kaons psi upsilon Baryons proton neutron lambda sigma xi omega .511 < 7 106 106 < 0.17 1777 < 24 140 135 494 498 3097 9460 938.3 939.6 1116 1189 1192 1315 1321 1672 stable stable? 2.2 106 stable? 2.9 1013 stable? 2.6 108 8.4 1017 1.2 108 9 1020 8 1021 1.3 1020 1031? 885* 2.6 1010 8 1011 7.4 1020 2.9 1010 1.6 1010 8.2 1011 *lifetime for a free neutron; neutrons in nuclei are stable Units for Subatomic Masses Note that Table 17.3 lists masses in units of MeV/c2. The kilogram is not always the most convenient unit for expressing the mass of subatomic particles. Physicists often deal with transformations between mass and energy using Einstein’s famous equation E mc2. Rearranging E . It follows that mass can be expressed in this equation gives m 2 c energy terms of units of speed of light squared . info BIT A pion will decay in the time it takes light to travel across a classroom. e MATH To better understand the relative sizes of subatomic particles, visit www.pearsoned.ca/school/ physicssource. Particle physicists find it convenient to use a factor of c2 to relate mass to electron volts, the traditional energy unit for particle physics. Conversion factors for such units are 1 eV/c2 1.7827 1036 kg 1 MeV/c2 1.7827 1030 kg 1 GeV/c2 1.7827 1027 kg Chapter 17 The development of models of the structure of matter is ongoing. 843 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 844 For example, the mass of a proton expressed in these units is mp 1.6726 1027 kg 938.23 MeV/c2 1 MeV/c2 1.7827 1030 kg The masses of the known subatomic particles range from 0.5 MeV/c2 to 10 GeV/c2, so exponent notation is usually not necessary with these units. Table 17.4 compares subatomic masses expressed in several common units. ▼ Table 17.4 Comparison of Mass Units for Subatomic Particles (to Five Significant Digits) Particle Mass (kg) Mass (u) Mass (MeV/c2) Electron Proton Neutron 9.1094 1031 5.4858 104 0.511 00 1.6726 1027 1.6749 1027 1.0073 1.0087 938.23 939.52 17.3 Check and Reflect 17.3 Check and Reflect Knowledge 1. Why do physicists require extremely highenergy particles for studying the structure of nucleons? 2. List two natural sources of energetic particles. 3. What is the advantage of high-altitude locations for performing experiments with cosmic rays? 8. (a) Find the energy equivalent of the mass of an electron. (b) The mass of a psi particle is 3.097 GeV/c2. Express this mass in kilograms. 9. Calculate the conversion factor between atomic mass units and MeV/c2. Give your answer to four significant digits. 4. List two uses of particle accelerators in Extensions 10. How has the development of superconducting electromagnets aided research into the structure of matter? 11. (a) What relativistic effect limits the energy of particles accelerated in an ordinary cyclotron? (b) Describe three different ways this limit can be overcome. e TEST To check your understanding of particle accelerators and subatomic particles, follow the eTest links at www.pearsoned.ca/school/physicssource. (a) medicine (b) industry 5. Identify a major difference that distinguishes (a) leptons from hadrons (b) mesons from baryons Applications 6. Can alpha particles from the radioactive decay of polonium be used to probe the nucleus? Explain your answer. 7. Calculate the momentum and kinetic energy of a proton that is accelerated to a speed of (a) 0.01c (b) 5.0 105 m/s 844 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 845 17.4 Quarks and the Standard Model By 1960, physicists faced a large and growing menagerie of subatomic particles. Since the leptons are small and there are only a few types of them, it seemed likely that they were fundamental particles. However, the number of hadrons was a puzzle: Could there really be a hundred or more fundamental particles? info BIT Gell-Mann coined the term “quark” from an obscure line in a novel by the Irish writer James Joyce. Zweig had called the new fundamental particles “aces.” The Quark Model In the 19th century, chemists studied the properties and reactions of the elements. The patterns observed in these properties led to the development of the periodic table and an understanding of the electron structure in atoms. Physicists searched for similar patterns in the properties and interactions of subatomic particles. In 1963, Americans Murray Gell-Mann (b. 1929) and George Zweig (b. 1937) independently proposed that all hadrons are composed of simpler particles, which Gell-Mann called quarks. By grouping the subatomic particles into distinct classes and families, Gell-Mann and Zweig showed that all the hadrons then known could be made from just three smaller particles and their antiparticles. These three particles are now called the up quark, the down quark, and the strange quark. This theory required that the quarks have fractional charges that are either one-third of the charge on an electron or two-thirds of the charge on a proton. Understandably, many physicists had trouble accepting this radical concept. Using the quark model, Gell-Mann accurately predicted not only the existence of the omega (Ω) particle, but also the exact method for producing it. The quark model also accurately predicted key aspects of electronpositron interactions. Stronger evidence for the quark theory came in 1967 when Jerome Friedman, Henry Kendall, and Richard Taylor used the powerful Stanford Linear Accelerator to beam extremely high-energy electrons at protons. The electrons scattered off the protons, somewhat like the alpha particles that scattered off the gold nuclei in Rutherford’s scattering experiment (Figure 17.15). The pattern of the scattered electrons suggested that the mass and charge of a proton are concentrated in three centres within the proton. Later experiments confirmed these results and showed a similar pattern for scattering from neutrons. In the quark model, protons and neutrons contain only up and down quarks. The strange quark accounts for the properties of strange particles, hadrons that decay via the weak nuclear force even though they originate from and decay into particles that can interact via the strong nuclear force. In 1974, the discovery of the psi meson confirmed the existence of a fourth quark, the charm quark. Then, in 1977, the heavy upsilon meson was detected and found to involve a fifth quark, the bottom quark. Since there are six leptons, physicists wondered if there might be an equal number of quarks. In 1995, a large team of researchers at Fermilab found evidence for the top quark. This discovery required a huge accelerator because the top quark is about 40 000 times heavier than the up quark. quark: any of the group of fundamental particles in hadrons info BIT Richard Taylor was born in Medicine Hat and became interested in experimental physics while studying at th
e University of Alberta. In 1990, he shared the Nobel Prize in physics with Friedman and Kendall ▲ Figure 17.15 Scattering of high-energy electrons from a proton strange particle: a particle that interacts primarily via the strong nuclear force yet decays only via the weak nuclear force e WEB To learn more about the strange particles, follow the links at www.pearsoned.ca/ school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 845 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 846 THEN, NOW, AND FUTURE Small Particles and Big Science In the summer of 1936, Carl Anderson and his first graduate student, Seth Neddermeyer, lugged a cloud chamber and photographic equipment to the summit of Pikes Peak in Colorado, about 4300 m above sea level. They chose this location because the cosmic rays at that altitude were then the only source of the high-energy particles needed for their research. The work was lonely, uncomfortable, and poorly funded. It was also highly successful. Anderson and Neddermeyer discovered the muon with the cloud chamber photographs they took during their summer on Pikes Peak. Flash forward 60 years. How particle physics has changed! In 1995, the Fermi National Accelerator Laboratory (Fermilab) near Chicago, Illinois, announced that a team of researchers there had discovered the elusive top quark. In all, 450 people worked on this project. A key member of the team was Melissa Franklin (Figure 17.16), who has worked for over 18 years on the collider detector at Fermilab, a machine for studying the interactions resulting from colliding highenergy protons and antiprotons. A graduate of the University of Toronto and Stanford University, she is now a professor of physics at Harvard. Franklin is seeking to understand the structure of matter at the smallest scale, as Anderson did. However, Franklin collaborates with physicists from around the world and uses particle accelerators and detectors costing hundreds of millions of dollars. Although her work centres on tiny particles, she practises science on a big scale! Questions 1. What other fields of scientific research require huge budgets and international cooperation? 2. What are some advantages and drawbacks of “big science”? 3. Teamwork skills are becoming increasingly important in many areas of research. What other skills would be useful for a career in science? ▲ Figure 17.16 Melissa Franklin Table 17.5 compares some properties of the six quarks. The mass of an individual quark cannot be measured directly. The masses given here were derived mainly by taking the total mass of various particles and subtracting estimates of the mass-energy the quarks gain from motion and interactions via the strong nuclear force within the particles. For each quark there is a corresponding antiquark with the opposite charge. ▼ Table 17.5 Some Properties of Quarks Generation Name Symbol Mass (MeV/c2) Charge First up down Second strange Third charm bottom (or beauty) top (or truth) u d s c b t 1.5–4* 4–8 80–130 1.15–1.35 103 4.1–4.9 103 1.7–1.9 104 Some physicists think the up quark may be essentially massless. info BIT To name quarks, physicists chose words that would not be mistaken for visible physical properties. In Europe, physicists commonly call the top quark “truth” and the bottom quark “beauty.” 846 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 847 Individual quarks probably cannot be observed. The strong nuclear force binds the quarks in a particle very tightly. The energy required to separate quarks is large enough to create new quarks or antiquarks that bind to the quark being separated before it can be observed on its own. Composition of Protons and Neutrons The proton and the neutron contain only first-generation quarks. As shown in Figure 17.17, the proton consists of a down quark and two e e 2 up quarks. The net charge of these three quarks is 2 3 3 e e. The other quantum properties of the quarks also sum to 1 3 those of a proton. Similarly, a neutron consists of two down quarks and an up quark. In this combination, the positive charge on the up quark exactly balances the negative charge on the two down quarks. Composition of Other Hadrons All of the hadrons discovered in the 20th century can be accounted for with a combination of either two or three quarks: • All the mesons consist of a quark and an antiquark. • All the baryons consist of three quarks. • All the antibaryons consist of three antiquarks. However, experiments in 2003 produced strong evidence that the recently discovered theta particle () consists of five quarks: two up quarks, two down quarks, and an antistrange quark. Table 17.6 gives some examples of quark combinations. ▼ Table 17.6 Some Quark Combinations e WEB To learn more about the difficulties in measuring quarks, follow the links at www.pearsoned.ca/ school/physicssource. 2 3 e u u 2 3 e Proton Neutron d 1 3 e ▲ Figure 17.17 The quarks making up protons and neutrons Meson Composition Baryon Composition Antibaryon Composition pion () pion (o) pion () kaon () ud uu ud us proton (p) neutron (n) sigma-plus () sigma-minus () uud udd uus dds antiproton (p) antineutron (n) uud udd Describing Beta Decay Using Quarks and Leptons Recall from section 16.2 that during beta decays of elements, the nuclei emit either an electron or a positron. Since both these particles are leptons, beta decay must proceed via the weak nuclear force. In decay of nuclei, a neutron transforms into a proton, an electron, and an antineutrino. Figure 17.18 shows that a neutron consists of an up quark and two down quarks while a proton consists of two up quarks and a down quark. So, the decay can be written as: udd → uud e e Charge is conserved since the difference between the 1 e charge on the 3 down quark and the 2 e charge on the new up quark equals the charge 3 on the electron emitted by the neutron. Physicists think that the down ▲ Figure 17.18 During decay, a down quark changes into an up quark. Chapter 17 The development of models of the structure of matter is ongoing. 847 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 848 quark emits a virtual W particle (a mediator for the weak nuclear force) that then decays into an electron and an antineutrino: d → u [W] e e e WEB To learn more about the decay of subatomic particles, follow the links at www.pearsoned.ca/ school/physicssource. The idea of mediating particles is essential to understanding beta decay and is a central idea in the standard model. Similarly, in decay of nuclei, an up quark in a proton turns into a down quark by emitting a virtual W particle that then decays into a positron and a neutrino: uud → udd [W] e e The Standard Model standard model: the current theory describing the nature of matter and the fundamental forces The term standard model now refers to a model originally proposed in 1978 to explain the nature of matter and the fundamental forces. Here are some key concepts of this model: electroweak force: a fundamental force that combines the electromagnetic force and the weak nuclear force colour: a quantum property related to the strong nuclear force quantum chromodynamics: quantum field theory that describes the strong nuclear force in terms of quantum colour • All matter is composed of 12 fundamental particles — the 6 leptons and the 6 quarks — plus their antiparticles. • The electromagnetic force and the weak nuclear force are both aspects of a single fundamental force. Sheldon Glashow, Abdus Salaam, and Steven Weinberg developed the theory for this electroweak force in the late 1960s. This theory accurately predicted the existence and masses of the W, W, and Z0 particles. • The electromagnetic and nuclear forces are mediated by virtual particles. As discussed in section 17.2, these mediating particles are the photon, the gluon, and the W, W, and Z0 particles. • All quarks have a quantum property, termed colour, which determines how the strong nuclear force acts between quarks. (Quantum colour is not related to visible colours at all.) The quantum field theory describing the strong nuclear force in this way is called quantum chromodynamics. It is analogous to quantum electrodynamics with colour instead of electric charge and gluons instead of photons. Table 17.7 summarizes the fundamental particles in the standard model. ▼ Table 17.7 Fundamental Particles in the Standard Model Matter Generation First Second Third Quarks Leptons Force Mediating particle(s) up down strange charm bottom top electron electron-neutrino muon muon-neutrino tau tau-neutrino Fundamental Forces Electromagnetic Weak Nuclear Strong Nuclear photon W, W, and Z0 gluon 848 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 849 What’s Next in Quantum Theory? Many theorists are working to combine quantum chromodynamics and the electroweak force theory into a grand unified theory. One such theory suggests that the electromagnetic, strong nuclear, and weak nuclear forces would blend into a single force at distances less than 1030 m, and leptons and quarks could transform from one into the other. However, it would take tremendously high energy to push particles so close together. Although of no relevance for everyday life, such theories could have a great effect on calculations about the origin of the universe. Another challenge is to develop a theory that unifies gravity with the other three forces. One of the most promising approaches is string theory, which treats all particles as exceedingly tiny vibrating strings of mass-energy. The vibration of the strings is quantized (like standing waves). The various kinds of particles are just different modes of vibration, with the graviton being the lowest mode. At present, these theories are highly speculative. The only thing known for sure is that the people who solve these problems will be in line for a Nobel Prize! grand unified theory: quantum theory unifying the electromagnetic, strong nuclear, an
d weak nuclear forces string theory: theory that treats particles as quantized vibrations of extremely small strings of mass-energy Project LINK For your unit project, you may want to describe theories that unify the fundamental forces. 17.4 Check and Reflect 17.4 Check and Reflect Knowledge 1. What experimental evidence suggests that the proton contains three smaller particles? 2. Why is it probably impossible to observe an individual quark on its own? 3. Compare the quark composition of a proton to that of a neutron. 4. Describe the difference between mesons and baryons in terms of quarks. 5. State two differences between leptons and hadrons. 6. List the 12 fundamental particles of matter in the standard model. 8. Is the beta decay → e possible? Justify your answer. e 9. Describe what happens in this decay process: uud → udd [W] e e Extension 10. Explain why a grand unified theory could have a great effect on speculations about the origin of the universe. Applications e TEST 7. (a) Using quark theory, write an equation for the beta decay of a neutron. (b) Show that charge is conserved in this decay process. To check your understanding of fundamental particles and the nature of matter, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 17 The development of models of the structure of matter is ongoing. 849 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 850 CHAPTER 17 SUMMARY Key Terms and Concepts cloud chamber bubble chamber fundamental particle quantum field theory mediating particle virtual particle quantum electrodynamics gluon graviton primary cosmic rays secondary cosmic rays Van de Graaff accelerator drift tube accelerator cyclotron synchrotron muon Key Equations Electron-positron annihilation: e e → 2 Mass units: 1 eV/c2 1.7827 1036 kg pion lepton hadron meson baryon spin fermion boson quark strange particle standard model electroweak force colour quantum chromodynamics grand unified theory string theory decay: udd → uud [W] decay: uud → udd [W] e e e e Conceptual Overview Summarize the chapter by copying and completing this concept map. Nature of Matter involves fundamental particles classed as have corresponding antiparticles leptons quarks generations generations first first second second quark & antiquark have integer spin third third bosons have half-integer spin interact via make up all gluons can interact via 3 quarks fundamental forces mediated by W W Z0 photons gravitons? unified in electroweak unified in 850 Unit VIII Atomic Physics 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 851 CHAPTER 17 REVIEW Knowledge 1. (17.1) How does a bubble chamber detect the path of a charged particle? 2. (17.1) Describe the technological advance in atomic physics made by (a) Charles Wilson (b) Marietta Blau (c) Donald Glaser 3. (17.2, 17.4) (a) In the early 1900s, which three subatomic particles were thought to be the fundamental building blocks of matter? (b) Which of these particles is still thought to be fundamental? 4. (17.2) What theory predicted the existence of antimatter? 5. (17.2) For each of these particles, list the corresponding antimatter particle and explain how it differs from the ordinary matter particle. (a) electron (b) proton 6. (17.2) How does quantum field theory account for fundamental forces acting over a distance? 7. (17.2, 17.3) Describe a similarity and a difference between a muon and a pion. 8. (17.3) What are two advantages of using units of MeV/c2 to express the mass of subatomic particles? 9. (17.3) Compare the size of an electron to that of a proton. 10. (17.4) Describe an experiment that provided evidence for the existence of quarks. 11. (17.4) Give two reasons why Millikan did not detect any quarks with his oil-drop experiment. 12. (17.4) (a) Why did physicists suspect that there might be a sixth quark? (b) What is the name of this sixth quark? (c) Why was a huge accelerator necessary for the discovery of this quark? 15. (17.4) How does the string theory explain the various kinds of subatomic particles? Applications 16. Sketch the paths that alpha, beta, and gamma radiation would follow when travelling perpendicular to a magnetic field directed out of the page. 17. The red tracks in this diagram show a high- speed proton colliding with a hydrogen atom in a bubble chamber, deflecting downward (toward the bottom of the page), and then colliding with another hydrogen atom. These tracks curve clockwise slightly. (a) In which direction is the magnetic field oriented? (b) What conclusions can you make about the mass, speed, and charge of the particles involved in the first collision? (c) What conclusion can you make about the mass, speed, and charge of the particles that made the small spiral tracks? 18. (a) Write a nuclear decay equation to show how fluorine-18 can produce positrons for use in positron-emission tomography. (b) Describe the role that quarks play in this decay process. (c) Write an equation to describe what happens to the positrons within a patient undergoing a PET scan. 13. (17.4) Compare the quark composition of 19. The mass of a top quark is about 176 GeV/c2. antiprotons and antineutrons. Express this mass in kilograms. 14. (17.4) Which two fundamental forces are united 20. (a) Determine the charge on a particle having in the standard model? the quark composition uus. (b) Estimate the mass of this particle. Chapter 17 The development of models of the structure of matter is ongoing. 851 17-PearsonPhys30-Chap17 7/24/08 4:34 PM Page 852 21. The diagram shows a particle track recorded in a bubble chamber at the CERN particle accelerator. There is good reason to suspect that the particle is either an electron or a positron. The magnetic field in the bubble chamber was 1.2 T directed out of the page. 0 2 4 6 8 10 cm (a) Does the particle have a positive or negative charge? Explain your reasoning. (b) Estimate the initial radius of the particle’s path. (c) Determine the initial momentum of the particle. Assume that the particle is an electron or a positron. (d) Why does the particle’s path spiral inward? (e) What could cause the short tracks that branch off from the large spiral track? Extension 22. Research Pauli’s exclusion principle at the library or on the Internet. Write a paragraph describing this principle and the classes of particles to which it applies. Consolidate Your Understanding 1. Describe three experiments that discovered new subatomic particles. Explain how these experiments changed physicists’ understanding of the nature of matter. 2. Give two examples of theories that accurately predicted the existence of previously unknown subatomic particles. 3. (a) Why was the quark theory first proposed? (b) Outline the experimental evidence that supports this theory. (c) Explain to a classmate why the standard model now includes six quarks instead of the three originally suggested by Gell-Mann and Zweig. Think About It Review your answers to the Think About It questions on page 829. How would you answer each question now? e TEST To check your understanding of the structure of matter, follow the eTest links at www.pearsoned.ca/school/physicssource. 852 Unit VIII Atomic Physics 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 853 UNIT VIII PROJECT How Atomic Physics Affects Science and Technology Scenario Atomic physics has enormously influenced the development of modern science and technology. Advances in atomic physics have profoundly changed scientists’ understanding of chemistry, biology, and medicine. A century ago, technology that is taken for granted now would have seemed impossible even in principle. In this project, you will work with two or three classmates to research how the concepts presented in this unit affected an aspect of science or technology. Planning Brainstorm with your classmates to make a list of possible topics. Look for branches of science or technology that apply concepts of atomic physics. Here are some starting points: • medical diagnostic technologies, such as X rays, magnetic resonance imaging (MRI), PET scans, and radioactive tracers • chemistry applications, such as understanding molecular bonds and using spectroscopy to identify compounds • biology topics such as quantum effects in photosynthesis and DNA replication • computer and electronic devices, such as microchips, tunnel diodes, and quantum computers • nanotechnologies such as nanotubes and atomic force microscopes • power technologies, such as nuclear reactors, tokamaks, and radioisotope thermoelectric generators (RTGs) Decide upon a topic to research. Often, you will find it easier to deal with a specific topic rather than a general one. For example, you could focus on nanotubes rather than trying to cover the whole field of nanotechnology. You may want to do some preliminary research on two or three promising ideas to see how much information is available. Consider the best way to present your findings. You might use a written report, an oral presentation, slides, a poster, a model, a video, or a combination of methods. Consult with your teacher on the format of your presentation. Assessing Results Assess the success of your project based on a rubric* designed in class that considers: research strategies clarity and thoroughness of the written report effectiveness of the team’s presentation Materials • library resources, including books and periodicals • Web browser and Internet connection Check with your teacher about any special resources, such as computer software, that you may need for your presentation. Procedure 1 Assign tasks for each group member. Each member should do part of the research as well as some of the preparation for the presentation, such as writing, preparing graphics, or building a model. Clearly identify who is responsible for each part of your project. 2 Once you have gathered basic information about your topic, consider what further research you need to do. For example, you may be able to interview an expert either in person, or by
telephone or e-mail. Many university departments have an outreach program that might suggest an expert you could consult. 3 Check whether the presentation method your group has chosen is suitable for the information you have found during your research. Consult with your teacher if you think you need to change to another type of presentation. If you will be making an oral presentation, practise. Having friends and family critique your presentation can often help you find ways to improve it. Thinking Further • Atomic physics often seems to be too abstract or theoretical to have any relevance to the “real world.” How has your investigation changed your understanding of the relationship between atomic physics and other disciplines? • Does the influence of atomic physics extend beyond science? Can you find ways in which atomic physics has influenced the arts or philosophy? • Science fiction often employs ideas from physics. List several science-fiction books, movies, or television series that use some of the concepts in this unit. How accurate is their treatment of atomic physics? *Note: Your instructor will assess the project using a similar assessment rubric. Unit VIII Atomic Physics 853 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 854 UNIT VIII SUMMARY Unit Concepts and Skills: Quick Reference Summary Resources and Skill Building Concepts Chapter 15 cathode rays Electric force and energy quantization determine atomic structure. 15.1 The Discovery of the Electron Thomson’s experiments showed that cathode rays are subatomic particles with a negative charge. charge-to-mass ratio You can use electric and magnetic fields to measure the charge-to-mass ratio of a particle. charge quantization Millikan’s experiment 15.2 Quantization of Charge Electric charge exists only in multiples of the fundamental unit of charge, e. Millikan’s oil-drop experiment measured the charge on an electron and showed that charge is quantized. 15-1 QuickLab Example 15.2 15-2 QuickLab Example 15.3 eSIM of Millikan’s oil-drop experiment classical model of the atom 15.3 The Discovery of the Nucleus Rutherford’s gold-foil experiment led to the solar-system model with electrons orbiting a tiny positively charged nucleus at the centre of the atom. 15-3 QuickLab Example 15.5 spectra Bohr model energy levels 15.4 The Bohr Model of the Atom Elements and compounds have characteristic emission and absorption line spectra. The Bohr model uses energy levels to account for stability of the atom and to explain line spectra. This model accurately predicts many properties of hydrogen, but has several serious failings. 15-4 Design a Lab Example 15.6 An electron in an atom can occupy only orbits that give the electron discrete, quantized amounts of energy that are inversely proportional to the square of the principal quantum number. Example 15.7 15-5 Inquiry Lab quantum mechanical model 15.5 The Quantum Model of the Atom The wave properties of electrons lead to a powerful new model based on probability distributions. Figures 15.24–15.25 Chapter 16 Nuclear reactions are among the most powerful energy sources in nature. nuclear structure 16.1 The Nucleus Nuclei contain protons and neutrons bound together by the strong nuclear force. mass-energy equivalence Mass and energy are equivalent, and the one can be transformed into the other. binding energy, mass defect The binding energy and mass defect of a nucleus indicate how tightly its nucleons are bound together. nuclear decay, transmutation 16.2 Radioactive Decay Some nuclei spontaneously transmute into a different element by emitting an alpha or beta particle. Nuclei can also give off gamma rays. All three types of radiation can be harmful. Examples 16.1 and 16.2 Example 16.3 Figure 16.3 Example 16.4 Examples 16.5–16.10 16-1 Inquiry Lab 16-2 Design a Lab half-life, activity nuclear reactions 16.3 Radioactive Decay Rates You can use the decay constant and the half-life of a radioisotope to predict the activity of a sample. Examples 16.12 and 16.13 16.4 Fission and Fusion Both the fission of a heavy nucleus into smaller nuclei and the fusion of light nuclei into a single heavier nucleus can release tremendous amounts of energy. Figures 16.17 and 16.18 Examples 16.14–16.16 Chapter 17 The development of models of the structure of matter is ongoing. particle tracks antimatter 17.1 Detecting and Measuring Subatomic Particles The existence and basic properties of subatomic particles can be determined by analyzing the paths of particles in magnetic and electric fields. 17.2 Quantum Theory and the Discovery of New Particles Quantum theory predicted the existence of antimatter, which was confirmed by Anderson’s discovery of the positron. Example 17.1 17-1 QuickLab, 17-2 Inquiry Lab Figure 17.11 quantum field theory According to this theory, the electromagnetic and nuclear forces are mediated by virtual particles. Figure 17.13 particle accelerators families of particles fundamental particles 17.3 Probing the Structure of Matter Particle accelerators produce high-energy particles, which are used to study the structure of matter. Hadrons interact via the strong nuclear force, whereas leptons do not. Bosons have integer spin and fermions have half-integer spin. 17.4 Quarks and the Standard Model The fundamental particles are the six leptons, the six quarks, and their antiparticles. All hadrons consist of a combination of quarks and/or antiquarks. Figure 17.14 Tables 17.2 and 17.3 Tables 17.5,17.6, and 17.7 854 Unit VIII Atomic Physics 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 855 UNIT VIII REVIEW Vocabulary 1. Using your own words, define these terms: absorption line spectrum activity (A) or decay rate alpha radiation antimatter atomic mass number atomic mass unit (u) atomic number baryon becquerel (Bq) beta () particle beta radiation beta-negative () decay beta-positive () decay binding energy Bohr radius boson bubble chamber cathode ray cloud chamber colour cyclotron daughter element decay constant drift tube accelerator electroweak force elementary unit of charge emission line spectrum energy level excited state femto fermion fission Fraunhofer line fundamental particle fusion gamma () decay gamma radiation gluon grand unified theory graviton gray (Gy) ground state hadron ) half-life ionization energy isotopes lepton mass defect mediating particle meson muon neutrino neutron neutron number nucleon nucleosynthesis orbital parent element pion planetary model positron (e or 0 1 primary cosmic rays principal quantum number proton proton-proton chain quantum chromodynamics quantum electrodynamics quantum field theory quark radioactive decay series radioisotope relative biological effectiveness (RBE) secondary cosmic rays sievert (Sv) spectrometer spectroscopy spin standard model stationary state strange particle string theory strong nuclear force supernova synchrotron transmute Van de Graaff accelerator virtual particle weak nuclear force Unit VIII Atomic Physics 855 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 856 Knowledge CHAPTER 15 2. Calculate the electrical charge carried by 1 kg of protons. Give your answer to two significant digits. 3. How many coulombs of charge are on a dust particle that has gained 10 electrons? 4. Calculate the force exerted by an electric field of strength 100 N/C [S] on a dust particle having a charge of 10e. 5. Explain how the path of this particle shows whether its charge is positive or negative. E CHAPTER 16 10. How many neutrons are in a nucleus of gallium 64 31Ga? How many protons? 11. Why is the mass of an atom always less than Zm1 1H Nmneutron? 12. Express the energy equivalent of 0.021 u of mass in electron volts. 13. Express the energy equivalent of 7.0 u in joules. 14. Calculate the binding energy for a nucleus that has a mass defect of 0.0072 u. 15. What is the activity of a sample that contains 1.5 1022 nuclei of an element with a decay rate of 1.5 1013 Bq? 16. Write the decay process for 18 9F, and identify the daughter element. 17. The half-life of sulfur-35 is 87.51 days. How much of a 25-g sample of this isotope will be left after a year? 18. Write the alpha decay process for 228 90Th and 6. Determine whether the charge on this particle is identify the daughter element. positive or negative. 19. Explain the difference between fission and fusion. CHAPTER 17 20. What is a positron? 21. What is a pion? 22. If an electron and positron collide, they annihilate each other and are converted into energy. (a) How much energy does the annihilation of a positron-electron pair produce? (b) Explain why the annihilation must produce two gamma rays with the same wavelength. (c) Estimate the wavelength of these gamma rays. Assume that the kinetic energy of the electron and positron was negligible. 23. What is a quark? 24. Describe this reaction in words: ve p → n e B 7. What is an alpha particle? 8. Here are four energy-level transitions for an electron in a hydrogen atom: 4 → nf ni ni 6 → nf ni ni (a) For which transition(s) does the atom 1 → nf 2 → nf 5 6 1 2 lose energy? (b) For which transition does the atom gain the most energy? 25. Identify the particle formed by each of these (c) Which transition emits the shortest wavelength photon? 9. You are comparing the energy released by two different atomic transitions in a mercury atom. Transition A produces a very bright green line while transition B produces a fainter violetcoloured line. Which of these transitions releases more energy? Explain. 856 Unit VIII Atomic Physics combinations of quarks: (a) uud (c) ud (b) us (d) dds 26. Use quarks to describe how a neutron decays into a proton and an electron. 18-PearsonPhys30-U8-Closer 7/28/08 10:48 AM Page 857 Applications 27. A beam of protons enters a vacuum chamber where the electric field strength is 40 kN/C and the magnetic field strength is 0.55 T. (a) Sketch an orientation of the electric and magnetic fields that could let the protons pass undeflected through the chamber. (b) What
speed must the protons have if they are not deflected by this orientation of the fields? 28. Find the magnetic field strength that will deflect a sodium ion (Na) in an arc of radius 0.50 m when the ion has a speed of 1.0 106 m/s. 29. This diagram shows an electron moving at 2.5 106 m/s through perpendicular electric and magnetic fields. B 0.50 T [out of the page] E 100 N/C [down] v (a) Calculate the electric and magnetic forces acting on the electron. (b) Calculate the net force acting on the particle. 30. An oil droplet with a mass of 1.6 1016 kg is suspended motionless in a uniform electric field of strength 981 N/C [down]. (a) Find the charge on this droplet. (b) How many electrons has the droplet either gained or lost? 31. (a) Find the wavelengths of the first four spectral lines produced by transitions into the n 3 energy level of a hydrogen atom. (b) What part of the electromagnetic spectrum are these lines in? 32. (a) Use the Bohr model to calculate the radius of the n 2 energy level in a hydrogen atom. (b) Find the de Broglie wavelength for an electron in this energy level. (c) Use the formula for the de Broglie wavelength to find the momentum of this electron. (d) Find the electron’s speed and kinetic energy. 33. Calculate the binding energy for 40 20Ca. 34. Identify the nucleus produced in each reaction. (a) (b) (c) 12C → ? 6 14N → ? n 7 206Tl → ? v 81 35. Explain why each of these reactions cannot occur. 5 15C → (a) 6 1H → 3 (b) 3 11Na n → 23 15B ve 2He ve 19F 9 (c) 36. How much energy is released by decay of 16 7N? 37. Some blood-flow tests use iodine-131 as a tracer. This isotope has a half-life of 8.04 days. Estimate the percentage of iodine-131 left after 30 days. 38. How much energy is given off in the alpha decay of neodymium isotope 144 element does this decay produce? 60Nd? What daughter 39. A radioactive sample has an activity of 0.50 MBq and a half-life of 6 h. What will the activity of the sample be after 3.0 days? 40. The proportion of carbon-14 in charcoal used in a cave painting is only 12.5% of the proportion in living trees nearby. Estimate the age of this cave painting. 41. Calculate the amount of energy released when a carbon-12 nucleus absorbs an alpha particle and transmutes into oxygen-16. 42. Calculate the energy released by the reaction 2 1H 2 1H → 3 1H 1 1H. 43. (a) What fundamental particles does a neutron contain, according to the standard model? (b) Show that this combination of particles has zero net charge. Unit VIII Atomic Physics 857 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 858 44. The size of a nucleus is in the order of 1 fm. (a) Calculate the electrostatic force of repulsion between two protons separated by 1 fm. (b) Determine the potential energy of this pair of protons. (c) What keeps a nucleus together despite the electrostatic repulsion between protons? Extensions 45. Two hydrogen atoms in the ground state collide head on and both ionize. Find the minimum speed at which the atoms could have been moving toward each other. 46. (a) Describe the reaction p → 0 p in words. (b) Calculate the minimum energy the photon must have to produce this reaction. 47. In decay, a proton becomes a neutron and the nucleus emits a positron and a neutrino. A proton has less mass than a neutron, and the positron and the neutrino carry away some mass and energy. Explain how such decays conserve mass-energy despite this apparent imbalance. 48. A 5.0-GeV photon creates, via pair-production, an electron and a positron. Calculate the total momentum of the two particles and sketch their motion relative to the path of the original photon. 51. A spill of radioactive material at an industrial site emits 1.25 mGy per hour, measured at a distance of 1.0 m from the spill. The relative biological effectiveness of this radiation is 2. (a) Compare the radiation dose from this spill to exposure from background radiation. (b) At what distance from the spill would the annual absorbed dose be less than 0.1 mSv? (c) A newspaper headline reads “Dangerous Spill at Local Factory.” Is this description fair? Explain why or why not. 52. (a) What is the fundamental difference between a fusion process and one that combines matter and antimatter? (b) Compare the energy released by the fusion of ordinary hydrogen into helium-4 with the energy released by combining two protons with two antiprotons. (c) Why can antimatter not be used for generating power or propelling a spaceship now? 53. Suppose that an electricity generator powered by 1H 3 the fusion reaction 2 overall efficiency of 20%. How much deuterium and tritium will this generator need to produce 10 MWh of electricity, the annual consumption of a typical home? 0n has an 2He 1 1H → 4 49. Imagine that protons and electrons were not Skills Practice charged but could still form a hydrogen atom through gravitational attraction. Calculate the radius of the ground state. (Hint: Assume that the electron travels in a circular orbit and has a total energy me.) Gm p of 2 r 50. A typical banana contains about 0.40 g of potassium. Naturally occurring potassium is mainly 39 isotope 40 1.8 1017 s1. The average atomic mass for natural potassium is 39.1 u. 19K, but 0.012% of it is the radioactive 19K, which has a decay constant of (a) Calculate the activity of a typical banana. (b) Does the radiation exposure from bananas outweigh their health benefit as a source of potassium, fibre, and vitamins A, B6, and C? Explain your reasoning. 54. After two years, 6% remains of the original radioisotope in a sample. Estimate the half-life of this isotope. 55. A nucleus of boron 10 5B absorbs an alpha particle and emits a proton. Use nuclear notation to write this reaction process, and identify the element that it produces. 56. Does an electron that moves from an energy level of 5.1 eV to an energy level of 6.7 eV emit or absorb a photon? Find the wavelength of the photon. 57. How much energy is produced by the conversion of 0.250 u of matter into energy? 58. Calculate the radius of a hydrogen atom in the n 2 state. 858 Unit VIII Atomic Physics 18-PearsonPhys30-U8-Closer 7/25/08 7:36 AM Page 859 59. An electron jumping from the n 3 to the n 2 state in a hydrogen atom emits a 656-nm photon. (a) Which state has the greater energy? (b) Find the energy difference between the Self-assessment 64. Outline how you would describe Rutherford’s gold-foil experiment to a friend. Explain why the results were startling for physicists in 1910. two states. 60. Calculate the binding energy for 24 12Mg. 61. Find the parent atom for this decay: ? → 14N e v 7 62. Calculate the electrical charge of a particle composed of the quarks uus. 63. Find the activity of a sample containing 1.5 1020 radioactive atoms with a decay constant of 3.5 1015 s1. 65. (a) Explain why classical physics predicts that hydrogen will always produce a continuous spectrum rather than discrete spectral lines. (b) How does the Bohr model explain spectral lines? 66. Draw a concept map of the atomic physics topics that you find the most difficult. If you have trouble completing this concept map, discuss the concepts with a classmate or your teacher. 67. Explain why pair annihilation, such as e e → 2, does not violate the law of conservation of mass. 68. List the four fundamental forces and explain which ones are involved in nuclear binding energy, decays, fission, and fusion. e TEST To check your understanding of atomic physics, follow the eTEST links at www.pearsoned.ca/school/physicssource. Unit VIII Atomic Physics 859 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 860 APPENDIX STUDENT REFERENCES Contents SR 1 SR 2 SR 3 SR 4 SR 5 SR 6 SR 7 Safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 861 The Inquiry Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864 The Problem-Solving Process: GRASP . . . . . . . . . . . . . . . . . . 867 Using Graphic Organizers . . . . . . . . . . . . . . . . . . . . . . . . . . . 869 Graphing Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 872 5.1 Graphing Techniques . . . . . . . . . . . . . . . . . . . . . . . . . 872 5.2 Using the Graphing Tools . . . . . . . . . . . . . . . . . . . . . . 874 Math Skills . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 875 6.1 Measurement: Accuracy and Precision . . . . . . . . . . . 875 6.2 Mathematical Operations with Data . . . . . . . . . . . . . . 876 6.3 Exponential Notation and Scientific Notation . . . . . 877 6.4 Unit Conversions (Unit Factor Method) . . . . . . . . . . 878 6.5 Trigonometry for Solving Right Triangles . . . . . . . . . 879 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 880 7.1 SI Prefixes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 880 7.2 Fundamental Quantities and Base Units . . . . . . . . . . 880 7.3 Derived Quantities and Units . . . . . . . . . . . . . . . . . . . 880 7.4 Numerical Constants . . . . . . . . . . . . . . . . . . . . . . . . . . 881 7.5 Atomic Masses of Selected Isotopes . . . . . . . . . . . . . 881 7.6 Masses of Subatomic Particles . . . . . . . . . . . . . . . . . . 881 860 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 861 SR 1 Safety In our modern society, safety has become much more than just protecting one’s well being. Issues around safety have become extremely important to industry, business, governments, and all kinds of institutions including educational institutes. The understanding and application of safety in a broad sense has become an industry in itself. Today, even to be considered for many jobs, one must take safety courses or have a variety of safety-training certificates. Now is the time for you to continue developing an attitude and awareness of safety. Safety is everyone’s responsibility. The Provincial Government, the local school board, your teachers, and you all have an important
role in keeping a safe environment. Alberta Education has prepared a detailed document, “Safety in the Science Classroom”, outlining safety roles and responsibilities, and providing extensive information on potential hazards and safety procedures. This document is available online; go to www.pearsoned.ca/school/physicssource and follow the link to Safety in the Science Classroom. Of particular interest to you in this physics course is Chapter 6: Physical Hazards. For more technical information on issues, follow the link to Health Canada Index and search the topic alphabetically. In general, your role in maintaining safety is to act responsibly by carefully following directions, learning how to recognize potential safety hazards, and how to respond to potentially unsafe situations and emergencies. If you are unsure about how to proceed, ask your teacher. The Canadian Hazardous Products Act requires chemical manufacturers to include all hazard symbols and the degree of hazard on product labels. You may recognize hazard symbols on many household products. These symbols may indicate hazard(s), precaution, and first-aid treatment. Hazardous Product and WHMIS Symbols Household hazardous product symbols indicate the type of danger and the degree of danger. They appear in either a triangle (which means “caution”), a diamond (which means “warning”), or an octagon (which means “danger”). There are also numerous laboratory and industry hazard symbols in use. Some symbols relevant to Physics 20 and 30 are shown below: Figure SR 1.1 Laboratory and industry hazard symbols Many of the chemical products used in Canadian schools are manufactured in the United States. To standardize the labelling systems, WHMIS (the Workplace Hazardous Materials Information System) was developed. The symbols belonging to this system appear on materials and products used both in workplaces and our schools. Appendix Student References 861 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 862 2. Precautions with Mechanical Hazards a) Rotating machinery or moving devices can catch loose clothing, fingers or hair; therefore, keep a safe distance away from moving parts. b) Strong magnets can snap on ferromagnetic materials and other magnets very quickly. Use caution to avoid pinching skin or cutting clothing. c) Projectile launchers should be used only with appropriate eye protection and a clear “line-of-fire.” Be aware of the potential for a misfire or backfire. d) Model rockets with air, water, or solid-fuel motors can be a hazard. Wear eye protection and stay well clear of the launch area and potential trajectory. Make sure everyone watches for rocket parts falling back to the ground. 3. Precautions with Electrical Sources a) Do not use 110-V AC equipment if it has a damaged plug (e.g., missing the ground pin) or a frayed cord. Always disconnect the cord from the socket by pulling the plug, not the cord. b) Keep water and wet hands away from electrical cords. c) Do not touch a person in contact with live electrical currents. Disconnect the power source first. Then give artificial respiration if necessary. Call for help and treat burns. d) Make sure electrical cords are not placed where someone could trip over them. e) Do not allow a short circuit connection to a dry cell or battery. Dangerous amounts of heat can be generated in the wires and in the cells themselves, potentially causing an explosion or fire. f) Never attempt to recharge a non-rechargeable battery. Never cut open batteries. Their contents can be corrosive and poisonous. g) Keep flammable liquids away from electrical equipment. Sparks, in a motor for example, could ignite flammable vapours. Figure SR 1.2 Laboratory Safety Approach all investigations, especially in the laboratory, with maturity. Before you begin, read the instructions carefully, noting all safety precautions. In addition, your teacher may provide other safety reminders and rules pertaining to the laboratory activity. It is your responsibility to inform your teacher of medical conditions such as possible allergies to materials used (e.g., latex) or by-products of the activity. Inform your teacher if you wear contact lenses. 1. General Precautions and Safety Equipment a) Identify all safety equipment in the laboratory. b) Know the location of and how to operate safety equipment, including the fire extinguisher, fire blankets, eyewash fountains, sand, and the first-aid kit. c) Know how and where to get help if needed. d) Wear appropriate laboratory apparel, which may include safety goggles, gloves, and/or lab aprons. e) Tie back long hair and secure any loose clothing. 862 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 863 b) Never look directly into the beam of an operating laser, even one with a low power. The eye focuses the laser light onto the retina, resulting in a power density of about 50 times that of direct sunlight. This can cause pinpoint burns to the retina. c) Guard against stray reflections and turn the laser off when not in use. d) Use radioactive sources only under the direction of your teacher. e) In all cases, the potential for harm from radiation increases with exposure. Exposure can be minimized by limiting the time of use and maximizing the distance away from the source. h) Spark timers create a very short but high voltage spark, which can give a minor electrical shock to anyone who touches a “live” part of the circuit. Although the shock itself is not dangerous, the surprise and sudden reaction can cause elbows to fly or objects to be dropped. i) Some high voltage devices can cause nasty shocks or skin burns. Be aware of the potential danger of charged capacitors, tesla coils, electrostatic generators, and transformers. Use only under the guidance of your teacher. j) When hooking up circuits, always have your teacher check the circuit before turning on the power. 4. Precautions with Electromagnetic Radiation a) Never look directly into an infrared (IR) or an intense, visible ultraviolet (UV) light source. Intense light can harm the retina. UV and IR radiation are absorbed by the cornea and eye contents, and can cause burning and overheating or other damage. Appendix Student References 863 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 864 SR 2 The Inquiry Process Inquire: to seek knowledge of, to ask about, to investigate, or to seek information by asking. Have you ever seen “sun dogs”? They are a common occurrence in Alberta, especially in the winter. Often they are coloured and associated with a coloured ring, or halo, around the sun. They seem to occur in thin clouds or in frosty air. You may have noticed that they are always the same distance from the sun and the colours are always in the same order: red closest to the sun and violet farther away. This set of observations can be the focus of an inquiry process. Figure SR 2.1 Prairie sun dogs The inquiry process is a model of learning incorporated in Alberta Education curricula; it is not a separate topic or option. The inquiry process is applicable to all learning, and is especially suited to learning physics. Learning an inquiry approach is more than a way to succeed in physics; it is a useful way to deal with problems and challenging situations throughout any future career. The inquiry process is non-linear (there may be some side-tracks or dead ends), flexible (you can bend the rules or the process), individual (you can develop your own process), and recursive (you will need to revisit or loop through parts of the process as you go). An inquiry process model contains six components that connect together, all around your own thinking or reflection on the process. These components are: • Planning • Retrieving • Processing • Creating • Sharing • Evaluating 864 Appendix Student References Planning In order to inquire, you must have something in mind about which to inquire. Normally, the planning stage involves recognizing a situation, event, topic, or occurrence for which there is some unknown component. This leads to questions. In the planning stage, you (or your teacher) will need to look at the situation at hand and ask a question to be investigated. You may have many questions, but part of the process is to reflect on the situation and narrow (or in some cases, broaden) your question so that it is something you can actually investigate. From there, you will need to develop your process to lead through the other stages of the inquiry process. One way of working through the planning stage is to ask yourself the following questions: • What do I want to know? • What do I think the result might be? • How can I find out? • What do I need to do to find out? • How will I know when I have found out? • What form will my final results take? • How can I best share my results with others? • How can I evaluate what I have done? In this stage, you will need to develop a clear inquiry question, propose a thesis or hypothesis, identify variables or related factors, create a data or information gathering process (experiment or research strategy), and recognize where your results may end up. Sometimes, this is the most difficult or lengthy part of the entire process, but very important. Retrieving Once you have an inquiry question, a hypothesis to test, and a plan to follow, you can begin the process of retrieving. This may involve experimentation or research. You may be gathering text information or numerical data from measurements. You may be using several different data sources including your own experimental data. In this stage, you may need to revisit Planning if you find difficulties or discrepancies during your information gathering. 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 865 Processing Evaluating The information or data you retrieve must be processed. You must evaluate the information, sort good data from poor data, decide what data is relevant, and perhaps re-identify variables. At this stage, you may find you
rself looping through parts of the Planning and Retrieving stages again. Once you have good information, you need to decide how to display and analyze the data. What combination of data tables, graphs or graphic organizers should you use? Is there some mathematical analysis such as finding a slope, modelling a curve with an equation, or some statistical calculations that will be useful? Again, you may want to loop back to Planning, or ahead to Creating and Sharing to decide what detail of analysis you want to perform on your information. Ultimately, you need to be able to answer the question, “What does this information mean?” Creating You have asked a question, retrieved your information, and processed it. Now you must put the package together by creating a final product. Remember to look ahead to Sharing. In this product, you must clearly indicate your initial inquiry, provide a summary of your data or information, explain the meaning of your data, and state some conclusions regarding your inquiry question. At this point, you may find that you have more questions. This could lead you to another inquiry, and another, and another. This is the essence of how scientific knowledge continues to grow. Sharing This stage in the inquiry process is often devalued, yet it is a crucial step in the process. In science, if a new discovery is not communicated, then it is lost. In education, communicating ideas to others is one of the best learning processes. You don’t really know or understand something until you can share the ideas with others. How you share will depend in part on how you created your final product. There are many modes of sharing: oral presentation, poster display, written report, demonstration, art work, working model, skit… Your job will be to choose a method that best fits you, your results, your classroom situation, and your audience. Again, you may need to loop back to Planning and Creating to get this in the most suitable form. Now is the time to look at the whole process (not just the results or answers). Ask yourself these questions: • What worked well? • What became a challenge? • Is there another or better way of doing any one of the stages in this process? • What parts of the process were easier or more difficult, or more or less effective? • How would I coach someone else to do this same inquiry in a more efficient way? • Are there other questions or situations that might be resolved by the process I followed? • What have I learned (about the inquiry question and about learning)? By critically evaluating what you have done, you will learn process skills beyond physics or science; you will develop skills to last a lifetime. A skeleton of a sample inquiry: Planning Situation: a solar halo display is very clear and colourful in the sky. Question: Where do the sun dog’s colours come from? Hypothesis: This is the same effect as the rainbow. If I spray water into sunlight, then I should be able to see a halo. What to do: Research rainbows and halos in print and online. Do an experiment with light rays and water drops to try to create a halo. How do I know I have the answer: I can create a halo and a rainbow and match them together. My final results: I will have a poster display with photos I have taken, and I will explain the results to my classmates. Retrieving Internet search informs me that halos and rainbows are different. From photos, I see that the colours are reversed. There are different types of halos. Sun dogs are part of ice crystal halos caused by refraction. Revisit planning Question: Does refraction of light through ice or other transparent solid crystals model the position of colours in sun dogs? Appendix Student References 865 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 866 Hypothesis: If I shine a white light beam on a transparent solid crystal, then I will be able to see colours located in different directions. Experiment: Shine light through ice crystals, prisms, or other transparent solids, and look for colours in different directions. Sharing I will give a short presentation to my class to explain my question and results, and model the path of light with white and coloured yarn through a foam block. Classmates will be able to see “colours” only at certain directions. Evaluating I will look critically at the entire process once I am finished. Processing Can I see colours produced by light passing through the crystals? Is there a way for me to quantify my observations, e.g., can I measure directions? To what degree of accuracy can I measure directions? Do I have enough data? Do I need to look at more variables such as the shape or material of the crystal? How can I record my observations in the most meaningful way for this context? Creating I will summarize my specific question and investigative process. My results include: data tables showing the approximate angles where I see different colours of light refracted through prisms of different shapes; three photos of my apparatus set-up and colours I could see; and internet photos of more sophisticated experiments. I will also create diagrams of what could be seen in my apparatus according to theory, and a poster display of the theory behind the experiment and actual halos. There are other questions arising from my work that I will pose on my poster. For example: What happens to the halo when an ice crystal tips on its side? 866 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 867 SR 3 The Problem-Solving Process: GRASP Solving Numerical Problems A significant amount of effort in physics is spent in solving problems that have a numerical component. Often these problems seem more difficult than they really are because they involve physics concepts, principles, and/or laws as well as mathematical operations. Research into the problemsolving abilities of professionals and novices shows that professionals have logical procedures they follow when solving problems while novices who are having difficulty do not. The more methods the problem-solver can apply, the more adept he/she is at problem solving. The approach used in numerical problems throughout the textbook follows four basic steps. These steps are easy to remember and apply because combined, the first letters of the key words spell GRASP. A description of the four steps is provided below. Step 1: List what is Given The first step in solving numerical problems involves answering the question, “What information are we given?” This is sometimes referred to as data extraction. To answer this question, read the problem carefully, study the information given, and represent physical quantities and numerical data with appropriate symbols, units, and directions (if necessary). Write the data in scientific notation to the correct number of significant digits (SR 6.3). Step 2: List what you are Required to find The second step involves answering the question, “What am I required to find?” To answer this question, identify what the problem is asking you to do. Be sure to note the units requested, if specified, and, for vector quantities, the direction. Answering this question will point you in the right direction and prevent you from being distracted by irrelevant information. Step 3: Analysis and Solution — Analyze the problem carefully and work out the Solution The third step requires a careful analysis of the problem before solving. To analyze the problem you must break it down into a series of logical steps. Begin by sketching a diagram. Many physics problems lend themselves to a diagram and the diagram often provides the key to solving the problem. Write down all the relationships you know involving the givens and the required. Also, write down any assumptions that must be made in order to solve the problem. An assumption is anything that must be taken for granted. Next, start with what you are trying to find, and answer the question, “What additional information do I need to calculate the unknown?” This may be a constant that you have to look up in a reference book or from a table of constants given in the textbook. Organize and sequence the information you have to form the solution. In physics, this often involves substituting appropriate data into an equation. It is good practice to rearrange an equation to solve for an unknown variable in terms of the other variables before substituting to obtain the final answer. Always be on the lookout for errors in the mathematical computations and check that the answer has the correct number of significant digits, and that appropriate units are included. Step 4: Paraphrase the solution The numerical answer should be stated in a form that answers the original question. Since the original question was a sentence, the statement of the final answer should also be a complete sentence. Physical quantities should include units and directions, if appropriate. You can use the following Numerical Problem Checklist to guide your work. Numerical Problem Checklist Given read problem carefully extract data represent physical quantities with appropriate symbols include units with physical quantities include directions where needed show the correct number of significant digits Required identify what the problem is asking for identify units of the final answer identify direction of the final answer (for vector quantities) Appendix Student References 867 Paraphrase write the final answer in a complete sentence check that units are included with numerals check accuracy of significant digits check accuracy of direction (if required) check that the original question has been answered and that the answer seems realistic 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 868 Analysis and Solution draw a sketch write down possible relationships list viable assumptions write an appropriate equation containing the needed unknown variable and only other variables that are known or can be found from the given data identify and look up const
ants needed identify inconsistent units and perform needed unit conversions identify “red herrings” (i.e., extra bits of numerical information not needed to solve the problem) note the least number of significant digits in the given data note directions (for vector quantities) rearrange the equation to solve for the unknown substitute data into the rearranged equation simplify the mathematics: solve for the numerical answer and simplify the units check the mathematical calculations check the number of significant digits check the direction (if required) 868 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 869 SR 4 Using Graphic Organizers Graphic organizers are effective tools that can help you learn. They enable you to solve problems and think critically, through analyzing similarities and differences, inferring sequences, and establishing cause-and-effect relationships. They generate discussion and negotiation of ideas, extend comprehension of a concept, theme, or topic, and lead to organized representation and presentation of understandings. You can use them to brainstorm, demonstrate what you know, and organize your thoughts before planning a presentation or writing a report or essay. The following chart outlines a number of graphic organizers, their intended purposes, and how to use them as you study science. Type of Graphic Organizer Purpose Method Concept Map connector descriptor concept Used to clarify relationships and linkages between concepts, events, or ideas Brainstorm ideas and link together from “big to small” with arrows or lines linking words. Venn Diagram Web Diagram Used to visualize similarities and differences between two or more ideas, topics, or concepts Brainstorm similarities, and list these in the overlapping section of the two circles. Then brainstorm differences and list these in the non- overlapping sections. Used to clarify concepts and ideas by clustering them Cluster words and/or information around a central object, concept, or idea. Pie Chart Used to estimate the relationship of parts to the whole Estimate/research the importance or amount of proportionate time of each aspect of an event in relation to the whole. Flowchart/Sequence Chart Used to map out your thinking about an issue or to organize ideas for an essay or report Brainstorm aspects of the whole event. Select important aspects and put them into sequential order. Appendix Student References 869 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 870 Type of Graphic Organizer Purpose Method Ranking Ladder Used to rank ideas in order of importance Brainstorm ideas and rank them in order from least important (bottom rung) to most important (top rung). Fishbone Diagram Used to identify cause-and-effect relationships Right-Angle Diagram Used to explore the consequences of an idea and the impact of its application Target Diagram Used to weigh the importance of facts and ideas Identify a problem to be solved. List the “effect” at the head of the fish. Brainstorm “possible causes” in each bone. Rank the causes and circle the most probable ones, justifying your choice. Briefly describe the idea you are exploring on the horizontal arrow. Brainstorm consequences of the idea, and list these to the right of the horizontal arrow. Expand on one consequence, and list details about it along the vertical arrow. Describe social impacts of that trait below the vertical arrow. Brainstorm facts and ideas. Rank their importance and place the most important facts or ideas centrally, and the least important ones toward the outer rings. 870 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 871 Type of Graphic Organizer Purpose Method Agree/Disagree Chart Used to organize data to support a position for or against an idea or decision List a series of statements relating to a topic or issue. Survey agree-disagreement before discussion. Survey again after discussion and research. PMI (Plus, Minus, Interesting) Chart Used to summarize the positive and negative aspects of a topic or issue, as well as identify interesting aspects of the topic for possible further research Sort ideas or information about a topic or issue in a three-column chart that has the following headings: Plus (), Minus (), and Interesting. Gathering Grid Used to make distinctions between ideas or events Concept Hierarchy Diagram Used to identify and sequence the subordinate concepts needed to understand a higher-order concept Gather information on a number of ideas or events and arrange it on a grid. Each idea or event is assigned to a separate row. Analyze the information according to selected criteria in each specific column. Place the higher-order concept at the top of a page. Then consider the question, “What concepts need to be understood before the higher-order concept above can be grasped?” The same question is then asked for each of the subordinate concepts identified and a hierarchy of connected concepts is created. Appendix Student References 871 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 872 SR 5 Graphing Data 5.1 Graphing Techniques Table SR 5.1 Total mass of system as sand is added to a beaker Total Volume of Sand Added V (mL) Total Mass of System m (g) 28 55 84 106 148 174 210 188 258 333 391 500 567 661 Physicists make extensive use of graphs to convey information and to help determine how one physical quantity is affected by another. To review simple graphical analysis techniques, as an example, use the data from a simple measurement experiment where students added given volumes of sand as measured in a graduated cylinder to a beaker on a balance, recording the total volume of sand in the beaker and the total mass of the system as shown on the balance. The Data Table A data table is the most practical way to record quantitative data. Table SR 5.1 above shows the data from the student experiment of adding sand to a beaker on a balance. Note that the name of each variable, the symbol, and the unit of measurement (in round brackets) are recorded at the top of each column. The Title of the Graph Figure SR 5.2 shows a sample graph for a student’s experiment. Every graph needs a title to describe what it is about. The title is placed at the top of the graph or in a box on a clear area above the graph. The Axes of the Graph Plot the independent variable on the horizontal x-axis and the dependent variable on the vertical y-axis. The variable that is changed intentionally is called the manipulated or independent variable. Volume of sand in the beaker was the manipulated or independent variable in the experiment 872 Appendix Student References 800 600 400 200 ) g ( m s s a M 0 0 Total Mass of System vs. Volume of Sand in Beaker m 640 g 165 g 475 g V 200 mL 20 mL 180 mL 100 200 Volume V (mL) 300 Figure SR 5.2 as students chose how much to add for each trial. The mass of the system depended on how much sand was added, thus the mass was the responding or dependent variable. Label each axis with the name, symbol, and unit of the variable being plotted, as shown in Figure SR 5.2. Scales are chosen for each axis to spread the measured values across the graph paper without making the plotting difficult. The maximum values in the data table determine the maximum numbers on the scales of the axes. To set a scale for an axis, analyze the data to be plotted and choose increments appropriate to the data. In Table SR 5.1, the maximum volume value was 210 mL, and the minimum volume value was 28 mL. Increments of 20 mL on the x-axis would be appropriate for the data. As well, the maximum total mass of system was 661 g, and the minimum total mass of system was 188 g. Increments of 50 g on the y-axis would be appropriate for the data. Plotting the Data and Drawing the Line of Best Fit Use a pencil to plot the data points as accurately as possible by making a small visible dot. Accuracy is important. Use the actual data values and make your best estimate of values within the scale grid. Once all of the data points have been plotted, a line of best fit is drawn. A line of best fit is a line that shows the trend of the points. Make the smoothest curve you can, balancing points that do not fit the curve evenly above and 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 873 below the curve. Do not try to have the curve or straight line go through all the dots since most data points have some error. The scatter of the data points from the smooth line indicates the extent of the errors in the data. Where a point is far off the line, a serious error may have been made. If this occurs, measure the data for that point again, if possible. If you believe an error was made, still plot the point, but ignore it while drawing your best fit curve. Interpolating from the Graph Interpolation is the process of estimating a value that is between two directly measured data points of a variable. To interpolate, first locate the point on the appropriate axis for the value of the variable in which you are interested. Next, draw a line perpendicular to this axis to intercept the line of best fit. From this point on the line of best fit, draw a second line perpendicular to the second axis. Read the value of the second variable from this axis. For example, in Figure SR 5.2, a volume of 70 mL of sand is interpolated to a total mass of 300 g (indicated by the small star). There is some risk of inaccuracy involved in interpolation, since it is assumed that the trend of the line continues between the measured points. This assumption is not always valid. Extrapolating from the Graph Extrapolation is the process of estimating the values of a data point beyond the limits of the known or measured values. However, there is a considerable risk of inaccuracy, because it is assumed that the trend of the curve continues outside the range of the data. When the line is extended, a dotted line is used to show that the extension is little more than guesswork. The arrow in Figu
re SR 5.2 shows the process of extrapolating the curve to find the mass of the system without any sand in the beaker. What is this value and what does it represent? How valid is the value? Calculating the Slope If the line of best fit is straight, the slope of the line can be found. The slope of the line is defined as the rate of change of one variable with respect to the other and is found by the ratio of the rise to the run. To find the slope, find two points far apart on the line of best fit whose values are easily readable from the scales on the axes. Note that the points used should not be data points (i.e. values that were measured and plotted to draw the line of best fit). On the graph, lightly draw a horizontal line from the lower point and a vertical line from the higher point so that they intersect. Use the axis scale to determine the change in vertical value (rise) and change in horizontal value (run) along these two line segments. In Figure SR 5.2, the rise is shown as 475 g and the run as 180 mL. The slope of this graph is thus: slope rise/run 475 g/180 mL 2.64 g/mL Notice that the slope in this example has units; it also has some physical significance: it represents the density of the sand. Writing the Equation of the Line If the trend of the curve is a straight line, then changes in the plotted variables are directly proportional to each other. As the change in one variable doubles, the change in the other doubles, and vice versa. The general equation for a straight line is y mx b, where y is the variable on the vertical axis, x is the variable on the horizontal axis, m is the slope of the line, and b is the vertical axis intercept. Figure SR 5.2 shows a straight line for volumes of sand at least up to 210 mL. For this range, the change in mass is directly proportional to the volume of sand added. The general equation for the linear graph in Figure SR 5.2 is: m V b where m is the total mass of the system, is the density of the sand, V is the total volume of sand in the beaker, and b is the mass of the empty beaker. The specific equation for this graph is: m (2.64 g/mL)V 115 g. Using the Equation of the Line It is often more convenient to extrapolate or interpolate from the specific equation than from the graph. For example, the total mass of the system could be determined if 800 mL of sand is added, even though our beaker may not be that big and our graph does not extend that far. To do this, substitute 800 mL for V in the equation and calculate m. The accuracy of this result depends on the accuracy of the equation, which in turn depends on the accuracy of the determination of slope and vertical axis intercept. Appendix Student References 873 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 874 5.2 Using the Graphing Tools Graphing calculators make the process of plotting and interpreting graphs easy and efficient. Data from an experiment can be entered into the calculator and displayed as a scatterplot. The calculator can be used to determine the function that best models a given scatterplot. The information provided for this function can also be used to write the equation that best describes the relationship between the two plotted variables. The graphing calculator can also be used to help us explore the graph of a given equation or relationship. It can be used to interpolate values between the plotted points or to extrapolate values beyond the plotted points. These uses make the graphing calculator a very powerful laboratory tool. Data can also be stored and plotted in a com- puter spreadsheet such as Microsoft Excel. The eMath activities in this textbook provide opportunities for you to use the graphing calculator or a computer spreadsheet. 874 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 875 SR 6 Math Skills 6.1 Measurement: Accuracy and Precision Measurement is a process of comparing some unknown characteristic, attribute, or quantity to some known or accepted scale or standard. Measurement involves tools and technique. If there is a problem with either, there can be problems with the quality of the measured data. A tape measure with the end broken off would give incorrect measurements if the problem was not noticed. Using a metre stick to measure the thickness of a hair would not work well. Forgetting to include the weight of the fuel when weighing in freight to load a plane could be disastrous. At best, a measurement is an estimate; there is always some amount of uncertainty to the value you record. To make a measurement, and to use measured values correctly, we must understand a number of important issues. Accuracy Accuracy is a means of describing the quality of measurements, or how closely a measurement agrees with the accepted or actual value of the quantity being measured. A broken tape measure will not give accurate values for length. The difference between an observed value (or the average of observed values) and the accepted value is called the deviation. The size of the deviation is an indication of the accuracy. Thus, the smaller the error, the greater is the accuracy. The percent deviation is determined by subtracting the accepted value from the measured value, dividing this by the accepted value, and multiplying by 100. Thus, percent deviation (measured value accepted value) accepted value 100% red-heads. The smaller the divisions of its scale, the less uncertainty there will be in reading values. Any measurement that falls between the smallest divisions on the measuring instrument is an estimate. We should always try to read any instrument by estimating tenths of the smallest division. For a ruler calibrated in centimetres, this means estimating to the nearest tenth of a centimetre, or to 1 mm. Using this procedure, the length of the object in Figure SR 6.1 is found to be 6.7 cm. We are certain of the 6, but the 0.7 is an estimate. In reality, it could easily be 0.6 or 0.8. It is however, unlikely that it would be 0.5 10 11 I 6.7 cm Figure SR 6.1 Figure SR 6.2 shows the measurement of the same object using a ruler calibrated in millimetres. The reading estimated to the nearest tenth of a millimetre appears closest to 6.74 cm. It might be tempting to record the length as either 6.7 cm or 6.8 cm. This would be wrong. We can tell that the length is between the two divisions. The estimated digit is always shown when recording the measurement. The estimated digit in this reading is 0.04 cm 10 11 Precision Precision is the degree of repeatability of measurements; it depends on your care and technique. If the same measurement is carefully made several times independently, we find that we may get variations in the last digit we read. This limitation defines the precision of the measurement. The precision of a measuring instrument depends on how finely the scale is divided. A ruler with mm divisions will not be useful in measuring the difference in hair thickness between blondes and Figure SR 6.2 Figure SR 6.3 shows a different object being measured with a ruler calibrated in centimetres. The length falls exactly on the 6-cm mark. Should the length be recorded as 6 cm or 6.0 cm? Remember that with a centimetre ruler we can estimate to tenths of a centimetre. With this ruler we can therefore distinguish readings of 5.9 cm and 6.1 cm. The object is right on a division mark, Appendix Student References 875 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 876 so the estimated digit is zero-tenths of a centimetre. Zero-tenths is indicated by 0.0 and the correct reading is 6.0 cm, not 6 cm. • All of the digits from one to nine (1, 2, 3, … 9) are significant, so 424.7 m or 0.4247 km each have four significant digits 10 11 Figure SR 6.3 Indicating the Precision of Measured Quantities The precision of a measurement is indicated by the number of decimal places. For example, 2.861 cm is more precise than 581.86 cm even though the latter contains more digits. This is because the three decimal places in 2.861 make it precise to the nearest one-thousandth of a centimetre, while the two decimal places in 581.86 make it precise to the nearest one-hundredth of a centimetre. In this physics textbook, angle measures will have a precision no better than 0.1° (that is, angle measures are either whole numbers or to one decimal place only). Significant Digits The accuracy of a measurement is indicated by the number of significant digits. Significant digits are the specific number of digits used to communicate the degree of uncertainty in a measurement. When we are expressing a physical quantity as a number, how many significant digits should we indicate? These rules should help you decide. • Numbers obtained by counting are considered to be exact and contain an infinite number of significant digits. For example, if there are 12 stopwatches in a classroom, there are not 11 stopwatches, or 13 stopwatches, or 12.35 stopwatches. There are exactly 12.000… stopwatches. The zeros may be extended to as many decimal places as necessary in calculations. • Numbers obtained from definitions are considered to be exact and contain an infinite number of significant digits. For example, 1 m 100 cm, and 1 kWh 3600 kJ, are definitions of equalities. ( 3.141 592 654) has an infinite number of decimal places, as do numbers in formulae such as P 4s, where P is perimeter of a square, and s is the length of a side. 876 Appendix Student References • All zeros to the left of the first non-zero digit are not significant. For example, 1.4 kg and 0.0014 kg each have two significant digits. • Zeros between other non-zero digits are significant. Therefore, 501.009 s has six significant digits. • Alberta Education Assessment Branch considers all trailing zeros to be significant. Any zero to the right of a non-zero digit is significant. Therefore, the mass of an object written as 2000 kg has four significant digits. If the mass is a stated value, not a measured value, we can indicate that we know it to f
our significant digits by using scientific notation and writing the mass as 2.000 103 kg. Good measurements have high degrees of both accuracy and precision. In order to assure better accuracy, instruments should be calibrated. Calibration involves making sure the scale divisions are spaced properly and that the zero reading is correct. Better precision is attained by using better tools, those with finer scales. Good technique, of course, is also necessary for both precision and accuracy. 6.2 Mathematical Operations with Data When doing calculations with measured values, never keep more digits in the final answer than in the least accurate number in the calculation. For example, 0.6 0.32 0.9, not 0.92. This procedure for using only meaningful digits is called rounding off. The procedure for rounding off digits is as follows: • When the first digit discarded is less than five, the last digit retained is left the same. Notice that we start rounding off at the digit immediately after the last digit we are retaining. For example, 14.248 kg rounded to three digits is 14.2 kg, since the fourth digit (4) is less than five. • When the first digit discarded is a five or greater, we increase the last digit retained by one. Therefore, 7.8361 km rounded to three digits is 7.84 km, and 4.255 01 s rounded to three digits is 4.26 s. • Consider numbers that are exact counts to be perfectly precise. For example, the average mass of three cars having masses of 1000 kg, 1250 kg, and 1165 kg is (1000 kg 1250 kg 1165 kg)/3 or 1138 kg. The denominator (3) in this example is an exact count, and therefore the answer includes four significant digits. 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 877 • Consider fractions and defined equalities to be perfectly precise. The fraction 1 in the equa2 1 mv2 does not influence rounding off. tion Ek 2 Neither does the defined equality 10 mm 1 cm. Note: In the examples in this textbook, intermediate steps are shown rounded off to one extra significant digit. In reality, all digits are carried through the calculations until the final answer is reached, at which point the final answer is rounded off appropriately. Rules for Significant Digits in Mathematical Operations Adding or subtracting when it is the only operation: the precision, as shown by the number of decimal places in the values being used, determines the number of significant digits in the answer. Round off the answer to the same precision as the least precise value used. For example, 11.2 kg 0.24 kg 0.336 kg 11.776 kg, is rounded off to 11.8 kg because the least precise value, 11.2 kg, is only given to the first decimal place. Multiplying or dividing when it is the only operation: the value with the least number of significant digits determines the number of significant digits in the answer. For example, a distance of 34.28 m is travelled in a time of 4.8 s. The average speed is calculated by (34.28 m)/(4.8 s) 7.141 666 m/s, rounded off to 7.1 m/s to two significant digits because the time of 4.8 s has only two significant digits. When a series of calculations is performed, each interim value should not be rounded before carrying out the next calculation. The final answer should then be rounded to the same number of significant digits as are contained in the quantity in the original data with the lowest number of significant digits. For example: In calculating (1.23)(4.321) ÷ (3.45 – 3.21), three steps are required: a. 3.45 – 3.21 = 0.24 b. (1.23)(4.321) = 5.314 83 c. 5.314 83 ÷ 0.24 = 22.145 125 The answer should be rounded to 22.1 since 3 is the lowest number of significant digits in the original data. The interim values are not used in determining the number of significant digits in the final answer. When calculations involve exact numbers (counted and defined values), the calculated answer should be rounded based upon the precision of the measured values. For example: 12 eggs 49.6 g/egg 595.2 g (to one decimal place) Note: In this textbook, in calculations involving angle measurements, the number of significant digits in the angle measurements is not taken into consideration when determining the number of significant digits for the final answer. The rules of operation apply to all other measurements. Answers for angle measurements should be no more precise than the least precise in the given angle data. 6.3 Exponential Notation and Scientific Notation Exponential Notation Exponential notation makes use of powers of ten to write large and small quantities and to convey the number of significant digits. The first part of the number is called the coefficient, and the power of ten is the exponent. The radius of Earth may be written in exponential notation to three significant digits as 638 104 m, 63.8 105 m, 6.38 106 m, or 0.638 107 m. The diameter of a typical atom may be expressed to one significant digit as 1 108 cm or 0.1 107 cm. Scientific Notation Any measurement that consists of a coefficient multiplied by a power of ten is expressed in exponential notation. Both 6.38 106 and 0.638 107 are in exponential notation. Scientific notation is a special kind of exponential notation. For a number to be in scientific notation, the coefficient must be greater than or equal to 1, and less than 10. This means that 6.38 106 is expressed in scientific notation and 0.638 107 is not. Scientific notation enables us to show the correct number of significant digits. Remember that any zero to the right of the decimal point is significant. Therefore, if all four digits in the measurement 3400 J are significant, then it would be written in scientific notation as 3.400 103 J. However, if only two digits are significant, it would be as 3.4 103 J. Sometimes you may be required to express the results of calculations in scientific notation. This involves moving the decimal point and changing the exponent until the coefficient is between 1 and 9. The exponent is decreased by one for each position the decimal point in the coefficient is moved to the right, and increased Appendix Student References 877 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 878 by one for each position the decimal point is moved to the left. For example, exponent in the answer remains the same as the largest exponent. For example, a) 500 104 5.00 102 104 5.00 1024 5.00 106 b) 0.068 103 6.8 102 103 6.8 102(3) 6.8 105 Exponential Notation and Mathematical Operations Multiplication The product of exponential numbers is determined by multiplying the coefficients and adding the exponents. For example, (3.0 102)(4.0 106) (3.0 4.0)(102 106) 12 102(6) 12 104 1.2 103 Division To divide numbers written in scientific notation, the coefficients are first divided, and then the exponent in the denominator is subtracted from the exponent in the numerator. For example, 3.3 105 6.6 102 .50 105(2) 0.50 1052 0.50 107 5.0 106 Addition and Subtraction When the Exponents are the Same When the exponents are the same, the coefficients are added or subtracted as in normal arithmetic. The exponent in the final answer remains the same. For example, (2 104) (3 104) (1 104) (2 3 1) 104 4 104 Addition and Subtraction When the Exponents are Different When the exponents are different, the numbers must first be converted to a form in which all exponents are the same. The decimal point is moved so that all have the same exponent as the largest number in the group. Then, the coefficients are added or subtracted accordingly. The 878 Appendix Student References (1.00 103) (2.00 104) (400 105) (1.00 103) (0.200 103) (4.00 103) (1.00 0.200 4.00) 103 4.80 103 6.4 Unit Conversions (Unit Factor Method) Conversions are often necessary in both math and science problems. Whether you are working on a problem involving the metric system or converting moles to grams, the unit factor method is a useful tool. The unit factor method is the sequential application of conversion factors expressed as fractions. They are arranged so that any dimensional unit appearing in both the numerator and denominator of any of the fractions can be cancelled out until only the desired set of dimensional units is obtained Example 1: Convert a speed of 25 m/s to km/h. The equivalent relationships are: 1000 m 1 km, 60 s 1 min, and 60 min 1 h. Multiplying by unit factors and carefully analyzing the units results in 0 6 k m m 25 1 m n i 0 0 s 10 60 m in m 90 km/h h 1 1 s Example 2: 1 ng 109 g, therefore, 9 g 10 n g 1 1 or 9 g n g 10 This type of relation can be used to convert units from one to another. Example 3: Convert 5.3 mL to L. Multiply by factors of 1 to remove the prefix “m” and to introduce “”. 3 L L 10 5.3 103 L 5.3 mL 6 L 10 m L mL “cancel” as do L, leaving the desired units of L. Subtract the exponents for division: 3 (6) 3 This skill is also useful for drawing scale diagrams. For example, a force of 840 N is to be drawn at a scale of 1 cm 50 N. The scale length, in centimetres, will be found thus, 1 840 N 50 cm 16.8 cm N 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 879 6.5 Trigonometry for Solving Right Triangles A right triangle is a special triangle with one right angle (90°). The side opposite the right angle is always the longest side and is called the hypotenuse. The other two sides are called the legs. There are several important relationships that allow us to solve right triangles as long as we know the lengths of any two sides, or the length of one side and the measure of one of the acute angles. φ a Figure SR 6.4 c b Pythagorean Theorem The square of the length of the hypotenuse is equal to the sum of the squares of the length of each of the legs. c2 a2 b2 Sine Ratio Sine of one of the acute angles is equal to the ratio of the length of the leg opposite the angle to the length of the hypotenuse. sin opp/hyp sin a/c Cosine Ratio Cosine of one of the acute angles is equal to the ratio of the length of the leg adjacent the angle to the length of the hypotenuse. cos adj/hyp cos b/c θ Tangent Ratio Tangent of one of the acute
angles is equal to the ratio of the length of the leg opposite the angle to the length of the leg adjacent to the angle. In general, the relationships are: Angle Sum The angles of a plane triangle add to 180°; the acute angles of a plane right triangle add to 90°. tan opp/adj tan a/b Appendix Student References 879 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 880 SR 7 Tables 7.1 SI Prefixes 7.2 Fundamental Quantities and Base Units Prefix Symbol Scientific Notation yotta- zetta- exa- peta- tera- giga- mega- kilo- hecto- deka- deci- centi- milli- micro- nano- pico- femto- atto- zepto- yocto da d c m n p f a z y 1024 1021 1018 1015 1012 109 106 103 102 101 101 102 103 106 109 1012 1015 1018 1021 1024 Quantity Length or Distance Mass Time Electrical current Thermodynamic temperature Quantity of matter Quantity Symbol m t I T n Unit metre kilogram second ampere kelvin mole Unit Symbol m kg s A K mol 7.3 Derived Quantities and Units Quantity Area Volume Speed, Velocity Acceleration Frequency Force Momentum Impulse Quantity Symbol A V v a f F p J Unit square metre cubic metre metre per second metre per second squared hertz newton kilogrammetre per second newtonsecond Energy, Work E, W Power Electric charge Electric potential Electric resistance Activity Absorbed dose Equivalent absorbed dose P q V R A D E joule watt coulomb volt ohm becquerel gray sievert Unit Symbol Expression in terms of SI Base Units m2 m3 m/s m/s2 Hz N — — — — s1 kgm/s2 kgm/s — Ns J W C V Bq Gy Sv kgm/s kgm2/s2 kgm2/s3 As kgm2/s3A kgm2/s3A2 s1 m2/s2 m2/s2 880 Appendix Student References 19-PearsonPhys-Appendix(SR) 7/25/08 7:35 AM Page 881 7.4 Numerical Constants 7.6 Masses of Subatomic Particles Name Symbol Value Particle Symbol Mass (u) Elementary unit of charge Gravitational constant Coulomb’s constant Atomic mass unit Rest mass of an electron Rest mass of a proton Rest mass of a neutron Speed of light Planck’s constant Rydberg’s constant (hydrogen) e G k u me mp mn c h RH 1.60 1019 C 6.67 1011 Nm2/kg2 8.99 109 Nm2/C2 Electron Proton Neutron me mp mn 5.485 799 104 1.007 276 1.008 665 1.66 1027 kg 9.11 1031 kg 1.67 1027 kg 1.67 1027 kg 3.00 108 m/s 6.63 1034 Js 1.097 107 m1 7.5 Atomic Masses of Selected Isotopes Isotope Symbol Atomic Mass (u) Isotope Symbol Atomic Mass (u) hydrogen deuterium tritium helium-3 helium-4 carbon-12 nitrogen-16 oxygen-16 neon-20 neon-22 sodium-22 sodium-23 magnesium-24 silicon-28 silicon-30 phosphorus-30 potassium-39 potassium-40 calcium-40 iron-56 iron-58 cobalt-60 nickel-60 bromine-87 krypton-92 zirconium-94 H H (or D) H (or T) He He C N O Ne Ne Na Na Mg Si Si P K K Ca Fe Fe Co Ni Br Kr Zr 1.007 825 2.014 102 3.016 049 3.016 029 4.002 603 12 (by definition) 16.006 102 15.994 915 19.992 440 21.991 385 21.994 436 22.989 769 23.985 042 27.976 927 29.973 770 29.978 314 38.963 707 39.963 998 39.962 591 55.934 938 57.933 276 59.933 817 59.930 786 86.920 711 91.926 156 93.906 315 tellurium-112 tellurium-139 cerium-140 cesium-140 barium-141 neodymium-144 lanthanum-146 lead-204 lead-208 polonium-208 lead-210 polonium-212 polonium-214 radon-222 radium-226 thorium-230 thorium-234 protactinium-234 uranium-235 uranium-238 Te Te Ce Cs Ba Nd La Pb Pb Po Pb Po Po Rn Ra Th Th Pa U U 111.917 010 138.934 700 139.905 439 139.917 282 140.914 412 143.910 087 145.925 791 203.973 044 207.976 652 207.981 246 209.984 189 211.988 868 213.995 201 222.017 578 226.025 410 230.033 134 234.043 601 234.043 308 235.043 930 238.050 788 Note: Measurements of the atomic mass of most stable isotopes are accurate to at least a millionth of an atomic mass unit. However, the masses of highly unstable isotopes are more difficult to measure. For such isotopes, measurement errors can be large enough that the last one or two digits listed for their masses are not known for certain. Appendix Student References 881 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 882 GLOSSARY Note: The number in parentheses at the end of each definition indicates the page number in this book where the term first appears. A acceleration vector quantity representing the change in velocity (magnitude or direction) per unit time; non-uniform motion (23) acceleration due to gravity constant acceleration of an object falling near Earth’s surface (54) activity (decay rate) number of nuclei in a sample that decay within a given time (812) alpha radiation emission of a helium nucleus; symbol is (797) altitude elevation of the ground above sea level (221) flow of 1 C of charge past ampere a point in a conductor in 1 s (602) amplitude of oscillation maximum displacement of a body from its equilibrium position during oscillatory motion (355) angle of diffraction angle formed between the perpendicular bisector and the straight line to a nodal or antinodal point on the interference pattern (689) angle of incidence angle formed between the incident ray and the normal line (654) angle of reflection angle formed between the reflected ray and the normal line (654) angle of refraction angle formed between the normal line and the refracted ray (666) annihilate energy (836) convert entirely into form of matter that antimatter has a key property, such as charge, opposite to that of ordinary matter (804) antinode point of interaction between waves on a spring or other medium at which only constructive interference occurs; in a standing wave, antinodes occur at intervals ; in an interference pattern, of 1 2 antinodes occur at path difference intervals of whole wavelengths (417) 882 Glossary fundamental armature (rotor) component of a simple DC electric motor consisting of a rotating loop of conducting wire on a shaft (608) artificial satellite artificially created object intended to orbit Earth or other celestial body to perform a variety of tasks; includes weather, communication, observation, science, broadcast, navigation, and military satellites (284) at rest not moving; stationary (13) atomic mass number number of nucleons in the nucleus, Z N; symbol is A (790) 1 of exactly atomic mass unit 1 2 the mass of the carbon-12 atom; symbol is u, where 1 u 1.660 539 1027 kg (791) atomic number number of protons in a nucleus; symbol is Z (790) axis of rotation imaginary line that passes through the centre of rotation perpendicular to circular motion (242) shaft on which a wheel axle rotates (242) B ballistic pendulum type of pendulum used to determine the speed of bullets before electronic timing devices were invented (483) baryon hadron with half-integer spin (842) becquerel unit of activity equal to 1 decay per second; unit is Bq (812) beta-negative decay nuclear decay involving emission of an electron; symbol is (802) beta-positive decay nuclear decay involving emission of a positron; symbol is (805) beta radiation emission of a highenergy electron; symbol is (797) binding energy net energy required to liberate all of the protons and neutrons in a nucleus (793) blackbody object that completely absorbs any light energy that falls on it (705) blackbody radiation curve of the intensity of light emitted versus wavelength for an object of a given temperature (705) graph Bohr radius orbit in a hydrogen atom (774) radius of the smallest boson particle with integer spin (842) bright fringe (antinodal line) of constructive interference (686) region bubble chamber device that uses trails of bubbles in a superheated liquid to show the paths of charged particles (831) C capacitor two conductors, holding equal amounts of opposite charges, placed near one another without touching (642) cathode ray free electrons emitted by a negative electrode (754) central maximum line of antinodes along the perpendicular bisector of the line joining the point sources (426) centre of curvature (C) point in space representing the centre of the sphere from which a curved mirror was cut (657) centre of mass point where the total mass of an object can be assumed to be concentrated (492) centripetal acceleration acceleration acting toward the centre of a circle (244) force acting centripetal force toward the centre of a circle causing an object to move in a circular path; centre-seeking acceleration (244) charge migration movement of electrons in a neutral object where one side of the object becomes positive and the other side becomes negative (520) charge shift movement of electrons in an atom where one side of an atom becomes positive and the other side becomes negative (521) charging by induction process of polarizing an object by induction while grounding it (521) closed pipe (closed tube) pipe closed at one end; the longest wavelength that can resonate in a closed pipe is four times the length of the pipe (419) closed-pipe (closed-tube) resonance if an antinode occurs at 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 883 the open end of a pipe, a point of resonance (resulting from constructive interference) occurs at the open end of the pipe, and the sound appears to be amplified (419) cloud chamber device that uses trails of droplets of condensed vapour to show the paths of charged particles (830) along the same straight collinear line, either in the same or in opposite directions (71) non-collinear not along the same straight line (80) collision interaction between two objects where each receives an impulse (469) elastic collision collision in which the total kinetic energy of Ekf ) a system is conserved (Eki (481) inelastic collision collision in which the total kinetic energy of a system is not conserved (Eki Ekf ) (483) colour quantum property related to the strong nuclear force (848) commutator fundamental component of a simple DC electric motor consisting of a mechanism for maintaining a properly polarized connection to the moving coil in a motor or generator (608) components perpendicular parts (Rx and Ry) into which a vector can be separated (77) Compton effect length of the scattered X-ray photon (721) change in wave- Compton scattering an X ray by an electron (721) scattering of conduction process of charging an object thr
ough the direct transfer of electrons when a charged object touches a neutral object (519) conductor material in which electrons in the outermost regions of the atom are free to move (513) forces that act conservative forces within systems but do not change their mechanical energy; includes gravity and elastic forces (314) non-conservative forces such as friction, and forces applied from outside a system, that cause the energy of the forces, system to change so that energy is not conserved (319) converging lens lens that refracts rays travelling parallel to the principal axis inward to intersect at the principal focus (677) converging mirror concave reflecting surface that causes parallel light rays to converge after being reflected (657) coulomb (C) SI unit for electric charge, equivalent to the charge on 6.25 1018 electrons or protons (517) Coulomb’s law magnitude of the force of electrostatic attraction or repulsion (⏐F e⏐) is directly proportional to the product of the two e⏐ q1q2) and charges q1 and q2 (⏐F inversely proportional to the square of the distance between their centres r (529) crest region where the medium rises above the equilibrium position (394) for any two media, critical angle the size of the incident angle for which the angle of refraction is 90° (672) current quantity of charge that flows through a wire in a given unit of time (602) cycle one complete back-and-forth motion or oscillation (249) cyclotron particle accelerator Particle accelerator in which the magnetic field perpendicular to the paths of the charged particles makes them follow circular paths within two hollow semicircular electrodes. An alternating voltage accelerates the charged particles each time they cross the gap between the two electrodes. The radius of each particle’s path increases with its speed, so the accelerated particles spiral toward the outer wall of the cyclotron. (841) D dark fringe (nodal line) destructive interference (686) region of daughter element duced by a decay process (799) element pro- decay constant probability of a nucleus decaying in a given time; symbol is (811) diffraction change in shape and direction of a wave front as a result of encountering a small opening or aperture in a barrier, or a corner (685) sheet of glass diffraction grating or plastic etched with a large number of parallel lines; when light is incident on the grating, each line or slit acts as one individual light source (692) diffuse (irregular) reflection behaviour describing parallel incident rays scattered in different directions when reflected from an irregular surface (653) dispersion separation of white light into its components (675) displacement straight line between initial and final positions; includes magnitude and direction (7) length of the path taken distance to move from one position to another (7) lens that refracts diverging lens rays travelling parallel to the principal axis outward to appear as though they have originated at a virtual principal focus (677) convex reflecting diverging mirror surface that causes parallel light rays to spread out after being reflected (657) diverging ray ray that spreads out as it moves away from the origin (397) domain region of a material in which the magnetic fields of most of the atoms are aligned (589) Doppler effect apparent change in frequency and wavelength of a wave that is perceived by an observer moving relative to the source of the wave (429) drift tube particle accelerator particle accelerator in which alternating voltage accelerates charged particles through a series of electrodes shaped like open tubes; particles are always attracted to the next tube in the line (841) Glossary 883 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 884 dynamics branch of mechanics dealing with the cause of motion (126) E eccentricity degree to which an ellipse is elongated; number between 0 and 1, with 0 being a perfect circle and 1 being a parabola (269) efficiency ratio of the energy output to the energy input of any system (324) elastic potential energy energy resulting from an object being altered from its standard shape, without permanent deformation (300) electric field lines lines drawn to represent the electric field; density of the lines represents the magnitude of the electric field (554) electric poten- electric potential tial energy stored per unit charge at a given point in an electric field; symbol is V (564) electric potential difference change in electric potential experienced by a charge moving between two points in an electric field (565) electric potential energy energy stored in a system of two charges a certain distance apart; change in electric potential energy equals work done to move a small charge (Ep electromagnet magnet having its magnetic field produced by electric current flowing through a coil of wire (588) W) (561) electromagnetic radiation (EMR) radiant energy in the form of a wave produced by the acceleration of electrons or other charged particles; does not require a material medium; can travel through a vacuum (636) electron volt change in energy of an electron when it moves through a potential difference of 1 V; unit is eV (564) electrostatics charges at rest (513) study of electric fundamental electroweak force force that combines the electromagnetic force and the weak nuclear force (848) 884 Glossary elementary unit of charge on a proton; symbol is e (762) charge elongated circle; consists ellipse of two foci, a major, and a minor axis (269) energy ability to do work (292) energy level discrete and quantized amount of energy (773) equilibrium position rest position or position of a medium from which the amplitude of a wave can be measured (394) excited state higher than the ground state (775) any energy level F femto metric prefix meaning 1015 (790) fermion particle with half-integer spin (842) ferromagnetic having magnetic properties, like those of iron (589) field three-dimensional region of influence surrounding an object (200) electric field three-dimensional region of electrostatic influence surrounding a charged object (641) gravitational field region of influence surrounding any object that has mass (200) magnetic field three-dimensional region of magnetic influence surrounding a magnet, in which other magnets are affected by magnetic forces (584) first order maximum line of antinodes resulting from a onewavelength phase shift (427) fission reaction in which a nucleus with A 120 splits into smaller nuclei that have greater binding energy per nucleon (818) focal length (f) distance from the vertex to the focal point, measured along the principal axis; related to the radius of curvature by f r/2 (657) force quantity measuring a push or a pull on an object; measured in newtons (127) action force object A on object B (160) force initiated by action-at-a-distance force force that acts even if the objects involved are not touching (200) force exerted applied force directly on an object by a app (130) person; symbol is F gravitational force attractive force between any two objects due to their masses; symbol is F g (196) net force vector sum of all the forces acting simultaneously on net (131) an object; symbol is F force on an object normal force that is perpendicular to a common contact surface; N (130) symbol is F reaction force object B on object A (160) force exerted by force acting restoring force opposite to the displacement to move an object back to its equilibrium position (353) strong nuclear force binds together the protons and neutrons in a nucleus (793) force that weak nuclear force tal force that acts on electrons and neutrinos (804) fundamen- forced frequency frequency at which an external force is applied to an oscillating object (382) Fraunhofer line dark line in the spectrum of the Sun (773) free-body diagram vector diagram of an object in isolation showing all the forces acting on it (129) situation in which the only free fall force acting on an object that has mass is the gravitational force (226) frequency number of cycles per second measured in hertz (Hz) (249) friction force that opposes either the motion of an object or the direction the object would be moving in if there were no friction; symbol is f (169) F fundamental forces basic forces of nature that physicists think underlie all interactions in the universe (194) fundamental frequency lowest frequency produced by a particular instrument; corresponds to the standing wave having a single antinode, with a node at each end of the string (422) fusion reaction in which two lowmass nuclei combine to form a single nucleus with A 60; the resulting nucleus is more tightly bound (818) 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 885 G gamma decay emission of a highenergy photon by a nucleus; symbol is (806) gamma radiation emission of a high-energy photon; symbol is (797) generator effect (electromagnetic induction) production of electricity by magnetism (611) gluon mediating particle for the strong nuclear force (838) grand unified theory quantum theory unifying the electromagnetic, strong nuclear, and weak nuclear forces (849) sensitive instrument gravimeter used to detect small variations in the magnitude of the gravitational field strength on Earth’s surface (222) gravitational field strength gravitational force per unit mass at a specific location (201) gravitational mass mass measurement based on comparing the known weight of one object to the unknown weight of another object (199) gravitational potential energy energy of an object due to its position relative to the surface of Earth (295) graviton hypothetical mediating particle for the gravitational force (838) gravity assist use of the gravitational force exerted by celestial bodies to reduce interplanetary travel times (214) gray dose of ionizing radiation that delivers 1 J of energy to each kilogram of material absorbing the radiation; unit is Gy (809) ground state energy level (774) lowest possible grounding process o
f transferring charge to and from Earth (521) H hadron subatomic particle that interacts via the strong nuclear force (842) half-life time it takes for half of the radioactive nuclei in a sample to decay (812) Heisenberg’s uncertainty principle it is impossible to know both the position and momentum of a particle with unlimited precision at the same time (735) high tide highest level of ocean water that occurs near Earth’s coastlines (211) Hooke’s Law relationship where the stretch produced by a force applied to a spring is proportional to the magnitude of the force (299) horsepower (hp) unit used to identify the power output of motors, mainly in the automotive industry (324) Huygens’ Principle model of wave theory, which predicted the motion of a wave front as being many small point sources propagating outward in a concentric circle at the same speed as the wave itself (684) I image attitude orientation characteristic of an image, whether erect or inverted (656) image position where the image forms relative to the surface of the mirror (656) image type distinction between real and virtual images (656) real image image from which light rays come; can be formed on a diffusely reflecting surface or screen (654) virtual image image from which light rays appear to come; cannot be formed on a non-reflective surface or screen (654) impulse product of the net force on an object and the time interval during an interaction. Impulse causes a change in the momentum of the object. (457) incandescent (704) glowing with heat induction movement of charge caused by an external charged object (520) inertia property of an object that resists acceleration (138) inertial mass mass measurement based on the ratio of a known net force on an object to the acceleration of the object (148) insulator material in which the electrons are tightly bound to the nucleus and are not free to move within the substance (513) interference (or two waves) crossing within a effect of two pulses medium; the medium takes on a shape that is different from the shape of either pulse alone (411) constructive interference overlap of pulses to create a pulse of greater amplitude (412) destructive interference overlap of pulses to create a pulse of lesser amplitude (412) interference fringes ence pattern of light and dark bands (686) fixed interfer- interference pattern pattern of maxima and minima resulting from the interaction of waves, as crests and troughs overlap while the waves move through each other (425) ionization energy energy required to remove an electron from an atom (775) atoms that have the same isotopes number of protons, but different numbers of neutrons (791) K kinematics branch of physics that describes motion (6) kinetic energy energy due to the motion of an object; symbol is Ek (302) kinetic friction force exerted on an object in motion that opposes the motion of the object as it slides on another object; symbol is F fkinetic (176) coefficient of kinetic friction proportionality constant relating F fkinetic and FN (183) L latitude south of the equator (221) angular distance north or law of conservation of charge net charge of an isolated system is conserved (517) law of conservation of energy within an isolated system, energy may be transferred from one object to another or transformed from one form to another, but it cannot be increased nor decreased (312) law of conservation of momentum momentum of an isolated system is constant (473) law of magnetism like magnetic ends repel and unlike ends attract each other Glossary 885 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 886 law of reflection angle of reflection is equal to the angle of incidence and is in the same plane (654) Lenz’s law direction of a magnetically induced current is such as to oppose the cause of the current (618) lepton subatomic particle that does not interact via the strong nuclear force (842) lowest level of ocean low tide water that occurs near Earth’s coastlines (211) M magnification relationship of the size of the image to the size of the object (656) mass defect difference between the sum of the masses of the separate nucleons and the mass of the nucleus; symbol is m (794) maximum (line of antinodes) of points linking antinodes that occur as the result of constructive interference between waves (426) line Maxwell’s Equations series of equations that summarized the relationships between electricity and magnetism, and predicted the existence of electromagnetic waves and their propagation through space (642) mechanical energy sum of potential and kinetic energies; symbol is Em (306) increase in mechanical resonance amplitude of oscillation of a system as a result of a periodic force whose frequency is equal or very close to the resonant frequency of the system (382) mechanics statics, and dynamics (306) study of kinematics, medium material, for example, air or water through which waves travel; the medium does not travel with the wave (394) meson hadron with integer spin (842) minimum (nodal line) points linking nodes that occur as the result of destructive interference between waves (426) line of mirror equation equation relating the focal length of a curved mirror to the image and object distances (662) 886 Glossary momentum product of the mass of an object and its velocity (449) sum of momentum (of a system) the momenta of all the objects in the system (470) motor effect force deflecting force acting on a charged particle moving in a magnetic field (593) muon unstable subatomic particle having many of the properties of an electron but a mass 207 times greater (842) N natural satellite naturally formed body that revolves around a planet (moon) (273) navigator method method commonly used to show direction for vector quantities in two dimensions; uses compass bearings north [N], south [S], east [E], and west [W] to identify vector directions (78) net charge charges in the system (517) sum of all electric neutrino extremely small neutral subatomic particle; symbol is v (804) neutron neutral particle found in nuclei (790) neutron number number of neutrons in the nucleus; symbol is N (790) Newton’s first law of motion an object will continue either being at rest or moving at constant velocity unless acted upon by an external non-zero net force (139) Newton’s law of universal gravitation Any two objects, A and B, in the universe exert gravitational forces of equal magnitude but opposite direction on each other. The forces are directed along the line joining the centres of both objects. (204) Newton’s second law of motion when an external non-zero net force acts on an object, the object accelerates in the direction of the net force; the magnitude of the acceleration is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object (148) Newton’s third law of motion if object A exerts a force on object B, then B exerts a force on A that is equal in magnitude and opposite in direction (161) node (nodal point) point on a spring or other medium at which only destructive interference occurs; in a standing wave, a point that never vibrates between maximum positive amplitude and maximum negative amplitude; in a standing wave, nodes occur at intervals of ; in an interference pattern, 1 2 nodes occur at path difference (417) intervals of 1 2 normal line perpendicular to the reflecting surface (654) nucleon proton or neutron (790) nucleosynthesis elements by the fusion of lighter elements (823) imaginary line drawn formation of O open pipe (open tube) pipe opened at both ends; the longest wavelength that can resonate in an open pipe is twice the length of the pipe (424) optical fibre central core of glass with a refractive index of approximately 1.5, surrounded by a cladding material of a slightly lower refractive index (671) orbital probability distribution of an electron in an atom (783) orbital period time required for a planet to make one full orbit; may be measured in Earth days (271) orbital perturbation irregularity or disturbance in the predicted orbit of a planet (282) orbital radius distance between the centre of the ellipse and the planet; average orbital radius corresponds to the semi-major axis (269) origin reference point (6) oscillation repetitive back-andforth motion (344) oscillatory motion motion in which the period of each cycle is constant (344) overtone any frequency of vibration of a string that may exist simultaneously with the fundamental frequency (423) P parent element original element in a decay process (799) 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 887 particle discrete unit of matter having mass, momentum, and the ability to carry an electric charge (639) alpha particle two neutrons bound together to form a stable particle (497) two protons and electron emitted beta particle by a nucleus; symbol is (802) fundamental particle particle that cannot be divided into smaller particles; an elementary particle (836) mediating particle virtual particle that carries one of the fundamental forces (837) strange particle particle that interacts primarily via the strong nuclear force yet decays only via the weak nuclear force (845) virtual particle particle that exists for such a short time that it is not detectable (837) particle model describes EMR as a stream of tiny particles radiating outward from a source (639) path length distance between a point source and a chosen point in space (688) difference in path length difference between two path lengths, each measured from a different origin and extending to a common point in space (688) period time required for an object to make one complete oscillation (cycle); measured in s/cycle (249) phase shift result of waves from one source having to travel farther to reach a particular point in the interference pattern than waves from another source (426) emission of photoelectric effect electrons when a metal is illuminated by short wavelengths of light (712) photoelectron
electron emitted from a metal because of the photoelectric effect (712) photon (from the Greek word meaning “light”) quantum of light; discrete packet of energy associated with an electromagnetic field (640) pion unstable subatomic particle with a mass roughly 270 times that of an electron (842) Planck’s formula light comes in quanta of energy that can be calculated using the equation E nhf (705) plane mirror ing surface (654) smooth, flat, reflect- light resultplane polarized light ing from polarization, in which only one plane of the electric field is allowed to pass through a filter (696) close to the principal axis converge, or appear to diverge from, after being reflected (657) principal quantum number quantum number that determines the size and energy of an orbit (774) planetary model that has electrons orbiting a nucleus (768) atomic model plasma highly ionized gas containing nearly equal numbers of free electrons and positive ions (522) point of incidence point at which the incident ray contacts a polished, reflecting surface and is reflected from the surface as the reflected ray (652) point source single point of disturbance that generates a circular wave (395) polar coordinates method method commonly used to show direction for vector quantities in two dimensions; the positive x-axis is at 0° and angles are measured by moving counterclockwise about the origin, or pole (78) polarization production of a state in which the plane of the electric field for each electromagnetic wave occurs only in one direction (696) polarizing filter filter that allows only one plane of the electric field to pass through it; plane polarized EMR emerges (696) position straight-line distance between the origin and an object’s location; includes magnitude and direction (6) positron antielectron; positively charged particle with its other properties the same as those of an (804) electron; symbol is e or 0 1 potential energy energy that is stored and held in readiness; includes gravitational and elastic potential energies; symbol is Ep rate of doing work (324) power primary cosmic rays high-energy particles that flow from space into Earth’s atmosphere (841) imaginary line principal axis (PA) drawn through the vertex, perpendicular to the surface of the curved mirror at this point (657) principal focal point (F) point where light rays parallel to and principle of superposition displacement of the combined pulse at each point of interference is the sum of the displacements of the individual pulses (412) projectile object released or thrown into the air (54) projectile motion motion in a vertical plane (54) proton positively charged particle found in all nuclei (790) proton–proton chain fusion process in which four hydrogen nuclei combine to form a helium nucleus (821) pulse disturbance of short duration in a medium; usually seen as the crest or trough of a wave (401) compression pulse where the coils of a spring are more tightly compressed (404) region rarefaction pulse the coils of a spring are more widely spaced (404) region where transverse pulse pulse in which the coils of the spring move at right angles to the direction of the pulse’s motion (401) Q quanta discrete units of energy (638) quantized limited to whole multiples of a basic amount (quantum) (705) quantum smallest amount or “bundle” of energy that a wavelength of light can possess (pl. quanta) (705) quantum chromodynamics quantum field theory that describes the strong nuclear force in terms of quantum colour (848) quantum electrodynamics quantum field theory dealing with the interactions of electromagnetic fields, charged particles, and photons (838) quantum field theory field theory developed using both quantum mechanics and relativity theory (837) Glossary 887 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 888 quantum indeterminacy probability of finding a particle at a particular location in a double-slit interference pattern (740) quantum model light and all other EMR are discrete bundles of energy, each of which has both particle and wave characteristics (640) quark any of the group of fundamental particles in hadrons (845) R radioactive decay series process of successive decays in which a radioactive nucleus decays into a daughter nucleus that is itself radioactive, and the daughter nucleus decays into another unstable nucleus until a stable nucleus is created (807) radioisotope radioactive (808) isotope that is radius of curvature (r) distance from the centre of curvature to the mirror surface (657) range distance a projectile travels horizontally over level ground (105) ray line that indicates only the direction of motion of the wave front at any point where the ray and the wave front intersect (397) ray diagram diagram showing the result of a light ray interacting with a surface (653) recomposition (of the spectrum) production of white light by a combination of light of all colours of the spectrum (674) rectilinear propagation movement of light in straight lines through a uniform medium (653) reference point point from which distances are measured (297) refracted ray path of a light ray after it has changed direction at an interface, due to a change in its speed (664) refraction change in the direction of a light wave due to a change in its speed as it passes from one medium to another (666) refractive index ratio comparing the speed of light in a vacuum to the measured speed of light in the medium (666) relative biological effectiveness (RBE) factor indicating how much arbitrarily chosen 888 Glossary a particular type of radiation affects the human body (809) any light ray passing through the boundary between two media. (667) relative motion motion measured with respect to an observer (91) solenoid electromagnet that operates a mechanical device (589) resonance increase in the amplitude of a wave due to a transfer of energy in phase with the natural frequency of the wave (418) resonant frequencies natural frequencies of vibration of an object that will produce a standing wave pattern; at a resonant frequency, energy added is in phase with existing oscillations (418) sum of a series of resultant vector vectors; drawn from the tail of the first vector to the tip of the last vector (71) revolution one complete cycle for an object moving in a circular path (249) rpm revolutions per minute; imperial unit used to measure frequency (249) S scalar quantity measurement that has magnitude only (6) secondary cosmic rays particles created by collisions between primary cosmic rays and atoms in the atmosphere (841) shower of semiconductor material that lies in the middle, between a good conductor and a good insulator; because of its nature, a semiconductor is a good conductor in certain situations, and a good insulator in other situations (514) absorbed dose of ionizing sievert radiation that has the same effect on a person as 1 Gy of photon radiation, such as X rays or gamma rays; absorbed dose in sieverts is equal to the dose in grays multiplied by the relative biological effectiveness (RBE); unit is Sv (809) simple harmonic motion (SHM) oscillatory motion where the restoring force is proportional to the displacement of the mass (355) simple harmonic oscillator object that moves with simple harmonic motion (355) Snell’s Law For any angle of incidence greater than zero, the ratio sin r is a constant for i/sin sound barrier term applied to the increase in aerodynamic resistance as an aircraft approaches the speed of sound (433) source charge duces an electric field (546) charge that pro- spectrometer device for measuring the wavelengths of light in a spectrum (773) spectroscopy study of the light emitted and absorbed by different materials (771) spectrum bands of colours making up white light; in order: red, orange, yellow, green, blue, and violet (675) absorption line spectrum pattern of dark lines produced when light passes through a gas at low pressure (772) electromagnetic spectrum all types of EMR considered in terms of frequency, wavelength, or energy (637) emission line spectrum pattern of bright lines produced by a hot gas at low pressure (772) specular (regular) reflection behaviour describing parallel incident rays reflected from a flat, smooth, reflecting surface as parallel reflected rays (653) spin quantum property resembling rotational angular momentum (842) constant of pro- spring constant portionality k which appears in Hooke’s Law for springs; represents the slope of the line and is measured in units of force per unit length; amount of stiffness of a spring (299) standard model describing the nature of matter and the fundamental forces (848) current theory static friction force exerted on an object at rest that prevents it from sliding; symbol is F fstatic (171) coefficient of static friction proportionality constant relating (Ffstatic )max and FN (182) stationary state a fixed energy level (773) stable state with 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 889 stator fundamental component of a simple DC electric motor consisting of a frame with a coil or permanent magnet to provide a magnetic field (608) stopping potential potential difference for which the kinetic energy of a photoelectron equals the work needed to move through a potential difference V (716) string theory theory that treats particles as quantized vibrations of extremely small strings of massenergy (849) superconductor has no measurable resistance at very low temperatures (515) conductor that supernova sudden, extremely powerful explosion of a massive star (823) synchrotron particle accelerator particle accelerator in which an advanced type of cyclotron increases the strength of the magnetic field as the particles’ energy increases, so that the particles travel in a circle rather than spiralling outward (841) system two or more objects that interact with each other (470) group of objects isolated system (in the context of energy) assumed to be isolated from all other objects in the uni
verse (311) isolated system (in the context of momentum) when mass of a system is constant and no external net force acts on the system (470) non-isolated system system in which there is an energy exchange with the surroundings (320) T tangent a curved-line graph at only one point (24) straight line that touches tension (in a rope) magnitude of T exerted by a rope on an a force F object at the point where the rope is attached to the object (132) test charge charge with a magnitude small enough that it does not disturb the charge on the source charge and thus change its electric field (546) thin lens equation equation that relates object distance, image dis- tance, and focal length of a curved lens (680) air velocity object’s velocity relative to still air (92) threshold frequency minimum frequency that a photon can have to cause photoemission from a metal; symbol is f0 (712) torsion balance device used to measure very small forces (205) total internal reflection reflection of all incident light back into an optically more dense medium due to inability to refract beyond the maximum angle of 90° (672) trajectory parabolic path or motion of a projectile (103) transmute element (806) trough region where the medium is lower than the equilibrium position (394) tuning (a musical instrument) changing the tension in the string of a musical instrument (424) change into a different U uniform circular motion motion in a circular path at a constant speed (242) uniform motion constant velocity (motion or rest) (13) non-uniform motion acceleration (23) uniformly accelerated motion constant change in velocity per unit time (25) universal gravitational constant constant in Newton’s law of universal gravitation that is equal to 6.67 1011 N•m2/kg2; symbol is G (204) universal wave equation relationship between the speed, frequency, and wavelength of a wave: v f (408) V Van de Graaff particle accelerator particle accelerator in which a moving belt transfers charge to a hollow, conductive sphere, building up a large potential difference that propels ions through an accelerator chamber (841) vector quantity measurement that has both magnitude and direction (6) ground velocity velocity relative to an observer on the ground (92) instantaneous velocity moment-to-moment measure of an object’s velocity (24) wind velocity velocity of the wind relative to the ground (92) vertex (V) curved mirror surface (657) geometric centre of the W wave disturbance that moves outward from its point of origin, transferring energy through a medium by means of vibrations (394) bow wave V-shaped wave produced as a boat moves through water or an airplane moves through the atmosphere (433) electromagnetic wave periodic variation in perpendicular electric and magnetic fields, propagating at right angles to both fields (643) incident wave wave front moving out from the point of origin toward a barrier (395) longitudinal wave wave with the motion of the medium being parallel to the motion of the wave (401) reflected wave wave front moving away from a barrier (395) strong compression shock wave wave produced as an aircraft exceeds the speed of sound (433) condition in a standing wave spring or other medium in which a wave seems to oscillate around stationary points called nodes; wavelength of a standing wave is the distance between alternate nodes or alternate antinodes (417) transverse wave wave with the motion of the medium being perpendicular to the motion of the wave (401) wave amplitude distance from the equilibrium position to the top of a crest or the bottom of a trough (395) velocity rate of change in position; includes magnitude (speed) and direction (12) imaginary line that wave front joins all points reached by the wave at the same instant (395) Glossary 889 20-PearsonPhys-Glossary 7/25/08 7:39 AM Page 890 waves out of phase occurs when a crest from one wave occupies the same point in the medium as a trough from a second wave; produces destructive interference (416) gravitational force exerted weight on an object by a celestial body; symbol is F g (198) apparent weight negative of the normal force acting on an object; symbol is w (224) true weight acting on an object that has mass (222) gravitational force true weightlessness which w 0 for an object and F g 0 on the object (228) situation in work measure of the amount of energy transferred when a force acts over a given displacement; calculated as the product of the magnitude of applied force and the displacement of the object in the direction of the force (293) work–energy theorem work done on a system is equal to the sum of the changes in the potential and kinetic energies of the system (307) work function minimum energy that a photon can have to cause photoemission from a metal; specific for every metal; symbol is W (713) wave model describes EMR as a stream of transverse waves radiating outward from a source (639) wave–particle duality light has both wave-like and particle-like properties (726) wave train series of waves forming a continuous series of crests and troughs (395) wavelength distance between two points on a wave that have identical status; usually measured from crest to crest or from trough to trough (395) waves in phase occurs when crests or troughs from two waves occupy the same point in the medium; produces constructive interference (395) 890 Glossary 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 891 NUMERICAL ANSWERS For some questions, answers may vary slightly depending on the method or data chosen for the solution. page 10, 1.1 Check and Reflect page 63, 1.6 Check and Reflect 4. (e) 13.0 m [right] 5. 45.0 km [W] 6. Distance 11.0 m; Displacement 5.0 m [right] 0.50 m [right] 0.75 m [left] groom best man maid of honour 1.25 m [right] flower girl 1.50 m [left] 7. d d d d page 20, 1.2 Check and Reflect 3. 72.3 m 4. 1.5 m 5. 1.61 m/s2 [down] 6. 2.8 m 7. (1) 26 m/s [down] (2) 33 m 8. 0.376 s 9. 47.9 m 10. 1.6 s 11. 6.22 s 12. 10.6 m 3. (i) D (ii) C (iii) A (iv) B 13. 1.4 s 4. 10 m; A is ahead 5. 7 km [W] 8. 2.5 m right 9. Insect B is ahead by 1.2 m. 11. Distance 26 m; Time 13 s 12. Time 20 s; Displacement 45 m [N] (2) 31 s 13. (1) 22 s (3) 14 s page 30, 1.3 Check and Reflect 1. (a) 5.6 m/s2 [forward] (b) 2.8 m/s2 [forward] (c) 0.30 m/s2 [forward] (iii) C (ii) B 3. (i) A 4. Time (s) Velocity (m/s [forward]) (iv) D 2.0 4.0 6.0 8.0 3.8 7.0 0.0 7.5 page 44, 1.4 Check and Reflect 7. 75 m [E] 8. 36 km [up] 13. 15 m/s2 [E] 15. Acceleration 1.25 m/s2 [W]; Time 8.00 s 16. (a) 81 km/h [N] 17. 0.33 m/s2 [right] page 53, 1.5 Check and Reflect 1. 20 cm [forward] 2. 75 m [right] 3. 59.9 m 4. 1.67 105 m/s2 [forward] 5. 11 s 6. 39.2 m 7. 0.41 m/s2 8. 0.24 m/s2 [forward] 9. 75.0 m/s2 [W] 10. 23.3 m/s2 11. 3.52 m/s2 [S] 12. 9.31 m/s2 13. 9.5 m 14. 0.064 m/s2 [N] 14. 2.4 m 15. (a) 4.0 m/s2 [up] (b) 5.0 103 m (d) 7.0 103 m 16. 9.68 m/s [down] page 65, Chapter 1 Review 3. (a) 1.0 m/s [backward] (b) 2.0 m/min [right] (c) 1.7 m/s [forward] 4. 27.0 km [W] 5. 42 min 6. (1) 3.75 m/s (2) 1.25 m/s [right] (3) 25.0 m [right] 8. 2.8 s 9. 7.0 s 10. 0 m/s 11. 1.9 102 s 12. 1.5 m/s [W] 14. 60 m 15. 7.2 107 m/s2 16. 18 km 17. 34.5 km 18. 72 times faster 19. 1.1 102 km/h, 31 m/s 20. (1) 2.81 m/s2 [downhill] (2) 22.5 m/s [downhill] 21. (1) 1.3 m/s2 [N] (2) 10 m/s [N] 22. (1) 16.0 m/s [S] (2) 22.0 m/s [S] (3) 0.267 m/s2 [S] 23. 35 m/s 24. (1) 12 s 26. (1) 20 km [right] (2) 39 m/s (2) 0 m/s2 27. 9.03 s 28. 0.467 s 29. 0.298 m/s2 30. 24 km/h [E] 32. 11.2 m/s2 [W] 33. 6.8 s 34. 1.17 s 35. 12 m 36. 9.29 m 37. 1.32 s 38. (1) 15.2 m (2) 17.3 m/s 39. (1) 20 m/s (2) 2.0 s page 75, 2.1 Check and Reflect 3. 24 cm 4. 1.0 cm : 20 km 5. 2.1 103 km 6. (b) 100 yards [down field] (c) 1000 yards 7. 100 km [S], 150 km [S], and 200 km [S] 9. (1) 36.0 m (2) 8.0 m [down] page 90, 2.2 Check and Reflect 28 cm 48 cm; dx 4. dy 5. (a) Distance 11.5 km; Displacement 3.4 km [27° N of W] (b) Distance 522 m; Displacement 522 m [17° E of N] (c) Distance 2.95 km; Displacement 1.45 m [270°] 6. 2.0 km [45° S of E] 7. 178 m/s 8. 2.65 km [48° S of E] 9. Total displacement 2.8 km [49° N of W]; Average velocity 3.0 km/h [49° N of W] 10. (a) 27 m in both the x and y directions (b) 54 m 11. 10.6 m [293°] page 101, 2.3 Check and Reflect 5. (a) 88 s (b) 79 s; 47 m 6. (a) 233 km/h [N] (b) 297 km/h [N] (c) 267 km/h [83.1° N of W] 7. 7.2 102 s 8. 8.4 102 km/h [3° N of E] 9. Ground velocity 2.8 m/s [35° S of E]; Time 19 min 10. 6.3 102 km/h [3° S of W] 11. (a) 4.7 m/s [32° W of N] (b) 2.0 102 s page 112, 2.4 Check and Reflect 5. 2.0 m 6. Initial horizontal velocity component 22.1 m/s; Initial vertical velocity component 15.5 m/s; Up 12.2 m; Out 69.8 m 7. 7.21 m/s 8. 10.8 m/s Numerical Answers 891 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 892 9. (a) 10.0 s (c) 20.5 s (e) Horizontal component 68.8 m/s; Vertical component 103 m/s (b) 542 m (d) 1.41 km page 114, Chapter 2 Review 2. x component 41 m; y component 37 m 5.0 m [N] 140.0 cm [E] 2.5 km [backward] 80.0 km [right] 3. 0 m/s 7. (a) d (b) d (c) d (d) d 8. 0 m/s 9. 3.9 102 km 10. Range 86.2 m; Maximum height 30.8 m 11. 2.23 102 m horizontally; 2.83 s 12. (a) 6.4 m/s [51° E of N] 28. 99 km/h [W] 29. 1.5 102 m 30. 1.2 m/s [210°]; 4.4 m/s [210°]; 6.7 m/s [210°] 31. 1.1 102 km/h [7° N of E] 32. 13.2 m/s2 33. 375 m [N] 35. 1.93 m 36. (a) 7 squares (b) 8.6 units [54°] 37. 1.6 m/s [N] 38. 320 m [51° S of E] 40. 5.98 m/s2 41. (a) 14.4 m/s [down] 42. (a) 347 m (b) 376 m (c) 82.6 m/s [down] 43. 13 m/s [30°] (b) 10.6 m 9. 1.2 m/s2 [uphill] 10. 16 m 11. 1.9 s page 192, Chapter 3 Review 1. 5.00 102 m/s2 [E] 3. 1850 N [W] 5. (a) 42 N [along rope connecting foot to pulley] 6. 1.2 103 N [forward] 3.7 106 N, R 7. L 8. 7.5 N [backward] 9. (a) 3.4 102 N 3.3 106 N (b) 1.4 103 N [forward] (c) 3.4 102 N [backward] 10. (a) F netA 36 N [W], F netB (b) (curler A) 0.71 m/s2 [W], (curler B) 0.26 m/s2 [E] 21 N [E] (b) 30 s page 136, 3.1 Check and Reflect 11. 0.38 13. (a) 29 m/s [down] (b) 23 m 3. (b) 300 N [forward] (c) 7.50 m/s [away from
cliff] (b) 3.5 m/s [N] 14. 6.5 m/s [76° up] 15. (a) 30° W of N (c) 1.4 102 s 16. 1.82 m 17. 39 s 18. 2.18 s (b) 180° 4. (a) 0° 5. 1.47 102 N [13°] 6. 42 N [153°] 7. F F 1.50 102 N [125°], 1.50 102 N [55°] T1 T2 page 139, 3.2 Concept Check 19. (a) 17° S of W (b) 95 km/h v 17 km/s [toward interstellar space] 20. [42.2°] 21. 5.0 m 22. 60.96 m 23. (a) 1.2 102 km/h 24. 2.92 s; Distance 59.9 m; Maximum height 10.5 m (b) east 25. [39.8°] 26. (a) 36 s (b) 1.5 km page 118, Unit I Review 3. (a) dx (b) vx 0 m; dy 15.0 m/s; vy 5.0 m 5.47 m/s 5. 9.8 km [30° S of E] 6. 2.70 m/s [W] 8. 3.03 s; Distance 78.0 m horizontally; Maximum height 11.3 m 13. 30 m/s 14. 32 km/h 16. (a) 30 m/s [90°] (b) 40 m/s [90°] 17. 12.1 km 18. 1.06 h 19. 5.7 m/s 20. 9.56 m/s2 21. 8.1 102 km/h [4° W of S] 22. 70.4 m/s2 23. 2.2 s 24. 58 m 25. (a) 51 s (b) 0.032 m/s2 26. 13.5 m 27. 63.4 m 892 Numerical Answers page 146, 3.3 Concept Check F app if F f F net 0 page 148, 3.3 Concept Check (a) a (b) a (c) a 1 6 1 page 158, 3.3 Check and Reflect (b) 15 m/s2 [up] 5. (a) 26 kg 6. 0.11 m/s2 [horizontally] 7. (a) 75 N [97°] 8. 0.75 m/s2 [right] 9. (a) (4.0-kg block) 1.3 m/s2 (b) 37 m/s2 [97°] [toward pulley], (2.0-kg block) 1.3 m/s2 [down] (b) 17 N page 168, 3.4 Check and Reflect 7. 10 N, 10 N [toward spring scale] 8. (a) F X on Y F Y on X (b) F X on Y F Y on X 12 N [right], 12 N [left] 12 N [right], 12 N [left] page 178, 3.5 Concept Check 90° page 190, 3.5 Check and Reflect 4. 2 N [backward] 5. 0.40 6. 2 103 N [backward] 7. 24° 12. 0.423 13. (a) 9.8 102 N (b) (i) (60 kg) 2.0 m/s2 [down], (40 kg) 2.0 m/s2 [up] (ii) 4.7 102 N (c) 9.4 102 N 14. 7.3 102 kg 15. 2 103 kg 18. (a) 6.9 m/s2 [backward] (b) 1.2 s page 201, 4.1 Concept Check (a) 16g (c) g (d) g (b) 1 g 4 (d) 0 page 202, 4.1 Check and Reflect 9. (b) 9.8 N/kg (c) gBanff page 205, 4.2 Concept Check 1 6 1 (c) (b) 4Fg (a) 4Fg Fg page 215, 4.2 Check and Reflect 3. (a) 1 Fg (b) 1 Fg 2 4 (ii) 978 N 4. (a) (i) 162 N 5. (b) (Deimos) 1.9 1014 N, (Phobos) 5.3 1015 N 6. (a) 132 N [toward Earth’s centre] (b) about 13.4 times greater in magnitude page 229, 4.3 Check and Reflect 5. 2.4 N/kg [toward Earth’s centre] 6. (b) (i) 1.6 N/kg (ii) 0.72 N/kg (iii) 0.40 N/kg (c) 10rMoon 7. 9.8 N/kg 8. 540 N [down] g) 4.9 102 N [down], 9. (b) (F (w) 2.9 103 N [down] 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 893 (c) 0 page 286, 5.3 Check and Reflect 10. (Figure 4.38) 2.45 m/s2 [up], (Figure 4.40) 1.23 m/s2 [down] page 231, Chapter 4 Review 2. (balance) 5.0 kg, (spring scale) 96 N 3. g Msource 4. 3.83 N/kg [toward Earth’s centre] 5. 1.4 s 7. (a) 222 N [toward centre of Mars] (b) 626 N [toward centre of Saturn] 8. 2.0 107 N 9. (a) 2.4 102 N (b) 7.1 102 N (c) 6.8 N 10. (a) 7.3 m/s2 [down] (b) 12 s (c) 4.9 102 N [down] page 234, Unit II Review 2. 86 N [350°] 6. 0.50 kg 7. (a) 4a 8. 40 N (b) 1 a 4 10. 15 N [up] 21. (mass m) Fg, (mass 2m) 2Fg 24. 306 N [357°] 25. 2.097 m/s2 [W] 27. 4.0 s 28. (b) 8.58 N [0°] 29. 0.56 m/s2 [in same direction as train A] 30. 8.0 104 N [up] 31. 2.2 m/s2 [forward] 32. 13 N [up] 33. (a) 5.3 m/s2 [right] (b) 4.8 102 N 34. 0.46 35. (a) 6 101 N [forward] (b) 2 101 N [forward] (c) 6 101 N [backward] 36. 3.2 m/s2 [downhill] 37. 1.5 m/s2 [up] 38. (a) 1.3 m/s2 [toward object A] (b) (string between A and B) 51 N, (string between B and C) 44 N 39. 8.00 N [toward Earth’s centre] 41. 8.57 N/kg [toward Earth’s centre] 42. 24.3 N less on Mars 4 3. (a) (i) and (ii) 5.9 102 N (iii) 8.8 102 N (iv) 3.9 102 N (b) (i) and (ii) (w) 5.9 102 N [down], g) 5.9 102 N [down] (F (iii) (w ) 8.8 102 N [down], g) 5.9 102 N [down] (F (iv) (w ) 3.9 102 N [down], g) 5.9 102 N [down] (F g) 5.9 102 N [down], 44. (F (a) 6.5 m/s2 [down] 45. (a) 9.4 m/s2 [down] (b) 11 N [up] (c) 11 N [down] 46. (Earth) 1.2 s, (Moon) 3.0 s 47. (a) 0.1 N [away from net] (b) 0.8 m/s2 [toward net] (c) 9 s (d) no 51. (a) 3.7 107 N [down] (b) 1.3 107 N [up] (c) 3.3 m/s2 [up] 52. Fg from person on you is 5.8 times greater page 268, 5.2 Check and Reflect 6. 0.2 s 7. 1.88 102 m/s 8. 26 m/s 9. 8.57 m/s2 10. 1.2 102 m/s2; 12 times 11. 1.21 m/s 12. 0.0337 m/s2 13. 1.7 Hz 14. 1.3 Hz or 80 rpm 9. 0.723 AU 10. 1.36 103 m/s 11. 3.547 d 12. 1.43 104 m/s 13. 1.98 1030 kg page 288, Chapter 5 Review 14. 4 greater 17. 2.0 101 m/s 18. 7.8 102 N 19. 44.3 m/s 20. 68 N 22. 7.83 102 m/s2 23. 3.0 103 rpm 25. 13.1 m/s 26. (a) 6.00 107 m/s (b) 1.20 1016 m/s2 27. 18.0 AU 28. 1.09 1030 kg 29. (a) 1.16 1018 s or 3.66 1010 a (b) 5.18 1036 kg (c) 6.71 1015 m/s2 30. (a) 7.91 103 m/s (b) 5.07 103 s 31. 2.73 103 m/s2; 2.00 1020 N 32. 0.430 d or 3.71 104 s 33. 5.51 103 m/s 34. (a) 4.01 1033 kg (b) 6.38 1011 m page 305, 6.1 Check and Reflect 5. (a) 3.60 104 J (b) 1.18 104 J 6. 2.76 104 J 7. (a) 6.18 104 J (b) 7.55 104 J (c) 1.37 105 J 8. (a) 3.25 J (b) 0.0732 m 9. (a) A 2.04 105 J; B 3.48 105 J 10. (a) 110 J (b) 33.8 J 11. (a) 160 J (b) 12.6 m/s page 310, 6.2 Check and Reflect 5. 1.79 106 J; Ek 6. 3.54 105 J 7. (a) 5.10 103 J (b) 1.25 103 J (c) 3.85 103 J (d) 31.0 m/s 8. (a) 288 J (c) 126 J (b) 288 J (d) 126 J page 323, 6.3 Check and Reflect 7. 12.4 J 10. (a) 0.482 m (b) 3.08 m/s (c) 2.13 m/s page 330, 6.4 Check and Reflect 4. 164 W 5. 1.50 106 W 6. 4.1 106 J 7. 380 N 8. 1.04 107 W page 332, Chapter 6 Review 5. 3.27 105 J 6. (a) increases by 4 7. (a) 5.12 m/s 8. 1.70 102 N/m 9. 12.7 m 10. 2.30 104 W (b) 5.59 m/s page 336, Unit III Review 16. 1 J 1 kg·m2/s2 29. 0.017 s 30. 62.5 Hz 31. 5.000 Hz, 0.2000 s 32. 1.02 103 m/s 33. 1.6 101 s 34. 3.09 m/s2 35. 1.40 102 m/s 36. 7.10 103 N 37. (a) 28.1 m (b) 4 greater; 113 m (b) 1.41 103 J 38. (a) 2.40 103 J 39. (a) 2.80 103 N [0°] (b) 1.54 105 J (c) 3.79 105 J (d) 19.5 m/s Numerical Answers 893 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 894 40. 1.03 m 41. (a) 1.20 103 J 42. (a) 35.9 J 44. (a) 2.60 J (b) 15.5 m/s (b) 0.814 m (b) 0.0651 m 45. 1.32 m/s 46. (a) 1.31 105 J (b) 917 m (c) 295 m/s 47. (a) 8.91 J 48. 3.68 103 W 49. 1.62 104 W 50. 60.0 m/s (216 km/h) (b) 1.78 N 51. 3.63 m/s 52. (a) 4.01 1030 kg (b) 4.27 1025 kg (c) 2.18 103 m/s 54. (a) 7.75 m/s (b) 4.38 m/s page 347, 7.1 Check and Reflect 7. 20.0 Hz 9. 2.50 102 Hz 10. 0.026 s 11. 0.01250 s 12. (a) 0.400 s (b) 1.50 102 wags page 365, 7.2 Check and Reflect 4. 6.0 N opposite to the displacement 5. 1.6 m 6. 19 N [away from equilibrium] 7. 1.5 N/m 8. 0.342 N [toward equilibrium] 9. 0.028 N/m page 380, 7.3 Check and Reflect 6. 1.5 m 7. 1.08 103 m/s2 [left] 8. 1.3 103 N/m 9. 11.0° 10. 3.49 s 11. (a) 1.88 s (b) 0.900 kg (c) 16.7 m/s2 12. 3.14 m/s 13. 7.99 cm [east] 14. 0.900 s page 390, Chapter 7 Review 11. (a) 2.5 N [toward equilibrium] (b) 0.85 N [toward equilibrium] 12. 2.5 102 Hz 13. 1.5 s 14. 10.0 Hz 15. 5.1 N [toward equilibrium] 16. 2.0 102 N/m 17. (a) 1.96 N/m (b) 0.392 N (c) 0.300 m 894 Numerical Answers 18. 0.750 m 20. 24.8 cm 21. 3.00 s 22. 0.120 m/s 23. 0.13 m/s2 [down] 24. 1.02 s 25. 0.65 N/kg 26. (a) l/g (b) 0.796 m 27. (a) 1.26 m/s (b) 0.993 s 28. (a) 0.566 Hz (b) 0.800 Hz 46. (a) 10 nodes and 9 antinodes (b) 2.22 Hz 47. (a) 193 m/s (b) 440 Hz 48. (a) 0.552 m (b) 1.66 m 49. 8.62 m/s [away from you] 50. (a) 1.28 103 Hz (b) 9.96 102 Hz page 449, 9.1 Concept Check (b) 1 p 3 (c) p [W] page 410, 8.2 Check and Reflect (a) 2p 5. 1.20 103 Hz 6. 2.07 m 7. 0.135 m 9. (a) 0.911 m (b) 3.91 m page 428, 8.3 Check and Reflect 7. 0.106 m 9. 0.60 cm page 434, 8.4 Check and Reflect 4. 748 Hz 5. 15.7 m/s (56.6 km/h) page 436, Chapter 8 Review 5. 0.133 s 7. 0.833 cm 8. 8.6 cm 9. 1.59 102 m 10. (b) 5.6 m/s 12. (a) 2.5 Hz (b) 0.50 Hz 13. (a) 435 m/s (b) 777 Hz 14. (a) 19.7 cm (b) 59.0 cm 16. 1.4 cm 17. 308 Hz 18. 86.9 km/h (24.1 m/s) 19. 1.26 103 km/h 20. 694 Hz 21. 175 m/s; f 2 original 3 page 440, Unit IV Review 11. 2.5, 3.5 28. 3.06 N 29. 7.9 g 32. 1.25 m/s2 33. 4.0 mN/cm 34. (a) 0.100 m (b) 2.24 m/s 35. 15 m/s 36. 6.73 s 37. 0.99 m/s 38. 15.9 m 41. 4.00 1014 Hz to 6.98 1014 Hz 42. 240 m 43. 0.294 m 44. 7.76 s 45. 120 m/s page 453, 9.1 Check and Reflect 8. 13 kg•m/s [S] 9. 1.2 102 m/s [N] 11. 0.16 kg 13. 75 m/s [S] 14. (a) 4.28 105 kg•m/s [W] (b) 3.85 105 kg•m/s [W] 15. 1.36 105 kg 16. 32.6 m/s [W] page 456, 9.2 Concept Check pi 0 page 467, 9.2 Check and Reflect 4. (a) 2 (impulse) (b) 1 (impulse) 3 7. (a) 2.3 N•s (b) 47 m/s [S] 8. 6.2 N•s 9. (a) 7.0 103 N•s (b) 11 m/s 10. 12 s 11. 560 N [W] 12. 545 N•s [W] page 486, 9.3 Check and Reflect 6. 0.018 m/s [away from bag] 7. 3.1 103 m/s [down] 8. 1.2 m/s [S] 9. 0.47 m/s [E] 10. (a) 1.11 m/s [right] (b) inelastic 11. (a) 274 kg (b) inelastic page 499, 9.4 Check and Reflect 5. 0.505 m/s [320°] 6. 0.625 m/s [48.1° N of W] 7. inelastic, 0.098 J 9. 0.603 m/s [49.6° S of W] 10. 27.4 m/s [37°] 11. 27.0 m/s [349°] 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 895 page 503, Unit V Review 37. 1.4 104 kg•m/s [N] 39. 1.85 104 kg•m/s [S] 40. 6.2 kg 41. 37 kg•m/s [W] 42. 2.4 kg•m/s [W] 43. (a) 7.5 N•s (b) 0.26 kg 44. (a) 3.02 N•s [210°] (b) 6.9 m/s [210°] 45. 2.7 103 N [toward drop-off] 46. (a) 5.50 103 N•s [W] (b) 7.75 s 47. 11.3 m/s [6.4° S of W] 48. (a) 7.62 103 N•s [forward] (b) 22 m/s [forward] 49. (a) 3.0 N•s (b) 11 m/s 50. 6.5 107 N 51. 8.7 102 N [up] 52. (a) 12 kg•m/s [toward pitcher] (b) 1.6 103 N [toward pitcher] 53. 2.1 m/s [225°] 54. (a) 1.58 m/s (b) 0.948 m/s 55. 0.750 m/s [backward] 56. 1.27 103 m/s [3.5°] 57. 3.5 m/s [65°] 58. 1.26 m/s [8.3° E of N] 59. (a) 1.8 m/s [1.8° E of N] (b) elastic 60. 1.3 m/s [downhill] 61. 15.6 m/s [34.1° S of W] 62. (a) 17 m/s 63. 4.90 m/s [2°] 64. (a) 3.04 m/s [0.0°] (b) elastic 65. 4.21 1025 kg•m/s [W] 66. (a) 3.46 m/s [4.7° W of N] (b) inelastic 67. (a) 40.0 kg (b) 14.0mT kg•m/s [S] 68. 33 m/s [72.5° N of W] 69. (a) 0.36 m/s (b) 18° S of E 70. (a) 5.95 m/s [E] (b) 2.76 s 71. 10 m/s [73.8° N of E] 72. 0.580 kg 73. 1.18 m/s [307.9°] 74. 0.641 m/s [S] 75. (a) S (b) NW (c) 248° 77. 10.5 m/s [46.4° W of S and 52.5° up] 80. (initial velocity of car) 90 km/h [W]; (initial velocity of truck) 88 km/h [N] 82. 6.1 102 m/s page 538, 10.2 Check and Reflect 5. (a) 20 N (b) 40 N 6. (b) 3.1 1010 electrons 7. (a) 6.74 N—repulsive (b) 6.80 N—repulsive
8. (a) 1.20 104 N toward charge B (b) 7.49 103 N toward charge A page 540, Chapter 10 Review 11. 1.69 N—repulsive 12. 6.70 103 m 13. (a) 1.60 102 N [left] (b) 1.24 102 N [left] (b) 6.7 N 22. (a) 160 N 23. Fe Fg Fe 8.22 108 N 3.63 1047 N 2.27 1039 Fg 24. X—2.43 N [90°]; Y—2.43 N [210°]; Z—2.43 N [330°] 25. (b) Fe varies as 1/r or 1/r2 (e) 0.0360 N•m2 (f ) kq1q2 (g) 2.00 106 C page 553, 11.1 Check and Reflect 7. (a) 4.50 105 N/C [right] (b) 8.99 103 N [right] 8. (a) 2.04 104 N/C [away from larger sphere] (b) 3.63 109 C 9. (a) 2.50 107 N/C [toward the 3.00 C charge] (b) 0.661 m [left of the 3.00 C charge] 10. 0.00 N/C page 569, 11.2 Check and Reflect 4. (a) 2.40 1017 N [downward] (b) 1.64 1026 N [downward] 6. 1.60 103 N/C 7. 7.2 105 N/C [toward the 6.4 C charge] 8. 2.00 103 J 9. 1.8 106 N/C [left] 10. (a) 2.00 104 V/m (b) 3.86 105 J page 575, 11.3 Check and Reflect 4. Speed of electron is 8.79 106 m/s; speed of proton is 2.05 105 m/s 5. 9.4 107 m/s 6. 6.00 1019 C 7. 2.40 104 J 8. (a) 1.38 106 m/s (b) 9.79 105 m/s 9. (a) 1.8 105 J (b) 1.8 105 J (c) 1.1 m/s 10. (b) 2.00 1017 N [down] (c) 2.20 1013 m/s2 [down] (d) 0.130 m page 578, Chapter 11 Review 12. Ep W 17. (a) 5.17 103 N/C [away] (b) 1.03 102 N [toward positive charge] 18. (a) 7.49 1010 N/C [toward the 5.00 C charge] (b) 0.735 m from the 5.00 C charge 19. 1.2 107 N/C [90.0°] (b) 3.0 J 20. (a) 3.0 J 21. 5.62 104 J 22. (a) 545 V/m (b) 20.5 V 23. 0 J 24. 7.0 105 V/m 25. 2.88 1013 J or 1.80 106 eV 26. 6.40 1015 J or 4.00 104 eV page 601, 12.2 Check and Reflect 6. (a) 6.40 1015 N (b) 3.67 1015 N 7. 1.3 106 m/s 8. 6.53 1026 N 9. 1.06 104 m 10. 8.31 109 T page 613, 12.3 Check and Reflect 9. 1.04 102 C 11. 8.4 102 N page 622, Chapter 12 Review 20. 9.86 1014 N 21. (a) 2.31 104 m/s (b) 4.44 1019 J or 2.77 eV 22. 4.08 103 m/s 23. (a) 4.13 1016 N (b) 2.06 1016 N (c) no force 24. 2.7 105 T 26. (b) B varies as 1/r or 1/r 2 28. 25.0 N page 626, Unit VI Review 39. (a) 1.4 108 V/m (b) 1.1 1019 J 41. 2.56 109 m 42. 5.06 107 N—repulsive 45. 15 N/C 46. 3.2 1015 J; maximum speed is 2.0 106 m/s 47. 0.268 m to the right of the first charge 48. 5.0 105 C 49. 0.73 A 50. (a) 2.6 106 N/C [left] (b) 1.6 N—attractive 52. 2.0 103 N 53. 1.67 104 m 54. (a) 3.63 1047 N (b) 8.22 108 N (c) 2.27 1039 times greater 57. 1.53 103 m/s; positive charge Numerical Answers 895 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 896 58. (a) 3.17 1012 N (b) 1.89 102 m 59. (a) 113 N [right] (b) 28.4 N [left] (c) 30.8 N [69.4°] 60. 6.6 103 V 61. (a) 2.50 104 V (b) 9.37 107 m/s (c) 2.25 1011 N 81. (b) B varies as 1/r (c) Plot B vs 1/r (f ) 0.79 T•m/A (g) 0.789 T•m/A page 647, 13.1 Check and Reflect 8. 2.5 103 Hz page 652, 13.2 Check and Reflect 4. 2.1 108 m/s 5. 6.67 102 s 6. 293 Hz 7. 2.7 103 Hz 8. 3.44 104 m 9. 1.86 103 Hz 10. 2.80 108 m/s 11. 2.85 108 m/s page 665, 13.3 Check and Reflect 8. 50 cm page 683, 13.4 Check and Reflect 9. 2.26 108 m/s 10. 23.0° 11. 32.4° 12. 41.1° 14. angle of refraction 15.9°; wavelength 351 nm 16. 2.25 102 m; inverted 18. (a) 648 cm (b) 140 cm (c) 8.2 cm page 697, 13.5 Check and Reflect 6. 5.7 106 m 7. 6.1° 8. 0.350 m 9. 3.76 103 m 10. wavelength 6.11 107 m; frequency 4.91 1014 Hz 11. wavelength 4.13 107 m; frequency 7.26 1014 Hz page 699, Chapter 13 Review 23. 3.41 104 m 24. 2.88 108 m/s 26. 13 cm 30. 5.00 cm high, 6.67 cm from the mirror 31. 1.82 cm 32. (a) 2.26 108 m/s (b) 2.19 108 m/s (c) 1.97 108 m/s 896 Numerical Answers (d) 2.04 108 m/s (e) 1.24 108 m/s 33. angle of refraction 23°; wavelength 415 nm 34. (a) no (b) yes (c) yes (d) no 35. 1.12, which is less than that of water 36. (a) 48.8° (c) 33.3° (b) 24.4° (d) 41.1° 37. 2.67 cm high, 3.33 cm from the lens. It is virtual, erect, and diminished. 38. 0.7°, 1.4°, 2.0° 39. 4.4 107 m; violet 40. 4.9 107 m; blue 41. 6.0 105 m page 710, 14.1 Check and Reflect 1. 4.42 1019 J 2. 8.29 108 m 2E600 3. E300 4. (a) 2.41 1019 Hz 5. 2.77 1021 photons 6. 4.93 1021 photons 7. 100 photons/s 15. 0.00388 nm 16. (a) 1.1 1034 m 18. 3.32 1030 photons/s 19. 0.52 eV, 2.1 eV, 4.7 eV 20. 5.33 1023 kg·m/s 22. (c) 9.3 N 23. 1.5 1017 page 746, Unit VII Review 26. 3.62 1019 J 29. 6.00 1014 Hz 33. 3.9 107 m 36. 2.21 1027 N·s 37. 0.243 nm 41. 8.3 1029 photons/s 43. 3.84 105 km 45. 4.84 104 m 46. 3.75 103 Hz 47. 1.39 104 s 49. refractive index 2.42, therefore the material is diamond 51. 2.91 108 m/s; error 2.93% 53. 7.5 cm from mirror. It is erect and page 715, 14.2 Concept Check Ephoton hf W page 720, 14.2 Check and Reflect 1. 3.11 eV 3. 9.82 1014 Hz 4. yes, hf W 5. 1.25 V 9. 4.1 1015 eV•s; this is close to Planck’s constant Page 725, 14.3 Check and Reflect 1. 1.33 1027 N•s 3. wavelength 1.11 1013 m; energy 1.80 1012 J 6. 0.0124 nm 7. 2.0 104 m/s page 736, 14.4 Check and Reflect 1. 36.4 nm 2. 1.33 1027 kg•m/s 3. 5.3 1026 kg•m/s 4. 2.1 1023 kg•m/s 5. (a) 3.36 1015 J (b) 8.47 1012 m 1.83 103 e/ 6. p page 742, Chapter 14 Review 3. 4.42 1019 J or 2.76 eV 4. Ex/Ev 7. 4.0 1019 J or 2.5 eV 9. It increases by 0.0024 nm. 100 10. 73 nm 11. 6.63 1027 N•s 13. 3.0 1018 photons 14. 3.62 1019 J or 2.26 eV 2.5 cm high. 4 A; pA 4pB 54. B 55. 9° 56. 7.0 1016 J 57. 10 cm from the lens 58. (b) 0.1 m/s 60. 0.0045 nm 62. 0.4 eV 63. for violet light 24.8°; for red light 40.5° 64. 2.63 102 m 67. Energy 0.2 eV; momentum 3 1025 N•s 68. 6.67 102 s 74. 2.5 eV 76. momentum 1.2 1022 kg•m/s; wavelength 5.5 1012 m 78. 6.6 1026 N•s page 760, 15.1 Check and Reflect 1. (a) 8.0 1015 N (b) 0 N 4. (b) 4.00 105 m/s 5. 1.8 107 m/s 6. 1.57 103 T 8. 8.00 104 m/s 9. (a) 8.0 1015 N [downward] 10. (b) 3 109 m/s page 765, 15.2 Check and Reflect 3. 6.40 1019 C 4. 8.00 1017 N [up] 5. (a) 0 N 6. (a) 1.6 1018 C, or 10e (b) gained 10 (c) downward 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 897 page 770, 15.3 Check and Reflect 7. 177.8 MeV 3. (a) 3.6 1016 J (b) 3.6 1012 J 6. (a) 2 1010 m (b) The gold atom is approximately 7000 times larger than the gold nucleus. 7. (b) 5.0 1015 m page 780, 15.4 Concept Check 634 nm, 654 nm page 780, 15.4 Check and Reflect 5. (b) 486 nm (c) 4.09 1019 J 6. 1.96 eV 7. (b) 1 1.9 eV; 2 2.1 eV; 3 1.5 eV; 4 3.9 eV 654 nm, visible, red; 592 nm, visible, 829 nm; (c) 1 2 yellow; 3 4 319 nm 9. (a) 3.03 1019 J (b) 2.66 1020 J 11. (a) 140 nm (b) 3.27 106 m (c) 15 page 784, 15.5 Check and Reflect 4. (a) 3.32 1010 m (b) 2.00 1024 kg·m/s (c) kinetic energy 2.19 1018 J; speed 2.19 106 m/s 5. (a) 2 2 1; 3 3 1 page 786, Chapter 15 Review 6. 3.68 1018 C 7. 1.76 1011 C 16. 4.76 1010 m 19. 2.4 1016 N 20. (a) 1.09 106 m/s 21. 4.1 104 N/C directed upward 22. (a) 1.06 1019 J; 1.55 1019 J; 1.82 1019 J (b) wavelengths: 1.88 106 m, 1.28 106 m, 1.09 106 m. Frequencies: 1.60 1014 Hz, 2.34 1014 Hz, 2.74 1014 Hz. (c) 1.55 1019 J 23. (b) 634 nm 24. 2.5 103 N/C [up] 25. (a) 2.17 1018 J (b) 9.01 1022 m/s2 (d) 4.72 1011 s page 796, 16.1 Check and Reflect 1. (a) 38 protons, 52 neutrons (b) 6 protons, 7 neutrons (c) 26 protons, 30 neutrons (d) 1 proton, 0 neutrons 2. 1.0 109 eV 1.0 GeV 3. 2.3 108 eV 4. 5.56 108 kg 8. (a) 0.366 642 u (b) 8.538 MeV/nucleon 9. (a) 100 MeV (b) 476 MeV (c) 1737 MeV 13. Nuclear radius 5.38 fm Approximate distance between nucleons 2 fm page 810, 16.2 Check and Reflect 2. 3 : 2 8. 1.819 MeV page 817, 16.3 Check and Reflect 1 1. 6 1 2. 3.7 1012 atoms/s 4. 1.2 million years 5. 1.7 103 of the original quantity 6. 1.1 104 a 7. 1.5 102 Bq 8. (a) 4 h (b) 3600 decays/min Page 824, 16.4 Check and Reflect 6. (b) 4.730 MeV 7. (b) 17.59 MeV 8. (a) 8.1 1019 (b) 1.0 103 kg Page 826, Chapter 16 Review 5. 60 neutrons, 55 protons 6. 8.0 1012 J 7. 9 1013 J 8. 3.4 1010 J 9. 0.322 u 10. 20 MeV 18. 1.8 108 decays/s 19. 2 1019 nuclei 24. (a) 7.074 MeV (b) 8.448 MeV (c) 8.792 MeV (d) 7.591 MeV 26. (b) 8.95 MeV 27. (b) 3.210 MeV 28. about 1.1 104 a 29. (a) 5.1 1015 atoms (b) 1.905 109 kg (c) 1.1 104 Bq 30. (a) 400 Bq (b) 4 h 31. 7.274 MeV 32. (b) 0.23 kg page 844, 17.3 Check and Reflect 7. (a) momentum 5 1021 kg•m/s; kinetic energy 8 1015 J (b) momentum 8.4 1022 kg•m/s; kinetic energy 2.1 1016 J 8. (a) 0.51100 MeV (b) 5.521 1027 kg 9. 931.5 page 851, Chapter 17 Review 19. 3.14 1025 kg 20. (a) e (b) 83138 MeV/c 2 21. (b) 10 cm (c) 1.9 1020 N•s page 855, Unit VIII Review 1 and 2 5 1 2. 9.6 107 C 3. 1.60 1018 C 4. 1.60 1016 N [N] 8. (a) ni ni (b) ni (c) ni 4 → nf 6 → nf 1 → nf 4 → nf 10. 33 neutrons, 31 protons 12. 2.0 107 eV 13. 1.0 109 J 14. 6.7 MeV 15. 2.3 109 decays/s 17. 1.4 g 22. (a) 1.022 MeV (c) 2.43 1012 m 25. (a) proton (b) kaon (c) pion (d) 27. (b) 7.3 104 m/s 28. 0.48 T 29. (a) electric force 1.6 1017 N [up] magnetic force 2.0 1013 N [up] (b) 2.0 1013 N [up] 30. (a) 1.6 1018 C (b) It has gained 10 electrons. 31. (a) 1005 nm, 1094 nm, 1282 nm, 1875 nm (b) infrared 32. (a) 2.12 1010 m (b) 6.65 1010 m (c) 9.97 1025 kg·m/s (d) speed 1.09 106 m/s; kinetic energy 5.46 1019 J 33. 342.1 MeV 36. 10.42 MeV 37. 7.5% 38. 1.905 MeV; cerium140 39. 1 102 Bq 40. 1.72 104 years old 41. 7.161 MeV 42. 4.033 MeV 44. (a) 2 102 N (b) 2 1013 J 45. 5.10 104 m/s 46. (b) 135 MeV 48. 2.7 1018 N•s 49. 2.33 1050 m 50. (a) 13 Bq Numerical Answers 897 21-PearsonPhys-AnswerKey 7/29/08 1:40 PM Page 898 51. (a) the level of 21.9 Sv is much higher than the background level of 400 Sv (b) 5 102 m 52. (b) The matter — anti-matter reaction is more than 150 times more powerful than nuclear fusion. 53. 0.22 g deuterium, 0.33 g tritium 54. 0.5 years 56. 7.8 107 m photon emitted 57. 3.73 1011 J 58. 2.12 1010 m 59. (a) The n 3 state (b) 1.89 eV 60. 198.3 MeV 62. 1 63. 5.3 105 Bq page 131, Example 3.1 Practice Problems 1. up down forward backward Ff Fair FN Fg 898 Numerical Answers 2. up down backward forward FN Fair F f Fg page 225, Example 4.7 Practice Problem 1. up down FN (force exerted by rocket on scale) 300 200 100 400 500 600 700 0 Fg (for scale) Fg (for astronaut) 22-PearsonPhys-Index 7/25/08 7:40 AM Page 899 INDEX Note: Bold-font numbers indicate page(s) where the term is defined; t indicates the term is in a Table; f indicates the term is in a Figure. A Absorption line spectrum, 772, 776 Acceleration, 10, 23–28, 31–44, 46, 47, 450
and mass, 147 and net force, 146 of a mass-spring system, 366–369 units of, 23, 26 Acceleration due to gravity, 54–62 Acceleration-time graph, 26, 42 Accelerometer, 366 Action force, 160–167 Action-at-a-distance force, 200 Activity (also Decay) rate, 812 Adams, Henry, 617 Adams, John, 282 Air velocity, 92–97 Air, refractive index of, 666, 667t, 670, 673 wavelength of light in, 669 Airbag systems, 140, 141 Alpha decay, 799–801 Alpha emission, 797 Alpha particle, 497, 767–769, 799–801 hazards of, 808t Alternating current, 638t Altitude, 221 Ammeter, 604 Ampere (A), 602, 607 Ampere, André-Marie, 602, 607 Ampere’s law, 642 Amplitude, 355f, 371, 378, 382, 386, 395, 408, 412 Amplitude modulation (AM), 646 Analog radio technology, 646 Anderson, Carl, 836, 842, 846 Angle of diffraction, 689–691 Angle of incidence, 654, 666f Angle of reflection, 654, 666f Angle of refraction, 666, 672 Anti-locking brake systems, 187 Antimatter, 804, 836, 837 Antineutrino, 804 Antinodal lines (also Bright fringes), 686–690, 692 Antinodes, 415, 417–419, 422f, 423f, 424f, 426 Apparent weight, 224–226 Applied force, 130, 171–189, 307 Arago, Dominique, 692 Aristotle, 634 Armature (also Rotor), 608 Artificial satellites, 284–286 Astronomical units (AU), 272, 273 At rest, 13 Atom, 513 Bohr model, 773, 774 energy levels, 774–776 nucleus, 766–769, 790–796 orbit size, 774 planetary model, 768 quantum model, 782, 783 raison-bun-model, 758 structure theories of, 752f subatomic particle models, 845–849 subatomic particles, 830–849 Atomic emission spectroscopy, 779 Atomic mass number, 790, 791, 798 Atomic mass unit, 791 Atomic number, 790, 791, 794, 795 Attitude of images, 656, 662, 664 Audio frequency signal, 646 Aurora borealis, 580, 597, 598f, 779 Avalanches, 448, 449 Average net force, 450 Average velocity, 12–18, 36–38 Axis of rotation, 242, 244f Axle, 242 B Background radiation, 637 Ballistic pendulum, 483–485 Balmer, Johann Jacob, 773 Balmer’s formula, 773, 774, 778 Balmer series, 776f Baryons, 842, 843t, 847 Bearing method, 78 Becker, Wilhelm, 497 Becquerel, Antoine Henri, 797 Becquerel (Bq), 812 Beta emission, 797 Beta particles, 802–805, 808t Beta-negative decay, 802–805, 847 Beta-positive decay, 805, 848 Bethe, Hans, 821 Binding energy, 793–796, 818 Binoculars, 674 Biomechanics, 29 Blackbody radiation curves, 704f, 705 Blackbody, 705 Blau, Marietta, 842 Bohr, Niels, 513, 771, 773, 774 Bohr model, 773, 774, 783 Bohr radius, 774 Born, Max, 739, 783 Bosons, 842 Bothe, Walther, 497 Bow wave, 433 Brahe, Tycho, 214, 269 Bright-light (also Emission line) spectrum, 772f Bubble chamber, 831, 833 C Capacitor, 642 Carbon dating, 815, 816 Cartesian plane, 127 Cassini, Giovanni, 359 Cathode ray, 593, 754, 755f–757 Cathode ray tube, 568, 593f Cavendish, Henry, 205, 524 Cell phones, 646 Central antinode, 686 Centre of curvature, 657, 658 Centre of mass, 492 Centrifugal force, 247 Centripetal acceleration, 243, 244, 247t, 252–255, 265–267 Centripetal force, 244, 246, 247t, 256, 258–267, 277–280 CERN, 831 CGS system, 642 Chadwick, James, 497, 790 “Change in” (Δ), 7 Charge, 593–600, 762, 798 determination of, 528, 529 magnitude of, 529–531 transfer of, 517–522 Charge migration, 520 Charge shift, 521 Charge-to-mass ratio, 755–758 Charging by induction, 521 Chemical energy vs nuclear energy, 820 Circular motion, 242–247 and Newton’s laws, 248–267 of satellites and celestial bodies, 269–286 Clocks, 383, 387 Closed-pipe (also Closed-tube) resonance, 419 Cloud chamber, 830, 831, 833, 842 Coaxial cables, 558 Coefficient of friction, 259 Coefficient of kinetic friction, 182, 183 Coefficient of static friction, 182, 183t, 186 Coils, electron flow in, 587f Collinear charges, 532 Collinear collisions (see Collisions in one dimension) Collinear forces, 132, 133 Collinear vectors, 71, 73, 75 Collisions, 469, 470 charge transfer during, 518, 519 impulse in, 458 Collisions in one dimension, 468–485 Collisions in two dimensions, 487–498 Colour of quarks, 848 Commutator, 608, 609 Compression, 404 Compton, Arthur, 721f Compton effect, 721–724 Compton equation, 722–724 Compton scattering, 721 Concave reflecting surface, 657f–659 Conduction, 519 Conductors, 513, 514 and magnetic fields, 602–611 electric field lines of, 555–558 Conservation of momentum, in discovery of subatomic particles, 497, 498 in one-dimensional collisions, 473–479 Index 899 22-PearsonPhys-Index 7/25/08 7:40 AM Page 900 in two-dimensional collisions, 489–495 Conservative forces, 314, 319, 320 Constant acceleration, 450 Constant mass, 450 Constructive interference, 412, 416f, 417, 425–427, 686–689, 782f Continuous spectrum, 772f Converging lens, 677, 678 Converging mirror, 657–659 Convex reflecting surface, 657f–659 Copernicus, Nicholas, 214 Cosmic radiation, 637, 638t Coulomb (C), 529 Coulomb’s constant, 774 Coulomb’s force (see Electrostatic force) Coulomb’s law, 529–531, 768 Coulomb’s proportionality constant (k), 529, 530 Crash test dummies, 29 Crest, 394, 395, 397f, 408, 430 Critical angle, 672, 673 Crookes, William, 754 Curie, Marie and Pierre, 797, 808 Current, 587, 602–609, 642 Curved mirrors, 657–662 Cycle, 249, 344, 355 Cyclotron particle accelerator, 841 D da Vinci, Leonardo, 180 Dalton, John, 754 Damping, 385 Daughter element, 799 Daughter nucleus decay, 807 Davisson, C. J., 729 Davisson-Germer experiment, 729f, 730f de Broglie, Louis, 726f, 782 de Broglie’s wave equation, 726–728, 730–733 de Broglie’s wavelength, 726 de Coulomb, Charles, 524, 528, 529 de Maricourt, Pierre, 582 Decay (also Activity) rate, 812 Decay constant, 811, 812 Defibrillators, 559 Deformation from colliding objects, 481 “Delta” symbol (Δ), 7 Destructive interference, 412, 413f, 416f, 417, 419, 425–427, 687, 689, 782f Diamond, 667t, 670, 672 Difference in path length, 688 Diffraction, 685, 690–694 Diffraction grating, 692, 693, 771 Diffuse (also Irregular) reflection, 653 Digital wireless technology, 646 Dirac, Paul Adrien Maurice, 582, 836 Direct current, 608 Dispersion, 675 Displacement, 7, 9, 56–58, 73–75, 80–89, 293–295 from velocity-time graphs, 33–35, 40, 48–51 Diverging lens, 677, 678 Diverging mirror, 657–659 Diverging rays, 397, 398 Domains, 589 Doping, 515 Doppler, Christian, 429 Doppler effect, 430–432 Doppler frequency, 431, 432 Doppler ultrasound, 434 Doppler wavelength, 431–434 Down quark, 845, 846t Drift tube particle accelerator, 841 Dynamics, 126 E Earth, magnetic field of, 591, 597, 598 orbital diameter, 648f, 649 Ebonite rod, 518, 520, 521, 529 Eddington, Arthur Stanley, 821 Efficiency, 324 Effluvium theory, 544 Einstein, Albert, 199, 640, 706, 713 Einstein’s mass-energy equation, 793, 794 Elastic (also Spring) constant, 299 Elastic collisions, 481, 482, 495–497 Elastic force, 128, 129 Elastic limit, 349 Elastic potential energy, 300–302, 307, 370, 402, 406 Electric charge (see Charge) Electric current (see Current) Electric field, 545–553, 554–559, 641–643, 756f between charged plates, 567, 568 in one dimension, 550 in two dimensions, 551, 552 magnitude and direction of, 546–553 Electric field lines, 554–559 Electric force on a charged particle, 755 Electric generators, 617 Electric motor, 602, 608, 609, 614, 617, 619 Electric potential (also Voltage), 564, 565 Electric potential difference, 565, 566 Electric potential energy, 561–564, 571 of an alpha particle, 769 of an electron, 774, 775 Electrical conductivity, 513, 514 Electrical interactions, 512–522, 524–537 and the Law of conservation of energy, 570–575 Electricity, 512–522 Electromagnetic force, 838t Electromagnetic induction (see Generator effect) Electromagnetic radiation (see also Light), 636–646 diffraction and interference, 684–696 models of, 639–643 photoelectric effect, 712–719 production of, 644, 645 speed of, 648–651 Distance, 6, 7, 106–111 Electromagnetic spectrum, 637, 638t, 708, 709f Electromagnets, 588, 589 Electromagnetic theory, 641–643 Electromagnetic wave, 643 Electromotive force, 611 Electron, 498, 513, 842 charge of, 529, 762 charge-to-mass ratio, 755–758 de Broglie’s wave equation for, 726, 727 discovery of, 754 energy level transition of, 776, 778, 779 in charge transfer, 517–522 in Compton scattering, 721, 722 in photoelectric effect, 712–719 kinetic energy of, 730–733 mass of, 792t, 844t wave nature of, 782, 783 wavelength of, 728, 836, 837 Electron microscope, 727 Electron-positron annihilations, 836 Electron volt (eV), 565, 843, 844 Electron-wave interference, 729 Electrostatic discharge, 537 Electrostatic (also Coulomb’s) force, 528–537 Electrostatics, 513–521 Electroweak force, 848 Elementary charge, 774 Elementary unit of charge, 762 Ellipse, 269 Emission line (also Bright-light) spectrum, 772, 776 Energy, 292–304, 306–309, 311–322, 324–329, 639 along a pulse, 402 in quanta, 705–709 Energy conservation, 497, 498 Energy level, 773–776 Energy-mass equation, 843 Equilibrium position, 394, 395 Erect image, 660, 662, 680 Euclid, 634 Excited state, 775 Explosion interaction, 476 Extrasolar planets, 283 F Faraday, Michael, 545, 584, 610, 611, 614 Faraday’s ice pail experiment, 559 Faraday’s law, 642 Femtometres, 790 Fermi, Enrico, 804 Fermilab Tevatron accelerator, 840f Fermions, 842 Ferromagnets, 589 Feynman, Richard, 739 Fibre optics, 674 Field (see also Electric fields), 200, 545 First order maximum, 427 Fission, 818–820 Fizeau, Armand, 649, 666 Fletcher, Harvey, 761 Focal length, 657, 658, 662, 667, 678 Force, 127–135, 293–295 900 Index 22-PearsonPhys-Index 7/25/08 7:40 AM Page 901 and centripetal acceleration, 252–255 effect on momentum and impulse, velocity-displacement, 368f, 369f velocity-time, 21f–28f, 62f 455–466 in collisions, 474 Forced frequency, 382, 383f, 385 Frame of reference, 13, 14 Franklin, Benjamin, 512, 513 Fraunhofer lines, 773 Free fall, 226, 227 Free-body diagrams, 129–135f, 146f, Gravimeters, 222 Gravitational acceleration, 217–219 Gravitational field, 200, 201, 545 Gravitational field strength, 201, 217, 220–222, 378, 379 Gravitational force, 128, 129, 131, 173–175, 196–201, 203–214, 216–228, 25
9f–264, 277, 524, 838t 150f, 151f, 161f–167f, 172f–189f, 197f, 198f, 209f, 222f–224f, 226f Frequency, 249, 265–267, 344, 345, 408, Gravitational mass, 199 Gravitational potential energy, 295–298, 307–309, 312–316, 321, 560, 562 409, 429–432, 636, 637 Frequency modulation (FM), 646 Fresnel, Augustin, 691, 692f Fresnel lens, 634f Friction, 130, 165, 169–189, 320 463 and momentum, 449 charging objects by, 518, 519 effects on motion, 184 in circular motion, 258 Friedman, Jerome, 845 Fuel cells, 329 Fundamental forces, 194 Fundamental frequency, 422 Fundamental particles, 836, 837 Fusion, 818, 821, 822 G g force, 216 Galilean telescope, 682f Galileo, 54, 103, 137, 138, 281, 648 Galle, Johann Gottfried, 282 Galvani, Luigi, 602 Galvanometer, 602f, 604f, 611f Gamma decay, 806 Gamma emission, 797, 806, 808t Gamma radiation, 636 Gamma rays, 637f, 638t Gauss (G), 586 Geiger, Hans, 767 Gell-Mann, Murray, 845 Generator effect (also Electromagnetic induction), 609–611 Generator effect, 615, 617–619 Genetic damage, 808 Geostationary satellites, 285, 286 Germer, L. H., 729 Gilbert, William, 512, 583 Glaser, Donald, 831 Glashow, Sheldon, 848 Glass, 666, 667t, 668, 673 Gluons, 838 Gramme, Zénoble Théopile, 617 Grand unified theory, 849 Graphs, acceleration-displacement, 367f, 368f, 369f acceleration-time, 26f, 42f force-displacement of a pendulum, 361f force-displacement of a spring, 351f net force-time, 459f–462f position-time, 11f–18f, 61f, 62f Graviton, 838 Gravity, 54–62 Gravity assist, 214 Gray (Gy), 809 Ground state, 774 Ground velocity, 92–99 Grounding, 521, 610 H Hadrons, 842, 843 Half-life, 812–814 Hansen, Hans Marius, 771 Harrison, John, 383 Heisenberg, Werner, 734f Heisenberg’s uncertainty principle, 735 Helium formation, 821, 822 Henry, Joseph, 610, 611 Herschel, William, 282 Hertz (Hz), 249, 250, 344, 345 Hertz, Heinrich, 644, 645, 648, 712 High tide, 210, 211 Hit-and-stick interaction, 477 Hollow conducting objects, 557, 558 Hooke, Robert, 299, 349f, 684 Hooke’s law, 299–301, 349–354, 366, 368 Horsepower (Hp), 324 Huygens, Christiaan, 359, 381, 634, 639, 648, 649, 684f Huygens’ Principle, 684, 685 Hydrogen, 773f Hydrogen fusion, 821–823 I Images, 653–665 formation in a curved mirror, 657–662 formation in a plane mirror, 654–656 Impulse, 457–466, 470 In phase, 395, 416, 425, 426 Incandescent, 704 Incident light, 653, 654f, 666 Incident wave, 395, 398 Inclines, 173, 174, 176–179, 188, 189, 308 Induced current, 611, 617–619 Induction, 520, 521 Induction coils, 615t Inelastic collisions, 483–485, 495–497 Inertia, 138–140, 147–157 Inertial mass, 148–157 Infinity, 562 Infrared radiation, 636 Infrared ray, 709f Infrared spectrum, 637, 638t Input energy, 324 Instantaneous momentum, 449 Instantaneous velocity, 24, 25 Insulators, 513, 514 Interference, 411–413, 416, 417, 685–690, 695, 696 Interference fringes, 686 Interference pattern, 425, 426f Interferometer, 649 Inverse square law, 529 Inverted image, 660, 662, 680 Ionization energy, 775 Ionization smoke detectors, 802 Irregular (also Diffuse) reflection, 653 Isolated systems, 311–316, 470, 475 Isotopes, 791, 796f, 806 J Jerk, 26 K Kammerlingh Onnes, Heike, 515 Kendall, Henry, 845 Kepler, Johannes, 214, 269 Kepler’s constant, 271–273 Kepler’s laws of planetary motion, 269–279 Keplerian telescope, 682f Kilogram-metres per second (kg•m/s), 449, 457 Kinematics, 6 Kinematics equations, 47–53, 56, 144 Kinetic energy, 302–304, 306–309, 312–316, 321, 370, 402, 480, 570, 571, 573 in alpha decay, 799, 801 in beta decay, 804 in elastic collisions, 481, 482, 495 in inelastic collisions, 483–485, 495 of alpha particles, 769 of electrons, 774, 775 of photoelectrons, 713–717 Kinetic friction, 176–182, 187, 258 Kirchhoff, Gustav, 705, 771, 772 L “Lambda” symbol (), 408, 409, 417, 419, 424, 427, 429–432 Latitude, 221 Law of charges, 512 Law of conservation of charge, 517 Law of conservation of energy, 312, 497, 498, 721, 722 and electrical interactions, 570–575 for the photoelectric effect, 714, 718 Law of conservation of momentum, 473, 721, 722 in ballistic pendulum system, 484 in two-dimensional collisions, 491, 492, 495 Law of magnetism, 582 Law of reflection, 654, 658 Le Verrier, Urbain, 282 Index 901 22-PearsonPhys-Index 7/25/08 7:40 AM Page 902 Left-hand rule, 834 for deflection of charged particles, 594, 595 for magnetic fields, 588 for magnetic force, 604, 611, 756f Lenses, 677–681 Lenz’s law, 618, 619 Leptons, 842, 843t, 848t Lewis, Gilbert, 706 Light (see also Electromagnetic radiation) from colliding objects, 481 quantum theory of, 705–709, 712–719, 721–724, 726–735 reflection, 653–665 refraction, 666–681 speed of, 636, 648–651, 669, 670 Light waves, 712–719 Lightning, 512, 513, 522, 529 Lightning rod, 554 Lodestone (also leading stone), 582, 583 Longitudinal waves, 401, 404–406, 695f Lord Kelvin (see William Thomson) Loudspeaker, 614t Low tide, 210, 211 Lyman series, 776f M Maglev, 587 Magnesium, 636f Magnetic deflection, 593–597, 603, 604 Magnetic field, 584–591, 593–600, 602–611, 641–643, 756f–758 left-hand rule for, 588 Magnetic field strengths, 586t Magnetic force, between two current carrying conductors, 607 calculation of, 598–600 on a current carrying conductor, 603–606 Magnetic monopoles, 582 Magnetic poles, 582, 585 Magnetism, 512, 587–591 Magnetization, 589, 590 Magnetohydrodynamic propulsion, 614t Magnetron, 599 Magnets, 583, 585, 586, 617, 618 Magnification, 656, 662, 664 Magnitude, 6 Major axis, 269f Marconi, Guglielmo, 646 Marsden, Ernest, 767 Mass, 220 and conservation of momentum, 474 due to gravity, 199 in momentum, 448–452 of celestial bodies, 218t, 280, 281 Mass defect, 794, 795 Mass spectrometer, 759 Mass-energy equation, 793, 794 Mass-spring systems, 354–360, 366–376 Matter waves, 726–735, 729 Maximum, 426 Maxwell, James Clerk, 641, 648, 695, 771 902 Index Maxwell’s equations, 642 Mechanical energy, 306, 309, 311–322 Mechanical resonance, 382 Mechanical waves (see Waves) Mechanics, 306 Mediating particles, 837, 838 Medium, 394, 395, 404, 406, 411 Mesons, 842, 843t, 845, 847 Michell, John, 205 Michelson, Albert, 650, 651 Microwaves, 637f, 638t Millikan, Robert Andrews, 713f–715, 761–763 Minor axis, 269f Mirage, 667 Mirror equation, 662, 664 Mirrors, 654–662 Momentum, 449–452, 470 and impulse, 454–466 and Newton’s Second Law, 450–452, 456 and non-linear net forces, 461, 462 conservation of, 473–479, 489–498 in one-dimensional collisions, 473–479 in two-dimensional collisions, 489–495 of a photon, 728 Motion, in one dimension, 6–62, 70–75 in two dimensions, 76–89 of projectiles, 102–111 sign conventions for, 8–10 Motor effect force, 593 Muons, 842, 843t N Nanotechnology, 619, 674 Navigator method, 77, 78 Neddermeyer, Seth, 842, 846 Negative acceleration, 28, 29, 44 Net charge, 517, 518 Net force, 131–135, 139, 146, 152, 155, 157, 164, 165, 171–189, 306, 307, 356, 366, 456–466, 473, 474, 489 in circular motion, 256 on momentum, 450 Neutrino, 804 Neutron, 497, 498, 790, 792t, 844t, 847 Neutron number, 790, 791, 794, 795 Newton (N), 127 Newton, Isaac, 194, 196, 214, 276–281, 545, 634, 639, 675 Newton’s first law of motion, 139, 140 Newton’s law of gravitation, 524 Newton’s law of universal gravitation, 204–214, 216 Newton’s second law of motion, 143–157, 366, 367 and horizontal motion, 149–151 and momentum, 450–452, 456 and single pulley system, 154, 155 and two-body systems, 153, 154 and two-pulley system, 156, 157 and vertical motion, 152, 153 on objects in systems, 470, 473 Newton’s third law of motion, 161–167, 459, 474 Newton-second (N•s), 457 Nodal lines (also Minimum; Dark fringes), 426, 686 Nodes (also Nodal points), 415, 417–419, 423f, 424f, 687 Non-collinear forces, 133, 134 Non-collinear vectors, 80–89 Non-conservative forces, 319 Non-isolated systems, 320–322 Non-linear net force, 461, 462 Non-uniform motion, 22–28, 31 Non-zero net force, 307, 450 Normal (also Perpendicular) force, 130, 131 Normal force, 258, 259, 260 Normal line (N), 653, 654f, 666f North pole, 582 Nuclear energy vs chemical energy, 820 Nuclear reactions, 818–823 Nucleon, 790, 795, 796 Nucleosynthesis, 823 Nucleus, 513, 766–769, 790–796 decay rates of, 811–816 reactions in, 818–823 size of, 768, 769 O Oersted, Hans Christian, 587 Oil-drop experiment, 761 Omega particle, 845 Open-pipe (also Open-tube) resonance, 424 Opsin, 636 Optical centre, 678f Optical fibres, 673 Orbital height, 279, 280 Orbital period, 271–289 Orbital perturbations (also Wobbles), 281, 282, 283 Orbital radius, 271–289 Orbital, 783 Origin, 6 Oscillation, 344 Oscillators, 640 Oscillatory motion, 344, 345, 348–364, 366–379, 381–387 Out of phase, 416 Output energy, 324 Overtones, 422, 423, 424 Oxygen, 779f P Parabola, 61, 62 Parabolic mirror, 661 Parallel plates, electric field between, 567 motion of charges between, 572, 573 Parallel-plate capacitors, 558, 559 electric field between, 574, 575 electric potential energy between, 563 Parent element, 799 Partially reflected/refracted rays, 666 22-PearsonPhys-Index 7/25/08 7:40 AM Page 903 Particle, 639 Particle accelerator, 840f, 841, 842 Particle model, 639, 640 Particle-in-a-box model, 730–735 Paschen series, 776f Path length, 688, 689 Pauli, Wolfgang, 498, 804 Payne-Gaposchkin, Cecilia, 821 Pendulum, 314–316, 359–362, 377–379, 381, 382, 483 Pendulum-bullet system (also Ballistic pendulum), 483–485 Period, 249, 251, 265–267, 344, 345, 408, 409 of a mass-spring system, 373–376 of a pendulum, 377–379 Perpendicular bisector, 686–691 Perrin, Jean Baptiste, 754 Phase shift, 426 Photoelectric effect, 712–719 Photoelectrons, 712 Photon, 640, 706–709, 722–724, 728, 776 Piezoelectric material, 387 Pions, 842, 843t Pitch, 430 Planck, Max, 640, 641, 705f, 752 Planck’s constant, 705, 706, 713–715, 726, 774 Planck’s formula, 705–708, 713 Plane mirror, 654 Plane polarized light, 696 Planetary (also Solar-system, Nuclear, or Rutherford) model, 768 Planetary motion, 269–289 Plasma, 522 Plates as conductors, 556 Point of incidence, 653, 654f, 666f Point source, 425, 426 Poisson, S
imon, 691 Poisson’s Bright Spot, 692 Polar coordinates method, 77, 78 Polarization, 696 Polarizing filters, 696 Porro prism, 674 Position, 6 Position-time graphs, 11–18, 61, 62 Position-time graphs, vs Velocity-time graphs, 25, 26, 28, 41–44 Positron, 804, 836, 837 Positron-emission tomography (PET), 837 Potential energy (see also Elastic potential energy; Gravitational potential energy), 570 Powell, Cecil Frank, 842 Power, 324–329 Pressure waves, 406, 433 Primary cosmic rays, 841 Principal axis, 657, 677, 678 Principal focal point, 657 Principal focus, 677, 678 Principal quantum number, 774 Principle of superposition, 412, 413 Prisms, 675–677, 771 Projectile, 54, 465 Projectile motion, 54, 102–112, 108–111 Propellar aircraft, 166 Proportionalities, 452 Proton, 497, 498, 513, 529, 790, 792t, 844t, 847 Proton-proton chain, 821 Ptolemy, 214 Pulse, 401–408, 411–413 Pythagorean theorem, 83, 85, 535–537 Q Quadratic equations, 56 Quanta, 640 Quantization of charge, 761–763 Quantized, 705 Quantum, 705 Quantum chromodynamics, 848 Quantum colour, 848 Quantum electrodynamics, 838 Quantum field theory, 837 Quantum indeterminancy, 740 Quantum model (also Quantum theory), 640, 641, 782, 783 Quantum physics, 830–849 Quarks, 845, 846t, 847, 848t Quartz crystals, 387 R Radar vectors, 7 Radial line, 242f–244 Radiation sickness, 808 Radiation, 811–816 Radio frequency signal, 646 Radio waves, 637f, 638t, 645, 646 Radioactive decay, 797–809, 812–814 Radioactive decay series, 807 Radioisotopes, 808, 812–814 Radiotherapy, 815 Radius of celestial bodies, 218t Radius of curvature, 657 Rainbows, 675 Raisin-bun model, 758 Range of a projectile, 105 Rarefaction, 404 Ray diagrams, 653, 654f, 657f–659f, 666f, 678f–681f Rays, 397, 398 Reaction force, 160–167 Real image, 654, 659, 660, 662, 680 Recomposition, 675f, 676 Rectilinear propagation, 653 Reference coordinates, 71 Reference point, 297, 298, 309 Reflected ray, 653, 654f, 666f Reflected wave, 395, 398, 404 Reflection, 653–665 Refracted ray, 666 Refraction, 666–681 Refractive index, 666, 667t, 672, 676 Regular (also Specular) reflection, 653 Relative biological effectiveness, 809 Relative motion, 91–100 Resonance, 418–420, 422–424 Resonant frequency, 381–383f, 385, 387, 418, 419 Restoring force, 353, 354, 356, 360, 362, 372, 375 Resultant vector, 71–73 Retinal, 636 Revolution, 249, 251 Revolutions per minute (rpm), 249, 250 Right-hand rule, 833f, 834 Rocket, 167, 459, 460, 475 Rømer, Olaus, 648 Rosette nanotube, 619f Rotational kinetic energy, 302 Rotor (also Armature), 608, 619 Rutherford, Ernest, 513, 767, 790, 797 Rutherford’s scattering experiment, 767 Rydberg, Johannes Robert, 773 Rydberg’s constant, 773, 778 S Safety devices, 463–465 Salaam, Abdus, 848 Satellites, 269–286 Scalar quantity, 6–10, 449 Scanning electron microscope, 727 Schrödinger, Erwin, 783 Secondary cosmic rays, 841 van de Graaf particle accelerator, 841 Semiconductors, 514, 515 Shock wave, 433 Sievert (Sv), 809 Simple harmonic motion, 355–364, 366–379, 381–387, 408 Simple harmonic oscillators, 355, 366–379 Snell, Willebrord, 667 Snell’s law, 667–669, 673 Solenoids, 589 Sonic boom, 433, 434 Sound, 418–420, 422–424, 429–434, 481 Sound barrier, 433 Source charge, 546, 547f, 548f South pole, 582 Spectrometers, 773 Spectroscopy, 771–773 Spectrum, 675, 676, 772 Specular (also Regular) reflection, 653 Speed, 10, 12, 52, 56, 106–111 and power, 327, 328 in circular motion, 242–244, 250, 251, 254, 255, 265–267 of a mass-spring system, 369–372, 374 of a satellite, 277 of a spring pulse, 406, 409 of light, 636 of sound, 420, 430, 431, 434 Spin, 842 Sports, 466 Spring constant, 350, 351, 357, 374 Spring scale, 128f Spring systems, 349–354 Spring waves, 401, 402 Standard model, 848 Standing wave, 415, 417, 418, 422f, 423f Index 903 22-PearsonPhys-Index 7/25/08 7:40 AM Page 904 Static equilibrium, 555 Static friction, 171–175, 180–183, 258, 259 Statics, 126 Stationary states, 773 Stator, 608 Step leader, 522 Stopping potential, 716, 717 Straight wire, 587 Strange particles, 845 Strange quark, 845, 846t Streamer, 522 String theory, 544, 849 Strong nuclear force, 793, 838t Subatomic models, 845–849 Subatomic particles, 497, 498, 830–849 Sudbury Neutrino Observatory, 498 Sudden infant death syndrome monitors, 615t Sun, 821, 822 Superconductors, 515 Supernova, 823 Superposition, 208–210 Superposition principle, 549 Surfaces, cushioning effects of, 455, 456, 496 electric field lines on, 556, 557 of mirrors, 657 Synchrotron particle accelerator, 841 System, 470 T Tail of a vector, 70–75 Tangent, 24, 25 Tangent function, 535–537 Tangential line, 242f–244 Taylor, Richard, 845 Technologies, 614, 615, 619, 646 Telescopes, 682f Telluric currents, 610 Tension, 132, 134, 135, 264, 356, 463 Terminal velocity, 61 Tesla (T), 585, 834 Test charge, 546, 547f, 548f Thales, 512 Thin lens equation, 680, 681 Thomson, George Paget, 729f Thomson, Joseph John, 754, 757, 758 Thomson, William (also Lord Kelvin), 702f, 703 Threshold frequency, 712, 713, 718 Time, 455–466 Tip of a vector, 70–75 Torsion balance, 205 Total internal reflection, 672 Trajectory, 103 Translational kinetic energy, 302 Transmission electron microscope, 727 Transmutation, 806 Transverse pulse, 401 Transverse waves, 401, 406, 408, 695f Triboelectric series, 518f Trough, 394, 395, 397f, 408 True weight, 222, 223 True weightlessness, 228 Tuned mass damper, 385, 386f Tuning, 424 Turnbull, Wallace Rupert, 166 U Ultraviolet catastrophe, 705 Ultraviolet rays, 637f, 638t Uniform circular motion, 242 Uniform motion, 13, 21–28, 31–44, 104 Uniformly accelerated motion, 25, 52, 104 Universal gravitational constant (G), 204–207, 524 Universal wave equation, 408, 424, 645, 648, 669, 670, 676 Up quark, 845, 846t Uranium-235 fission, 819f V Vacuum, 666, 667t, 668 Van Allen belt, 597 Van de Graaff generator, 560 Vectors, 6–10, 70–75, 77, 78, 83, 84, 127, 449, 451 adding, 71–73, 80–82, 86–89, 489–495, 550, 552 analyzing electrostatic forces, 532–537 subtracting, 73 Velocity, 10, 12–18, 21–28, 31–44, 46–51, 53, 56–62, 91–100, 139 in circular motion, 242–244, 247t in momentum, 448–452, 474 of a mass-spring system, 367, 368 Velocity-time graphs, 21–28, 31–44, 46–50, 62 Vertex, 657, 658 Vertical inversion, 656 Vertical projectiles, 58–62 Virtual image, 654, 659, 660, 662, 680 Virtual particles, 837 Visible ray, 709f Visible spectrum, 637, 638t Volt, 564 Volta, Alessandro, 564 Voltage (also Electric potential), 564, 565 Voltmeter, 604 von Fraunhofer, Josef, 771 Voyager missions, 213, 214 W Water, 666–669 Watt (W), 324 Watt, James, 324 Wave, 394, 395–409, 411–427, 428–434, 639 Wave (also Point) source, 395, 397f, 398 Wave equation, 726, 727 Wave front, 395, 398, 684 Wave model, 639, 640, 669, 684–698 Wave patterns, 731 Wave train, 395, 417 Wavelength, 395, 397f, 408, 409, 424, 427, 429–432, 636, 637, 676t and angle of diffraction, 690, 691 and Snell’s law, 669, 670 Wavelet, 684 Wave-particle duality, 726, 737–740 Weak nuclear force, 804, 838t Weight (see also Apparent weight; True weight), 128, 197, 198, 216, 219, 220 Weightlessness, 228 Weinberg, Steven, 848 Wheel, 242f, 243 White light, 675–677 Wilson, Charles Thomas Rees, 831 Wind velocity, 92–97 Work function, 712t, 713, 718 Work, 293–304, 306–309, 320, 321, 324, 560–562 and electric potential energy, 563, 567 Work-energy theorem, 307, 308 X X rays, 637f, 638t, 709f, 721, 722 x vector component, 77, 86–89, 105–111, 134, 135 Y y vector component, 77, 86–89, 105–111, 134, 135 Young, Thomas, 639f, 685, 640, 691 Young’s double-slit experiment, 685, 686, 738–740 Z Zero electric potential energy, 562 Zero net forces, 307 Zero reference point, 561, 562 Zweig, George, 845 904 Index
ing is changing. Q: In each picture in the Figure 1.1, what is moving and how is its position changing? A: The train and all its passengers are speeding straight down a track to the next station. The man and his bike are racing along a curving highway. The geese are flying over their wetland environment. The meteor is shooting through the atmosphere toward Earth, burning up as it goes. Frame of Reference There’s more to motion than objects simply changing position. You’ll see why when you consider the following example. Assume that the school bus pictured in the Figure 1.2 passes by you as you stand on the sidewalk. It’s obvious to you that the bus is moving, but what about to the children inside the bus? The bus isn’t moving relative to them, and if they look at the other children sitting on the bus, they won’t appear to be moving either. If the ride 1 www.ck12.org FIGURE 1.1 FIGURE 1.2 is really smooth, the children may only be able to tell that the bus is moving by looking out the window and seeing you and the trees whizzing by. This example shows that how we perceive motion depends on our frame of reference. Frame of reference refers to something that is not moving with respect to an observer that can be used to detect motion. For the children on the bus, if they use other children riding the bus as their frame of reference, they do not appear to be moving. But if they use objects outside the bus as their frame of reference, they can tell they are moving. The video at the URL below illustrates other examples of how frame of reference is related to motion. http://www.youtube.com/watch?v=7FYBG5GSklU MEDIA Click image to the left for more content. Q: What is your frame of reference if you are standing on the sidewalk and see the bus go by? How can you tell that the bus is moving? A: Your frame of reference might be the trees and other stationary objects across the street. As the bus goes by, it momentarily blocks your view of these objects, and this helps you detect the bus’ motion. 2 www.ck12.org Summary Chapter 1. Motion • Motion is defined as a change of position. • How we perceive motion depends on our frame of reference. Frame of reference refers to something that is not moving with respect to an observer that can be used to detect motion. Vocabulary • frame of reference: Something that is not moving with respect to an observer that can be used to detect motion. • motion: Change in position. Practice Do the frame of reference activity at the following URL. Watch the introduction and then do the nine trials. Repeat any trial you answer incorrectly until you get the correct answer. http://www.amnh.org/learn/pd/physical_science/week2/frame_reference.html Review 1. How is motion defined in science? 2. Describe an original example that shows how frame of reference influences the perception of motion. References 1. Train: John H. Gray; Bike: Flickr:DieselDemon; Geese: Don McCullough; Meteor: Ed Sweeney (Flickr:Navicore). . CC BY 2.0 2. Bus: Flickr:torbakhopper; Children: Flickr:woodleywonderworks. . CC BY 2.0 3 CHAPTER 2 Position and Displacement www.ck12.org • Define and give an example of a frame of reference. • Describe the difference between distance and displacement. • Identify the position, distance, and displacements in various descriptions of motions. In stockcar races, the winners frequently travel a distance of 500 miles but at the end of the race, their displacement is only a few feet from where they began. Position, Distance, and Displacement In order to study how something moves, we must know where it is. For straight line motion, it is easy to visualize the object on a number line. The object may be placed at any point on the number line either in the positive numbers or the negative numbers. In making the zero mark the reference point, you have chosen a frame of reference. The position of an object is the separation between the object and the reference point. It is common to choose the original position of the object to be on the zero mark. When an object moves, we often refer to the amount it moves as the distance. Distance does not need a reference point and does not need a direction. If an automobile moves 50 kilometers, the distance traveled is 50 kilometers regardless of the starting point or the direction of movement. If we wish to find the final position of the automobile, however, just having the distance traveled will not allow us to determine the final position. In order to find the final position of the object, we need to know the starting point and the direction of the motion. The change in the position of the object is called its displacement. The displacement must include a direction because the final position may be either in the positive or negative direction along the number line from the initial position. The displacement is a vector quantity and vectors are discussed in another section. Summary • The length traveled by an object moving in any direction or even changing direction is called distance. 4 www.ck12.org Chapter 2. Position and Displacement • The location of an object in a frame of reference is called position. • For straight line motion, positions can be shown using a number line. • The separation between original and final position is called displacement. Practice The following url is for a discussion of the difference between distance and displacement. http://www.tutorvista.com/content/physics/physics-i/motion/distance-and-displacement.php Use this resource to answer the questions that follow. MEDIA Click image to the left for more content. 1. What is the vector equivalent of the scalar “distance”? 2. What is the vector equivalent of the scalar “speed”? Review 1. Explain the difference between distance and displacement in your own words. 2. Suppose that John lives on a square block that is 180 yards per side, and in the evenings, he walks with his dog around the block after dinner for a little exercise. (a) If John walks once around the block, what distance does he travel? (b) If John walks around the block, what is his displacement at the end? 3. Joanna’s house is 8000 feet due west of her school. If her house is assigned the position of zero and her school is assigned the possition of +8000, what would Joanna’s position be if she walked 100 feet west of her house? • distance: The space between two objects but this is not adequate when considering the distance travelled. The distance travelled cannot be negative and can never get smaller –in this sense, distance is the total length of path traversed by the moving body irrespective of direction. • displacement: The vector from the initial position to a subsequent position assumed by a body. References 1. Image copyright Action Sports Photography, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 5 CHAPTER 3 • Distinguish between velocity and speed. • Represent velocity with vector arrows. • Describe objects that have different velocities. • Show how to calculate average velocity when direction is constant. www.ck12.org Velocity Ramey and her mom were driving down this highway at 45 miles per hour, which is the speed limit on this road. As they approached this sign, Ramey’s mom put on the brakes and started to slow down so she could safely maneuver the upcoming curves in the road. This speed limit sign actually represents two components of motion: speed and direction. Speed and Direction Speed tells you only how fast or slow an object is moving. It doesn’t tell you the direction the object is moving. The measure of both speed and direction is called velocity. Velocity is a vector. A vector is measurement that includes both size and direction. Vectors are often represented by arrows. When using an arrow to represent velocity, the length of the arrow stands for speed, and the way the arrow points indicates the direction. If you’re still not sure of the difference between speed and velocity, watch the cartoon at this URL: http://www.youtube.com/watch?v=mDcaeO0WxBI MEDIA Click image to the left for more content. 6 www.ck12.org Chapter 3. Velocity Using Vector Arrows to Represent Velocity The arrows in the Figure 3.1 represent the velocity of three different objects. Arrows A and B are the same length but point in different directions. They represent objects moving at the same speed but in different directions. Arrow C is shorter than arrow A or B but points in the same direction as arrow A. It represents an object moving at a slower speed than A or B but in the same direction as A. FIGURE 3.1 Differences in Velocity Objects have the same velocity only if they are moving at the same speed and in the same direction. Objects moving at different speeds, in different directions, or both have different velocities. Look again at arrows A and B from the Figure 3.1. They represent objects that have different velocities only because they are moving in different directions. A and C represent objects that have different velocities only because they are moving at different speeds. Objects represented by B and C have different velocities because they are moving in different directions and at different speeds. Q: Jerod is riding his bike at a constant speed. As he rides down his street he is moving from east to west. At the end of the block, he turns right and starts moving from south to north, but he’s still traveling at the same speed. Has his velocity changed? A: Although Jerod’s speed hasn’t changed, his velocity has changed because he is moving in a different direction. Q: How could you use vector arrows to represent Jerod’s velocity and how it changes? A: The arrows might look like this (see Figure 3.2): FIGURE 3.2 7 Calculating Average Velocity You can calculate the average velocity of a moving object that is not changing direction by dividing the distance the object travels by the time it takes to travel that distance. You would use this formula: www.ck12.org velocity = distance time This is the same formula tha
t is used for calculating average speed. It represents velocity only if the answer also includes the direction that the object is traveling. Let’s work through a sample problem. Toni’s dog is racing down the sidewalk toward the east. The dog travels 36 meters in 18 seconds before it stops running. The velocity of the dog is: velocity = distance time 36 m 18 s = 2 m/s east = Note that the answer is given in the SI unit for velocity, which is m/s, and it includes the direction that the dog is traveling. Q: What would the dog’s velocity be if it ran the same distance in the opposite direction but covered the distance in 24 seconds? A: In this case, the velocity would be: velocity = = distance time 36 m 24 s = 1:5 m/s west Summary • Velocity is a measure of both speed and direction of motion. Velocity is a vector, which is a measurement that includes both size and direction. • Velocity can be represented by an arrow, with the length of the arrow representing speed and the way the arrow points representing direction. • Objects have the same velocity only if they are moving at the same speed and in the same direction. Objects moving at different speeds, in different directions, or both have different velocities. • The average velocity of an object moving in a constant direction is calculated with the formula: velocity = distance time . The SI unit for velocity is m/s, plus the direction the object is traveling. Vocabulary • vector: Measure such as velocity that includes both size and direction; may be represented by an arrow. • velocity: Measure of both speed and direction of motion. 8 www.ck12.org Practice Chapter 3. Velocity At the following URL, review how to calculate speed and velocity, and work through the sample problems. Then solve the 10 practice problems. http://www2.franciscan.edu/academic/mathsci/mathscienceintegation/MathScienc eIntegation-827.htm Review 1. What is velocity? 2. How does velocity differ from speed? Why is velocity a vector? 3. Explain how an arrow can be used to represent velocity. 4. Use vector arrows to represent the velocity of a car that travels north at 50 mi/h and then travels east at 25 mi/h. 5. Another car travels northwest for 2 hours and covers a distance of 90 miles. What is the average velocity of the car? References 1. Christopher Auyeung (CK-12 Foundation); Compass: Seamus McGill. . CC BY-NC 3.0; Compass: Public Domain 2. . . CC BY-NC 3.0 9 www.ck12.org Average Velocity CHAPTER 4 • Explain the difference between speed and velocity. • Define the concept of average velocity. • Given displacement and time, calculate average velocity. • Solve for any variable in the equation Vave = Dx Dt Test Pilot Neil Armstrong (later to become a famous astronaut) is seen here next to the X-15 ship after a research flight. Armstrong made his first X-15 flight on November 30, 1960. This was the first X-15 flight to use the ball nose, which provided accurate measurement of air speed at hypersonic speeds. The servo-actuated ball nose can be seen in this photo in front of Armstrong’s right hand. The X-15 employed a non-standard landing gear. It had a nose gear with a wheel and tire, but the main landing consisted of skids mounted at the rear of the vehicle. In the photo, the left skid is visible, as are marks on the lakebed from both skids. Because of the skids, the rocket-powered aircraft could only land on a dry lakebed, not on a concrete runway. The X-15 weighed about 14,000 lb empty and approximately 34,000 lb at launch. The X-15 was flown over a period of nearly 10 years – June 1959 to Oct. 1968 – and set the world’s unofficial speed record of 4,520 mph. Average Velocity The terms speed and velocity are used interchangeably in ordinary language. In physics, however, we make a distinction between the two. Essentially both words refer to how fast an object is moving. Speed is the number we read off the speedometer of a car. It indicates how fast the car is moving at any instant but given no indication of the direction it is moving. For a particular time interval, average speed would be calculated by dividing the distance travelled by the time interval of travel. Velocity, in physics, is a vector, meaning that it must have a direction as well as a magnitude. Furthermore, the average velocity is defined in terms of displacement rather than distance. Average 10 www.ck12.org Chapter 4. Average Velocity velocity would be calculated by dividing the displacement by the time interval where displacement is the change in position of the object. To show the distinction, we could calculate the average speed and the average velocity of a person who walks 50 m to the east, then turns around and walks 50 m to the west. The total time interval is 20 seconds. The distance traveled in this trip is 100 m but the displacement is zero. The average speed would be calculated by dividing 100 m by 20 s with a result of 5 m/s. The average velocity, on the other hand, would be calculated by dividing 0 m by 20 s giving a result of 0 m/s. Neither average speed nor average velocity implies a constant rate of motion. That is to say, an object might travel at 10 m/s for 10 s and then travel at 20 m/s for 5 s and then travel at 100 m/s for 5s. This motion would cover a distance of 700 m in 20 s and the average speed would be 35 m/s. We would report the average speed during this 20 s interval to be 35 m/s and yet at no time during the interval was the speed necessarily 35 m/s. The concept of constant velocity is very different from average velocity. If an object traveled at 35 m/s for 20 s, it would travel the same distance in the same time as the previous example but in the second case, the speed of the object would always be 35 m/s. Example: The position of a runner as a function of time is plotted as moving along the x-axis of a coordinate system. During a 3.00 s time interval, the runner’s position changes from x1 = 50:0 m to x2 = 30:5 m. What was the runner’s average velocity. Solution: Displacement = 30:5 m 50:0 m = 19:5 m (the object was traveling back toward zero) Dt = 3:00 s vave = Dx Dt = 19:5 m 3:00 s = 6:50 m/s Summary • Average speed is distance divided by time. • Average velocity is displacement divided by time. Practice The url below is a physics classroom discussion of speed versus velocity with a short animation. http://www.physicsclassroom.com/Class/1DKin/U1L1d.cfm Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=BWP1tN7PZps MEDIA Click image to the left for more content. 1. The velocity versus time graph in the video is divided into six sections. In how many of these sections is the velocity constant? 2. In how many sections of the graph is the velocity zero? 3. What does the area under the curve of a velocity versus time graph represent? 11 www.ck12.org Review 1. On a one day vacation, Jane traveled 340 miles in 8.0 hours. What was her average speed? 2. An object on a number line moved from x = 12 m to x = 124 m and moved back to x = 98 m. The time interval for all the motion was 10. s. What was the average velocity of the object? 3. An object on a number line moved from x = 15 cm to x = 165 cm and then moved back to x = 25 cm all in a time of 100 seconds. (a) What was the average velocity of the object? (b) What was the average speed of the object? • speed: Distance travelled per unit time. • velocity: A vector measurement of the rate and direction of motion or, in other terms, the rate and direction of the change in the position of an object. References 1. Courtesy of NASA. http://commons.wikimedia.org/wiki/File:Pilot_Neil_Armstrong_and_X-15.jpg. Public Domain 12 www.ck12.org Chapter 5. Instantaneous Velocity CHAPTER 5 Instantaneous Velocity • Define instantaneous velocity. • Plot and interpret position vs time graphs. • Determine the slope of a curve on a position vs time graph. In a footrace such as the one shown here, the initial velocity of a runner is zero. The runner increases his velocity out of the starting blocks and his velocity continues to increase as the race proceeds. For the well-trained athlete, his highest velocity is maintained through the finish line. Instantaneous Velocity The instantaneous velocity of an object is the precise velocity at a given moment. It is a somewhat difficult quantity to determine unless the object is moving with constant velocity. If the object is moving with constant velocity, then the instantaneous velocity at every instant, the average velocity, and the constant velocity are all exactly the same. Position vs Time Graphs Consider a position versus time graph for an object starting at t = 0 and x = 0 that has a constant velocity of 80. m/s. 13 www.ck12.org The velocity of an object can be found from a position vs time graph. On a position vs time graph, the displacement The ratio of is the vertical separation between two points and the time interval is the horizontal separation. displacement to time interval is the average velocity. The ratio of the vertical separation to the horizontal separation is also the slope of the line. Therefore, the slope of straight line is the average velocity. The slope at any given time is the instantaneous velocity. For the motion pictured above, slope = rise run = Dd Dt = 400: m 5:0 s = 80: m/s For accelerated motion (the velocity is constantly changing), the position vs time graph will be a curved line. The slope of the curved line at any point will be the instantaneous velocity at that time. If we were using calculus, the slope of a curved line could be calculated. Since we are not using calculus, we can only approximate the slope of curved line by laying a straight edge along the curved line and guessing at the slope. 14 www.ck12.org Chapter 5. Instantaneous Velocity In the image above, the red line is the position vs time graph and the blue line is an approximated slope for the line at t = 2:5 seconds. The rise for this slope is approximately 170 m and the time interval (run) is 4.0 seconds. Therefor
e, the approximated slope is 43 m/s. Summary • The slope of a position versus time graph is the velocity. For a constant velocity motion, the slope gives the constant velocity, the average velocity, and the instantaneous velocity at every point. For constant acceleration motion, the slope of the position versus time curve gives only the instantaneous velocity at that point. Practice Draw a velocity versus time graph for an object whose constant velocity is 15 m/s and whose position starts at x = 0 when t = 0. Graph the motion for the first 5.0 seconds. Use this resource to answer the questions that follow. http://www.youtube.com/watch?v=sujsb5ZlM8o MEDIA Click image to the left for more content. 1. In the graph on the video, what is graphed on the vertical axis? 2. What is graphed on the horizontal axis. 3. What does the slope of this graph represent? Review 1. For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 1 to 3 seconds? 2. For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 3 to 4 seconds? 3. For the motion graphed in the position versus time graph shown above, what is the average velocity in the time interval 5 to 6 seconds? 15 • instantaneous velocity: The velocity of an object at any given instant. www.ck12.org References 1. Image copyright Denis Kuvaev, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 4. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 16 www.ck12.org Chapter 6. Average Acceleration CHAPTER 6 Average Acceleration • Define average acceleration. • Given initial velocity, final velocity and time, calculate acceleration. • Given three of initial velocity, acceleration, time, and final velocity, calculate the fourth. End of an era. The Space Shuttle Atlantis blasts off on mission STS-125, the final mission to service and upgrade the Hubble Space Telescope, one of NASA’s greatest legacies and triumphs. Canceled in the wake of the Columbia tragedy and then reinstated, the only mission not to go to the international space station post-accident will see seven astronauts undertake one of the most ambitious shuttle missions in history, with five spacewalks to install new and replace old components on Hubble. It will be the closing chapter in one of the original purposes of the shuttle. Average Acceleration An object whose velocity is changing is said to be accelerating. Average acceleration, a is defined as the rate of change of velocity, or the change in velocity per unit time. A symbol with a bar over it is read as average –so a-bar is average acceleration. Example: A car accelerates along a straight road from rest to 60. km/h in 5.0 s. What is the magnitude of its average acceleration? Solution: This is read as kilometers per hour per second. In general, it is undesirable to have to different units for the same quantity in a unit expression. For example, in this case, it is undesirable to have two different units for time (hours and seconds) in the same unit expression. To eliminate this problem, we would convert the hour units to seconds. If we converted the original 60. km/h to m/s, it would be 17 m/s. Then the acceleration would be 3.4 m/s2. Example: An automobile is moving along a straight highway in the positive direction and the driver puts on If the initial velocity is 15.0 m/s and 5.0 s is required to slow down to 5.0 m/s, what was the car’s the brakes. acceleration? 17 www.ck12.org Solution: a = Dv Dt = 10: m/s 5:0 s = 2:0 m/s/s Summary • Average acceleration is the rate of change of velocity, or the change in velocity per unit time. Practice The following url has a lesson on the difference between average and instantaneous acceleration and practice calculating average acceleration. http://www.brighthubeducation.com/homework-math-help/102434-definition-and-how-to-calculate-acceleration/ Review 1. The velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. What was the average acceleration? 2. If an automobile slows from 26 m/s to 18 m/s in a period of 4.0 s, what was the average acceleration? 3. If a runner increases his velocity from 0 m/s to 20 m/s in 2.0 s, what was his average acceleration? 4. If a runner decreases his velocity from 20 m/s to 10 m/s in 2.0 s, what was his average acceleration? • average acceleration: The change in velocity over the change in time. References 1. Courtesy of NASA. http://spaceflight.nasa.gov/gallery/images/shuttle/sts-125/html/sts125-s-025.html. Public Domain 18 www.ck12.org Chapter 7. Uniform Acceleration CHAPTER 7 Uniform Acceleration • Define uniform acceleration. • Given initial velocity, acceleration, and time, calculate final velocity. Wingtip vortices are often thought to be a type of contrail but are actually produced from a different process. During very specific weather conditions you may see vapor trails form at the rear of the wingtips of jet aircraft on takeoff or landing. This phenomenon occurs due to a decrease in pressure and temperature as the wing generates lift. The image is an F-35 departing from Elgin Air Force Base in Florida. Uniform Acceleration Acceleration that does not change in time is called uniform or constant acceleration. The velocity at the beginning of the time interval is called initial velocity, vi, and the velocity at the end of the time interval is called final velocity, v f . In a velocity versus time graph for uniform acceleration, the slope of the line is the acceleration. The equation that describes the curve is v f = vi + at. Example: velocity? If an automobile with a velocity of 4.0 m/s accelerates at a rate of 4.0 m/s2 for 2.5 s, what is the final Solution: v f = vi + at = 4:0 m/s + (4:0 m/s2)(2:5 s) = 4:0 m/s + 10: m/s = 14 m/s Example: If a cart slows from 22.0 m/s to 4.0 m/s with an acceleration of -2.0 m/s2, how long does it require? Solution: t = v f vi a = 18 m/s 2:0 m/s2 = 9:0 s 19 www.ck12.org Summary • Acceleration that does not change in time is uniform or constant acceleration. • The equation relating initial velocity, final velocity, time, and acceleration is v f = vi + at. Practice The following url has instruction in one dimensional uniformly accelerated motion and it also has a series of practice problems. http://dallaswinwin.com/Motion_in_One_Dimension/uniform_accelerated_motion.htm Review 1. If an object has zero acceleration, does that mean it has zero velocity? Give an example. 2. If an object has zero velocity, does that mean it has zero acceleration? Give an example. 3. If the acceleration of a motorboat is 4.0 m/s2, and the motorboat starts from rest, what is its velocity after 6.0 s? 4. The friction of the water on a boat produces an acceleration of -10. m/s2. If the boat is traveling at 30. m/s and the motor is shut off, how long it take the boat to slow down to 5.0 m/s? • uniform acceleration: Acceleration that does not change in time is uniform or constant acceleration. References 1. Courtesy of Senior Airman Julianne Showalter/U.S. Air Force. http://www.af.mil/news/story.asp?id=1232 61835. Public Domain 20 www.ck12.org Chapter 8. Displacement During Constant Acceleration CHAPTER 8 Displacement During Constant Acceleration • Plot and interpret a velocity vs time graph. • Find the area under a curve on a velocity vs time graph and calculate the displacement from such a graph. • Calculate the displacement of an object undergoing uniform acceleration when given two of the three quanti- ties acceleration, time, velocity. Long distance runners try to maintain constant velocity with very little acceleration or deceleration because acceleration requires more energy than simply maintaining velocity. Displacement During Constant Acceleration When the acceleration is constant, there are three equations that relate displacement to two of the other three It is absolutely vital that you do NOT quantities we use to describe motion –time, velocity, and acceleration. try to use these equations when the acceleration is NOT constant. Fortunately, there are quite a few cases of motion where the acceleration is constant. One of the most common, if we ignore air resistance, are objects falling due to gravity. When an object is moving with constant velocity, the displacement can be found by multiplying the velocity by the time interval. d = vt If the object is moving with constant acceleration, the velocity in that equation is replaced with the average velocity. The average velocity for a uniformly accelerated object can be found by adding the beginning and final velocities and dividing by 2. vave = 1 2 (v f + vi) The distance, then, for uniformly accelerating motion can be found by multiplying the average velocity by the time. 21 d = 1 2 (v f + vi)(t) (Equation 1) We know that the final velocity for constantly accelerated motion can be found by multiplying the acceleration times time and adding the result to the initial velocity, v f = vi + at. The second equation that relates, displacement, time, initial velocity, and final velocity is generated by substituting into equation 1. www.ck12.org d = 1 2 (v f + vi)(t vit but v f = vi + at and substituting for v f yields d = 1 2 vit + d = vit + 1 2 1 2 at2 (t)(vi + at) = 1 2 vit + 1 2 vit + at2 1 2 (Equation 2) The third equation is formed by combining v f = vi + at and d = 1 and then substitute into the second equation, we get 2 (v f + vi)(t). If we solve the first equation for t d = 1 2 (v f + vi) v f vi a = 1 2 v2 f v2 i a And solving for v f 2 yields v f 2 = vi 2 + 2ad (Equation 3) Keep in mind that these three equations are only valid when acceleration is constant. velocity can be set to zero and that simplifies the three equations considerably. In many cases, the initial With both constant acceleration and initial velocity of zero = 2ad: d = at2 and v f Example: Su
ppose a planner is designing an airport for small airplanes. Such planes must reach a speed of 56 m/s before takeoff and can accelerate at 12.0 m/s2. What is the minimum length for the runway of this airport? Solution: The acceleration in this problem is constant and the initial velocity of the airplane is zero, therefore, we can use the equation v f 2a = (56 m/s)2 d = v f (2)(12:0 m/s2 2 = 2ad and solve for d. = 130 m ) 2 Example: How long does it take a car to travel 30.0 m if it accelerates from rest at a rate of 2.00 m/s2? Solution: The acceleration in this problem is constant and the initial velocity is zero, therefore, we can use d = 1 solved for t. r s 2 at2 t = 2d a = (2)(30:0 m) 2:00 m/s2 = 5:48 s Example: A baseball pitcher throws a fastball with a speed of 30.0 m/s. Assuming the acceleration is uniform and the distance through which the ball is accelerated is 3.50 m. What is the acceleration? 22 www.ck12.org Chapter 8. Displacement During Constant Acceleration Solution: Since the acceleration is uniform and the initial velocity is zero, we can use v f 2 = 2ad solved for a. a = v2 f 2d = (30:0 m/s)2 (2)(3:50 m) = 900: m2=s2 7:00 m = 129 m/s2 Suppose we plot the velocity versus time graph for an object undergoing uniform acceleration. In this first case, we will assume the object started from rest. If the object has a uniform acceleration of 6.0 m/s2 and started from rest, then each succeeding second, the velocity will increase by 6.0 m/s. Here is the table of values and the graph. In displacement versus time graphs, the slope of the line is the velocity of the object and in this case of velocity versus time graph, the slope of the line is the acceleration. If you take any segment of this line and determine the Dy to Dx ratio, you will get 6.0 m/s2 which we know to be the constant acceleration of this object. The area of a triangle is calculated by multiplying one-half the base times the height. The area under the curve in the image above is the area of the triangle shown below. 23 www.ck12.org The area of this triangle would be calculated by area = 1 2 The distance traveled by an object accelerating uniformly from rest at 6.0 m/s2 would be displacement = 1 (6:0 s)(36 m/s) = 108 m. 2 at2. Therefore, the displacement of this object in the first 6 seconds of travel would be displacement = 1 2 (6:0 m/s2)(6:0 s)2 = 108 m. In fact, the area underneath the curve in a velocity versus time graph is always equal to the displacement that occurs during that time interval. Summary • There are three equations relating displacement to two of the other three quantities we use to describe motion –time, velocity, and acceleration: – d = 1 2 – d = vit + 1 2 = vi – v f (v f + vi)(t) (Equation 1) 2 at2 (Equation 2) 2 + 2ad (Equation 3) • When the initial velocity of the object is zero, these three equations become: (v f )(t) (Equation 1’) – d = 1 2 – d = 1 2 at2 (Equation 2’) 2 = 2ad (Equation 3’) – v f • The slope of a velocity versus time graph is the acceleration of the object. • The area under the curve of a velocity versus time graph is the displacement that occurs during the given time interval. Practice Use this resource to answer the questions that follow. MEDIA Click image to the left for more content. 1. For the example in the video, what acceleration is used? 2. What time period is used in the example? 3. What does the slope of the line in the graph represent? 4. What does the area under the curve of the line represent? Review 1. An airplane accelerates with a constant rate of 3.0 m/s2 starting at a velocity of 21 m/s. If the distance traveled during this acceleration was 535 m, what is the final velocity? 2. An car is brought to rest in a distance of 484 m using a constant acceleration of -8.0 m/s2. What was the velocity of the car when the acceleration first began? 3. An airplane starts from rest and accelerates at a constant 3.00 m/s2 for 20.0 s. What is its displacement in this time? 4. A driver brings a car to a full stop in 2.0 s. 24 www.ck12.org Chapter 8. Displacement During Constant Acceleration (a) If the car was initially traveling at 22 m/s, what was the acceleration? (b) How far did the car travel during braking? References 1. Image copyright Maridav, 2013. http://www.shutterstock.com. Used under license from Shutterstock.com 2. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 25 CHAPTER 9 Acceleration Due to Gravity www.ck12.org • Solve problems of the motion of objects uniformly accelerated by gravity. In the absence of air resistance, all objects fall toward the earth with the same acceleration. Man, however, make maximum use of air resistance in the construction of parachutes for both entertainment and military use. The image at left was taken during a 2008 Graduation demonstration jump by the U.S. Army Parachute Team. The 2008 team contained the first amputee member and the largest number of females in history. Acceleration Due to Gravity One of the most common examples of uniformly accelerated motion is that an object allowed to fall vertically to the earth. In treating falling objects as uniformly accelerated motion, we must ignore air resistance. Galileo’s original statement about the motion of falling objects is: At a given location on the earth and in the absence of air resistance, all objects fall with the same uniform acceleration. We call this acceleration due to gravity on the earth and we give it the symbol g. The value of g is 9.80 m/s2. All of the equations involving constant acceleration can be used for falling bodies but we insert g wherever “a” appeared and the value of g is always 9.80 m/s2. Example: A rock is dropped from a tower 70.0 m high. How far will the rock have fallen after 1.00 s, 2.00 s, and 3.00 s? Assume the distance is positive downward. 26 www.ck12.org Chapter 9. Acceleration Due to Gravity Solution: We are looking for displacement and we have time and acceleration. Therefore, we can use d = 1 Displacement after 1:00 s = 1 2 Displacement after 2:00 s = 1 2 Displacement after 3:00 s = 1 2 (9:80 m/s2)(1:00 s)2 = 4:90 m (9:80 m/s2)(2:00 s)2 = 19:6 m (9:80 m/s2)(3:00 s)2 = 44:1 m 2 at2. Example: (a) A person throws a ball upward into the air with an initial velocity of 15.0 m/s. How high will it go before it comes to rest? (b) How long will the ball be in the air before it returns to the person’s hand? Solution: In part (a), we know the initial velocity (15.0 m/s), the final velocity (0 m/s), and the acceleration -9.80 m/s2. We wish to solve for the displacement, so we can use v f d = v f 2 + 2ad and solve for d. 2a = (0 m/s)2(15:0 m/s)2 2vi = 11:5 m 2 = vi 2 (2)(9:80 m/s2 ) There are a number of methods by which we can solve part (b). Probably the easiest is to divide the distance traveled by the average velocity to get the time going up and then double this number since the motion is symmetrical –that is, time going up equals the time going down. The average velocity is half of 15.0 m/s or 7.5 m/s and dividing this into the distance of 11.5 m yields 1.53 seconds. This is the time required for the ball to go up and the time for the ball to come down will also be 1.53 s, so the total time for the trip up and down is 3.06 seconds. Example: A car accelerates with uniform acceleration from 11.1 m/s to 22.2 m/s in 5.0 s. acceleration and (b) how far did it travel during the acceleration? (a) What was the Dt = 22:2 m/s11:1 m/s Solution: (a) a = Dv (b) We can find the distance traveled by d = vit + 1 average velocity and multiply it by the time. = 2:22 m/s2 5:0 s 2 at2 and we can also find the distance traveled by determining the d = vit + at2 1 2 = (11:1 m/s)(5:0 s) + = 55:5 m + 27:8 m = 83 m 1 2 (2:22 m/s2)(5:0 s)2 d = (vave)(t) = (16:6 m/s)(5:0 s) = 83 m Example: A stone is dropped from the top of a cliff. It is seen to hit the ground after 5.5 s. How high is the cliff? Solution: d = vit + 1 2 at2 = (0 m/s)(5:5 s) + 1 2 (9:80 m/s2)(5:5 s)2 = 150 m Summary • At a given location on the earth and in the absence of air resistance, all objects fall with the same uniform acceleration. • We call this acceleration the acceleration due to gravity on the earth and we give it the symbol g. • The value of g is 9.80 m/s2. 27 Practice This url shows a video of a discussion and demonstration of the acceleration due to gravity. http://www.youtube.com/watch?v=izXGpivLvgY www.ck12.org MEDIA Click image to the left for more content. Review 1. A baseball is thrown vertically into the air with a speed of 24.7 m/s. (a) How high does it go? (b) How long does the round trip up and down require? 2. A salmon jumps up a waterfall 2.4 m high. With what minimum speed did the salmon leave the water below to reach the top? 3. A kangaroo jumps to a vertical height of 2.8 m. How long will it be in the air before returning to earth? • acceleration due to gravity: The acceleration experienced by a body in free fall in a gravitational field. References 1. Courtesy of Donna Dixon/U.S. Military. http://commons.wikimedia.org/wiki/File:Flickr_-_The_U.S._Arm y_-_U.S._Army_Parachute_Team_graduates_first_wounded_warrior_and_largest_female_class_%282%29.jp g. Public Domain 28 www.ck12.org Chapter 10. Graphing Motion CHAPTER 10 Graphing Motion Students will learn how to graph motion vs time. Specifically students will learn how to take the slope of a graph and relate that to the instantaneous velocity or acceleration for position or velocity graphs, respectively. Finally students will learn how to take the area of a velocity vs time graph in order to calculate the displacement. Students will learn how to graph motion vs time. Specifically students will learn how to take the slope of a graph and relate that to the instantaneous velocity or acceleration for position or velocity graphs, respectively. Finally students will learn how to take the area of a velocity vs time graph in order to calculate the displacement. Key Equations For a graph of position vs. time. Th
e slope is the rise over the run, where the rise is the displacement and the run is the time. thus, Slope = vavg = Dx Dt Note: Slope of the tangent line for a particular point in time = the instantaneous velocity For a graph of velocity vs. time. The slope is the rise over the run, where the rise is the change in velocity and the run is the time. thus, Slope = aavg = Dv Dt Note: Slope of the tangent line for a particular point in time = the instantaneous acceleration Guidance • One must first read a graph correctly. For example on a position vs. time graph (thus the position is graphed on the y-axis and the time on the x-axis) for a given a data point, go straight down from it to get the time and straight across to get the position. • If there is constant acceleration the graph x vs. t produces a parabola. The slope of the x vs. t graph equals the instantaneous velocity. The slope of a v vs. t graph equals the acceleration. • The slope of the graph v vs. t can be used to find acceleration; the area of the graph v vs. t can be used to find displacement. Welcome to calculus! What is a Graph Watch this Explanation MEDIA Click image to the left for more content. 29 www.ck12.org MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. MEDIA Click image to the left for more content. Time for Practice 1. The position graph below is of the movement of a fast turtle who can turn on a dime. a. Sketch the velocity vs. time graph of the turtle below. 30 www.ck12.org Chapter 10. Graphing Motion b. Explain what the turtle is doing (including both speed and direction) from: i) 0-2s. ii) 2-3s. iii) 3-4s. c. How much distance has the turtle covered after 4s? d. What is the turtle’s displacement after 4s? 2. Draw the position vs. time graph that corresponds to the velocity vs. time graph below. You may assume a starting position x0 = 0. Label the yaxis with appropriate values. 3. The following velocity-time graph represents 10 seconds of actress Halle Berry’s drive to work (it’s a rough morning). 31 a. Fill in the tables below –remember that displacement and position are not the same thing! www.ck12.org 32 www.ck12.org Chapter 10. Graphing Motion TABLE 10.1: Displacement (m) Acceleration(m=s2) Instantaneous Time (s) 0 sec Position (m) 0 m 2 sec 4 sec 5 sec 9 sec 10 sec Interval (s) 0-2 sec 2-4 sec 4-5 sec 5-9 sec 9-10 sec b. On the axes below, draw an acceleration-time graph for the car trip. Include numbers on your acceleration axis. c. On the axes below, draw a position-time graph for the car trip. Include numbers on your position axis. Be sure to note that some sections of this graph are linear and some curve –why? 4. Two cars are drag racing down El Camino. At time t = 0, the yellow Maserati starts from rest and accelerates at 10 m=s2. As it starts to move it’s passed by a ’63 Chevy Nova (cherry red) traveling at a constant velocity of 30 m/s. a. On the axes below, show a line for each car representing its speed as a function of time. Label each line. 33 www.ck12.org b. At what time will the two cars have the same speed (use your graph)? (or curve) for each car representing its position as a function of time. Label each curve. c. On the axes below, draw a line d. At what time would the two cars meet (other than at the start)? Answers: 1c. 25 m 1d. -5 m 2. discuss in class 3. discuss in class 4b. 3 sec 4d. 6 sec 34 Physics Unit 3 (Vectors) Patrick Marshall Ck12 Science Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org AUTHORS Patrick Marshall Ck12 Science CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform®. Copyright © 2014 CK-12 Foundation, www.ck12.org The names “CK-12” and “CK12” and associated logos and the terms “FlexBook®” and “FlexBook Platform®” (collectively “CK-12 Marks”) are trademarks and service marks of CK-12 Foundation and are protected by federal, state, and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link http://www.ck12.org/saythanks (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution-Non-Commercial 3.0 Unported (CC BY-NC 3.0) License (http://creativecommons.org/ licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated herein by this reference. Complete terms can be found at http://www.ck12.org/terms. Printed: February 4, 2014 iii Contents www.ck12.org Contents 1 Graphical Methods of Vector Addition 2 Vector Addition 1 5 iv www.ck12.org Chapter 1. Graphical Methods of Vector Addition CHAPTER 1 Graphical Methods of Vector Addition • Differentiate between scalars and vectors. • Graphically add vectors in one dimension by placing the vectors head to toe on a number line. • Define resultant. • Graphically add vectors in two dimensions by placing them head to toe on a two-dimensional coordinate system. Successfully shooting a basketball requires a subconscious understanding of the vectors involved in how the basketball moves through the air. The vertical and horizontal vectors must be perfectly organized if the ball is to pass through the basket. Graphical Methods Vector Addition In physics, a quantity, such as mass, length, or speed, that is completely specified by its magnitude and has no direction is called a scalar. A vector, on the other hand, is a quantity possessing both magnitude and direction. A vector quantity can be represented by an arrow-tipped line segment. The length of the line, drawn to scale, represents the magnitude of the quantity. The direction of the arrow indicates the direction of the vector. Not only can vectors be represented graphically, but they can also be added graphically. For one dimensional vector addition, the first vector is placed on a number line with the tail of the vector on the origin. The second vector is placed with its tail exactly on the arrow head of the first vector. The sum of the two vectors is the vector that begins at the origin and ends at the arrow head of the final added vector. Consider the following two vectors. 1 www.ck12.org The red vector has a magnitude of 11 in the positive direction on the number line. The blue vector has a magnitude of -3 in the negative direction on the number line. In order to add these two vectors, we place one of the vectors on a number line and then the second vector is placed on the same number line such that its origin is on the arrow head of the first vector. The sum of these two vectors is the vector that begins at the origin of the first vector (the red one) and ends at the arrow head of the blue vector. So the sum of these two vectors is the purple vector as shown below. The vector sum of the first two vectors is a vector that begins at the origin and has a magnitude of 8 units in the positive direction. If we were adding three or four vectors all in one dimension, we would continue to place them head to toe in sequence on the number line. The sum would be the vector that begins at the beginning of the first vector and goes to the ending of the final vector. Adding Vectors in Two Dimensions In the following image, vectors A and B represent the two displacements of a person who walked 90. m east and then 50. m north. We want to add these two vectors to get the vector sum of the two movements. The graphical process for adding vectors in two dimensions is to place the tail of the second vector on the arrow head of the first vector as shown above. The sum of the two vectors is the vector that begins at the origin of the first vector and goes to the ending of the second vector, as shown below. 2 www.ck12.org Chapter 1. Graphical Methods of Vector Addition If we are using totally graphic means of adding these vectors, the magnitude of the sum would be determined by measuring the length of the sum vector and comparing it to the original standard. We would also use a compass to measure the angle of the summation vector. If we are using calculation means, we can determine the inverse tangent of 50 units divided by 90 units and get the angle of 29° north of east. The length of the sum vector can also be determined mathematically by the Pythagorean theorem, a2 + b2 = c2. In this case, the length of the hypotenuse would be the square root of (8100 + 2500) or 103 units. If three or four vectors are to be added by graphical means, we would continue to place each new vector head to toe with the vectors to be added until all the vectors were in the coordinate system and then the sum vector would be the vector goes from the origin of the first vector to the arrowhead of the last vector. The magnitude and direction of the sum vector would be measured. Summary • Scalars are quantities, such as mass, length, or speed, that are completely specified by magnitude and has no direction. • Vectors are quantities possessing both magnitude and direction and can be represented by an arrow; the direction of the arrow indicates the direction of the quantity and the length of the arrow is proportional to the magnitude. • Vectors that are in one dimension can be added arithmetically. • Vectors that are in two dimensions are added geometrically. • When vectors are added graphically, graphs must be done to scale and answers are only as accurate as the graph
ing. Practice Video on the graphical method of adding vectors. http://www.youtube.com/watch?v=_Vppxdho6JU MEDIA Click image to the left for more content. 3 Review 1. On the following number line, add the vector 7.5 m/s and the vector -2.0 m/s. www.ck12.org 2. On a sheet of graph paper, add a vector that is 4.0 N due east and a vector that is 3.0 N due north. • scalar: A quantity, such as mass, length, or speed, that is completely specified by its magnitude and has no direction. • vector: A quantity possessing both magnitude and direction, represented by an arrow the direction ofwhich indicates the direction of the quantity and the length of which is proportional to the magnitude. • vector addition: The process of finding one vector that is equivalent to the result of the successive application of two or more given vectors. References 1. Official White House photo by Pete Souza. http://commons.wikimedia.org/wiki/File:Barack_Obama_play ing_basketball.jpg. Public Domain 2. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 4. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 5. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 6. CK-12 Foundation - CC-BY-NC-SA 3.0. . CC-BY-NC-SA 3.0 7. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 4 www.ck12.org Chapter 2. Vector Addition CHAPTER 2 Vector Addition • Describe the independence of perpendicular vectors. • Resolve vectors into axial components. • Define resultant. • Add vectors using geometric and trigonometric methods. Vector Addition Adding Vectors in Two Dimensions In the following image, vectors A and B represent the two displacements of a person who walked 90. m east and then 50. m north. We want to add these two vectors to get the vector sum of the two movements. The graphical process for adding vectors in two dimensions is to place the tail of the second vector on the arrow head of the first vector as shown above. The sum of the two vectors is the vector that begins at the origin of the first vector and goes to the ending of the second vector, as shown below. If we are using totally graphic means of adding these vectors, the magnitude of the sum would be determined by measuring the length of the sum vector and comparing it to the original standard. We would also use a compass to measure the angle of the summation vector. If we are using calculation means, we can determine the inverse tangent of 50 units divided by 90 units and get the angle of 29° north of east. The length of the sum vector can also be determined mathematically by the Pythagorean 5 www.ck12.org theorem, a2 + b2 = c2. In this case, the length of the hypotenuse would be the square root of (8100 + 2500) or 103 units. If three or four vectors are to be added by graphical means, we would continue to place each new vector head to toe with the vectors to be added until all the vectors were in the coordinate system and then the sum vector would be the vector goes from the origin of the first vector to the arrowhead of the last vector. The magnitude and direction of the sum vector would be measured. Mathematical Methods of Vector Addition We can add vectors mathematically using trig functions, the law of cosines, or the Pythagorean theorem. If the vectors to be added are at right angles to each other, we would assign them to the sides of a right triangle and calculate the sum as the hypotenuse of the right triangle. We would also calculate the direction of the sum vector by using an inverse sin or some other trig function. Suppose, however, that we wish to add two vectors that are not at right angles to each other. Let’s consider the vectors in the following images. The two vectors we are to add is a force of 65 N at 30° north of east and a force of 35 N at 60° north of west. We know that vectors in the same dimension can be added by regular arithmetic. Therefore, we can resolve each of these vectors into components that lay on the axes –pictured below. We can resolve each of the vectors into two components. The components are on the axes lines. The resolution of vectors reduces each vector to a component on the north-south axis and a component on the east-west axis. We can now mathematically determine the magnitude of the components and add then arithmetically because they are in the same dimension. Once we have added the components, we will once again have only two vectors that are perpendicular to each other and can be the legs of a right triangle. The east-west component of the first vector is (65 N)(cos 30°) = (65 N)(0.866) = 56.3 N north The north-south component of the first vector is (65 N)(sin 30°) = (65 N)(0.500) = 32.5 N north 6 www.ck12.org Chapter 2. Vector Addition The east-west component of the second vector is (35 N)(cos 60°) = (35 N)(0.500) = 17.5 N west The north-south component of the second vector is (35 N)(sin 60°) = (35 N)(0.866) = 30.3 N north The sum of the two east-west components is 56.3 N - 17.5 N = 38.8 N east The sum of the two north-south components is 32.5 N + 30.3 N = 62.8 N north We can now consider those two vectors to be the sides of a right triangle and use the Pythagorean Theorem to find the length of the hypotenuse and use a trig function to find its direction. p c = sin x = 62:8 38:82 + 62:82 = 74 N 74 so x = sin1 0:84 so x = 58 The direction of the sum vector is 74 N at 58° north of east. Perpendicular vectors have no components in the other direction. For example, if a boat is floating down a river due south, and you are paddling the boat due east, the eastward vector has no component in the north-south direction and therefore, has no effect on the north-south motion. If the boat is floating down the river at 5 miles/hour south and you paddle the boat eastward at 5 miles/hour, the boat continues to float southward at 5 miles/hour. The eastward motion has absolutely no effect on the southward motion. Perpendicular vectors have NO effect on each other. Example Problem: A motorboat heads due east at 16 m/s across a river that flows due north at 9.0 m/s. (a) What is the resultant velocity of the boat? (b) If the river is 135 m wide, how long does it take the boat to reach the other side? (c) When the boat reaches the other side, how far downstream will it be? Solution: Sketch: (a) Since the two motions are perpendicular to each other, they can be assigned to the legs of a right triangle and the hypotenuse (resultant) calculated. c = p a2 + b2 = q (16 m/s)2 + (9:0 m/s)2 = 18 m/s sin q = 9:0 18 = 0:500 and therefore q = 30 The resultant is 18 m/s at 30° north of east. 7 www.ck12.org (b) The boat is traveling across the river at 16 m/s due to the motor. The current is perpendicular and therefore has no effect on the speed across the river. The time required for the trip can be determined by dividing the distance by the velocity. t = d = 8:4 s v = 135 m 16 m/s (c) The boat is traveling across the river for 8.4 seconds and therefore, it is also traveling downstream for 8.4 seconds. We can determine the distance downstream the boat will travel by multiplying the speed downstream by the time of the trip. ddownstream = (vdownstream)(t) = (9:0 m/s)(8:4 s) = 76 m Summary • Vectors can be added mathematically using geometry and trigonometry. • Vectors that are perpendicular to each other have no effect on each other. • Vector addition can be accomplished by resolving into axial components those vectors that are to be added, adding up the axial components, and then combining the two axial components. Practice A video demonstrating the component method of vector addition. http://www.youtube.com/watch?v=nFDzRWw08Ew MEDIA Click image to the left for more content. Review 1. A hiker walks 11 km due north from camp and then turns and walks 11 km due east. (a) What is the total distance walked by the hiker? (b) What is the displacement (on a straight line) of the hiker from the camp? 2. While flying due east at 33 m/s, an airplane is also being carried due north at 12 m/s by the wind. What is the plane’s resultant velocity? 3. Two students push a heavy crate across the floor. John pushes with a force of 185 N due east and Joan pushes with a force of 165 N at 30° north of east. What is the resultant force on the crate? 4. An airplane flying due north at 90. km/h is being blown due west at 50. km/h. What is the resultant velocity of the plane? 5. A golf ball is struck with a golf club and travels in a parabolic curve. The horizontal distance traveled by the golf ball is 240 meters and the time of flight is 4.00 seconds. What was the initial velocity magnitude and direction? • axial component: A component situated in or on an axis. • resolution of vectors: Any vector directed at an angle to the horizontal (or the vertical) can be thought of as having two parts (or components) that lie on the axes (one horizontal and one vertical). The process of identifying these two components is known as the resolution of the vector. 8 Chapter 2. Vector Addition www.ck12.org References 1. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 2. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 3. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 4. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 5. CK-12 Foundation - Richard Parsons. . CC-BY-NC-SA 3.0 9 Physics Unit 4 (Two Dimensional Motion) Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required) AUTHORS Patrick Marshall Jean Brainard, Ph.D. Ck12 Science James Dann, Ph.D. CONTRIBUTORS Chris Addiego Antonio De Jesus López www.ck12.org To access a customizable version of this book, as well as other interactive content, visit www.ck12.org CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook®, CK-12 intends to pioneer t