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zing 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 v
2 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” in 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 shorter 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 telescope (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 photoemission 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! 721 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 –––––––– v
2 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 resonance 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 particle, 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 sl
its. 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 photon 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 Brogl
ie’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 the 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 magnitude 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 cath
ode 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 science, 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 nucleus 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 spect
ra 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 clearly. 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 Em
ission 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 Zeeman 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 2r
1 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
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

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