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-slit diffraction pattern, where λis the wavelength of light, Dis the slit width, is the angle between a line from the slit to a minimum and a line perpendicular to the screen, Lis the distance from the slit to the screen, yis the distance from the center of the pattern to the minimum, and mis a nonzero integer indicating the order of the minimum. Below we summarize the equations needed for the calculations to follow. The speed of light in a vacuum, c, the wavelength of the light, , and its frequency, f, are related as follows. The wavelength of light in a medium, , compared to its wavelength in a vacuum, , is given by To calculate the positions of constructive interference for a double slit, the path-length difference must be an integral multiple, m, of the wavelength. 17.1 where dis the distance between the slits and is the angle between a line from the slits to the maximum and a line perpendicular to the barrier in which the slits are located. To calculate the positions of destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength: For a single-slit diffraction pattern, the width of the slit, D, the distance of the first (m= 1) destructive interference minimum, y, the distance from the slit to the screen, L, and the wavelength, , are given by Also, for single-slit diffraction, is the angle between a line from the slit to the minimum and a line perpendicular to the screen, and mis the order of where the minimum. WORKED EXAMPLE Two-Slit Interference Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm, and you find that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam. What is the wavelength of the light? STRATEGY The third bright line is due to third-order constructive interference, which means that m= 3. You are given d= 0.0100 mm and = 10.95º. The wavelength can thus be found using the equation for constructive interference. Access for free at openstax.org. 17.1 • Understanding Diffraction and Interference 531 Solution The equation is . Solving for the wavelength, , gives Substituting known values yields 17.2 17.3 Discussion To three digits, 633 nm is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did that for visible wavelengths. His analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with , so spectra (measurements of intensity versus wavelength) can be obtained. WORKED EXAMPLE Single-Slit Diffraction Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0° relative to the incident direction of the light. What is the width of the slit? STRATEGY From the given information, and assuming the screen is far away from the slit, you can use the equation D. to find Solution Quantities given are = 550 nm, m= 2, and gives = 45.0°. Solving the equation for Dand substituting known values Discussion You see that the slit is narrow (it is only a few times greater than the wavelength of light). That is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects, such as this single-slit diffraction pattern. Practice Problems 1. Monochromatic light from a laser passes through two slits separated by . The third bright line on a screen is 17.4 relative to the incident beam. What is the wavelength of the light? formed at an angle of a. b. c. d. 2. What is the width of a single slit through which 610-nm orange light passes to form a first diffraction minimum at an angle of 30.0°? a. 0.863 µm b. 0.704 µm c. 0.610 µm d. 1.22 µm 532 Chapter 17 • Diffraction and Interference Check Your Understanding 3. Which aspect of a beam of monochromatic light changes when it passes from a vacuum into water, and how does it change? a. The wavelength first decreases and then increases. b. The wavelength first increases and then decreases. c. The wavelength increases. d. The wavelength decreases. 4. Go outside in the sunlight and observe your shadow. It has fuzzy edges, even if you do not. Is this a diffraction effect? Explain. a. This is a diffraction effect. Your whole body acts as the origin for a new wavefront. b. This is a diffraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront. c. This is a refraction effect. Your whole body acts as the origin for a new wavefront. d. This is a refraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront. 5. Which aspect of monochromatic green light changes when it passes from a vacuum into diamond, and how does it change? a. The wavelength first decreases and then increases. b. The wavelength first increases and then decreases. c. The wavelength increases. d. The wavelength decreases. 17.2 Applications of Diffraction, Interference, and Coherence Section Learning Objectives By the end of this section, you will be able to do the following: • Explain behaviors of waves, including reflection, refraction, diffraction, interference, and coherence, and describe applications based on these behaviors • Perform calculations related to applications based on wave properties of light Section Key Terms differential interference contrast (DIC) diffraction grating iridescence laser monochromator Rayleigh criterion resolution Wave-Based Applications of Light In 1917, Albert Einstein was thinking about photons and excited atoms. He considered an atom excited by a certain amount of energy and what would happen if that atom were hit by a photon with the same amount of energy. He suggested that the atom would emit a photon with that amount of energy, and it would be accompanied by the original photon. The exciting part is that you would have twophotons with the same energy andthey would be in phase. Those photons could go on to hit other excited atoms, and soon you would have a stream of in-phase photons. Such a light stream is said to be coherent. Some four decades later, Einstein’s idea found application in a process called, light amplification by stimulated emission of radiation. Take the first letters of all the words (except byand “of”) and write them in order. You get the word laser(see (a)), which is the name of the device that produces such a beam of light. Laser beams are directional, very intense, and narrow (only about 0.5 mm in diameter). These properties lead to a number of applications in industry and medicine. The following are just a few examples: • This chapter began with a picture of a compact disc (see ). Those audio and data-storage devices began replacing cassette tapes during the 1990s. CDs are read by interpreting variations in reflections of a laser beam from the surface. • Some barcode scanners use a laser beam. • Lasers are used in industry to cut steel and other metals. • Lasers are bounced off reflectors that astronauts left on the Moon. The time it takes for the light to make the round trip can be used to make precise calculations of the Earth-Moon distance. • Laser beams are used to produce holograms. The name hologram means entire picture(from the Greek holo-, as in Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 533 holistic), because the image is three-dimensional. A viewer can move around the image and see it from different perspectives. Holograms take advantage of the wave properties of light, as opposed to traditional photography which is based on geometric optics. A holographic image is produced by constructive and destructive interference of a split laser beam. • One of the advantages of using a laser as a surgical tool is that it is accompanied by very little bleeding. • Laser eye surgery has improved the vision of many people, without the need for corrective lenses. A laser beam is used to change the shape of the lens of the eye, thus changing its focal length. Virtual Physics Lasers Click to view content (https://www.openstax.org/l/28lasers) This animation allows you to examine the workings of a laser. First view the picture of a real laser. Change the energy of the incoming photons, and see if you can match it to an excitation level that will produce pairs of coherent photons. Change the excitation level and try to match it to the incoming photon energy. In the animation there is only one excited atom. Is that the case for a real laser? Explain. a. No, a laser would have two excited atoms. b. No, a laser would have several million excited atoms. c. Yes, a laser would have only one excited atom. d. No, a laser would have on the order of 1023 excited atoms. An interesting thing happens if you pass light through a large number of evenly-spaced parallel slits. Such an arrangement of slits is called a diffraction grating. An interference pattern is created that is very similar to the one formed by double-slit diffraction (see and ). A diffraction grating can be manufactured by scratching glass with a sharp tool to form a number of precisely positioned parallel lines, which act like slits. Diffraction gratings work both for transmission of light, as in Figure 17.13, and for reflection of light, as on the butterfly wings or the Australian opal shown in Figure 17.14, or the CD pictured in the opening illustration of this chapter. In addition to their use as novelty items, diffraction gratings are commonly used for spectroscopic dispersion and analysis of light. What makes them particularly useful is the fact that they form a sharper pattern than do double slits. That is, their bright regions are narrower and brighter, while their dark regions are darker. Figure 17.15 shows idealized graphs dem |
onstrating the sharper pattern. Natural diffraction gratings occur in the feathers of certain birds. Tiny, fingerlike structures in regular patterns act as reflection gratings, producing constructive interference that gives the feathers colors not solely due to their pigmentation. The effect is called iridescence. Figure 17.13 A diffraction grating consists of a large number of evenly-spaced parallel slits. (a) Light passing through the grating is diffracted in a pattern similar to a double slit, with bright regions at various angles. (b) The pattern obtained for white light incident on a grating. The central maximum is white, and the higher-order maxima disperse white light into a rainbow of colors. 534 Chapter 17 • Diffraction and Interference Figure 17.14 (a) This Australian opal and (b) the butterfly wings have rows of reflectors that act like reflection gratings, reflecting different colors at different angles. (credit: (a) Opals-On-Black.com, via Flickr (b) whologwhy, Flickr) Figure 17.15 Idealized graphs of the intensity of light passing through a double slit (a) and a diffraction grating (b) for monochromatic light. Maxima can be produced at the same angles, but those for the diffraction grating are narrower, and hence sharper. The maxima become narrower and the regions between become darker as the number of slits is increased. Snap Lab Diffraction Grating • A CD (compact disc) or DVD • A measuring tape • Sunlight near a white wall Instructions Procedure 1. Hold the CD in direct sunlight near the wall, and move it around until a circular rainbow pattern appears on the wall. 2. Measure the distance from the CD to the wall and the distance from the center of the circular pattern to a color in the rainbow. Use those two distances to calculate . Find . . 3. Look up the wavelength of the color you chose. That is 4. Solve 5. Compare your answer to the usual spacing between CD tracks, which is 1,600 nm (1.6 μm). for d. How do you know what number to use for m? a. Count the rainbow rings preceding the chosen color. Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 535 b. Calculate mfrom the frequency of the light of the chosen color. c. Calculate mfrom the wavelength of the light of the chosen color. d. The value of mis fixed for every color. FUN IN PHYSICS CD Players Can you see the grooves on a CD or DVD (see Figure 17.16)? You may think you can because you know they are there, but they are extremely narrow—1,600 in a millimeter. Because the width of the grooves is similar to wavelengths of visible light, they form a diffraction grating. That is why you see rainbows on a CD. The colors are attractive, but they are incidental to the functions of storing and retrieving audio and other data. Figure 17.16 For its size, this CD holds a surprising amount of information. Likewise, the CD player it is in houses a surprising number of electronic devices. The grooves are actually one continuous groove that spirals outward from the center. Data are recorded in the grooves as binary code (zeroes and ones) in small pits. Information in the pits is detected by a laser that tracks along the groove. It gets even more complicated: The speed of rotation must be varied as the laser tracks toward the circumference so that the linear speed along the groove remains constant. There is also an error correction mechanism to prevent the laser beam from getting off track. A diffraction grating is used to create the first two maxima on either side of the track. If those maxima are not the same distance from the track, an error is indicated and then corrected. The pits are reflective because they have been coated with a thin layer of aluminum. That allows the laser beam to be reflected back and directed toward a photodiode detector. The signal can then be processed and converted to the audio we hear. The longest wavelength of visible light is about 780 nm . How does that compare to the distance between CD grooves? a. The grooves are about 3 times the longest wavelength of visible light. b. The grooves are about 2 times the longest wavelength of visible light. c. The grooves are about 2 times the shortest wavelength of visible light. d. The grooves are about 3 times the shortest wavelength of visible light. LINKS TO PHYSICS Biology: DIC Microscopy If you were completely transparent, it would be hard to recognize you from your photograph. The same problem arises when using a traditional microscope to view or photograph small transparent objects such as cells and microbes. Microscopes using differential interference contrast (DIC) solve the problem by making it possible to view microscopic objects with enhanced contrast, as shown in Figure 17.17. 536 Chapter 17 • Diffraction and Interference Figure 17.17 This aquatic organism was photographed with a DIC microscope. (credit: Public Library of Science) A DIC microscope separates a polarized light source into two beams polarized at right angles to each other and coherent with each other, that is, in phase. After passing through the sample, the beams are recombined and realigned so they have the same plane of polarization. They then create an interference pattern caused by the differences in their optical path and the refractive indices of the parts of the sample they passed through. The result is an image with contrast and shadowing that could not be observed with traditional optics. Where are diffraction gratings used? Diffraction gratings are key components of monochromators—devices that separate the various wavelengths of incoming light and allow a beam with only a specific wavelength to pass through. Monochromators are used, for example, in optical imaging of particular wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength of light emitted by molecules in diseased cells in a biopsy sample, or to help excite strategic molecules in the sample with a selected frequency of light. Another important use is in optical fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings is available for selecting specific wavelengths for such use. Diffraction gratings are used in spectroscopes to separate a light source into its component wavelengths. When a material is heated to incandescence, it gives off wavelengths of light characteristic of the chemical makeup of the material. A pure substance will produce a spectrum that is unique, thus allowing identification of the substance. Spectroscopes are also used to measure wavelengths both shorter and longer than visible light. Such instruments have become especially useful to astronomers and chemists. Figure 17.18 shows a diagram of a spectroscope. Figure 17.18 The diagram shows the function of a diffraction grating in a spectroscope. Light diffracts as it moves through space, bending around obstacles and interfering constructively and destructively. While diffraction allows light to be used as a spectroscopic tool, it also limits the detail we can obtain in images. Figure 17.19 (a) shows the effect of passing light through a small circular aperture. Instead of a bright spot with sharp edges, a spot with a fuzzy edge surrounded by circles of light is obtained. This pattern is caused by diffraction similar to that produced by a single slit. Light from different parts of the circular aperture interferes constructively and destructively. The effect is most noticeable when the aperture is small, but the effect is there for large apertures, too. Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 537 Figure 17.19 (a) Monochromatic light passed through a small circular aperture produces this diffraction pattern. (b) Two point light sources that are close to one another produce overlapping images because of diffraction. (c) If they are closer together, they cannot be resolved, that is, distinguished. How does diffraction affect the detail that can be observed when light passes through an aperture? Figure 17.19 (b) shows the diffraction pattern produced by two point light sources that are close to one another. The pattern is similar to that for a single point source, and it is just barely possible to tell that there are two light sources rather than one. If they are closer together, as in Figure 17.19 (c), you cannot distinguish them, thus limiting the detail, or resolution, you can obtain. That limit is an inescapable consequence of the wave nature of light. There are many situations in which diffraction limits the resolution. The acuity of vision is limited because light passes through the pupil, the circular aperture of the eye. Be aware that the diffraction-like spreading of light is due to the limited diameter of a light beam, not the interaction with an aperture. Thus light passing through a lens with a diameter of Dshows the diffraction effect and spreads, blurring the image, just as light passing through an aperture of diameter Ddoes. Diffraction limits the resolution of any system having a lens or mirror. Telescopes are also limited by diffraction, because of the finite diameter, D, of their primary mirror. Why are diffraction gratings used in spectroscopes rather than just two slits? a. The bands produced by diffraction gratings are dimmer but sharper than the bands produced by two slits. b. The bands produced by diffraction gratings are brighter, though less sharp, than the bands produced by two slits. c. The bands produced by diffraction gratings are brighter and sharper than the bands produced by two slits. d. The bands produced by diffraction gratings are dimmer and less sharp, but more widely dispersed, than the bands produced by two slits. Calculations Involving Diffraction Gratings and Resolution Early in the chapter, it was mentioned that when light passes from one medium to another, its speed and wavelength change, |
but its frequency remains constant. The equation , is related to the wavelength in a vacuum, shows how to the wavelength in a given medium, , and the refractive index, n, of the medium. The equation is useful for calculating the change in wavelength of a monochromatic laser beam in various media. The analysis of a diffraction grating is very similar to that for a double slit. As you know from the discussion of double slits in Young’s double-slit experiment, light is diffracted by, and spreads out after passing through, each slit. Rays travel at an angle relative to the incident direction. Each ray travels a different distance to a common point on a screen far away. The rays start in phase, and they can be in or out of phase when they reach a screen, depending on the difference in the path lengths traveled. Each ray travels a distance that differs by equals an integral number of wavelengths, the rays all arrive in phase, and constructive interference (a maximum) is obtained. Thus, the condition necessary to obtain constructive interference for a diffraction grating is from that of its neighbor, where dis the distance between slits. If where dis the distance between slits in the grating, is the wavelength of the light, and mis the order of the maximum. Note that this is exactly the same equation as for two slits separated by d. However, the slits are usually closer in diffraction gratings than in double slits, producing fewer maxima at larger angles. 538 Chapter 17 • Diffraction and Interference WATCH PHYSICS Diffraction Grating This video (https://www.openstax.org/l/28diffraction) explains the geometry behind the diffraction pattern produced by a diffraction grating. Click to view content (https://www.openstax.org/l/28diffraction) The equation that gives the points of constructive interference produced by a diffraction grating is equation look familiar? a. b. c. d. It is the same as the equation for destructive interference for a double-slit diffraction pattern. It is the same as the equation for constructive interference for a double-slit diffraction pattern. It is the same as the equation for constructive interference for a single-slit diffraction pattern. It is the same as the equation for destructive interference for a single-slit diffraction pattern. . Why does that Just what is the resolution limit of an aperture or lens? To answer that question, consider the diffraction pattern for a circular aperture, which, similar to the diffraction pattern of light passing through a slit, has a central maximum that is wider and brighter than the maxima surrounding it (see Figure 17.19 (a)). It can be shown that, for a circular aperture of diameter D, the first minimum in the diffraction pattern occurs at wavelength of light, which is the case for most optical instruments. The accepted criterion for determining the diffraction limit to resolution based on diffraction was developed by Lord Rayleigh in the 19th century. The Rayleigh criterion for the diffraction limit to resolution states that two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other. See Figure 17.20 (b). The first minimum is at an angle of , so that two point objects are just resolvable if they are separated by the angle , provided that the aperture is large compared with the is the wavelength of the light (or other electromagnetic radiation) and Dis the diameter of the aperture, lens, mirror, where etc., with which the two objects are observed. In the expression above, has units of radians. Figure 17.20 (a) Graph of intensity of the diffraction pattern for a circular aperture. Note that, similar to a single slit, the central maximum is wider and brighter than those to the sides. (b) Two point objects produce overlapping diffraction patterns. Shown here is the Rayleigh criterion for their being just resolvable. The central maximum of one pattern lies on the first minimum of the other. Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 539 Snap Lab Resolution • A sheet of white paper • A black pen or pencil • A measuring tape Instructions Procedure 1. Draw two lines several mm apart on a white sheet of paper. 2. Move away from the sheet as it is held upright, and measure the distance at which you can just distinguish (resolve) the lines as separate. 3. Use to calculate Dthe diameter of your pupil. Use the distance between the lines and the maximum distance at which they were resolved to calculate . Use the average wavelength for visible light as the value for . 4. Compare your answer to the average pupil diameter of 3 mm. Describe resolution in terms of minima and maxima of diffraction patterns. a. The limit for resolution is when the minimum of the pattern for one of the lines is directly over the first minimum of the pattern for the other line. b. The limit for resolution is when the maximum of the pattern for one of the lines is directly over the first minimum of the pattern for the other line. c. The limit for resolution is when the maximum of the pattern for one of the lines is directly over the second minimum of the pattern for the other line. d. The limit for resolution is when the minimum of the pattern for one of the lines is directly over the second maximum of the pattern for the other line. WORKED EXAMPLE Change of Wavelength A monochromatic laser beam of green light with a wavelength of 550 nm passes from air to water. The refractive index of water is 1.33. What will be the wavelength of the light after it enters the water? STRATEGY You can assume that the refractive index of air is the same as that of light in a vacuum because they are so close. You then have all the information you need to solve for . Solution 17.5 Discussion The refractive index of air is 1.0003, so the approximation holds for three significant figures. You would not see the light change color, however. Color is determined by frequency, not wavelength. WORKED EXAMPLE Diffraction Grating A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum form for green light with a wavelength of 520 nm? STRATEGY You are given enough information to calculate d, and you are given the values of and m. You will have to find the arcsin of a 540 Chapter 17 • Diffraction and Interference number to find . Solution First find d. Rearrange the equation for constructive interference conditions for a diffraction grating, and substitute the known values. 17.6 Discussion This angle seems reasonable for the first maximum. Recall that the meaning of sin‒1 (or arcsin) is the angle with a sine that is (the unknown). Remember that the value of will not be greater than 1 for any value of . WORKED EXAMPLE Resolution What is the minimum angular spread of a 633-nm-wavelength He-Ne laser beam that is originally 1.00 mm in diameter? STRATEGY The diameter of the beam is the same as if it were coming through an aperture of that size, so D= 1.00 mm. You are given , and you must solve for . Solution 17.7 Discussion The conversion factor for radians to degrees is 1.000 radian = 57.3°. The spread is very small and would not be noticeable over short distances. The angle represents the angular separation of the central maximum and the first minimum. Practice Problems 6. A beam of yellow light has a wavelength of 600 nm in a vacuum and a wavelength of 397 nm in Plexiglas. What is the refractive index of Plexiglas? a. 1.51 b. 2.61 3.02 c. 3.77 d. 7. What is the angle between two just-resolved points of light for a 3.00 mm diameter pupil, assuming an average wavelength of 550 nm? a. 224 rad 183 rad b. 1.83 × 10–4 rad c. d. 2.24 × 10–4 rad Check Your Understanding 8. How is an interference pattern formed by a diffraction grating different from the pattern formed by a double slit? Access for free at openstax.org. 17.2 • Applications of Diffraction, Interference, and Coherence 541 a. The pattern is colorful. b. The pattern is faded. c. The pattern is sharper. d. The pattern is curved. 9. A beam of light always spreads out. Why can a beam not be produced with parallel rays to prevent spreading? a. Light is always polarized. b. Light is always reflected. c. Light is always refracted. d. Light is always diffracted. 10. Compare interference patterns formed by a double slit and by a diffraction grating in terms of brightness and narrowness of bands. a. The pattern formed has broader and brighter bands. b. The pattern formed has broader and duller bands. c. The pattern formed has narrower and duller bands. d. The pattern formed has narrower and brighter bands. 11. Describe the slits in a diffraction grating in terms of number and spacing, as compared to a two-slit diffraction setup. a. The slits in a diffraction grating are broader, with space between them that is greater than the separation of the two slits in two-slit diffraction. b. The slits in a diffraction grating are broader, with space between them that is the same as the separation of the two slits in two-slit diffraction. c. The slits in a diffraction grating are narrower, with space between them that is the same as the separation of the two slits in two-slit diffraction. d. The slits in a diffraction grating are narrower, with space between them that is greater than the separation of the two slits in two-slit diffraction. 542 Chapter 17 • Key Terms KEY TERMS differential interference contrast (DIC) separating a tangent to all of the wavelets. polarized light source into two beams polarized at right angles to each other and coherent with each other then, after passing through the sample, recombining and realigning the beams so they have the same plane of polarization, and then creating an interference pattern caused by the differences in their optical path and the refractive indices of the parts of the sample they passed through; the result is an image with contrast and shadowing that c |
ould not be observed with traditional optics diffraction bending of a wave around the edges of an opening or an obstacle diffraction grating many of evenly spaced slits having dimensions such that they produce an interference pattern Huygens’s principle Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself; the new wavefront is a line SECTION SUMMARY 17.1 Understanding Diffraction and Interference iridescence the effect that occurs when tiny, fingerlike structures in regular patterns act as reflection gratings, producing constructive interference that gives feathers colors not solely due to their pigmentation laser acronym for a device that produces light amplification by stimulated emission of radiation monochromatic one color monochromator device that separates the various wavelengths of incoming light and allows a beam with only a specific wavelength to pass through Rayleigh criterion two images are just resolvable when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other resolution degree to which two images can be distinguished from one another, which is limited by diffraction wavefront points on a wave surface that all share an identical, constant phase 17.2 Applications of Diffraction, Interference, and Coherence • The wavelength of light varies with the refractive index • The focused, coherent radiation emitted by lasers has of the medium. many uses in medicine and industry. • Slits produce a diffraction pattern if their width and • Characteristics of diffraction patterns produced with separation are similar to the wavelength of light passing through them. Interference bands of a single-slit diffraction pattern can be predicted. Interference bands of a double-slit diffraction pattern can be predicted. • • KEY EQUATIONS 17.1 Understanding Diffraction and Interference speed of light, frequency, and wavelength change of wavelength with index of refraction two-slit constructive interference , for m = 0, 1, −1, 2, −2, … diffraction gratings can be determined. • Diffraction gratings have been incorporated in many instruments, including microscopes and spectrometers. • Resolution has a limit that can be predicted. two-slit destructive interference , for m= 0, 1, −1, 2, −2, … one-slit, first-order destructive interference; wavelength related to dimensions one-slit destructive interference Access for free at openstax.org. Chapter 17 • Chapter Review 543 17.2 Applications of Diffraction, Interference, and Coherence wavelength change with change in medium diffraction grating constructive interference resolution CHAPTER REVIEW Concept Items 17.1 Understanding Diffraction and Interference 1. Which behavior of light is indicated by an interference pattern? a. ray behavior b. particle behavior c. d. wave behavior corpuscular behavior 2. Which behavior of light is indicated by diffraction? a. wave behavior b. particle behavior ray behavior c. corpuscular behavior d. 17.2 Applications of Diffraction, Interference, and Coherence 3. There is a principle related to resolution that is expressed by this equation. 17.8 What is that principle stated in full? Critical Thinking Items 17.1 Understanding Diffraction and Interference 6. Describe a situation in which bodies of water and a line of rocks could create a diffraction pattern similar to light passing through double slits. Include the arrangement of the rocks, the positions of the bodies of water, and the location of the diffraction pattern. Note the dimensions that are necessary for the production of the pattern. a. When waves from a small body of water pass through two widely separated openings and enter a larger body of water, a diffraction pattern is produced that is similar tothe diffraction pattern formed by light passing through two slits. The width of each opening is larger than the size of the wavelength of the waves. b. When waves from a large body of water pass 4. A principle related to resolution states, “Two images are just resolved when the center of the diffraction pattern of one is directly over the first minimum of the diffraction pattern of the other.” Write the equation that expresses that principle. a. b. c. d. 5. Which statement completes this resolution? Two images are just resolved when — a. The center of the diffraction pattern of one image is directly over the central maximum of the diffraction pattern of the other. b. The center of the diffraction pattern of one image is directly over the central minimum of the diffraction pattern of the other c. The center of the diffraction pattern of one image is directly over the first minimum of the diffraction pattern of the other d. The center of the diffraction pattern of one is directly over the first maximum of the diffraction pattern of the other through two narrow openings and enter a smaller body of water, a diffraction pattern is produced that is similar to the diffraction pattern formed by light passing through two slits. The widths and separation of the openings are similar to the size of the wavelength of the waves. c. When waves from a small body of water pass through two wide openings and enter a larger body of water, a diffraction pattern is produced that is similar tothe diffraction pattern formed by light passing through two slits. The separation between the openings is similar to the size of the wavelength of the waves. d. When waves from a large body of water pass through two wide openings and enter a smaller body of water, a diffraction pattern is produced that is similar to the diffraction pattern formed by light passing through two slits. The widths and 544 Chapter 17 • Chapter Review separation of the openings are larger than the size of the wavelength of the waves. 17.2 Applications of Diffraction, Interference, and Coherence 7. For what type of electromagnetic radiation would a grating with spacing greater than 800 nm be useful as a spectroscopic tool? a. It can be used to analyze spectra only in the infrared portion of the spectrum. It can be used to analyze spectra in the entire visible portion of the electromagnetic spectrum. b. Problems 17.1 Understanding Diffraction and Interference 9. What is the distance between two slits that produce a diffraction pattern with the first minimum at an angle of 45.0° when 410-nm violet light passes through the slits? a. 2,030 nm b. 1,450 nm c. 410 nm d. 290 nm 10. A breakwater at the entrance to a harbor consists of a rock barrier with a 50.0 − m -wide opening. Ocean waves with a 20.0-m wavelength approach the opening straight on. At what angle to the incident direction are the boats inside the harbor most protected against wave action? 11.5° a. 7.46° b. c. 5.74° d. 23.6° Performance Task 17.2 Applications of Diffraction, Interference, and Coherence 13. In this performance task you will create one- and two- slit diffraction and observe the interference patterns that result. • A utility knife (a knife with a razor blade-like cutting edge) • Aluminum foil • A straight edge • A strong, small light source or a laser pointer • A tape measure • A white wall Access for free at openstax.org. c. d. It can only be used to analyze spectra in the short‐wavelength visible. It can only be used to analyze spectra in the short‐wavelength visible and ultraviolet. 8. A beam of green light has a wavelength of in a in Plexiglas. What vacuum and a wavelength of is the refractive index of Plexiglas? a. b. c. d. 17.2 Applications of Diffraction, Interference, and Coherence 11. A 500-nm beam of light passing through a diffraction grating creates its second band of constructive interference at an angle of 1.50°. How far apart are the slits in the grating? 38,200 nm a. b. 19,100 nm c. 667 nm 333 nm d. 12. The range of the visible-light spectrum is 380 nm to 780 nm. What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light? a. 26,300 lines/cm b. 13,200 lines/cm c. 6,410 lines/cm d. 12,820 lines/cm Procedure 1. Cut a piece of aluminum foil about 15 cm × 15 cm. 2. Use the utility knife and the straight edge to cut a straight slit several cm long in the center of the foil square. 3. With the room darkened, one partner shines the light through the slit and toward the wall. The other partner observes the pattern on the wall. The partner with the light changes the distance from the foil to the wall and the distance from the light to the foil. 4. When the sharpest, brightest pattern possible is obtained, the partner who is not holding the foil and light makes measurements. 5. Measure the perpendicular (shortest) distance from the slit to the wall, the distance from the center of the pattern to several of the dark bands, and the distance from the slit to the same dark bands. 6. Carefully make a second slit parallel to the first slit and 1 mm or less away. 7. Repeat steps 2 through 5, only this time measure the distances to bright bands. NOTE—In your calculations, use 580 nm for if you used white light. If you used a colored laser pointer, look up the wavelength of the color. You from its tangent may find it easier to calculate TEST PREP Multiple Choice 17.1 Understanding Diffraction and Interference 14. Which remains unchanged when a monochromatic beam of light passes from air into water? a. b. c. d. the speed of the light the direction of the beam the frequency of the light the wavelength of the light 15. Two slits are separated by a distance of 3500 nm . If light with a wavelength of 500 nm passes through the slits and produces an interference pattern, the m = ________ order minimum appears at an angle of 30.0°. a. 0 b. 1 c. 2 3 d. 16. In the sunlight, the shadow of a building has fuzzy edges even if the building does not. Is this a refraction effect? Explain. a. Yes, this is a refraction effect, where every point on the building acts as the origin for |
a new wavefront. b. Yes, this is a refraction effect, where the whole building acts as the origin for a new wavefront. c. No, this is a diffraction effect, where every point on the edge of the building’s shadow acts as the origin for a new wavefront. d. No, this is a diffraction effect, where the whole building acts as the origin for a new wavefront. 17.2 Applications of Diffraction, Interference, and Coherence 17. Two images are just resolved when the center of the diffraction pattern of one is directly over ________ of the diffraction pattern of the other. a. the center Chapter 17 • Test Prep 545 rather than from its sine. a. Which experiment gave the most distinct pattern—one or two slits? b. What was the width of the single slit? Compare the calculated distance with the measured distance. c. What was the distance between the two slits? Compare the calculated distance with the measured distance. b. c. d. the first minimum the first maximum the last maximum 18. Two point sources of light are just resolvable as . What is the they pass through a small hole. The angle to the first minimum of one source is diameter of the hole? a. b. c. d. 19. Will a beam of light shining through a 1-mm hole behave any differently than a beam of light that is 1 mm wide as it leaves its source? Explain.? a. Yes, the beam passing through the hole will spread out as it travels, because it is diffracted by the edges of the hole, whereas the 1 -mm beam, which encounters no diffracting obstacle, will not spread out. b. Yes, the beam passing through the hole will be made more parallelby passing through the hole, and so will not spread out as it travels, whereas the unaltered wavefronts of the 1-mm beam will cause the beam to spread out as it travels. c. No, both beams will remain the same width as they travel, and they will not spread out. d. No, both beams will spread out as they travel. 20. A laser pointer emits a coherent beam of parallel light rays. Does the light from such a source spread out at all? Explain. a. Yes, every point on a wavefront is not a source of wavelets, which prevent the spreading of light waves. b. No, every point on a wavefront is not a source of wavelets, so that the beam behaves as a bundles of rays that travel in their initial direction. c. No, every point on a wavefront is a source of 546 Chapter 17 • Test Prep wavelets, which keep the beam from spreading. d. Yes, every point on a wavefront is a source of wavelets, which cause the beam to spread out steadily as it moves forward. Short Answer 17.1 Understanding Diffraction and Interference 21. Light passing through double slits creates a diffraction pattern. How would the spacing of the bands in the pattern change if the slits were closer together? a. The bands would be closer together. b. The bands would spread farther apart. c. The bands would remain stationary. d. The bands would fade and eventually disappear. 22. A beam of light passes through a single slit to create a diffraction pattern. How will the spacing of the bands in the pattern change if the width of the slit is increased? a. The width of the spaces between the bands will remain the same. spaced parallel lines that produces an interference pattern that is similar to but sharper and better dispersed than that of a double slit. b. A diffraction grating is a large collection of randomly spaced parallel lines that produces an interference pattern that is similar to but less sharp or well-dispersed as that of a double slit. c. A diffraction grating is a large collection of randomly spaced intersecting lines that produces an interference pattern that is similar to but sharper and better dispersed than that of a double slit. d. A diffraction grating is a large collection of evenly spaced intersecting lines that produces an interference pattern that is similar to but less sharp or well-dispersed as that of a double slit. b. The width of the spaces between the bands will 26. Suppose pure-wavelength light falls on a diffraction increase. c. The width of the spaces between the bands will decrease. d. The width of the spaces between the bands will first decrease and then increase. grating. What happens to the interference pattern if the same light falls on a grating that has more lines per centimeter? a. The bands will spread farther from the central maximum. b. The bands will come closer to the central 23. What is the wavelength of light falling on double slits maximum. if the third-order maximum is at c. The bands will not spread farther from the first maximum. d. The bands will come closer to the first maximum. 27. How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for 473 nm blue light at an angle of 25.0°? a. b. c. 851 lines/cm d. 8,934 lines/cm 529,000 lines/cm 50,000 lines/cm 28. What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angle of 60.0°? a. 2.28 × 104 nm 3.29 × 102 nm b. c. 2.53 × 101 nm 1.76 × 103 nm d. ? separated by an angle of a. b. c. d. 24. What is the longest wavelength of light passing through a single slit of width 1.20 μm for which there is a firstorder minimum? 1.04 µm a. b. 0.849 µm c. 0.600 µm d. 2.40 µm 17.2 Applications of Diffraction, Interference, and Coherence 25. Describe a diffraction grating and the interference pattern it produces. a. A diffraction grating is a large collection of evenly Access for free at openstax.org. Chapter 17 • Test Prep 547 grating. a. All three interference pattern produce identical bands. b. A double slit produces the sharpest and most distinct bands. c. A single slit produces the sharpest and most distinct bands. d. The diffraction grating produces the sharpest and most distinct bands. 32. An electric current through hydrogen gas produces several distinct wavelengths of visible light. The light is projected onto a diffraction grating having per centimeter. What are the wavelengths of the hydrogen spectrum if the light forms first-order maxima at angles of a. , and ? , , lines b. c. d. Extended Response 17.1 Understanding Diffraction and Interference 29. Suppose you use a double slit to perform Young’s double-slit experiment in air, and then repeat the experiment with the same double slit in water. Does the color of the light change? Do the angles to the same parts of the interference pattern get larger or smaller? Explain. a. No, the color is determined by frequency. The magnitude of the angle decreases. b. No, the color is determined by wavelength. The magnitude of the angle decreases. c. Yes, the color is determined by frequency. The magnitude of the angle increases. d. Yes, the color is determined by wavelength. The magnitude of the angle increases. 30. A double slit is located at a distance xfrom a screen, with the distance along the screen from the center given by y. When the distance dbetween the slits is relatively large, there will be numerous bright bands. For small angles (where sinθ= θ, with θin radians), what is the distance between fringes? a. b. c. d. 17.2 Applications of Diffraction, Interference, and Coherence 31. Compare the interference patterns of single-slit diffraction, double-slit diffraction, and a diffraction 548 Chapter 17 • Test Prep Access for free at openstax.org. CHAPTER 18 Static Electricity Figure 18.1 This child’s hair contains an imbalance of electrical charge (commonly called static electricity), which causes it to stand on end. The sliding motion stripped electrons away from the child’s body, leaving him with an excess of positive charges, which repel each other along each strand of hair. (credit: Ken Bosma, Wikimedia Commons) Chapter Outline 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 18.2 Coulomb's law 18.3 Electric Field 18.4 Electric Potential 18.5 Capacitors and Dielectrics You may have been introduced to static electricity like the child sliding down the slide in the opening INTRODUCTION photograph (Figure 18.1). The zapthat he is likely to receive if he touches a playmate or parent tends to bring home the lesson. But static electricity is more than just fun and games—it is put to use in many industries. The forces between electrically charged particles are used in technologies such as printers, pollution filters, and spray guns used for painting cars and trucks. Static electricityis the study of phenomena that involve an imbalance of electrical charge. Although creating this imbalance typically requires moving charge around, once the imbalance is created, it often remains static for a long time. The study of charge in motion is called electromagnetismand will be covered in a later chapter. What is electrical charge, how is it associated 550 Chapter 18 • Static Electricity with objects, and what forces does it create? These are just some of the questions that this chapter addresses. 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge Section Learning Objectives By the end of this section, you will be able to do the following: • Describe positive and negative electric charges • Use conservation of charge to calculate quantities of charge transferred between objects • Characterize materials as conductors or insulators based on their electrical properties • Describe electric polarization and charging by induction Section Key Terms conduction conductor electron induction insulator law of conservation of charge polarization proton Electric Charge You may know someone who has an electricpersonality, which usually means that other people are attracted to this person. This saying is based on electric charge, which is a property of matter that causes objects to attract or repel each other. Electric charge comes in two varieties, which we call positiveand negative.Like charges repel each other, and unlike charges attract each other. Thus, two positive charges repel each other, as do two negative charges. A positive charge and a negative charge at |
tract each other. How do we know there are two types of electric charge? When various materials are rubbed together in controlled ways, certain combinations of materials always result in a net charge of one type on one material and a net charge of the opposite type on the other material. By convention, we call one type of charge positive and the other type negative. For example, when glass is rubbed with silk, the glass becomes positively charged and the silk negatively charged. Because the glass and silk have opposite charges, they attract one another like clothes that have rubbed together in a dryer. Two glass rods rubbed with silk in this manner will repel one another, because each rod has positive charge on it. Similarly, two silk cloths rubbed in this manner will repel each other, because both cloths have negative charge. Figure 18.2 shows how these simple materials can be used to explore the nature of the force between charges. Figure 18.2 A glass rod becomes positively charged when rubbed with silk, whereas the silk becomes negatively charged. (a) The glass rod is attracted to the silk, because their charges are opposite. (b) Two similarly charged glass rods repel. (c) Two similarly charged silk cloths repel. It took scientists a long time to discover what lay behind these two types of charges. The word electricitself comes from the Greek word elektronfor amber, because the ancient Greeks noticed that amber, when rubbed by fur, attracts dry straw. Almost 2,000 years later, the English physicist William Gilbert proposed a model that explained the effect of electric charge as being due to a mysterious electrical fluid that would pass from one object to another. This model was debated for several hundred years, but it was finally put to rest in 1897 by the work of the English physicist J. J. Thomson and French physicist Jean Perrin. Along with many others, Thomson and Perrin were studying the mysterious cathode raysthat were known at the time to consist of particles smaller than the smallest atom. Perrin showed that cathode rays actually carried negative electrical charge. Later, Thomson’s work led him to declare, “I can see no escape from the conclusion that [cathode rays] are charges of negative Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 551 electricity carried by particles of matter.” It took several years of further experiments to confirm Thomson’s interpretation of the experiments, but science had in fact discovered the particle that carries the fundamental unit of negative electrical charge. We now know this particle as the electron. Atoms, however, were known to be electrically neutral, which means that they carry the same amount of positive and negative charge, so their net charge is zero. Because electrons are negative, some other part of the atom must contain positive charge. Thomson put forth what is called the plum pudding model, in which he described atoms as being made of thousands of electrons swimming around in a nebulous mass of positive charge, as shown by the left-side image of Figure 18.3. His student, Ernest Rutherford, originally believed that this model was correct and used it (along with other models) to try to understand the results of his experiments bombarding gold foils with alphaparticles (i.e., helium atoms stripped of their electrons). The results, however, did not confirm Thomson’s model but rather destroyed it! Rutherford found that most of the space occupied by the gold atoms was actually empty and that almost all of the matter of each atom was concentrated into a tiny, extremely dense nucleus, as shown by the right-side image of Figure 18.3. The atomic nucleus was later found to contain particles called protons, each of which carries a unit of positive electric charge.1 Figure 18.3 The left drawing shows Thompson’s plum-pudding model, in which the electrons swim around in a nebulous mass of positive charge. The right drawing shows Rutherford’s model, in which the electrons orbit around a tiny, massive nucleus. Note that the size of the nucleus is vastly exaggerated in this drawing. Were it drawn to scale with respect to the size of the electron orbits, the nucleus would not be visible to the naked eye in this drawing. Also, as far as science can currently detect, electrons are point particles, which means that they have no size at all! Protons and electrons are thus the fundamental particles that carry electric charge. Each proton carries one unit of positive charge, and each electron carries one unit of negative charge. To the best precision that modern technology can provide, the charge carried by a proton is exactlythe opposite of that carried by an electron. The SI unit for electric charge is the coulomb (abbreviated as “C”), which is named after the French physicist Charles Augustin de Coulomb, who studied the force between charged objects. The proton carries protons required to make +1.00 C is and the electron carries . The number nof The same number of electrons is required to make −1.00 C of electric charge. The fundamental unit of charge is often represented as e. Thus, the charge on a proton is e, and the charge on an electron is −e. Mathematically, 18.1 LINKS TO PHYSICS Measuring the Fundamental Electric Charge The American physicist Robert Millikan (1868–1953) and his student Harvey Fletcher (1884–1981) were the first to make a relatively accurate measurement of the fundamental unit of charge on the electron. They designed what is now a classic 1Protons were later found to contain sub particles called quarks, which have fractional electric charge. But that is another story that we leave for subsequent physics courses. 552 Chapter 18 • Static Electricity experiment performed by students. The Millikan oil-drop experiment is shown in Figure 18.4. The experiment involves some concepts that will be introduced later, but the basic idea is that a fine oil mist is sprayed between two plates that can be charged with a known amount of opposite charge. Some oil drops accumulate some excess negative charge when being sprayed and are attracted to the positive charge of the upper plate and repelled by the negative charge on the lower plate. By tuning the charge on these plates until the weight of the oil drop is balanced by the electric forces, the net charge on the oil drop can be determined quite precisely. Figure 18.4 The oil-drop experiment involved spraying a fine mist of oil between two metal plates charged with opposite charges. By knowing the mass of the oil droplets and adjusting the electric charge on the plates, the charge on the oil drops can be determined with precision. Millikan and Fletcher found that the drops would accumulate charge in discrete units of about within 1 percent of the modern value of times greater than the possible error Millikan reported for his results! which is Although this difference may seem quite small, it is actually five Because the charge on the electron is a fundamental constant of nature, determining its precise value is very important for all of science. This created pressure on Millikan and others after him that reveals some equally important aspects of human nature. First, Millikan took sole credit for the experiment and was awarded the 1923 Nobel Prize in physics for this work, although his student Harvey Fletcher apparently contributed in significant ways to the work. Just before his death in 1981, Fletcher divulged that Millikan coerced him to give Millikan sole credit for the work, in exchange for which Millikan promoted Fletcher’s career at Bell Labs. Another great scientist, Richard Feynman, points out that many scientists who measured the fundamental charge after Millikan were reluctant to report values that differed much from Millikan’s value. History shows that later measurements slowly crept up from Millikan’s value until settling on the modern value. Why did they not immediately find the error and correct the value, asks Feynman. Apparently, having found a value higher than the much-respected value found by Millikan, scientists would look for possible mistakes that might lower their value to make it agree better with Millikan’s value. This reveals the important psychological weight carried by preconceived notions and shows how hard it is to refute them. Scientists, however devoted to logic and data they may be, are apparently just as vulnerable to this aspect of human nature as everyone else. The lesson here is that, although it is good to be skeptical of new results, you should not discount them just because they do not agree with conventional wisdom. If your reasoning is sound and your data are reliable, the conclusion demanded by the data must be seriously considered, even if that conclusion disagrees with the commonly accepted truth. GRASP CHECK Suppose that Millikan observed an oil drop carrying three fundamental units of charge. What would be the net charge on this oil drop? a. −4.81 × 10−19 C b. −1.602 × 10−19 C 1.602 × 10−19 C c. d. 4.81 × 10−19 C Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 553 Snap Lab Like and Unlike Charges This activity investigates the repulsion and attraction caused by static electrical charge. • Adhesive tape • Nonconducting surface, such as a plastic table or chair Instructions Procedure for Part (a) 1. Prepare two pieces of tape about 4 cm long. To make a handle, double over about 0.5 cm at one end so that the sticky side sticks together. 2. Attach the pieces of tape side by side onto a nonmetallic surface, such as a tabletop or the seat of a chair, as shown in Figure 18.5(a). 3. Peel off both pieces of tape and hang them downward, holding them by the handles, as shown in Figure 18.5(b). If the tape bends upward and sticks to your hand, try using a shorter piece of tape, or simply shake the tape so that it no longer sticks to your hand. 4. Now slowly bring t |
he two pieces of tape together, as shown in Figure 18.5(c). What happens? Figure 18.5 Procedure for Part (b) 5. Stick one piece of tape on the nonmetallic surface, and stick the second piece of tape on top of the first piece, as shown in Figure 18.6(a). 6. Slowly peel off the two pieces by pulling on the handle of the bottom piece. 7. Gently stroke your finger along the top of the second piece of tape (i.e., the nonsticky side), as shown in Figure 18.6(b). 8. Peel the two pieces of tape apart by pulling on their handles, as shown in Figure 18.6(c). 9. Slowly bring the two pieces of tape together. What happens? Figure 18.6 GRASP CHECK In step 4, why did the two pieces of tape repel each other? In step 9, why did they attract each other? a. Like charges attract, while unlike charges repel each other. b. Like charges repel, while unlike charges attract each other. c. Tapes having positive charge repel, while tapes having negative charge attract each other. d. Tapes having negative charge repel, while tapes having positive charge attract each other. 554 Chapter 18 • Static Electricity Conservation of Charge Because the fundamental positive and negative units of charge are carried on protons and electrons, we would expect that the total charge cannot change in any system that we define. In other words, although we might be able to move charge around, we cannot create or destroy it. This should be true provided that we do not create or destroy protons or electrons in our system. In the twentieth century, however, scientists learned how to create and destroy electrons and protons, but they found that charge is still conserved. Many experiments and solid theoretical arguments have elevated this idea to the status of a law. The law of conservation of charge says that electrical charge cannot be created or destroyed. The law of conservation of charge is very useful. It tells us that the net charge in a system is the same before and after any interaction within the system. Of course, we must ensure that no external charge enters the system during the interaction and that no internal charge leaves the system. Mathematically, conservation of charge can be expressed as 18.2 where is the net charge of the system before the interaction, and is the net charge after the interaction. WORKED EXAMPLE What is the missing charge? Figure 18.7 shows two spheres that initially have +4 C and +8 C of charge. After an interaction (which could simply be that they touch each other), the blue sphere has +10 C of charge, and the red sphere has an unknown quantity of charge. Use the law of conservation of charge to find the final charge on the red sphere. Strategy The net initial charge of the system is . The net final charge of the system is is the final charge on the red sphere. Conservation of charge tells us that we can solve for . , where , so 18.3 Solution Equating and and solving for gives The red sphere has +2 C of charge. Figure 18.7 Two spheres, one blue and one red, initially have +4 C and +8 C of charge, respectively. After the two spheres interact, the blue sphere has a charge of +10 C. The law of conservation of charge allows us to find the final charge on the red sphere. Discussion Like all conservation laws, conservation of charge is an accounting scheme that helps us keep track of electric charge. Practice Problems 1. Which equation describes conservation of charge? a. qinitial = qfinal = constant b. qinitial = qfinal = 0 c. qinitial − qfinal = 0 Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 555 d. qinitial/qfinal = constant 2. An isolated system contains two objects with charges and . If object 1 loses half of its charge, what is the final charge on object 2? a. b. c. d. Conductors and Insulators Materials can be classified depending on whether they allow charge to move. If charge can easily move through a material, such as metals, then these materials are called conductors. This means that charge can be conducted (i.e., move) through the material rather easily. If charge cannot move through a material, such as rubber, then this material is called an insulator. Most materials are insulators. Their atoms and molecules hold on more tightly to their electrons, so it is difficult for electrons to move between atoms. However, it is not impossible. With enough energy, it is possible to force electrons to move through an insulator. However, the insulator is often physically destroyed in the process. In metals, the outer electrons are loosely bound to their atoms, so not much energy is required to make electrons move through metal. Such metals as copper, silver, and aluminum are good conductors. Insulating materials include plastics, glass, ceramics, and wood. The conductivity of some materials is intermediate between conductors and insulators. These are called semiconductors. They can be made conductive under the right conditions, which can involve temperature, the purity of the material, and the force applied to push electrons through them. Because we can control whether semiconductors are conductors or insulators, these materials are used extensively in computer chips. The most commonly used semiconductor is silicon. Figure 18.8 shows various materials arranged according to their ability to conduct electrons. Figure 18.8 Materials can be arranged according to their ability to conduct electric charge. The slashes on the arrow mean that there is a very large gap in conducting ability between conductors, semiconductors, and insulators, but the drawing is compressed to fit on the page. The numbers below the materials give their resistivityin Ω•m (which you will learn about below). The resistivity is a measure of how hard it is to make charge move through a given material. What happens if an excess negative charge is placed on a conducting object? Because like charges repel each other, they will push against each other until they are as far apart as they can get. Because the charge can move in a conductor, it moves to the outer surfaces of the object. Figure 18.9(a) shows schematically how an excess negative charge spreads itself evenly over the outer surface of a metal sphere. What happens if the same is done with an insulating object? The electrons still repel each other, but they are not able to move, because the material is an insulator. Thus, the excess charge stays put and does not distribute itself over the object. Figure 18.9(b) shows this situation. 556 Chapter 18 • Static Electricity Figure 18.9 (a) A conducting sphere with excess negative charge (i.e., electrons). The electrons repel each other and spread out to cover the outer surface of the sphere. (b) An insulating sphere with excess negative charge. The electrons cannot move, so they remain in their original positions. Transfer and Separation of Charge Most objects we deal with are electrically neutral, which means that they have the same amount of positive and negative charge. However, transferring negative charge from one object to another is fairly easy to do. When negative charge is transferred from one object to another, an excess of positive charge is left behind. How do we know that the negative charge is the mobile charge? The positive charge is carried by the proton, which is stuck firmly in the nucleus of atoms, and the atoms are stuck in place in solid materials. Electrons, which carry the negative charge, are much easier to remove from their atoms or molecules and can therefore be transferred more easily. Electric charge can be transferred in several manners. One of the simplest ways to transfer charge is charging by contact, in which the surfaces of two objects made of different materials are placed in close contact. If one of the materials holds electrons more tightly than the other, then it takes some electrons with it when the materials are separated. Rubbing two surfaces together increases the transfer of electrons, because it creates a closer contact between the materials. It also serves to present freshmaterial with a full supply of electrons to the other material. Thus, when you walk across a carpet on a dry day, your shoes rub against the carpet, and some electrons are removed from the carpet by your shoes. The result is that you have an excess of negative charge on your shoes. When you then touch a doorknob, some of your excess of electrons transfer to the neutral doorknob, creating a small spark. Touching the doorknob with your hand demonstrates a second way to transfer electric charge, which is charging by conduction. This transfer happens because like charges repel, and so the excess electrons that you picked up from the carpet want to be as far away from each other as possible. Some of them move to the doorknob, where they will distribute themselves over the outer surface of the metal. Another example of charging by conduction is shown in the top row of Figure 18.10. A metal sphere with 100 excess electrons touches a metal sphere with 50 excess electrons, so 25 electrons from the first sphere transfer to the second sphere. Each sphere finishes with 75 excess electrons. The same reasoning applies to the transfer of positive charge. However, because positive charge essentially cannot move in solids, it is transferred by moving negative charge in the opposite direction. For example, consider the bottom row of Figure 18.10. The first metal sphere has 100 excess protons and touches a metal sphere with 50 excess protons, so the second sphere transfers 25 electrons to the first sphere. These 25 extra electrons will electrically cancel 25 protons so that the first metal sphere is left with 75 excess protons. This is shown in the bottom row of Figure 18.10. The second metal sphere lost 25 electrons so it has 25 more excess protons, for a total of 75 excess protons. The end result is the same if we consider that the first ball transferred a net positive charge equal to that of 25 protons to the |
first ball. Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 557 Figure 18.10 In the top row, a metal sphere with 100 excess electrons transfers 25 electrons to a metal sphere with an excess of 50 electrons. After the transfer, both spheres have 75 excess electrons. In the bottom row, a metal sphere with 100 excess protons receives 25 electrons from a ball with 50 excess protons. After the transfer, both spheres have 75 excess protons. In this discussion, you may wonder how the excess electrons originally got from your shoes to your hand to create the spark when you touched the doorknob. The answer is that noelectrons actually traveled from your shoes to your hands. Instead, because like charges repel each other, the excess electrons on your shoe simply pushed away some of the electrons in your feet. The electrons thus dislodged from your feet moved up into your leg and in turn pushed away some electrons in your leg. This process continued through your whole body until a distribution of excess electrons covered the extremities of your body. Thus your head, your hands, the tip of your nose, and so forth all received their doses of excess electrons that had been pushed out of their normal positions. All this was the result of electrons being pushed out of your feet by the excess electrons on your shoes. This type of charge separation is called polarization. As soon as the excess electrons leave your shoes (by rubbing off onto the floor or being carried away in humid air), the distribution of electrons in your body returns to normal. Every part of your body is again electrically neutral (i.e., zero excess charge). The phenomenon of polarization is seen in . The child has accumulated excess positive charge by sliding on the slide. This excess charge repels itself and so becomes distributed over the extremities of the child’s body, notably in his hair. As a result, the hair stands on end, because the excess negative charge on each strand repels the excess positive charge on neighboring strands. Polarization can be used to charge objects. Consider the two metallic spheres shown in Figure 18.11. The spheres are electrically neutral, so they carry the same amounts of positive and negative charge. In the top picture (Figure 18.11(a)), the two spheres are touching, and the positive and negative charge is evenly distributed over the two spheres. We then approach a glass rod that carries an excess positive charge, which can be done by rubbing the glass rod with silk, as shown in Figure 18.11(b). Because opposite charges attract each other, the negative charge is attracted to the glass rod, leaving an excess positive charge on the opposite side of the right sphere. This is an example of charging by induction, whereby a charge is created by approaching a charged object with a second object to create an unbalanced charge in the second object. If we then separate the two spheres, as shown in Figure 18.11(c), the excess charge is stuck on each sphere. The left sphere now has an excess negative charge, and the right sphere has an excess positive charge. Finally, in the bottom picture, the rod is removed, and the opposite charges attract each other, so they move as close together as they can get. 558 Chapter 18 • Static Electricity Figure 18.11 (a) Two neutral conducting spheres are touching each other, so the charge is evenly spread over both spheres. (b) A positively charged rod approaches, which attracts negative charges, leaving excess positive charge on the right sphere. (c) The spheres are separated. Each sphere now carries an equal magnitude of excess charge. (d) When the positively charged rod is removed, the excess negative charge on the left sphere is attracted to the excess positive charge on the right sphere. FUN IN PHYSICS Create a Spark in a Science Fair Van de Graaff generators are devices that are used not only for serious physics research but also for demonstrating the physics of static electricity at science fairs and in classrooms. Because they deliver relatively little electric current, they can be made safe for use in such environments. The first such generator was built by Robert Van de Graaff in 1931 for use in nuclear physics research. Figure 18.12 shows a simplified sketch of a Van de Graaff generator. Van de Graaff generators use smooth and pointed surfaces and conductors and insulators to generate large static charges. In the version shown in Figure 18.12, electrons are “sprayed” from the tips of the lower comb onto a moving belt, which is made of an insulating material like, such as rubber. This technique of charging the belt is akin to charging your shoes with electrons by walking across a carpet. The belt raises the charges up to the upper comb, where they transfer again, akin to your touching the doorknob and transferring your charge to it. Because like charges repel, the excess electrons all rush to the outer surface of the globe, which is made of metal (a conductor). Thus, the comb itself never accumulates too much charge, because any charge it gains is quickly depleted by the charge moving to the outer surface of the globe. Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 559 Figure 18.12 Van de Graaff generators transfer electrons onto a metallic sphere, where the electrons distribute themselves uniformly over the outer surface. Van de Graaff generators are used to demonstrate many interesting effects caused by static electricity. By touching the globe, a person gains excess charge, so his or her hair stands on end, as shown in Figure 18.13. You can also create mini lightning bolts by moving a neutral conductor toward the globe. Another favorite is to pile up aluminum muffin tins on top of the uncharged globe, then turn on the generator. Being made of conducting material, the tins accumulate excess charge. They then repel each other and fly off the globe one by one. A quick Internet search will show many examples of what you can do with a Van de Graaff generator. Figure 18.13 The man touching the Van de Graaff generator has excess charge, which spreads over his hair and repels hair strands from his neighbors. (credit: Jon “ShakataGaNai” Davis) GRASP CHECK Why don’t the electrons stay on the rubber belt when they reach the upper comb? a. The upper comb has no excess electrons, and the excess electrons in the rubber belt get transferred to the comb by contact. b. The upper comb has no excess electrons, and the excess electrons in the rubber belt get transferred to the comb by conduction. c. The upper comb has excess electrons, and the excess electrons in the rubber belt get transferred to the comb by conduction. d. The upper comb has excess electrons, and the excess electrons in the rubber belt get transferred to the comb by contact. Virtual Physics Balloons and Static Electricity Click to view content (http://www.openstax.org/l/28balloons) 560 Chapter 18 • Static Electricity This simulation allows you to observe negative charge accumulating on a balloon as you rub it against a sweater. You can then observe how two charged balloons interact and how they cause polarization in a wall. GRASP CHECK Click the reset button, and start with two balloons. Charge a first balloon by rubbing it on the sweater, and then move it toward the second balloon. Why does the second balloon not move? a. The second balloon has an equal number of positive and negative charges. b. The second balloon has more positive charges than negative charges. c. The second balloon has more negative charges than positive charges. d. The second balloon is positively charged and has polarization. Snap Lab Polarizing Tap Water This lab will demonstrate how water molecules can easily be polarized. • Plastic object of small dimensions, such as comb or plastic stirrer • Source of tap water Instructions Procedure 1. Thoroughly rub the plastic object with a dry cloth. 2. Open the faucet just enough to let a smooth filament of water run from the tap. 3. Move an edge of the charged plastic object toward the filament of running water. What do you observe? What happens when the plastic object touches the water filament? Can you explain your observations? GRASP CHECK Why does the water curve around the charged object? a. The charged object induces uniform positive charge on the water molecules. b. The charged object induces uniform negative charge on the water molecules. c. The charged object attracts the polarized water molecules and ions that are dissolved in the water. d. The charged object depolarizes the water molecules and the ions dissolved in the water. WORKED EXAMPLE Charging Ink Droplets Electrically neutral ink droplets in an ink-jet printer pass through an electron beam created by an electron gun, as shown in Figure 18.14. Some electrons are captured by the ink droplet, so that it becomes charged. After passing through the electron . How many electrons are captured by the ink droplet? beam, the net charge of the ink droplet is Access for free at openstax.org. 18.1 • Electrical Charges, Conservation of Charge, and Transfer of Charge 561 Figure 18.14 Electrons from an electron guncharge a passing ink droplet. STRATEGY A single electron carries a charge of of a single electron will give the number of electrons captured by the ink droplet. . Dividing the net charge of the ink droplet by the charge Solution The number nof electrons captured by the ink droplet are 18.4 Discussion This is almost a billion electrons! It seems like a lot, but it is quite small compared to the number of atoms in an ink droplet, which number about Thus, each extra electron is shared between about atoms. Practice Problems 3. How many protons are needed to make 1 nC of charge? 1 nC = 10−9 C 1.6 × 10−28 a. 1.6 × 10−10 b. 3 × 109 c. d. 6 × 109 4. In a physics lab, you charge up three metal spheres, two with and one with . When you bring all three sph |
eres together so that they all touch one another, what is the total charge on the three spheres? a. b. c. d. Check Your Understanding 5. How many types of electric charge exist? a. one type two types b. three types c. four types d. 6. Which are the two main electrical classifications of materials based on how easily charges can move through them? 562 Chapter 18 • Static Electricity a. b. c. d. conductor and insulator semiconductor and insulator conductor and superconductor conductor and semiconductor 7. True or false—A polarized material must have a nonzero net electric charge. a. b. true false 8. Describe the force between two positive point charges that interact. a. The force is attractive and acts along the line joining the two point charges. b. The force is attractive and acts tangential to the line joining the two point charges. c. The force is repulsive and acts along the line joining the two point charges. d. The force is repulsive and acts tangential to the line joining the two point charges. 9. How does a conductor differ from an insulator? a. Electric charges move easily in an insulator but not in a conducting material. b. Electric charges move easily in a conductor but not in an insulator. c. A conductor has a large number of electrons. d. More charges are in an insulator than in a conductor. 10. True or false—Charging an object by polarization requires touching it with an object carrying excess charge. a. b. true false 18.2 Coulomb's law Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Coulomb’s law verbally and mathematically • Solve problems involving Coulomb’s law Section Key Terms Coulomb’s law inverse-square law More than 100 years before Thomson and Rutherford discovered the fundamental particles that carry positive and negative electric charges, the French scientist Charles-Augustin de Coulomb mathematically described the force between charged objects. Doing so required careful measurements of forces between charged spheres, for which he built an ingenious device called a torsion balance. This device, shown in Figure 18.15, contains an insulating rod that is hanging by a thread inside a glass-walled enclosure. At one end of the rod is the metallic sphere A. When no charge is on this sphere, it touches sphere B. Coulomb would touch the spheres with a third metallic ball (shown at the bottom of the diagram) that was charged. An unknown amount of charge would distribute evenly between spheres A and B, which would then repel each other, because like charges repel. This force would cause sphere A to rotate away from sphere B, thus twisting the wire until the torsion in the wire balanced the electrical force. Coulomb then turned the knob at the top, which allowed him to rotate the thread, thus bringing sphere A closer to sphere B. He found that bringing sphere A twice as close to sphere B required increasing the torsion by a factor of four. Bringing the sphere three times closer required a ninefold increase in the torsion. From this type of measurement, he deduced that the electrical force between the spheres was inversely proportional to the distance squared between the spheres. In other words, where ris the distance between the spheres. An electrical charge distributes itself equally between two conducting spheres of the same size. Knowing this allowed Coulomb to divide an unknown charge in half. Repeating this process would produce a sphere with one quarter of the initial charge, and 18.5 Access for free at openstax.org. so on. Using this technique, he measured the force between spheres A and B when they were charged with different amounts of charge. These measurements led him to deduce that the force was proportional to the charge on each sphere, or 18.6 where is the charge on sphere A, and is the charge on sphere B. 18.2 • Coulomb's law 563 Figure 18.15 A drawing of Coulomb’s torsion balance, which he used to measure the electrical force between charged spheres. (credit: Charles-Augustin de Coulomb) Combining these two proportionalities, he proposed the following expression to describe the force between the charged spheres. This equation is known as Coulomb’s law, and it describes the electrostatic force between charged objects. The constant of proportionality kis called Coulomb’s constant. In SI units, the constant khas the value 18.7 The direction of the force is along the line joining the centers of the two objects. If the two charges are of opposite signs, Coulomb’s law gives a negative result. This means that the force between the particles is attractive. If the two charges have the same signs, Coulomb’s law gives a positive result. This means that the force between the particles is repulsive. For example, if and both negative charge and This is shown in Figure 18.16(b). is a is a positive charge (or vice versa), then the charges are different, so the force between them is attractive. are negative or if both are positive, the force between them is repulsive. This is shown in Figure 18.16(a). If Figure 18.16 The magnitude of the electrostatic force Fbetween point charges q1 and q2 separated by a distance ris given by Coulomb’s law. Note that Newton’s third law (every force exerted creates an equal and opposite force) applies as usual—the force (F1,2) on q1 is equal in magnitude and opposite in direction to the force (F2,1) it exerts on q2. (a) Like charges. (b) Unlike charges. Note that Coulomb’s law applies only to charged objects that are not moving with respect to each other. The law says that the force is proportional to the amount of charge on each object and inversely proportional to the square of the distance between the objects. If we double the charge the force between them decreasesby a factor of spherical objects or to objects that are much smaller than the distance between the objects (in which case, the objects can be approximated as spheres). , for instance, then the force is doubled. If we double the distance between the objects, then . Although Coulomb’s law is true in general, it is easiest to apply to 564 Chapter 18 • Static Electricity Coulomb’s law is an example of an inverse-square law, which means the force depends on the square of the denominator. Another inverse-square law is Newton’s law of universal gravitation, which is they differ in two important respects: (i) The gravitational constant Gis much, much smaller than k ( differences explain why gravity is so much weaker than the electrostatic force and why gravity is only attractive, whereas the electrostatic force can be attractive or repulsive. ); and (ii) only one type of mass exists, whereas two types of electric charge exist. These two . Although these laws are similar, Finally, note that Coulomb measured the distance between the spheres from the centers of each sphere. He did not explain this assumption in his original papers, but it turns out to be valid. From outsidea uniform spherical distribution of charge, it can be treated as if all the charge were located at the center of the sphere. WATCH PHYSICS Electrostatics (part 1): Introduction to charge and Coulomb's law This video explains the basics of Coulomb’s law. Note that the lecturer uses dfor the distance between the center of the particles instead of r. Click to view content (https://www.openstax.org/l/28coulomb) GRASP CHECK True or false—If one particle carries a positive charge and another carries a negative charge, then the force between them is attractive. true a. false b. Snap Lab Hovering plastic In this lab, you will use electrostatics to hover a thin piece of plastic in the air. • Balloon • Light plastic bag (e.g., produce bag from grocery store) Instructions Inflate the balloon. Procedure 1. Cut the plastic bag to make a plastic loop about 2 inches wide. 2. 3. Charge the balloon by rubbing it on your clothes. 4. Charge the plastic loop by placing it on a nonmetallic surface and rubbing it with a cloth. 5. Hold the balloon in one hand, and in the other hand hold the plastic loop above the balloon. If the loop clings too much to your hand, recruit a friend to hold the strip above the balloon with both hands. Now let go of the plastic loop, and maneuver the balloon under the plastic loop to keep it hovering in the air above the balloon. GRASP CHECK How does the balloon keep the plastic loop hovering? a. The balloon and the loop are both negatively charged. This will help the balloon keep the plastic loop hovering. b. The balloon is charged, while the plastic loop is neutral.This will help the balloon keep the plastic loop hovering. c. The balloon and the loop are both positively charged. This will help the balloon keep the plastic loop hovering. d. The balloon is positively charged, while the plastic loop is negatively charged. This will help the balloon keep the plastic loop hovering. Access for free at openstax.org. WORKED EXAMPLE 18.2 • Coulomb's law 565 Using Coulomb’s law to find the force between charged objects between the two charged spheres when they are separated by 5.0 cm. By Suppose Coulomb measures a force of turning the dial at the top of the torsion balance, he approaches the spheres so that they are separated by 3.0 cm. Which force does he measure now? STRATEGY Apply Coulomb’s law to the situation before and after the spheres are brought closer together. Although we do not know the charges on the spheres, we do know that they remain the same. We call these unknown but constant charges Because these charges appear as a product in Coulomb’s law, they form a single unknown. We thus have two equations and two unknowns, which we can solve. The first unknown is the force (which we call second is ) when the spheres are 3.0 cm apart, and the and . . Use the following notation: When the charges are 5.0 cm apart, the force is where the subscript imeans initial. Once the charges are brought closer together, we know the subscript fmeans final. and , , where Solution Coulomb’s law app |
lied to the spheres in their initial positions gives Coulomb’s law applied to the spheres in their final positions gives Dividing the second equation by the first and solving for the final force leads to Inserting the known quantities yields 18.8 18.9 18.10 18.11 The force acts along the line joining the centers of the spheres. Because the same type of charge is on each sphere, the force is repulsive. Discussion As expected, the force between the charges is greater when they are 3.0 cm apart than when they are 5.0 cm apart. Note that although it is a good habit to convert cm to m (because the constant kis in SI units), it is not necessary in this problem, because the distances cancel out. We can also solve for the second unknown . By using the first equation, we find 566 Chapter 18 • Static Electricity 18.12 Note how the units cancel in the second-to-last line. Had we not converted cm to m, this would not occur, and the result would be incorrect. Finally, because the charge on each sphere is the same, we can further deduce that 18.13 WORKED EXAMPLE Using Coulomb’s law to find the distance between charged objects An engineer measures the force between two ink drops by measuring their acceleration and their diameter. She finds that each member of a pair of ink drops exerts a repulsive force of on its partner. If each ink drop carries a charge , how far apart are the ink drops? STRATEGY We know the force and the charge on each ink drop, so we can solve Coulomb’s law for the distance rbetween the ink drops. Do not forget to convert the force into SI units: Solution The charges in Coulomb’s law are so the numerator in Coulomb’s law takes the form . Inserting this into Coulomb’s law and solving for the distance rgives 18.14 or 130 microns (about one-tenth of a millimeter). Discussion The plus-minus sign means that we do not know which ink drop is to the right and which is to the left, but that is not important, because both ink drops are the same. Practice Problems 11. A charge of −4 × 10−9 C is a distance of 3 cm from a charge of 3 × 10−9 C . What is the magnitude and direction of the force between them? 1.2 × 10−4 N, and the force is attractive a. 1.2 × 1014 N, and the force is attractive b. c. 6.74 × 1023 N, and the force is attractive d. −ŷ, and the force is attractive 12. Two charges are repelled by a force of 2.0 N. If the distance between them triples, what is the force between the charges? a. 0.22 N b. 0.67 N c. 2.0 N 18.0 N d. Access for free at openstax.org. 18.3 • Electric Field 567 Check Your Understanding 13. How are electrostatic force and charge related? a. The force is proportional to the product of two charges. b. The force is inversely proportional to the product of two charges. c. The force is proportional to any one of the charges between which the force is acting. d. The force is inversely proportional to any one of the charges between which the force is acting. 14. Why is Coulomb’s law called an inverse-square law? a. because the force is proportional to the inverse of the distance squared between charges b. because the force is proportional to the product of two charges c. because the force is proportional to the inverse of the product of two charges d. because the force is proportional to the distance squared between charges 18.3 Electric Field Section Learning Objectives By the end of this section, you will be able to do the following: • Calculate the strength of an electric field • Create and interpret drawings of electric fields Section Key Terms electric field test charge You may have heard of a force fieldin science fiction movies, where such fields apply forces at particular positions in space to keep a villain trapped or to protect a spaceship from enemy fire. The concept of a fieldis very useful in physics, although it differs somewhat from what you see in movies. A fieldis a way of conceptualizing and mapping the force that surrounds any object and acts on another object at a distance without apparent physical connection. For example, the gravitational field surrounding Earth and all other masses represents the gravitational force that would be experienced if another mass were placed at a given point within the field. Michael Faraday, an English physicist of the nineteenth century, proposed the concept of an electric field. If you know the electric field, then you can easily calculate the force (magnitude and direction) applied to any electric charge that you place in the field. An electric field is generated by electric charge and tells us the force per unit charge at all locations in space around a charge distribution. The charge distribution could be a single point charge; a distribution of charge over, say, a flat plate; or a more complex distribution of charge. The electric field extends into space around the charge distribution. Now consider placing a test charge in the field. A test charge is a positive electric charge whose charge is so small that it does not significantly disturb the charges that create the electric field. The electric field exerts a force on the test charge in a given direction. The force exerted is proportional to the charge of the test charge. For example, if we double the charge of the test charge, the force exerted on it doubles. Mathematically, saying that electric field is the force per unit charge is written as 18.15 where we are considering only electric forces. Note that the electric field is a vector field that points in the same direction as the force on the positive test charge. The units of electric field are N/C. If the electric field is created by a point charge or a sphere of uniform charge, then the magnitude of the force between this point charge Qand the test charge is given by Coulomb’s law where the absolute value is used, because we only consider the magnitude of the force. The magnitude of the electric field is then 568 Chapter 18 • Static Electricity 18.16 This equation gives the magnitude of the electric field created by a point charge Q. The distance rin the denominator is the distance from the point charge, Q, or from the center of a spherical charge, to the point of interest. If the test charge is removed from the electric field, the electric field still exists. To create a three-dimensional map of the electric field, imagine placing the test charge in various locations in the field. At each location, measure the force on the charge, and use the vector equation to calculate the electric field. Draw an arrow at each point where you place the test charge to represent the strength and the direction of the electric field. The length of the arrows should be proportional to the strength of the electric field. If you join together these arrows, you obtain lines. Figure 18.17 shows an image of the threedimensional electric field created by a positive charge. Figure 18.17 Three-dimensional representation of the electric field generated by a positive charge. Just drawing the electric field lines in a plane that slices through the charge gives the two-dimensional electric-field maps shown in Figure 18.18. On the left is the electric field created by a positive charge, and on the right is the electric field created by a negative charge. Notice that the electric field lines point away from the positive charge and toward the negative charge. Thus, a positive test charge placed in the electric field of the positive charge will be repelled. This is consistent with Coulomb’s law, which says that like charges repel each other. If we place the positive charge in the electric field of the negative charge, the positive charge is attracted to the negative charge. The opposite is true for negative test charges. Thus, the direction of the electric field lines is consistent with what we find by using Coulomb’s law. says that the electric field gets stronger as we approach the charge that generates it. For example, at The equation 2 cm from the charge Q(r= 2 cm), the electric field is four times stronger than at 4 cm from the charge (r= 4 cm). Looking at Figure 18.17 and Figure 18.18 again, we see that the electric field lines become denser as we approach the charge that generates it. In fact, the density of the electric field lines is proportional to the strength of the electric field! Figure 18.18 Electric field lines from two point charges. The red point on the left carries a charge of +1 nC, and the blue point on the right carries a charge of –1 nC. The arrows point in the direction that a positive test charge would move. The field lines are denser as you approach the point charge. Electric-field maps can be made for several charges or for more complicated charge distributions. The electric field due to multiple charges may be found by adding together the electric field from each individual charge. Because this sum can only be a single number, we know that only a single electric-field line can go through any given point. In other words, electric-field lines cannotcross each other. Figure 18.19(a) shows a two-dimensional map of the electric field generated by a charge of +qand a nearby charge of −q. The three-dimensional version of this map is obtained by rotating this map about the axis that goes through both charges. A positive Access for free at openstax.org. test charge placed in this field would experience a force in the direction of the field lines at its location. It would thus be repelled from the positive charge and attracted to the negative charge. Figure 18.19(b) shows the electric field generated by two charges of −q. Note how the field lines tend to repel each other and do not overlap. A positive test charge placed in this field would be attracted to both charges. If you are far from these two charges, where far means much farther than the distance between the charges, the electric field looks like the electric field from a single charge of −2q. 18.3 • Electric Field 569 Figure 18.19 (a) The electric field generated by a positive point ch |
arge (left) and a negative point charge of the same magnitude (right). (b) The electric field generated by two equal negative charges. Virtual Physics Probing an Electric Field Click to view content (http://www.openstax.org/l/28charge-field) This simulation shows you the electric field due to charges that you place on the screen. Start by clicking the top checkbox in the options panel on the right-hand side to show the electric field. Drag charges from the buckets onto the screen, move them around, and observe the electric field that they form. To see more precisely the magnitude and direction of the electric field, drag an electric-field sensor, or E-fieldsensor from the bottom bucket, and move it around the screen. GRASP CHECK Two positive charges are placed on a screen. Which statement describes the electric field produced by the charges? a. b. c. d. It is constant everywhere. It is zero near each charge. It is zero halfway between the charges. It is strongest halfway between the charges. WATCH PHYSICS Electrostatics (part 2): Interpreting electric field This video explains how to calculate the electric field of a point charge and how to interpret electric-field maps in general. Note 570 Chapter 18 • Static Electricity that the lecturer uses dfor the distance between particles instead of r. Note that the point charges are infinitesimally small, so all their charges are focused at a point. When larger charged objects are considered, the distance between the objects must be measured between the center of the objects. Click to view content (https://www.youtube.com/embed/0YOGrTNgGhE) GRASP CHECK True or false—If a point charge has electric field lines that point into it, the charge must be ositive. a. b. true false WORKED EXAMPLE What is the charge? Look at the drawing of the electric field in Figure 18.20. What is the relative strength and sign of the three charges? Figure 18.20 Map of electric field due to three charged particles. STRATEGY We know the electric field extends out from positive charge and terminates on negative charge. We also know that the number of electric field lines that touch a charge is proportional to the charge. Charge 1 has 12 fields coming out of it. Charge 2 has six field lines going into it. Charge 3 has 12 field lines going into it. Solution The electric-field lines come out of charge 1, so it is a positive charge. The electric-field lines go into charges 2 and 3, so they are negative charges. The ratio of the charges is of charge 2. . Thus, magnitude of charges 1 and 3 is twice that Discussion Although we cannot determine the precise charge on each particle, we can get a lot of information from the electric field regarding the magnitude and sign of the charges and where the force on a test charge would be greatest (or least). WORKED EXAMPLE Electric field from doorknob A doorknob, which can be taken to be a spherical metal conductor, acquires a static electricity charge of the electric field 1.0 cm in front of the doorknob? The diameter of the doorknob is 5.0 cm. STRATEGY Because the doorknob is a conductor, the entire charge is distributed on the outside surface of the metal. In addition, because the doorknob is assumed to be perfectly spherical, the charge on the surface is uniformly distributed, so we can treat the doorknob as if all the charge were located at the center of the doorknob. The validity of this simplification will be proved in a to indicate the outward direction later physics course. Now sketch the doorknob, and define your coordinate system. Use What is Access for free at openstax.org. perpendicular to the door, with at the center of the doorknob (as shown in the figure below). 18.3 • Electric Field 571 If the diameter of the doorknob is 5.0 cm, its radius is 2.5 cm. We want to know the electric field 1.0 cm from the surface of the doorknob, which is a distance from the center of the doorknob. We can use the equation to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge, which is negative in this case. Solution Inserting the charge gives and the distance into the equation 18.17 Because the charge is negative, the electric-field lines point toward the center of the doorknob. Thus, the electric field at is . Discussion This seems like an enormous electric field. Luckily, it takes an electric field roughly 100 times stronger ( air to break down and conduct electricity. Also, the weight of an adult is about feel a force on the protons in your hand as you reach for the doorknob? The reason is that your hand contains an equal amount of negative charge, which repels the negative charge in the doorknob. A very small force might develop from polarization in your hand, but you would never notice it. ) to cause , so why don’t you Practice Problems 15. What is the magnitude of the electric field from 20 cm from a point charge of q= 33 nC? 7.4 × 103 N/C 1.48 × 103 N / C 7.4 × 1012 N / C a. b. c. d. 0 16. A −10 nC charge is at the origin. In which direction does the electric field from the charge point at x+ 10 cm ? a. The electric field points away from negative charges. b. The electric field points toward negative charges. c. The electric field points toward positive charges. d. The electric field points away from positive charges. Check Your Understanding 17. When electric field lines get closer together, what does that tell you about the electric field? 572 Chapter 18 • Static Electricity a. The electric field is inversely proportional to the density of electric field lines. b. The electric field is directly proportional to the density of electric field lines. c. The electric field is not related to the density of electric field lines. d. The electric field is inversely proportional to the square root of density of electric field lines. 18. If five electric-field lines come out of a +5 nC charge, how many electric-field lines should come out of a +20 nC charge? a. five field lines 10 field lines b. c. 15 field lines d. 20 field lines 18.4 Electric Potential Section Learning Objectives By the end of this section, you will be able to do the following: • Explain the similarities and differences between electric potential energy and gravitational potential energy • Calculate the electric potential difference between two point charges and in a uniform electric field Section Key Terms electric potential electric potential energy As you learned in studying gravity, a mass in a gravitational field has potential energy, which means it has the potential to accelerate and thereby increase its kinetic energy. This kinetic energy can be used to do work. For example, imagine you want to use a stone to pound a nail into a piece of wood. You first lift the stone high above the nail, which increases the potential energy of the stone-Earth system—because Earth is so large, it does not move, so we usually shorten this by saying simply that the potential energy of the stone increases. When you drop the stone, gravity converts the potential energy into kinetic energy. When the stone hits the nail, it does work by pounding the nail into the wood. The gravitational potential energy is the work that a mass can potentially do by virtue of its position in a gravitational field. Potential energy is a very useful concept, because it can be used with conservation of energy to calculate the motion of masses in a gravitational field. Electric potential energy works much the same way, but it is based on the electric field instead of the gravitational field. By virtue of its position in an electric field, a charge has an electric potential energy. If the charge is free to move, the force due to the electric field causes it to accelerate, so its potential energy is converted to kinetic energy, just like a mass that falls in a gravitational field. This kinetic energy can be used to do work. The electric potential energy is the work that a charge can do by virtue of its position in an electric field. The analogy between gravitational potential energy and electric potential energy is depicted in Figure 18.21. On the left, the ballEarth system gains gravitational potential energy when the ball is higher in Earth's gravitational field. On the right, the twocharge system gains electric potential energy when the positive charge is farther from the negative charge. Access for free at openstax.org. 18.4 • Electric Potential 573 Figure 18.21 On the left, the gravitational field points toward Earth. The higher the ball is in the gravitational field, the higher the potential energy is of the Earth-ball system. On the right, the electric field points toward a negative charge. The farther the positive charge is from the negative charge, the higher the potential energy is of the two-charge system. Let’s use the symbol potential energy decreases. Conservation of energy tells us that the work done by the gravitational field to make the mass accelerate must equal the loss of potential energy of the mass. If we use the symbol to denote gravitational potential energy. When a mass falls in a gravitational field, its gravitational to denote this work, then where the minus sign reflects the fact that the potential energy of the ball decreases. The work done by gravity on the mass is 18.18 18.19 where Fis the force due to gravity, and are the initial and final positions of the ball, respectively. The negative sign is because gravity points down, which we consider to be the negative direction. For the constantgravitational field near Earth’s surface, . The change in gravitational potential energy of the mass is and 18.20 . in electric Note that is just the negative of the height hfrom which the mass falls, so we usually just write We now apply the same reasoning to a charge in an electric field to find the electric potential energy. The change potential energy is the work done by the electric field to move a charge qfrom an initial position ( ). The definition of |
work does not change, except that now the work is done by the electric field: to a final position electric field is . The change in electric potential energy of the charge is thus . For a charge that falls through a constantelectric field E, the force applied to the charge by the or 18.21 18.22 This equation gives the change in electric potential energy of a charge qwhen it moves from position constantelectric field E. to position in a Figure 18.22 shows how this analogy would work if we were close to Earth’s surface, where gravity is constant. The top image shows a charge accelerating due to a constant electric field. Likewise, the round mass in the bottom image accelerates due to a constant gravitation field. In both cases, the potential energy of the particle decreases, and its kinetic energy increases. 574 Chapter 18 • Static Electricity Figure 18.22 In the top picture, a mass accelerates due to a constant electric field. In the bottom picture, the mass accelerates due to a constant gravitational field. WATCH PHYSICS Analogy between Gravity and Electricity This video discusses the analogy between gravitational potential energy and electric potential energy. It reviews the concepts of work and potential energy and shows the connection between a mass in a uniform gravitation field, such as on Earth’s surface, and an electric charge in a uniform electric field. Click to view content (https://www.openstax.org/l/28grav-elec) If the electric field is not constant, then the equation energy becomes more involved. For example, consider the electric potential energy of an assembly of two point charges is not valid, and deriving the electric potential and of the same sign that are initially very far apart. We start by placing charge at the origin of our coordinate system. This in from very far away to a distance rfrom the center of charge takes no electrical energy, because there is no electric field at the origin (because charge charge of charge . The energy it takes to assemble these two charges can be recuperated if we let them fly apart again. Thus, the charges have potential energy when they are a distance rapart. It turns out that the electric a distance rapart is potential energy of a pair of point charges . This requires some effort, because the electric field applies a repulsive force on charge is very far away). We then bring and To recap, if charges are free to move, they can accumulate kinetic energy by flying apart, and this kinetic energy can be used to do work. The maximum amount of work the two charges can do (if they fly infinitely far from each other) is given by the equation above. and Notice that if the two charges have opposite signs, then the potential energy is negative. This means that the charges have more potential to do work when they are farapart than when they are at a distance rapart. This makes sense: Opposite charges attract, so the charges can gain more kinetic energy if they attract each other from far away than if they start at only a short distance apart. Thus, they have more potential to do work when they are far apart. Figure 18.23 summarizes how the electric potential energy depends on charge and separation. 18.23 Access for free at openstax.org. 18.4 • Electric Potential 575 Figure 18.23 The potential energy depends on the sign of the charges and their separation. The arrows on the charges indicate the direction in which the charges would move if released. When charges with the same sign are far apart, their potential energy is low, as shown in the top panel for two positive charges. The situation is the reverse for charges of opposite signs, as shown in the bottom panel. Electric Potential Recall that to find the force applied by a fixed charge Qon any arbitrary test charge q, it was convenient to define the electric field, which is the force per unit charge applied by Qon any test charge that we place in its electric field. The same strategy is used here with electric potential energy: We now define the electric potential V, which is the electric potential energy per unit charge. 18.24 Normally, the electric potential is simply called the potentialor voltage. The units for the potential are J/C, which are given the name volt(V) after the Italian physicist Alessandro Volta (1745–1827). From the equation distance rfrom a point charge , the electric potential a is 18.25 This equation gives the energy required per unit charge to bring a charge Mathematically, this is written as from infinity to a distance rfrom a point charge ∞ 18.26 Note that this equation actually represents a differencein electric potential. However, because the second term is zero, it is normally not written, and we speak of the electric potential instead of the electric potential difference, or we just say the potential difference, or voltage). Below, when we consider the electric potential energy per unit charge between two points not infinitely far apart, we speak of electric potential differenceexplicitly. Just remember that electric potential and electric potential difference are really the same thing; the former is used just when the electric potential energy is zero in either the initial or final charge configuration. Coming back now to the electric potential a distance rfrom a point charge we can drop the subscripts and simply write , note that can be any arbitrary point charge, so Now consider the electric potential near a group of charges q1, q2, and q3, as drawn in Figure 18.24. The electric potential is 18.27 576 Chapter 18 • Static Electricity derived by considering the electric field. Electric fields follow the principle of superposition and can be simply added together, so the electric potential from different charges also add together. Thus, the electric potential of a point near a group of charges is where 18.24. are the distances from the center of charges to the point of interest, as shown in Figure 18.28 Figure 18.24 The potential at the red point is simply the sum of the potentials due to each individual charge. Now let’s consider the electric potential in a uniform electric field. From the equation potential difference in going from to in a uniform electric field Eis , we see that the TIPS FOR SUCCESS Notice from the equation that the electric field can be written as 18.29 18.30 which means that the electric field has units of V/m. Thus, if you know the potential difference between two points, calculating the electric field is very simple—you simply divide the potential difference by the distance! Notice that a positive charge in a region with high potential will experience a force pushing it toward regions of lower potential. In this sense, potential is like pressure for fluids. Imagine a pipe containing fluid, with the fluid at one end of the pipe under high pressure and the fluid at the other end of the pipe under low pressure. If nothing prevents the fluid from flowing, it will flow from the high-pressure end to the low-pressure end. Likewise, a positive charge that is free to move will move from a region with high potential to a region with lower potential. WATCH PHYSICS Voltage This video starts from electric potential energy and explains how this is related to electric potential (or voltage). The lecturer calculates the electric potential created by a uniform electric field. Click to view content (https://www.openstax.org/l/28voltage) GRASP CHECK What is the voltage difference between the positions a. and in an electric field of ? Access for free at openstax.org. b. c. d. 18.4 • Electric Potential 577 LINKS TO PHYSICS Electric Animals Many animals generate and/or detect electric fields. This is useful for activities such as hunting, defense, navigation, communication, and mating. Because salt water is a relatively good conductor, electric fish have evolved in all the world’s oceans. These fish have intrigued humans since the earliest times. In the nineteenth century, parties were even organized where the main attraction was getting a jolt from an electric fish! Scientists also studied electric fish to learn about electricity. Alessandro Volta based his research that led to batteries in 1799 on electric fish. He even referred to batteries as artificial electric organs, because he saw them as imitations of the electric organs of electric fish. Animals that generate electricity are called electrogenicand those that detect electric fields are called electroreceptive. Most fish that are electrogenic are also electroreceptive. One of the most well-known electric fish is the electric eel (see Figure 18.25), which is both electrogenic and electroreceptive. These fish have three pairs of organs that produce the electric charge: the main organ, Hunter’s organ, and Sach’s organ. Together, these organs account for more than 80percent of the fish’s body. Electric eels can produce electric discharges of much greater voltage than what you would get from a standard wall socket. These discharges can stun or even kill their prey. They also use low-intensity discharges to navigate. The electric fields they generate reflect off nearby obstacles or animals and are then detected by electroreceptors in the eel’s skin. The three organs that produce electricity contain electrolytes, which are substances that ionize when dissolved in water (or other liquids). An ionized atom or molecule is one that has lost or gained at least one electron, so it carries a net charge. Thus, a liquid solution containing an electrolyte conducts electricity, because the ions in the solution can move if an electric field is applied. To produce large discharges, the main organ is used. It contains approximately 6,000 rows of electroplaques connected in a long chain. Connected this way, the voltage between electroplaques adds up, creating a large final voltage. Each electroplaque consists of a column of cells controlled by an excitor nerve. When triggered by the excitor nerve, the electroplaques allow ionized sodium to flow thro |
ugh them, creating a potential difference between electroplaques. These potentials add up, and a large current can flow through the electrolyte. This geometry is reflected in batteries, which also use stacks of plates to produce larger potential differences. Figure 18.25 An electric eel in its natural environment. (credit: Steven G. Johnson) GRASP CHECK If an electric eel produces 1,000 V, which voltage is produced by each electroplaque in the main organ? a. 0.17 mV b. c. d. 1.7 mV 17 mV 170 mV 578 Chapter 18 • Static Electricity WORKED EXAMPLE X-ray Tube Dentists use X-rays to image their patients’ teeth and bones. The X-ray tubes that generate X-rays contain an electron source separated by about 10 cm from a metallic target. The electrons are accelerated from the source to the target by a uniform electric field with a magnitude of about 100 kN/C, as drawn in Figure 18.26. When the electrons hit the target, X-rays are produced. (a) What is the potential difference between the electron source and the metallic target? (b) What is the kinetic energy of the electrons when they reach the target, assuming that the electrons start at rest? Figure 18.26 In an X-ray tube, a large current flows through the electron source, causing electrons to be ejected from the electron source. The ejected electrons are accelerated toward the target by the electric field. When they strike the target, X-rays are produced. STRATEGY FOR (A) Use the equation as point in the negative xdirection. This way, the force qand Eare negative. Thus, and the target position as . to find the potential difference given a constant electric field. Define the source position . To accelerate the electrons in the positive xdirection, the electric field must on the electrons will point in the positive xdirection, because both Solution for (a) and Using electron source and the target is , the equation tells us that the potential difference between the Discussion for (a) The potential difference is positive, so the energy per unit positive charge is higher at the target than at the source. This means that free positive charges would fall from the target to the source. However, electrons are negative charges, so they accelerate from the source toward the target, gaining kinetic energy as they go. STRATEGY FOR (B) Apply conservation of energy to find the final kinetic energy of the electrons. In going from the source to the target, the change The change in in electric potential energy plus the change in kinetic energy of the electrons must be zero, so electric potential energy for moving through a constant electric field is given by the equation 18.31 where the electric field is the change in kinetic energy is simply their final kinetic energy, so . . Because the electrons start at rest, their initial kinetic energy is zero. Thus, Solution for (b) Again gives and . The charge of an electron is . Conservation of energy 18.32 Access for free at openstax.org. Inserting the known values into the right-hand side of this equation gives 18.4 • Electric Potential 579 18.33 Discussion for (b) This is a very small energy. However, electrons are very small, so they are easy to accelerate, and this energy is enough to make . The result is that an electron go extremely fast. You can find their speed by using the definition of kinetic energy, the electrons are moving at more than 100 million miles per hour! WORKED EXAMPLE Electric Potential Energy of Doorknob and Dust Speck Consider again the doorknob from the example in the previous section. The doorknob is treated as a spherical conductor with a on its surface. What is the electric potential energy between the doorknob and a speck of uniform static charge dust carrying a charge at 1.0 cm from the front surface of the doorknob? The diameter of the doorknob is 5.0 cm. STRATEGY As we did in the previous section, we treat the charge as if it were concentrated at the center of the doorknob. Again, as you will be able to validate in later physics classes, we can make this simplification, because the charge is uniformly distributed over the surface of the spherical object. Make a sketch of the situation and define a coordinate system, as shown in the image below. We at the center of the doorknob. If the diameter of use the doorknob is 5.0 cm, its radius is 2.5 cm. Thus, the speck of dust 1.0 cm from the surface of the doorknob is a distance to indicate the outward direction perpendicular to the door, with from the center of the doorknob. To solve this problem, use the equation . Solution The charge on the doorknob is gives . The distance . Inserting these values into the equation and the charge on the speck of dust is Discussion The energy is negative, which means that the energy will decrease that is, get even morenegative as the speck of dust approaches the doorknob. This helps explain why dust accumulates on objects that carry a static charge. However, note that insulators normally collect more static charge than conductors, because any charge that accumulates on insulators cannot move about on the insulator to find a way to escape. They must simply wait to be removed by some passing moist speck of dust or other host. 18.34 580 Chapter 18 • Static Electricity Practice Problems 19. What is the electric potential 10 cm from a −10 nC charge? a. 9.0 × 102 V b. 9.0 × 103 V c. 9.0 × 104 V d. 9.0 × 105 V 20. An electron accelerates from 0 to 10 × 104 m/s in an electric field. Through what potential difference did the electron travel? The mass of an electron is 9.11 × 10–31 kg, and its charge is −1.60 × 10–19 C. a. 29 mV b. 290 mV c. 2,900 mV d. 29 V Check Your Understanding 21. Gravitational potential energy is the potential for two masses to do work by virtue of their positions with respect to each other. What is the analogous definition of electric potential energy? a. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to the origin point. b. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to infinity. c. Electric potential energy is the potential for two charges to do work by virtue of their positions with respect to each other. d. Electric potential energy is the potential for single charges to do work by virtue of their positions with respect to their final positions. 22. A negative charge is 10 m from a positive charge. Where would you have to move the negative charge to increase the potential energy of the system? a. The negative charge should be moved closer to the positive charge. b. The negative charge should be moved farther away from the positive charge. c. The negative charge should be moved to infinity. d. The negative charge should be placed just next to the positive charge. 18.5 Capacitors and Dielectrics Section Learning Objectives By the end of this section, you will be able to do the following: • Calculate the energy stored in a charged capacitor and the capacitance of a capacitor • Explain the properties of capacitors and dielectrics Section Key Terms capacitor dielectric Capacitors Consider again the X-ray tube discussed in the previous sample problem. How can a uniform electric field be produced? A single positive charge produces an electric field that points away from it, as in . This field is not uniform, because the space between the lines increases as you move away from the charge. However, if we combine a positive and a negative charge, we obtain the electric field shown in (a). Notice that, between the charges, the electric field lines are more equally spaced. What happens if we place, say, five positive charges in a line across from five negative charges, as in Figure 18.27? Now the region between the lines of charge contains a fairly uniform electric field. Access for free at openstax.org. 18.5 • Capacitors and Dielectrics 581 Figure 18.27 The red dots are positive charges, and the blue dots are negative charges. The electric-field direction is shown by the red arrows. Notice that the electric field between the positive and negative dots is fairly uniform. We can extend this idea even further and into two dimensions by placing two metallic plates face to face and charging one with positive charge and the other with an equal magnitude of negative charge. This can be done by connecting one plate to the positive terminal of a battery and the other plate to the negative terminal, as shown in Figure 18.28. The electric field between these charged plates will be extremely uniform. 582 Chapter 18 • Static Electricity Figure 18.28 Two parallel metal plates are charged with opposite charge, by connecting the plates to the opposite terminals of a battery. The magnitude of the charge on each plate is the same. Let’s think about the work required to charge these plates. Before the plates are connected to the battery, they are neutral—that is, they have zero net charge. Placing the first positive charge on the left plate and the first negative charge on the right plate requires very little work, because the plates are neutral, so no opposing charges are present. Now consider placing a second positive charge on the left plate and a second negative charge on the right plate. Because the first two charges repel the new arrivals, a force must be applied to the two new charges over a distance to put them on the plates. This is the definition of work, which means that, compared with the first pair, more work is required to put the second pair of charges on the plates. To place the third positive and negative charges on the plates requires yet more work, and so on. Where does this work come from? The battery! Its chemical potential energy is converted into the work required to separate the positive and negative charges. Although the battery does work, this work remains within the battery-plate system. Therefore, conservation of energy tells us that, if the potential energy of the battery decreases t |
o separate charges, the energy of another part of the system must increase by the same amount. In fact, the energy from the battery is stored in the electric field between the plates. This idea is analogous to considering that the potential energy of a raised hammer is stored in Earth’s gravitational field. If the gravitational field were to disappear, the hammer would have no potential energy. Likewise, if no electric field existed between the plates, no energy would be stored between them. If we now disconnect the plates from the battery, they will hold the energy. We could connect the plates to a lightbulb, for example, and the lightbulb would light up until this energy was used up. These plates thus have the capacity to store energy. For this reason, an arrangement such as this is called a capacitor. A capacitor is an arrangement of objects that, by virtue of their geometry, can store energy an electric field. Various real capacitors are shown in Figure 18.29. They are usually made from conducting plates or sheets that are separated by Access for free at openstax.org. an insulating material. They can be flat or rolled up or have other geometries. 18.5 • Capacitors and Dielectrics 583 The capacity of a capacitor is defined by its capacitance C, which is given by Figure 18.29 Some typical capacitors. (credit: Windell Oskay) 18.35 where Qis the magnitudeof the charge on each capacitor plate, and Vis the potential difference in going from the negative plate to the positive plate. This means that both Qand Vare always positive, so the capacitance is always positive. We can see from the equation for capacitance that the units of capacitance are C/V, which are called farads (F) after the nineteenth-century English physicist Michael Faraday. makes sense: A parallel-plate capacitor (like the one shown in Figure 18.28) the size of a football field The equation could hold a lot of charge without requiring too much work per unit charge to push the charge into the capacitor. Thus, Qwould be large, and Vwould be small, so the capacitance Cwould be very large. Squeezing the same charge into a capacitor the size of a fingernail would require much more work, so Vwould be very large, and the capacitance would be much smaller. Although the equation makes it seem that capacitance depends on voltage, in fact it does not. For a given capacitor, the ratio of the charge stored in the capacitor to the voltage difference between the plates of the capacitor always remains the same. Capacitance is determined by the geometry of the capacitor and the materials that it is made from. For a parallel-plate capacitor with nothing between its plates, the capacitance is given by 18.36 where Ais the area of the plates of the capacitor and dis their separation. We use nothing between its plates (in the next section, we’ll see what happens when this is not the case). The constant zerois called the permittivity of free space, and its value is instead of C, because the capacitor has read epsilon 18.37 Coming back to the energy stored in a capacitor, we can ask exactly how much energy a capacitor stores. If a capacitor is charged by putting a voltage Vacross it for example, by connecting it to a battery with voltage V—the electrical potential energy stored in the capacitor is 18.38 Notice that the form of this equation is similar to that for kinetic energy, . WATCH PHYSICS Where does Capacitance Come From? This video shows how capacitance is defined and why it depends only on the geometric properties of the capacitor, not on voltage or charge stored. In so doing, it provides a good review of the concepts of work and electric potential. Click to view content (https://www.openstax.org/l/28capacitance) 584 Chapter 18 • Static Electricity GRASP CHECK If you increase the distance between the plates of a capacitor, how does the capacitance change? a. Doubling the distance between capacitor plates will reduce the capacitance four fold. b. Doubling the distance between capacitor plates will reduce the capacitance two fold. c. Doubling the distance between capacitor plates will increase the capacitance two times. d. Doubling the distance between capacitor plates will increase the capacitance four times. Virtual Physics Charge your Capacitor Click to view content (http://www.openstax.org/l/28charge-cap) For this simulation, choose the tab labeled Introductionat the top left of the screen. You are presented with a parallel-plate capacitor connected to a variable-voltage battery. The battery is initially at zero volts, so no charge is on the capacitor. Slide the battery slider up and down to change the battery voltage, and observe the charges that accumulate on the plates. Display the capacitance, top-plate charge, and stored energy as you vary the battery voltage. You can also display the electric-field lines in the capacitor. Finally, probe the voltage between different points in this circuit with the help of the voltmeter, and probe the electric field in the capacitor with the help of the electric-field detector. GRASP CHECK True or false— In a capacitor, the stored energy is always positive, regardless of whether the top plate is charged with negative or positive charge. a. b. false true WORKED EXAMPLE Capacitance and Charge Stored in a Parallel Plate Capacitor (a) What is the capacitance of a parallel-plate capacitor with metal plates, each of area 1.00 m2, separated by 0.0010 m? (b) What charge is stored in this capacitor if a voltage of 3.00 × 103 V is applied to it? STRATEGY FOR (A) Use the equation . Solution for (a) Entering the given values into this equation for the capacitance of a parallel-plate capacitor yields 18.39 Discussion for (a) This small value for the capacitance indicates how difficult it is to make a device with a large capacitance. Special techniques help, such as using very-large-area thin foils placed close together or using a dielectric (to be discussed below). STRATEGY FOR (B) Knowing C, find the charge stored by solving the equation , for the charge Q. Access for free at openstax.org. Solution for (b) The charge Qon the capacitor is 18.5 • Capacitors and Dielectrics 585 18.40 Discussion for (b) This charge is only slightly greater than typical static electricity charges. More charge could be stored by using a dielectric between the capacitor plates. WORKED EXAMPLE What battery is needed to charge a capacitor? Your friend provides you with a STRATEGY Use the equation capacitor. To store to find the voltage needed to charge the capacitor. on this capacitor, what voltage battery should you buy? Solution Solving for the voltage gives gives . Inserting and Discussion Such a battery should be easy to procure. There is still a question of whether the battery contains enough energy to provide the desired charge. The equation allows us to calculate the required energy. 18.42 18.41 A typical commercial battery can easily provide this much energy. Practice Problems 23. What is the voltage on a 35 μF with 25 nC of charge? a. 8.75 × 10−13 V b. 0.71 × 10−3 V 1.4 × 10−3 V c. 1.4 × 103 V d. 24. Which voltage is across a 100 μF capacitor that stores 10 J of energy? a. −4.5 × 102 V b. 4.5 × 102 V c. ±4.5 × 102 V d. ±9 × 102 V Dielectrics Before working through some sample problems, let’s look at what happens if we put an insulating material between the plates of a capacitor that has been charged and then disconnected from the charging battery, as illustrated in Figure 18.30. Because the material is insulating, the charge cannot move through it from one plate to the other, so the charge Qon the capacitor does not change. An electric field exists between the plates of a charged capacitor, so the insulating material becomes polarized, as shown in the lower part of the figure. An electrically insulating material that becomes polarized in an electric field is called a dielectric. 586 Chapter 18 • Static Electricity Figure 18.30 shows that the negative charge in the molecules in the material shifts to the left, toward the positive charge of the capacitor. This shift is due to the electric field, which applies a force to the left on the electrons in the molecules of the dielectric. The right sides of the molecules are now missing a bit of negative charge, so their net charge is positive. Figure 18.30 The top and bottom capacitors carry the same charge Q. The top capacitor has no dielectric between its plates. The bottom capacitor has a dielectric between its plates. The molecules in the dielectric are polarized by the electric field of the capacitor. All electrically insulating materials are dielectrics, but some are betterdielectrics than others. A good dielectric is one whose molecules allow their electrons to shift strongly in an electric field. In other words, an electric field pulls their electrons a fair bit away from their atom, but they do not escape completely from their atom (which is why they are insulators). Figure 18.31 shows a macroscopic view of a dielectric in a charged capacitor. Notice that the electric-field lines in the capacitor with the dielectric are spaced farther apart than the electric-field lines in the capacitor with no dielectric. This means that the electric field in the dielectric is weaker, so it stores less electrical potential energy than the electric field in the capacitor with no dielectric. Access for free at openstax.org. 18.5 • Capacitors and Dielectrics 587 Where has this energy gone? In fact, the molecules in the dielectric act like tiny springs, and the energy in the electric field goes into stretching these springs. With the electric field thus weakened, the voltage difference between the two sides of the capacitor is smaller, so it becomes easier to put more charge on the capacitor. Placing a dielectric in a capacitor before charging it therefore allows more charge and potential energy to be stored in the capacitor. A parallel plate with a dielectric has a capacitance of 18.43 (kappa) is a dimensionless const |
ant called the dielectric constant. Because where capacitance increases when a dielectric is placed between the capacitor plates. The dielectric constant of several materials is shown in Table 18.1. is greater than 1 for dielectrics, the Material Dielectric Constant ( ) Vacuum 1.00000 Air 1.00059 Fused quartz 3.78 Neoprene rubber 6.7 Nylon Paper 3.4 3.7 Polystyrene 2.56 Pyrex glass Silicon oil 5.6 2.5 Strontium titanate 233 Teflon Water 2.1 80 Table 18.1 Dielectric Constants for Various Materials at 20 °C 588 Chapter 18 • Static Electricity Figure 18.31 The top and bottom capacitors carry the same charge Q. The top capacitor has no dielectric between its plates. The bottom capacitor has a dielectric between its plates. Because some electric-field lines terminate and start on polarization charges in the dielectric, the electric field is less strong in the capacitor. Thus, for the same charge, a capacitor stores less energy when it contains a dielectric. WORKED EXAMPLE Capacitor for Camera Flash A typical flash for a point-and-shoot camera uses a capacitor of about capacitor plates is 100 V—that is, 100 V is placed “across the capacitor,” how much energy is stored in the capacitor? (b) If the dielectric used in the capacitor were a 0.010-mm-thick sheet of nylon, what would be the surface area of the capacitor plates? STRATEGY FOR (A) . (a) If the potential difference between the , we can use the equation , to find the electric potential energy Given that stored in the capacitor. and Access for free at openstax.org. Solution for (a) Inserting the given quantities into gives 18.5 • Capacitors and Dielectrics 589 18.44 Discussion for (a) This is enough energy to lift a 1-kg ball about 1 m up from the ground. The flash lasts for about 0.001 s, so the power delivered by the capacitor during this brief time is power, this is not bad for a little capacitor! . Considering that a car engine delivers about 100 kW of STRATEGY FOR (B) Because the capacitor plates are in contact with the dielectric, we know that the spacing between the capacitor plates is . From the previous table, the dielectric constant of nylon is . We can now use the equation to find the area Aof the capacitor. Solution (b) Solving the equation for the area Aand inserting the known quantities gives 18.45 Discussion for (b) This is much too large an area to roll into a capacitor small enough to fit in a handheld camera. This is why these capacitors don’t use simple dielectrics but a more advanced technology to obtain a high capacitance. Practice Problems 25. With 12 V across a capacitor, it accepts 10 mC of charge. What is its capacitance? a. 0.83 μF b. 83 μF 120 μF c. d. 830 μF 26. A parallel-plate capacitor has an area of 10 cm2 and the plates are separated by 100 μm . If the capacitor contains paper between the plates, what is its capacitance? a. b. c. d. 3.3 × 10−10 F 3.3 × 10−8 F 3.3 × 10−6 F 3.3 × 10−4 F Check Your Understanding 27. If the area of a parallel-plate capacitor doubles, how is the capacitance affected? a. The capacitance will remain same. b. The capacitance will double. c. The capacitance will increase four times. d. The capacitance will increase eight times. 28. If you double the area of a parallel-plate capacitor and reduce the distance between the plates by a factor of four, how is the capacitance affected? 590 Chapter 18 • Static Electricity a. b. c. d. It will increase by a factor of two. It will increase by a factor of four. It will increase by a factor of six. It will increase by a factor of eight. Access for free at openstax.org. Chapter 18 • Key Terms 591 KEY TERMS capacitor arrangement of objects that can store electrical energy by virtue of their geometry conductor material through which electric charge can easily move, such as metals Coulomb’s law describes the electrostatic force between charged objects, which is proportional to the charge on each object and inversely proportional to the square of the distance between the objects of negative electric charge induction creating an unbalanced charge distribution in an object by moving a charged object toward it (but without touching) insulator material through which a charge does not move, such as rubber inverse-square law law that has the form of a ratio, with the denominator being the distance squared dielectric electrically insulating material that becomes law of conservation of charge states that total charge is polarized in an electric field constant in any process electric field defines the force per unit charge at all locations in space around a charge distribution polarization separation of charge induced by nearby excess charge electric potential the electric potential energy per unit proton subatomic particle that carries the same magnitude charge charge as the electron, but its charge is positive electric potential energy the work that a charge can do by test charge positive electric charge whose with a charge virtue of its position in an electric field electron subatomic particle that carries one indivisible unit magnitude so small that it does not significantly perturb any nearby charge distribution SECTION SUMMARY 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge • Electric charge is a conserved quantity, which means it can be neither created nor destroyed. • Electric charge comes in two varieties, which are called positiveand negative. • Charges with the same sign repel each other. Charges with opposite signs attract each other. • Charges can move easily in conducting material. Charges cannot move easily in an insulating material. • Objects can be charged in three ways: by contact, by conduction, and by induction. • Although a polarized object may be neutral, its electrical charge is unbalanced, so one side of the object has excess negative charge and the other side has an equal magnitude of excess positive charge. 18.2 Coulomb's law • Coulomb’s law is an inverse square law and describes the electrostatic force between particles. • The electrostatic force between charged objects is proportional to the charge on each object and inversely proportional to the distance squared between the objects. If Coulomb’s law gives a negative result, the force is attractive; if the result is positive, the force is repulsive. • 18.3 Electric Field • The electric field defines the force per unit charge in the space around a charge distribution. • For a point charge or a sphere of uniform charge, the electric field is inversely proportional to the distance from the point charge or from the center of the sphere. • Electric-field lines never cross each other. • More force is applied to a charge in a region with many electric field lines than in a region with few electric field lines. • Electric field lines start at positive charges and point away from positive charges. They end at negative charges and point toward negative charges. 18.4 Electric Potential • Electric potential energy is a concept similar to gravitational potential energy: It is the potential that charges have to do work by virtue of their positions relative to each other. • Electric potential is the electric potential energy per unit charge. • The potential is always measured between two points, where one point may be at infinity. • Positive charges move from regions of high potential to regions of low potential. • Negative charges move from regions of low potential to regions of high potential. 18.5 Capacitors and Dielectrics • The capacitance of a capacitor depends only on the geometry of the capacitor and the materials from which it is made. It does not depend on the voltage across the capacitor. • Capacitors store electrical energy in the electric field between their plates. 592 Chapter 18 • Key Equations • A dielectric material is an insulator that is polarized in • Putting a dielectric between the plates of a capacitor an electric field. increases the capacitance of the capacitor. KEY EQUATIONS 18.2 Coulomb's law Coulomb’s law 18.3 Electric Field electric field magnitude of electric field of point charge 18.4 Electric Potential change in electric potential energy for a charge that moves in a constant electric field electric potential energy of a charge a distance rfrom a point charge or sphere of uniform charge definition of electric potential change in electric potential for a charge that moves in a constant electric field electric potential of a charge a distance rfrom a point charge or sphere of uniform charge 18.5 Capacitors and Dielectrics capacitance energy stored in a capacitor capacitance of a parallel-plate capacitor CHAPTER REVIEW Concept Items 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 1. There are very large numbers of charged particles in most objects. Why, then, don’t most objects exhibit static electric effects? a. Most objects are neutral. b. Most objects have positive charge only. c. Most objects have negative charge only. d. Most objects have excess protons. 2. Can an insulating material be used to charge a conductor? If so, how? If not, why not? a. No, an insulator cannot charge a conductor by induction. b. No, an insulating material cannot charge a conductor. Access for free at openstax.org. c. Yes, an uncharged insulator can charge a conductor by induction. d. Yes, a charged insulator can charge a conductor upon contact. 3. True or false—A liquid can be an insulating material. a. b. true false 18.2 Coulomb's law 4. Two plastic spheres with uniform charge repel each other with a force of 10 N . If you remove the charge from one sphere, what will be the force between the spheres? a. The force will be 15 N. b. The force will be 10 N. c. The force will be 5 N. d. The force will be zero. 5. What creates a greater magnitude of force, two charges +qa distance r apart or two charges – qthe same distance apart? a. Two charges +qa distance raway b. Two charges −qa distance raway c. The magnitudes of forces are equal. 6. In Newton’s law of universal |
gravitation, the force between two masses is proportional to the product of the two masses. What plays the role of mass in Coulomb’s law? a. b. c. d. the electric charge the electric dipole the electric monopole the electric quadruple 18.3 Electric Field 7. Why can electric fields not cross each other? a. Many electric-field lines can exist at any given point in space. Chapter 18 • Chapter Review 593 b. true 10. True or false—The characteristics of an electric field make it analogous to the gravitational field near the surface of Earth. false a. true b. 11. An electron moves in an electric field. Does it move toward regions of higher potential or lower potential? Explain. a. It moves toward regions of higher potential because its charge is negative. It moves toward regions of lower potential because its charge is negative It moves toward regions of higher potential because its charge is positive. It moves toward regions of lower potential because its charge is positive. b. c. d. b. No electric-field lines can exist at any given point in 18.5 Capacitors and Dielectrics space. c. Only a single electric-field line can exist at any given point in space. d. Two electric-field lines can exist at the same point in space. 8. A constant electric field is (4.5 × 105 N/C)ŷ. In which direction is the force on a −20 nC charge placed in this field? a. The direction of the force is in the direction. direction. b. The direction of the force is in the c. The direction of the force is in the −ŷ direction. d. The direction of the force is in the +ŷ direction. 18.4 Electric Potential 9. True or false—The potential from a group of charges is the sum of the potentials from each individual charge. a. false Critical Thinking Items 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 15. If you dive into a pool of seawater through which an equal amount of positively and negatively charged particles is moving, will you receive an electric shock? a. Yes, because negatively charged particles are moving. b. No, because positively charged particles are moving. c. Yes, because positively and negatively charged particles are moving. 12. You insert a dielectric into an air-filled capacitor. How does this affect the energy stored in the capacitor? a. Energy stored in the capacitor will remain same. b. Energy stored in the capacitor will decrease. c. Energy stored in the capacitor will increase. d. Energy stored in the capacitor will increase first, and then it will decrease. 13. True or false— Placing a dielectric between the plates of a capacitor increases the energy of the capacitor. a. b. false true 14. True or false— The electric field in an air-filled capacitor is reduced when a dielectric is inserted between the plates. a. b. false true d. No, because equal amounts of positively and negatively charged particles are moving. 16. True or false—The high-voltage wires that you see connected to tall metal-frame towers are held aloft by insulating connectors, and these wires are wrapped in an insulating material. a. b. true false 17. By considering the molecules of an insulator, explain how an insulator can be overall neutral but carry a surface charge when polarized. a. Inside the insulator, the oppositely charged ends of 594 Chapter 18 • Chapter Review b. the molecules cancel each other. Inside the insulator, the oppositely charged ends of the molecules do not cancel each other. c. The electron distribution in all the molecules shifts in every possible direction, leaving an excess of positive charge on the opposite end of each molecule. d. The electron distribution in all the molecules shifts in a given direction, leaving an excess of negative charge on the opposite end of each molecule. 18.2 Coulomb's law 18. In terms of Coulomb’s law, why are water molecules attracted by positive and negative charges? a. Water molecules are neutral. b. Water molecules have a third type of charge that is attracted by positive as well as negative charges. c. Water molecules are polar. d. Water molecule have either an excess of electrons or an excess of protons. 19. A negative lightning strike occurs when a negatively charged cloud discharges its excess electrons to the positively charged ground. If you observe a cloud-tocloud lightning strike, what can you say about the charge on the area of the cloud struck by lightning? a. The area of the cloud that was struck by lightning had a positive charge. b. The area of the cloud that was struck by lightning had a negative charge. repelled or attracted. b. No, because an electrically neutral body can be attracted but not repelled. c. Yes, because an electrically neutral body can be repelled or attracted. d. Yes, because an electrically neutral body can be repelled. 18.4 Electric Potential 22. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential? a. Voltage is the energy per unit mass at some point in space. b. Voltage is the energy per unit length in space. c. Voltage is the energy per unit charge at some point in space. d. Voltage is the energy per unit area in space. 23. Three parallel plates are stacked above each other, with a separation between each plate. If the potential difference between the first two plates is ΔV1 and the potential between the second two plates is ΔV2, what is the potential difference between the first and the third plates? a. ΔV3 = ΔV2 + ΔV1 b. ΔV3 = ΔV2 − ΔV1 c. ΔV3 = ΔV2 / ΔV1 d. ΔV3 = ΔV2×ΔV1 c. The area of the cloud that was struck by lightning is 18.5 Capacitors and Dielectrics neutral. d. The area of the cloud that was struck by lightning had a third type of charge. 18.3 Electric Field 20. An arbitrary electric field passes through a box-shaped volume. There are no charges in the box. If 11 electricfield lines enter the box, how many electric-field lines must exit the box? a. nine electric field lines 10 electric field lines b. 11 electric field lines c. 12 electric field lines d. 21. In a science-fiction movie, a villain emits a radial electric field to repulse the hero. Knowing that the hero is electrically neutral, is this possible? Explain your reasoning. a. No, because an electrically neutral body cannot be 24. When you insert a dielectric into a capacitor, the energy stored in the capacitor decreases. If you take the dielectric out, the energy increases again. Where does this energy go in the former case, and where does the energy come from in the latter case? a. Energy is utilized to remove the dielectric and is released when the dielectric is introduced between the plates. b. Energy is released when the dielectric is added and is utilized when the dielectric is introduced between the plates. c. Energy is utilized to polarize the dielectric and is released when the dielectric is introduced between the plates. d. Energy is released to polarize the dielectric and is utilized when dielectric is introduced between the plates. Access for free at openstax.org. Problems 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 25. A dust particle acquires a charge of −13 nC. How many excess electrons does it carry? a. 20.8 × 10−28 electrons b. 20.8 × −19 electrons c. 8.1 × 1010 electrons d. 8.1 × 1019 electrons 26. Two identical conducting spheres are charged with a net charge of +5.0 qon the first sphere and a net charge of −8.0 qon the second sphere. The spheres are brought together, allowed to touch, and then separated. What is the net charge on each sphere now? a. −3.0q b. −1.5q c. +1.5q d. +3.0q 18.2 Coulomb's law 27. Two particles with equal charge experience a force of 10 nN when they are 30 cm apart. What is the magnitude of the charge on each particle? -5.8 × 10-10 C a. -3.2 × 10-10 C b. c. +3.2 × 10-10 C d. +1.4 × 10-5 C 28. Three charges are on a line. The left charge is q1 = 2.0 nC . The middle charge is q2 = 5.0 nC . The right charge is q3 = − 3.0 nC . The left and right charges are 2.0 cm from the middle charge. What is the force on the middle charge? a. −5.6 × 10−4 N to the left b. −1.12 × 10−4 N to the left c. +1.12 × 10−4 N to the right 5.6 × 10−4 N to the right d. 18.3 Electric Field 29. An electric field (15 N/C)ẑapplies a force (− 3 × 10–6 N)ẑ on a particle. What is the charge on the particle? a. −2.0 × 10–7 C b. 2.0 × 10–7 C Chapter 18 • Chapter Review 595 c. 2.0 × 10–8 C d. 2.0 × 10–9 C 30. Two uniform electric fields are superimposed. The first electric field is . The second electric . With respect to the positive field is xaxis, at which angle will a positive test charge accelerate in this combined field? a. 27° 54° b. c. 90° 108° d. 18.4 Electric Potential 31. You move a charge qfrom ri = 20 cm to rf = 40 cm from a fixed charge Q= 10 nC. What is the difference in potential for these two positions? a. −2.2 × 102 V b. −1.7 × 103 V c. −2.2 × 104 V d. −1.7 × 102 V 32. How much work is required from an outside agent to ? move an electron from xi = 0 to xf = 20 cm in an electric field a. b. c. d. 1.6 × 10−15 J 1.6 × 10−16 J 1.6 × 10−20 J 1.6 × 10−18 J 18.5 Capacitors and Dielectrics 33. A 4.12 µF parallel-plate capacitor has a plate area of 2,000 cm2 and a plate separation of 10 µm . What dielectric is between the plates? a. b. 466, the dielectric is strontium c. 699, the dielectric is strontium nitrate d. 1,000, the dielectric is strontium chloride 1, the dielectric is strontium titanate 34. What is the capacitance of a metal sphere of radius ? a. b. c. d. Performance Task 18.5 Capacitors and Dielectrics 35. Newton’s law of universal gravitation is where . This describes the gravitational force between two point masses m1 and m2. Coulomb’s law is 18.46 18.47 596 Chapter 18 • Test Prep . This describes the where electric force between two point charges q1 and q2. (a) Describe how the force in each case depends on the distance rbetween the objects. How do the forces change if the distance is reduced by half? If the distance is doubled? (b) D |
escribe the similarities and differences between the two laws. Consider the signs of the quantities that create the interaction (i.e., mass and charge), the constants G and k, and their dependence on separation r. (c) Given that the electric force is much stronger than TEST PREP Multiple Choice 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 36. A neutral hydrogen atom has one proton and one electron. If you remove the electron, what will be the leftover sign of the charge? a. negative b. positive c. zero d. neutral 37. What is the charge on a proton? a. +8.99 × 10–9 C b. −8.99 × 10–9 C c. + 1.60 × 10–19 C d. −1.60 × 10–19 C 38. True or false—Carbon is more conductive than pure water. a. b. true false 39. True or false—Two insulating objects are polarized. To cancel the polarization, it suffices to touch them together. true a. false b. 40. How is the charge of the proton related to the charge of the electron? a. The magnitudes of charge of the proton and the electron are equal, but the charge of the proton is positive, whereas the charge of the electron is negative. b. The magnitudes of charge of the proton and the electron are unequal, but the charge of the proton is positive, whereas the charge of the electron is negative. c. The magnitudes of charge of the proton and the electron are equal, but the charge of the proton is Access for free at openstax.org. the gravitational force, discuss why the law for gravitational force was discovered much earlier than the law for electric force. (d) Consider a hydrogen atom, which is a single proton orbited by a single electron. The electric force holds the electron and proton together so that the hydrogen atom has a radius of about force between electron and proton does not change, what would be the approximate radius of the hydrogen atom if . Assuming the ? negative, whereas the charge of the electron is positive. d. The magnitudes of charge of the proton and the electron are unequal, but the charge of the proton is negative, whereas the charge of the electron is positive. 18.2 Coulomb's law 41. If you double the distance between two point charges, by which factor does the force between the particles change? 1/2 a. b. 2 c. 4 d. 1/4 42. The combined charge of all the electrons in a dime is hundreds of thousands of coulombs. Because like charges repel, what keeps the dime from exploding? a. The dime has an equal number of protons, with positive charge. b. The dime has more protons than electrons, with positive charge. c. The dime has fewer protons than electrons, with positive charge. d. The dime is polarized, with electrons on one side and protons on the other side. 43. How can you modify the charges on two particles to quadruple the force between them without moving them? a. Increase the distance between the charges by a factor of two. Increase the distance between the charges by a factor of four. Increase the product of the charges by a factor of two Increase the product of the charges by a factor of four. b. c. d. 18.3 Electric Field 44. What is the magnitude of the electric field 12 cm from a charge of 1.5 nC ? a. 9.4 × 107 N/ C 1.1 × 102 N/C b. c. 9.4 × 102 N/C d. 9.4 × 10–2 N/C 45. A charge distribution has electric field lines pointing into it. What sign is the net charge? a. positive b. neutral c. final d. negative 46. If five electric field lines come out of point charge q1 and 10 electric-field lines go into point charge q2, what is the ratio q1/q2? a. –2 b. –1 c. –1/2 d. 0 47. True or false—The electric-field lines from a positive point charge spread out radially and point outward. a. b. false true 18.4 Electric Potential 48. What is the potential at 1.0 m from a point charge Q= − 25 nC? a. 6.6 × 102 V b. −2.3 × 102 V c. −6.6 × 102 V d. 2.3 × 102 V 49. Increasing the distance by a factor of two from a point charge will change the potential by a factor of how Short Answer 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 54. Compare the mass of the electron with the mass of the proton. a. The mass of the electron is about 1,000 times that of the proton. b. The mass of the proton is about 1,000 times that of the electron. c. The mass of the electron is about 1,836 times that of the proton. d. The mass of the proton is about 1,836 times that of the electron. Chapter 18 • Test Prep 597 much? a. 2 b. 4 c. d. 1/2 1/4 50. True or false—Voltageis the common word for potential difference, because this term is more descriptive than potential difference. a. b. false true 18.5 Capacitors and Dielectrics 51. Which magnitude of charge is stored on each plate of a 12 µF capacitor with 12 V applied across it? a. –1.0 × 10–6 C 1.0 × 10–6 C b. c. –1.4 × 10–4 C 1.4 × 10–4 C d. 52. What is the capacitance of a parallel-plate capacitor with an area of 200 cm2, a distance of 0.20 mm between the plates, and polystyrene as a dielectric? a. 2.3 nC b. 0.89 nC c. 23 nC d. 8.9 nC 53. Which factors determine the capacitance of a device? a. Capacitance depends only on the materials that make up the device. b. Capacitance depends on the electric field surrounding the device. c. Capacitance depends on the geometric and material parameters of the device. d. Capacitance depends only on the mass of the capacitor 55. The positive terminal of a battery is connected to one connection of a lightbulb, and the other connection of the lightbulb is connected to the negative terminal of the battery. The battery pushes charge through the circuit but does not become charged itself. Does this violate the law of conservation of charge? Explain. a. No, because this is a closed circuit. b. No, because this is an open circuit. c. Yes, because this is a closed circuit. d. Yes, because this is an open circuit. 56. Two flat pieces of aluminum foil lay one on top of the other. What happens if you add charge to the top piece of aluminum foil? a. The charge will distribute over the top of the top 598 Chapter 18 • Test Prep piece. b. The charge will distribute to the bottom of the bottom piece. c. The inner surfaces will have excess charge of the opposite sign. d. The inner surfaces will have excess charge of the same sign. 57. The students in your class count off consecutively so each student has a number. The odd-numbered students are told to act as negative charge, and the evennumbered students are told to act as positive charge. How would you organize them to represent a polarized material? a. The even-numbered and odd-numbered students will be arranged one after the other. b. Two even-numbered will be followed by two odd- numbered, and so on. c. Even-numbered students will be asked to come to the front, whereas odd-numbered students will be asked to go to the back of the class. d. Half even-numbered and odd-numbered will come to the front, whereas half even-numbered and oddnumbered will go to the back. 58. An ion of iron contains 56 protons. How many electrons must it contain if its net charge is +5e? a. five electrons 51 electrons b. c. 56 electrons d. 61 electrons 59. An insulating rod carries of charge. After rubbing it with a material, you find it carries charge. How much charge was transferred to it? a. b. c. d. of 60. A solid cube carries a charge of +8e. You measure the charge on each face of the cube and find that each face carries +0.5eof charge. Is the cube made of conducting or insulating material? Explain. a. The cube is made of insulating material, because all the charges are on the surface of the cube. b. The cube is made of conducting material, because some of the charges are inside the cube. c. The cube is made of insulating material, because all the charges are on the surface of the cube. d. The cube is made of insulating material, because some of the charges are inside the cube. 61. You have four neutral conducting spheres and a charging device that allows you to place charge qon any neutral object. You want to charge one sphere with a Access for free at openstax.org. charge q/2 and the other three with a charge q6 . How do you proceed? a. Charge one sphere with charge q. Touch it simultaneously to the three remaining neutral spheres. b. Charge one sphere with charge q. Touch it to one other sphere to produce two spheres with charge . Touch one of these spheres to one other neutral sphere. c. Charge one sphere with charge q. Touch it to one other sphere to produce two spheres with charge . Touch one of these spheres simultaneously to the two remaining neutral spheres. d. Charge one sphere with charge q. Touch it simultaneously to two other neutral spheres to produce three spheres with charge q/3. Touch one of these spheres to one other neutral sphere. 18.2 Coulomb's law 62. Why does dust stick to the computer screen? a. The dust is neutral. b. The dust is polarized. c. The dust is positively charged. d. The dust is negatively charged. 63. The force between two charges is 4 × 10–9 N . If the magnitude of one charge is reduced by a factor of two and the distance between the charges is reduced by a factor of two, what is the new force between the charges? a. 2 × 10–9 N b. 4 × 10–9 N c. 6 × 10–9 N d. 8 × 10–9 N 64. True or false—Coulomb’s constant is k= 8.99 × 109 N·m2/C2. Newton’s gravitational constant is G= 6.67 × 10−11 m3/kg⋅s2. This tells you about the relative strength of the electrostatic force versus that of gravity. a. b. true false 65. An atomic nucleus contains 56 protons, for iron. Which force would this nucleus apply on an electron at a distance of 10×10–12 m? a. 0.65 × 10–4 N b. 0.02 × 10–4 N 1.3 × 10–4 N c. 72.8 × 10–4 N d. 18.3 Electric Field 66. The electric field a distance of 10 km from a storm cloud is 1,000 N/C . What is the approximate charge in the cloud? a. 0.0011 C 11 C b. 110 C c. 1,100 C d. Chapter 18 • Test Prep 599 d. +400 mC 72. Given the potential difference between two points and the distance between the points, explain how to obtain the electric field between the points. a. Add the electric potential to the distance to |
obtain 67. Which electric field would produce a 10 N force in the the electric field. +x- direction on a charge of – 10 nC ? a. − 1.0 × 109 N/C 1.0 × 109 N/C b. 1.0 × 1010 N/C c. 1.0 × 1011 N/C d. 68. A positive charge is located at x= 0 . When a negative charge is placed at x= 10 cm, what happens to the electric field lines between the charges? a. The electric field lines become denser between the charges. b. The electric field lines become denser between the charges. c. The electric field lines remains same between the charges. d. The electric field lines will be zero between the charges. 18.4 Electric Potential 69. The energy required to bring a charge q= − 8.8 nC from far away to 5.5 cm from a point charge Q is 13 mJ. What is the potential at the final position of q? a. −112 MV b. −1.5 MV c. −0.66 MV d. +1.5 MV b. Divide the electric potential by the distance to obtain the electric field. c. Multiply the electric potential and the distance to obtain the electric field. d. Subtract the electric potential from the distance to obtain the electric field. 18.5 Capacitors and Dielectrics 73. If you double the voltage across the plates of a capacitor, how is the stored energy affected? a. Stored energy will decrease two times. b. Stored energy will decrease four times. c. Stored energy will increase two times. d. Stored energy will increase four times. 74. A capacitor with neoprene rubber as the dielectric stores 0.185 mJ of energy with a voltage of 50 V across the plates. If the area of the plates is 500 cm2, what is the plate separation? a. 20 µm b. 20 m c. 80 µm d. 80 m 75. Explain why a storm cloud before a lightning strike is like a giant capacitor. a. The storm cloud acts as a giant charged capacitor, 70. How is electric potential related to electric potential as it can store a large amount of charge. energy? a. Electric potential is the electric potential energy per unit mass at a given position in space. b. Electric potential is the electric potential energy per unit length at a given position in space. This relation is not dimensionally correct. c. Electric potential is the electric potential energy per unit area in space. d. Electric potential is the electric potential energy per unit charge at a given position in space. 71. If it takes 10 mJ to move a charge qfrom xi = 25 cm to xf = − 25 cm in an electric field of charge q? a. −1.0 mC b. +0.25 mC c. + 1.0 mC what is the b. The storm cloud acts as a giant charged capacitor, as it contains a high amount of excess charges. c. The storm cloud acts as a giant charged capacitor, as it splits in two capacitor plates with equal and opposite charge. d. The storm cloud acts as a giant charged capacitor, as it splits in two capacitor plates with unequal and opposite charges. 76. A storm cloud is 2 km above the surface of Earth. The lower surface of the cloud is approximately 2 km2 in area. What is the approximate capacitance of this storm cloud-Earth system? a. 9 × 10–15 F b. 9 × 10-9 F c. d. 17.7 × 10-15 F 17.7 × 10-9 F 600 Chapter 18 • Test Prep Extended Response 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge 77. Imagine that the magnitude of the charge on the electron differed very slightly from that of the proton. How would this affect life on Earth and physics in general? a. Many macroscopic objects would be charged, so we would experience the enormous force of electricity on a daily basis. b. Many macroscopic objects would be charged, so we would experience the small force of electricity on a daily basis. c. Many macroscopic objects would be charged, but it would not affect life on Earth and physics in general. d. Macroscopic objects would remain neutral, so it would not affect life on Earth and physics in general. 78. True or false—Conservation of charge is like balancing a budget. a. b. true false 79. True or false—Although wood is an insulator, lightning can travel through a tree to reach Earth. a. b. true false touched to a second small metal sphere that is initially neutral. The spheres are then placed 20 cm apart. What is the force between the spheres? 1.02 × 10−7 N a. b. 2.55 × 10−7 N 5.1 × 10−7 N c. d. 20.4 × 10−7 N 18.3 Electric Field 83. Point charges are located at each corner of a square with sides of 5.0 cm . The top-left charge is q1 = 8.0 nC The top right charge is q2 = 4.0 nC. The bottom-right charge is q3 = 4.0 nC. The bottom-left charge is q4 = 8.0 nC. What is the electric field at the point midway between charges q2 and q3? a. b. c. d. 84. A long straight wire carries a uniform positive charge distribution. Draw the electric field lines in a plane containing the wire at a location far from the ends of the wire. Do not worry about the magnitude of the charge on the wire. a. Take the wire on the x-axis, and draw electric-field lines perpendicular to it. b. Take the wire on the x-axis, and draw electric-field lines parallel to it. 80. True or false—An eccentric inventor attempts to levitate c. Take the wire on the y-axis, and draw electric-field by first placing a large negative charge on himself and then putting a large positive charge on the ceiling of his workshop. Instead, while he attempts to place a large negative charge on himself, his clothes fly off. a. b. true false 18.2 Coulomb's law 81. Electrostatic forces are enormous compared to gravitational force. Why do you not notice electrostatic forces in everyday life, whereas you do notice the force due to gravity? a. Because there are two types of charge, but only one type of mass exists. b. Because there is only one type of charge, but two types of mass exist. c. Because opposite charges cancel each other, while gravity does not cancel out. d. Because opposite charges do not cancel each other, while gravity cancels out. 82. A small metal sphere with a net charge of 3.0 nC is lines along it. d. Take the wire on the z-axis, and draw electric-field lines along it. 18.4 Electric Potential 85. A square grid has charges of Q= 10 nC are each corner. The sides of the square at 10 cm . How much energy does it require to bring a q= 1.0 nC charge from very far away to the point at the center of this square? 1.3 × 10−6 J a. b. 2.5 × 10−6 J 3.8 × 10−6 J c. 5.1 × 10−6 J d. 86. How are potential difference and electric-field strength related for a constant electric field? a. The magnitude of electric-field strength is equivalent to the potential divided by the distance. b. The magnitude of electric-field strength is equivalent to the product of the electric potential and the distance. c. The magnitude of electric-field strength is Access for free at openstax.org. Chapter 18 • Test Prep 601 equivalent to the difference between magnitude of the electric potential and the distance. d. The magnitude of electric-field strength is equivalent to the sum of the magnitude of the electric potential and the distance. 88. Explain why capacitance should be inversely proportional to the separation between the plates of a capacitor. a. Capacitance is directly proportional to the electric field, which is inversely proportional to the distance between the capacitor plates. 18.5 Capacitors and Dielectrics b. Capacitance is inversely proportional to the electric 87. A 12 μF air-filled capacitor has 12 V across it. If the surface charge on each capacitor plate is σ= 7.2 mC / m2, what is the attractive force of one capacitor plate toward the other? a. 0.81 × 105 N b. 0.81 × 106 N 1.2 × 105 N c. 1.2 × 106 N d. field, which is inversely proportional to the distance between the capacitor plates. c. Capacitance is inversely proportional to the electric field, which is directly proportional to the distance between the capacitor plates. d. Capacitance is directly proportional to the electric field, which is directly proportional to the distance between the capacitor plates. 602 Chapter 18 • Test Prep Access for free at openstax.org. CHAPTER 19 Electrical Circuits Figure 19.1 Electric energy in massive quantities is transmitted from this hydroelectric facility, the Srisailam power station located along the Krishna River in India, by the movement of charge—that is, by electric current. (credit: Chintohere, Wikimedia Commons) Chapter Outline 19.1 Ohm's law 19.2 Series Circuits 19.3 Parallel Circuits 19.4 Electric Power The flicker of numbers on a handheld calculator, nerve impulses carrying signals of vision to the brain, an INTRODUCTION ultrasound device sending a signal to a computer screen, the brain sending a message for a baby to twitch its toes, an electric train pulling into a station, a hydroelectric plant sending energy to metropolitan and rural users—these and many other examples of electricity involve electric current, which is the movement of charge. Humanity has harnessed electricity, the basis of this technology, to improve our quality of life. Whereas the previous chapter concentrated on static electricity and the fundamental force underlying its behavior, the next two chapters will be devoted to electric and magnetic phenomena involving current. In addition to exploring applications of electricity, we shall gain new insights into the workings of nature. 604 Chapter 19 • Electrical Circuits 19.1 Ohm's law Section Learning Objectives By the end of this section, you will be able to do the following: • Describe how current is related to charge and time, and distinguish between direct current and alternating current • Define resistance and verbally describe Ohm’s law • Calculate current and solve problems involving Ohm’s law Section Key Terms alternating current ampere conventional current direct current electric current nonohmic ohmic Ohm’s law resistance Direct and Alternating Current Just as water flows from high to low elevation, electrons that are free to move will travel from a place with low potential to a place with high potential. A battery has two terminals that are at different potentials. If the terminals are connected by a conducting wire, an electric current (charges) will flo |
w, as shown in Figure 19.2. Electrons will then move from the low-potential terminal of the battery (the negativeend) through the wire and enter the high-potential terminal of the battery (the positiveend). Figure 19.2 A battery has a wire connecting the positive and negative terminals, which allows electrons to move from the negative terminal to the positive terminal. Electric current is the rate at which electric charge moves. A large current, such as that used to start a truck engine, moves a large amount very quickly, whereas a small current, such as that used to operate a hand-held calculator, moves a small amount of charge more slowly. In equation form, electric current Iis defined as is the amount of charge that flows past a given area and where The SI unit for electric current is the ampere (A), which is named in honor of the French physicist André-Marie Ampère (1775–1836). One ampere is one coulomb per second, or is the time it takes for the charge to move past the area. Electric current moving through a wire is in many ways similar to water current moving through a pipe. To define the flow of water through a pipe, we can count the water molecules that flow past a given section of the pipe. As shown in Figure 19.3, electric current is very similar. We count the number of electrical charges that flow past a section of a conductor; in this case, a wire. Access for free at openstax.org. 19.1 • Ohm's law 605 Figure 19.3 The electric current moving through this wire is the charge that moves past the cross-section A divided by the time it takes for this charge to move past the section A. Assume each particle qin Figure 19.3 carries a charge , in which case the total charge shown would be . If these charges move past the area Ain a time , then the current would be 19.1 Note that we assigned a positive charge to the charges in Figure 19.3. Normally, negative charges—electrons—are the mobile charge in wires, as indicated in Figure 19.2. Positive charges are normally stuck in place in solids and cannot move freely. However, because a positive current moving to the right is the same as a negative current of equal magnitude moving to the left, as shown in Figure 19.4, we define conventional current to flow in the direction that a positive charge would flow if it could move. Thus, unless otherwise specified, an electric current is assumed to be composed of positive charges. Also note that one Coulomb is a significant amount of electric charge, so 5 A is a very large current. Most often you will see current on the order of milliamperes (mA). Figure 19.4 (a) The electric field points to the right, the current moves to the right, and positive charges move to the right. (b) The equivalent situation but with negative charges moving to the left. The electric field and the current are still to the right. Snap Lab Vegetable Current This lab helps students understand how current works. Given that particles confined in a pipe cannot occupy the same space, pushing more particles into one end of the pipe will force the same number of particles out of the opposite end. This creates a current of particles. Find a straw and dried peas that can move freely in the straw. Place the straw flat on a table and fill the straw with peas. When you push one pea in at one end, a different pea should come out of the other end. This demonstration is a model for 606 Chapter 19 • Electrical Circuits an electric current. Identify the part of the model that represents electrons and the part of the model that represents the supply of electrical energy. For a period of 30 s, count the number of peas you can push through the straw. When finished, calculate the pea currentby dividing the number of peas by the time in seconds. Note that the flow of peas is based on the peas physically bumping into each other; electrons push each other along due to mutually repulsive electrostatic forces. GRASP CHECK Suppose four peas per second pass through a straw. If each pea carried a charge of be through the straw? a. The electric current would be the pea charge multiplied by b. The electric current would be the pea current calculated in the lab multiplied by c. The electric current would be the pea current calculated in the lab. d. The electric current would be the pea charge divided by time. . , what would the electric current . The direction of conventional current is the direction that positive charge would flow. Depending on the situation, positive charges, negative charges, or both may move. In metal wires, as we have seen, current is carried by electrons, so the negative charges move. In ionic solutions, such as salt water, both positively charged and negatively charged ions move. This is also true in nerve cells. Pure positive currents are relatively rare but do occur. History credits American politician and scientist Benjamin Franklin with describing current as the direction that positive charges flow through a wire. He named the type of charge associated with electrons negative long before they were known to carry current in so many situations. As electrons move through a metal wire, they encounter obstacles such as other electrons, atoms, impurities, etc. The electrons scatter from these obstacles, as depicted in Figure 19.5. Normally, the electrons lose energy with each interaction. 1 To keep the electrons moving thus requires a force, which is supplied by an electric field. The electric field in a wire points from the end of the wire at the higher potential to the end of the wire at the lower potential. Electrons, carrying a negative charge, move on average (or drift) in the direction opposite the electric field, as shown in Figure 19.5. Figure 19.5 Free electrons moving in a conductor make many collisions with other electrons and atoms. The path of one electron is shown. The average velocity of free electrons is in the direction opposite to the electric field. The collisions normally transfer energy to the conductor, so a constant supply of energy is required to maintain a steady current. So far, we have discussed current that moves constantly in a single direction. This is called direct current, because the electric charge flows in only one direction. Direct current is often called DCcurrent. Many sources of electrical power, such as the hydroelectric dam shown at the beginning of this chapter, produce alternating current, in which the current direction alternates back and forth. Alternating current is often called AC current. Alternating current moves back and forth at regular time intervals, as shown in Figure 19.6. The alternating current that comes from a normal wall socket does not suddenly switch directions. Rather, it increases smoothly up to a maximum current and then smoothly decreases back to zero. It then grows again, but in the opposite direction until it has reached the same maximum value. After that, it decreases smoothly back to zero, and the cycle starts over again. 1This energy is transferred to the wire and becomes thermal energy, which is what makes wires hot when they carry a lot of current. Access for free at openstax.org. 19.1 • Ohm's law 607 Figure 19.6 With alternating current, the direction of the current reverses at regular time intervals. The graph on the top shows the current versus time. The negative maxima correspond to the current moving to the left. The positive maxima correspond to current moving to the right. The current alternates regularly and smoothly between these two maxima. Devices that use AC include vacuum cleaners, fans, power tools, hair dryers, and countless others. These devices obtain the power they require when you plug them into a wall socket. The wall socket is connected to the power grid that provides an alternating potential (AC potential). When your device is plugged in, the AC potential pushes charges back and forth in the circuit of the device, creating an alternating current. Many devices, however, use DC, such as computers, cell phones, flashlights, and cars. One source of DC is a battery, which provides a constant potential (DC potential) between its terminals. With your device connected to a battery, the DC potential pushes charge in one direction through the circuit of your device, creating a DC current. Another way to produce DC current is by using a transformer, which converts AC potential to DC potential. Small transformers that you can plug into a wall socket are used to charge up your laptop, cell phone, or other electronic device. People generally call this a chargeror a battery, but it is a transformer that transforms AC voltage into DC voltage. The next time someone asks to borrow your laptop charger, tell them that you don’t have a laptop charger, but that they may borrow your converter. WORKED EXAMPLE Current in a Lightning Strike A lightning strike can transfer as many as average electric current in the lightning? STRATEGY electrons from the cloud to the ground. If the strike lasts 2 ms, what is the Use the definition of current, . The charge from electrons is , where is the number of electrons and is the charge on the electron. This gives The time is the duration of the lightning strike. Solution The current in the lightning strike is Discussion 19.2 19.3 608 Chapter 19 • Electrical Circuits The negative sign reflects the fact that electrons carry the negative charge. Thus, although the electrons flow from the cloud to the ground, the positive current is defined to flow from the ground to the cloud. WORKED EXAMPLE Average Current to Charge a Capacitor In a circuit containing a capacitor and a resistor, it takes 1 min to charge a 16 μF capacitor by using a 9-V battery. What is the average current during this time? STRATEGY We can determine the charge on the capacitor by using the definition of capacitance: . This gives a charge of a 9-V battery, the voltage across the capacitor will be . When the capacitor is charged by By inserting this expression for charge into the equation |
for current, , we can find the average current. Solution The average current is 19.4 19.5 Discussion This small current is typical of the current encountered in circuits such as this. Practice Problems 1. 10 nC of charge flows through a circuit in 3.0 × 10−6 s . What is the current during this time? a. The current passes through the circuit is 3.3 × 10−3 A. b. The current passes through the circuit is 30 A. c. The current passes through the circuit is 33 A. d. The current passes through the circuit is 0.3 A. 2. How long would it take a current to charge a capacitor with ? a. b. c. d. Resistance and Ohm’s Law As mentioned previously, electrical current in a wire is in many ways similar to water flowing through a pipe. The water current that can flow through a pipe is affected by obstacles in the pipe, such as clogs and narrow sections in the pipe. These obstacles slow down the flow of current through the pipe. Similarly, electrical current in a wire can be slowed down by many factors, including impurities in the metal of the wire or collisions between the charges in the material. These factors create a resistance to the electrical current. Resistance is a description of how much a wire or other electrical component opposes the flow of charge through it. In the 19th century, the German physicist Georg Simon Ohm (1787–1854) found experimentally that current through a conductor is proportional to the voltage drop across a current-carrying conductor. Access for free at openstax.org. 19.1 • Ohm's law 609 The constant of proportionality is the resistance Rof the material, which leads to This relationship is called Ohm’s law. It can be viewed as a cause-and-effect relationship, with voltage being the cause and the current being the effect. Ohm’s law is an empirical law like that for friction, which means that it is an experimentally observed phenomenon. The units of resistance are volts per ampere, or V/A. We call a V/A an ohm, which is represented by the uppercase Greek letter omega ( ). Thus, Ohm’s law holds for most materials and at common temperatures. At very low temperatures, resistance may drop to zero (superconductivity). At very high temperatures, the thermal motion of atoms in the material inhibits the flow of electrons, increasing the resistance. The many substances for which Ohm’s law holds are called ohmic. Ohmic materials include good conductors like copper, aluminum, and silver, and some poor conductors under certain circumstances. The resistance of ohmic materials remains essentially the same for a wide range of voltage and current. WATCH PHYSICS Introduction to Electricity, Circuits, Current, and Resistance This video presents Ohm’s law and shows a simple electrical circuit. The speaker uses the analogy of pressure to describe how electric potential makes charge move. He refers to electric potential as electric pressure. Another way of thinking about electric potential is to imagine that lots of particles of the same sign are crowded in a small, confined space. Because these charges have the same sign (they are all positive or all negative), each charge repels the others around it. This means that lots of charges are constantly being pushed towards the outside of the space. A complete electric circuit is like opening a door in the small space: Whichever particles are pushed towards the door now have a way to escape. The higher the electric potential, the harder each particle pushes against the others. GRASP CHECK , two resistors each with resistance If, instead of a single resistor what can you say about the current through the circuit? a. The amount of current through the circuit must decrease by half. b. The amount of current through the circuit must increase by half. c. The current must remain the same through the circuit. d. The amount of current through the circuit would be doubled. are drawn in the circuit diagram shown in the video, Virtual Physics Ohm’s Law Click to view content (http://www.openstax.org/l/28ohms_law) This simulation mimics a simple circuit with batteries providing the voltage source and a resistor connected across the batteries. See how the current is affected by modifying the resistance and/or the voltage. Note that the resistance is modeled as an element containing small scattering centers. These represent impurities or other obstacles that impede the passage of the current. GRASP CHECK In a circuit, if the resistance is left constant and the voltage is doubled (for example, from current change? Does this conform to Ohm’s law? a. The current will get doubled. This conforms to Ohm’s law as the current is proportional to the voltage. b. The current will double. This does not conform to Ohm’s law as the current is proportional to the voltage. to ), how does the 610 Chapter 19 • Electrical Circuits c. The current will increase by half. This conforms to Ohm’s law as the current is proportional to the voltage. d. The current will decrease by half. This does not conform to Ohm’s law as the current is proportional to the voltage. WORKED EXAMPLE Resistance of a Headlight What is the resistance of an automobile headlight through which 2.50 A flows when 12.0 V is applied to it? STRATEGY Ohm’s law tells us the battery, headlight. . The voltage drop in going through the headlight is just the voltage rise supplied by . We can use this equation and rearrange Ohm’s law to find the resistance of the Solution Solving Ohm’s law for the resistance of the headlight gives 19.6 Discussion This is a relatively small resistance. As we will see below, resistances in circuits are commonly measured in kW or MW. WORKED EXAMPLE Determine Resistance from Current-Voltage Graph Suppose you apply several different voltages across a circuit and measure the current that runs through the circuit. A plot of your results is shown in Figure 19.7. What is the resistance of the circuit? Access for free at openstax.org. 19.1 • Ohm's law 611 Figure 19.7 The line shows the current as a function of voltage. Notice that the current is given in milliamperes. For example, at 3 V, the current is 0.003 A, or 3 mA. STRATEGY The plot shows that current is proportional to voltage, which is Ohm’s law. In Ohm’s law ( proportionality is the resistance R. Because the graph shows current as a function of voltage, we have to rearrange Ohm’s law in that form: Figure 19.7, we can calculate the resistance R. . This shows that the slope of the line of Iversus Vis . Thus, if we find the slope of the line in ), the constant of Solution The slope of the line is the risedivided by the run. Looking at the lower-left square of the grid, we see that the line rises by 1 mA (0.001 A) and runs over a voltage of 1 V. Thus, the slope of the line is Equating the slope with and solving for Rgives or 1 k-ohm. Discussion 19.7 19.8 This resistance is greater than what we found in the previous example. Resistances such as this are common in electric circuits, as we will discover in the next section. Note that if the line in Figure 19.7 were not straight, then the material would not be ohmic and we would not be able to use Ohm’s law. Materials that do not follow Ohm’s law are called nonohmic. Practice Problems 3. If you double the voltage across an ohmic resistor, how does the current through the resistor change? a. The current will double. b. The current will increase by half. c. The current will decrease by half. d. The current will decrease by a factor of two. 4. The current through a resistor is . What is the voltage drop across the resistor? a. b. c. d. 612 Chapter 19 • Electrical Circuits Check Your Understanding 5. What is electric current? a. Electric current is the electric charge that is at rest. b. Electric current is the electric charge that is moving. c. Electric current is the electric charge that moves only from the positive terminal of a battery to the negative terminal. d. Electric current is the electric charge that moves only from a region of lower potential to higher potential. 6. What is an ohmic material? a. An ohmic material is a material that obeys Ohm’s law. b. An ohmic material is a material that does not obey Ohm’s law. c. An ohmic material is a material that has high resistance. d. An ohmic material is a material that has low resistance. 7. What is the difference between direct current and alternating current? a. Direct current flows continuously in every direction whereas alternating current flows in one direction. b. Direct current flows continuously in one direction whereas alternating current reverses its direction at regular time intervals. c. Both direct and alternating current flow in one direction but the magnitude of direct current is fixed whereas the magnitude of alternating current changes at regular intervals of time. d. Both direct and alternating current changes its direction of flow but the magnitude of direct current is fixed whereas the magnitude of alternating current changes at regular intervals of time. 19.2 Series Circuits Section Learning Objectives By the end of this section, you will be able to do the following: • Interpret circuit diagrams and diagram basic circuit elements • Calculate equivalent resistance of resistors in series and apply Ohm’s law to resistors in series and apply Ohm’s law to resistors in series Section Key Terms circuit diagram electric circuit equivalent resistance in series resistor steady state Electric Circuits and Resistors Now that we understand the concept of electric current, let’s see what we can do with it. As you are no doubt aware, the modern lifestyle relies heavily on electrical devices. These devices contain ingenious electric circuits, which are complete, closed pathways through which electric current flows. Returning to our water analogy, an electric circuit is to electric charge like a network of pipes is to water: The electric circuit guides electric charge from one point to the next, running the charge through various devices along the way to extract work or info |
rmation. Electric circuits are made from many materials and cover a huge range of sizes, as shown in Figure 19.8. Computers and cell phones contain electric circuits whose features can be as small as roughly a billionth of a meter (a nanometer, or pathways that guide the current in these devices are made by ultraprecise chemical treatments of silicon or other semiconductors. Large power systems, on the other hand, contain electric circuits whose features are on the scale of meters. These systems carry such large electric currents that their physical dimensions must be relatively large. ). The Access for free at openstax.org. 19.2 • Series Circuits 613 Figure 19.8 The photo on the left shows a chipthat contains complex integrated electric circuitry. Chips such as this are at the heart of devices such as computers and cell phones. The photograph on the right shows some typical electric circuitry required for high-power electric power transmission. The pathways that form electric circuits are made from a conducting material, normally a metal in macroscopic circuits. For example, copper wires inside your school building form the electrical circuits that power lighting, projectors, screens, speakers, etc. To represent an electric circuit, we draw circuit diagrams. We use lines and symbols to represent the elements in the circuit. A simple electric circuit diagram is shown on the left side of Figure 19.9. On the right side is an analogous water circuit, which we discuss below. Figure 19.9 On the left is a circuit diagram showing a battery (in red), a resistor (black zigzag element), and the current I. On the right is the analogous water circuit. The pump is like the battery, the sand filter is like the resistor, the water current is like the electrical current, and the reservoir is like the ground. There are many different symbols that scientists and engineers use in circuit diagrams, but we will focus on four main symbols: the wire, the battery or voltage source, resistors, and the ground. The thin black lines in the electric circuit diagram represent the pathway that the electric charge must follow. These pathways are assumed to be perfect conductors, so electric charge can move along these pathways without losing any energy. In reality, the wires in circuits are not perfect, but they come close enough for our purposes. The zigzag element labeled Ris a resistor, which is a circuit element that provides a known resistance. Macroscopic resistors are often color coded to indicate their resistance, as shown in Figure 19.10. The red element in Figure 19.9 is a battery, with its positive and negative terminals indicated; the longer line represents the positive terminal of the battery, and the shorter line represents the negative terminal. Note that the battery icon is not always colored red; this is done in Figure 19.9 just to make it easy to identify. Finally, the element labeled groundon the lower left of the circuit indicates that the circuit is connected to Earth, which is a large, essentially neutral object containing an infinite amount of charge. Among other things, the ground determines the potential of the negative terminal of the battery. Normally, the potential of the ground is defined to be zero: . This 614 Chapter 19 • Electrical Circuits means that the entire lower wire in Figure 19.10 is at a voltage of zero volts. Figure 19.10 Some typical resistors. The color bands indicate the value of the resistance of each resistor. The electric current in Figure 19.9 is indicted by the blue line labeled I. The arrow indicates the direction in which positive charge would flow in this circuit. Recall that, in metals, electrons are mobile charge carriers, so negative charges actually flow in the opposite direction around this circuit (i.e., counterclockwise). However, we draw the current to show the direction in which positive charge would move. On the right side of Figure 19.9 is an analogous water circuit. Water at a higher pressure leaves the top of the pump, which is like charges leaving the positive terminal of the battery. The water travels through the pipe, like the charges traveling through the wire. Next, the water goes through a sand filter, which heats up as the water squeezes through. This step is like the charges going through the resistor. When charges flow through a resistor, they do work to heat up the resistor. After flowing through the sand filter, the water has converted its potential energy into heat, so it is at a lower pressure. Likewise, the charges exiting the resistor have converted their potential energy into heat, so they are at a lower voltage. Recall that voltage is just potential energy per charge. Thus, water pressure is analogous to electric potential energy (i.e., voltage). Coming back to the water circuit again, we see that the water returns to the bottom of the pump, which is like the charge returning to the negative terminal of the battery. The water pump uses a source of energy to pump the water back up to a high pressure again, giving it the pressure required to go through the circuit once more. The water pump is like the battery, which uses chemical energy to increase the voltage of the charge up to the level of the positive terminal. The potential energy per charge at the positive terminal of the battery is the voltage rating of the battery. This voltage is like water pressure in the upper pipe. Just like a higher pressure forces water to move toward a lower pressure, a higher voltage forces electric charge to flow toward a lower voltage. The pump takes water at low pressure and does work on it, ejecting water at a higher pressure. Likewise, a battery takes charge at a low voltage, does work on it, and ejects charge at a higher voltage. Note that the current in the water circuit of Figure 19.9 is the same throughout the circuit. In other words, if we measured the number of water molecules passing a cross-section of the pipe per unit time at any point in the circuit, we would get the same answer no matter where in the circuit we measured. The same is true of the electrical circuit in the same figure. The electric current is the same at all points in this circuit, including inside the battery and in the resistor. The electric current neither speeds up in the wires nor slows down in the resistor. This would create points where too much or too little charge would be bunched up. Thus, the current is the same at all points in the circuit shown in Figure 19.9. Although the current is the same everywhere in both the electric and water circuits, the voltage or water pressure changes as you move through the circuits. In the water circuit, the water pressure at the pump outlet stays the same until the water goes through the sand filter, assuming no energy loss in the pipe. Likewise, the voltage in the electrical circuit is the same at all points in a given wire, because we have assumed that the wires are perfect conductors. Thus, as indicated by the constant red color of the upper wire in Figure 19.11, the voltage throughout this wire is constant at through the resistor, but once you reach the blue wire, the voltage stays at its new level of terminal of the battery (i.e., the blue terminal of the battery). . The voltage then drops as you go all the way to the negative Access for free at openstax.org. 19.2 • Series Circuits 615 Figure 19.11 The voltage in the red wire is constant at from the positive terminal of the battery to the top of the resistor. The voltage in the blue wire is constant at from the bottom of the resistor to the negative terminal of the battery. If we go from the blue wire through the battery to the red wire, the voltage increases from we go from the blue wire up through the resistor to the red wire, the voltage also goes from Ohm’s law, we can write to to . Likewise, if . Thus, using is measured from the bottom of the resistor to the top, meaning that the top of the resistor is at a higher Note that voltage than the bottom of the resistor. Thus, current flows from the top of the resistor or higher voltage to the bottom of the resistor or lower voltage. Virtual Physics Battery-Resistor Circuit Click to view content (http://www.openstax.org/l/21batteryresist) Use this simulation to better understand how resistance, voltage, and current are related. The simulation shows a battery with a resistor connected between the terminals of the battery, as in the previous figure. You can modify the battery voltage and the resistance. The simulation shows how electrons react to these changes. It also shows the atomic cores in the resistor and how they are excited and heat up as more current goes through the resistor. Draw the circuit diagram for the circuit, being sure to draw an arrow indicating the direction of the current. Now pick three spots along the wire. Without changing the settings, allow the simulation to run for 20 s while you count the number of electrons passing through that spot. Record the number on the circuit diagram. Now do the same thing at each of the other two spots in the circuit. What do you notice about the number of charges passing through each spot in 20 s? Remember that that current is defined as the rate that charges flow through the circuit. What does this mean about the current through the entire circuit? GRASP CHECK With the voltage slider, give the battery a positive voltage. Notice that the electrons are spaced farther apart in the left wire than they are in the right wire. How does this reflect the voltage in the two wires? a. The voltage between static charges is directly proportional to the distance between them. b. The voltage between static charges is directly proportional to square of the distance between them. c. The voltage between static charges is inversely proportional to the distance between them. d. The voltage between static charges is inversely proportional to square of the distance between them. Other possible circuit elements include |
capacitors and switches. These are drawn as shown on the left side of Figure 19.12. A switch is a device that opens and closes the circuit, like a light switch. It is analogous to a valve in a water circuit, as shown on the right side of Figure 19.12. With the switch open, no current passes through the circuit. With the switch closed, it becomes part of the wire, so the current passes through it with no loss of voltage. The capacitor is labeled C on the left of Figure 19.12. A capacitor in an electrical circuit is analogous to a flexible membrane in a 616 Chapter 19 • Electrical Circuits water circuit. When the switch is closed in the circuit of Figure 19.12, the battery forces electrical current to flow toward the capacitor, charging the upper capacitor plate with positive charge. As this happens, the voltage across the capacitor plates increases. This is like the membrane in the water circuit: When the valve is opened, the pump forces water to flow toward the membrane, making it stretch to store the excess water. As this happens, the pressure behind the membrane increases. Now if we open the switch, the capacitor holds the voltage between its plates because the charges have nowhere to go. Likewise, if we close the valve, the water has nowhere to go and the membrane maintains the water pressure in the pipe between itself and the valve. If the switch is closed for a long time in the electric circuit or if the valve is open for a long time in the water circuit, the current will eventually stop flowing because the capacitor or the membrane will have become completely charged. Each circuit is now in the steady state, which means that its characteristics do not change over time. In this case, the steady state is characterized by zero current, and this does not change as long as the switch or valve remains in the same position. In the steady state, no electrical current passes through the capacitor, and no water current passes through the membrane. The voltage difference between the capacitor plates will be the same as the battery voltage. In the water circuit, the pressure behind the membrane will be the same as the pressure created by the pump. Although the circuit in Figure 19.12 may seem a bit pointless because all that happens when the switch is closed is that the capacitor charges up, it does show the capacitor’s ability to store charge. Thus, the capacitor serves as a reservoir for charge. This property of capacitors is used in circuits in many ways. For example, capacitors are used to power circuits while batteries are being charged. In addition, capacitors can serve as filters. To understand this, let’s go back to the water analogy. Suppose you have a water hose and are watering your garden. Your friend thinks he’s funny, and kinksthe hose. While the hose is kinked, you experience no water flow. When he lets go, the water starts flowing again. If he does this really fast, you experience water-no water-water-no water, and that’s really no way to water your garden. Now imagine that the hose is filling up a big bucket, and you are watering from the bottom of the bucket. As long as you had water in your bucket to begin with and your friend doesn’t kink the water hose for too long, you would be able to water your garden without the interruptions. Your friend kinking the water hose is filteredby the big bucket’s supply of water, so it does not impact your ability to water the garden. We can think of the interruptions in the current (be it water or electrical current) as noise. Capacitors act in an analogous way as the water bucket to help filter out the noise. Capacitors have so many uses that it is very rare to find an electronic circuit that does not include some capacitors. Figure 19.12 On the left is an electrical circuit containing a battery, a switch, and a capacitor. On the left is the analogous water circuit with a pump, a valve, and a stretchable membrane. The pump is like the battery, the valve is like the switch, and the stretchable membrane is like the capacitor. When the switch is closed, electrical current flows as the capacitor charges and its voltage increases. Likewise in the water circuit, when the valve is open, water current flows as the stretchable membrane stretches and the water pressure behind it increases. Access for free at openstax.org. 19.2 • Series Circuits 617 WORK IN PHYSICS What It Takes to be an Electrical Engineer Physics is used in a wide variety of fields. One field that requires a very thorough knowledge of physics is electrical engineering. An electrical engineer can work on anything from the large-scale power systems that provide power to big cities to the nanoscale electronic circuits that are found in computers and cell phones (Figure 19.13). In working with power companies, you can be responsible for maintaining the power grid that supplies electrical power to large areas. Although much of this work is done from an office, it is common to be called in for overtime duty after storms or other natural events. Many electrical engineers enjoy this part of the job, which requires them to race around the countryside repairing high-voltage transformers and other equipment. However, one of the more unpleasant aspects of this work is to remove the carcasses of unfortunate squirrels or other animals that have wandered into the transformers. Other careers in electrical engineering can involve designing circuits for cell phones, which requires cramming some 10 billion transistors into an electronic chip the size of your thumbnail. These jobs can involve much work with computer simulations and can also involve fields other than electronics. For example, the 1-m-diameter lenses that are used to make these circuits (as of 2015) are so precise that they are shipped from the manufacture to the chip fabrication plant in temperature-controlled trucks to ensure that they are held within a certain temperature range. If they heat up or cool down too much, they deform ever so slightly, rendering them useless for the ultrahigh precision photolithography required to manufacture these chips. In addition to a solid knowledge of physics, electrical engineers must above all be practical. Consider, for example, how one corporation managed to launch some anti-ballistic missiles at the White Sands Missile Test Range in New Mexico in the 1960s. Before launch, the skin of the missile had to be at the same voltage as the rail from which it was launched. The rail was connected to the ground by a large copper wire connected to a stake driven into the sandy earth. The missile, however, was connected by an umbilical cord to the equipment in the control shed a few meters away, which was grounded via a different grounding circuit. Before launching the missile, the voltage difference between the missile skin and the rail had to be less than 2.5 V. After an especially dry spell of weather, the missile could not be launched because the voltage difference stood at 5 V. A group of electrical engineers, including the father of your author, stood around pondering how to reduce the voltage difference. The situation was resolved when one of the engineers realized that urine contains electrolytes and conducts electricity quite well. With that, the four engineers quickly resolved the problem by urinating on the rail spike. The voltage difference immediately dropped to below 2.5 V and the missile was launched on schedule. Figure 19.13 The systems that electrical engineers work on range from microprocessor circuits (left)] to missile systems (right). Virtual Physics Click to view content (http://www.openstax.org/l/21phetcirconstr) Amuse yourself by building circuits of all different shapes and sizes. This simulation provides you with various standard circuit elements, such as batteries, AC voltage sources, resistors, capacitors, light bulbs, switches, etc. You can connect these in any configuration you like and then see the result. Build a circuit that starts with a resistor connected to a capacitor. Connect the free side of the resistor to the positive terminal of a battery and the free side of the capacitor to the negative terminal of the battery. Click the reset dynamics button to see how the current flows starting with no charge on the capacitor. Now right click on the resistor to change its 618 Chapter 19 • Electrical Circuits value. When you increase the resistance, does the circuit reach the steady state more rapidly or more slowly? GRASP CHECK When the circuit has reached the steady state, how does the voltage across the capacitor compare to the voltage of the battery? What is the voltage across the resistor? a. The voltage across the capacitor is greater than the voltage of the battery. In the steady state, no current flows through this circuit, so the voltage across the resistor is zero. b. The voltage across the capacitor is smaller than the voltage of the battery. In the steady state, finite current flows through this circuit, so the voltage across the resistor is finite. c. The voltage across the capacitor is the same as the voltage of the battery. In the steady state, no current flows through this circuit, so the voltage across the resistor is zero. d. The voltage across the capacitor is the same as the voltage of the battery. In the steady state, finite current flows through this circuit, so the voltage across the resistor is finite. Resistors in Series and Equivalent Resistance Now that we have a basic idea of how electrical circuits work, let’s see what happens in circuits with more than one circuit element. In this section, we look at resistors in series. Components connected in series are connected one after the other in the same branch of a circuit, such as the resistors connected in series on the left side of Figure 19.14. Figure 19.14 On the left is an electric circuit with three resistors R1, R2, and R3 connected in series. On the right is an electric circuit with one resistor Requiv that is equivalen |
t to the combination of the three resistors R1, R2, and R3. We will now try to find a single resistance that is equivalent to the three resistors in series on the left side of Figure 19.14. An equivalent resistor is a resistor that has the same resistance as the combined resistance of a set of other resistors. In other words, the same current will flow through the left and right circuits in Figure 19.14 if we use the equivalent resistor in the right circuit. where Iis the current in According to Ohm’s law, the voltage drop Vacross a resistor when a current flows through it is amperes (A) and Ris the resistance in ohms ( ). Another way to think of this is that Vis the voltage necessary to make a current Iflow through a resistance R. Applying Ohm’s law to each resistor on the left circuit of Figure 19.14, we find that the voltage drop across , that across voltage output of the battery, that is . The sum of these voltages equals the , and that across is is is You may wonder why voltages must add up like this. One way to understand this is to go once around the circuit and add up the successive changes in voltage. If you do this around a loop and get back to the starting point, the total change in voltage should be zero, because you end up at the same place that you started. To better understand this, consider the analogy of going for a stroll through some hilly countryside. If you leave your car and walk around, then come back to your car, the total height you gained in your stroll must be the same as the total height you lost, because you end up at the same place as you started. Thus, the gravitational potential energy you gain must be the same as the gravitational potential energy you lose. The same reasoning holds for voltage in going around an electric circuit. Let’s apply this reasoning to the left circuit in Figure 19.14. We start just below the battery and move up through the battery, which contributes a voltage gainof resistors. The voltage dropsby . Next, we got through the in going through resistor in going through resistor in going through , and by , by 19.9 Access for free at openstax.org. 19.2 • Series Circuits 619 resistor . After going through resistor and set the sum equal to zero. This gives , we arrive back at the starting point, so we add up these four changes in voltage which is the same as the previous equation. Note that the minus signs in front of drops, whereas is a voltage rise. 19.10 are because these are voltage Ohm’s law tells us that gives , , and . Inserting these values into equation Applying this same logic to the right circuit in Figure 19.14 gives Dividing the equation by , we get 19.11 19.12 19.13 This shows that the equivalent resistance for a series of resistors is simply the sum of the resistances of each resistor. In general, Nresistors connected in series can be replaced by an equivalent resistor with a resistance of WATCH PHYSICS Resistors in Series This video discusses the basic concepts behind interpreting circuit diagrams and then shows how to calculate the equivalent resistance for resistors in series. Click to view content (https://www.openstax.org/l/02resistseries) GRASP CHECK True or false—In a circuit diagram, we can assume that the voltage is the same at every point in a given wire. a. b. false true WORKED EXAMPLE Calculation of Equivalent Resistance In the left circuit of the previous figure, suppose the voltage rating of the battery is 12 V, and the resistances are STRATEGY FOR (A) Use the equation for the equivalent resistance of resistors connected in series. Because the circuit has three resistances, we only need to keep three terms, so it takes the form . (a) What is the equivalent resistance? (b) What is the current through the circuit? 19.14 Solution for (a) Inserting the given resistances into the equation above gives 620 Chapter 19 • Electrical Circuits Discussion for (a) We can thus replace the three resistors with a single 20- resistor. STRATEGY FOR (B) Apply Ohm’s law to the circuit on the right side of the previous figure with the equivalent resistor of 20 . Solution for (b) The voltage drop across the equivalent resistor must be the same as the voltage rise in the battery. Thus, Ohm’s law gives 19.15 19.16 Discussion for (b) To check that this result is reasonable, we calculate the voltage drop across each resistor and verify that they add up to the voltage rating of the battery. The voltage drop across each resistor is Adding these voltages together gives which is the voltage rating of the battery. WORKED EXAMPLE 19.17 19.18 Determine the Unknown Resistance The circuit shown in figure below contains three resistors of known value and a third element whose resistance Given that the equivalent resistance for the entire circuit is 150 , what is the resistance ? is unknown. STRATEGY The four resistances in this circuit are connected in series, so we know that they must add up to give the equivalent resistance. We can use this to find the unknown resistance . Solution For four resistances in series, the equation for the equivalent resistance of resistors in series takes the form 19.19 Access for free at openstax.org. Solving for R3 and inserting the known values gives 19.3 • Parallel Circuits 621 19.20 Discussion The equivalent resistance of a circuit can be measured with an ohmmeter. This is sometimes useful for determining the effective resistance of elements whose resistance is not marked on the element. Check your Understanding 8. Figure 19.15 What circuit element is represented in the figure below? a. a battery b. a resistor c. a capacitor d. an inductor 9. How would a diagram of two resistors connected in series appear? a. b. c. d. 19.3 Parallel Circuits Section Learning Objectives By the end of this section, you will be able to do the following: • Interpret circuit diagrams with parallel resistors • Calculate equivalent resistance of resistor combinations containing series and parallel resistors Section Key Terms in parallel Resistors in Parallel In the previous section, we learned that resistors in series are resistors that are connected one after the other. If we instead combine resistors by connecting them next to each other, as shown in Figure 19.16, then the resistors are said to be connected in parallel. Resistors are in parallel when both ends of each resistor are connected directly together. Note that the tops of the resistors are all connected to the same wire, so the voltage at the top of the each resistor is the same. Likewise, the bottoms of the resistors are all connected to the same wire, so the voltage at the bottom of each resistor is the same. This means that the voltage drop across each resistor is the same. In this case, the voltage drop is the voltage rating Vof the battery, because the top and bottom wires connect to the positive and negative terminals of the battery, respectively. Although the voltage drop across each resistor is the same, we cannot say the same for the current running through each resistor. Thus, are not necessarily the same, because the resistors do not necessarily have the same 622 Chapter 19 • Electrical Circuits resistance. Note that the three resistors in Figure 19.16 provide three different paths through which the current can flow. This means that the equivalent resistance for these three resistors must be less than the smallest of the three resistors. To understand this, imagine that the smallest resistor is the only path through which the current can flow. Now add on the alternate paths by connecting other resistors in parallel. Because the current has more paths to go through, the overall resistance (i.e., the equivalent resistance) will decrease. Therefore, the equivalent resistance must be less than the smallest resistance of the parallel resistors. Figure 19.16 The left circuit diagram shows three resistors in parallel. The voltage Vof the battery is applied across all three resistors. The currents that flow through each branch are not necessarily equal. The right circuit diagram shows an equivalent resistance that replaces the three parallel resistors. To find the equivalent resistance voltage drop across each resistor is V, we obtain of the three resistors , we apply Ohm’s law to each resistor. Because the or 19.21 19.22 We also know from conservation of charge that the three currents through the battery. If this were not true, current would have to be mysteriously created or destroyed somewhere in the circuit, which is physically impossible. Thus, we have must add up to give the current Ithat goes Inserting the expressions for into this equation gives or This formula is just Ohm’s law, with the factor in parentheses being the equivalent resistance. Thus, the equivalent resistance for three resistors in parallel is 19.23 19.24 19.25 19.26 19.27 The same logic works for any number of resistors in parallel, so the general form of the equation that gives the equivalent resistance of Nresistors connected in parallel is 19.28 Access for free at openstax.org. 19.3 • Parallel Circuits 623 WORKED EXAMPLE Find the Current through Parallel Resistors The three circuits below are equivalent. If the voltage rating of the battery is the circuit and what current runs through the circuit? , what is the equivalent resistance of STRATEGY The three resistors are connected in parallel and the voltage drop across them is Vbattery. Thus, we can apply the equation for the equivalent resistance of resistors in parallel, which takes the form 19.29 The circuit with the equivalent resistance is shown below. Once we know the equivalent resistance, we can use Ohm’s law to find the current in the circuit. Solution Inserting the given values for the resistance into the equation for equivalent resistance gives The current through the circuit is thus 19.30 19.31 Discussion Although 0.62 A flows through the entire circuit, note that this current does not flow through each resistor. However, because 624 Chapter 19 • Electrical |
Circuits electric charge must be conserved in a circuit, the sum of the currents going through each branch of the circuit must add up to the current going through the battery. In other words, we cannot magically create charge somewhere in the circuit and add this new charge to the current. Let’s check this reasoning by using Ohm’s law to find the current through each resistor. 19.32 As expected, these currents add up to give 0.62 A, which is the total current found going through the equivalent resistor. Also, note that the smallest resistor has the largest current flowing through it, and vice versa. WORKED EXAMPLE Reasoning with Parallel Resistors Without doing any calculation, what is the equivalent resistance of three identical resistors Rin parallel? STRATEGY Three identical resistors Rin parallel make three identical paths through which the current can flow. Thus, it is three times easier for the current to flow through these resistors than to flow through a single one of them. Solution If it is three times easier to flow through three identical resistors Rthan to flow through a single one of them, the equivalent resistance must be three times less: R/3. Discussion Let’s check our reasoning by calculating the equivalent resistance of three identical resistors Rin parallel. The equation for the equivalent resistance of resistors in parallel gives 19.33 Thus, our reasoning was correct. In general, when more paths are available through which the current can flow, the equivalent resistance decreases. For example, if we have identical resistors Rin parallel, the equivalent resistance would be R/10. Practice Problems 10. Three resistors, 10, 20, and 30 Ω, are connected in parallel. What is the equivalent resistance? a. The equivalent resistance is 5.5 Ω b. The equivalent resistance is 60 Ω c. The equivalent resistance is 6 × 103 Ω d. The equivalent resistance is 6 × 104 Ω 11. If a drop occurs across a. Voltage drop across is b. Voltage drop across is c. Voltage drop across is d. Voltage drop across is , and . . . . is connected in parallel to , what is the voltage drop across ? Resistors in Parallel and in Series More complex connections of resistors are sometimes just combinations of series and parallel. Combinations of series and parallel resistors can be reduced to a single equivalent resistance by using the technique illustrated in Figure 19.17. Various parts are identified as either series or parallel, reduced to their equivalents, and further reduced until a single resistance is left. The Access for free at openstax.org. process is more time consuming than difficult. 19.3 • Parallel Circuits 625 Figure 19.17 This combination of seven resistors has both series and parallel parts. Each is identified and reduced to an equivalent resistance, and these are further reduced until a single equivalent resistance is reached. Let’s work through the four steps in Figure 19.17 to reduce the seven resistors to a single equivalent resistor. To avoid distracting . In step 1, we reduce the two sets of parallel resistors circled by the blue dashed loop. algebra, we’ll assume each resistor is 10 The upper set has three resistors in parallel and will be reduced to a single equivalent resistor resistors in parallel and will be reduced to a single equivalent resistor resistors in parallel, we obtain . Using the equation for the equivalent resistance of . The lower set has two These two equivalent resistances are encircled by the red dashed loop following step 1. They are in series, so we can use the 19.34 626 Chapter 19 • Electrical Circuits equation for the equivalent resistance of resistors in series to reduce them to a single equivalent resistance step 2, with the result being . This is done in The equivalent resistor pair can be replaced by the equivalent resistor , which is given by appears in the green dashed loop following step 2. This resistor is in parallel with resistor 19.35 , so the 19.36 This is done in step 3. The resistor These two resistors are combined in the final step to form the final equivalent resistor , as shown in the purple dashed loop following step 3. , which is is in series with the resistor 19.37 Thus, the entire combination of seven resistors may be replaced by a single resistor with a resistance of about 14.5 . That was a lot of work, and you might be asking why we do it. It’s important for us to know the equivalent resistance of the entire circuit so that we can calculate the current flowing through the circuit. Ohm’s law tells us that the current flowing through a circuit depends on the resistance of the circuit and the voltage across the circuit. But to know the current, we must first know the equivalent resistance. Here is a general approach to find the equivalent resistor for any arbitrary combination of resistors: Identify a group of resistors that are only in parallel or only in series. 1. 2. For resistors in series, use the equation for the equivalent resistance of resistors in series to reduce them to a single equivalent resistance. For resistors in parallel, use the equation for the equivalent resistance of resistors in parallel to reduce them to a single equivalent resistance. 3. Draw a new circuit diagram with the resistors from step 1 replaced by their equivalent resistor. 4. If more than one resistor remains in the circuit, return to step 1 and repeat. Otherwise, you are finished. FUN IN PHYSICS Robot Robots have captured our collective imagination for over a century. Now, this dream of creating clever machines to do our dirty work, or sometimes just to keep us company, is becoming a reality. Roboticshas become a huge field of research and development, with some technology already being commercialized. Think of the small autonomous vacuum cleaners, for example. Figure 19.18 shows just a few of the multitude of different forms robots can take. The most advanced humanoid robots can walk, pour drinks, even dance (albeit not very gracefully). Other robots are bio-inspired, such as the dogbotshown in the middle photograph of Figure 19.18. This robot can carry hundreds of pounds of load over rough terrain. The photograph on the right in Figure 19.18 shows the inner workings of an M-block,developed by the Massachusetts Institute of Technology. These simplelooking blocks contain inertial wheels and electromagnets that allow them to spin and flip into the air and snap together in a variety of shapes. By communicating wirelessly between themselves, they self-assemble into a variety of shapes, such as desks, chairs, and someday maybe even buildings. All robots involve an immense amount of physics and engineering. The simple act of pouring a drink has only recently been mastered by robots, after over 30 years of research and development! The balance and timing that we humans take for granted is in fact a very tricky act to follow, requiring excellent balance, dexterity, and feedback. To master this requires sensors to detect balance, computing power to analyze the data and communicate the appropriate compensating actions, and joints and actuators to implement the required actions. In addition to sensing gravity or acceleration, robots can contain multiple different sensors to detect light, sound, temperature, smell, taste, etc. These devices are all based on the physical principles that you are studying in this text. For example, the optics used for robotic vision are similar to those used in your digital cameras: pixelated semiconducting detectors in which light is Access for free at openstax.org. converted into electrical signals. To detect temperature, simple thermistors may be used, which are resistors whose resistance changes depending on temperature. Building a robot today is much less arduous than it was a few years ago. Numerous companies now offer kits for building robots. These range in complexity something suitable for elementary school children to something that would challenge the best professional engineers. If interested, you may find these easily on the Internet and start making your own robot today. 19.3 • Parallel Circuits 627 Figure 19.18 Robots come in many shapes and sizes, from the classic humanoidtype to dogbotsto small cubes that self-assemble to perform a variety of tasks. WATCH PHYSICS Resistors in Parallel This video shows a lecturer discussing a simple circuit with a battery and a pair of resistors in parallel. He emphasizes that electrons flow in the direction opposite to that of the positive current and also makes use of the fact that the voltage is the same at all points on an ideal wire. The derivation is quite similar to what is done in this text, but the lecturer goes through it well, explaining each step. Click to view content (https://www.openstax.org/l/28resistors) GRASP CHECK True or false—In a circuit diagram, we can assume that the voltage is the same at every point in a given wire. a. b. false true WATCH PHYSICS Resistors in Series and in Parallel This video shows how to calculate the equivalent resistance of a circuit containing resistors in parallel and in series. The lecturer uses the same approach as outlined above for finding the equivalent resistance. Click to view content (https://www.openstax.org/l/28resistorssp) GRASP CHECK Imagine connected Nidentical resistors in parallel. Each resistor has a resistance of R. What is the equivalent resistance for this group of parallel resistors? a. The equivalent resistance is (R)N. 628 Chapter 19 • Electrical Circuits b. The equivalent resistance is NR. c. The equivalent resistance is d. The equivalent resistance is WORKED EXAMPLE Find the Current through a Complex Resistor Circuit The battery in the circuit below has a voltage rating of 10 V. What current flows through the circuit and in what direction? STRATEGY Apply the strategy for finding equivalent resistance to replace all the resistors with a single equivalent resistance, then use Ohm’s law to find the current through the equivalent resistor. Soluti |
on The resistor combination and can be reduced to an equivalent resistance of 19.38 by 19.39 Replacing and with this equivalent resistance gives the circuit below. We now replace the two upper resistors their equivalent resistor and . These resistors are in series, so we add them together to find the equivalent resistance. and the two lower resistors by the equivalent resistor and Replacing the relevant resistors with their equivalent resistor gives the circuit below. Access for free at openstax.org. 19.3 • Parallel Circuits 629 Now replace the two resistors , which are in parallel, with their equivalent resistor . The resistance of is Updating the circuit diagram by replacing with this equivalent resistance gives the circuit below. Finally, we combine resistors , which are in series. The equivalent resistance is The final circuit is shown below. We now use Ohm’s law to find the current through the circuit. 19.40 19.41 The current goes from the positive terminal of the battery to the negative terminal of the battery, so it flows clockwise in this circuit. Discussion This calculation may seem rather long, but with a little practice, you can combine some steps. Note also that extra significant digits were carried through the calculation. Only at the end was the final result rounded to two significant digits. 630 Chapter 19 • Electrical Circuits WORKED EXAMPLE Strange-Looking Circuit Diagrams Occasionally, you may encounter circuit diagrams that are not drawn very neatly, such as the diagram shown below. This circuit diagram looks more like how a real circuit might appear on the lab bench. What is the equivalent resistance for the resistors in this diagram, assuming each resistor is 10 and the voltage rating of the battery is 12 V. STRATEGY Let’s redraw this circuit diagram to make it clearer. Then we’ll apply the strategy outlined above to calculate the equivalent resistance. Solution To redraw the diagram, consider the figure below. In the upper circuit, the blue resistors constitute a path from the positive terminal of the battery to the negative terminal. In parallel with this circuit are the red resistors, which constitute another path from the positive to negative terminal of the battery. The blue and red paths are shown more cleanly drawn in the lower circuit diagram. Note that, in both the upper and lower circuit diagrams, the blue and red paths connect the positive terminal of the battery to the negative terminal of the battery. Now it is easier to see that turn is in parallel with the series combination of . The equivalent resistance is are in parallel, and the parallel combination is in series with . This combination in . First, we calculate the blue branch, which contains 19.42 where we show the contribution from the parallel combination of resistors and from the series combination of resistors. We now calculate the equivalent resistance of the red branch, which is 19.43 Inserting these equivalent resistors into the circuit gives the circuit below. Access for free at openstax.org. 19.3 • Parallel Circuits 631 These two resistors are in parallel, so they can be replaced by a single equivalent resistor with a resistance of 19.44 The final equivalent circuit is show below. Discussion Finding the equivalent resistance was easier with a clear circuit diagram. This is why we try to make clear circuit diagrams, where the resistors in parallel are lined up parallel to each other and at the same horizontal position on the diagram. We can now use Ohm’s law to find the current going through each branch to this circuit. Consider the circuit diagram with and is . The voltage across each of these branches is 12 V (i.e., the voltage rating of the battery). The current in the blue branch 19.45 19.46 The current across the red branch is The current going through the battery must be the sum of these two currents (can you see why?), or 1.4 A. Practice Problems 12. What is the formula for the equivalent resistance of two parallel resistors with resistance R1 and R2? a. Equivalent resistance of two parallel resistors b. Equivalent resistance of two parallel resistors c. Equivalent resistance of two parallel resistors d. Equivalent resistance of two parallel resistors 632 Chapter 19 • Electrical Circuits 13. Figure 19.19 What is the equivalent resistance for the two resistors shown below? a. The equivalent resistance is 20 Ω b. The equivalent resistance is 21 Ω c. The equivalent resistance is 90 Ω d. The equivalent resistance is 1,925 Ω Check Your Understanding 14. The voltage drop across parallel resistors is ________. a. the same for all resistors b. greater for the larger resistors less for the larger resistors c. d. greater for the smaller resistors 15. Consider a circuit of parallel resistors. The smallest resistor is 25 Ω . What is the upper limit of the equivalent resistance? a. The upper limit of the equivalent resistance is 2.5 Ω. b. The upper limit of the equivalent resistance is 25 Ω. c. The upper limit of the equivalent resistance is 100 Ω. d. There is no upper limit. 19.4 Electric Power Section Learning Objectives By the end of this section, you will be able to do the following: • Define electric power and describe the electric power equation • Calculate electric power in circuits of resistors in series, parallel, and complex arrangements Section Key Terms electric power Power is associated by many people with electricity. Every day, we use electric power to run our modern appliances. Electric power transmission lines are visible examples of electricity providing power. We also use electric power to start our cars, to run our computers, or to light our homes. Power is the rate at which energy of any type is transferred; electric power is the rate at which electric energy is transferred in a circuit. In this section, we’ll learn not only what this means, but also what factors determine electric power. To get started, let’s think of light bulbs, which are often characterized in terms of their power ratings in watts. Let us compare a 25-W bulb with a 60-W bulb (see Figure 19.20). Although both operate at the same voltage, the 60-W bulb emits more light intensity than the 25-W bulb. This tells us that something other than voltage determines the power output of an electric circuit. Access for free at openstax.org. Incandescent light bulbs, such as the two shown in Figure 19.20, are essentially resistors that heat up when current flows through them and they get so hot that they emit visible and invisible light. Thus the two light bulbs in the photo can be considered as two different resistors. In a simple circuit such as a light bulb with a voltage applied to it, the resistance determines the current by Ohm’s law, so we can see that current as well as voltage must determine the power. 19.4 • Electric Power 633 Figure 19.20 On the left is a 25-W light bulb, and on the right is a 60-W light bulb. Why are their power outputs different despite their operating on the same voltage? The formula for power may be found by dimensional analysis. Consider the units of power. In the SI system, power is given in watts (W), which is energy per unit time, or J/s Recall now that a voltage is the potential energy per unit charge, which means that voltage has units of J/C We can rewrite this equation as and substitute this into the equation for watts to get 19.47 19.48 But a Coulomb per second (C/s) is an electric current, which we can see from the definition of electric current, Qis the charge in coulombs and tis time in seconds. Thus, equation above tells us that electric power is voltage times current, or , where This equation gives the electric power consumed by a circuit with a voltage drop of Vand a current of I. For example, consider the circuit in Figure 19.21. From Ohm’s law, the current running through the circuit is Thus, the power consumed by the circuit is Where does this power go? In this circuit, the power goes primarily into heating the resistor in this circuit. 19.49 19.50 Figure 19.21 A simple circuit that consumes electric power. In calculating the power in the circuit of Figure 19.21, we used the resistance and Ohm’s law to find the current. Ohm’s law gives the current: , which we can insert into the equation for electric power to obtain 634 Chapter 19 • Electrical Circuits This gives the power in terms of only the voltage and the resistance. We can also use Ohm’s law to eliminate the voltage in the equation for electric power and obtain an expression for power in terms of just the current and the resistance. If we write Ohm’s law as and use this to eliminate Vin the equation , we obtain This gives the power in terms of only the current and the resistance. Thus, by combining Ohm’s law with the equation terms of voltage and resistance and one in terms of current and resistance. Note that only resistance (not capacitance or anything else), current, and voltage enter into the expressions for electric power. This means that the physical characteristic of a circuit that determines how much power it dissipates is its resistance. Any capacitors in the circuit do not dissipate electric power—on the contrary, capacitors either store electric energy or release electric energy back to the circuit. for electric power, we obtain two more expressions for power: one in To clarify how voltage, resistance, current, and power are all related, consider Figure 19.22, which shows the formula wheel. The quantities in the center quarter circle are equal to the quantities in the corresponding outer quarter circle. For example, to express a potential V in terms of power and current, we see from the formula wheel that . Figure 19.22 The formula wheel shows how volts, resistance, current, and power are related. The quantities in the inner quarter circles equal the quantities in the corresponding outer quarter circles. WORKED EXAMPLE Find the Resistance of a Lightbulb A typical older incandescent lightbulb was 60 W. Assuming that 120 V is ap |
plied across the lightbulb, what is the current through the lightbulb? STRATEGY We are given the voltage and the power output of a simple circuit containing a lightbulb, so we can use the equation find the current Ithat flows through the lightbulb. to Solution Solving for the current and inserting the given values for voltage and power gives 19.51 Thus, a half ampere flows through the lightbulb when 120 V is applied across it. Access for free at openstax.org. 19.4 • Electric Power 635 Discussion This is a significant current. Recall that household power is AC and not DC, so the 120 V supplied by household sockets is an alternating power, not a constant power. The 120 V is actually the time-averaged power provided by such sockets. Thus, the average current going through the light bulb over a period of time longer than a few seconds is 0.50 A. WORKED EXAMPLE Boot Warmers To warm your boots on cold days, you decide to sew a circuit with some resistors into the insole of your boots. You want 10 W of heat output from the resistors in each insole, and you want to run them from two 9-V batteries (connected in series). What total resistance should you put in each insole? STRATEGY We know the desired power and the voltage (18 V, because we have two 9-V batteries connected in series), so we can use the equation to find the requisite resistance. Solution Solving for the resistance and inserting the given voltage and power, we obtain 19.52 Thus, the total resistance in each insole should be 32 Discussion Let’s see how much current would run through this circuit. We have 18 V applied across a resistance of 32 , so Ohm’s law gives 19.53 All batteries have labels that say how much charge they can deliver (in terms of a current multiplied by a time). A typical 9-V alkaline battery can deliver a charge of 565 ), so this heating system would function for a time of (so two 9 V batteries deliver 1,130 19.54 WORKED EXAMPLE Power through a Branch of a Circuit Each resistor in the circuit below is 30 . What power is dissipated by the middle branch of the circuit? STRATEGY The middle branch of the circuit contains resistors the equivalent resistance in this branch, and then use in series. The voltage across this branch is 12 V. We will first find to find the power dissipated in the branch. Solution The equivalent resistance is circuit is . The power dissipated by the middle branch of the 636 Chapter 19 • Electrical Circuits Discussion Let’s see if energy is conserved in this circuit by comparing the power dissipated in the circuit to the power supplied by the battery. First, the equivalent resistance of the left branch is 19.55 The power through the left branch is 19.56 19.57 The right branch contains only , so the equivalent resistance is . The power through the right branch is The total power dissipated by the circuit is the sum of the powers dissipated in each branch. The power provided by the battery is 19.58 19.59 19.60 where Iis the total current flowing through the battery. We must therefore add up the currents going through each branch to obtain I. The branches contributes currents of The total current is and the power provided by the battery is 19.61 19.62 19.63 This is the same power as is dissipated in the resistors of the circuit, which shows that energy is conserved in this circuit. Practice Problems 16. What is the formula for the power dissipated in a resistor? a. The formula for the power dissipated in a resistor is b. The formula for the power dissipated in a resistor is c. The formula for the power dissipated in a resistor is P= IV. d. The formula for the power dissipated in a resistor is P= I2V. 17. What is the formula for power dissipated by a resistor given its resistance and the voltage across it? a. The formula for the power dissipated in a resistor is b. The formula for the power dissipated in a resistor is c. The formula for the power dissipated in a resistor is d. The formula for the power dissipated in a resistor is Access for free at openstax.org. 19.4 • Electric Power 637 Check your Understanding 18. Which circuit elements dissipate power? a. b. c. d. capacitors inductors ideal switches resistors 19. Explain in words the equation for power dissipated by a given resistance. a. Electric power is proportional to current through the resistor multiplied by the square of the voltage across the resistor. b. Electric power is proportional to square of current through the resistor multiplied by the voltage across the resistor. c. Electric power is proportional to current through the resistor divided by the voltage across the resistor. d. Electric power is proportional to current through the resistor multiplied by the voltage across the resistor. 638 Chapter 19 • Key Terms KEY TERMS alternating current electric current whose direction alternates back and forth at regular intervals ampere unit for electric current; one ampere is one coulomb per second ( ) circuit diagram schematic drawing of an electrical circuit including all circuit elements, such as resistors, capacitors, batteries, and so on conventional current flows in the direction that a positive charge would flow if it could move same as the combined resistance of a group of resistors in parallel when a group of resistors are connected side by side, with the top ends of the resistors connected together by a wire and the bottom ends connected together by a different wire in series when elements in a circuit are connected one after the other in the same branch of the circuit nonohmic material that does not follow Ohm’s law Ohm’s law electric current is proportional to the voltage direct current electric current that flows in a single applied across a circuit or other path direction electric circuit physical network of paths through which electric current can flow electric current electric charge that is moving electric power in a circuit equivalent resistor rate at which electric energy is transferred resistance of a single resistor that is the ohmic material that obeys Ohm’s law resistance how much a circuit element opposes the passage of electric current; it appears as the constant of proportionality in Ohm’s law circuit element that provides a known resistance resistor steady state when the characteristics of a system do not change over time SECTION SUMMARY 19.1 Ohm's law 19.3 Parallel Circuits • Direct current is constant over time; alternating current • The equivalent resistance of a group of Nidentical alternates smoothly back and forth over time. resistors Rconnected in parallel is R/N. • Electrical resistance causes materials to extract work • from the current that flows through them. In ohmic materials, voltage drop along a path is proportional to the current that runs through the path. 19.2 Series Circuits • Circuit diagrams are schematic representations of electric circuits. • Resistors in series are resistors that are connected head to tail. • The same current runs through all resistors in series; however, the voltage drop across each resistor can be different. • The voltage is the same at every point in a given wire. • Connecting resistors in parallel provides more paths for the current to go through, so the equivalent resistance is always less than the smallest resistance of the parallel resistors. • The same voltage drop occurs across all resistors in parallel; however, the current through each resistor can differ. 19.4 Electric Power • Electric power is dissipated in the resistances of a circuit. Capacitors do not dissipate electric power. • Electric power is proportional to the voltage and the current in a circuit. • Ohm’s law provides two extra expressions for electric power: one that does not involve current and one that does not involve voltage. KEY EQUATIONS 19.1 Ohm's law electric current Iis the charge passes a plane per unit time that an ampere is the coulombs per unit time that pass a plane Access for free at openstax.org. Ohm’s law: the current Iis proportional to the voltage V, with the resistance Rbeing the constant of proportionality 19.2 Series Circuits 19.4 Electric Power Chapter 19 • Chapter Review 639 the equivalent resistance of N resistors connected in series 19.3 Parallel Circuits the equivalent resistance of Nresistors connected in parallel CHAPTER REVIEW Concept Items 19.1 Ohm's law 1. You connect a resistor across a battery. In which direction do the electrons flow? a. The electrons flow from the negative terminal of the battery to the positive terminal of the battery. b. The electrons flow from the positive terminal of the battery to the negative terminal of the battery. 2. How does current depend on resistance in Ohm’s law? a. Current is directly proportional to the resistance. b. Current is inversely proportional to the resistance. c. Current is proportional to the square of the resistance. d. Current is inversely proportional to the square of the resistance. 3. In the context of electricity, what is resistance? a. Resistance is the property of materials to resist the passage of voltage. b. Resistance is the property of materials to resist the passage of electric current. c. Resistance is the property of materials to increase the passage of voltage. d. Resistance is the property of materials to increase the passage of electric current. 4. What is the mathematical formula for Ohm’s law? a. b. c. d. 19.2 Series Circuits 5. In which circuit are all the resistors connected in series? for a given current Iflowing through a potential difference V, the electric power dissipated for a given current Iflowing through a resistance R, the electric power dissipated for a given voltage difference Vacross a resistor R, the electric power dissipated a. b. c. d. 6. What is the voltage and current through the capacitor in the circuit below a long time after the switch is closed? a. 0 V, 0 A b. 0 V, 10 A 10 V, 0 A c. 10 V, 10 A d. decreases the overall resistance. d. Adding resistors in parallel gives the current longer path through wh |
ich it can flow hence decreases the overall resistance. 19.4 Electric Power 9. To draw the most power from a battery, should you connect a small or a large resistance across its terminals? Explain. a. Small resistance, because smaller resistance will lead to the largest power b. Large resistance, because smaller resistance will lead to the largest power 10. If you double the current through a resistor, by what factor does the power dissipated by the resistor change? a. Power increases by a factor of two. b. Power increases by a factor of four. c. Power increases by a factor of eight. d. Power increases by a factor of 16. material is ohmic. 19.2 Series Circuits 13. Given three batteries (5V, 9V, 12V) and five resistors (10, 20, 30, 40, 50Ω) to choose from, what can you choose to form a circuit diagram with a current of 0.175A? You do not need to use all of the components. a. Batteries (5V, 9V) and resistors (30Ω, 50Ω) connected in series b. Batteries (5V4, 12V) and resistors (10Ω, 20Ω, 40Ω, and 50Ω) connected in series. c. Batteries (5V, 9V, and 12V) and resistors (10Ω, 20Ω, and 30Ω) connected in series. 14. What is the maximum resistance possible given a and a resistor of ? resistor of a. b. c. d. 15. Rank the points A, B, C, and D in the circuit diagram from lowest voltage to highest voltage. 640 Chapter 19 • Chapter Review 19.3 Parallel Circuits 7. If you remove resistance from a circuit, does the total resistance of the circuit always decrease? Explain. a. No, because for parallel combination of resistors, the resistance through the remaining circuit increases. b. Yes, because for parallel combination of resistors, the resistance through the remaining circuit increases. 8. Explain why the equivalent resistance of a parallel combination of resistors is always less than the smallest of the parallel resistors. a. Adding resistors in parallel gives the current a shorter path through which it can flow hence decreases the overall resistance. b. Adding resistors in parallel gives the current another path through which it can flow hence decreases the overall resistance. c. Adding resistors in parallel reduce the number of paths through which the current can flow hence Critical Thinking Items 19.1 Ohm's law 11. An accelerator accelerates He nuclei (change = 2e) to a speed of v= 2 × 106 m/s. What is the current if the linear density of He nuclei is λ= 108 m–1? a. I= 9.6 × 10–5 A b. I= 3.2 × 10–5 A c. I= 12.8 × 10–5 A d. I= 6.4 × 10–5 A 12. How can you verify whether a certain material is ohmic? a. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is proportional to the voltage, then the material is ohmic. b. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is inversely proportional to the voltage, then the material is ohmic. c. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is proportional to the square of the voltage, then the material is ohmic. d. Make a resistor from this material and measure the current going through this resistor for several different voltages. If the current is inversely proportional to the square of the voltage, then the Access for free at openstax.org. Chapter 19 • Chapter Review 641 a. The equivalent resistance of the circuit 14 Ω. b. The equivalent resistance of the circuit 16.7 Ω. c. The equivalent resistance of the circuit 140 Ω. d. The equivalent resistance of the circuit 195 Ω. 19.4 Electric Power 18. Two lamps have different resistances. (a) If the lamps are connected in parallel, which one is brighter, the lamp with greater resistance or the lamp with less resistance? (b) If the lamps are connected in series, which one is brighter? Note that the brighter lamp dissipates more power. a. (a) lamp with greater resistance; (b) lamp with less resistance (a) lamp with greater resistance; (b) lamp with greater resistance (a) lamp with less resistance; (b) lamp with less resistance (a) lamp with less resistance; (b) lamp with greater resistance b. c. d. 19. To measure the power consumed by your laptop computer, you place an ammeter (a device that measures electric current) in series with its DC power supply. When the screen is off, the computer draws 0.40 A of current. When the screen is on at full brightness, it draws 0.90 A of current. Knowing the DC power supply delivers 16 V, how much power is used by the screen? a. The power used by the screen is −8.0 W. b. The power used by the screen is 0.3 W. c. The power used by the screen is 3.2 W. d. The power used by the screen is 8.0 W. which of the following functions would be best to fit the data? Assume that a, b, and care nonzero constants adjusted to fit the data. a. b. c. d. 22. A battery of unknown voltage is attached across a resistor in series with . You add a second battery with so that the voltage across is now and measure of current through resistor . You add a third battery with series with the first two batteries so that the voltage of across current through ? . What is the resistance of and measure is in a. A, B, C, D b. B, C, A, D c. C, B, A, D d. D, A, B, C 19.3 Parallel Circuits 16. Can all resistor combinations be reduced to series and parallel combinations? a. No, all practical resistor circuits cannot be reduced to series and parallel combinations. b. Yes, all practical resistor circuits can be reduced to series and parallel combinations. 17. What is the equivalent resistance of the circuit shown below? Figure 19.23 Problems 19.1 Ohm's law 20. What voltage is needed to make 6 C of charge traverse a 100-Ω resistor in 1 min? a. The required voltage is 1 × 10−3 V. b. The required voltage is 10 V. c. The required voltage is 1,000 V. d. The required voltage is 10,000 V. 21. Resistors typically obey Ohm’s law at low currents, but show deviations at higher currents because of heating. Suppose you were to conduct an experiment measuring the voltage, V, across a resistor as a function of current, I, including currents whose deviations from Ohm’s law start to become apparent. For a data plot of Vversus I, 642 Chapter 19 • Chapter Review a. b. c. d. 19.2 Series Circuits 23. What is the voltage drop across two 80-Ω resistors connected in series with 0.15 A flowing through them? a. 12 V b. 24 V c. 36 V d. 48 V 24. In this circuit, the voltage drop across the upper resistor is 4.5 V. What is the battery voltage? a. 4.5V 7.5V b. 12V c. 18V d. 19.3 Parallel Circuits 25. What is the equivalent resistance of this circuit? Performance Task 19.4 Electric Power 29. 1. An incandescent light bulb (i.e., an old-fashioned light bulb with a little wire in it) 2. A lightbulb socket to hold the light bulb 3. A variable voltage source 4. An ammeter Procedure • Screw the lightbulb into its socket. Connect the Access for free at openstax.org. a. The equivalent resistance of the circuit is 32.7 Ω. b. The equivalent resistance of the circuit is 100 Ω. c. The equivalent resistance of the circuit is 327 Ω. d. The equivalent resistance of the circuit is 450 Ω. 26. What is the equivalent resistance of the circuit shown below? a. The equivalent resistance is 25 Ω. b. The equivalent resistance is 50 Ω. c. The equivalent resistance is 75 Ω. d. The equivalent resistance is 100 Ω. 19.4 Electric Power 27. When 12 V are applied across a resistor, it dissipates 120 W of power. What is the current through the resistor? a. The current is 1,440 A. b. The current is 10 A. c. The current is 0.1 A. d. The current is 0.01 A. 28. Warming 1 g of water requires 1 J of energy per . How long would it take to warm 1 L of water from 20 to 40 °C if you immerse in the water a 1-kW resistor connected across a 9.0-V batteries aligned in series? a. 10 min b. 20 min c. 30 min d. 40 min positive terminal of the voltage source to the input of the ammeter. Connect the output of the ammeter to one connection of the socket. Connect the other connection of the socket to the negative terminal of the voltage source. Ensure that the voltage source is set to supply DC voltage and that the ammeter is set to measure DC amperes. The desired circuit is shown below. Chapter 19 • Test Prep 643 10 rows. Label the left column voltsand the right column current.Adjust the voltage source so that it supplies from between 1 and 10 volts DC. For each voltage, write the voltage in the volts column and the corresponding amperage measured by the ammeter in the current column. Make a plot of volts versus current, that is, a plot with volts on the vertical axis and current on the horizontal axis. Use this data and the plot to answer the following questions: 1. What is the resistance of the lightbulb? 2. What is the range of possible error in your result 3. for the resistance? In a single word, how would you describe the curve formed by the data points? • On a piece of paper, make a two-column table with TEST PREP Multiple Choice 19.1 Ohm's law 30. What are the SI units for electric current? a. a battery b. a capacitor the ground c. d. a switch a. b. c. d. 31. What is the SI unit for resistance? a. b. c. d. 32. The equivalent unit for an ohm is a ________. a. V/A b. C/m c. d. V/s 33. You put DC across resistor and measure the current through it. With the same voltage across resistor , you measure twice the current. What is the ratio 1 ? a. b. c. 4 d. 2 19.2 Series Circuits 34. What does the circuit element shown represent? 35. How many 10-Ω resistors must be connected in series to make an equivalent resistance of 80 Ω? a. 80 b. 8 c. 20 d. 40 36. Which two circuit elements are represented in the circuit diagram? a. a battery connected in series with an inductor b. a capacitor connected in series with a resistor c. a resistor connected in series with a battery d. an inductor connected in series with a resistor 37. How much current will flow through a 10-V battery with a |
100-Ω resistor connected across its terminals? a. 0.1 A 1.0 A b. c. 0 d. 1,000 A 19.3 Parallel Circuits 38. A 10-Ω resistor is connected in parallel to another resistor R . The equivalent resistance of the pair is 8 Ω. What is the resistance R? 10 Ω a. b. 20 Ω 30 Ω c. 644 Chapter 19 • Test Prep d. 40 Ω 39. Are the resistors shown connected in parallel or in series? Explain. a. The resistors are connected in parallel because the same current flows through all three resistors. b. The resistors are connected in parallel because different current flows through all three resistors. c. The resistors are connected in series because the same current flows through all three resistors. d. The resistors are connected in series because different current flows through all three resistors. 19.4 Electric Power 40. Which equation below for electric power is incorrect? a. b. Short Answer 19.1 Ohm's law 44. True or false—it is possible to produce nonzero DC current by adding together AC currents. a. b. false true 45. What type of current is used in cars? a. alternating current indirect current b. c. direct current d. straight current 46. If current were represented by , voltage by , and resistance by , what would the mathematical expression be for Ohm’s law? a. b. c. d. 47. Give a verbal expression for Ohm’s law. a. Ohm’s law says that the current through a resistor equals the voltage across the resistor multiplied by the resistance of the resistor. b. Ohm’s law says that the voltage across a resistor equals the current through the resistor multiplied by the resistance of the resistor. c. Ohm’s law says that the resistance of the resistor equals the current through the resistor multiplied Access for free at openstax.org. c. d. 41. What power is dissipated in a circuit through which flows across a potential drop of ? a. b. c. Voltage drop across is d. . 42. How does a resistor dissipate power? a. A resistor dissipates power in the form of heat. b. A resistor dissipates power in the form of sound. c. A resistor dissipates power in the form of light. d. A resistor dissipates power in the form of charge. 43. What power is dissipated in a circuit through which 0.12 A flows across a potential drop of 3.0 V? a. 0.36 W b. 0.011 W c. 5 V d. 2.5 W by the voltage across a resistor. d. Ohm’s law says that the voltage across a resistor equals the square of the current through the resistor multiplied by the resistance of the resistor. 48. What is the current through a 100-Ω resistor with 12 V across it? a. 0 b. 0.12 A c. 8.33 A d. 1,200 A 49. What resistance is required to produce from a battery? a. 1 b. c. 60 120 d. 19.2 Series Circuits 50. Given a circuit with one 9-V battery and with its negative terminal connected to ground. The two paths are connected to ground from the positive terminal: the right path with a 20-Ω and a 100-Ω resistor and the left path with a 50-Ω resistor. How much current will flow in the right branch? a. b. Chapter 19 • Test Prep 645 c. d. a. b. c. d. 51. Through which branch in the circuit below does the most current flow? 19.3 Parallel Circuits 54. Ten 100-Ω resistors are connected in series. How can you increase the total resistance of the circuit by about 40 percent? a. Adding two 10-Ω resistors increases the total resistance of the circuit by about 40 percent. b. Removing two 10-Ω resistors increases the total resistance of the circuit by about 40 percent. c. Adding four 10-Ω resistors increases the total resistance of the circuit by about 40 percent. d. Removing four 10-Ω resistors increases the total resistance of the circuit by about 40 percent. 55. Two identical resistors are connected in parallel across the terminals of a battery. If you increase the resistance of one of the resistors, what happens to the current through and the voltage across the other resistor? a. The current and the voltage remain the same. b. The current decreases and the voltage remains the a. All of the current flows through the left branch due to the open switch. b. All of the current flows through the right branch due to the open switch in the left branch. c. All of the current flows through the middle branch due to the open switch in the left branch d. There will be no current in any branch of the circuit due to the open switch. 52. What current flows through the resistor in the same. circuit below? c. The current and the voltage increases. d. The current increases and the voltage remains the same. 56. a. b. c. d. 53. What is the equivalent resistance for the circuit below if and ? In the circuit below, through which resistor(s) does the most current flow? Through which does the least flow? Explain. a. The most current flows through the 15-Ω resistor because all the current must pass through this resistor. b. The most current flows through the 20-Ω resistor because all the current must pass through this resistor. c. The most current flows through the 25-Ω resistor because it is the highest resistance. d. The same current flows through the all the resistor because all the current must pass through each of the resistors. 19.4 Electric Power 57. You want to increase the power dissipated in a circuit. 646 Chapter 19 • Test Prep You have the choice between doubling the current or doubling the resistance, with the voltage remaining constant. Which one would you choose? a. doubling the resistance b. doubling the current battery? a. The power dissipated is 2430 W. b. The power dissipated is 270 W. c. The power dissipated is 2.7 W. d. The power dissipated is 0.37 W. 58. You want to increase the power dissipated in a circuit. You have the choice between reducing the voltage or reducing the resistance, with the current remaining constant. Which one would you choose? a. b. reduce the voltage to increase the power reduce the resistance to increase the power 59. What power is dissipated in the circuit consisting of 310-Ω resistors connected in series across a 9.0-V Extended Response 19.1 Ohm's law 61. Describe the relationship between current and charge. Include an explanation of how the direction of the current is defined. a. Electric current is the charge that passes through a conductor per unit time. The direction of the current is defined to be the direction in which positive charge would flow. b. Electric current is the charges that move in a conductor. The direction of the current is defined to be the direction in which positive charge would flow. c. Electric current is the charge that passes through a conductor per unit time. The direction of the current is defined to be the direction in which negative charge would flow. d. Electric current is the charges that move in a conductor. The direction of the current is defined to be the direction in which negative charge would flow. 62. What could cause Ohm’s law to break down? a. b. c. d. If small amount of current flows through a resistor, the resistor will heat up so much that it will change state, in violation of Ohm’s law. If excessive amount of current flows through a resistor, the resistor will heat up so much that it will change state, in violation of Ohm’s law. If small amount of current flows through a resistor, the resistor will not heat up so much and it will not change its state, in violation of Ohm’s law. If excessive amount of current flows through a resistor, the resistor will heat up so much that it will not change its state, in violation of Ohm’s law. 63. You connect a single resistor across a battery and find that flows through the circuit. You add Access for free at openstax.org. 60. What power is dissipated in a circuit consisting of three 10-Ω resistors connected in parallel across a 9.0-V battery? a. The power dissipated is 270 W. b. The power dissipated is 30 W. c. The power dissipated is 24 W. d. The power dissipated is 1/24 W. another resistor after the first resistor and find that flows through the circuit. If you have connected in a line one after the other, what resistors would be their total resistance? a. b. c. d. 19.2 Series Circuits 64. Explain why the current is the same at all points in the circuit below. a. b. c. If the current were not constant, the mobile charges would bunch up in places, which means that the voltage would decrease at that point. A lower voltage at some point would push the current in the direction that further decreases the voltage. If the current were not constant, the mobile charges would bunch up in places, which means that the voltage would increase at that point. But a higher voltage at some point would push the current in the direction that decreases the voltage. If the current were not constant, the mobile charges would bunch up in places, which mean that the voltage would increase at that point. A higher voltage at some point would push the current in the direction that further increases the d. voltage. If the current were not constant, the mobile charges would bunch up in places, which mean that the voltage would decrease at that point. But a lower voltage at some point would push the current in the direction that increases the voltage. 65. What is the current through each resistor in the circuit? a. Current through resistors R1, R2, R3, and R4 is 0.48 A, 0.30 A, 1.2 A, and 0.24 A, respectively. b. Current through resistors R1, R2, R3, and R4 is 1200 A, 1920 A, 480 A, and 2400 A, respectively. c. Current through resistors R1, R2, R3, and is R4 2.08 A, 3.34 A, 0.833 A, and 4.17 A, respectively. d. The same amount of current, 0.096 A, flows through all of the resistors. 19.3 Parallel Circuits 66. In a house, a single incoming wire at a high potential with respect to the ground provides electric power. How are the appliances connected between this wire and the Chapter 19 • Test Prep 647 ground, in parallel or in series? Explain. a. The appliances are connected in parallel to provide different voltage differences across each appliance. b. The appliances are connected in parallel to provide the same voltage difference across each app |
liance. c. The appliances are connected in series to provide the same voltage difference across each appliance. d. The appliances are connected in series to provide different voltage differences across each appliance. 19.4 Electric Power 67. A single resistor is connected across the terminals of a battery When you attach a second resistor in parallel with the first, does the power dissipated by the system change? a. No, the power dissipated remain same. b. Yes, the power dissipated increases. c. Yes, the power dissipated decreases. 68. In a flashlight, the batteries are normally connected in series. Why are they not connected in parallel? a. Batteries are connected in series for higher voltage and power output. b. Batteries are connected in series for lower voltage and power output. c. Batteries are connected in series so that power output is a much lower for the same amount of voltage. d. Batteries are connected in series to reduce the overall loss of energy from the circuit. 648 Chapter 19 • Test Prep Access for free at openstax.org. CHAPTER 20 Magnetism Figure 20.1 The magnificent spectacle of the Aurora Borealis, or northern lights, glows in the northern sky above Bear Lake near Eielson Air Force Base, Alaska. Shaped by Earth’s magnetic field, this light is produced by radiation spewed from solar storms. (credit: Senior Airman Joshua Strang, Flickr) Chapter Outline 20.1 Magnetic Fields, Field Lines, and Force 20.2 Motors, Generators, and Transformers 20.3 Electromagnetic Induction You may have encountered magnets for the first time as a small child playing with magnetic toys or INTRODUCTION refrigerator magnets. At the time, you likely noticed that two magnets that repulse each other will attract each other if you flip one of them around. The force that acts across the air gaps between magnets is the same force that creates wonders such as the Aurora Borealis. In fact, magnetic effects pervade our lives in myriad ways, from electric motors to medical imaging and computer memory. In this chapter, we introduce magnets and learn how they work and how magnetic fields and electric currents interact. 650 Chapter 20 • Magnetism 20.1 Magnetic Fields, Field Lines, and Force Section Learning Objectives By the end of this section, you will be able to do the following: • Summarize properties of magnets and describe how some nonmagnetic materials can become magnetized • Describe and interpret drawings of magnetic fields around permanent magnets and current-carrying wires • Calculate the magnitude and direction of magnetic force in a magnetic field and the force on a current- carrying wire in a magnetic field Section Key Terms Curie temperature domain electromagnet electromagnetism ferromagnetic magnetic dipole magnetic field magnetic pole magnetized north pole permanent magnet right-hand rule solenoid south pole Magnets and Magnetization People have been aware of magnets and magnetism for thousands of years. The earliest records date back to ancient times, particularly in the region of Asia Minor called Magnesia—the name of this region is the source of words like magnet. Magnetic rocks found in Magnesia, which is now part of western Turkey, stimulated interest during ancient times. When humans first discovered magnetic rocks, they likely found that certain parts of these rocks attracted bits of iron or other magnetic rocks more strongly than other parts. These areas are called the polesof a magnet. A magnetic pole is the part of a magnet that exerts the strongest force on other magnets or magnetic material, such as iron. For example, the poles of the bar magnet shown in Figure 20.2 are where the paper clips are concentrated. Figure 20.2 A bar magnet with paper clips attracted to the two poles. If a bar magnet is suspended so that it rotates freely, one pole of the magnet will always turn toward the north, with the opposite pole facing south. This discovery led to the compass, which is simply a small, elongated magnet mounted so that it can rotate freely. An example of a compass is shown Figure 20.3. The pole of the magnet that orients northward is called the north pole, and the opposite pole of the magnet is called the south pole. Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 651 Figure 20.3 A compass is an elongated magnet mounted in a device that allows the magnet to rotate freely. The discovery that one particular pole of a magnet orients northward, whereas the other pole orients southward allowed people to identify the north and south poles of any magnet. It was then noticed that the north poles of two different magnets repel each other, and likewise for the south poles. Conversely, the north pole of one magnet attracts the south pole of other magnets. This situation is analogous to that of electric charge, where like charges repel and unlike charges attract. In magnets, we simply replace charge with pole: Like poles repel and unlike poles attract. This is summarized in Figure 20.4, which shows how the force between magnets depends on their relative orientation. Figure 20.4 Depending on their relative orientation, magnet poles will either attract each other or repel each other. Consider again the fact that the pole of a magnet that orients northward is called the north pole of the magnet. If unlike poles attract, then the magnetic pole of Earth that is close to the geographic North Pole must be a magnetic south pole! Likewise, the magnetic pole of Earth that is close to the geographic South Pole must be a magnetic north pole. This situation is depicted in Figure 20.5, in which Earth is represented as containing a giant internal bar magnet with its magnetic south pole at the geographic North Pole and vice versa. If we were to somehow suspend a giant bar magnet in space near Earth, then the north pole of the space magnet would be attracted to the south pole of Earth’s internal magnet. This is in essence what happens with a compass needle: Its magnetic north pole is attracted to the magnet south pole of Earth’s internal magnet. 652 Chapter 20 • Magnetism Figure 20.5 Earth can be thought of as containing a giant magnet running through its core. The magnetic south pole of Earth’s magnet is at the geographic North Pole, so the north pole of magnets is attracted to the North Pole, which is how the north pole of magnets got their name. Likewise, the south pole of magnets is attracted to the geographic South Pole of Earth. What happens if you cut a bar magnet in half? Do you obtain one magnet with two south poles and one magnet with two north poles? The answer is no: Each half of the bar magnet has a north pole and a south pole. You can even continue cutting each piece of the bar magnet in half, and you will always obtain a new, smaller magnet with two opposite poles. As shown in Figure 20.6, you can continue this process down to the atomic scale, and you will find that even the smallest particles that behave as magnets have two opposite poles. In fact, no experiment has ever found any object with a single magnetic pole, from the smallest subatomic particle such as electrons to the largest objects in the universe such as stars. Because magnets always have two poles, they are referred to as magnetic dipoles—dimeans two. Below, we will see that magnetic dipoles have properties that are analogous to electric dipoles. Figure 20.6 All magnets have two opposite poles, from the smallest, such as subatomic particles, to the largest, such as stars. WATCH PHYSICS Introduction to Magnetism This video provides an interesting introduction to magnetism and discusses, in particular, how electrons around their atoms contribute to the magnetic effects that we observe. Click to view content (https://www.openstax.org/l/28_intro_magn) GRASP CHECK Toward which magnetic pole of Earth is the north pole of a compass needle attracted? Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 653 a. The north pole of a compass needle is attracted to the north magnetic pole of Earth, which is located near the geographic North Pole of Earth. b. The north pole of a compass needle is attracted to the south magnetic pole of Earth, which is located near the geographic North Pole of Earth. c. The north pole of a compass needle is attracted to the north magnetic pole of Earth, which is located near the geographic South Pole of Earth. d. The north pole of a compass needle is attracted to the south magnetic pole of Earth, which is located near the geographic South Pole of Earth. Only certain materials, such as iron, cobalt, nickel, and gadolinium, exhibit strong magnetic effects. Such materials are called ferromagnetic, after the Latin word ferrumfor iron. Other materials exhibit weak magnetic effects, which are detectable only with sensitive instruments. Not only do ferromagnetic materials respond strongly to magnets—the way iron is attracted to magnets—but they can also be magnetized themselves—that is, they can be induced to be magnetic or made into permanent magnets (Figure 20.7). A permanent magnet is simply a material that retains its magnetic behavior for a long time, even when exposed to demagnetizing influences. Figure 20.7 An unmagnetized piece of iron is placed between two magnets, heated, and then cooled, or simply tapped when cold. The iron becomes a permanent magnet with the poles aligned as shown: Its south pole is adjacent to the north pole of the original magnet, and its north pole is adjacent to the south pole of the original magnet. Note that attractive forces are created between the central magnet and the outer magnets. When a magnet is brought near a previously unmagnetized ferromagnetic material, it causes local magnetization of the material with unlike poles closest, as in the right side of Figure 20.7. This causes an attractive force, which is why unmagnetized iron is attracted to a magnet. What happens on a microscopic scale is illustrated in Figure 7(a). Regions within the mate |
rial called domains act like small bar magnets. Within domains, the magnetic poles of individual atoms are aligned. Each atom acts like a tiny bar magnet. Domains are small and randomly oriented in an unmagnetized ferromagnetic object. In response to an external magnetic field, the domains may grow to millimeter size, aligning themselves, as shown in Figure 7(b). This induced magnetization can be made permanent if the material is heated and then cooled, or simply tapped in the presence of other magnets. Figure 20.8 (a) An unmagnetized piece of iron—or other ferromagnetic material—has randomly oriented domains. (b) When magnetized by an external magnet, the domains show greater alignment, and some grow at the expense of others. Individual atoms are aligned within 654 Chapter 20 • Magnetism domains; each atom acts like a tiny bar magnet. Conversely, a permanent magnet can be demagnetized by hard blows or by heating it in the absence of another magnet. Increased thermal motion at higher temperature can disrupt and randomize the orientation and size of the domains. There is a well-defined temperature for ferromagnetic materials, which is called the Curie temperature, above which they cannot be magnetized. The Curie temperature for iron is 1,043 K (770 elements and alloys that have Curie temperatures much lower than room temperature and are ferromagnetic only below those temperatures. ), which is well above room temperature. There are several Snap Lab Refrigerator Magnets We know that like magnetic poles repel and unlike poles attract. See if you can show this for two refrigerator magnets. Will the magnets stick if you turn them over? Why do they stick to the refrigerator door anyway? What can you say about the magnetic properties of the refrigerator door near the magnet? Do refrigerator magnets stick to metal or plastic spoons? Do they stick to all types of metal? GRASP CHECK You have one magnet with the north and south poles labeled. How can you use this magnet to identify the north and south poles of other magnets? a. If the north pole of a known magnet is repelled by a pole of an unknown magnet on bringing them closer, that pole of unknown magnet is its north pole; otherwise, it is its south pole. If the north pole of known magnet is attracted to a pole of an unknown magnet on bringing them closer, that pole of unknown magnet is its north pole; otherwise, it is its south pole. b. Magnetic Fields We have thus seen that forces can be applied between magnets and between magnets and ferromagnetic materials without any contact between the objects. This is reminiscent of electric forces, which also act over distances. Electric forces are described using the concept of the electric field, which is a force field around electric charges that describes the force on any other charge placed in the field. Likewise, a magnet creates a magnetic field around it that describes the force exerted on other magnets placed in the field. As with electric fields, the pictorial representation of magnetic field lines is very useful for visualizing the strength and direction of the magnetic field. As shown in Figure 20.9, the direction of magnetic field lines is defined to be the direction in which the north pole of a compass needle points. If you place a compass near the north pole of a magnet, the north pole of the compass needle will be repelled and point away from the magnet. Thus, the magnetic field lines point away from the north pole of a magnet and toward its south pole. Figure 20.9 The black lines represent the magnetic field lines of a bar magnet. The field lines point in the direction that the north pole of a small compass would point, as shown at left. Magnetic field lines never stop, so the field lines actually penetrate the magnet to form complete loops, as shown at right. Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 655 Magnetic field lines can be mapped out using a small compass. The compass is moved from point to point around a magnet, and at each point, a short line is drawn in the direction of the needle, as shown in Figure 20.10. Joining the lines together then reveals the path of the magnetic field line. Another way to visualize magnetic field lines is to sprinkle iron filings around a magnet. The filings will orient themselves along the magnetic field lines, forming a pattern such as that shown on the right in Figure 20.10. Virtual Physics Using a Compass to Map Out the Magnetic Field Click to view content (http://www.openstax.org/l/28magcomp) This simulation presents you with a bar magnet and a small compass. Begin by dragging the compass around the bar magnet to see in which direction the magnetic field points. Note that the strength of the magnetic field is represented by the brightness of the magnetic field icons in the grid pattern around the magnet. Use the magnetic field meter to check the field strength at several points around the bar magnet. You can also flip the polarity of the magnet, or place Earth on the image to see how the compass orients itself. GRASP CHECK With the slider at the top right of the simulation window, set the magnetic field strength to 100 percent . Now use the magnetic field meter to answer the following question: Near the magnet, where is the magnetic field strongest and where is it weakest? Don’t forget to check inside the bar magnet. a. The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet. b. The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are least dense at the center and densest between the two poles just outside the bar magnet. c. The magnetic field is weakest at the center and strongest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet. d. The magnetic field is weakest at the center and strongest between the two poles just outside the bar magnet and the magnetic field lines are least dense at the center and densest between the two poles just outside the bar magnet. Figure 20.10 Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle. Joining the points together reveals the path of the magnetic field lines. Another way to visualize magnetic field lines is to sprinkle iron filings around a magnet, as shown at right. When two magnets are brought close together, the magnetic field lines are perturbed, just as happens for electric field lines when two electric charges are brought together. Bringing two north poles together—or two south poles—will cause a repulsion, and the magnetic field lines will bend away from each other. This is shown in Figure 20.11, which shows the magnetic field lines created by the two closely separated north poles of a bar magnet. When opposite poles of two magnets are brought together, the 656 Chapter 20 • Magnetism magnetic field lines join together and become denser between the poles. This situation is shown in Figure 20.11. Figure 20.11 (a) When two north poles are approached together, the magnetic field lines repel each other and the two magnets experience a repulsive force. The same occurs if two south poles are approached together. (b) If opposite poles are approached together, the magnetic field lines become denser between the poles and the magnets experience an attractive force. Like the electric field, the magnetic field is stronger where the lines are denser. Thus, between the two north poles in Figure 20.11, the magnetic field is very weak because the density of the magnetic field is almost zero. A compass placed at that point would essentially spin freely if we ignore Earth’s magnetic field. Conversely, the magnetic field lines between the north and south poles in Figure 20.11 are very dense, indicating that the magnetic field is very strong in this region. A compass placed here would quickly align with the magnetic field and point toward the south pole on the right. Note that magnets are not the only things that make magnetic fields. Early in the nineteenth century, people discovered that electrical currents cause magnetic effects. The first significant observation was by the Danish scientist Hans Christian Oersted (1777–1851), who found that a compass needle was deflected by a current-carrying wire. This was the first significant evidence that the movement of electric charges had any connection with magnets. An electromagnet is a device that uses electric current to make a magnetic field. These temporarily induced magnets are called electromagnets. Electromagnets are employed for everything from a wrecking yard crane that lifts scrapped cars to controlling the beam of a 90-km-circumference particle accelerator to the magnets in medical-imaging machines (see Figure 20.12). Figure 20.12 Instrument for magnetic resonance imaging (MRI). The device uses a cylindrical-coil electromagnet to produce for the main magnetic field. The patient goes into the tunnelon the gurney. (credit: Bill McChesney, Flickr) The magnetic field created by an electric current in a long straight wire is shown in Figure 20.13. The magnetic field lines form concentric circles around the wire. The direction of the magnetic field can be determined using the right-hand rule. This rule shows up in several places in the study of electricity and magnetism. Applied to a straight current-carrying wire, the right-hand rule says that, with your right thumb pointed in the direction of the current, the magnetic field will be in the direction in which your right fingers curl, as shown in Figure 20.13. If the wire is very long compared to the distance rfrom the wire, the |
strength B of the magnetic field is given by where Iis the current in the wire in amperes. The SI unit for magnetic field is the tesla (T). The symbol —read “mu-zero”—is a constant called the “permeability of free space” and is given by 20.2 20.1 Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 657 Figure 20.13 This image shows how to use the right-hand rule to determine the direction of the magnetic field created by current flowing through a straight wire. Point your right thumb in the direction of the current, and the magnetic field will be in the direction in which your fingers curl. WATCH PHYSICS Magnetic Field Due to an Electric Current This video describes the magnetic field created by a straight current-carrying wire. It goes over the right-hand rule to determine the direction of the magnetic field, and presents and discusses the formula for the strength of the magnetic field due to a straight current-carrying wire. Click to view content (https://www.openstax.org/l/28magfield) GRASP CHECK A long straight wire is placed on a table top and electric current flows through the wire from right to left. If you look at the wire end-on from the left end, does the magnetic field go clockwise or counterclockwise? a. By pointing your right-hand thumb in the direction opposite of current, the right-hand fingers will curl counterclockwise, so the magnetic field will be in the counterclockwise direction. b. By pointing your right-hand thumb in the direction opposite of current, the right-hand fingers will curl clockwise, so the magnetic field will be in the clockwise direction. c. By pointing your right-hand thumb in the direction of current, the right-hand fingers will curl counterclockwise, so the magnetic field will be in the counterclockwise direction. d. By pointing your right-hand thumb in the direction of current, the right-hand fingers will curl clockwise, so the magnetic field will be in the clockwise direction. Now imagine winding a wire around a cylinder with the cylinder then removed. The result is a wire coil, as shown in Figure 20.14. This is called a solenoid. To find the direction of the magnetic field produced by a solenoid, apply the right-hand rule to several points on the coil. You should be able to convince yourself that, inside the coil, the magnetic field points from left to right. In fact, another application of the right-hand rule is to curl your right-hand fingers around the coil in the direction in which the current flows. Your right thumb then points in the direction of the magnetic field inside the coil: left to right in this case. Figure 20.14 A wire coil with current running through as shown produces a magnetic field in the direction of the red arrow. Each loop of wire contributes to the magnetic field inside the solenoid. Because the magnetic field lines must form closed loops, the field lines close the loop outside the solenoid. The magnetic field lines are much denser inside the solenoid than outside the solenoid. The resulting magnetic field looks very much like that of a bar magnet, as shown in Figure 20.15. The magnetic field strength deep inside a solenoid is 658 Chapter 20 • Magnetism where Nis the number of wire loops in the solenoid and is the length of the solenoid. 20.3 Figure 20.15 Iron filings show the magnetic field pattern around (a) a solenoid and (b) a bar magnet. The fields patterns are very similar, especially near the ends of the solenoid and bar magnet. Virtual Physics Electromagnets Click to view content (http://www.openstax.org/l/28elec_magnet) Use this simulation to visualize the magnetic field made from a solenoid. Be sure to click on the tab that says Electromagnet. You can drive AC or DC current through the solenoid by choosing the appropriate current source. Use the field meter to measure the strength of the magnetic field and then change the number of loops in the solenoid to see how this affects the magnetic field strength. GRASP CHECK Choose the battery as current source and set the number of wire loops to four. With a nonzero current going through the solenoid, measure the magnetic field strength at a point. Now decrease the number of wire loops to two. How does the magnetic field strength change at the point you chose? a. There will be no change in magnetic field strength when number of loops reduces from four to two. b. The magnetic field strength decreases to half of its initial value when number of loops reduces from four to two. c. The magnetic field strength increases to twice of its initial value when number of loops reduces from four to two. d. The magnetic field strength increases to four times of its initial value when number of loops reduces from four to two. Magnetic Force If a moving electric charge, that is electric current, produces a magnetic field that can exert a force on another magnet, then the reverse should be true by Newton’s third law. In other words, a charge moving through the magnetic field produced by another object should experience a force—and this is exactly what we find. As a concrete example, consider Figure 20.16, which shows a Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 659 charge qmoving with velocity force experienced by this charge is through a magnetic field between the poles of a permanent magnet. The magnitude Fof the where is the angle between the velocity of the charge and the magnetic field. The direction of the force may be found by using another version of the right-hand rule: First, we join the tails of the velocity vector and a magnetic field vector, as shown in step 1 of Figure 20.16. We then curl our right fingers from to in step (2) of Figure 20.16. The direction in which the right thumb points is the direction of the force. For the charge in Figure 20.16, we find that the force is directed into the page. , as indicated 20.4 Note that the factor magnetic field because the magnetic field, because in the equation and and means that zero force is applied on a charge that moves parallel to a . The maximum force a charge can experience is when it moves perpendicular to Figure 20.16 (a) An electron moves through a uniform magnetic field. (b) Using the right-hand rule, the force on the electron is found to be directed into the page. LINKS TO PHYSICS Magnetohydrodynamic Drive In Tom Clancy’s Cold War novel “The Hunt for Red October,” the Soviet Union built a submarine (see Figure 20.17) with a magnetohydrodynamic drive that was so silent it could not be detected by surface ships. The only conceivable purpose to build such a submarine was to give the Soviet Union first-strike capability, because this submarine could sneak close to the coast of the United States and fire its ballistic missiles, destroying key military and government installations to prevent an American counterattack. Figure 20.17 A Typhoon-class Russian ballistic-missile submarine on which the fictional submarine Red October was based. A magnetohydrodynamic drive is supposed to be silent because it has no moving parts. Instead, it uses the force experienced by charged particles that move in a magnetic field. The basic idea behind such a drive is depicted in Figure 20.18. Salt water flows through a channel that runs from the front to the back of the submarine. A magnetic field is applied horizontally across the channel, and a voltage is applied across the electrodes on the top and bottom of the channel to force a downward electric current through the water. The charge carriers are the positive sodium ions and the negative chlorine ions of salt. Using the right-hand 660 Chapter 20 • Magnetism rule, the force on the charge carriers is found to be toward the rear of the vessel. The accelerated charges collide with water molecules and transfer their momentum, creating a jet of water that is propelled out the rear of the channel. By Newton’s third law, the vessel experiences a force of equal magnitude, but in the opposite direction. Figure 20.18 A schematic drawing of a magnetohydrodynamic drive showing the water channel, the current direction, the magnetic field direction, and the resulting force. Fortunately for all involved, it turns out that such a propulsion system is not very practical. Some back-of-the-envelope calculations show that, to power a submarine, either extraordinarily high magnetic fields or extraordinarily high electric currents would be required to obtain a reasonable thrust. In addition, prototypes of magnetohydrodynamic drives show that they are anything but silent. Electrolysis caused by running a current through salt water creates bubbles of hydrogen and oxygen, which makes this propulsion system quite noisy. The system also leaves a trail of chloride ions and metal chlorides that can easily be detected to locate the submarine. Finally, the chloride ions are extremely reactive and very quickly corrode metal parts, such as the electrode or the water channel itself. Thus, the Red October remains in the realm of fiction, but the physics involved is quite real. GRASP CHECK If the magnetic field is downward, in what direction must the current flow to obtain rearward-pointing force? a. The current must flow vertically from up to down when viewed from the rear of the boat. b. The current must flow vertically from down to up when viewed from the rear of the boat. c. The current must flow horizontally from left to right when viewed from the rear of the boat. d. The current must flow horizontally from right to left when viewed from the rear of the boat. Instead of a single charge moving through a magnetic field, consider now a steady current Imoving through a straight wire. If we place this wire in a uniform magnetic field, as shown in Figure 20.19, what is the force on the wire or, more precisely, on the electrons in the wire? An electric current involves charges that move. If the charges qmove a distance speed is Inserting this into the equation in a time t, then their gi |
ves The factor q/tin this equation is nothing more than the current in the wire. Thus, using , we obtain 20.5 20.6 This equation gives the force on a straight current-carrying wire of length angle between the current vector and the magnetic field vector. Note that for which as shown in Figure 20.19. in a magnetic field of strength B. The angle is the is the length of wire that is in the magnetic field and The direction of the force is determined in the same way as for a single charge. Curl your right fingers from the vector for Ito the vector for B, and your right thumb will point in the direction of the force on the wire. For the wire shown in Figure 20.19, the force is directed into the page. Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 661 Figure 20.19 A straight wire carrying current Iin a magnetic field B. The force exerted on the wire is directed into the page. The length is the length of the wire that is inthe magnetic field. Throughout this section, you may have noticed the symmetries between magnetic effects and electric effects. These effects all fall under the umbrella of electromagnetism, which is the study of electric and magnetic phenomena. We have seen that electric charges produce electric fields, and moving electric charges produce magnetic fields. A magnetic dipole produces a magnetic field, and, as we will see in the next section, moving magnetic dipoles produce an electric field. Thus, electricity and magnetism are two intimately related and symmetric phenomena. WORKED EXAMPLE Trajectory of Electron in Magnetic Field A proton enters a region of constant magnetic field, as shown in Figure 20.20. The magnetic field is coming out of the page. If the electron is moving at and the magnetic field strength is 2.0 T, what is the magnitude and direction of the force on the proton? Figure 20.20 A proton enters a region of uniform magnetic field. The magnetic field is coming out of the page—the circles with dots represent vector arrow heads coming out of the page. STRATEGY Use the equation and the velocity vector of the proton is Solution The charge of the proton is the equation gives to find the magnitude of the force on the proton. The angle between the magnetic field vectors The direction of the force may be found by using the right-hand rule. . Entering this value and the given velocity and magnetic field strength into 20.7 To find the direction of the force, first join the velocity vector end to end with the magnetic field vector, as shown in Figure 20.21. Now place your right hand so that your fingers point in the direction of the velocity and curl them upward toward the magnetic field vector. The force is in the direction in which your thumb points. In this case, the force is downward in the plane of the paper in the -direction, as shown in Figure 20.21. 662 Chapter 20 • Magnetism Figure 20.21 The velocity vector and a magnetic field vector from Figure 20.20 are placed end to end. A right hand is shown with the fingers curling up from the velocity vector toward the magnetic field vector. The thumb points in the direction of the resulting force, which is the -direction in this case. Thus, combining the magnitude and the direction, we find that the force on the proton is Discussion This seems like a very small force. However, the proton has a mass of , so its acceleration is , or about ten thousand billion times the acceleration due to gravity! We found that the proton’s initial acceleration as it enters the magnetic field is downward in the plane of the page. Notice that, as the proton accelerates, its velocity remains perpendicular to the magnetic field, so the magnitude of the force does not change. In addition, because of the right-hand rule, the direction of the force remains perpendicular to the velocity. This force is nothing more than a centripetal force: It has a constant magnitude and is always perpendicular to the velocity. Thus, the magnitude of the velocity does not change, and the proton executes circular motion. The radius of this circle may be found by using the kinematics relationship. 20.8 The path of the proton in the magnetic field is shown in Figure 20.22. Figure 20.22 When traveling perpendicular to a constant magnetic field, a charged particle will execute circular motion, as shown here for a proton. WORKED EXAMPLE Wire with Current in Magnetic Field Now suppose we run a wire through the uniform magnetic field from the previous example, as shown. If the wire carries a current of 1.0 A in the -direction, and the region with magnetic field is 4.0 cm long, what is the force on the wire? Access for free at openstax.org. 20.1 • Magnetic Fields, Field Lines, and Force 663 STRATEGY to find the magnitude of the force on the wire. The length of the wire inside the magnetic field is Use equation 4.0 cm, and the angle between the current direction and the magnetic field direction is 90°. To find the direction of the force, use the right-hand rule as explained just after the equation Solution Insert the given values into equation to find the magnitude of the force 20.9 To find the direction of the force, begin by placing the current vector end to end with a vector for the magnetic field. The result is as shown in the figure in the previous Worked Example with and your right thumb points down the page, again as shown in the figure in the previous Worked Example. Thus, the direction of the force is in the . Curl your right-hand fingers from to -direction. The complete force is thus replaced by . Discussion The direction of the force is the same as the initial direction of the force was in the previous example for a proton. However, because the current in a wire is confined to a wire, the direction in which the charges move does not change. Instead, the entire wire accelerates in the -direction. The force on a current-carrying wire in a magnetic field is the basis of all electrical motors, as we will see in the upcoming sections. Practice Problems 1. What is the magnitude of the force on an electron moving at 1.0 × 106 m/s perpendicular to a 1.0-T magnetic field? a. 0.8 × 10–13 N 1.6 × 10–14 N b. c. 0.8 × 10–14 N 1.6 × 10–13 N d. 2. A straight 10 cm wire carries 0.40 A and is oriented perpendicular to a magnetic field. If the force on the wire is 0.022 N, what is the magnitude of the magnetic field? 1.10 × 10–2 T a. b. 0.55 × 10–2 T c. 1.10 T d. 0.55 T Check Your Understanding 3. If two magnets repel each other, what can you conclude about their relative orientation? a. Either the south pole of magnet 1 is closer to the north pole of magnet 2 or the north pole of magnet 1 is closer to the south pole of magnet 2. b. Either the south poles of both the magnet 1 and magnet 2 are closer to each other or the north poles of both the magnet 1 664 Chapter 20 • Magnetism and magnet 2 are closer to each other. 4. Describe methods to demagnetize a ferromagnet. a. by cooling, heating, or submerging in water b. by heating, hammering, and spinning it in external magnetic field c. by hammering, heating, and rubbing with cloth d. by cooling, submerging in water, or rubbing with cloth 5. What is a magnetic field? a. The directional lines present inside and outside the magnetic material that indicate the magnitude and direction of the magnetic force. b. The directional lines present inside and outside the magnetic material that indicate the magnitude of the magnetic force. c. The directional lines present inside the magnetic material that indicate the magnitude and the direction of the magnetic force. d. The directional lines present outside the magnetic material that indicate the magnitude and the direction of the magnetic force. 6. Which of the following drawings is correct? a. b. c. d. Access for free at openstax.org. 20.2 • Motors, Generators, and Transformers 665 20.2 Motors, Generators, and Transformers Section Learning Objectives By the end of this section, you will be able to do the following: • Explain how electric motors, generators, and transformers work • Explain how commercial electric power is produced, transmitted, and distributed Section Key Terms electric motor generator transformer Electric Motors, Generators, and Transformers As we learned previously, a current-carrying wire in a magnetic field experiences a force—recall motors, which convert electrical energy into mechanical energy, are the most common application of magnetic force on currentcarrying wires. Motors consist of loops of wire in a magnetic field. When current is passed through the loops, the magnetic field exerts a torque on the loops, which rotates a shaft. Electrical energy is converted to mechanical work in the process. Figure 20.23 shows a schematic drawing of an electric motor. . Electric Figure 20.23 Torque on a current loop. A vertical loop of wire in a horizontal magnetic field is attached to a vertical shaft. When current is passed through the wire loop, torque is exerted on it, making it turn the shaft. Let us examine the force on each segment of the loop in Figure 20.23 to find the torques produced about the axis of the vertical shaft—this will lead to a useful equation for the torque on the loop. We take the magnetic field to be uniform over the rectangular loop, which has width wand height loop. To determine the direction of the force, we use the right-hand rule. The current goes from left to right into the page, and the magnetic field goes from left to right in the plane of the page. Curl your right fingers from the current vector to the magnetic field vector and your right thumb points down. Thus, the force on the top segment is downward, which produces no torque on the shaft. Repeating this analysis for the bottom segment—neglect the small gap where the lead wires go out—shows that the force on the bottom segment is upward, again producing no torque on the shaft. as shown in the figure. First, consider the force on the top segment of the Consider now the left vertical segme |
nt of the loop. Again using the right-hand rule, we find that the force exerted on this segment is perpendicular to the magnetic field, as shown in Figure 20.23. This force produces a torque on the shaft. Repeating this analysis on the right vertical segment of the loop shows that the force on this segment is in the direction opposite that of the force on the left segment, thereby producing an equal torque on the shaft. The total torque on the shaft is thus twice the toque on one of the vertical segments of the loop. To find the magnitude of the torque as the wire loop spins, consider Figure 20.24, which shows a view of the wire loop from where Fis the applied force, ris the distance from the pivot to where the above. Recall that torque is defined as force is applied, and θis the angle between rand F. Notice that, as the loop spins, the current in the vertical loop segments is always perpendicular to the magnetic field. Thus, the equation segment as The distance rfrom the shaft to where this force is applied is w/2, so the torque created by this force is gives the magnitude of the force on each vertical Because there are two vertical segments, the total torque is twice this, or 20.10 20.11 If we have a multiple loop with Nturns, we get Ntimes the torque of a single loop. Using the fact that the area of the loop is the expression for the torque becomes 666 Chapter 20 • Magnetism This is the torque on a current-carrying loop in a uniform magnetic field. This equation can be shown to be valid for a loop of any shape. 20.12 Figure 20.24 View from above of the wire loop from Figure 20.23. The magnetic field generates a force Fon each vertical segment of the wire loop, which generates a torque on the shaft. Notice that the currents have the same magnitude because they both represent the current flowing in the wire loop, but flows into the page and flows out of the page. From the equation a maximum positive torque of to From will oscillate back and forth. we see that the torque is zero when when The torque then decreases back to zero as the wire loop rotates to the torque is negative. Thus, the torque changes sign every half turn, so the wire loop As the wire loop rotates, the torque increases to For the coil to continue rotating in the same direction, the current is reversed as the coil passes through using automatic switches called brushes, as shown in Figure 20.25. Figure 20.25 (a) As the angular momentum of the coil carries it through the brushes reverse the current and the torque remains clockwise. (b) The coil rotates continuously in the clockwise direction, with the current reversing each half revolution to maintain the clockwise torque. Consider now what happens if we run the motor in reverse; that is, we attach a handle to the shaft and mechanically force the coil to rotate within the magnetic field, as shown in Figure 20.26. As per the equation is the angle —where between the vectors and in the wires of the loop experience a magnetic force because they are moving in a magnetic field. Again using the right-hand rule, where we curl our fingers from vector the top and bottom segments feel a force perpendicular to the wire, which does not cause a current. However, charges in the vertical wires experience forces parallel to the wire, causing a current to flow through the wire and through an external circuit if one is connected. A device such as this that converts mechanical energy into electrical energy is called a generator. , we find that charges in to vector Access for free at openstax.org. 20.2 • Motors, Generators, and Transformers 667 Figure 20.26 When this coil is rotated through one-fourth of a revolution, the magnetic flux Φ changes from its maximum to zero, inducing an emf, which drives a current through an external circuit. Because current is induced only in the side wires, we can find the induced emf by only considering these wires. As explained in Induced Current in a Wire, motional emf in a straight wire moving at velocity vthrough a magnetic field Bis where the velocity is perpendicular to the magnetic field. In the generator, the velocity makes an angle with B(see Figure 20.27), so the velocity component perpendicular to Bis Thus, in this case, the emf induced on each vertical wire segment is and they are in the same direction. The total emf around the loop is then 20.13 Although this expression is valid, it does not give the emf as a function of time. To find how the emf evolves in time, we assume that the coil is rotated at a constant angular velocity The angle is related to the angular velocity by so that Recall that tangential velocity vis related to angular velocity by Here, , so that and Noting that the area of the loop is and allowing for Nwire loops, we find that 20.14 20.15 20.16 is the emf induced in a generator coil of Nturns and area Arotating at a constant angular velocity B. This can also be expressed as in a uniform magnetic field where is the maximum (peak) emf. 20.17 20.18 668 Chapter 20 • Magnetism Figure 20.27 The instantaneous velocity of the vertical wire segments makes an angle with the magnetic field. The velocity is shown in the figure by the green arrow, and the angle is indicated. to a negative maximum of Figure 20.28 shows a generator connected to a light bulb and a graph of the emf vs. time. Note that the emf oscillates from a positive maximum of current flows through the light bulb at these times. Thus, the light bulb actually flickers on and off at a frequency of 2f, because there are two zero crossings per period. Since alternating current such as this is used in homes around the world, why do we not notice the lights flickering on and off? In the United States, the frequency of alternating current is 60 Hz, so the lights flicker on and off at a frequency of 120 Hz. This is faster than the refresh rate of the human eye, so you don’t notice the flicker of the lights. Also, other factors prevent various different types of light bulbs from switching on and off so fast, so the light output is smoothed outa bit. In between, the emf goes through zero, which means that zero Figure 20.28 The emf of a generator is sent to a light bulb with the system of rings and brushes shown. The graph gives the emf of the generator as a function of time. is the peak emf. The period is where fis the frequency at which the coil is rotated in the magnetic field. Virtual Physics Generator Click to view content (http://www.openstax.org/l/28gen) Use this simulation to discover how an electrical generator works. Control the water supply that makes a water wheel turn a magnet. This induces an emf in a nearby wire coil, which is used to light a light bulb. You can also replace the light bulb with a voltmeter, which allows you to see the polarity of the voltage, which changes from positive to negative. GRASP CHECK Set the number of wire loops to three, the bar-magnet strength to about 50 percent, and the loop area to 100 percent. Note the maximum voltage on the voltmeter. Assuming that one major division on the voltmeter is 5V, what is the maximum voltage when using only a single wire loop instead of three wire loops? a. b. 5 V 15 V Access for free at openstax.org. c. d. 125 V 53 V 20.2 • Motors, Generators, and Transformers 669 In real life, electric generators look a lot different than the figures in this section, but the principles are the same. The source of mechanical energy that turns the coil can be falling water—hydropower—steam produced by the burning of fossil fuels, or the kinetic energy of wind. Figure 20.29 shows a cutaway view of a steam turbine; steam moves over the blades connected to the shaft, which rotates the coil within the generator. Figure 20.29 Steam turbine generator. The steam produced by burning coal impacts the turbine blades, turning the shaft which is connected to the generator. (credit: Nabonaco, Wikimedia Commons) Another very useful and common device that exploits magnetic induction is called a transformer. Transformers do what their name implies—they transform voltages from one value to another; the term voltage is used rather than emf because transformers have internal resistance. For example, many cell phones, laptops, video games, power tools, and small appliances have a transformer built into their plug-in unit that changes 120 V or 240 V AC into whatever voltage the device uses. Figure 20.30 shows two different transformers. Notice the wire coils that are visible in each device. The purpose of these coils is explained below. Figure 20.30 On the left is a common laminated-core transformer, which is widely used in electric power transmission and electrical appliances. On the right is a toroidal transformer, which is smaller than the laminated-core transformer for the same power rating but is more expensive to make because of the equipment required to wind the wires in the doughnut shape. Figure 20.31 shows a laminated-coil transformer, which is based on Faraday’s law of induction and is very similar in construction to the apparatus Faraday used to demonstrate that magnetic fields can generate electric currents. The two wire coils are called the primary and secondary coils. In normal use, the input voltage is applied across the primary coil, and the secondary produces the transformed output voltage. Not only does the iron core trap the magnetic field created by the primary coil, but also its magnetization increases the field strength, which is analogous to how a dielectric increases the electric field strength in a capacitor. Since the input voltage is AC, a time-varying magnetic flux is sent through the secondary coil, inducing an AC output voltage. 670 Chapter 20 • Magnetism Figure 20.31 A typical construction of a simple transformer has two coils wound on a ferromagnetic core. The magnetic field created by the primary coil is mostly confined to and increased by the core, which transmits it to the secondary coil. Any change in current in the primary coil induc |
es a current in the secondary coil. LINKS TO PHYSICS Magnetic Rope Memory To send men to the moon, the Apollo program had to design an onboard computer system that would be robust, consume little power, and be small enough to fit onboard the spacecraft. In the 1960s, when the Apollo program was launched, entire buildings were regularly dedicated to housing computers whose computing power would be easily outstripped by today’s most basic handheld calculator. To address this problem, engineers at MIT and a major defense contractor turned to magnetic rope memory, which was an offshoot of a similar technology used prior to that time for creating random access memories. Unlike random access memory, magnetic rope memory was read-only memory that contained not only data but instructions as well. Thus, it was actually more than memory: It was a hard-wired computer program. The components of magnetic rope memory were wires and iron rings—which were called cores. The iron cores served as transformers, such as that shown in the previous figure. However, instead of looping the wires multiple times around the core, individual wires passed only a single time through the cores, making these single-turn transformers. Up to 63 wordwires could pass through a single core, along with a single bitwire. If a word wire passed through a given core, a voltage pulse on this wire would induce an emf in the bit wire, which would be interpreted as a one. If the word wire did not pass through the core, no emf would be induced on the bit wire, which would be interpreted as a zero. Engineers would create programs that would be hard wired into these magnetic rope memories. The wiring process could take as long as a month to complete as workers painstakingly threaded wires through some cores and around others. If any mistakes were made either in the programming or the wiring, debugging would be extraordinarily difficult, if not impossible. These modules did their job quite well. They are credited with correcting an astronaut mistake in the lunar landing procedure, thereby allowing Apollo 11 to land on the moon. It is doubtful that Michael Faraday ever imagined such an application for magnetic induction when he discovered it. GRASP CHECK If the bit wire were looped twice around each core, how would the voltage induced in the bit wire be affected? a. b. c. d. If number of loops around the wire is doubled, the emf is halved. If number of loops around the wire is doubled, the emf is not affected. If number of loops around the wire is doubled, the emf is also doubled. If number of loops around the wire is doubled, the emf is four times the initial value. For the transformer shown in Figure 20.31, the output voltage voltage the secondary coil gives its induced output voltage to be across the primary coil and the number of loops in the primary and secondary coils. Faraday’s law of induction for from the secondary coil depends almost entirely on the input Access for free at openstax.org. 20.19 20.2 • Motors, Generators, and Transformers 671 is the number of loops in the secondary coil and where equals the induced emf sectional area of the coils is the same on each side, as is the magnetic field strength, and so input primary voltage is the rate of change of magnetic flux. The output voltage provided coil resistance is small—a reasonable assumption for transformers. The cross- is also related to changing flux by is the same on each side. The 20.20 20.21 Taking the ratio of these last two equations yields the useful relationship This is known as the transformer equation. It simply states that the ratio of the secondary voltage to the primary voltage in a transformer equals the ratio of the number of loops in secondary coil to the number of loops in the primary coil. Transmission of Electrical Power Transformers are widely used in the electric power industry to increase voltages—called step-uptransformers—before longdistance transmission via high-voltage wires. They are also used to decrease voltages—called step-downtransformers—to deliver power to homes and businesses. The overwhelming majority of electric power is generated by using magnetic induction, whereby a wire coil or copper disk is rotated in a magnetic field. The primary energy required to rotate the coils or disk can be provided by a variety of means. Hydroelectric power plants use the kinetic energy of water to drive electric generators. Coal or nuclear power plants create steam to drive steam turbines that turn the coils. Other sources of primary energy include wind, tides, or waves on water. Once power is generated, it must be transmitted to the consumer, which often means transmitting power over hundreds of kilometers. To do this, the voltage of the power plant is increased by a step-up transformer, that is stepped up, and the current decreases proportionally because 20.22 The lower current flow. This heating is caused by the small, but nonzero, resistance environment through this heat is in the transmission wires reduces the Joule losses, which is heating of the wire due to a current of the transmission wires. The power lost to the which is proportional to the current squaredin the transmission wire. This is why the transmitted current small as possible and, consequently, the voltage must be large to transmit the power 20.23 must be as Voltages ranging from 120 to 700 kV are used for transmitting power over long distances. The voltage is stepped up at the exit of the power station by a step-up transformer, as shown in Figure 20.32. Figure 20.32 Transformers change voltages at several points in a power distribution system. Electric power is usually generated at greater than 10 kV, and transmitted long distances at voltages ranging from 120 kV to 700 kV to limit energy losses. Local power distribution to neighborhoods or industries goes through a substation and is sent short distances at voltages ranging from 5 to 13 kV. This is reduced to 120, 240, or 480 V for safety at the individual user site. Once the power has arrived at a population or industrial center, the voltage is stepped down at a substation to between 5 and 30 672 Chapter 20 • Magnetism kV. Finally, at individual homes or businesses, the power is stepped down again to 120, 240, or 480 V. Each step-up and stepdown transformation is done with a transformer designed based on Faradays law of induction. We’ve come a long way since Queen Elizabeth asked Faraday what possible use could be made of electricity. Check Your Understanding 7. What is an electric motor? a. An electric motor transforms electrical energy into mechanical energy. b. An electric motor transforms mechanical energy into electrical energy. c. An electric motor transforms chemical energy into mechanical energy. d. An electric motor transforms mechanical energy into chemical energy. 8. What happens to the torque provided by an electric motor if you double the number of coils in the motor? a. The torque would be doubled. b. The torque would be halved. c. The torque would be quadrupled. d. The torque would be tripled. 9. What is a step-up transformer? a. A step-up transformer decreases the current to transmit power over short distance with minimum loss. b. A step-up transformer increases the current to transmit power over short distance with minimum loss. c. A step-up transformer increases voltage to transmit power over long distance with minimum loss. d. A step-up transformer decreases voltage to transmit power over short distance with minimum loss. 10. What should be the ratio of the number of output coils to the number of input coil in a step-up transformer to increase the voltage fivefold? a. The ratio is five times. b. The ratio is 10 times. c. The ratio is 15 times. d. The ratio is 20 times. 20.3 Electromagnetic Induction Section Learning Objectives By the end of this section, you will be able to do the following: • Explain how a changing magnetic field produces a current in a wire • Calculate induced electromotive force and current Section Key Terms emf induction magnetic flux Changing Magnetic Fields In the preceding section, we learned that a current creates a magnetic field. If nature is symmetrical, then perhaps a magnetic field can create a current. In 1831, some 12 years after the discovery that an electric current generates a magnetic field, English scientist Michael Faraday (1791–1862) and American scientist Joseph Henry (1797–1878) independently demonstrated that magnetic fields can produce currents. The basic process of generating currents with magnetic fields is called induction; this process is also called magnetic induction to distinguish it from charging by induction, which uses the electrostatic Coulomb force. When Faraday discovered what is now called Faraday’s law of induction, Queen Victoria asked him what possible use was electricity. “Madam,” he replied, “What good is a baby?” Today, currents induced by magnetic fields are essential to our technological society. The electric generator—found in everything from automobiles to bicycles to nuclear power plants—uses magnetism to generate electric current. Other devices that use magnetism to induce currents include pickup coils in electric guitars, transformers of every size, certain microphones, airport security gates, and damping mechanisms on sensitive chemical balances. Access for free at openstax.org. 20.3 • Electromagnetic Induction 673 One experiment Faraday did to demonstrate magnetic induction was to move a bar magnet through a wire coil and measure the resulting electric current through the wire. A schematic of this experiment is shown in Figure 20.33. He found that current is induced only when the magnet moves with respect to the coil. When the magnet is motionless with respect to the coil, no current is induced in the coil, as in Figure 20.33. In addition, moving the magnet in the opposite direction (compare Figure 20.33 with Figure 20.33) or reversing the poles of the magnet (compare |
Figure 20.33 with Figure 20.33) results in a current in the opposite direction. Figure 20.33 Movement of a magnet relative to a coil produces electric currents as shown. The same currents are produced if the coil is moved relative to the magnet. The greater the speed, the greater the magnitude of the current, and the current is zero when there is no motion. The current produced by moving the magnet upward is in the opposite direction as the current produced by moving the magnet downward. Virtual Physics Faraday’s Law Click to view content (http://www.openstax.org/l/faradays-law) Try this simulation to see how moving a magnet creates a current in a circuit. A light bulb lights up to show when current is flowing, and a voltmeter shows the voltage drop across the light bulb. Try moving the magnet through a four-turn coil and through a two-turn coil. For the same magnet speed, which coil produces a higher voltage? GRASP CHECK With the north pole to the left and moving the magnet from right to left, a positive voltage is produced as the magnet enters the coil. What sign voltage will be produced if the experiment is repeated with the south pole to the left? a. The sign of voltage will change because the direction of current flow will change by moving south pole of the magnet to the left. b. The sign of voltage will remain same because the direction of current flow will not change by moving south pole of the magnet to the left. c. The sign of voltage will change because the magnitude of current flow will change by moving south pole of the magnet to the left. d. The sign of voltage will remain same because the magnitude of current flow will not change by moving south pole of the magnet to the left. Induced Electromotive Force If a current is induced in the coil, Faraday reasoned that there must be what he called an electromotive forcepushing the charges through the coil. This interpretation turned out to be incorrect; instead, the external source doing the work of moving the magnet adds energy to the charges in the coil. The energy added per unit charge has units of volts, so the electromotive force is actually a potential. Unfortunately, the name electromotive force stuck and with it the potential for confusing it with a real force. For this reason, we avoid the term electromotive forceand just use the abbreviation emf, which has the mathematical symbol The emf may be defined as the rate at which energy is drawn from a source per unit current flowing through a circuit. Thus, emf is the energy per unit charge addedby a source, which contrasts with voltage, which is the energy per unit charge 674 Chapter 20 • Magnetism releasedas the charges flow through a circuit. To understand why an emf is generated in a coil due to a moving magnet, consider Figure 20.34, which shows a bar magnet moving downward with respect to a wire loop. Initially, seven magnetic field lines are going through the loop (see left-hand image). Because the magnet is moving away from the coil, only five magnetic field lines are going through the loop after a short time (see right-hand image). Thus, when a change occurs in the number of magnetic field lines going through the area defined by the wire loop, an emf is induced in the wire loop. Experiments such as this show that the induced emf is proportional to the rate of changeof the magnetic field. Mathematically, we express this as 20.24 where is the change in the magnitude in the magnetic field during time and Ais the area of the loop. Figure 20.34 The bar magnet moves downward with respect to the wire loop, so that the number of magnetic field lines going through the loop decreases with time. This causes an emf to be induced in the loop, creating an electric current. Note that magnetic field lines that lie in the plane of the wire loop do not actually pass through the loop, as shown by the leftmost loop in Figure 20.35. In this figure, the arrow coming out of the loop is a vector whose magnitude is the area of the loop and whose direction is perpendicular to the plane of the loop. In Figure 20.35, as the loop is rotated from the contribution of the magnetic field lines to the emf increases. Thus, what is important in generating an emf in the wire loop is the component of the magnetic field that is perpendicularto the plane of the loop, which is to This is analogous to a sail in the wind. Think of the conducting loop as the sail and the magnetic field as the wind. To maximize the force of the wind on the sail, the sail is oriented so that its surface vector points in the same direction as the winds, as in the right-most loop in Figure 20.35. When the sail is aligned so that its surface vector is perpendicular to the wind, as in the leftmost loop in Figure 20.35, then the wind exerts no force on the sail. Thus, taking into account the angle of the magnetic field with respect to the area, the proportionality becomes 20.25 Figure 20.35 The magnetic field lies in the plane of the left-most loop, so it cannot generate an emf in this case. When the loop is rotated so that the angle of the magnetic field with the vector perpendicular to the area of the loop increases to (see right-most loop), the magnetic field contributes maximally to the emf in the loop. The dots show where the magnetic field lines intersect the plane defined by the loop. Access for free at openstax.org. 20.3 • Electromagnetic Induction 675 Another way to reduce the number of magnetic field lines that go through the conducting loop in Figure 20.35 is not to move the magnet but to make the loop smaller. Experiments show that changing the area of a conducting loop in a stable magnetic field induces an emf in the loop. Thus, the emf produced in a conducting loop is proportional to the rate of change of the productof the perpendicular magnetic field and the loop area 20.26 is the perpendicular magnetic field and Ais the area of the loop. The product where proportional to the number of magnetic field lines that pass perpendicularly through a surface of area A. Going back to our sail analogy, it would be proportional to the force of the wind on the sail. It is called the magnetic flux and is represented by is very important. It is . The unit of magnetic flux is the weber (Wb), which is magnetic field per unit area, or T/m2. The weber is also a volt second (Vs). The induced emf is in fact proportional to the rate of change of the magnetic flux through a conducting loop. 20.27 20.28 Finally, for a coil made from Nloops, the emf is Ntimes stronger than for a single loop. Thus, the emf induced by a changing magnetic field in a coil of Nloops is The last question to answer before we can change the proportionality into an equation is “In what direction does the current flow?” The Russian scientist Heinrich Lenz (1804–1865) explained that the current flows in the direction that creates a magnetic field that tries to keep the flux constant in the loop. For example, consider again Figure 20.34. The motion of the bar magnet causes the number of upward-pointing magnetic field lines that go through the loop to decrease. Therefore, an emf is generated in the loop that drives a current in the direction that creates more upward-pointing magnetic field lines. By using the righthand rule, we see that this current must flow in the direction shown in the figure. To express the fact that the induced emf acts to counter the change in the magnetic flux through a wire loop, a minus sign is introduced into the proportionality , which gives Faraday’s law of induction. 20.29 Lenz’s law is very important. To better understand it, consider Figure 20.36, which shows a magnet moving with respect to a wire coil and the direction of the resulting current in the coil. In the top row, the north pole of the magnet approaches the coil, so the magnetic field lines from the magnet point toward the coil. Thus, the magnetic field right increases in the coil. According to Lenz’s law, the emf produced in the coil will drive a current in the direction that creates a pointing to the magnetic field inside the coil pointing to the left. This will counter the increase in magnetic flux pointing to the right. To see which way the current must flow, point your right thumb in the desired direction of the magnetic field and the current will flow in the direction indicated by curling your right fingers. This is shown by the image of the right hand in the top row of Figure 20.36. Thus, the current must flow in the direction shown in Figure 4(a). In Figure 4(b), the direction in which the magnet moves is reversed. In the coil, the right-pointing magnetic field the moving magnet decreases. Lenz’s law says that, to counter this decrease, the emf will drive a current that creates an due to additional right-pointing magnetic field magnetic field, and the current will flow in the direction indicate by curling your right fingers (Figure 4(b)). in the coil. Again, point your right thumb in the desired direction of the Finally, in Figure 4(c), the magnet is reversed so that the south pole is nearest the coil. Now the magnetic field toward the magnet instead of toward the coil. As the magnet approaches the coil, it causes the left-pointing magnetic field in the coil to increase. Lenz’s law tells us that the emf induced in the coil will drive a current in the direction that creates a magnetic field pointing to the right. This will counter the increasing magnetic flux pointing to the left due to the magnet. Using the righthand rule again, as indicated in the figure, shows that the current must flow in the direction shown in Figure 4(c). points 676 Chapter 20 • Magnetism Figure 20.36 Lenz’s law tells us that the magnetically induced emf will drive a current that resists the change in the magnetic flux through a circuit. This is shown in panels (a)–(c) for various magnet orientations and velocities. The right hands at right show how to apply the right- hand rule to find in which direction the induced current |
flows around the coil. Virtual Physics Faraday’s Electromagnetic Lab Click to view content (http://www.openstax.org/l/Faraday-EM-lab) This simulation proposes several activities. For now, click on the tab Pickup Coil, which presents a bar magnet that you can move through a coil. As you do so, you can see the electrons move in the coil and a light bulb will light up or a voltmeter will indicate the voltage across a resistor. Note that the voltmeter allows you to see the sign of the voltage as you move the magnet about. You can also leave the bar magnet at rest and move the coil, although it is more difficult to observe the results. GRASP CHECK Orient the bar magnet with the north pole facing to the right and place the pickup coil to the right of the bar magnet. Now move the bar magnet toward the coil and observe in which way the electrons move. This is the same situation as depicted below. Does the current in the simulation flow in the same direction as shown below? Explain why or why not. Access for free at openstax.org. 20.3 • Electromagnetic Induction 677 a. Yes, the current in the simulation flows as shown because the direction of current is opposite to the direction of flow of electrons. b. No, current in the simulation flows in the opposite direction because the direction of current is same to the direction of flow of electrons. WATCH PHYSICS Induced Current in a Wire This video explains how a current can be induced in a straight wire by moving it through a magnetic field. The lecturer uses the cross product, which a type of vector multiplication. Don’t worry if you are not familiar with this, it basically combines the righthand rule for determining the force on the charges in the wire with the equation Click to view content (https://www.openstax.org/l/induced-current) GRASP CHECK through a uniform magnetic field What emf is produced across a straight wire 0.50 m long moving at a velocity of (1.5 m/s) (0.30 T)ẑ? The wire lies in the ŷ-direction. Also, which end of the wire is at the higher potential—let the lower end of the wire be at y= 0 and the upper end at y= 0.5 m)? a. 0.15 V and the lower end of the wire will be at higher potential b. 0.15 V and the upper end of the wire will be at higher potential c. 0.075 V and the lower end of the wire will be at higher potential d. 0.075 V and the upper end of the wire will be at higher potential WORKED EXAMPLE EMF Induced in Conducing Coil by Moving Magnet Imagine a magnetic field goes through a coil in the direction indicated in Figure 20.37. The coil diameter is 2.0 cm. If the magnetic field goes from 0.020 to 0.010 T in 34 s, what is the direction and magnitude of the induced current? Assume the coil has a resistance of 0.1 678 Chapter 20 • Magnetism Figure 20.37 A coil through which passes a magnetic field B. STRATEGY Use the equation solenoid, we find it has 16 loops, so to find the induced emf in the coil, where . Counting the number of loops in the Use the equation to calculate the magnetic flux 20.30 where dis the diameter of the solenoid and we have used in the magnetic of the flux through the solenoid is Because the area of the solenoid does not vary, the change 20.31 Once we find the emf, we can use Ohm’s law, to find the current. Finally, Lenz’s law tells us that the current should produce a magnetic field that acts to oppose the decrease in the applied magnetic field. Thus, the current should produce a magnetic field to the right. Solution Combining equations and gives Solving Ohm’s law for the current and using this result gives 20.32 20.33 Lenz’s law tells us that the current must produce a magnetic field to the right. Thus, we point our right thumb to the right and curl our right fingers around the solenoid. The current must flow in the direction in which our fingers are pointing, so it enters at the left end of the solenoid and exits at the right end. Discussion Let’s see if the minus sign makes sense in Faraday’s law of induction. Define the direction of the magnetic field to be the positive direction. This means the change in the magnetic field is negative, as we found above. The minus sign in Faraday’s law of induction negates the negative change in the magnetic field, leaving us with a positive current. Therefore, the current must flow in the direction of the magnetic field, which is what we found. Now try defining the positive direction to be the direction opposite that of the magnetic field, that is positive is to the left in Figure 20.37. In this case, you will find a negative current. But since the positive direction is to the left, a negative current must flow to the right, which again agrees with what we found by using Lenz’s law. WORKED EXAMPLE Magnetic Induction due to Changing Circuit Size The circuit shown in Figure 20.38 consists of a U-shaped wire with a resistor and with the ends connected by a sliding conducting rod. The magnetic field filling the area enclosed by the circuit is constant at 0.01 T. If the rod is pulled to the right at speed what current is induced in the circuit and in what direction does the current flow? Access for free at openstax.org. 20.3 • Electromagnetic Induction 679 Figure 20.38 A slider circuit. The magnetic field is constant and the rod is pulled to the right at speed v. The changing area enclosed by the circuit induces an emf in the circuit. STRATEGY We again use Faraday’s law of induction, although this time the magnetic field is constant and the area enclosed by the circuit changes. The circuit contains a single loop, so The rate of change of the area is Thus the rate of change of the magnetic flux is 20.34 where we have used the fact that the angle between the area vector and the magnetic field is 0°. Once we know the emf, we can find the current by using Ohm’s law. To find the direction of the current, we apply Lenz’s law. Solution Faraday’s law of induction gives Solving Ohm’s law for the current and using the previous result for emf gives 20.35 20.36 As the rod slides to the right, the magnetic flux passing through the circuit increases. Lenz’s law tells us that the current induced will create a magnetic field that will counter this increase. Thus, the magnetic field created by the induced current must be into the page. Curling your right-hand fingers around the loop in the clockwise direction makes your right thumb point into the page, which is the desired direction of the magnetic field. Thus, the current must flow in the clockwise direction around the circuit. Discussion Is energy conserved in this circuit? An external agent must pull on the rod with sufficient force to just balance the force on a current-carrying wire in a magnetic field—recall that be balanced by the rate at which the circuit dissipates power. Using constant speed vis the force required to pull the wire at a The rate at which this force does work on the rod should where we used the fact that the angle between the current and the magnetic field is the current into this equation gives Inserting our expression above for 20.37 20.38 The power contributed by the agent pulling the rod is 680 Chapter 20 • Magnetism The power dissipated by the circuit is 20.39 20.40 We thus see that agent that pulls the rod. Thus, energy is conserved in this system. which means that power is conserved in the system consisting of the circuit and the Practice Problems 11. The magnetic flux through a single wire loop changes from 3.5 Wb to 1.5 Wb in 2.0 s. What emf is induced in the loop? a. –2.0 V b. –1.0 V c. +1.0 V d. +2.0 V 12. What is the emf for a 10-turn coil through which the flux changes at 10 Wb/s? a. –100 V b. –10 V c. +10 V d. +100 V Check Your Understanding 13. Given a bar magnet, how can you induce an electric current in a wire loop? a. An electric current is induced if a bar magnet is placed near the wire loop. b. An electric current is induced if wire loop is wound around the bar magnet. c. An electric current is induced if a bar magnet is moved through the wire loop. d. An electric current is induced if a bar magnet is placed in contact with the wire loop. 14. What factors can cause an induced current in a wire loop through which a magnetic field passes? a. b. c. d. Induced current can be created by changing the size of the wire loop only. Induced current can be created by changing the orientation of the wire loop only. Induced current can be created by changing the strength of the magnetic field only. Induced current can be created by changing the strength of the magnetic field, changing the size of the wire loop, or changing the orientation of the wire loop. Access for free at openstax.org. Chapter 20 • Key Terms 681 KEY TERMS Curie temperature well-defined temperature for the magnetic force ferromagnetic materials above which they cannot be magnetized domain region within a magnetic material in which the magnetic flux component of the magnetic field perpendicular to the surface area through which it passes and multiplied by the area magnetic poles of individual atoms are aligned magnetic pole part of a magnet that exerts the strongest electric motor device that transforms electrical energy into force on other magnets or magnetic material mechanical energy magnetized material that is induced to be magnetic or that electromagnet device that uses electric current to make a is made into a permanent magnet magnetic field north pole part of a magnet that orients itself toward the electromagnetism study of electric and magnetic geographic North Pole of Earth phenomena emf rate at which energy is drawn from a source per unit current flowing through a circuit ferromagnetic material such as iron, cobalt, nickel, or gadolinium that exhibits strong magnetic effects generator device that transforms mechanical energy into electrical energy permanent magnet material that retains its magnetic behavior for a long time, even when exposed to demagnetizing influences right-hand rule rule involving curling the right-hand fingers from one vector to |
another; the direction in which the right thumb points is the direction of the resulting vector induction rate at which energy is drawn from a source per unit current flowing through a circuit magnetic dipole term that describes magnets because they solenoid uniform cylindrical coil of wire through which electric current is passed to produce a magnetic field south pole part of a magnet that orients itself toward the always have two poles: north and south geographic South Pole of Earth magnetic field directional lines around a magnetic transformer device that transforms voltages from one material that indicates the direction and magnitude of value to another SECTION SUMMARY 20.1 Magnetic Fields, Field Lines, and Force • All magnets have two poles: a north pole and a south pole. If the magnet is free to move, its north pole orients itself toward the geographic North Pole of Earth, and the south pole orients itself toward the geographic South Pole of Earth. • A repulsive force occurs between the north poles of two magnets and likewise for two south poles. However, an attractive force occurs between the north pole of one magnet and the south pole of another magnet. • A charged particle moving through a magnetic field experiences a force whose direction is determined by the right-hand rule. • An electric current generates a magnetic field. • Electromagnets are magnets made by passing a current through a system of wires. 20.2 Motors, Generators, and Transformers • Electric motors contain wire loops in a magnetic field. Current is passed through the wire loops, which forces them to rotate in the magnetic field. The current is reversed every half rotation so that the torque on the loop is always in the same direction. • Electric generators contain wire loops in a magnetic field. An external agent provides mechanical energy to force the loops to rotate in the magnetic field, which produces an AC voltage that drives an AC current through the loops. • Transformers contain a ring made of magnetic material and, on opposite sides of the ring, two windings of wire wrap around the ring. A changing current in one wire winding creates a changing magnetic field, which is trapped in the ring and thus goes through the second winding and induces an emf in the second winding. The voltage in the second winding is proportional to the ratio of the number of loops in each winding. • Transformers are used to step up and step down the voltage for power transmission. • Over long distances, electric power is transmitted at high voltage to minimize the current and thereby minimize the Joule losses due to resistive heating. 20.3 Electromagnetic Induction • Faraday’s law of induction states that a changing magnetic flux that occurs within an area enclosed by a conducting loop induces an electric current in the loop. • Lenz’ law states that an induced current flows in the direction such that it opposes the change that induced it. 682 Chapter 20 • Key Equations KEY EQUATIONS 20.1 Magnetic Fields, Field Lines, and Force the magnitude of the force on an electric charge the force on a wire carrying current the magnitude of the magnetic field created by a long, straight current-carrying wire CHAPTER REVIEW Concept Items 20.1 Magnetic Fields, Field Lines, and Force 1. If you place a small needle between the north poles of two bar magnets, will the needle become magnetized? a. Yes, the magnetic fields from the two north poles will point in the same directions. b. Yes, the magnetic fields from the two north poles will point in opposite directions. c. No, the magnetic fields from the two north poles will point in opposite directions. d. No, the magnetic fields from the two north poles will point in the same directions. 2. If you place a compass at the three points in the figure, at which point will the needle experience the greatest torque? Why? the magnitude of the magnetic field inside a solenoid 20.3 Electromagnetic Induction magnetic flux emf greatest torque at B. b. The density of the magnetic field is minimized at C, so the magnetic compass needle will experience the greatest torque at C. c. The density of the magnetic field is maximized at B, so the magnetic compass needle will experience the greatest torque at B. d. The density of the magnetic field is maximized at A, so the magnetic compass needle will experience the greatest torque at A. 3. In which direction do the magnetic field lines point near the south pole of a magnet? a. Outside the magnet the direction of magnetic field lines is towards the south pole of the magnet. b. Outside the magnet the direction of magnetic field lines is away from the south pole of the magnet. 20.2 Motors, Generators, and Transformers 4. Consider the angle between the area vector and the and magnetic field in an electric motor. At what angles is the torque on the wire loop the greatest? a. b. c. d. and and and 5. What is a voltage transformer? a. A transformer is a device that transforms current to voltage. b. A transformer is a device that transforms voltages from one value to another. c. A transformer is a device that transforms resistance of wire to voltage. 6. Why is electric power transmitted at high voltage? a. The density of the magnetic field is minimized at B, so the magnetic compass needle will experience the Access for free at openstax.org. Chapter 20 • Chapter Review 683 a. To increase the current for the transmission b. To reduce energy loss during transmission c. To increase resistance during transmission d. To reduce resistance during transmission 20.3 Electromagnetic Induction 7. Yes or no—Is an emf induced in the coil shown when it is stretched? If so, state why and give the direction of the induced current. a. b. c. d. If induced current flows, its direction is such that it adds to the changes which induced it. If induced current flows, its direction is such that it opposes the changes which induced it. If induced current flows, its direction is always clockwise to the changes which induced it. If induced current flows, its direction is always counterclockwise to the changes which induced it. 9. Explain how magnetic flux can be zero when the a. No, because induced current does not depend upon the area of the coil. b. Yes, because area of the coil increases; the direction of the induced current is counterclockwise. c. Yes, because area of the coil increases; the direction of the induced current is clockwise. d. Yes, because the area of the coil does not change; the direction of the induced current is clockwise. 8. What is Lenz’s law? Critical Thinking Items 20.1 Magnetic Fields, Field Lines, and Force 10. True or false—It is not recommended to place credit cards with magnetic strips near permanent magnets. a. b. false true 11. True or false—A square magnet can have sides that alternate between north and south poles. a. b. false true 12. You move a compass in a circular plane around a planar magnet. The compass makes four complete revolutions. How many poles does the magnet have? two poles a. b. four poles c. eight poles 12 poles d. 20.2 Motors, Generators, and Transformers 13. How can you maximize the peak emf from a generator? a. The peak emf from a generator can be maximized only by maximizing number of turns. b. The peak emf from a generator can be maximized only by maximizing area of the wired loop. magnetic field is not zero. a. If angle between magnetic field and area vector is 0°, then its sine is also zero, which means that there is zero flux. If angle between magnetic field and area vector is 45°, then its sine is also zero, which means that there is zero flux. If angle between magnetic field and area vector is 60°, then its cosine is also zero, which means that there is zero flux. If the angle between magnetic field and area vector is 90°, then its cosine is also zero, which means that there is zero flux. b. c. d. c. The peak emf from a generator can be maximized only by maximizing frequency. d. The peak emf from a generator can be maximized by maximizing number of turns, maximizing area of the wired loop or maximizing frequency. 14. Explain why power is transmitted over long distances at high voltages. a. Plost = Itransmitted Vtransmitted, so to maximize current, the voltage must be maximized b. Ptransmitted = Itransmitted Vtransmitted, so to maximize current, the voltage must be maximized c. Plost = Itransmitted Vtransmitted, so to minimize current, the voltage must be maximized d. Ptransmitted = Itransmitted Vtransmitted, so to minimize current, the voltage must be maximized 20.3 Electromagnetic Induction 15. To obtain power from the current in the wire of your vacuum cleaner, you place a loop of wire near it to obtain an induced emf. How do you place and orient the loop? a. A loop of wire should be placed nearest to the vacuum cleaner wire to maximize the magnetic flux through the loop. b. A loop of wire should be placed farthest to the vacuum cleaner wire to maximize the magnetic flux through the loop. 684 Chapter 20 • Test Prep c. A loop of wire should be placed perpendicular to the vacuum cleaner wire to maximize the magnetic flux through the loop. d. A loop of wire should be placed at angle greater than 90° to the vacuum cleaner wire to maximize the magnetic flux through the loop. that is proportional to the rate of change of the magnetic flux. b. The magnetic field in the coil changes rapidly due to spinning of magnet which creates an emf in the coil that is proportional to the rate of change of the magnetic flux. 16. A magneto is a device that creates a spark across a gap 17. If you drop a copper tube over a bar magnet with its by creating a large voltage across the gap. To do this, the device spins a magnet very quickly in front of a wire coil, with the ends of the wires forming the gap. Explain how this creates a sufficiently large voltage to produce a spark. a. The electric field in the coil increases rapidly due to spinning of magnet which creates an emf in the coil nor |
th pole up, is a current induced in the copper tube? If so, in what direction? Consider when the copper tube is approaching the bar magnet. a. Yes, the induced current will be produced in the clockwise direction when viewed from above. b. No, the induced current will not be produced. Problems 20.1 Magnetic Fields, Field Lines, and Force 18. A straight wire segment carries 0.25 A. What length would it need to be to exert a 4.0-mN force on a magnet that produces a uniform magnetic field of 0.015 T that is perpendicular to the wire? a. 0.55 m b. 1.10 m c. 2.20 m d. 4.40 m 20.3 Electromagnetic Induction 19. What is the current in a wire loop of resistance 10 Ω through which the magnetic flux changes from zero to Performance Task 20.2 Motors, Generators, and Transformers 21. Your family takes a trip to Cuba, and rents an old car to drive into the countryside to see the sights. Unfortunately, the next morning you find yourself deep in the countryside and the car won’t start because the battery is too weak. Wanting to jump-start the car, you open the hood and find that you can’t tell which battery TEST PREP Multiple Choice 20.1 Magnetic Fields, Field Lines, and Force 22. For a magnet, a domain refers to ______. a. b. c. the region between the poles of the magnet the space around the magnet that is affected by the magnetic field the region within the magnet in which the Access for free at openstax.org. 10 Wb in 1.0 s? a. –100 A b. –2.0 A c. –1.0 A d. +1.0 A 20. An emf is induced by rotating a 1,000 turn, 20.0 cm diameter coil in Earth’s 5.00 × 10–5 T magnetic field. What average emf is induced, given the plane of the coil is originally perpendicular to Earth’s field and is rotated to be parallel to the field in 10.0 ms? a. –1.6 × 10-4 V b. +1.6 × 10-4 V c. +1.6 × 10-1 V d. –1.6 × 10-1 V terminal is positive and which is negative. However, you do have a bar magnet with the north and south poles labeled and you manage to find a short wire. How do you use these to determine which terminal is which? For starters, how do you determine the direction of a magnetic field around a current-carrying wire? And in which direction will the force be on another magnet placed in this field? Do you need to worry about the sign of the mobile charge carriers in the wire? d. magnetic poles of individual atoms are aligned the region from which the magnetic material is mined 23. In the region just outside the south pole of a magnet, the magnetic field lines ______. a. point away from the south pole b. go around the south pole c. are less concentrated than at the north pole d. point toward the south pole 24. Which equation gives the force for a charge moving through a magnetic field? a. b. c. d. 25. Can magnetic field lines cross each other? Explain why or why not. a. Yes, magnetic field lines can cross each other because that point of intersection indicates two possible directions of magnetic field, which is possible. b. No, magnetic field lines cannot cross each other because that point of intersection indicates two possible directions of magnetic field, which is not possible. 26. True or false—If a magnet shatters into many small pieces, all the pieces will have north and south poles a. b. true false 20.2 Motors, Generators, and Transformers 27. An electrical generator ________. is a generator powered by electricity a. b. must be turned by hand c. converts other sources of power into electrical power d. uses magnetism to create electrons 28. A step-up transformer increases the a. voltage from power lines for use in homes b. c. current from the power lines for use in homes current from the electrical generator for transmission along power lines d. voltage from the electrical power plant for transmission along power lines Chapter 20 • Test Prep 685 b. Torque is doubled. c. Torque is quadrupled. d. Torque is halved. 30. Why are the coils of a transformer wrapped around a loop of ferrous material? a. The magnetic field from the source coil is trapped and also increased in strength. b. The magnetic field from the source coil is dispersed and also increased in strength. c. The magnetic field from the source coil is trapped and also decreased in strength. d. Magnetic field from the source coil is dispersed and also decreased in strength. 20.3 Electromagnetic Induction 31. What does emfstand for? a. electromotive force b. electro motion force c. electromagnetic factor d. electronic magnetic factor 32. Which formula gives magnetic flux? a. b. c. d. 33. What is the relationship between the number of coils in a solenoid and the emf induced in it by a change in the magnetic flux through the solenoid? a. The induced emf is inversely proportional to the number of coils in a solenoid. b. The induced emf is directly proportional to the number of coils in a solenoid. c. The induced emf is inversely proportional to the square of the number of coils in a solenoid. d. The induced emf is proportional to square of the number of coils in a solenoid. 29. What would be the effect on the torque of an electric 34. True or false—If you drop a bar magnet through a motor of doubling the width of the current loop in the motor? a. Torque remains the same. copper tube, it induces an electric current in the tube. a. b. false true Short Answer 20.1 Magnetic Fields, Field Lines, and Force 35. Given a bar magnet, a needle, a cork, and a bowl full of water, describe how to make a compass. a. Magnetize the needle by holding it perpendicular to a bar magnet’s north pole and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself along the magnetic field lines of Earth. b. Magnetize the needle by holding it perpendicular to a bar magnet’s north pole and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself perpendicular to the Hence, it will attract the south pole of other magnet. c. The needle will magnetize and the point of a needle kept closer to the north pole will act as a north pole. Hence, it will repel the south pole of the other magnet. d. The needle will magnetize and the point of needle kept closer to the north pole will act as a north pole. Hence, it will attract the south pole of other magnet. 39. Using four solenoids of the same size, describe how to orient them and in which direction the current should flow to make a magnet with two opposite-facing north poles and two opposite-facing south poles. 686 Chapter 20 • Test Prep magnetic field lines of Earth. c. Magnetize the needle by holding its axis parallel to the axis of a bar magnet and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself along the magnetic field lines of Earth. d. Magnetize the needle by holding its axis parallel to the axis of a bar magnet and pierce the cork along its longitudinal axis by the needle and place the needle-cork combination in the water. The needle now orients itself perpendicular to the magnetic field lines of Earth. 36. Give two differences between electric field lines and magnetic field lines. a. Electric field lines begin and end on opposite charges and the electric force on a charge is in the direction of field, while magnetic fields form a loop and the magnetic force on a charge is perpendicular to the field. b. Electric field lines form a loop and the electric force on a charge is in the direction of field, while magnetic fields begin and end on opposite charge and the magnetic force on a charge is perpendicular to the field. c. Electric field lines begin and end on opposite charges and the electric force on a charge is in the perpendicular direction of field, while magnetic fields form a loop and the magnetic force on a charge is in the direction of the field. d. Electric field lines form a loop and the electric force on a charge is in the perpendicular direction of field, while magnetic fields begin and end on opposite charge and the magnetic force on a charge is in the direction of the field. 37. To produce a magnetic field of 0.0020 T, what current is required in a 500-turn solenoid that is 25 cm long? a. 0.80 A b. 1.60 A c. 80 A 160 A d. 38. You magnetize a needle by aligning it along the axis of a bar magnet and just outside the north pole of the magnet. Will the point of the needle that was closest to the bar magnet then be attracted to or repelled from the south pole of another magnet? a. The needle will magnetize and the point of needle kept closer to the north pole will act as a south pole. Hence, it will repel the south pole of other magnet. b. The needle will magnetize and the point of needle kept closer to the north pole will act as a south pole. Access for free at openstax.org. Chapter 20 • Test Prep 687 a. The output emf will be doubled. b. The output emf will be halved. c. The output emf will be quadrupled. d. The output emf will be tripled. 44. In a hydroelectric dam, what is used to power the electrical generators that provide electric power? Explain. a. The electric potential energy of stored water is used to produce emf with the help of a turbine. b. The electric potential energy of stored water is used to produce resistance with the help of a turbine. c. Gravitational potential energy of stored water is used to produce resistance with the help of a turbine. d. Gravitational potential energy of stored water is used to produce emf with the help of a turbine. 20.3 Electromagnetic Induction 45. A uniform magnetic field is perpendicular to the plane of a wire loop. If the loop accelerates in the direction of the field, will a current be induced in the loop? Explain why or why not. a. No, because magnetic flux through the loop remains constant. b. No, because magnetic flux through the loop changes continuously. c. Yes, because magnetic flux through the loop remains constant. d. Yes, because magnetic flux th |
rough the loop changes continuously. 46. The plane of a square wire circuit with side 4.0 cm long is at an angle of 45° with respect to a uniform magnetic field of 0.25 T. The wires have a resistance per unit length of 0.2. If the field drops to zero in 2.5 s, what magnitude current is induced in the square circuit? 35 µA a. b. 87.5 µA 3.5 mA c. 35 A d. 47. Yes or no—If a bar magnet moves through a wire loop as shown in the figure, is a current induced in the loop? Explain why or why not. 40. How far from a straight wire carrying 0.45 A is the magnetic field strength 0.040 T? a. 0.23 µm b. 0.72 µm c. 2.3 µm 7.2 µm d. 20.2 Motors, Generators, and Transformers 41. A laminated-coil transformer has a wire coiled 12 times around one of its sides. How many coils should you wrap around the opposite side to get a voltage output that is one half of the input voltage? Explain. a. six output coils because the ratio of output to input voltage is the same as the ratio of number of output coils to input coils 12 output coils because the ratio of output to input voltage is the same as the ratio of number of output coils to input coils b. c. 24 output coils because the ratio of output to input voltage is half the ratio of the number of output coils to input coils 36 output coils because the ratio of output to input voltage is three times the ratio of the number of output coils to input coils d. 42. Explain why long-distance electrical power lines are designed to carry very high voltages. a. Ptransmitted = Itransmitted> 2 Rwire and Plost = Itransmitted Vtransmitted, so Vmust be low to make the current transmitted as high as possible. b. Ptransmitted = Itransmitted> 2 Rwire and Plost = Ilost Vlost, so Vmust be low to make the current transmitted as high as possible. c. Ptransmitted = Itransmitted> 2 Rwire and Plost = Itransmitted Vtransmitted, so Vmust be high to make the current transmitted as low as possible d. Plost = Itransmitted 2 Rwire and Ptransmitted = Itransmitted Vtransmitted, so Vmust be high to make the current transmitted as low as possible. a. No, because the net magnetic field passing through the loop is zero. 43. How is the output emf of a generator affected if you b. No, because the net magnetic field passing through double the frequency of rotation of its coil? 688 Chapter 20 • Test Prep the loop is nonzero. c. Yes, because the net magnetic field passing through the loop is zero. d. Yes, because the net magnetic field line passing through the loop is nonzero. 48. What is the magnetic flux through an equilateral Extended Response 20.1 Magnetic Fields, Field Lines, and Force 49. Summarize the properties of magnets. a. A magnet can attract metals like iron, nickel, etc., but cannot attract nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic south pole of Earth. b. A magnet can attract metals like iron, nickel, etc., but cannot attract nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic north pole of Earth. c. A magnet can attract metals like iron, nickel, etc., and nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic south pole of Earth. d. A magnet can attract metals like iron, nickel, etc., and nonmetals like piece of plastic or wood, etc. If free to rotate, an elongated magnet will orient itself so that its north pole will face the magnetic north pole of Earth. 50. The magnetic field shown in the figure is formed by current flowing in two rings that intersect the page at the dots. Current flows into the page at the dots with crosses (right side) and out of the page at the dots with points (left side). Access for free at openstax.org. triangle with side 60 cm long and whose plane makes a 60° angle with a uniform magnetic field of 0.33 T? a. 0.045 Wb b. 0.09 Wb c. 0.405 Wb d. 4.5 Wb Where is the field strength the greatest and in what direction do the magnetic field lines point? a. The magnetic field strength is greatest where the magnetic field lines are less dense; magnetic field lines points up the page. b. The magnetic field strength is greatest where the magnetic field lines are most dense; magnetic field lines points up the page. c. The magnetic field strength is greatest where the magnetic field lines are most dense; magnetic field lines points down the page. d. The magnetic field strength is greatest where the magnetic field lines are less dense; magnetic field lines points down the page. 51. The forces shown below are exerted on an electron as it moves through the magnetic field. In each case, what direction does the electron move? a. b. c. d. (a) left to right, (b) out of the page, (c) upwards (a) left to right, (b) into the page, (c) downwards (a) right to left, (b) out of the page, (c) upwards (a) right to left, (b) into the page, (c) downwards 20.2 Motors, Generators, and Transformers 52. Explain why increasing the frequency of rotation of the coils in an electrical generator increases the output emf. a. The induced emf is proportional to the rate of change of magnetic flux with respect to distance. b. The induced emf is inversely proportional to the rate of change of magnetic flux with respect to distance. c. The induced emf is inversely proportional to the rate of change of magnetic flux with respect to time. d. The induced emf is proportional to the rate of change of magnetic flux with respect to time. 53. Your friend tells you that power lines must carry a maximum current because P= I2R, where R is the resistance of the transmission line. What do you tell her? a. Ptransmitted = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be high to reduce power lost due to transmission. b. Plost = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be high to reduce power lost due to transmission. c. Ptransmitted = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be low to reduce power lost due to transmission. d. Plost = Itransmitted 2Rwire and Plost = Itransmitted Vtransmitted, so Imust be low to reduce power lost due to transmission. 20.3 Electromagnetic Induction 54. When you insert a copper ring between the poles of two bar magnets as shown in the figure, do the magnets exert an attractive or repulsive force on the ring? Explain your reasoning. a. Magnets exert an attractive force, because magnetic field due to induced current is repulsed by the magnetic field of the magnets. b. Magnets exert an attractive force, because Chapter 20 • Test Prep 689 magnetic field due to induced current is attracted by the magnetic field of the magnets. c. Magnets exert a repulsive force, because magnetic field due to induced current is repulsed by the magnetic field of the magnets. d. Magnets exert a repulsive force, because magnetic field due to induced current is attracted by the magnetic field of the magnets. 55. The figure shows a uniform magnetic field passing through a closed wire circuit. The wire circuit rotates at an angular frequency of about the axis shown by the dotted line in the figure. What is an expression for the magnetic flux through the circuit as a function of time? a. expression for the magnetic flux through the circuit Φ(t) = BAcos ωt b. expression for the magnetic flux through the circuit c. expression for the magnetic flux through the circuit d. expression for the magnetic flux through the circuit Φ(t) = 2BAcos ωt 690 Chapter 20 • Test Prep Access for free at openstax.org. CHAPTER 21 The Quantum Nature of Light Figure 21.1 In Lewis Carroll’s classic text Alice’s Adventures in Wonderland, Alice follows a rabbit down a hole into a land of curiosity. While many of her interactions in Wonderland are of surprising consequence, they follow a certain inherent logic. (credit: modification of work by John Tenniel, Wikimedia Commons) Chapter Outline 21.1 Planck and Quantum Nature of Light 21.2 Einstein and the Photoelectric Effect 21.3 The Dual Nature of Light At first glance, the quantum nature of light can be a strange and bewildering concept. Between light acting as INTRODUCTION discrete chunks, massless particles providing momenta, and fundamental particles behaving like waves, it may often seem like something out of Alice in Wonderland. For many, the study of this branch of physics can be as enthralling as Lewis Carroll’s classic novel. Recalling the works of legendary characters and brilliant scientists such as Einstein, Planck, and Compton, the study of light’s quantum nature will provide you an interesting tale of how a clever interpretation of some small details led to the most important discoveries of the past 150 years. From the electronics revolution of the twentieth century to our future progress in solar energy and space exploration, the quantum nature of light should yield a rabbit hole of curious consequence, within which lie some of the most fascinating truths of our time. 692 Chapter 21 • The Quantum Nature of Light 21.1 Planck and Quantum Nature of Light Section Learning Objectives By the end of this section, you will be able to do the following: • Describe blackbody radiation • Define quantum states and their relationship to modern physics • Calculate the quantum energy of lights • Explain how photon energies vary across divisions of the electromagnetic spectrum Section Key Terms blackbody quantized quantum ultraviolet catastrophe Blackbodies Our first story of curious significance begins with a T-shirt. You are likely aware that wearing a tight black T-shirt outside on a hot day provides a significantly less comfortable experience than wearing a white shirt. Black shirts, as well as all other black objects, will absorb and re-emit a significantly greater amount of radiation from the sun. This shirt is a good approximation of what is called a bl |
ackbody. A perfect blackbody is one that absorbs and re-emits all radiated energy that is incident upon it. Imagine wearing a tight shirt that did this! This phenomenon is often modeled with quite a different scenario. Imagine carving a small hole in an oven that can be heated to very high temperatures. As the temperature of this container gets hotter and hotter, the radiation out of this dark hole would increase as well, re-emitting all energy provided it by the increased temperature. The hole may even begin to glow in different colors as the temperature is increased. Like a burner on your stove, the hole would glow red, then orange, then blue, as the temperature is increased. In time, the hole would continue to glow but the light would be invisible to our eyes. This container is a good model of a perfect blackbody. It is the analysis of blackbodies that led to one of the most consequential discoveries of the twentieth century. Take a moment to carefully examine Figure 21.2. What relationships exist? What trends can you see? The more time you spend interpreting this figure, the closer you will be to understanding quantum physics! Figure 21.2 Graphs of blackbody radiation (from an ideal radiator) at three different radiator temperatures. The intensity or rate of radiation emission increases dramatically with temperature, and the peak of the spectrum shifts toward the visible and ultraviolet parts of the spectrum. The shape of the spectrum cannot be described with classical physics. TIPS FOR SUCCESS When encountering a new graph, it is best to try to interpret the graph before you read about it. Doing this will make the following text more meaningful and will help to remind yourself of some of the key concepts within the section. Access for free at openstax.org. 21.1 • Planck and Quantum Nature of Light 693 Understanding Blackbody Graphs Figure 21.2 is a plot of radiation intensity against radiated wavelength. In other words, it shows how the intensity of radiated light changes when a blackbody is heated to a particular temperature. It may help to just follow the bottom-most red line labeled 3,000 K, red hot. The graph shows that when a blackbody acquires a temperature of 3,000 K, it radiates energy across the electromagnetic spectrum. However, the energy is most intensely emitted at a wavelength of approximately 1000 nm. This is in the infrared portion of the electromagnetic spectrum. While a body at this temperature would appear red-hotto our eyes, it would truly appear ‘infrared-hot’ if we were able to see the entire spectrum. A few other important notes regarding Figure 21.2: • As temperature increases, the total amount of energy radiated increases. This is shown by examining the area underneath each line. • Regardless of temperature, all red lines on the graph undergo a consistent pattern. While electromagnetic radiation is emitted throughout the spectrum, the intensity of this radiation peaks at one particular wavelength. • As the temperature changes, the wavelength of greatest radiation intensity changes. At 4,000 K, the radiation is most intense in the yellow-green portion of the spectrum. At 6,000 K, the blackbody would radiate white hot,due to intense radiation throughout the visible portion of the electromagnetic spectrum. Remember that white light is the emission of all visible colors simultaneously. • As the temperature increases, the frequency of light providing the greatest intensity increases as well. Recall the equation Because the speed of light is constant, frequency and wavelength are inversely related. This is verified by the leftward movement of the three red lines as temperature is increased. While in science it is important to categorize observations, theorizing as to why the observations exist is crucial to scientific advancement. Why doesn’t a blackbody emit radiation evenly across all wavelengths? Why does the temperature of the body change the peak wavelength that is radiated? Why does an increase in temperature cause the peak wavelength emitted to decrease? It is questions like these that drove significant research at the turn of the twentieth century. And within the context of these questions, Max Planck discovered something of tremendous importance. Planck’s Revolution The prevailing theory at the time of Max Planck’s discovery was that intensity and frequency were related by the equation This equation, derived from classical physics and using wave phenomena, infers that as wavelength increases, the intensity of energy provided will decrease with an inverse-squared relationship. This relationship is graphed in Figure 21.3 and shows a troubling trend. For starters, it should be apparent that the graph from this equation does not match the blackbody graphs found experimentally. Additionally, it shows that for an object of any temperature, there should be an infinite amount of energy quickly emitted in the shortest wavelengths. When theory and experimental results clash, it is important to re-evaluate both models. The disconnect between theory and reality was termed the ultraviolet catastrophe. 694 Chapter 21 • The Quantum Nature of Light Figure 21.3 The graph above shows the true spectral measurements by a blackbody against those predicted by the classical theory at the time. The discord between the predicted classical theory line and the actual results is known as the ultraviolet catastrophe. Due to concerns over the ultraviolet catastrophe, Max Planck began to question whether another factor impacted the relationship between intensity and wavelength. This factor, he posited, should affect the probability that short wavelength light would be emitted. Should this factor reduce the probability of short wavelength light, it would cause the radiance curve to not progress infinitely as in the classical theory, but would instead cause the curve to precipitate back downward as is shown in the 5,000 K, 4,000 K, and 3,000 K temperature lines of the graph in Figure 21.3. Planck noted that this factor, whatever it may be, must also be dependent on temperature, as the intensity decreases at lower and lower wavelengths as the temperature increases. The determination of this probability factorwas a groundbreaking discovery in physics, yielding insight not just into light but also into energy and matter itself. It would be the basis for Planck’s 1918 Nobel Prize in Physics and would result in the transition of physics from classical to modern understanding. In an attempt to determine the cause of the probability factor,Max Planck constructed a new theory. This theory, which created the branch of physics called quantum mechanics, speculated that the energy radiated by the blackbody could exist only in specific numerical, or quantum, states. This theory is described by the where nis any nonnegative integer (0, 1, 2, 3, …) and his Planck’s constant, given by equation and fis frequency. Through this equation, Planck’s probability factor can be more clearly understood. Each frequency of light provides a specific quantized amount of energy. Low frequency light, associated with longer wavelengths would provide a smaller amount of energy, while high frequency light, associated with shorter wavelengths, would provide a larger amount of energy. For specified temperatures with specific total energies, it makes sense that more low frequency light would be radiated than high frequency light. To a degree, the relationship is like pouring coins through a funnel. More of the smaller pennies would be able to pass through the funnel than the larger quarters. In other words, because the value of the coin is somewhat related to the size of the coin, the probability of a quarter passing through the funnel is reduced! Furthermore, an increase in temperature would signify the presence of higher energy. As a result, the greater amount of total blackbody energy would allow for more of the high frequency, short wavelength, energies to be radiated. This permits the peak of the blackbody curve to drift leftward as the temperature increases, as it does from the 3,000 K to 4,000 K to 5,000 K values. Furthering our coin analogy, consider a wider funnel. This funnel would permit more quarters to pass through and allow for a reduction in concern about the probability factor. In summary, it is the interplay between the predicted classical model and the quantum probability that creates the curve depicted in Figure 21.3. Just as quarters have a higher currency denomination than pennies, higher frequencies come with larger Access for free at openstax.org. 21.1 • Planck and Quantum Nature of Light 695 amounts of energy. However, just as the probability of a quarter passing through a fixed diameter funnel is reduced, so is the probability of a high frequency light existing in a fixed temperature object. As is often the case in physics, it is the balancing of multiple incredible ideas that finally allows for better understanding. Quantization It may be helpful at this point to further consider the idea of quantum states. Atoms, molecules, and fundamental electron and proton charges are all examples of physical entities that are quantized—that is, they appear only in certain discrete values and do not have every conceivable value. On the macroscopic scale, this is not a revolutionary concept. A standing wave on a string allows only particular harmonics described by integers. Going up and down a hill using discrete stair steps causes your potential energy to take on discrete values as you move from step to step. Furthermore, we cannot have a fraction of an atom, or part of an electron’s charge, or 14.33 cents. Rather, everything is built of integral multiples of these substructures. That said, to discover quantum states within a phenomenon that science had always considered continuous would certainly be surprising. When Max Planck was able to use quantization to correctly describe the experimentally known shape of the blackbody spectrum |
, it was the first indication that energy was quantized on a small scale as well. This discovery earned Planck the Nobel Prize in Physics in 1918 and was such a revolutionary departure from classical physics that Planck himself was reluctant to accept his own idea. The general acceptance of Planck’s energy quantization was greatly enhanced by Einstein’s explanation of the photoelectric effect (discussed in the next section), which took energy quantization a step further. Figure 21.4 The German physicist Max Planck had a major influence on the early development of quantum mechanics, being the first to recognize that energy is sometimes quantized. Planck also made important contributions to special relativity and classical physics. (credit: Library of Congress, Prints and Photographs Division, Wikimedia Commons) WORKED EXAMPLE How Many Photons per Second Does a Typical Light Bulb Produce? Assuming that 10 percent of a 100-W light bulb’s energy output is in the visible range (typical for incandescent bulbs) with an average wavelength of 580 nm, calculate the number of visible photons emitted per second. Strategy The number of visible photons per second is directly related to the amount of energy emitted each second, also known as the bulb’s power. By determining the bulb’s power, the energy emitted each second can be found. Since the power is given in watts, which is joules per second, the energy will be in joules. By comparing this to the amount of energy associated with each photon, the number of photons emitted each second can be determined. Solution The power in visible light production is 10.0 percent of 100 W, or 10.0 J/s. The energy of the average visible photon is found by substituting the given average wavelength into the formula By rearranging the above formula to determine energy per photon, this produces 21.1 The number of visible photons per second is thus 696 Chapter 21 • The Quantum Nature of Light Discussion This incredible number of photons per second is verification that individual photons are insignificant in ordinary human experience. However, it is also a verification of our everyday experience—on the macroscopic scale, photons are so small that quantization becomes essentially continuous. WORKED EXAMPLE How does Photon Energy Change with Various Portions of the EM Spectrum? Refer to the Graphs of Blackbody Radiation shown in the first figure in this section. Compare the energy necessary to radiate one photon of infrared light and one photon of visible light. Strategy To determine the energy radiated, it is necessary to use the equation frequency for infrared light and visible light. It is also necessary to find a representative Solution According to the first figure in this section, one representative wavelength for infrared light is 2000 nm (2.000 × 10-6 m). The associated frequency of an infrared light is Using the equation , the energy associated with one photon of representative infrared light is The same process above can be used to determine the energy associated with one photon of representative visible light. According to the first figure in this section, one representative wavelength for visible light is 500 nm. 21.2 21.3 21.4 21.5 Discussion This example verifies that as the wavelength of light decreases, the quantum energy increases. This explains why a fire burning with a blue flame is considered more dangerous than a fire with a red flame. Each photon of short-wavelength blue light emitted carries a greater amount of energy than a long-wavelength red light. This example also helps explain the differences in the 3,000 K, 4,000 K, and 6,000 K lines shown in the first figure in this section. As the temperature is increased, more energy is available for a greater number of short-wavelength photons to be emitted. Practice Problems 1. An AM radio station broadcasts at a frequency of 1,530 kHz . What is the energy in Joules of a photon emitted from this station? a. b. c. d. 10.1 × 10-26 J 1.01 × 10-28 J 1.01 × 10-29 J 1.01 × 10-27 J 2. A photon travels with energy of 1.0 eV. What type of EM radiation is this photon? a. visible radiation Access for free at openstax.org. 21.1 • Planck and Quantum Nature of Light 697 b. microwave radiation infrared radiation c. d. ultraviolet radiation Check Your Understanding 3. Do reflective or absorptive surfaces more closely model a perfect blackbody? reflective surfaces a. b. absorptive surfaces 4. A black T-shirt is a good model of a blackbody. However, it is not perfect. What prevents a black T-shirt from being considered a perfect blackbody? a. The T-shirt reflects some light. b. The T-shirt absorbs all incident light. c. The T-shirt re-emits all the incident light. d. The T-shirt does not reflect light. 5. What is the mathematical relationship linking the energy of a photon to its frequency? a. b. c. d. 6. Why do we not notice quantization of photons in everyday experience? a. because the size of each photon is very large b. because the mass of each photon is so small c. because the energy provided by photons is very large d. because the energy provided by photons is very small 7. Two flames are observed on a stove. One is red while the other is blue. Which flame is hotter? a. The red flame is hotter because red light has lower frequency. b. The red flame is hotter because red light has higher frequency. c. The blue flame is hotter because blue light has lower frequency. d. The blue flame is hotter because blue light has higher frequency. 8. Your pupils dilate when visible light intensity is reduced. Does wearing sunglasses that lack UV blockers increase or decrease the UV hazard to your eyes? Explain. Increase, because more high-energy UV photons can enter the eye. a. b. Increase, because less high-energy UV photons can enter the eye. c. Decrease, because more high-energy UV photons can enter the eye. d. Decrease, because less high-energy UV photons can enter the eye. 9. The temperature of a blackbody radiator is increased. What will happen to the most intense wavelength of light emitted as this increase occurs? a. The wavelength of the most intense radiation will vary randomly. b. The wavelength of the most intense radiation will increase. c. The wavelength of the most intense radiation will remain unchanged. d. The wavelength of the most intense radiation will decrease. 698 Chapter 21 • The Quantum Nature of Light 21.2 Einstein and the Photoelectric Effect Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Einstein’s explanation of the photoelectric effect • Describe how the photoelectric effect could not be explained by classical physics • Calculate the energy of a photoelectron under given conditions • Describe use of the photoelectric effect in biological applications, photoelectric devices and movie soundtracks Section Key Terms electric eye photoelectric effect photoelectron photon The Photoelectric Effect Teacher Support [EL]Ask the students what they think the term photoelectricmeans. How does the term relate to its definition? When light strikes certain materials, it can eject electrons from them. This is called the photoelectric effect, meaning that light (photo) produces electricity. One common use of the photoelectric effect is in light meters, such as those that adjust the automatic iris in various types of cameras. Another use is in solar cells, as you probably have in your calculator or have seen on a rooftop or a roadside sign. These make use of the photoelectric effect to convert light into electricity for running different devices. Figure 21.5 The photoelectric effect can be observed by allowing light to fall on the metal plate in this evacuated tube. Electrons ejected by the light are collected on the collector wire and measured as a current. A retarding voltage between the collector wire and plate can then be adjusted so as to determine the energy of the ejected electrons. (credit: P. P. Urone) Revolutionary Properties of the Photoelectric Effect When Max Planck theorized that energy was quantized in a blackbody radiator, it is unlikely that he would have recognized just how revolutionary his idea was. Using tools similar to the light meter in Figure 21.5, it would take a scientist of Albert Einstein’s stature to fully discover the implications of Max Planck’s radical concept. Through careful observations of the photoelectric effect, Albert Einstein realized that there were several characteristics that could be explained only if EM radiation is itself quantized. While these characteristics will be explained a bit later in this section, you can already begin to appreciate why Einstein’s idea is very important. It means that the apparently continuous stream of energy in an EM wave is actually not a continuous stream at all. In fact, the EM wave itself is actually composed of tiny quantum packets of energy called photons. In equation form, Einstein found the energy of a photon or photoelectron to be where Eis the energy of a photon of frequency fand his Planck’s constant. A beam from a flashlight, which to this point had been considered a wave, instead could now be viewed as a series of photons, each providing a specific amount of energy see Figure 21.6. Furthermore, the amount of energy within each individual photon is based upon its individual frequency, as Access for free at openstax.org. dictated by frequency-dependent photon energies added together. As a result, the total amount of energy provided by the beam could now be viewed as the sum of all 21.2 • Einstein and the Photoelectric Effect 699 Figure 21.6 An EM wave of frequency fis composed of photons, or individual quanta of EM radiation. The energy of each photon is , where his Planck’s constant and fis the frequency of the EM radiation. Higher intensity means more photons per unit area per second. The flashlight emits large numbers of photons of many different frequencies, hence others have energy , and so on. Jus |
t as with Planck’s blackbody radiation, Einstein’s concept of the photon could take hold in the scientific community only if it could succeed where classical physics failed. The photoelectric effect would be a key to demonstrating Einstein’s brilliance. Consider the following five properties of the photoelectric effect. All of these properties are consistent with the idea that individual photons of EM radiation are absorbed by individual electrons in a material, with the electron gaining the photon’s energy. Some of these properties are inconsistent with the idea that EM radiation is a simple wave. For simplicity, let us consider what happens with monochromatic EM radiation in which all photons have the same energy hf. Figure 21.7 Incident radiation strikes a clean metal surface, ejecting multiple electrons from it. The manner in which the frequency and intensity of the incoming radiation affect the ejected electrons strongly suggests that electromagnetic radiation is quantized. This event, called the photoelectric effect, is strong evidence for the existence of photons. 1. If we vary the frequency of the EM radiation falling on a clean metal surface, we find the following: For a given material, there is a threshold frequency f0 for the EM radiation below which no electrons are ejected, regardless of intensity. Using the photon model, the explanation for this is clear. Individual photons interact with individual electrons. Thus if the energy of an individual photon is too low to break an electron away, no electrons will be ejected. However, if EM radiation were a simple wave, sufficient energy could be obtained simply by increasing the intensity. 2. Once EM radiation falls on a material, electrons are ejected without delay. As soon as an individual photon of sufficiently high frequency is absorbed by an individual electron, the electron is ejected. If the EM radiation were a simple wave, several minutes would be required for sufficient energy to be deposited at the metal surface in order to eject an electron. 3. The number of electrons ejected per unit time is proportional to the intensity of the EM radiation and to no other 4. characteristic. High-intensity EM radiation consists of large numbers of photons per unit area, with all photons having the same characteristic energy, hf. The increased number of photons per unit area results in an increased number of electrons per unit area ejected. If we vary the intensity of the EM radiation and measure the energy of ejected electrons, we find the following: The maximum kinetic energy of ejected electrons is independent of the intensity of the EM radiation. Instead, as noted in point 3 above, increased intensity results in more electrons of the same energy being ejected. If EM radiation were a simple wave, a higher intensity could transfer more energy, and higher-energy electrons would be ejected. 5. The kinetic energy KE of an ejected electron equals the photon energy minus the binding energy BE of the electron in the 700 Chapter 21 • The Quantum Nature of Light specific material. An individual photon can give all of its energy to an electron. The photon’s energy is partly used to break the electron away from the material. The remainder goes into the ejected electron’s kinetic energy. In equation form, this is given by 21.6 is the maximum kinetic energy of the ejected electron, where electron to the particular material. This equation explains the properties of the photoelectric effect quantitatively and demonstrates that BE is the minimum amount of energy necessary to eject an electron. If the energy supplied is less than BE, the electron cannot be ejected. The binding energy can also be written as particular material. Figure 21.8 shows a graph of maximum particular material. versus the frequency of incident EM radiation falling on a is the photon’s energy, and BE is the binding energy of the is the threshold frequency for the where Figure 21.8 A graph of the kinetic energy of an ejected electron, KEe, versus the frequency of EM radiation impinging on a certain material. There is a threshold frequency below which no electrons are ejected, because the individual photon interacting with an individual electron has insufficient energy to break it away. Above the threshold energy, KEe increases linearly with f, consistent with KEe = hf− BE. The slope of this line is h, so the data can be used to determine Planck’s constant experimentally. TIPS FOR SUCCESS The following five pieces of information can be difficult to follow without some organization. It may be useful to create a table of expected results of each of the five properties, with one column showing the classical wave model result and one column showing the modern photon model result. The table may look something like Table 21.1 Classical Wave Model Modern Photon Model Threshold Frequency Electron Ejection Delay Intensity of EM Radiation Speed of Ejected Electrons Relationship between Kinetic Energy and Binding Energy Table 21.1 Table of Expected Results Virtual Physics Photoelectric Effect Click to view content (http://www.openstax.org/l/28photoelectric) Access for free at openstax.org. In this demonstration, see how light knocks electrons off a metal target, and recreate the experiment that spawned the field of quantum mechanics. 21.2 • Einstein and the Photoelectric Effect 701 GRASP CHECK In the circuit provided, what are the three ways to increase the current? a. decrease the intensity, decrease the frequency, alter the target b. decrease the intensity, decrease the frequency, don’t alter the target increase the intensity, increase the frequency, alter the target c. increase the intensity, increase the frequency, alter the target d. WORKED EXAMPLE Photon Energy and the Photoelectric Effect: A Violet Light (a) What is the energy in joules and electron volts of a photon of 420-nm violet light? (b) What is the maximum kinetic energy of electrons ejected from calcium by 420 nm violet light, given that the binding energy of electrons for calcium metal is 2.71 eV? Strategy To solve part (a), note that the energy of a photon is given by is a straightforward application of to find the ejected electron’s maximum kinetic energy, since BE is given. . For part (b), once the energy of the photon is calculated, it Solution for (a) Photon energy is given by Since we are given the wavelength rather than the frequency, we solve the familiar relationship yielding for the frequency, Combining these two equations gives the useful relationship Now substituting known values yields Converting to eV, the energy of the photon is Solution for (b) Finding the kinetic energy of the ejected electron is now a simple application of the equation the photon energy and binding energy yields 21.7 21.8 21.9 21.10 . Substituting 21.11 Discussion The energy of this 420 nm photon of violet light is a tiny fraction of a joule, and so it is no wonder that a single photon would be difficult for us to sense directly—humans are more attuned to energies on the order of joules. But looking at the energy in electron volts, we can see that this photon has enough energy to affect atoms and molecules. A DNA molecule can be broken with about 1 eV of energy, for example, and typical atomic and molecular energies are on the order of eV, so that the photon in this example could have biological effects, such as sunburn. The ejected electron has rather low energy, and it would not travel far, 702 Chapter 21 • The Quantum Nature of Light except in a vacuum. The electron would be stopped by a retarding potential of only 0.26 eV, a slightly larger KE than calculated above. In fact, if the photon wavelength were longer and its energy less than 2.71 eV, then the formula would give a negative kinetic energy, an impossibility. This simply means that the 420 nm photons with their 2.96 eV energy are not much above the frequency threshold. You can see for yourself that the threshold wavelength is 458 nm (blue light). This means that if calcium metal were used in a light meter, the meter would be insensitive to wavelengths longer than those of blue light. Such a light meter would be completely insensitive to red light, for example. Practice Problems 10. What is the longest-wavelength EM radiation that can eject a photoelectron from silver, given that the bonding energy is 4.73 eV ? Is this radiation in the visible range? a. 2.63 × 10−7 m; No, the radiation is in microwave region. b. 2.63 × 10−7 m; No, the radiation is in visible region. c. 2.63 × 10−7 m; No, the radiation is in infrared region. d. 2.63 × 10-7 m; No, the radiation is in ultraviolet region. 11. What is the maximum kinetic energy in eV of electrons ejected from sodium metal by 450-nm EM radiation, given that the binding energy is 2.28 eV? a. 0.48 V b. 0.82 eV c. 1.21 eV d. 0.48 eV Technological Applications of the Photoelectric Effect While Einstein’s understanding of the photoelectric effect was a transformative discovery in the early 1900s, its presence is ubiquitous today. If you have watched streetlights turn on automatically in response to the setting sun, stopped elevator doors from closing simply by putting your hands between them, or turned on a water faucet by sliding your hands near it, you are familiar with the electric eye, a name given to a group of devices that use the photoelectric effect for detection. All these devices rely on photoconductive cells. These cells are activated when light is absorbed by a semi-conductive material, knocking off a free electron. When this happens, an electron void is left behind, which attracts a nearby electron. The movement of this electron, and the resultant chain of electron movements, produces a current. If electron ejection continues, further holes are created, thereby increasing the electrical conductivity of the cell. This current can turn switches on and off and activate various familiar mechanisms. One such mechanism takes place where you may not expect |
it. Next time you are at the movie theater, pay close attention to the sound coming out of the speakers. This sound is actually created using the photoelectric effect! The audiotape in the projector booth is a transparent piece of film of varying width. This film is fed between a photocell and a bright light produced by an exciter lamp. As the transparent portion of the film varies in width, the amount of light that strikes the photocell varies as well. As a result, the current in the photoconductive circuit changes with the width of the filmstrip. This changing current is converted to a changing frequency, which creates the soundtrack commonly heard in the theater. WORK IN PHYSICS Solar Energy Physicist According to the U.S. Department of Energy, Earth receives enough sunlight each hour to power the entire globe for a year. While converting all of this energy is impossible, the job of the solar energy physicist is to explore and improve upon solar energy conversion technologies so that we may harness more of this abundant resource. The field of solar energy is not a new one. For over half a century, satellites and spacecraft have utilized photovoltaic cells to create current and power their operations. As time has gone on, scientists have worked to adapt this process so that it may be used in homes, businesses, and full-scale power stations using solar cells like the one shown in Figure 21.9. Access for free at openstax.org. 21.2 • Einstein and the Photoelectric Effect 703 Figure 21.9 A solar cell is an example of a photovoltaic cell. As light strikes the cell, the cell absorbs the energy of the photons. If this energy exceeds the binding energy of the electrons, then electrons will be forced to move in the cell, thereby producing a current. This current may be used for a variety of purposes. (credit: U.S. Department of Energy) Solar energy is converted to electrical energy in one of two manners: direct transfer through photovoltaic cells or thermal conversion through the use of a CSP, concentrating solar power, system. Unlike electric eyes, which trip a mechanism when current is lost, photovoltaic cells utilize semiconductors to directly transfer the electrons released through the photoelectric effect into a directed current. The energy from this current can then be converted for storage, or immediately used in an electric process. A CSP system is an indirect method of energy conversion. In this process, light from the Sun is channeled using parabolic mirrors. The light from these mirrors strikes a thermally conductive material, which then heats a pool of water. This water, in turn, is converted to steam, which turns a turbine and creates electricity. While indirect, this method has long been the traditional means of large-scale power generation. There are, of course, limitations to the efficacy of solar power. Cloud cover, nightfall, and incident angle strike at high altitudes are all factors that directly influence the amount of light energy available. Additionally, the creation of photovoltaic cells requires rare-earth minerals that can be difficult to obtain. However, the major role of a solar energy physicist is to find ways to improve the efficiency of the solar energy conversion process. Currently, this is done by experimenting with new semi conductive materials, by refining current energy transfer methods, and by determining new ways of incorporating solar structures into the current power grid. Additionally, many solar physicists are looking into ways to allow for increased solar use in impoverished, more remote locations. Because solar energy conversion does not require a connection to a large-scale power grid, research into thinner, more mobile materials will permit remote cultures to use solar cells to convert sunlight collected during the day into stored energy that can then be used at night. Regardless of the application, solar energy physicists are an important part of the future in responsible energy growth. While a doctoral degree is often necessary for advanced research applications, a bachelor's or master's degree in a related science or engineering field is typically enough to gain access into the industry. Computer skills are very important for energy modeling, including knowledge of CAD software for design purposes. In addition, the ability to collaborate and communicate with others is critical to becoming a solar energy physicist. GRASP CHECK What role does the photoelectric effect play in the research of a solar energy physicist? a. The understanding of photoelectric effect allows the physicist to understand the generation of light energy when using photovoltaic cells. b. The understanding of photoelectric effect allows the physicist to understand the generation of electrical energy when using photovoltaic cells. c. The understanding of photoelectric effect allows the physicist to understand the generation of electromagnetic energy when using photovoltaic cells. d. The understanding of photoelectric effect allows the physicist to understand the generation of magnetic energy when using photovoltaic cells. 704 Chapter 21 • The Quantum Nature of Light Check Your Understanding 12. How did Einstein’s model of photons change the view of a beam of energy leaving a flashlight? a. A beam of light energy is now considered a continual stream of wave energy, not photons. b. A beam of light energy is now considered a collection of photons, each carrying its own individual energy. 13. True or false—Visible light is the only type of electromagnetic radiation that can cause the photoelectric effect. a. b. false true 14. Is the photoelectric effect a direct consequence of the wave character of EM radiation or the particle character of EM radiation? a. The photoelectric effect is a direct consequence of the particle nature of EM radiation. b. The photoelectric effect is a direct consequence of the wave nature of EM radiation. c. The photoelectric effect is a direct consequence of both the wave and particle nature of EM radiation. d. The photoelectric effect is a direct consequence of neither the wave nor the particle nature of EM radiation. 15. Which aspects of the photoelectric effect can only be explained using photons? a. aspects 1, 2, and 3 b. aspects 1, 2, and 4 c. aspects 1, 2, 4 and 5 d. aspects 1, 2, 3, 4 and 5 16. In a photovoltaic cell, what energy transformation takes place? a. Solar energy transforms into electric energy. b. Solar energy transforms into mechanical energy. c. Solar energy transforms into thermal energy. d. In a photovoltaic cell, thermal energy transforms into electric energy. 17. True or false—A current is created in a photoconductive cell, even if only one electron is expelled from a photon strike. a. b. false true 18. What is a photon and how is it different from other fundamental particles? a. A photon is a quantum packet of energy; it has infinite mass. b. A photon is a quantum packet of energy; it is massless. c. A photon is a fundamental particle of an atom; it has infinite mass. d. A photon is a fundamental particle of an atom; it is massless. 21.3 The Dual Nature of Light Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the Compton effect • Calculate the momentum of a photon • Explain how photon momentum is used in solar sails • Explain the particle-wave duality of light Section Key Terms Compton effect particle-wave duality photon momentum Photon Momentum Do photons abide by the fundamental properties of physics? Can packets of electromagnetic energy possibly follow the same rules as a ping-pong ball or an electron? Although strange to consider, the answer to both questions is yes. Despite the odd nature of photons, scientists prior to Einstein had long suspected that the fundamental particle of Access for free at openstax.org. electromagnetic radiation shared properties with our more macroscopic particles. This is no clearer than when considering the photoelectric effect, where photons knock electrons out of a substance. While it is strange to think of a massless particle exhibiting momentum, it is now a well-established fact within the scientific community. Figure 21.10 shows macroscopic evidence of photon momentum. 21.3 • The Dual Nature of Light 705 Figure 21.10 The tails of the Hale-Bopp comet point away from the Sun, evidence that light has momentum. Dust emanating from the body of the comet forms this tail. Particles of dust are pushed away from the Sun by light reflecting from them. The blue, ionized gas tail is also produced by photons interacting with atoms in the comet material. (credit: Geoff Chester, U.S. Navy, via Wikimedia Commons) Figure 21.10 shows a comet with two prominent tails. Comet tails are composed of gases and dust evaporated from the body of the comet and ionized gas. What most people do not know about the tails is that they always point awayfrom the Sun rather than trailing behind the comet. This can be seen in the diagram. Why would this be the case? The evidence indicates that the dust particles of the comet are forced away from the Sun when photons strike them. Evidently, photons carry momentum in the direction of their motion away from the Sun, and some of this momentum is transferred to dust particles in collisions. The blue tail is caused by the solar wind, a stream of plasma consisting primarily of protons and electrons evaporating from the corona of the Sun. Momentum, The Compton Effect, and Solar Sails Momentum is conserved in quantum mechanics, just as it is in relativity and classical physics. Some of the earliest direct experimental evidence of this came from the scattering of X-ray photons by electrons in substances, a phenomenon discovered by American physicist Arthur H. Compton (1892–1962). Around 1923, Compton observed that X-rays reflecting from materials had decreased energy and correctly interpreted this as being due to the scattering of the X-ray photons by elec |
trons. This phenomenon could be handled as a collision between two particles—a photon and an electron at rest in the material. After careful observation, it was found that both energy and momentum were conserved in the collision. See Figure 21.11. For the discovery of this conserved scattering, now known as the Compton effect, Arthur Compton was awarded the Nobel Prize in 1929. Shortly after the discovery of Compton scattering, the value of the photon momentum, was determined by Louis de Broglie. In this equation, called the de Broglie relation, hrepresents Planck’s constant and λis the photon wavelength. Figure 21.11 The Compton effect is the name given to the scattering of a photon by an electron. Energy and momentum are conserved, resulting in a reduction of both for the scattered photon. 706 Chapter 21 • The Quantum Nature of Light We can see that photon momentum is small, since observe photon momentum. Our mirrors do not recoil when light reflects from them, except perhaps in cartoons. Compton saw the effects of photon momentum because he was observing X-rays, which have a small wavelength and a relatively large momentum, interacting with the lightest of particles, the electron. and his very small. It is for this reason that we do not ordinarily WORKED EXAMPLE Electron and Photon Momentum Compared (a) Calculate the momentum of a visible photon that has a wavelength of 500 nm. (b) Find the velocity of an electron having the same momentum. (c) What is the energy of the electron, and how does it compare with the energy of the photon? Strategy Finding the photon momentum is a straightforward application of its definition: small, we can assume that an electron with the same momentum will be nonrelativistic, making it easy to find its velocity and kinetic energy from the classical formulas. If we find the photon momentum is Solution for (a) Photon momentum is given by the de Broglie relation. Entering the given photon wavelength yields 21.12 21.13 Solution for (b) Since this momentum is indeed small, we will use the classical expression momentum. Solving for vand using the known value for the mass of an electron gives to find the velocity of an electron with this Solution for (c) The electron has kinetic energy, which is classically given by Thus, Converting this to eV by multiplying by yields The photon energy Eis 21.14 21.15 21.16 21.17 21.18 which is about five orders of magnitude greater. Discussion Even in huge numbers, the total momentum that photons carry is small. An electron that carries the same momentum as a 500-nm photon will have a 1,460 m/s velocity, which is clearly nonrelativistic. This is borne out by the experimental observation that it takes far less energy to give an electron the same momentum as a photon. That said, for high-energy photons interacting with small masses, photon momentum may be significant. Even on a large scale, photon momentum can have an effect if there Access for free at openstax.org. 21.3 • The Dual Nature of Light 707 are enough of them and if there is nothing to prevent the slow recoil of matter. Comet tails are one example, but there are also proposals to build space sails that use huge low-mass mirrors (made of aluminized Mylar) to reflect sunlight. In the vacuum of space, the mirrors would gradually recoil and could actually accelerate spacecraft within the solar system. See the following figure. TIPS FOR SUCCESS When determining energies in particle physics, it is more sensible to use the unit eV instead of Joules. Using eV will help you to recognize differences in magnitude more easily and will make calculations simpler. Also, eV is used by scientists to describe the binding energy of particles and their rest mass, so using eV will eliminate the need to convert energy quantities. Finally, eV is a convenient unit when linking electromagnetic forces to particle physics, as one eV is the amount energy given to an electron when placed in a field of 1-V potential difference. Practice Problems 19. Find the momentum of a 4.00-cm wavelength microwave photon. a. 0.83 × 10−32 kg ⋅ m/s 1.66 × 10−34 kg ⋅ m/s b. c. 0.83 × 10−34 kg ⋅ m/s 1.66 × 10-32 kg ⋅ m/s d. 20. Calculate the wavelength of a photon that has the same momentum of a proton moving at 1.00 percent of the speed of light. a. 2.43 × 10−10 m b. 2.43 × 10−12 m 1.32 × 10−15 m c. 1.32 × 10−13 m d. Figure 21.12 (a) Space sails have been proposed that use the momentum of sunlight reflecting from gigantic low-mass sails to propel spacecraft about the solar system. A Russian test model of this (the Cosmos 1) was launched in 2005, but did not make it into orbit due to a rocket failure. (b) A U.S. version of this, labeled LightSail-1, is scheduled for trial launches in 2016. It will have a 40 m2 sail. (credit: Kim Newton/NASA) LINKS TO PHYSICS LightSail-1 Project “Provide ships or sails adapted to the heavenly breezes, and there will be some who will brave even that void.” — Johannes Kepler (in a letter to Galileo Galilei in 1608) 708 Chapter 21 • The Quantum Nature of Light Figure 21.13 NASA’s NanoSail-D, a precursor to LightSail-1, with its sails deployed. The Planetary Society will be launching LightSail-1 in early 2016. (credit: NASA/MSFC/D, Wikimedia Commons) Traversing the Solar System using nothing but the Sun’s power has long been a fantasy of scientists and science fiction writers alike. Though physicists like Compton, Einstein, and Planck all provided evidence of light’s propulsive capacity, it is only recently that the technology has become available to truly put these visions into motion. In 2016, by sending a lightweight satellite into space, the LightSail-1 project is designed to do just that. A citizen-funded project headed by the Planetary Society, the 5.45-million-dollar LightSail-1 project is set to launch two crafts into orbit around the Earth. Each craft is equipped with a 32-square-meter solar sail prepared to unfurl once a rocket has launched it to an appropriate altitude. The sails are made of large mirrors, each a quarter of the thickness of a trash bag, which will receive an impulse from the Sun’s reflecting photons. Each time the Sun’s photon strikes the craft’s reflective surface and bounces off, it will provide a momentum to the sail much greater than if the photon were simply absorbed. Attached to three tiny satellites called CubeSats, whose combined volume is no larger than a loaf of bread, the received momentum from the Sun’s photons should be enough to record a substantial increase in orbital speed. The intent of the LightSail-1 mission is to prove that the technology behind photon momentum travel is sound and can be done cheaply. A test flight in May 2015 showed that the craft’s Mylar sails could unfurl on command. With another successful result in 2016, the Planetary Society will be planning future versions of the craft with the hopes of eventually achieving interplanetary satellite travel. Though a few centuries premature, Kepler’s fantastic vision may not be that far away. If eventually set into interplanetary launch, what will be the effect of continual photon bombardment on the motion of a craft similar to LightSail-1? a. b. c. d. It will result in continual acceleration of the craft. It will first accelerate and then decelerate the craft. It will first decelerate and then accelerate the craft. It will result in the craft moving at constant velocity. Particle-Wave Duality We have long known that EM radiation is like a wave, capable of interference and diffraction. We now see that light can also be modeled as particles—massless photons of discrete energy and momentum. We call this twofold nature the particle-wave duality, meaning that EM radiation has properties of both particles and waves. This may seem contradictory, since we ordinarily deal with large objects that never act like both waves and particles. An ocean wave, for example, looks nothing like a grain of sand. However, this so-called duality is simply a term for properties of the photon analogous to phenomena we can observe directly, on a macroscopic scale. See Figure 21.14. If this term seems strange, it is because we do not ordinarily observe details on the quantum level directly, and our observations yield either particle-like orwave-like properties, but never both simultaneously. Access for free at openstax.org. 21.3 • The Dual Nature of Light 709 Figure 21.14 (a) The interference pattern for light through a double slit is a wave property understood by analogy to water waves. (b) The properties of photons having quantized energy and momentum and acting as a concentrated unit are understood by analogy to macroscopic particles. Since we have a particle-wave duality for photons, and since we have seen connections between photons and matter in that both have momentum, it is reasonable to ask whether there is a particle-wave duality for matter as well. If the EM radiation we once thought to be a pure wave has particle properties, is it possible that matter has wave properties? The answer, strangely, is yes. The consequences of this are tremendous, as particle-wave duality has been a constant source of scientific wonder during the twentieth and twenty-first centuries. Check Your Understanding 21. What fundamental physics properties were found to be conserved in Compton scattering? a. energy and wavelength b. energy and momentum c. mass and energy d. energy and angle 22. Why do classical or relativistic momentum equations not work in explaining the conservation of momentum that occurs in Compton scattering? a. because neither classical nor relativistic momentum equations utilize mass as a variable in their equations b. because relativistic momentum equations utilize mass as a variable in their formulas but classical momentum equations do not c. because classical momentum equations utilize mass as a variable in their formulas but relativistic momentum equations do not d. because both classical and relativistic momentum equations ut |
ilize mass as a variable in their formulas 23. If solar sails were constructed with more massive materials, how would this influence their effectiveness? a. The effect of the momentum would increase due to the decreased inertia of the sails. b. The effect of the momentum would reduce due to the decreased inertia of the sails. c. The effect of the momentum would increase due to the increased inertia of the sails. d. The effect of the momentum would be reduced due to the increased inertia of the sails. 24. True or false—It is possible to propel a solar sail craft using just particles within the solar wind. a. b. true false 25. True or false—Photon momentum more directly supports the wave model of light. a. b. false true 710 Chapter 21 • The Quantum Nature of Light 26. True or false—wave-particle duality exists for objects on the macroscopic scale. a. b. false true 27. What type of electromagnetic radiation was used in Compton scattering? a. visible light b. ultraviolet radiation c. d. X-rays radio waves Access for free at openstax.org. Chapter 21 • Key Terms 711 KEY TERMS blackbody object that absorbs all radiated energy that strikes it and also emits energy across all wavelengths of the electromagnetic spectrum material by a photon of light photon a quantum, or particle, of electromagnetic radiation Compton effect phenomenon whereby X-rays scattered photon momentum amount of momentum of a photon, from materials have decreased energy calculated by electric eye group of devices that use the photoelectric effect for detection particle-wave duality property of behaving like either a particle or a wave; the term for the phenomenon that all particles have wave-like characteristics and waves have particle-like characteristics photoelectric effect phenomenon whereby some materials eject electrons when exposed to light photoelectron electron that has been ejected from a SECTION SUMMARY 21.1 Planck and Quantum Nature of Light • A blackbody will radiate energy across all wavelengths of the electromagnetic spectrum. • Radiation of a blackbody will peak at a particular wavelength, dependent on the temperature of the blackbody. • Analysis of blackbody radiation led to the field of quantum mechanics, which states that radiated energy can only exist in discrete quantum states. 21.2 Einstein and the Photoelectric Effect • The photoelectric effect is the process in which EM radiation ejects electrons from a material. • Einstein proposed photons to be quanta of EM where fis the radiation having energy frequency of the radiation. • All EM radiation is composed of photons. As Einstein KEY EQUATIONS 21.1 Planck and Quantum Nature of Light quantum energy quantized the fact that certain physical entities exist only with particular discrete values and not every conceivable value quantum discrete packet or bundle of a physical entity such as energy ultraviolet catastrophe misconception that blackbodies would radiate high frequency energy at a much higher rate than energy radiated at lower frequencies explained, all characteristics of the photoelectric effect are due to the interaction of individual photons with individual electrons. • The maximum kinetic energy KEeof ejected electrons (photoelectrons) is given by is the photon energy and BE is the binding energy (or work function) of the electron in the particular material. where hf 21.3 The Dual Nature of Light • Compton scattering provided evidence that photon- electron interactions abide by the principles of conservation of momentum and conservation of energy. • The momentum of individual photons, quantified by , can be used to explain observations of comets and may lead to future space technologies. • Electromagnetic waves and matter have both wave-like and particle-like properties. This phenomenon is defined as particle-wave duality. maximum kinetic energy of a photoelectron binding energy of an electron 21.2 Einstein and the Photoelectric Effect 21.3 The Dual Nature of Light energy of a photon momentum of a photon (deBroglie relation) 712 Chapter 21 • Chapter Review CHAPTER REVIEW Concept Items 21.1 Planck and Quantum Nature of Light 1. What aspect of the blackbody spectrum forced Planck to propose quantization of energy levels in atoms and molecules? a. Radiation occurs at a particular frequency that does not change with the energy supplied. b. Certain radiation occurs at a particular frequency that changes with the energy supplied. c. Maximum radiation would occur at a particular frequency that does not change with the energy supplied. d. Maximum radiation would occur at a particular frequency that changes with the energy supplied. 2. Two lasers shine red light at 650 nm. One laser is twice as bright as the other. Explain this difference using photons and photon energy. a. The brighter laser emits twice the number of photons and more energy per photon. b. The brighter laser emits twice the number of photons and less energy per photon. c. Both lasers emit equal numbers of photons and equivalent amounts of energy per photon. d. The brighter laser emits twice the number of photons but both lasers emit equivalent amounts of energy per photon. 3. Consider four stars in the night sky: red, yellow, orange, and blue. The photons of which star will carry the greatest amount of energy? a. blue b. orange c. red d. yellow 4. A lightbulb is wired to a variable resistor. What will happen to the color spectrum emitted by the bulb as the resistance of the circuit is increased? a. The bulb will emit greener light. b. The bulb will emit bluer light. c. The bulb will emit more ultraviolet light. d. The bulb will emit redder light. 21.2 Einstein and the Photoelectric Effect 5. Light is projected onto a semi-conductive surface. However, no electrons are ejected. What will happen when the light intensity is increased? a. An increase in light intensity decreases the number of photons. However, no electrons are ejected. Access for free at openstax.org. b. Increase in light intensity increases the number of photons, so electrons with higher kinetic energy are ejected. c. An increase in light intensity increases the number of photons, so electrons will be ejected. d. An increase in light intensity increases the number of photons. However, no electrons are ejected. 6. True or false—The concept of a work function (or binding energy) is permissible under the classical wave model. a. b. false true 7. Can a single microwave photon cause cell damage? a. No, there is not enough energy associated with a single microwave photon to result in cell damage. b. No, there is zero energy associated with a single microwave photon, so it does not result in cell damage. c. Yes, a single microwave photon causes cell damage because it does not have high energy. d. Yes, a single microwave photon causes cell damage because it has enough energy. 21.3 The Dual Nature of Light 8. Why don’t we feel the momentum of sunlight when we are on the beach? a. The momentum of a singular photon is incredibly small. b. The momentum is not felt because very few photons strike us at any time, and not all have momentum. c. The momentum of a singular photon is large, but very few photons strike us at any time. d. A large number of photons strike us at any time, and so their combined momentum is incredibly large. 9. If a beam of helium atoms is projected through two slits and onto a screen, will an interference pattern emerge? a. No, an interference pattern will not emerge because helium atoms will strike a variety of locations on the screen. b. No, an interference pattern will not emerge because helium atoms will strike at certain locations on the screen. c. Yes, an interference pattern will emerge because helium atoms will strike a variety of locations on the screen. d. Yes, an interference pattern will emerge because helium atoms will strike at certain locations on the screen. Chapter 21 • Chapter Review 713 Critical Thinking Items 21.1 Planck and Quantum Nature of Light 13. Why is it assumed that a perfect absorber of light (like a blackbody) must also be a perfect emitter of light? a. To achieve electrostatic equilibrium with its 10. Explain why the frequency of a blackbody does not surroundings double when the temperature is doubled. a. Frequency is inversely proportional to temperature. b. Frequency is directly proportional to temperature. c. Frequency is directly proportional to the square of temperature. b. To achieve thermal equilibrium with its surroundings c. To achieve mechanical equilibrium with its surroundings d. To achieve chemical equilibrium with its d. Frequency is directly proportional to the fourth surroundings power of temperature. 11. Why does the intensity shown in the blackbody radiation graph decrease after its peak frequency is achieved? a. Because after reaching the peak frequency, the photons created at a particular frequency are too many for energy intensity to continue to decrease. 21.2 Einstein and the Photoelectric Effect 14. Light is projected onto a semi-conductive surface. If the intensity is held constant but the frequency of light is increased, what will happen? a. As frequency is increased, electrons will stop being ejected from the surface. b. As frequency is increased, electrons will begin to be ejected from the surface. c. As frequency is increased, it will have no effect on the electrons being ejected as the intensity is the same. d. As frequency is increased, the rate at which the electrons are being ejected will increase. 15. Why is it important to consider what material to use when designing a light meter? Consider the worked example from Section 21-2 for assistance. a. A light meter should contain material that responds only to high frequency light. b. A light meter should contain material that responds b. Because after reaching the peak frequency, the to low frequency light. photons created at a particular frequency are too few for energy intensity to continue to decrease. c. A light meter should |
contain material that has high binding energy. c. Because after reaching the peak frequency, the d. A light meter should contain a material that does photons created at a particular frequency are too many for energy intensity to continue to increase. d. Because after reaching the peak frequency, the photons created at a particular frequency are too few for energy intensity to continue to increase. not show any photoelectric effect. 16. Why does overexposure to UV light often result in sunburn when overexposure to visible light does not? This is why you can get burnt even on a cloudy day. a. UV light carries less energy than visible light and 12. Shortly after the introduction of photography, it was can penetrate our body. found that photographic emulsions were more sensitive to blue and violet light than they were to red light. Explain why this was the case. a. Blue-violet light contains greater amount of energy than red light. b. UV light carries more energy than visible light, so it cannot break bonds at the cellular level. c. UV light carries more energy than visible light and can break bonds at the cellular level. d. UV light carries less energy than visible light and b. Blue-violet light contains lower amount of energy cannot penetrate the human body. than red light. c. Both blue-violet light and red light have the same frequency but contain different amounts of energy. d. Blue-violet light frequency is lower than the frequency of red light. 17. If you pick up and shake a piece of metal that has electrons in it free to move as a current, no electrons fall out. Yet if you heat the metal, electrons can be boiled off. Explain both of these facts as they relate to the amount and distribution of energy involved with shaking the 714 Chapter 21 • Chapter Review object as compared with heating it. a. Thermal energy is added to the metal at a much higher rate than energy added due to shaking. b. Thermal energy is added to the metal at a much lower rate than energy added due to shaking. If the thermal energy added is below the binding energy of the electrons, they may be boiled off. If the mechanical energy added is below the binding energy of the electrons, they may be boiled off. d. c. 21.3 The Dual Nature of Light 18. In many macroscopic collisions, a significant amount of kinetic energy is converted to thermal energy. Explain why this is not a concern for Compton scattering. a. Because, photons and electrons do not exist on the molecular level, all energy of motion is considered kinetic energy. b. Because, photons exist on the molecular level while electrons do not exist on the molecular level, all energy of motion is considered kinetic energy. c. Because, electrons exist on the molecular level while photons do not exist on the molecular level, all energy of motion is considered kinetic energy. d. Because, photons and electrons exist on the molecular level, all energy of motion is considered kinetic energy. Problems 21.1 Planck and Quantum Nature of Light 22. How many X-ray photons per second are created by an X-ray tube that produces a flux of X-rays having a power of 1.00 W? Assume the average energy per photon is 75.0 keV. a. 8.33 × 1015 photons b. 9.1 × 107 photons c. 9.1 × 108 photons d. 8.33 × 1013 photons 23. What is the frequency of a photon produced in a CRT using a 25.0-kV accelerating potential? This is similar to the layout as in older color television sets. a. 6.04 × 10−48 Hz b. 2.77 × 10−48 Hz 3.02 × 1018 Hz c. d. 6.04 × 1018 Hz 21.2 Einstein and the Photoelectric Effect 24. What is the binding energy in eV of electrons in magnesium, if the longest-wavelength photon that can eject electrons is 337 nm? Access for free at openstax.org. 19. In what region of the electromagnetic spectrum will photons be most effective in accelerating a solar sail? a. ultraviolet rays b. infrared rays c. X-rays d. gamma rays 20. True or false—Electron microscopes can resolve images that are smaller than the images resolved by light microscopes. false a. true b. 21. How would observations of Compton scattering change if ultraviolet light were used in place of X-rays? a. Ultraviolet light carries less energy than X-rays. As a result, Compton scattering would be easier to detect. b. Ultraviolet light carries less energy than X-rays. As a result, Compton scattering would be more difficult to detect. c. Ultraviolet light carries more energy than X-rays. As a result, Compton scattering would be easier to detect. d. Ultraviolet light has higher energy than X-rays. As a result, Compton scattering would be more difficult to detect. a. b. c. d. 7.44 × 10−19 J 7.44 × 10−49 J 5.90 × 10−17 J 5.90 × 10−19 J 25. Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 420-nm photons. Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device? a. 8.5 × 10−6 s 3.5 × 10−7 s b. c. 43.5 × 10−9 s d. 8.5 × 10−8 s 21.3 The Dual Nature of Light 26. What is the momentum of a 0.0100-nm-wavelength photon that could detect details of an atom? a. 6.626 × 10−27 kg ⋅ m/s b. 6.626 × 10−32 kg ⋅ m/s c. 6.626 × 10−34 kg ⋅ m/s d. 6.626 × 10-23 kg ⋅ m/s 27. The momentum of light is exactly reversed when reflected straight back from a mirror, assuming negligible recoil of the mirror. Thus the change in Chapter 21 • Test Prep 715 momentum is twice the initial photon momentum. Suppose light of intensity 1.00 kW/m2 reflectsfrom a mirror of area 2.00 m2 each second. Using the most general form of Newton’s second law, what is the force on the mirror? a. b. c. d. 1.33 × 10-5 N 1.33 × 10−6 N 1.33 × 10−7 N 1.33 × 10−8 N Performance Task 21.3 The Dual Nature of Light 28. Our scientific understanding of light has changed over time. There is evidence to support the wave model of light, just as there is evidence to support the particle model of light. 1. Construct a demonstration that supports the wave model of light. Note—One possible method is to use a piece of aluminum foil, razor blade, and laser to demonstrate wave interference. Can you arrange these materials to create an effective demonstration? In writing, explain how evidence TEST PREP Multiple Choice 21.1 Planck and Quantum Nature of Light 29. A perfect blackbody is a perfect absorber of energy transferred by what method? a. b. c. d. conduction convection induction radiation 30. Which of the following is a physical entity that is quantized? a. electric charge of an ion frequency of a sound b. speed of a car c. 31. Find the energy in joules of photons of radio waves that leave an FM station that has a 90.0-MHz broadcast frequency. a. b. c. d. 1.8 × 10−25 J 1.11 × 10−25 J 7.1 × 10−43 J 5.96 × 10-26 J 32. Which region of the electromagnetic spectrum will provide photons of the least energy? infrared light a. b. radio waves c. ultraviolet light d. X-rays 33. A hot, black coffee mug is sitting on a kitchen table in a dark room. Because it cannot be seen, one assumes that from your demonstration supports the wave model of light. 2. Construct a demonstration that supports the particle model of light. Note—One possible method is to use a negatively charged electroscope, zinc plate, and three light sources of different frequencies. A red laser, a desk lamp, and ultraviolet lamp are typically used. Can you arrange these materials to demonstrate the photoelectric effect? In writing, explain how evidence from your demonstration supports the particle model of light. it is not emitting energy in the form of light. Explain the fallacy in this logic. a. Not all heat is in the form of light energy. b. Not all light energy falls in the visible portion of the electromagnetic spectrum. c. All heat is in the form of light energy. d. All light energy falls in the visible portion of the electromagnetic spectrum. 34. Given two stars of equivalent size, which will have a greater temperature: a red dwarf or a yellow dwarf? Explain. Note—Our sun is considered a yellow dwarf. a. a yellow dwarf, because yellow light has lower frequency b. a red dwarf, because red light has lower frequency c. a red dwarf, because red light has higher frequency d. a yellow dwarf, because yellow light has higher frequency 21.2 Einstein and the Photoelectric Effect 35. What is a quantum of light called? a. electron b. neutron c. photon d. proton 36. Which of the following observations from the photoelectric effect is not a violation of classical physics? a. Electrons are ejected immediately after impact from light. b. Light can eject electrons from a semi-conductive 716 Chapter 21 • Test Prep material. protons from a surface. c. Light intensity does not influence the kinetic d. UV, X-rays, and gamma rays are capable of ejecting energy of ejected electrons. electrons from a surface. d. No electrons are emitted if the light frequency is too low. 21.3 The Dual Nature of Light 37. If of energy is supplied to an electron with a 41. What two particles interact in Compton scattering? binding energy of the electron be launched? a. b. c. d. , with what kinetic energy will a. photon and electron b. proton and electron c. neutron and electron d. proton and neutron 38. Which of the following terms translates to light- producing voltage? a. photoelectric b. quantum mechanics c. photoconductive d. photovoltaic 39. Why is high frequency EM radiation considered more dangerous than long wavelength EM radiation? a. Long wavelength EM radiation photons carry less energy and therefore have greater ability to disrupt materials through the photoelectric effect. b. Long wavelength EM radiation photons carry more energy and therefore have greater ability to disrupt materials through the photoelectric effect. c. High frequency EM radiation photons carry less energy and therefore have lower ability to disrupt materials through the photoelectric effect. d. High frequency EM radiation photons carry more energy and therefore have greater ability to disrupt materials through the photoelectric effect. 40. Why are UV, X-rays, and gam |
ma rays considered ionizing radiation? a. UV, X-rays, and gamma rays are capable of ejecting photons from a surface. b. UV, X-rays, and gamma rays are capable of ejecting neutrons from a surface. c. UV, X-rays, and gamma rays are capable of ejecting Short Answer 21.1 Planck and Quantum Nature of Light 42. What is the momentum of a 500-nm photon? a. 8.35 × 10−26 kg ⋅ m/s 3.31 × 10−40 kg ⋅ m/s b. 7.55 × 1026 kg ⋅ m/s c. 1.33 × 10-27 kg ⋅ m/s d. 43. The conservation of what fundamental physics principle charge is behind the technology of solar sails? a. b. mass c. momentum d. angular momentum 44. Terms like frequency, amplitude, and period are tied to what component of wave-particle duality? a. neither the particle nor the wave model of light b. both the particle and wave models of light c. d. the particle model of light the wave model of light 45. Why was it beneficial for Compton to scatter electrons using X-rays and not another region of light like microwaves? a. because X-rays are more penetrating than microwaves b. because X-rays have lower frequency than microwaves c. because microwaves have shorter wavelengths than X-rays d. because X-rays have shorter wavelength than microwaves elliptical path. c. The blackbody radiation curve would look like a vertical line. 46. Scientists once assumed that all frequencies of light d. The blackbody radiation curve would look like a were emitted with equal probability. Explain what the blackbody radiation curve would look like if this were the case. a. The blackbody radiation curve would look like a circular path. b. The blackbody radiation curve would look like an horizontal line. 47. Because there are more gradations to high frequency radiation than low frequency radiation, scientists also thought it possible that a curve titled the ultraviolet catastrophewould occur. Explain what the blackbody radiation curve would look like if this were the case. Access for free at openstax.org. a. The curve would steadily increase in intensity with increasing frequency. b. The curve would steadily decrease in intensity with increasing frequency. c. The curve would be much steeper than in the blackbody radiation graph. d. The curve would be much flatter than in the blackbody radiation graph. 48. Energy provided by a light exists in the following quantities: 150 J, 225 J, 300 J. Define one possible quantum of energy and provide an energy state that cannot exist with this quantum. a. 65 J; 450 J cannot exist 70 J; 450 J cannot exist b. 75 J; 375 J cannot exist c. 75 J; 100 J cannot exist d. 49. Why is Planck’s recognition of quantum particles considered the dividing line between classical and modern physics? a. Planck recognized that energy is quantized, which was in sync with the classical physics concepts but not in agreement with modern physics concepts. b. Planck recognized that energy is quantized, which was in sync with modern physics concepts but not in agreement with classical physics concepts. Chapter 21 • Test Prep 717 a. b. c. d. radio, microwave, infrared, visible, ultraviolet, Xray, gamma radio, infrared, microwave, ultraviolet, visible, Xray, gamma radio, visible, microwave, infrared, ultraviolet, Xray, gamma radio, microwave, infrared, visible, ultraviolet, gamma, X-ray 53. Why are photons of gamma rays and X-rays able to penetrate objects more successfully than ultraviolet radiation? a. Photons of gamma rays and X-rays carry with them less energy. b. Photons of gamma rays and X-rays have longer wavelengths. c. Photons of gamma rays and X-rays have lower frequencies. d. Photons of gamma rays and X-rays carry with them more energy. 21.2 Einstein and the Photoelectric Effect 54. According to wave theory, what is necessary to eject electrons from a surface? a. Enough energy to overcome the binding energy of the electrons at the surface c. Prior to Planck’s hypothesis, all the classical b. A frequency that is higher than that of the electrons physics calculations were valid for subatomic particles, but quantum physics calculations were not valid. d. Prior to Planck’s hypothesis, all the classical physics calculations were not valid for macroscopic particles, but quantum physics calculations were valid. 50. How many 500-mm microwave photons are needed to supply the 8 kJ of energy necessary to heat a cup of water by 10 degrees Celsius? a. 8.05 × 1028 photons b. 8.05 × 1026 photons c. 2.01 × 1026 photons d. 2.01 × 1028 photons 51. What is the efficiency of a 100-W, 550-nm lightbulb if a photometer finds that 1 × 1020 photons are emitted each second? a. b. c. d. 101 percent 72 percent 18 percent 36 percent 52. Rank the following regions of the electromagnetic spectrum by the amount of energy provided per photon: gamma, infrared, microwave, ultraviolet, radio, visible, X-ray. at the surface c. Energy that is lower than the binding energy of the electrons at the surface d. A very small number of photons 55. What is the wavelength of EM radiation that ejects 2.00-eV electrons from calcium metal, given that the binding energy is 2.71 eV? 16.1 × 105 m a. b. 6.21 × 10−5 m c. 9.94 × 10−26 m d. 2.63 × 10-7 m 56. Find the wavelength of photons that eject . electrons from potassium, given that the binding energy is a. b. c. d. 57. How do solar cells utilize the photoelectric effect? a. A solar cell converts all photons that it absorbs to electrical energy using the photoelectric effect. b. A solar cell converts all electrons that it absorbs to electrical energy using the photoelectric effect. c. A solar cell absorbs the photons with energy less 718 Chapter 21 • Test Prep than the energy gap of the material of the solar cell and converts it to electrical energy using the photoelectric effect. d. A solar cell absorbs the photons with energy greater than the energy gap of the material of the solar cell and converts it to electrical energy using the photoelectric effect. 58. Explain the advantages of the photoelectric effect to other forms of energy transformation. a. The photoelectric effect is able to work on the Sun’s natural energy. b. The photoelectric effect is able to work on energy generated by burning fossil fuels. c. The photoelectric effect can convert heat energy into electrical energy. d. The photoelectric effect can convert electrical energy into light energy. 21.3 The Dual Nature of Light 5.18 × 105 m/s c. d. 4.18 × 105 m/s 63. When a photon strikes a solar sail, what is the direction of impulse on the photon? a. parallel to the sail b. perpendicular to the sail c. tangential to the sail d. opposite to the sail 64. What is a fundamental difference between solar sails and sails that are used on sailboats? a. Solar sails rely on disorganized strikes from light particles, while sailboats rely on disorganized strikes from air particles. b. Solar sails rely on disorganized strikes from air particles, while sailboats rely on disorganized strikes from light particles. c. Solar sails rely on organized strikes from air particles, while sailboats rely on organized strikes from light particles. 59. Upon collision, what happens to the frequency of a d. Solar sails rely on organized strikes from light photon? a. The frequency of the photon will drop to zero. b. The frequency of the photon will remain the same. c. The frequency of the photon will increase. d. The frequency of the photon will decrease. particles, while sailboats rely on organized strikes from air particles. 65. The wavelength of a particle is called the de Broglie wavelength, and it can be found with the equation . 60. How does the momentum of a photon compare to the momentum of an electron of identical energy? a. Momentum of the photon is greater than the momentum of an electron. b. Momentum of the photon is less than the momentum of an electron. c. Momentum of the photon is equal to the momentum of an electron. d. Momentum of the photon is zero due to zero rest mass but the momentum of an electron is finite. 61. A 500-nm photon strikes an electron and loses 20 percent of its energy. What is the new momentum of the photon? a. 4.24 × 10−27 kg ⋅ m/s 3.18 × 10−27 kg ⋅ m/s b. c. 2.12 × 10−27 kg ⋅ m/s 1.06 × 10−27 kg ⋅ m/s d. 62. A 500-nm photon strikes an electron and loses 20 percent of its energy. What is the speed of the recoiling electron? 7.18 × 105 m/s a. b. 6.18 × 105 m/s Yes or no—Can the wavelength of an electron match that of a proton? a. Yes, a slow-moving electron can achieve the same momentum as a slow-moving proton. b. No, a fast-moving electron cannot achieve the same momentum, and hence the same wavelength, as a proton. c. No, an electron can achieve the same momentum, and hence not the same wavelength, as a proton. d. Yes, a fast-moving electron can achieve the same momentum, and hence have the same wavelength, as a slow-moving proton. 66. Large objects can move with great momentum. Why then is it difficult to see their wave-like nature? a. Their wavelength is equal to the object’s size. b. Their wavelength is very small compared to the object’s size. c. Their wavelength is very large compared to the object’s size. d. Their frequency is very small compared to the object’s size. Access for free at openstax.org. Extended Response 21.1 Planck and Quantum Nature of Light 67. Some television tubes are CRTs. They use an approximately 30-kV accelerating potential to send electrons to the screen, where the electrons stimulate phosphors to emit the light that forms the pictures we watch. Would you expect X-rays also to be created? Explain. a. No, because the full spectrum of EM radiation is not emitted at any temperature. b. No, because the full spectrum of EM radiation is not emitted at certain temperatures. c. Yes, because the full spectrum of EM radiation is emitted at any temperature. d. Yes, because the full spectrum of EM radiation is emitted at certain temperatures. 68. If Planck’s constant were large, say times greater than it is, we would observe macroscopic entities to be quantized. Describe the motion of a chi |
ld’s swing under such circumstances. a. The child would not be able to swing with particular energies. b. The child could be released from any height. c. The child would be able to swing with constant velocity. d. The child could be released only from particular heights. 69. What is the accelerating voltage of an X-ray tube that produces X-rays with the shortest wavelength of 0.0103 nm? 1.21 × 1010 V a. b. 2.4 × 105 V c. d. 3.0 × 10−33 V 1.21 × 105 V 70. Patients in a doctor’s office are rightly concerned about receiving a chest X-ray. Yet visible light is also a form of electromagnetic radiation and they show little concern about sitting under the bright lights of the waiting room. Explain this discrepancy. a. X-ray photons carry considerably more energy so they can harm the patients. b. X-ray photons carry considerably less energy so they can harm the patients. c. X-ray photons have considerably longer wavelengths so they cannot harm the patients. d. X-ray photons have considerably lower frequencies so they can harm the patients. 21.2 Einstein and the Photoelectric Effect 71. When increasing the intensity of light shining on a Chapter 21 • Test Prep 719 metallic surface, it is possible to increase the current created on that surface. Classical theorists would argue that this is evidence that intensity causes charge to move with a greater kinetic energy. Argue this logic from the perspective of a modern physicist. a. The increased intensity increases the number of ejected electrons. The increased current is due to the increase in the number of electrons. b. The increased intensity decreases the number of ejected electrons. The increased current is due to the decrease in the number of electrons ejected. c. The increased intensity does not alter the number of electrons ejected. The increased current is due to the increase in the kinetic energy of electrons. d. The increased intensity alters the number of electrons ejected, but an increase in the current is due to an increase in the kinetic energy of electrons. 72. What impact does the quantum nature of electromagnetic radiation have on the understanding of speed at the particle scale? a. Speed must also be quantized at the particle scale. b. Speed will not be quantized at the particle scale. c. Speed must be zero at the particle scale. d. Speed will be infinite at the particle scale. 73. A 500 nm photon of light strikes a semi-conductive surface with a binding energy of 2 eV. With what velocity will an electron be emitted from the semi-conductive surface? a. 8.38 × 105 m/s b. 9.33 × 105 m/s 3 × 108 m/s c. d. 4.11 × 105 m/s 74. True or false—Treating food with ionizing radiation helps keep it from spoiling. a. b. true false 21.3 The Dual Nature of Light 75. When testing atomic bombs, scientists at Los Alamos recognized that huge releases of energy resulted in problems with power and communications systems in the area surrounding the blast site. Explain the possible tie to Compton scattering. a. The release of light energy caused large-scale emission of electrons. b. The release of light energy caused large-scale emission of protons. c. The release of light energy caused large-scale emission of neutrons. d. The release of light energy caused large-scale 720 Chapter 21 • Test Prep emission of photons. 76. Sunlight above the Earth’s atmosphere has an intensity of 1.30 kW/m2 . If this is reflected straight back from a mirror that has only a small recoil, the light’s momentum is exactly reversed, giving the mirror twice the incident momentum. If the mirror were attached to a solar sail craft, how fast would the craft be moving after 24 hr? Note—The average mass per square meter of the craft is 0.100 kg. a. 8.67 × 10−5 m/s2 b. 8.67 × 10−6 m/s2 c. 94.2 m/s 7.49 m/s d. 77. Consider the counter-clockwise motion of LightSail-1 around Earth. When will the satellite move the fastest? a. point A b. point B c. point C d. point D 78. What will happen to the interference pattern created by electrons when their velocities are increased? a. There will be more zones of constructive interference and fewer zones of destructive interference. b. There will be more zones of destructive interference and fewer zones of constructive interference. c. There will be more zones of constructive and destructive interference. d. There will be fewer zones of constructive and destructive interference. Access for free at openstax.org. CHAPTER 22 The Atom Figure 22.1 Individual carbon atoms are visible in this image of a carbon nanotube made by a scanning tunneling electron microscope. (credit: Taner Yildirim, National Institute of Standards and Technology, Wikimedia Commons) Chapter Outline 22.1 The Structure of the Atom 22.2 Nuclear Forces and Radioactivity 22.3 Half Life and Radiometric Dating 22.4 Nuclear Fission and Fusion 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation From childhood on, we learn that atoms are a substructure of all things around us, from the air we breathe to INTRODUCTION the autumn leaves that blanket a forest trail. Invisible to the eye, the atoms have properties that are used to explain many phenomena—a theme found throughout this text. In this chapter, we discuss the discovery of atoms and their own substructures. We will then learn about the forces that keep them together and the tremendous energy they release when we break them apart. Finally, we will see how the knowledge and manipulation of atoms allows us to better understand geology, biology, and the world around us. 22.1 The Structure of the Atom Section Learning Objectives By the end of this section, you will be able to do the following: • Describe Rutherford’s experiment and his model of the atom • Describe emission and absorption spectra of atoms • Describe the Bohr model of the atom • Calculate the energy of electrons when they change energy levels • Calculate the frequency and wavelength of emitted photons when electrons change energy levels • Describe the quantum model of the atom 722 Chapter 22 • The Atom Section Key Terms energy-level diagram excited state Fraunhofer lines ground state Heisenberg Uncertainty Principle hydrogen-like atoms planetary model of the atom Rutherford scattering Rydberg constant How do we know that atoms are really there if we cannot see them with our own eyes? While often taken for granted, our knowledge of the existence and structure of atoms is the result of centuries of contemplation and experimentation. The earliest known speculation on the atom dates back to the fifth century B.C., when Greek philosophers Leucippus and Democritus contemplated whether a substance could be divided without limit into ever smaller pieces. Since then, scientists such as John Dalton (1766–1844), Amadeo Avogadro (1776–1856), and Dmitri Mendeleev (1834–1907) helped to discover the properties of that fundamental structure of matter. While much could be written about any number of important scientific philosophers, this section will focus on the role played by Ernest Rutherford (1871–1937). Though his understanding of our most elemental matter is rooted in the success of countless prior investigations, his surprising discovery about the interior of the atom is most fundamental in explaining so many well-known phenomena. Rutherford’s Experiment In the early 1900’s, the plum puddingmodel was the accepted model of the atom. Proposed in 1904 by J. J. Thomson, the model suggested that the atom was a spherical ball of positive charge, with negatively charged electrons scattered evenly throughout. In that model, the positive charges made up the pudding, while the electrons acted as isolated plums. During its short life, the model could be used to explain why most particles were neutral, although with an unbalanced number of plums, electrically charged atoms could exist. When Ernest Rutherford began his gold foil experiment in 1909, it is unlikely that anyone would have expected that the plum pudding model would be challenged. However, using a radioactive source, a thin sheet of gold foil, and a phosphorescent screen, Rutherford would uncover something so great that he would later call it “the most incredible event that has ever happened to me in my life”[James, L. K. (1993). Nobel Laureates in Chemistry, 1901–1992. Washington, DC: American Chemical Society.] The experiment that Rutherford designed is shown in Figure 22.2. As you can see in, a radioactive source was placed in a lead container with a hole in one side to produce a beam of positively charged helium particles, called alpha particles. Then, a thin gold foil sheet was placed in the beam. When the high-energy alpha particles passed through the gold foil, they were scattered. The scattering was observed from the bright spots they produced when they struck the phosphor screen. Figure 22.2 Rutherford’s experiment gave direct evidence for the size and mass of the nucleus by scattering alpha particles from a thin gold foil. The scattering of particles suggests that the gold nuclei are very small and contain nearly all of the gold atom’s mass. Particularly significant in showing the size of the nucleus are alpha particles that scatter to very large angles, much like a soccer ball bouncing off a goalie’s head. The expectation of the plum pudding model was that the high-energy alpha particles would be scattered only slightly by the presence of the gold sheet. Because the energy of the alpha particles was much higher than those typically associated with atoms, the alpha particles should have passed through the thin foil much like a supersonic bowling ball would crash through a Access for free at openstax.org. 22.1 • The Structure of the Atom 723 few dozen rows of bowling pins. Any deflection was expected to be minor, and due primarily to the electrostatic Coulomb force between the alpha particles and the foil’s interior electric charges. However, the true result was nothing of the sort. While the majority of alpha particles passed |
through the foil unobstructed, Rutherford and his collaborators Hans Geiger and Ernest Marsden found that alpha particles occasionally were scattered to large angles, and some even came back in the direction from which they came! The result, called Rutherford scattering, implied that the gold nuclei were actually very small when compared with the size of the gold atom. As shown in Figure 22.3, the dense nucleus is surrounded by mostly empty space of the atom, an idea verified by the fact that only 1 in 8,000 particles was scattered backward. Figure 22.3 An expanded view of the atoms in the gold foil in Rutherford’s experiment. Circles represent the atoms that are about 10−10 m in diameter, while the dots represent the nuclei that are about 10−15 m in diameter. To be visible, the dots are much larger than scale—if the nuclei were actually the size of the dots, each atom would have a diameter of about five meters! Most alpha particles crash through but are relatively unaffected because of their high energy and the electron’s small mass. Some, however, strike a nucleus and are scattered straight back. A detailed analysis of their interaction gives the size and mass of the nucleus. Although the results of the experiment were published by his colleagues in 1909, it took Rutherford two years to convince himself of their meaning. Rutherford later wrote: “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. On consideration, I realized that this scattering backwards ... [meant] ... the greatest part of the mass of the atom was concentrated in a tiny nucleus.” In 1911, Rutherford published his analysis together with a proposed model of the atom, which was in part based on Geiger’s work from the previous year. As a result of the paper, the size of the nucleus was determined to be about cm3, much greater than any macroscopic matter. m, or 100,000 times smaller than the atom. That implies a huge density, on the order of g/ Based on the size and mass of the nucleus revealed by his experiment, as well as the mass of electrons, Rutherford proposed the planetary model of the atom. The planetary model of the atom pictures low-mass electrons orbiting a large-mass nucleus. The sizes of the electron orbits are large compared with the size of the nucleus, and most of the atom is a vacuum. The model is analogous to how low-mass planets in our solar system orbit the large-mass Sun. In the atom, the attractive Coulomb force is analogous to gravitation in the planetary system (see Figure 22.4). Figure 22.4 Rutherford’s planetary model of the atom incorporates the characteristics of the nucleus, electrons, and the size of the atom. The model was the first to recognize the structure of atoms, in which low-mass electrons orbit a very small, massive nucleus in orbits much larger than the nucleus. The atom is mostly empty and is analogous to our planetary system. 724 Chapter 22 • The Atom Virtual Physics Rutherford Scattering Click to view content (https://www.openstax.org/l/28rutherford) How did Rutherford figure out the structure of the atom without being able to see it? Explore the answer through this simulation of the famous experiment in which he disproved the plum pudding model by observing alpha particles bouncing off atoms and determining that they must have a small core. TIPS FOR SUCCESS As you progress through the model of the atom, consider the effect that experimentation has on the scientific process. Ask yourself the following: What would our model of the atom be without Rutherford’s gold foil experiment? What further understanding of the atom would not have been gained? How would that affect our current technologies? Though often confusing, experiments taking place today to further understand composition of the atom could perhaps have a similar effect. Absorption and Emission Spectra In 1900, Max Planck recognized that all energy radiated from a source is emitted by atoms in quantum states. How would that radical idea relate to the interior of an atom? The answer was first found by investigating the spectrum of light or emission spectrum produced when a gas is highly energized. Figure 22.5 shows how to isolate the emission spectrum of one such gas. The gas is placed in the discharge tube at the left, where it is energized to the point at which it begins to radiate energy or emit light. The radiated light is channeled by a thin slit and then passed through a diffraction grating, which will separate the light into its constituent wavelengths. The separated light will then strike the photographic film on the right. The line spectrum shown in part (b) of Figure 22.5 is the output shown on the film for excited iron. Note that this spectrum is not continuous but discrete. In other words, only particular wavelengths are emitted by the iron source. Why would that be the case? Figure 22.5 Part (a) shows, from left to right, a discharge tube, slit, and diffraction grating producing a line spectrum. Part (b) shows the emission spectrum for iron. The discrete lines imply quantized energy states for the atoms that produce them. The line spectrum for each element is unique, providing a powerful and much-used analytical tool, and many line spectra were well known for many years before they could be explained with physics. (credit:(b) Yttrium91, Wikimedia Commons) The spectrum of light created by excited iron shows a variety of discrete wavelengths emitted within the visible spectrum. Each element, when excited to the appropriate degree, will create a discrete emission spectrum as in part (b) of Figure 22.5. However, the wavelengths emitted will vary from element to element. The emission spectrum for iron was chosen for Figure 22.5 solely Access for free at openstax.org. because a substantial portion of its emission spectrum is within the visible spectrum. Figure 22.6 shows the emission spectrum for hydrogen. Note that, while discrete, a large portion of hydrogen emission takes place in the ultraviolet and infrared regions. 22.1 • The Structure of the Atom 725 Figure 22.6 A schematic of the hydrogen spectrum shows several series named for those who contributed most to their determination. Part of the Balmer series is in the visible spectrum, while the Lyman series is entirely in the ultraviolet, and the Paschen series and others are in the infrared. Values of nf and ni are shown for some of the lines. Their importance will be described shortly. Just as an emission spectrum shows all discrete wavelengths emitted by a gas, an absorption spectrum will show all light that is absorbed by a gas. Black lines exist where the wavelengths are absorbed, with the remainder of the spectrum lit by light is free to pass through. What relationship do you think exists between the black lines of a gas’s absorption spectrum and the colored lines of its emission spectrum? Figure 22.7 shows the absorption spectrum of the Sun. The black lines are called Fraunhofer lines, and they correspond to the wavelengths absorbed by gases in the Sun’s exterior. Figure 22.7 The absorption spectrum of the Sun. The black lines appear at wavelengths absorbed by the Sun’s gas exterior. The energetic photons emitted from the Sun’s interior are absorbed by gas in its exterior and reemitted in directions away from the observer. That results in dark lines within the absorption spectrum. The lines are called Fraunhofer lines, in honor of the German physicist who discovered them. Lines similar to those are used to determine the chemical composition of stars well outside our solar system. Bohr’s Explanation of the Hydrogen Spectrum To tie the unique signatures of emission spectra to the composition of the atom itself would require clever thinking. Niels Bohr (1885–1962), a Danish physicist, did just that, by making immediate use of Rutherford’s planetary model of the atom. Bohr, shown in Figure 22.8, became convinced of its validity and spent part of 1912 at Rutherford’s laboratory. In 1913, after returning to Copenhagen, he began publishing his theory of the simplest atom, hydrogen, based on Rutherford’s planetary model. Figure 22.8 Niels Bohr, Danish physicist, used the planetary model of the atom to explain the atomic spectrum and size of the hydrogen 726 Chapter 22 • The Atom atom. His many contributions to the development of atomic physics and quantum mechanics, his personal influence on many students and colleagues, and his personal integrity, especially in the face of Nazi oppression, earned him a prominent place in history. (credit: Unknown Author, Wikimedia Commons) Bohr was able to derive the formula for the hydrogen spectrum using basic physics, the planetary model of the atom, and some very important new conjectures. His first conjecture was that only certain orbits are allowed: In other words, in an atom, the orbits of electrons are quantized. Each quantized orbit has a different distinct energy, and electrons can move to a higher orbit by absorbing energy or drop to a lower orbit by emitting energy. Because of the quantized orbits, the amount of energy emitted or absorbed must also be quantized, producing the discrete spectra seen in Figure 22.5 and Figure 22.7. In equation form, the amount of energy absorbed or emitted can be found as refers to the energy of the initial quantized orbit, and where wavelength emitted can be found using the equation refers to the energy of the final orbits. Furthermore, the and relating the wavelength to the frequency found using the equation , where vcorresponds to the speed of light. It makes sense that energy is involved in changing orbits. For example, a burst of energy is required for a satellite to climb to a higher orbit. What is not expected is that atomic orbits should be quantized. Quantization is not observed for satellites or planets, which can have any orbit, given the proper energy (see Figure 22.9). 22.2 22.1 Figure 22.9 The planetary model of the atom, as modif |
ied by Bohr, has the orbits of the electrons quantized. Only certain orbits are allowed, explaining why atomic spectra are discrete or quantized. The energy carried away from an atom by a photon comes from the electron dropping from one allowed orbit to another and is thus quantized. The same is true for atomic absorption of photons. Figure 22.10 shows an energy-level diagram, a convenient way to display energy states. Each of the horizontal lines corresponds to the energy of an electron in a different orbital. Energy is plotted vertically with the lowest or ground state at the bottom and with excited states above. The vertical arrow downwards shows energy being emitted out of the atom due to an electron dropping from one excited state to another. That would correspond to a line shown on the atom’s emission spectrum. The Lyman series shown in Figure 22.6 results from electrons dropping to the ground state, while the Balmer and Paschen series result to electrons dropping to the n = 2and n = 3states, respectively. Access for free at openstax.org. 22.1 • The Structure of the Atom 727 Figure 22.10 An energy-level diagram plots energy vertically and is useful in visualizing the energy states of a system and the transitions between them. This diagram is for the hydrogen-atom electrons, showing a transition between two orbits having energies and . The energy transition results in a Balmer series line in an emission spectrum. Energy and Wavelength of Emitted Hydrogen Spectra The energy associated with a particular orbital of a hydrogen atom can be found using the equation 22.3 where ncorresponds to the orbital value from the atom’s nucleus. The negative value in the equation is based upon a baseline energy of zero when the electron is infinitely far from the atom. As a result, the negative value shows that energy is necessary to free the electron from its orbital state. The minimum energy to free the electron is also referred to as its binding energy. The equation is only valid for atoms with single electrons in their orbital shells (like hydrogen). For ionized atoms similar to hydrogen, the following formula may be used. 22.4 corresponds to –13.6 eV, as mentioned earlier. Additionally, Please note that studied. The atomic number is the number of protons in the nucleus—it is different for each element. The above equation is derived from some basic physics principles, namely conservation of energy, conservation of angular momentum, Coulomb’s law, and centripetal force. There are three derivations that result in the orbital energy equations, and they are shown below. While you can use the energy equations without understanding the derivations, they will help to remind you of just how valuable those fundamental concepts are. refers to the atomic number of the element Derivation 1 (Finding the Radius of an Orbital) One primary difference between the planetary model of the solar system and the planetary model of the atom is the cause of the circular motion. While gravitation causes the motion of orbiting planets around an interior star, the Coulomb force is responsible for the circular shape of the electron’s orbit. The magnitude of the centripetal force is , while the magnitude of the Coulomb force is . The assumption here is that the nucleus is more massive than the stationary electron, and the electron orbits about it. That is consistent with the planetary model of the atom. Equating the Coulomb force and the centripetal force, 22.5 which yields 728 Chapter 22 • The Atom Derivation 2 (Finding the Velocity of the Orbiting Electron) Bohr was clever enough to find a way to calculate the electron orbital energies in hydrogen. That was an important first step that has been improved upon, but it is well worth repeating here, because it does correctly describe many characteristics of hydrogen. Assuming circular orbits, Bohr proposed that the angular momentum Lof an electron in its orbit is also quantized, that is, it has only specific, discrete values. The value for Lis given by the formula 22.7 22.6 where Lis the angular momentum, meis the electron’s mass, rnis the radius of the nth orbit, and his Planck’s constant. Note that angular momentum is . For a small object at a radius r, , so that , and Bohr himself did not know why angular momentum should be quantized, but by using that assumption, he was able to calculate the energies in the hydrogen spectrum, something no one else had done at the time. Quantization says that the value of mvrcan only be equal to h/ 2, 2h/ 2, 3h/ 2, etc. At the time, Derivation 3 (Finding the Energy of the Orbiting Electron) To get the electron orbital energies, we start by noting that the electron energy is the sum of its kinetic and potential energy. 22.8 Kinetic energy is the familiar electron is electrical, or , assuming the electron is not moving at a relativistic speed. Potential energy for the , where Vis the potential due to the nucleus, which looks like a point charge. The nucleus has a positive charge ; thus, , recalling an earlier equation for the potential due to a point charge from the chapter on Electricity and Magnetism. Since the electron’s charge is negative, we see that Substituting the expressions for KE and PE, Now we solve for rnand vusing the equation for angular momentum , giving and Substituting the expression for rnand vinto the above expressions for energy (KE and PE), and performing algebraic manipulation, yields 22.9 22.10 22.11 22.12 for the orbital energies of hydrogen-like atoms. Here, Eois the ground-state energy (n= 1) for hydrogen (Z= 1) and is given by Thus, for hydrogen, The relationship between orbital energies and orbital states for the hydrogen atom can be seen in Figure 22.11. 22.13 22.14 Access for free at openstax.org. 22.1 • The Structure of the Atom 729 Figure 22.11 Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. The orbital energies are calculated using the above equation, first derived by Bohr. WORKED EXAMPLE A hydrogen atom is struck by a photon. How much energy must be absorbed from the photon to raise the electron of the hydrogen atom from its ground state to its second orbital? Strategy The hydrogen atom has an atomic number of Z= 1. Raising the electron from the ground state to its second orbital will increase its orbital level from n= 1 to n= 2. The energy determined will be measured in electron-volts. Solution The amount of energy necessary to cause the change in electron state is the difference between the final and initial energies of the electron. The final energy state of the electron can be found using Knowing the nand Zvalues for the hydrogen atom, and knowing that Eo= –13.6 eV, the result is The original amount of energy associated with the electron is equivalent to the ground state orbital, or 22.15 22.16 22.17 730 Chapter 22 • The Atom The amount of energy necessary to change the orbital state of the electron can be found by determining the electron’s change in energy. 22.18 Discussion The energy required to change the orbital state of the electron is positive. That means that for the electron to move to a state with greater energy, energy must be added to the atom. Should the electron drop back down to its original energy state, a change of –10.2 eV would take place, and 10.2 eV of energy would be emitted from the atom. Just as only quantum amounts of energy may be absorbed by the atom, only quantum amounts of energy can be emitted from the atom. That helps to explain many of the quantum light effects that you have learned about previously. WORKED EXAMPLE Characteristic X-Ray Energy Calculate the approximate energy of an X-ray emitted for an n= 2 to n= 1 transition in a tungsten anode in an X-ray tube. Strategy How do we calculate energies in a multiple-electron atom? In the case of characteristic X-rays, the following approximate calculation is reasonable. Characteristic X-rays are produced when an inner-shell vacancy is filled. Inner-shell electrons are nearer the nucleus than others in an atom and thus feel little net effect from the others. That is similar to what happens inside a charged conductor, where its excess charge is distributed over the surface so that it produces no electric field inside. It is reasonable to assume the inner-shell electrons have hydrogen-like energies, as given by For tungsten, Z= 74, so that the effective charge is 73. Solution The amount of energy given off as an X-ray is found using where and Thus, 22.19 22.20 22.21 22.22 22.23 Discussion This large photon energy is typical of characteristic X-rays from heavy elements. It is large compared with other atomic emissions because it is produced when an inner-shell vacancy is filled, and inner-shell electrons are tightly bound. Characteristic X-ray energies become progressively larger for heavier elements because their energy increases approximately as Z2. Significant accelerating voltage is needed to create such inner-shell vacancies, because other shells are filled and you cannot simply bump one electron to a higher filled shell. You must remove it from the atom completely. In the case of tungsten, at least 72.5 kV is needed. Tungsten is a common anode material in X-ray tubes; so much of the energy of the impinging electrons is absorbed, raising its temperature, that a high-melting-point material like tungsten is required. The wavelength of light emitted by an atom can also be determined through basic derivations. Let us consider the energy of a photon emitted from a hydrogen atom in a downward transition, given by the equation Access for free at openstax.org. Substituting , we get Dividing both sides of the equation by hcgives us an expression for , It can be shown that where Ris the Rydberg constant. Simplified, the formula for determining emitted wavelength can now be written as 22.1 • The Structure of the Atom 731 22.24 22.25 22.26 22.27 22.28 WORKED EXAMPLE What wavelength of light is emitted by an elec |
tron dropping from the third orbital to the ground state of a hydrogen atom? Strategy The ground state of a hydrogen atom is considered the first orbital of the atom. As a result, nf= 1 and ni= 3. The Rydberg constant has already been determined and will be constant regardless of atom chosen. Solution For the equation above, calculate wavelength based on the known energy states. Rearranging the equation for wavelength yields 22.29 22.30 22.31 Discussion This wavelength corresponds to light in the ultraviolet spectrum. As a result, we would not be able to see the photon of light emitted when an electron drops from its third to first energy state. However, it is worth noting that by supplying light of wavelength precisely 102.6 nm, we can cause the electron in hydrogen to move from its first to its third orbital state. Limits of Bohr’s Theory and the Quantum Model of the Atom There are limits to Bohr’s theory. It does not account for the interaction of bound electrons, so it cannot be fully applied to multielectron atoms, even one as simple as the two-electron helium atom. Bohr’s model is what we call semiclassical. The orbits are quantized (nonclassical) but are assumed to be simple circular paths (classical). As quantum mechanics was developed, it became clear that there are no well-defined orbits; rather, there are clouds of probability. Additionally, Bohr’s theory did not explain that some spectral lines are doublets or split into two when examined closely. While we shall examine a few of those aspects of quantum mechanics in more detail, it should be kept in mind that Bohr did not fail. Rather, he made very important steps along the path to greater knowledge and laid the foundation for all of atomic physics that has since evolved. 732 Chapter 22 • The Atom DeBroglie’s Waves Following Bohr’s initial work on the hydrogen atom, a decade was to pass before Louis de Broglie proposed that matter has wave properties. The wave-like properties of matter were subsequently confirmed by observations of electron interference when scattered from crystals. Electrons can exist only in locations where they interfere constructively. How does that affect electrons in atomic orbits? When an electron is bound to an atom, its wavelength must fit into a small space, something like a standing wave on a string (see Figure 22.12). Orbits in which an electron can constructively interfere with itself are allowed. All orbits in which constructive interference cannot occur are not able to exist. Thus, only certain orbits are allowed. The wave nature of an electron, according to de Broglie, is why the orbits are quantized! Figure 22.12 (a) Standing waves on a string have a wavelength related to the length of the string, allowing them to interfere constructively. (b) If we imagine the string formed into a closed circle, we get a rough idea of how electrons in circular orbits can interfere constructively. (c) If the wavelength does not fit into the circumference, the electron interferes destructively; it cannot exist in such an orbit. For a circular orbit, constructive interference occurs when the electron’s wavelength fits neatly into the circumference, so that wave crests always align with crests and wave troughs align with troughs, as shown in Figure 22.12(b). More precisely, when an integral multiple of the electron’s wavelength equals the circumference of the orbit, constructive interference is obtained. In equation form, the condition for constructive interference and an allowed electron orbit is where of a hydrogen atom. is the electron’s wavelength and rnis the radius of that circular orbit. Figure 22.13 shows the third and fourth orbitals 22.32 Figure 22.13 The third and fourth allowed circular orbits have three and four wavelengths, respectively, in their circumferences. Access for free at openstax.org. 22.1 • The Structure of the Atom 733 Heisenberg Uncertainty How does determining the location of an electron change its trajectory? The answer is fundamentally important—measurement affects the system being observed. It is impossible to measure a physical quantity exactly, and greater precision in measuring one quantity produces less precision in measuring a related quantity. It was Werner Heisenberg who first stated that limit to knowledge in 1929 as a result of his work on quantum mechanics and the wave characteristics of all particles (see Figure 22.14). Figure 22.14 Werner Heisenberg was the physicist who developed the first version of true quantum mechanics. Not only did his work give a description of nature on the very small scale, it also changed our view of the availability of knowledge. Although he is universally recognized for the importance of his work by receiving the Nobel Prize in 1932, for example, Heisenberg remained in Germany during World War II and headed the German effort to build a nuclear bomb, permanently alienating himself from most of the scientific community. (credit: Unknown Author, Wikimedia Commons) For example, you can measure the position of a moving electron by scattering light or other electrons from it. However, by doing so, you are giving the electron energy, and therefore imparting momentum to it. As a result, the momentum of the electron is affected and cannot be determined precisely. This change in momentum could be anywhere from close to zero up to the relative momentum of the electron ( particle. ). Note that, in this case, the particle is an electron, but the principle applies to any Viewing the electron through the model of wave-particle duality, Heisenberg recognized that, because a wave is not located at one fixed point in space, there is an uncertainty associated with any electron’s position. That uncertainty in position, approximately equal to the wavelength of the particle. That is, momentum. The uncertainty in an electron’s position can be reduced by using a shorter-wavelength electron, since But shortening the wavelength increases the uncertainty in momentum, since momentum can be reduced by using a longer-wavelength electron, but that increases the uncertainty in position. Mathematically, you can express the trade-off by multiplying the uncertainties. The wavelength cancels, leaving . There is an interesting trade-off between position and . Conversely, the uncertainty in ,is . Therefore, if one uncertainty is reduced, the other must increase so that their product is mathematics, Heisenberg showed that the best that can be done in a simultaneous measurement of position and momentum is .With the use of advanced 22.33 That relationship is known as the Heisenberg uncertainty principle. The Quantum Model of the Atom Because of the wave characteristic of matter, the idea of well-defined orbits gives way to a model in which there is a cloud of probability, consistent with Heisenberg’s uncertainty principle. Figure 22.15 shows how the principle applies to the ground state of hydrogen. If you try to follow the electron in some well-defined orbit using a probe that has a wavelength small enough to 734 Chapter 22 • The Atom measure position accurately, you will instead knock the electron out of its orbit. Each measurement of the electron’s position will find it to be in a definite location somewhere near the nucleus. Repeated measurements reveal a cloud of probability like that in the figure, with each speck the location determined by a single measurement. There is not a well-defined, circular-orbit type of distribution. Nature again proves to be different on a small scale than on a macroscopic scale. Figure 22.15 The ground state of a hydrogen atom has a probability cloud describing the position of its electron. The probability of finding the electron is proportional to the darkness of the cloud. The electron can be closer or farther than the Bohr radius, but it is very unlikely to be a great distance from the nucleus. Virtual Physics Models of the Hydrogen Atom Click to view content (https://www.openstax.org/l/28atom_model) How did scientists figure out the structure of atoms without looking at them? Try out different models by shooting light at the atom. Use this simulation to see how the prediction of the model matches the experimental results. Check Your Understanding 1. Alpha particles are positively charged. What influence did their charge have on the gold foil experiment? a. The positively charged alpha particles were attracted by the attractive electrostatic force from the positive nuclei of the gold atoms. b. The positively charged alpha particles were scattered by the attractive electrostatic force from the positive nuclei of the gold atoms. c. The positively charged alpha particles were scattered by the repulsive electrostatic force from the positive nuclei of the gold atoms. d. The positively charged alpha particles were attracted by the repulsive electrostatic force from the positive nuclei of the gold atoms. 22.2 Nuclear Forces and Radioactivity Section Learning Objectives By the end of this section, you will be able to do the following: • Describe the structure and forces present within the nucleus • Explain the three types of radiation • Write nuclear equations associated with the various types of radioactive decay Section Key Terms alpha decay atomic number beta decay gamma decay Geiger tube isotope mass number nucleons radioactive radioactive decay Access for free at openstax.org. 22.2 • Nuclear Forces and Radioactivity 735 radioactivity scintillator strong nuclear force transmutation There is an ongoing quest to find the substructures of matter. At one time, it was thought that atoms would be the ultimate substructure. However, just when the first direct evidence of atoms was obtained, it became clear that they have a substructure and a tiny nucleus. The nucleus itself has spectacular characteristics. For example, certain nuclei are unstable, and their decay emits radiations with energies millions of times greater than atomic energies. Some of the mysteries of nature, suc |
h as why the core of Earth remains molten and how the Sun produces its energy, are explained by nuclear phenomena. The exploration of radioactivity and the nucleus has revealed new fundamental particles, forces, and conservation laws. That exploration has evolved into a search for further underlying structures, such as quarks. In this section, we will explore the fundamentals of the nucleus and nuclear radioactivity. The Structure of the Nucleus At this point, you are likely familiar with the neutron and proton, the two fundamental particles that make up the nucleus of an atom. Those two particles, collectively called nucleons, make up the small interior portion of the atom. Both particles have nearly the same mass, although the neutron is about two parts in 1,000 more massive. The mass of a proton is equivalent to 1,836 electrons, while the mass of a neutron is equivalent to that of 1,839 electrons. That said, each of the particles is significantly more massive than the electron. When describing the mass of objects on the scale of nucleons and atoms, it is most reasonable to measure their mass in terms of atoms. The atomic mass unit (u) was originally defined so that a neutral carbon atom would have a mass of exactly 12 u. Given that protons and neutrons are approximately the same mass, that there are six protons and six neutrons in a carbon atom, and that the mass of an electron is minuscule in comparison, measuring this way allows for both protons and neutrons to have masses close to 1 u. Table 22.1 shows the mass of protons, neutrons, and electrons on the new scale. TIPS FOR SUCCESS For most conceptual situations, the difference in mass between the proton and neutron is insubstantial. In fact, for calculations that require fewer than four significant digits, both the proton and neutron masses may be considered equivalent to one atomic mass unit. However, when determining the amount of energy released in a nuclear reaction, as in Equation 22.40, the difference in mass cannot be ignored. Another other useful mass unit on the atomic scale is the one uses the equation , as will be addressed later in this text. . While rarely used in most contexts, it is convenient when Proton Mass Neutron Mass Electron Mass Kilograms (kg) Atomic mass units (u) Table 22.1 Atomic Masses for Multiple Units To more completely characterize nuclei, let us also consider two other important quantities: the atomic number and the mass number. The atomic number, Z, represents the number of protons within a nucleus. That value determines the elemental quality of each atom. Every carbon atom, for instance, has a Zvalue of 6, whereas every oxygen atom has a Zvalue of 8. For clarification, only oxygen atoms may have a Zvalue of 8. If the Zvalue is not 8, the atom cannot be oxygen. The mass number, A, represents the total number of protons and neutrons, or nucleons, within an atom. For an ordinary carbon atom the mass number would be 12, as there are typically six neutrons accompanying the six protons within the atom. In the case of carbon, the mass would be exactly 12 u. For oxygen, with a mass number of 16, the atomic mass is 15.994915 u. Of course, the difference is minor and can be ignored for most scenarios. Again, because the mass of an electron is so small compared to the nucleons, the mass number and the atomic mass can be essentially equivalent. Figure 22.16 shows an example of Lithium-7, which has an atomic number of 3 and a mass number of 7. How does the mass number help to differentiate one atom from another? If each atom of carbon has an atomic number of 6, 736 Chapter 22 • The Atom then what is the value of including the mass number at all? The intent of the mass number is to differentiate between various isotopes of an atom. The term isotope refers to the variation of atoms based upon the number of neutrons within their nucleus. While it is most common for there to be six neutrons accompanying the six protons within a carbon atom, it is possible to find carbon atoms with seven neutrons or eight neutrons. Those carbon atoms are respectively referred to as carbon-13 and carbon-14 atoms, with their mass numbers being their primary distinction. The isotope distinction is an important one to make, as the number of neutrons within an atom can affect a number of its properties, not the least of which is nuclear stability. Figure 22.16 Lithium-7 has three protons and four neutrons within its nucleus. As a result, its mass number is 7, while its atomic number is 3. The actual mass of the atom is 7.016 u. Lithium 7 is an isotope of lithium. To more easily identify various atoms, their atomic number and mass number are typically written in a form of representation called the nuclide. The nuclide form appears as follows: neutrons. , where Xis the atomic symbol and Nrepresents the number of Let us look at a few examples of nuclides expressed in the notation. The nucleus of the simplest atom, hydrogen, is a single (the zero for no neutrons is often omitted). To check the symbol, refer to the periodic table—you see that the proton, or atomic number Zof hydrogen is 1. Since you are given that there are no neutrons, the mass number Ais also 1. There is a scarce form of hydrogen found in nature called deuterium; its nucleus has one proton and one neutron and, hence, twice the mass of common hydrogen. The symbol for deuterium is, thus, tritium, since it has a single proton and two neutrons, and it is written identical chemistries, but the nuclei differ greatly in mass, stability, and other characteristics. Again, the different nuclei are referred to as isotopes of the same element. . An even rarer—and radioactive—form of hydrogen is called . The three varieties of hydrogen have nearly There is some redundancy in the symbols A, X, Z, and N. If the element Xis known, then Zcan be found in a periodic table. If both Aand Xare known, then Ncan also be determined by first finding Z; then, N= A– Z. Thus the simpler notation for nuclides is 22.34 which is sufficient and is most commonly used. For example, in this simpler notation, the three isotopes of hydrogen are , , and . For , should we need to know, we can determine that Z= 92 for uranium from the periodic table, and thus, N = 238 − 92 = 146. Radioactivity and Nuclear Forces In 1896, the French physicist Antoine Henri Becquerel (1852–1908) noticed something strange. When a uranium-rich mineral called pitchblende was placed on a completely opaque envelope containing a photographic plate, it darkened spots on the photographic plate.. Becquerel reasoned that the pitchblende must emit invisible rays capable of penetrating the opaque material. Stranger still was that no light was shining on the pitchblende, which means that the pitchblende was emitting the invisible rays continuously without having any energy input! There is an apparent violation of the law of conservation of energy, one that scientists can now explain using Einstein’s famous equation originate in the nuclei of the atoms and have other unique characteristics. It was soon evident that Becquerel’s rays To this point, most reactions you have studied have been chemical reactions, which are reactions involving the electrons Access for free at openstax.org. 22.2 • Nuclear Forces and Radioactivity 737 surrounding the atoms. However, two types of experimental evidence implied that Becquerel’s rays did not originate with electrons, but instead within the nucleus of an atom. First, the radiation is found to be only associated with certain elements, such as uranium. Whether uranium was in the form of an element or compound was irrelevant to its radiation. In addition, the presence of radiation does not vary with temperature, pressure, or ionization state of the uranium atom. Since all of those factors affect electrons in an atom, the radiation cannot come from electron transitions, as atomic spectra do. The huge energy emitted during each event is the second piece of evidence that the radiation cannot be atomic. Nuclear radiation has energies on the order of 106 eV per event, which is much greater than typical atomic energies that are a few eV, such as those observed in spectra and chemical reactions, and more than ten times as high as the most energetic X-rays. But why would reactions within the nucleus take place? And what would cause an apparently stable structure to begin emitting energy? Was there something special about Becquerel’s uranium-rich pitchblende? To answer those questions, it is necessary to look into the structure of the nucleus. Though it is perhaps surprising, you will find that many of the same principles that we observe on a macroscopic level still apply to the nucleus. Nuclear Stability A variety of experiments indicate that a nucleus behaves something like a tightly packed ball of nucleons, as illustrated in Figure 22.17. Those nucleons have large kinetic energies and, thus, move rapidly in very close contact. Nucleons can be separated by a large force, such as in a collision with another nucleus, but strongly resist being pushed closer together. The most compelling evidence that nucleons are closely packed in a nucleus is that the radius of a nucleus, r, is found to be approximately where 1.2 femtometer (fm) and Ais the mass number of the nucleus. Note that . Since many nuclei are spherical, and the volume of a sphere is , we see that —that is, the volume of a nucleus is proportional to the number of nucleons in it. That is what you expect if you pack nucleons so close that there is no empty space between them. 22.35 Figure 22.17 Nucleons are held together by nuclear forces and resist both being pulled apart and pushed inside one another. The volume of the nucleus is the sum of the volumes of the nucleons in it, here shown in different colors to represent protons and neutrons. So what forces hold a nucleus together? After all, the nucleus is very small and its protons, being positive, should exert tremendous repulsive forc |
es on one another. Considering that, it seems that the nucleus would be forced apart, not together! The answer is that a previously unknown force holds the nucleus together and makes it into a tightly packed ball of nucleons. This force is known as the strong nuclear force. The strong force has such a short range that it quickly fall to zero over a distance of only 10–15 meters. However, like glue, it is very strong when the nucleons get close to one another. The balancing of the electromagnetic force with the nuclear forces is what allows the nucleus to maintain its spherical shape. If, for any reason, the electromagnetic force should overcome the nuclear force, components of the nucleus would be projected outward, creating the very radiation that Becquerel discovered! Understanding why the nucleus would break apart can be partially explained using Table 22.2. The balance between the strong nuclear force and the electromagnetic force is a tenuous one. Recall that the attractive strong nuclear force exists between any two nucleons and acts over a very short range while the weaker repulsive electromagnetic force only acts between protons, although over a larger range. Considering the interactions, an imperfect balance between neutrons and protons can result in a nuclear reaction, with the result of regaining equilibrium. 738 Chapter 22 • The Atom Range of Force Direction Nucleon Interaction Magnitude of Force Electromagnetic Force Long range, though decreasing by 1/r2 Repulsive Proton –proton repulsion Relatively small Strong Nuclear Force Very short range, essentially zero at 1 femtometer Attractive Attraction between any two nucleons 100 times greater than the electromagnetic force Table 22.2 Comparing the Electromagnetic and Strong Forces The radiation discovered by Becquerel was due to the large number of protons present in his uranium-rich pitchblende. In short, the large number of protons caused the electromagnetic force to be greater than the strong nuclear force. To regain stability, the nucleus needed to undergo a nuclear reaction called alpha (α) decay. The Three Types of Radiation Radioactivity refers to the act of emitting particles or energy from the nucleus. When the uranium nucleus emits energetic nucleons in Becquerel’s experiment, the radioactive process causes the nucleus to alter in structure. The alteration is called radioactive decay. Any substance that undergoes radioactive decay is said to be radioactive. That those terms share a root with the term radiationshould not be too surprising, as they all relate to the transmission of energy. Alpha Decay Alpha decay refers to the type of decay that takes place when too many protons exist in the nucleus. It is the most common type of decay and causes the nucleus to regain equilibrium between its two competing internal forces. During alpha decay, the nucleus ejects two protons and two neutrons, allowing the strong nuclear force to regain balance with the repulsive electromagnetic force. The nuclear equation for an alpha decay process can be shown as follows. 22.36 Figure 22.18 A nucleus undergoes alpha decay. The alpha particle can be seen as made up of two neutrons and two protons, which constitute a helium-4 atom. Three things to note as a result of the above equation: 1. By ejecting an alpha particle, the original nuclide decreases in atomic number. That means that Becquerel’s uranium nucleus, upon decaying, is actually transformed into thorium, two atomic numbers lower on the periodic table! The process of changing elemental composition is called transmutation. 2. Note that the two protons and two neutrons ejected from the nucleus combine to form a helium nucleus. Shortly after decay, the ejected helium ion typically acquires two electrons to become a stable helium atom. 3. Finally, it is important to see that, despite the elemental change, physical conservation still takes place. The mass number of the new element and the alpha particle together equal the mass number of the original element. Also, the net charge of all particles involved remains the same before and after the transmutation. Beta Decay Like alpha decay, beta ( For beta decay, however, a neutron is transformed into a proton and electron or vice versa. The transformation allows for the total mass number of the atom to remain the same, although the atomic number will increase by one (or decrease by one). Once again, the transformation of the neutron allows for a rebalancing of the strong nuclear and electromagnetic forces. The nuclear ) decay also takes place when there is an imbalance between neutrons and protons within the nucleus. Access for free at openstax.org. 22.2 • Nuclear Forces and Radioactivity 739 equation for a beta decay process is shown below. in the equation above stands for a high-energy particle called the neutrino. A nucleus may also emit a positron, The symbol and in that case Zdecreases and Nincreases. It is beyond the scope of this section and will be discussed in further detail in the chapter on particles. It is worth noting, however, that the mass number and charge in all beta-decay reactions are conserved. Figure 22.19 A nucleus undergoes beta decay. The neutron splits into a proton, electron, and neutrino. This particular decay is called decay. Gamma Decay Gamma decay is a unique form of radiation that does not involve balancing forces within the nucleus. Gamma decay occurs when a nucleus drops from an excited state to the ground state. Recall that such a change in energy state will release energy from the nucleus in the form of a photon. The energy associated with the photon emitted is so great that its wavelength is shorter than that of an X-ray. Its nuclear equation is as follows. 22.37 Figure 22.20 A nucleus undergoes gamma decay. The nucleus drops in energy state, releasing a gamma ray. WORKED EXAMPLE Creating a Decay Equation Write the complete decay equation in Strategy Beta decay results in an increase in atomic number. As a result, the original (or parent) nucleus, must have an atomic number of one fewer proton. . Refer to the periodic table for values of Z. notation for beta decay producing Solution The equation for beta decay is as follows Considering that barium is the product (or daughter) nucleus and has an atomic number of 56, the original nucleus must be of an atomic number of 55. That corresponds to cesium, or Cs. 22.38 22.39 740 Chapter 22 • The Atom The number of neutrons in the parent cesium and daughter barium can be determined by subtracting the atomic number from the mass number (137 – 55 for cesium, 137 – 56 for barium). Substitute those values for the Nand N – 1subscripts in the above equation. 22.40 Discussion The terms parentand daughternucleus refer to the reactants and products of a nuclear reaction. The terminology is not just used in this example, but in all nuclear reaction examples. The cesium-137 nuclear reaction poses a significant health risk, as its chemistry is similar to that of potassium and sodium, and so it can easily be concentrated in your cells if ingested. WORKED EXAMPLE Alpha Decay Energy Found from Nuclear Masses Find the energy emitted in the decay of 239Pu. Strategy Nuclear reaction energy, such as released in decay, can be found using the equation difference in mass between the parent nucleus and the products of the decay. The mass of pertinent particles is as follows . We must first find , the 239Pu: 239.052157 u 235U: 235.043924 u 4He: 4.002602 u. Solution The decay equation for 239Pu is Determine the amount of mass lost between the parent and daughter nuclei. Now we can find Eby entering into the equation. And knowing that , we can find that 22.41 22.42 22.43 22.44 Discussion The energy released in this decay is in the MeV range, about 106 times as great as typical chemical reaction energies, consistent with previous discussions. Most of the energy becomes kinetic energy of the particle (or 4He nucleus), which moves away at high speed. The energy carried away by the recoil of the 235U nucleus is much smaller, in order to conserve momentum. The 235U nucleus can be left in an excited state to later emit photons ( rays). The decay is spontaneous and releases energy, because the products have less mass than the parent nucleus. Properties of Radiation The charges of the three radiated particles differ. Alpha particles, with two protons, carry a net charge of +2. Beta particles, with one electron, carry a net charge of –1. Meanwhile, gamma rays are solely photons, or light, and carry no charge. The difference Access for free at openstax.org. 22.2 • Nuclear Forces and Radioactivity 741 in charge plays an important role in how the three radiations affect surrounding substances. Alpha particles, being highly charged, will quickly interact with ions in the air and electrons within metals. As a result, they have a short range and short penetrating distance in most materials. Beta particles, being slightly less charged, have a larger range and larger penetrating distance. Gamma rays, on the other hand, have little electric interaction with particles and travel much farther. Two diagrams below show the importance of difference in penetration. Table 22.3 shows the distance of radiation penetration, and Figure 22.21 shows the influence various factors have on radiation penetration distance. Type of Radiation Range particles A sheet of paper, a few cm of air, fractions of a millimeter of tissue particles A thin aluminum plate, tens of cm of tissue rays Several cm of lead, meters of concrete Table 22.3 Comparing Ranges of Radioactive Decay Figure 22.21 The penetration or range of radiation depends on its energy, the material it encounters, and the type of radiation. (a) Greater energy means greater range. (b) Radiation has a smaller range in materials with high electron density. (c) Alphas have the smallest range, betas have a greater range, and gammas have the greatest range. LINKS TO PHYSICS Radiation Detectors The |
first direct detection of radiation was Becquerel’s darkened photographic plate. Photographic film is still the most common detector of ionizing radiation, being used routinely in medical and dental X-rays. Nuclear radiation can also be captured on film, as seen in Figure 22.22. The mechanism for film exposure by radiation is similar to that by photons. A quantum of energy from a radioactive particle interacts with the emulsion and alters it chemically, thus exposing the film. Provided the radiation has more than the few eV of energy needed to induce the chemical change, the chemical alteration will occur. The amount of film darkening is related to the type of radiation and amount of exposure. The process is not 100 percent efficient, since not all incident radiation interacts and not all interactions produce the chemical change. 742 Chapter 22 • The Atom Figure 22.22 Film badges contain film similar to that used in this dental X-ray film. It is sandwiched between various absorbers to determine the penetrating ability of the radiation as well as the amount. Film badges are worn to determine radiation exposure. (credit: Werneuchen, Wikimedia Commons) Another very common radiation detector is the Geiger tube. The clicking and buzzing sound we hear in dramatizations and documentaries, as well as in our own physics labs, is usually an audio output of events detected by a Geiger counter. These relatively inexpensive radiation detectors are based on the simple and sturdy Geiger tube, shown schematically in Figure 22.23. A conducting cylinder with a wire along its axis is filled with an insulating gas so that a voltage applied between the cylinder and wire produces almost no current. Ionizing radiation passing through the tube produces free ion pairs that are attracted to the wire and cylinder, forming a current that is detected as a count. Not every particle is detected, since some radiation can pass through without producing enough ionization. However, Geiger counters are very useful in producing a prompt output that reveals the existence and relative intensity of ionizing radiation. Figure 22.23 (a) Geiger counters such as this one are used for prompt monitoring of radiation levels, generally giving only relative intensity and not identifying the type or energy of the radiation. (credit: Tim Vickers, Wikimedia Commons) (b) Voltage applied between the cylinder and wire in a Geiger tube affects ions and electrons produced by radiation passing through the gas-filled cylinder. Ions move toward the cylinder and electrons toward the wire. The resulting current is detected and registered as a count. Another radiation detection method records light produced when radiation interacts with materials. The energy of the radiation is sufficient to excite atoms in a material that may fluoresce, such as the phosphor used by Rutherford’s group. Materials called scintillators use a more complex process to convert radiation energy into light. Scintillators may be liquid or solid, and they can Access for free at openstax.org. 22.3 • Half Life and Radiometric Dating 743 be very efficient. Their light output can provide information about the energy, charge, and type of radiation. Scintillator light flashes are very brief in duration, allowing the detection of a huge number of particles in short periods of time. Scintillation detectors are used in a variety of research and diagnostic applications. Among those are the detection of the radiation from distant galaxies using satellite-mounted equipment and the detection of exotic particles in accelerator laboratories. Virtual Physics Beta Decay Click to view content (https://www.openstax.org/l/21betadecayvid) Watch beta decay occur for a collection of nuclei or for an individual nucleus. With this applet, individuals or groups of students can compare half-lives! Check Your Understanding 2. What leads scientists to infer that the nuclear strong force exists? a. A strong force must hold all the electrons outside the nucleus of an atom. b. A strong force must counteract the highly attractive Coulomb force in the nucleus. c. A strong force must hold all the neutrons together inside the nucleus. d. A strong force must counteract the highly repulsive Coulomb force between protons in the nucleus. 22.3 Half Life and Radiometric Dating Section Learning Objectives By the end of this section, you will be able to do the following: • Explain radioactive half-life and its role in radiometric dating • Calculate radioactive half-life and solve problems associated with radiometric dating Section Key Terms activity becquerel carbon-14 dating decay constant half-life radioactive dating Half-Life and the Rate of Radioactive Decay Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by Marie and Pierre Curie, decay faster than uranium. That means they have shorter lifetimes, producing a greater rate of decay. Here we will explore half-life and activity, the quantitative terms for lifetime and rate of decay. Why do we use the term like half-liferather than lifetime? The answer can be found by examining Figure 22.24, which shows how the number of radioactive nuclei in a sample decreases with time. The time in which half of the original number of nuclei decay is defined as the half-life, . After one half-life passes, half of the remaining nuclei will decay in the next half-life. Then, half of that amount in turn decays in the following half-life. Therefore, the number of radioactive nuclei decreases from Nto N/ 2 in one half-life, to N/ 4 in the next, to N/ 8 in the next, and so on. Nuclear decay is an example of a purely statistical process. TIPS FOR SUCCESS A more precise definition of half-life is that each nucleus has a 50 percent chance of surviving for a time equal to one halflife. If an individual nucleus survives through that time, it still has a 50 percent chance of surviving through another half-life. Even if it happens to survive hundreds of half-lives, it still has a 50 percent chance of surviving through one more. Therefore, the decay of a nucleus is like random coin flipping. The chance of heads is 50 percent, no matter what has happened before. The probability concept aligns with the traditional definition of half-life. Provided the number of nuclei is reasonably large, half of the original nuclei should decay during one half-life period. 744 Chapter 22 • The Atom Figure 22.24 Radioactive decay reduces the number of radioactive nuclei over time. In one half-life ( ), the number decreases to half of its original value. Half of what remains decays in the next half-life, and half of that in the next, and so on. This is exponential decay, as seen in the graph of the number of nuclei present as a function of time. The following equation gives the quantitative relationship between the original number of nuclei present at time zero the number at a later time t and 22.45 where e= 2.71828... is the base of the natural logarithm, and is the decay constant for the nuclide. The shorter the half-life, the larger is the value of equation decreases with time. The decay constant can be found with the , and the faster the exponential Activity, the Rate of Decay What do we mean when we say a source is highly radioactive? Generally, it means the number of decays per unit time is very high. We define activity Rto be the rate of decay expressed in decays per unit time. In equation form, this is 22.46 where is the number of decays that occur in time . Activity can also be determined through the equation 22.47 22.48 which shows that as the amount of radiative material (N) decreases, the rate of decay decreases as well. The SI unit for activity is one decay per second and it is given the name becquerel (Bq) in honor of the discoverer of radioactivity. That is, Activity Ris often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the curie (Ci), defined to be the activity of 1 g of 226Ra, in honor of Marie Curie’s work with radium. The definition of the curie is 22.49 Access for free at openstax.org. or decays per second. Radiometric Dating 22.3 • Half Life and Radiometric Dating 745 Radioactive dating or radiometric dating is a clever use of naturally occurring radioactivity. Its most familiar application is carbon-14 dating. Carbon-14 is an isotope of carbon that is produced when solar neutrinos strike atmosphere. Radioactive carbon has the same chemistry as stable carbon, and so it mixes into the biosphere, where it is consumed and becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon, so if you know the number of carbon nuclei in an object (perhaps determined by mass and Avogadro’s number), you can multiply that number by back to with the environment ceases, and wrappings, with the normal abundance in living tissue, it is possible to determine the artifact’s age (or time since death). Carbon-14 dating can be used for biological tissues as old as 50 or 60 thousand years, but is most accurate for younger samples, since the abundance of with a half-life of 5,730 years (note that this is an example of beta decay). When an organism dies, carbon exchange nuclei within the object. Over time, carbon-14 will naturally decay is not replenished. By comparing the abundance of in an artifact, such as mummy nuclei in them is greater. to find the number of particles within the One of the most famous cases of carbon-14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus (see Figure 22.25). This relic was first displayed in Turin in 1354 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the then-accepted image of Jesus. As a result, the relic has been remained controversial throughout the centuries. Carbon-14 dating was |
not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece found in from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92 percent of the living tissues, allowing the shroud to be dated (see Equation 22.57). Figure 22.25 Part of the Shroud of Turin, which shows a remarkable negative imprint likeness of Jesus complete with evidence of crucifixion wounds. The shroud first surfaced in the 14th century and was only recently carbon-14 dated. It has not been determined how the image was placed on the material. (credit: Butko, Wikimedia Commons) WORKED EXAMPLE Carbon-11 Decay Carbon-11 has a half-life of 20.334 min. (a) What is the decay constant for carbon-11? If 1 kg of carbon-11 sample exists at the beginning of an hour, (b) how much material will remain at the end of the hour and (c) what will be the decay activity at that time? Strategy Since refers to the amount of carbon-11 at the start, then after one half-life, the amount of carbon-11 remaining will be The decay constant is equivalent to the probability that a nucleus will decay each second. As a result, the half-life will need to be converted to seconds. Solution (a) Since half of the carbon-11 remains after one half-life, . 22.50 22.51 746 Chapter 22 • The Atom Take the natural logarithm of each side to isolate the decay constant. Convert the 20.334 min to seconds. (b) The amount of material after one hour can be found by using the equation with tconverted into seconds and NOwritten as 1,000 g (c) The decay activity after one hour can be found by using the equation for the mass value after one hour. 22.52 22.53 22.54 22.55 22.56 22.57 Discussion (a) The decay constant shows that 0.0568 percent of the nuclei in a carbon-11 sample will decay each second. Another way of considering the decay constant is that a given carbon-11 nuclei has a 0.0568 percent probability of decaying each second. The decay of carbon-11 allows it to be used in positron emission topography (PET) scans; however, its 20.334 min half-life does pose challenges for its administration. (b) One hour is nearly three full half-lives of the carbon-11 nucleus. As a result, one would expect the amount of sample remaining to be approximately one eighth of the original amount. The 129.4 g remaining is just a bit larger than one-eighth, which is sensible given a half-life of just over 20 min. (c) Label analysis shows that the unit of Becquerel is sensible, as there are 0.0735 g of carbon-11 decaying each second. That is smaller amount than at the beginning of the hour, when g of carbon-11 were decaying each second. WORKED EXAMPLE How Old is the Shroud of Turin? Calculate the age of the Shroud of Turin given that the amount of Strategy Because 92 percent of the know that the half-life of assume that the decrease in remains, is 5,730 years, and so once is solely due to nuclear decay. found in it is 92 percent of that in living tissue. . Therefore, the equation can be used to find . We also is known, we can find and then find tas requested. Here, we Solution Solving the equation for gives Access for free at openstax.org. 22.58 22.4 • Nuclear Fission and Fusion 747 Thus, Taking the natural logarithm of both sides of the equation yields so that Rearranging to isolate tgives Now, the equation can be used to find for . Solving for and substituting the known half-life gives We enter that value into the previous equation to find t. 22.59 22.60 22.61 22.62 22.63 22.64 Discussion This dates the material in the shroud to 1988–690 = 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of 1320 ± 60. That uncertainty is typical of carbon-14 dating and is due to the small amount of 14 C in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). That said, is it notable that the carbon-14 date is consistent with the first record of the shroud’s existence and certainly inconsistent with the period in which Jesus lived. There are other noncarbon forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of The decay series for been since the rock solidified. Knowledge of the years ago. about , so the ratio of those nuclides in a rock can be used an indication of how long it has half-life has shown, for example, that the oldest rocks on Earth solidified ends with Virtual Physics Radioactive Dating Game Click to view content (https://www.openstax.org/l/02radioactive_dating_game) Learn about different types of radiometric dating, such as carbon dating. Understand how decay and half-life work to enable radiometric dating to work. Play a game that tests your ability to match the percentage of the dating element that remains to the age of the object. 22.4 Nuclear Fission and Fusion Section Learning Objectives By the end of this section, you will be able to do the following: • Explain nuclear fission • Explain nuclear fusion • Describe how the processes of fission and fusion work in nuclear weapons and in generating nuclear power 748 Chapter 22 • The Atom Section Key Terms chain reaction critical mass liquid drop model nuclear fission nuclear fusion proton-proton cycle The previous section dealt with naturally occurring nuclear decay. Without human intervention, some nuclei will change composition in order to achieve a stable equilibrium. This section delves into a less-natural process. Knowing that energy can be emitted in various forms of nuclear change, is it possible to create a nuclear reaction through our own intervention? The answer to this question is yes. Through two distinct methods, humankind has discovered multiple ways of manipulating the atom to release its internal energy. Nuclear Fission In simplest terms, nuclear fission is the splitting of an atomic bond. Given that it requires great energy separate two nucleons, it may come as a surprise to learn that splitting a nucleus can releasevast potential energy. And although it is true that huge amounts of energy can be released, considerable effort is needed to do so in practice. An unstable atom will naturally decay, but it may take millions of years to do so. As a result, a physical catalyst is necessary to produce useful energy through nuclear fission. The catalyst typically occurs in the form of a free neutron, projected directly at the nucleus of a high-mass atom. As shown in Figure 22.26, a neutron strike can cause the nucleus to elongate, much like a drop of liquid water. This is why the model is known as the liquid drop model. As the nucleus elongates, nucleons are no longer so tightly packed, and the repulsive electromagnetic force can overcome the short-range strong nuclear force. The imbalance of forces can result in the two ends of the drop flying apart, with some of the nuclear binding energy released to the surroundings. Figure 22.26 Neutron-induced fission is shown. First, energy is put into a large nucleus when it absorbs a neutron. Acting like a struck liquid drop, the nucleus deforms and begins to narrow in the middle. Since fewer nucleons are in contact, the repulsive Coulomb force is able to break the nucleus into two parts with some neutrons also flying away. As you can imagine, the consequences of the nuclei splitting are substantial. When a nucleus is split, it is not only energy that is released, but a small number of neutrons as well. Those neutrons have the potential to cause further fission in other nuclei, especially if they are directed back toward the other nuclei by a dense shield or neutron reflector (see part (d) of Figure 22.26). Access for free at openstax.org. 22.4 • Nuclear Fission and Fusion 749 However, not every neutron produced by fission induces further fission. Some neutrons escape the fissionable material, while others interact with a nucleus without making it split. We can enhance the number of fissions produced by neutrons by having a large amount of fissionable material as well as a neutron reflector. The minimum amount necessary for self-sustained fission of a given nuclide is called its critical mass. Some nuclides, such as 239Pu, produce more neutrons per fission than others, such as 235U. Additionally, some nuclides are easier to make fission than others. In particular, 235U and 239Pu are easier to fission than the much more abundant 238U. Both factors affect critical mass, which is smallest for 239Pu. The self-sustained fission of nuclei is commonly referred to as a chain reaction, as shown in Figure 22.27. Figure 22.27 A chain reaction can produce self-sustained fission if each fission produces enough neutrons to induce at least one more fission. This depends on several factors, including how many neutrons are produced in an average fission and how easy it is to make a particular type of nuclide fission. A chain reaction can have runaway results. If each atomic split results in two nuclei producing a new fission, the number of nuclear reactions will increase exponentially. One fission will produce two atoms, the next round of fission will create four atoms, the third round eight atoms, and so on. Of course, each time fission occurs, more energy will be emitted, further increasing the power of the atomic reaction. And that is just if two neutrons create fission reactions each round. Perhaps you can now see why so many people consider atomic energy to be an exciting energy source! To make a self-sustained nuclear fission reactor with 235U, it is necessary to slow down the neutrons. Water is very effective at this, since neutrons collide with protons in water molecules and lose energy. Figure 22.28 shows a schematic of a react |
or design called the pressurized water reactor. 750 Chapter 22 • The Atom Figure 22.28 A pressurized water reactor is cleverly designed to control the fission of large amounts of 235U, while using the heat produced in the fission reaction to create steam for generating electrical energy. Control rods adjust neutron flux so that it is self-sustaining. In case the reactor overheats and boils the water away, the chain reaction terminates, because water is needed to slow down the neutrons. This inherent safety feature can be overwhelmed in extreme circumstances. Control rods containing nuclides that very strongly absorb neutrons are used to adjust neutron flux. To produce large amounts of power, reactors contain hundreds to thousands of critical masses, and the chain reaction easily becomes self-sustaining. Neutron flux must be carefully regulated to avoid an out-of-control exponential increase in the rate of fission. Control rods help prevent overheating, perhaps even a meltdown or explosive disassembly. The water that is used to slow down neutrons, necessary to get them to induce fission in 235U, and achieve criticality, provides a negative feedback for temperature increase. In case the reactor overheats and boils the water to steam or is breached, the absence of water kills the chain reaction. Considerable heat, however, can still be generated by the reactor’s radioactive fission products. Other safety features, thus, need to be incorporated in the event of a loss of coolant accident, including auxiliary cooling water and pumps. Energies in Nuclear Fission The following are two interesting facts to consider: • The average fission reaction produces 200 MeV of energy. • If you were to measure the mass of the products of a nuclear reaction, you would find that their mass was slightly less than the mass of the original nucleus. How are those things possible? Doesn’t the fission reaction’s production of energy violate the conservation of energy? Furthermore, doesn’t the loss in mass in the reaction violate the conservation of mass? Those are important questions, and they can both be answered with one of the most famous equations in scientific history. 22.65 Recall that, according to Einstein’s theory, energy and mass are essentially the same thing. In the case of fission, the mass of the products is less than that of the reactants because the missing mass appears in the form of the energy released in the reaction, with a constant value of c2 Joules of energy converted for each kilogram of material. The value of c2 is substantial—from Einstein’s equation, the amount of energy in just 1 gram of mass would be enough to support the average U.S. citizen for more than 270 years! The example below will show you how a mass-energy transformation of this type takes place. Access for free at openstax.org. 22.4 • Nuclear Fission and Fusion 751 WORKED EXAMPLE Calculating Energy from a Kilogram of Fissionable Fuel Calculate the amount of energy produced by the fission of 1.00 kg of , given the average fission reaction of 22.66 Strategy The total energy produced is the number of number of atoms in 1.00 kg. atoms times the given energy per fission. We should therefore find the Solution The number of g; thus, there are atoms in 1.00 kg is Avogadro’s number times the number of moles. One mole of has a mass of 235.04 . The number of atoms is therefore So the total energy released is Discussion The result is another impressively large amount of energy, equivalent to about 14,000 barrels of crude oil or 600,000 gallons of gasoline. But, it is only one fourth the energy produced by the fusion of a kilogram of a mixture of deuterium and tritium. Even though each fission reaction yields about ten times the energy of a fusion reaction, the energy per kilogram of fission fuel is less, because there are far fewer moles per kilogram of the heavy nuclides. Fission fuel is also much scarcer than fusion fuel, and less than 1 percent of uranium (the 235 U) is readily usable. 22.67 Virtual Physics Nuclear Fission Click to view content (https://www.openstax.org/l/16fission) Start a chain reaction, or introduce nonradioactive isotopes to prevent one. Use the applet to control energy production in a nuclear reactor! Nuclear Fusion Nuclear fusion is defined as the combining, or fusing, of two nuclei and, the combining of nuclei also results in an emission of energy. For many, the concept is counterintuitive. After all, if energy is released when a nucleus is split, how can it also be released when nucleons are combined together? The difference between fission and fusion, which results from the size of the nuclei involved, will be addressed next. Remember that the structure of a nucleus is based on the interplay of the compressive nuclear strong force and the repulsive electromagnetic force. For nuclei that are less massive than iron, the nuclear force is actually stronger than that of the Coulomb force. As a result, when a low-mass nucleus absorbs nucleons, the added neutrons and protons bind the nucleus more tightly. The increased nuclear strong force does work on the nucleus, and energy is released. Once the size of the created nucleus exceeds that of iron, the short-ranging nuclear force does not have the ability to bind a nucleus more tightly, and the emission of energy ceases. In fact, for fusion to occur for elements of greater mass than iron, energy must be added to the system! Figure 22.29 shows an energy-mass curve commonly used to describe nuclear reactions. Notice the location of iron (Fe) on the graph. All low-mass nuclei to the left of iron release energy through fusion, while all highmass particles to the right of iron produce energy through fission. 752 Chapter 22 • The Atom Figure 22.29 Fusion of light nuclei to form medium-mass nuclei converts mass to energy, because binding energy per nucleon ( ) is greater for the product nuclei. The larger is, the less mass per nucleon, and so mass is converted to energy and released in such fusion reactions. TIPS FOR SUCCESS Just as it is not possible for the elements to the left of iron in the figure to naturally fission, it is not possible for elements to the right of iron to naturally undergo fusion, as that process would require the addition of energy to occur. Furthermore, notice that elements commonly discussed in fission and fusion are elements that can provide the greatest change in binding energy, such as uranium and hydrogen. Iron’s location on the energy-mass curve is important, and explains a number of its characteristics, including its role as an elemental endpoint in fusion reactions in stars. The major obstruction to fusion is the Coulomb repulsion force between nuclei. Since the attractive nuclear force that can fuse nuclei together is short ranged, the repulsion of like positive charges must be overcome in order to get nuclei close enough to induce fusion. Figure 22.30 shows an approximate graph of the potential energy between two nuclei as a function of the distance between their centers. The graph resembles a hill with a well in its center. A ball rolled to the left must have enough kinetic energy to get over the hump before it falls into the deeper well with a net gain in energy. So it is with fusion. If the nuclei are given enough kinetic energy to overcome the electric potential energy due to repulsion, then they can combine, release energy, and fall into a deep well. One way to accomplish that end is to heat fusion fuel to high temperatures so that the kinetic energy of thermal motion is sufficient to get the nuclei together. Figure 22.30 Potential energy between two light nuclei graphed as a function of distance between them. If the nuclei have enough kinetic energy to get over the Coulomb repulsion hump, they combine, release energy, and drop into a deep attractive well. You might think that, in our Sun, nuclei are constantly coming into contact and fusing. However, this is only partially true. Only at the Sun’s core are the particles close enough and the temperature high enough for fusion to occur! In the series of reactions below, the Sun produces energy by fusing protons, or hydrogen nuclei ( , by far the Sun’s most Access for free at openstax.org. 22.4 • Nuclear Fission and Fusion 753 abundant nuclide) into helium nuclei cycle . The principal sequence of fusion reactions forms what is called the proton-proton 22.68 stands for a positron and where the first two reactions must occur twice for the third to be possible, so the cycle consumes six protons ( Furthermore, the two positrons produced will find two electrons and annihilate to form four more overall cycle is thus is an electron neutrino. The energy in parentheses is releasedby the reaction. Note that ) but gives back two. rays, for a total of six. The where the 26.7 MeV includes the annihilation energy of the positrons and electrons and is distributed among all the reaction products. The solar interior is dense, and the reactions occur deep in the Sun where temperatures are highest. It takes about 32,000 years for the energy to diffuse to the surface and radiate away. However, the neutrinos can carry their energy out of the Sun in less than two seconds, because they interact so weakly with other matter. Negative feedback in the Sun acts as a thermostat to regulate the overall energy output. For instance, if the interior of the Sun becomes hotter than normal, the reaction rate increases, producing energy that expands the interior. The expansion cools it and lowers the reaction rate. Conversely, if the interior becomes too cool, it contracts, increasing the temperature and therefore the reaction rate (see Figure 22.31). Stars like the Sun are stable for billions of years, until a significant fraction of their hydrogen has been depleted. Figure 22.31 Nuclear fusion in the Sun converts hydrogen nuclei into helium; fusion occurs primarily at the boundary of the helium core, where the temperature is highest and sufficie |
nt hydrogen remains. Energy released diffuses slowly to the surface, with the exception of neutrinos, which escape immediately. Energy production remains stable because of negative-feedback effects. Nuclear Weapons and Nuclear Power The world was in political turmoil when fission was discovered in 1938. Compounding the troubles, the possibility of a selfsustained chain reaction was immediately recognized by leading scientists the world over. The enormous energy known to be in nuclei, but considered inaccessible, now seemed to be available on a large scale. Within months after the announcement of the discovery of fission, Adolf Hitler banned the export of uranium from newly occupied Czechoslovakia. It seemed that the possible military value of uranium had been recognized in Nazi Germany, and that a serious effort to build a nuclear bomb had begun. Alarmed scientists, many of whom fled Nazi Germany, decided to take action. None was more famous or revered than Einstein. It was felt that his help was needed to get the American government to make a serious effort at constructing nuclear weapons as a matter of survival. Leo Szilard, a Hungarian physicist who had emigrated to America, took a draft of a letter to Einstein, who, although a pacifist, signed the final version. The letter was for President Franklin Roosevelt, warning of the German potential to build extremely powerful bombs of a new type. It was sent in August of 1939, just before the German invasion of Poland that marked the start of World War II. It was not until December 6, 1941, the day before the Japanese attack on Pearl Harbor, that the United States made a massive commitment to building a nuclear bomb. The top secret Manhattan Project was a crash program aimed at beating the Germans. 754 Chapter 22 • The Atom It was carried out in remote locations, such as Los Alamos, New Mexico, whenever possible, and eventually came to cost billions of dollars and employ the efforts of more than 100,000 people. J. Robert Oppenheimer (1904–1967), a talented physicist, was chosen to head the project. The first major step was made by Enrico Fermi and his group in December 1942, when they completed the first self-sustaining nuclear reactor. This first atomic pile, built in a squash court at the University of Chicago, proved that a fission chain reaction was possible. Plutonium was recognized as easier to fission with neutrons and, hence, a superior fission material very early in the Manhattan Project. Plutonium availability was uncertain, and so a uranium bomb was developed simultaneously. Figure 22.32 shows a guntype bomb, which takes two subcritical uranium masses and shoots them together. To get an appreciable yield, the critical mass must be held together by the explosive charges inside the cannon barrel for a few microseconds. Since the buildup of the uranium chain reaction is relatively slow, the device to bring the critical mass together can be relatively simple. Owing to the fact that the rate of spontaneous fission is low, a neutron source is at the center the assembled critical mass. Figure 22.32 A gun-type fission bomb for utilizes two subcritical masses forced together by explosive charges inside a cannon barrel. The energy yield depends on the amount of uranium and the time it can be held together before it disassembles itself. Plutonium’s special properties necessitated a more sophisticated critical mass assembly, shown schematically in Figure 22.33. A spherical mass of plutonium is surrounded by shaped charges (high explosives that focus their blast) that implode the plutonium, crushing it into a smaller volume to form a critical mass. The implosion technique is faster and more effective, because it compresses three-dimensionally rather than one-dimensionally as in the gun-type bomb. Again, a neutron source is included to initiate the chain reaction. Figure 22.33 An implosion created by high explosives compresses a sphere of 239Pu into a critical mass. The superior fissionability of plutonium has made it the preferred bomb material. Access for free at openstax.org. Owing to its complexity, the plutonium bomb needed to be tested before there could be any attempt to use it. On July 16, 1945, the test named Trinity was conducted in the isolated Alamogordo Desert in New Mexico, about 200 miles south of Los Alamos (see Figure 22.34). A new age had begun. The yield of the Trinity device was about 10 kilotons (kT), the equivalent of 5,000 of the largest conventional bombs. 22.4 • Nuclear Fission and Fusion 755 Figure 22.34 Trinity test (1945), the first nuclear bomb (credit: U.S. Department of Energy) Although Germany surrendered on May 7, 1945, Japan had been steadfastly refusing to surrender for many months, resulting large numbers of civilian and military casualties. Invasion plans by the Allies estimated a million casualties of their own and untold losses of Japanese lives. The bomb was viewed as a way to end the war. The first bomb used was a gun-type uranium bomb dropped on Hiroshima on August 6 by the United States. Its yield of about 15 kT destroyed the city and killed an estimated 80,000 people, with 100,000 more being seriously injured. The second bomb was an implosion-type plutonium bomb dropped on Nagasaki only three days later. Its 20-kT yield killed at least 50,000 people, something less than Hiroshima because of the hilly terrain and the fact that it was a few kilometers off target. The Japanese were told that one bomb a week would be dropped until they surrendered unconditionally, which they did on August 14. In actuality, the United States had only enough plutonium for one more bomb, as yet unassembled. Knowing that fusion produces several times more energy per kilogram of fuel than fission, some scientists pursued the idea of constructing a fusion bomb. The first such bomb was detonated by the United States several years after the first fission bombs, on October 31, 1952, at Eniwetok Atoll in the Pacific Ocean. It had a yield of 10 megatons (MT), about 670 times that of the fission bomb that destroyed Hiroshima. The Soviet Union followed with a fusion device of its own in August 1953, and a weapons race, beyond the aim of this text to discuss, continued until the end of the Cold War. Figure 22.35 shows a simple diagram of how a thermonuclear bomb is constructed. A fission bomb is exploded next to fusion fuel in the solid form of lithium deuteride. Before the shock wave blows it apart, create tritium through the reaction . Additional fusion and fission fuels are enclosed in a dense shell of rays heat and compress the fuel, and neutrons . At the same time that the uranium shell reflects the neutrons back into the fuel to enhance its fusion, the fast-moving neutrons cause the plentiful and inexpensive to fission, part of what allows thermonuclear bombs to be so large. 756 Chapter 22 • The Atom Figure 22.35 This schematic of a fusion bomb (H-bomb) gives some idea of how the fission trigger is used to ignite fusion fuel. Neutrons and γrays transmit energy to the fusion fuel, create tritium from deuterium, and heat and compress the fusion fuel. The outer shell of serves to reflect some neutrons back into the fuel, causing more fusion, and it boosts the energy output by fissioning itself when neutron energies become high enough. Of course, not all applications of nuclear physics are as destructive as the weapons described above. Hundreds of nuclear fission power plants around the world attest to the fact that controlled fission is both practical and economical. Given growing concerns over global warming, nuclear power is often seen as a viable alternative to energy derived from fossil fuels. BOUNDLESS PHYSICS Fusion Reactors For decades, fusion reactors have been deemed the energy of the future. A safer, cleaner, and more abundant potential source of energy than its fission counterpart, images of the fusion reactor have been conjured up each time the need for a renewable, environmentally friendly resource is discussed. Now, after more than half a century of speculating, some scientists believe that fusion reactors are nearly here. In creating energy by combining atomic nuclei, the fusion reaction holds many advantages over fission. First, fusion reactions are more efficient, releasing 3 to 4 times more energy than fission per gram of fuel. Furthermore, unlike fission reactions that require heavyelements like uranium that are difficult to obtain, fusion requires lightelements that are abundant in nature. The greatest advantage of the fusion reaction, however, is in its ability to be controlled. While traditional nuclear reactors create worries about meltdowns and radioactive waste, neither is a substantial concern with the fusion reaction. Consider that fusion reactions require a large amount of energy to overcome the repulsive Coulomb force and that the byproducts of a fusion reaction are largely limited to helium nuclei. In order for fusion to occur, hydrogen isotopes of deuterium and tritium must be acquired. While deuterium can easily be gathered from ocean water, tritium is slightly more difficult to come by, though it can be manufactured from Earth’s abundant lithium. Once acquired, the hydrogen isotopes are injected into an empty vessel and subjected to temperature and pressure great enough to mimic the conditions at the core of our Sun. Using carefully controlled high-frequency radio waves, the hydrogen isotopes are broken into plasma and further controlled through an electromagnetic field. As the electromagnetic field continues to exert pressure on the hydrogen plasma, enough energy is supplied to cause the hydrogen plasma to fuse into helium. Access for free at openstax.org. 22.5 • Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 757 Figure 22.36 Tokamak confinement of nuclear fusion plasma. The magnetic field lines are used to confine the high-temperature plasma (purple). Research is currently being done to increas |
e the efficiency of the tokamak confinement model. Once the plasma fuses, high-velocity neutrons are ejected from the newly formed helium atoms. Those high velocity neutrons, carrying the excess energy stored within bonds of the original hydrogen, are able to travel unaffected by the applied magnetic field. In doing so, they strike a barrier around the nuclear reactor, transforming their excess energy to heat. The heat is then harvested to make steam that drives turbines. Hydrogen’s tremendous power is now usable! The historical concern with nuclear fusion reactors is that the energy required to control the electromagnetic field is greater than the energy harvested from the hydrogen atoms. However, recent research by both Lockheed Martin engineers and scientists at the Lawrence Livermore National Laboratory has yielded exciting theoretical improvements in efficiency. At the time of this writing, a test facility called ITER (International Thermonuclear Experimental Reactor) is being constructed in southern France. A joint venture of the European Union, the United States, Japan, Russia, China, South Korea, and India, ITER is designed for further study into the future of nuclear fusion energy production. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation Section Learning Objectives By the end of this section, you will be able to do the following: • Describe how nuclear imaging works (e.g., radioisotope imaging, PET) • Describe the ionizing effects of radiation and how they can be used for medical treatment Section Key Terms Anger camera rad radiopharmaceutical therapeutic ratio relative biological effectiveness (RBE) roentgen equivalent man (rem) tagged Medical Applications of Nuclear Physics Applications of nuclear physics have become an integral part of modern life. From the bone scan that detects one cancer to the radioiodine treatment that cures another, nuclear radiation has diagnostic and therapeutic effects on medicine. Medical Imaging A host of medical imaging techniques employ nuclear radiation. What makes nuclear radiation so useful? First, radiation can 758 Chapter 22 • The Atom easily penetrate tissue; hence, it is a useful probe to monitor conditions inside the body. Second, nuclear radiation depends on the nuclide and not on the chemical compound it is in, so that a radioactive nuclide can be put into a compound designed for specific purposes. When that is done, the compound is said to be tagged. A tagged compound used for medical purposes is called a radiopharmaceutical. Radiation detectors external to the body can determine the location and concentration of a radiopharmaceutical to yield medically useful information. For example, certain drugs are concentrated in inflamed regions of the body, and their locations can aid diagnosis and treatment as seen in Figure 22.37. Another application utilizes a radiopharmaceutical that the body sends to bone cells, particularly those that are most active, to detect cancerous tumors or healing points. Images can then be produced of such bone scans. Clever use of radioisotopes determines the functioning of body organs, such as blood flow, heart muscle activity, and iodine uptake in the thyroid gland. For instance, a radioactive form of iodine can be used to monitor the thyroid, a radioactive thallium salt can be used to follow the blood stream, and radioactive gallium can be used for cancer imaging. Figure 22.37 A radiopharmaceutical was used to produce this brain image of a patient with Alzheimer’s disease. Certain features are computer enhanced. (credit: National Institutes of Health) Once a radioactive compound has been ingested, a device like that shown in Figure 22.38 is used to monitor nuclear activity. The device, called an Anger camera or gamma camera uses a piece of lead with holes bored through it. The gamma rays are redirected through the collimator to narrow their beam, and are then interpreted using a device called a scintillator. The computer analysis of detector signals produces an image. One of the disadvantages of this detection method is that there is no depth information (i.e., it provides a two-dimensional view of the tumor as opposed to a three-dimensional view), because radiation from any location under that detector produces a signal. Figure 22.38 An Anger or gamma camera consists of a lead collimator and an array of detectors. Gamma rays produce light flashes in the scintillators. The light output is converted to an electrical signal by the photomultipliers. A computer constructs an image from the detector output. Single-photon-emission computer tomography (SPECT) used in conjunction with a CT scanner improves on the process carried out by the gamma camera. Figure 22.39 shows a patient in a circular array of SPECT detectors that may be stationary or rotated, with detector output used by a computer to construct a detailed image. The spatial resolution of this technique is poor, but the three-dimensional image created results in a marked improvement in contrast. Access for free at openstax.org. 22.5 • Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 759 Figure 22.39 SPECT uses a rotating camera to form an image of the concentration of a radiopharmaceutical compound. (credit: Woldo, Wikimedia Commons) Positron emission tomography (or PET) scans utilize images produced by encounters an electron, mutual annihilation occurs, producing two γrays. Those energy comes from the destruction of an electron or positron mass) and they move directly away from each other, allowing detectors to determine their point of origin accurately (as shown in Figure 22.40). It requires detectors on opposite sides to simultaneously (i.e., at the same time) detect photons of 0.511 MeV energy and utilizes computer imaging techniques similar to those in SPECT and CT scans. PET is used extensively for diagnosing brain disorders. It can note decreased metabolism in certain regions that accompany Alzheimer’s disease. PET can also locate regions in the brain that become active when a person carries out specific activities, such as speaking, closing his or her eyes, and so on. rays have identical 0.511 MeV energies (the emitters. When the emitted positron β+ Figure 22.40 A PET system takes advantage of the two identical -ray photons produced by positron-electron annihilation. The rays are emitted in opposite directions, so that the line along which each pair is emitted is determined. Various events detected by several pairs of detectors are then analyzed by the computer to form an accurate image. Ionizing Radiation on the Body We hear many seemingly contradictory things about the biological effects of ionizing radiation. It can cause cancer, burns, and hair loss, and yet it is used to treat and even cure cancer. How do we understand such effects? Once again, there is an underlying simplicity in nature, even in complicated biological organisms. All the effects of ionizing radiation on biological tissue can be understood by knowing that ionizing radiation affects molecules within cells, particularly DNA molecules. Let us take a brief look at molecules within cells and how cells operate. Cells have long, double-helical DNA molecules containing chemical patterns called genetic codes that govern the function and processes undertaken by the cells. Damage to DNA consists of breaks in chemical bonds or other changes in the structural features of the DNA chain, leading to changes in the genetic code. In human cells, we can have as many as a million individual instances of damage to DNA per cell per day. The repair ability of DNA is vital for maintaining the integrity of the genetic code and for the normal functioning of the entire organism. A cell with a damaged ability to repair DNA, which could have been induced by ionizing radiation, can do one of the following: • The cell can go into an irreversible state of dormancy, known as senescence. • The cell can commit suicide, known as programmed cell death. • The cell can go into unregulated cell division, leading to tumors and cancers. 760 Chapter 22 • The Atom Since ionizing radiation damages the DNA, ionizing radiation has its greatest effect on cells that rapidly reproduce, including most types of cancer. Thus, cancer cells are more sensitive to radiation than normal cells and can be killed by it easily. Cancer is characterized by a malfunction of cell reproduction, and can also be caused by ionizing radiation. There is no contradiction to say that ionizing radiation can be both a cure and a cause. Radiotherapy Radiotherapy is effective against cancer because cancer cells reproduce rapidly and, consequently, are more sensitive to radiation. The central problem in radiotherapy is to make the dose for cancer cells as high as possible while limiting the dose for normal cells. The ratio of abnormal cells killed to normal cells killed is called the therapeutic ratio, and all radiotherapy techniques are designed to enhance that ratio. Radiation can be concentrated in cancerous tissue by a number of techniques. One of the most prevalent techniques for well-defined tumors is a geometric technique shown in Figure 22.41. A narrow beam of radiation is passed through the patient from a variety of directions with a common crossing point in the tumor. The technique concentrates the dose in the tumor while spreading it out over a large volume of normal tissue. Figure 22.41 The 60Co source of -radiation is rotated around the patient so that the common crossing point is in the tumor, concentrating the dose there. This geometric technique works for well-defined tumors. Another use of radiation therapy is through radiopharmaceuticals. Cleverly, radiopharmaceuticals are used in cancer therapy by tagging antibodies with radioisotopes. Those antibodies are extracted from the patient, cultured, loaded with a radioisotope, and then returned to the patient. The antibodies are then concentrated almost entirel |
y in the tissue they developed to fight, thus localizing the radiation in abnormal tissue. This method is used with radioactive iodine to fight thyroid cancer. While the therapeutic ratio can be quite high for such short-range radiation, there can be a significant dose for organs that eliminate radiopharmaceuticals from the body, such as the liver, kidneys, and bladder. As with most radiotherapy, the technique is limited by the tolerable amount of damage to the normal tissue. Radiation Dosage To quantitatively discuss the biological effects of ionizing radiation, we need a radiation dose unit that is directly related to those effects. To do define such a unit, it is important to consider both the biological organism and the radiation itself. Knowing that the amount of ionization is proportional to the amount of deposited energy, we define a radiation dose unit called the rad. It 1/100 of a joule of ionizing energy deposited per kilogram of tissue, which is For example, if a 50.0-kg person is exposed to ionizing radiation over her entire body and she absorbs 1.00 J, then her wholebody radiation dose is 22.69 If the same 1.00 J of ionizing energy were absorbed in her 2.00-kg forearm alone, then the dose to the forearm would be 22.70 22.71 and the unaffected tissue would have a zero rad dose. When calculating radiation doses, you divide the energy absorbed by the mass of affected tissue. You must specify the affected region, such as the whole body or forearm in addition to giving the numerical dose in rads. Although the energy per kilogram in 1 rad is small, it can still have significant effects. Since only a few eV cause ionization, just 0.01 J of ionizing energy can create a huge number of ion pairs and have an effect at the cellular level. The effects of ionizing radiation may be directly proportional to the dose in rads, but they also depend on the type of radiation and the type of tissue. That is, for a given dose in rads, the effects depend on whether the radiation is , X-ray, or some , , Access for free at openstax.org. 22.5 • Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 761 other type of ionizing radiation. The relative biological effectiveness (RBE) relates to the amount of biological damage that can occur from a given type of radiation and is given in Table 22.4 for several types of ionizing radiation. Type and energy of radiation RBE X-rays rays rays greater than 32 keV rays less than 32 keV 1 1 1 1.7 Neutrons, thermal to slow (< 20 keV) 2–5 Neutrons, fast (1–10 MeV) 10 (body), 32 (eyes) Protons (1–10 MeV) 10 (body), 32 (eyes) rays from radioactive decay 10–20 Heavy ions from accelerators 10–20 Table 22.4 Relative Biological Effectiveness TIPS FOR SUCCESS The RBEs given in Table 22.4 are approximate, but they yield certain valuable insights. • The eyes are more sensitive to radiation, because the cells of the lens do not repair themselves. • Though both are neutral and have large ranges, neutrons cause more damage than rays because neutrons often cause secondary radiation when they are captured. • Short-range particles such as rays have a severely damaging effect to internal anatomy, as their damage is concentrated and more difficult for the biological organism to repair. However, the skin can usually block alpha particles from entering the body. Can you think of any other insights from the table? A final dose unit more closely related to the effect of radiation on biological tissue is called the roentgen equivalent man, or rem. A combination of all factors mentioned previously, the roentgen equivalent man is defined to be the dose in rads multiplied by the relative biological effectiveness. The large-scale effects of radiation on humans can be divided into two categories: immediate effects and long-term effects. Table 22.5 gives the immediate effects of whole-body exposures received in less than one day. If the radiation exposure is spread out over more time, greater doses are needed to cause the effects listed. Any dose less than 10 rem is called a low dose, a dose 10 to 100 rem is called a moderate dose, and anything greater than 100 rem is called a high dose. 22.72 Dose (rem) Effect 0–10 No observable effect 10–100 Slight to moderate decrease in white blood cell counts Table 22.5 Immediate Effects of Radiation (Adults, Whole Body, Single Exposure) 762 Chapter 22 • The Atom Dose (rem) Effect 50 Temporary sterility 100–200 Significant reduction in blood cell counts, brief nausea, and vomiting; rarely fatal 200–500 Nausea, vomiting, hair loss, severe blood damage, hemorrhage, fatalities 450 LD50/32; lethal to 50% of the population within 32 days after exposure if untreated 500–2,000 Worst effects due to malfunction of small intestine and blood systems; limited survival > 2,000 Fatal within hours due to collapse of central nervous system Table 22.5 Immediate Effects of Radiation (Adults, Whole Body, Single Exposure) WORK IN PHYSICS Health Physicist Are you interested in learning more about radiation? Are you curious about studying radiation dosage levels and ensuring the safety of the environment and people that are most closely affected by it? If so, you may be interested in becoming a health physicist. The field of health physics draws from a variety of science disciplines with the central aim of mitigating radiation concerns. Those that work as health physicists have a diverse array of potential jobs available to them, including those in research, industry, education, environmental protection, and governmental regulation. Furthermore, while the term health physicistmay lead many to think of the medical field, there are plenty of applications within the military, industrial, and energy fields as well. As a researcher, a health physicist can further environmental studies on the effects of radiation, design instruments for more accurate measurements, and assist in establishing valuable radiation standards. Within the energy field, a health physicsist often acts as a manager, closely tied to all operations at all levels, from procuring appropirate equipment to monitoring health data. Within industry, the health physicist acts as a consultant, assisting industry management in important decisions, designing facilities, and choosing appropriate detection tools. The health physicist possesses a unique knowledge base that allows him or her to operate in a wide variety of interesting disciplines! To become a health physicist, it is necessary to have a background in the physical sciences. Understanding the fields of biology, physiology, biochemistry, and genetics are all important as well. The ability to analyze and solve new problems is critical, and a natural aptitude for science and mathematics will assist in the continued necessary training. There are two possible certifications for health physicists: from the American Board of Health Physicists (ABHP) and the National Registry of Radiation Protection Technologists (NRRPT). Access for free at openstax.org. Chapter 22 • Key Terms 763 KEY TERMS activity rate of decay for radioactive nuclides alpha decay type of radioactive decay in which an atomic nucleus emits an alpha particle anger camera common medical imaging device that uses a scintillator connected to a series of photomultipliers atomic number number of protons in a nucleus becquerel SI unit for rate of decay of a radioactive material beta decay type of radioactive decay in which an atomic nucleus emits a beta particle nuclear fusion reaction in which two nuclei are combined, or fused, to form a larger nucleus nucleons particles found inside nuclei planetary model of the atom model of the atom that shows electrons orbiting like planets about a Sun-like nucleus proton-proton cycle combined reactions and that begins with carbon-14 dating radioactive dating technique based on the hydrogen and ends with helium radioactivity of carbon-14 rad amount of ionizing energy deposited per kilogram of chain reaction self-sustaining sequence of events, tissue exemplified by the self-sustaining nature of a fission reaction at critical mass radioactive substance or object that emits nuclear radiation critical mass minimum amount necessary for self- radioactive dating application of radioactive decay in sustained fission of a given nuclide decay constant quantity that is inversely proportional to the half-life and that is used in the equation for number of nuclei as a function of time energy-level diagram a diagram used to analyze the energy levels of electrons in the orbits of an atom excited state any state beyond the n= 1 orbital in which the electron stores energy Fraunhofer lines black lines shown on an absorption spectrum that show the wavelengths absorbed by a gas gamma decay type of radioactive decay in which an atomic nucleus emits a gamma ray which the age of a material is determined by the amount of radioactivity of a particular type that occurs radioactive decay process by which an atomic nucleus of an unstable atom loses mass and energy by emitting ionizing particles radioactivity emission of rays from the nuclei of atoms radiopharmaceutical compound used for medical imaging relative biological effectiveness (RBE) number that expresses the relative amount of damage that a fixed amount of ionizing radiation of a given type can inflict on biological tissues roentgen equivalent man (rem) dose unit more closely Geiger tube very common radiation detector that usually related to effects in biological tissue gives an audio output ground state the n=1 orbital of an electron half-life time in which there is a 50 percent chance that a nucleus will decay Heisenberg uncertainty principle fundamental limit to the precision with which pairs of quantities such as momentum and position can be measured hydrogen-like atom any atom with only a single electron isotope nuclei having the same Zand different N’s liquid drop model model of the atomic nucleus (useful only to understand some of its featu |
res) in which nucleons in a nucleus act like atoms in a drop mass number number of nucleons in a nucleus nuclear fission reaction in which a nucleus splits SECTION SUMMARY 22.1 The Structure of the Atom • Rutherford’s gold foil experiment provided evidence that the atom is composed of a small, dense nucleus with electrons occupying the mostly empty space around it. • Analysis of emission spectra shows that energy is emitted from energized gas in discrete quantities. Rutherford scattering scattering of alpha particles by gold nuclei in the gold foil experiment Rydberg constant a physical constant related to atomic spectra, with an established value of scintillator radiation detection method that records light produced when radiation interacts with materials strong nuclear force attractive force that holds nucleons together within the nucleus tagged having a radioactive substance attached (to a chemical compound) therapeutic ratio the ratio of abnormal cells killed to normal cells killed transmutation process of changing elemental composition • The Bohr model of the atom describes electrons existing in discrete orbits, with discrete energies emitted and absorbed as the electrons decrease and increase in orbital energy. • The energy emitted or absorbed by an electron as it changes energy state can be determined with the equation , where 764 Chapter 22 • Key Equations . • The wavelength of energy absorbed or emitted by an electron as it changes energy state can be determined by the equation , where . • Described as an electron cloud, the quantum model of the atom is the result of de Broglie waves and Heisenberg’s uncertainty principle. 22.2 Nuclear Forces and Radioactivity • The structure of the nucleus is defined by its two nucleons, the neutron and proton. • Atomic numbers and mass numbers are used to differentiate between various atoms and isotopes. Those numbers can be combined into an easily recognizable form called a nuclide. • The size and stability of the nucleus is based upon two forces: the electromagnetic force and strong nuclear force. • Radioactive decay is the alteration of the nucleus through the emission of particles or energy. • Alpha decay occurs when too many protons exist in the nucleus. It results in the ejection of an alpha particle, as described in the equation . • Beta decay occurs when too many neutrons (or protons) exist in the nucleus. It results in the transmutation of a neutron into a proton, electron, and neutrino. The decay is expressed through the equation . (Beta decay may also transform a proton into a neutron.) • Gamma decay occurs when a nucleus in an excited state move to a more stable state, resulting in the release of a photon. Gamma decay is represented with the equation . • The penetration distance of radiation depends on its energy, charge, and type of material it encounters. 22.3 Half Life and Radiometric Dating • Radioactive half-life is the time it takes a sample of nuclei to decay to half of its original amount. • The rate of radioactive decay is defined as the sample’s KEY EQUATIONS 22.1 The Structure of the Atom energy of hydrogen electron in an orbital Access for free at openstax.org. activity, represented by the equation . • Knowing the half-life of a radioactive isotope allows for the process of radioactive dating to determine the age of a material. If the half-life of a material is known, the age of the material can be found using the equation . • • The age of organic material can be determined using the decay of the carbon-14 isotope, while the age of rocks can be determined using the decay of uranium-238. 22.4 Nuclear Fission and Fusion • Nuclear fission is the splitting of an atomic bond, releasing a large amount of potential energy previously holding the atom together. The amount of energy released can be determined through the equation . • Nuclear fusion is the combining, or fusing together, of two nuclei. Energy is also released in nuclear fusion as the combined nuclei are closer together, resulting in a decreased strong nuclear force. • Fission was used in two nuclear weapons at the conclusion of World War II: the gun-type uranium bomb and the implosion-type plutonium bomb. • While fission has been used in both nuclear weapons and nuclear reactors, fusion is capable of releasing more energy per reaction. As a result, fusion is a wellresearched, if not yet well-controlled, energy source. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation • Medical imaging occurs when a radiopharmaceutical placed in the body provides information to an array of radiation detectors outside the body. • Devices utilizing medical imaging include the Anger • camera, SPECT detector, and PET scan. Ionizing radiation can both cure and cause cancer through the manipulation of DNA molecules. • Radiation dosage and its effect on the body can be measured using the quantities radiation dose unit (rad), relative biological effectiveness (RBE), and the roentgen equivalent man (rem). energy of any hydrogen-like electron in orbital wavelength of light emitted by an electron changing states wavelength of an orbital heisenberg’s uncertainty principle 22.2 Nuclear Forces and Radioactivity alpha decay equation beta decay equation gamma decay equation CHAPTER REVIEW Concept Items 22.1 The Structure of the Atom 1. A star emits light from its core. One observer views the emission unobstructed while a second observer views the emission while obstructed by a cloud of hydrogen gas. Describe the difference between their observations. a. b. c. Frequencies will be absorbed from the spectrum. d. Frequencies will be added to the spectrum. Intensity of the light in the spectrum will increase. Intensity of the light in the spectrum will decrease. 2. How does the orbital energy of a hydrogen-like atom change as it increases in atomic number? Critical Thinking Items 22.1 The Structure of the Atom 4. How would the gold foil experiment have changed if electrons were used in place of alpha particles, assuming that the electrons hit the gold foil with the same force as the alpha particles? a. Being less massive, the electrons might have been scattered to a greater degree than the alpha particles. b. Being less massive, the electrons might have been scattered to a lesser degree than the alpha particles. Chapter 22 • Chapter Review 765 22.3 Half Life and Radiometric Dating radioactive half-life 22.4 Nuclear Fission and Fusion energy–mass conversion protonproton chain 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation roentgen equivalent man a. The orbital energy will increase. b. The orbital energy will decrease. c. The orbital energy will remain constant. d. The orbital energy will be halved. 22.4 Nuclear Fission and Fusion 3. Aside from energy yield, why are nuclear fusion reactors more desirable than nuclear fission reactors? a. Nuclear fusion reactors have a low installation cost. b. Radioactive waste is greater for a fusion reactor. c. Nuclear fusion reactors are easy to design and build. d. A fusion reactor produces less radioactive waste. c. Being more massive, the electrons would have been scattered to a greater degree than the alpha particles. d. Being more massive, the electrons would have been scattered to a lesser degree than the alpha particles. 5. Why does the emission spectrum of an isolated gas differ from the emission spectrum created by a white light? a. White light and an emission spectrum are different varieties of continuous distribution of frequencies. b. White light and an emission spectrum are different series of discrete frequencies. 766 Chapter 22 • Chapter Review c. White light is a continuous distribution of frequencies, and an emission spectrum is a series of discrete frequencies. d. White light is a series of discrete frequencies, and an emission spectrum is a continuous distribution of frequencies. 6. Why would it most likely be difficult to observe quantized orbital states for satellites orbiting the earth? a. On a macroscopic level, the orbital states do exist for satellites orbiting Earth but are too closely spaced for us to see. d. While the alpha particle has a greater charge than a beta particle, the electron density in lead is much higher than that in air. 10. What influence does the strong nuclear force have on the electrons in an atom? a. b. c. The strong force makes electrons revolve around It attracts them toward the nucleus. It repels them away from the nucleus. the nucleus. It does not have any influence. d. b. On a macroscopic level, the orbital states do not 22.3 Half Life and Radiometric Dating exist for satellites orbiting Earth. c. On a macroscopic level, we cannot control the amount of energy that we give to an artificial satellite and thus control its orbital altitude. d. On a macroscopic level, we cannot control the amount of energy that we give to an artificial satellite but we can control its orbital altitude. 7. Do standing waves explain why electron orbitals are quantized? a. no b. yes 8. Some terms referring to the observation of light include emission spectrumand absorption spectrum. Based on these definitions, what would a reflection spectrum describe? a. The reflection spectrum would describe when incident waves are selectively reflected by a substance. b. The reflection spectrum would describe when incident waves are completely reflected by a substance. c. The reflection spectrum would describe when incident waves are not absorbed by a substance. d. The reflection spectrum would describe when incident waves are completely absorbed by a substance. 22.2 Nuclear Forces and Radioactivity 9. Explain why an alpha particle can have a greater range in air than a beta particle in lead. a. While the alpha particle has a lesser charge than a beta particle, the electron density in lead is much less than that in air. b. While the alpha particle has a greater charge than a beta particle, the electron density in lead is |
much lower than that in air. c. While the alpha particle has a lesser charge than a beta particle, the electron density in lead is much greater than that in air. Access for free at openstax.org. 11. Provide an example of something that decreases in a manner similar to radioactive decay. a. The potential energy of an object falling under the influence of gravity b. The kinetic energy of a ball that is dropped from a building to the ground c. Theh charge transfer from an ebonite rod to fur d. The heat transfer from a hot to a cold object 12. A sample of radioactive material has a decay constant of 0.05 s–1. Why is it wrong to presume that the sample will take just 20 seconds to fully decay? a. The decay constant varies with the mass of the sample. b. The decay constant results vary with the amount of the sample. c. The decay constant represents a percentage of the sample that cannot decay. d. The decay constant represents only the fraction of a sample that decays in a unit of time, not the decay of the entire sample. 22.4 Nuclear Fission and Fusion 13. What is the atomic number of the most strongly bound nuclide? a. b. c. d. 14. Why are large electromagnets necessary in nuclear fusion reactors? a. Electromagnets are used to slow down the movement of charge hydrogen plasma. b. Electromagnets are used to decrease the temperature of hydrogen plasma. c. Electromagnets are used to confine the hydrogen plasma. d. Electromagnets are used to stabilize the temperature of the hydrogen plasma. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 15. Why are different radiopharmaceuticals used to image different parts of the body? a. The different radiopharmaceuticals travel through different blood vessels. b. The different radiopharmaceuticals travel to different parts of the body. c. The different radiopharmaceuticals are used to treat different diseases of the body. d. The different radiopharmaceuticals produce different amounts of ionizing radiation. 16. Why do people think carefully about whether to receive a diagnostic test such as a CT scan? a. The radiation from a CT scan is capable of creating cancerous cells. Performance Task 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 18. On the Environmental Protection Agency’s website, a helpful tool exists to allow you to determine your average annual radiation dose. Use the tool to determine whether the radiation level you have been exposed to is dangerous and to compare your radiation dosage to other radiative events. 1. Visit the webpage (http://www.openstax.org/l/ 28calculate) and answer the series of questions provided to determine the average annual radiation dosage that you receive. 2. Table 22.5 shows the immediate effects of a radiation dosage. Using the table, explain what you would experience if your yearly dosage of radiation was received all over the course of one day. Also, determine whether your dosage is considered a low, moderate, or high. 3. Using the information input into the webpage, what percentage of your dosage comes from TEST PREP Multiple Choice 22.1 The Structure of the Atom 19. If electrons are negatively charged and the nucleus is positively charged, why do they not attract and collide with each other? Chapter 22 • Test Prep 767 b. The radiation from a CT scan is capable of destroying cancerous cells. c. The radiation from a CT scan is capable of creating diabetic cells. d. The radiation from a CT scan is capable of destroying diabetic cells. 17. Sometimes it is necessary to take a PET scan very soon after ingesting a radiopharmaceutical. Why is that the case? a. The radiopharmaceutical may have a short half-life. b. The radiopharmaceutical may have a long half-life. c. The radiopharmaceutical quickly passes through the digestive system. d. The radiopharmaceutical can become lodged in the digestive system. natural sources? The average percentage of radiation from natural sources for an individual is around 85 percent. 4. Research radiation dosages for evacuees from events like the Chernobyl and Fukushima meltdowns. How does your annual radiation exposure rate compare to the net dosage for evacuees of each event. Use numbers to support your answer. 5. The U.S. Department of Labor limits the amount of radiation that a given worker may receive in a 12 month period. a. Research the present maximum value and compare your annual exposure rate to that of a radiation worker. Use numbers to support your answer. b. What types of work are likely to cause an increase in the radiation exposure of a particular worker? Provide one question based upon the information gathered on the EPA website. a. The pull from the nucleus provides a centrifugal force, which is not strong enough to draw the electrons into the nucleus. b. The pull from the nucleus provides a centripetal force, which is not strong enough to draw the electrons into the nucleus. 768 Chapter 22 • Test Prep c. The pull from the nucleus provides a helical motion. d. The pull from the nucleus provides a cycloid b. Coulomb force between protons only c. Strong nuclear force between all nucleons and motion. 22.4 Nuclear Fission and Fusion 20. If a nucleus elongates due to a neutron strike, which of the following forces will decrease? a. Nuclear force between neutrons only Coulomb force between protons, but the strong force will decrease more d. Strong nuclear force between neutrons and Coulomb force between protons, but Coulomb force will decrease more Short Answer 22.1 The Structure of the Atom a. The radioactive activity decreases exponentially. b. The radioactive activity undergoes linear decay. c. The radioactive activity undergoes logarithmic 21. Why do Bohr’s calculations for electron energies not decay. work for all atoms? a. In atoms with more than one electron is an atomic shell, the electrons will interact. That requires a more complex formula than Bohr’s calculations accounted for. In atoms with 10 or more electorns in an atomic shell, the electrons will interact. That requires a more complex formula than Bohr’s calculations accounted for. In atoms with more than one electron in an atomic shell, the electrons will not interact. That requires a more complex formula than Bohr’s calculations accounted for. In atoms with 10 or more electrons in an atomic shell, the electrons will not interact. That requires a more complex formula than Bohr’s calculations accounted for. b. c. d. 22.2 Nuclear Forces and Radioactivity 22. Does transmutation occur within chemical reactions? a. no b. yes 22.3 Half Life and Radiometric Dating 23. How does the radioactive activity of a sample change with time? Extended Response 22.1 The Structure of the Atom 26. Compare the standing wavelength of an orbital to the standing wavelength of an a. The standing wavelength of an orbital. orbital is greater than the standing wavelength of an orbital. b. The standing wavelength of an than the standing wavelength of an orbital is less orbital. Access for free at openstax.org. d. The radioactive activity will not change with time. 22.4 Nuclear Fission and Fusion 24. Why does fission of heavy nuclei result in the release of neutrons? a. Heavy nuclei require more neutrons to achieve stability. b. Heavy nuclei require more neutrons to balance charge. c. Light nuclei require more neutrons to achieve stability. d. Light nuclei require more neutrons to balance charge. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 25. Why is radioactive iodine used to monitor the thyroid? a. Radioactive iodine can be used by the thyroid while absorbing information about the thyroid. b. Radioactive iodine can be used by the thyroid while emitting information about the thyroid. c. Radioactive iodine can be secreted by the thyroid while absorbing information about the thyroid. d. Radioactive iodine can be secreted by the thyroid while emitting information about the thyroid. c. There is no relation between the standing wavelength of an wavelength of an orbital and the standing orbital. d. The standing wavelength of an orbital is the same as the standing wavelength of an orbital. 27. Describe the shape of the electron cloud, based on total energy levels, for an atom with electrons in multiple orbital states. a. There are multiple regions of high electron Chapter 22 • Test Prep 769 probability of various shapes surrounding the nucleus. b. There is a single solid spherical region of high electron probability surrounding the nucleus. brought together and then rapidly increase once a minimum is reached. b. The potential energy will decrease as the nuclei are brought together. c. There are multiple concentric shells of high c. The potential energy will increase as the nuclei are electron probability surrounding the nucleus. brought together. d. There is a single spherical shell of high electron probability surrounding the nucleus. 22.2 Nuclear Forces and Radioactivity 28. How did Becquerel’s observations of pitchblende imply the existence of radioactivity? a. A chemical reaction occurred on the photographic plate without any external source of energy. b. Bright spots appeared on the photographic plate due to an external source of energy. c. Energy from the Sun was absorbed by the pitchblende and reflected onto the photographic plate. d. Dark spots appeared on the photographic plate due to an external source of energy. 22.4 Nuclear Fission and Fusion 29. Describe the potential energy of two nuclei as they approach each other. a. The potential energy will decrease as the nuclei are d. The potential energy will increase as the nuclei are brought together and then rapidly decrease once a maximum is reached. 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation 30. Why do X-rays and gamma rays have equivalent RBE values if they provide different amounts of energy to the body? a. The penetration distance, which depends on energy, is short for both X-rays and gamm |
a rays. b. The penetration distance, which depends on energy, is long for both X-rays and gamma rays. c. The penetration distance, as determined by their high mass, is different for both X-rays and gamma rays. d. The penetration distance, as determined by their low mass, is the same for both X-rays and gamma rays. 770 Chapter 22 • Test Prep Access for free at openstax.org. CHAPTER 23 Particle Physics Figure 23.1 Part of the Large Hadron Collider (LHC) at CERN, on the border of Switzerland and France. The LHC is a particle accelerator, designed to study fundamental particles. (credit: Image Editor, Flickr) Chapter Outline 23.1 The Four Fundamental Forces 23.2 Quarks 23.3 The Unification of Forces Following ideas remarkably similar to those of the ancient Greeks, we continue to look for smaller and INTRODUCTION smaller structures in nature, hoping ultimately to find and understand the most fundamental building blocks that exist. Atomic physics deals with the smallest units of elements and compounds. In its study, we have found a relatively small number of atoms with systematic properties, and these properties have explained a tremendous range of phenomena. Nuclear physics is concerned with the nuclei of atoms and their substructures. Here, a smaller number of components—the proton and neutron—make up all nuclei. Exploring the systematic behavior of their interactions has revealed even more about matter, forces, and energy. Particle physics deals with the substructures of atoms and nuclei and is particularly aimed at finding those truly fundamental particles that have no further substructure. Just as in atomic and nuclear physics, we have found a complex array of particles and properties with systematic characteristics analogous to the periodic table and the chart of nuclides. An underlying structure is apparent, and there is some reason to think that we arefinding particles that have no substructure. Of course, we have been in similar situations before. For example, atoms were once thought to be the ultimate substructures. It is possible that we could continue to find deeper and deeper structures without ever discovering the ultimate substructure—in science there is never complete certainty. See Figure 23.2. The properties of matter are based on substructures called molecules and atoms. Each atom has the substructure of a nucleus surrounded by electrons, and their interactions explain atomic properties. Protons and neutrons—and the interactions between them—explain the stability and abundance of elements and form the substructure of nuclei. Protons and neutrons are not fundamental—they are composed of quarks. Like electrons and a few other particles, quarks may be the fundamental building blocks of all matter, lacking any further substructure. But the story is not complete because quarks and electrons may have substructures smaller than details that are presently observable. 772 Chapter 23 • Particle Physics Figure 23.2 A solid, a molecule, an atom, a nucleus, a nucleon (a particle that makes up the nucleus—either a proton or a neutron), and a quark. This chapter covers the basics of particle physics as we know it today. An amazing convergence of topics is evolving in particle physics. We find that some particles are intimately related to forces and that nature on the smallest scale may have its greatest influence on the large scale character of the universe. It is an adventure exceeding the best science fiction because it is not only fantastic but also real. 23.1 The Four Fundamental Forces Section Learning Objectives By the end of the section, you will be able to do the following: • Define, describe, and differentiate the four fundamental forces • Describe the carrier particles and explain how their exchange transmits force • Explain how particle accelerators work to gather evidence about particle physics Section Key Terms carrier particle colliding beam cyclotron Feynman diagram graviton particle physics pion quantum electrodynamics synchrotron boson boson weak nuclear force boson Despite the apparent complexity within the universe, there remain just four basic forces. These forces are responsible for all interactions known to science: from the very small to the very large to those that we experience in our day-to-day lives. These forces describe the movement of galaxies, the chemical reactions in our laboratories, the structure within atomic nuclei, and the cause of radioactive decay. They describe the true cause behind familiar terms like friction and the normal force. These four basic forces are known as fundamental because they alone are responsible for all observations of forces in nature. The four fundamental forces are gravity, electromagnetism, weak nuclear force, and strong nuclear force. Understanding the Four Forces The gravitational force is most familiar to us because it describes so many of our common observations. It explains why a dropped ball falls to the ground and why our planet orbits the Sun. It gives us the property of weight and determines much about the motion of objects in our daily lives. Because gravitational force acts between all objects of mass and has the ability to act over large distances, the gravitational force can be used to explain much of what we observe and can even describe the motion of objects on astronomical scales! That said, gravity is incredibly weak compared to the other fundamental forces and is the weakest of all of the fundamental forces. Consider this: The entire mass of Earth is needed to hold an iron nail to the ground. Yet with a simple magnet, the force of gravity can be overcome, allowing the nail to accelerate upward through space. The electromagnetic force is responsible for both electrostatic interactions and the magnetic force seen between bar magnets. When focusing on the electrostatic relationship between two charged particles, the electromagnetic force is known as the coulomb force. The electromagnetic force is an important force in the chemical and biological sciences, as it is responsible for molecular connections like ionic bonding and hydrogen bonding. Additionally, the electromagnetic force is behind the common physics forces of friction and the normal force. Like the gravitational force, the electromagnetic force is an inverse square law. However, the electromagnetic force does not exist between any two objects of mass, only those that are charged. When considering the structure of an atom, the electromagnetic force is somewhat apparent. After all, the electrons are held in place by an attractive force from the nucleus. But what causes the nucleus to remain intact? After all, if all protons are positive, it Access for free at openstax.org. 23.1 • The Four Fundamental Forces 773 makes sense that the coulomb force between the protons would repel the nucleus apart immediately. Scientists theorized that another force must exist within the nucleus to keep it together. They further theorized that this nuclear force must be significantly stronger than gravity, which has been observed and measured for centuries, and also stronger than the electromagnetic force, which would cause the protons to want to accelerate away from each other. The strong nuclear force is an attractive force that exists between all nucleons. This force, which acts equally between protonproton connections, proton-neutron connections, and neutron-neutron connections, is the strongest of all forces at short ranges. However, at a distance of 10–13 cm, or the diameter of a single proton, the force dissipates to zero. If the nucleus is large (it has many nucleons), then the distance between each nucleon could be much larger than the diameter of a single proton. The weak nuclear force is responsible for beta decay, as seen in the equation decay is when a beta particle is ejected from an atom. In order to accelerate away from the nucleus, the particle must be acted on by a force. Enrico Fermi was the first to envision this type of force. While this force is appropriately labeled, it remains stronger than the gravitational force. However, its range is even smaller than that of the strong force, as can be seen in Table 23.1. The weak nuclear force is more important than it may appear at this time, as will be addressed when we discuss quarks. Recall that beta Force Approximate Relative Strength[1] Range Gravity Weak Electromagnetic Strong 1 ∞ ∞ [1]Relative strength is based on the strong force felt by a proton–proton pair. Table 23.1 Relative strength and range of the four fundamental forces Transmitting the Four Fundamental Forces Just as it troubled Einstein prior to formulating the gravitational field theory, the concept of forces acting over a distance had greatly troubled particle physicists. That is, how does one proton knowthat another exists? Furthermore, what causes one proton to make a second proton repel? Or, for that matter, what is it about a proton that causes a neutron to attract? These mysterious interactions were first considered by Hideki Yukawa in 1935 and laid the foundation for much of what we now understand about particle physics. Hideki Yukawa’s focus was on the strong nuclear force and, in particular, its incredibly short range. His idea was a blend of particles, relativity, and quantum mechanics that was applicable to all four forces. Yukawa proposed that the nuclear force is actually transmitted by the exchange of particles, called carrier particles, and that what we commonly refer to as the force’s field consists of these carrier particles. Specifically for the strong nuclear force, Yukawa proposed that a previously unknown particle, called a pion, is exchanged between nucleons, transmitting the force between them. Figure 23.3 illustrates how a pion would carry a force between a proton and a neutron. Figure 23.3 The strong nuclear force is transmitted between a proton and neutron by the creation and exchange of a pion. The pion, created through a temporary vio |
lation of conservation of mass-energy, travels from the proton to the neutron and is recaptured. It is not 774 Chapter 23 • Particle Physics directly observable and is called a virtual particle. Note that the proton and neutron change identity in the process. The range of the force is limited by the fact that the pion can exist for only the short time allowed by the Heisenberg uncertainty principle. Yukawa used the finite range of the strong nuclear force to estimate the mass of the pion; the shorter the range, the larger the mass of the carrier particle. In Yukawa’s strong force, the carrier particle is assumed to be transmitted at the speed of light and is continually transferred between the two nucleons shown. The particle that Yukawa predicted was finally discovered within cosmic rays in 1947. Its name, the pion, stands for pi meson, where meson means medium mass; it’s a medium mass because it is smaller than a nucleon but larger than an electron. Yukawa launched the field that is now called quantum chromodynamics, and the carrier particles are now called gluons due to their strong binding power. The reason for the change in the particle name will be explained when quarks are discussed later in this section. As you may assume, the strong force is not the only force with a carrier particle. Nuclear decay from the weak force also requires a particle transfer. In the weak force are the following three: the weak negative carrier, W–; the weak positive carrier, W+; and the zero charge carrier, Z0. As we will see, Fermi inferred that these particles must carry mass, as the total mass of the products of nuclear decay is slightly larger than the total mass of all reactants after nuclear decay. The carrier particle for the electromagnetic force is, not surprisingly, the photon. After all, just as a lightbulb can emit photons from a charged tungsten filament, the photon can be used to transfer information from one electrically charged particle to another. Finally, the graviton is the proposed carrier particle for gravity. While it has not yet been found, scientists are currently looking for evidence of its existence (see Boundless Physics: Searching for the Graviton). So how does a carrier particle transmit a fundamental force? Figure 23.4 shows a virtual photon transmitted from one positively charged particle to another. The transmitted photon is referred to as a virtual particle because it cannot be directly observed while transmitting the force. Figure 23.5 shows a way of graphing the exchange of a virtual photon between the two positively charged particles. This graph of time versus position is called a Feynman diagram, after the brilliant American physicist Richard Feynman (1918–1988), who developed it. Figure 23.4 The image in part (a) shows the exchange of a virtual photon transmitting the electromagnetic force between charges, just as virtual pion exchange carries the strong nuclear force between nucleons. The image in part (b) shows that the photon cannot be directly observed in its passage because this would disrupt it and alter the force. In this case, the photon does not reach the other charge. The Feynman diagram should be read from the bottom up to show the movement of particles over time. In it, you can see that the left proton is propelled leftward from the photon emission, while the right proton feels an impulse to the right when the photon is received. In addition to the Feynman diagram, Richard Feynman was one of the theorists who developed the field of quantum electrodynamics (QED), which further describes electromagnetic interactions on the submicroscopic scale. For this work, he shared the 1965 Nobel Prize with Julian Schwinger and S.I. Tomonaga. A Feynman diagram explaining the strong force interaction hypothesized by Yukawa can be seen in Figure 23.6. Here, you can see the change in particle type due to the exchange of the pi meson. Access for free at openstax.org. 23.1 • The Four Fundamental Forces 775 Figure 23.5 The Feynman diagram for the exchange of a virtual photon between two positively charged particles illustrates how electromagnetic force is transmitted on a quantum mechanical scale. Time is graphed vertically, while the distance is graphed horizontally. The two positively charged particles are seen to repel each other by the photon exchange. Figure 23.6 The image shows a Feynman diagram for the exchange of a π+ (pion) between a proton and a neutron, carrying the strong nuclear force between them. This diagram represents the situation shown more pictorially in Figure 23.3. The relative masses of the listed carrier particles describe something valuable about the four fundamental forces, as can be seen bosons) and Z bosons ( in Table 23.2. W bosons (consisting of nearly 1,000 times more massive than pions, carriers of the strong nuclear force. Simultaneously, the distance that the weak nuclear force can be transmitted is approximately times the strong force transmission distance. Unlike carrier particles, bosons), carriers of the weak nuclear force, are and which have a limited range, the photon is a massless particle that has no limit to the transmission distance of the electromagnetic force. This relationship leads scientists to understand that the yet-unfound graviton is likely massless as well. Force Carrier Particle Range Relative Strength[1] Gravity Graviton (theorized) Weak W and Z bosons Electromagnetic Photon ∞ ∞ Strong Pi mesons or pions (now known as gluons) 1 [1]Relative strength is based on the strong force felt by a proton-proton pair. Table 23.2 Carrier particles and their relative masses compared to pions for the four fundamental forces 776 Chapter 23 • Particle Physics BOUNDLESS PHYSICS Searching for the Graviton From Newton’s Universal Law of Gravitation to Einstein’s field equations, gravitation has held the focus of scientists for centuries. Given the discovery of carrier particles during the twentieth century, the importance of understanding gravitation has yet again gained the interest of prominent physicists everywhere. With carrier particles discovered for three of the four fundamental forces, it is sensible to scientists that a similar particle, titled the graviton, must exist for the gravitational force. While evidence of this particle is yet to be uncovered, scientists are working diligently to discover its existence. So what do scientists think about the unfound particle? For starters, the graviton (like the photon) should be a massless particle traveling at the speed of light. This is assumed because, like the electromagnetic force, gravity is an inverse square law, . Scientists also theorize that the graviton is an electrically neutral particle, as an empty space within the influence of gravity is chargeless. However, because gravity is such a weak force, searching for the graviton has resulted in some unique methods. LIGO, the Laser Interferometer Gravitational-Wave Observatory, is one tool currently being utilized (see Figure 23.7). While searching for a gravitational wave to find a carrier particle may seem counterintuitive, it is similar to the approach taken by Planck and Einstein to learn more about the photon. According to wave-particle duality, if a gravitational wave can be found, the graviton should be present along with it. Predicted by Einstein’s theory of general relativity, scientists have been monitoring binary star systems for evidence of these gravitational waves. Figure 23.7 In searching for gravitational waves, scientists are using the Laser Interferometer Gravitational-Wave Observatory (LIGO). Here we see the control room of LIGO in Hanford, Washington. Particle accelerators like the Large Hadron Collider (LHC) are being used to search for the graviton through high-energy collisions. While scientists at the LHC speculate that the particle may not exist long enough to be seen, evidence of its prior existence, like footprints in the sand, can be found through gaps in projected energy and momentum. Some scientists are even searching the remnants of the Big Bang in an attempt to find the graviton. By observing the cosmic background radiation, they are looking for anomalies in gravitational waves that would provide information about the gravity particles that existed at the start of our universe. Regardless of the method used, scientists should know the graviton once they find it. A massless, chargeless particle with a spin of 2 and traveling at the speed of light—there is no other particle like it. Should it be found, its discovery would surely be considered by future generations to be on par with those of Newton and Einstein. GRASP CHECK Why are binary star systems used by LIGO to find gravitational waves? a. Binary star systems have high temperature. b. Binary star systems have low density. c. Binary star systems contain a large amount of mass, but because they are orbiting each other, the gravitational field between the two is much less. d. Binary star systems contain a large amount of mass. As a result, the gravitational field between the two is great. Access for free at openstax.org. 23.1 • The Four Fundamental Forces 777 Accelerators Create Matter From Energy Before looking at all the particles that make up our universe, let us first examine some of the machines that create them. The fundamental process in creating unknown particles is to accelerate known particles, such as protons or electrons, and direct a beam of them toward a target. Collisions with target nuclei provide a wealth of information, such as information obtained by Rutherford in the gold foil experiment. If the energy of the incoming particles is large enough, new matter can even be created in the collision. The more energy input or ΔE, the more matter mcan be created, according to mass energy equivalence . Limitations are placed on what can occur by known conservation laws, such as conservation of mass-energy, momentum, and charge. Even more interesting are the unkn |
own limitations provided by nature. While some expected reactions do occur, others do not, and still other unexpected reactions may appear. New laws are revealed, and the vast majority of what we know about particle physics has come from accelerator laboratories. It is the particle physicist’s favorite indoor sport. Our earliest model of a particle accelerator comes from the Van de Graaff generator. The relatively simple device, which you have likely seen in physics demonstrations, can be manipulated to produce potentials as great as 50 million volts. While these machines do not have energies large enough to produce new particles, analysis of their accelerated ions was instrumental in exploring several aspects of the nucleus. Another equally famous early accelerator is the cyclotron, invented in 1930 by the American physicist, E.O. Lawrence (1901–1958). Figure 23.8 is a visual representation with more detail. Cyclotrons use fixed-frequency alternating electric fields to accelerate particles. The particles spiral outward in a magnetic field, making increasingly larger radius orbits during acceleration. This clever arrangement allows the successive addition of electric potential energy with each loop. As a result, greater particle energies are possible than in a Van de Graaff generator. Figure 23.8 On the left is an artist’s rendition of the popular physics demonstration tool, the Van de Graaff generator. A battery (A) supplies excess positive charge to a pointed conductor, the points of which spray the charge onto a moving insulating belt near the bottom. The pointed conductor (B) on top in the large sphere picks up the charge. (The induced electric field at the points is so large that it removes the charge from the belt.) This can be done because the charge does not remain inside the conducting sphere but moves to its outer surface. An ion source inside the sphere produces positive ions, which are accelerated away from the positive sphere to high velocities. On the right is a cyclotron. Cyclotrons use a magnetic field to cause particles to move in circular orbits. As the particles pass between the plates of the Dees, the voltage across the gap is oscillated to accelerate them twice in each orbit. A synchrotron is a modification of the cyclotron in which particles continually travel in a fixed-radius orbit, increasing speed each time. Accelerating voltages are synchronized with the particles to accelerate them, hence the name. Additionally, magnetic field strength is increased to keep the orbital radius constant as energy increases. A ring of magnets and accelerating tubes, as shown in Figure 23.9, are the major components of synchrotrons. High-energy particles require strong magnetic fields to steer 778 Chapter 23 • Particle Physics them, so superconducting magnets are commonly employed. Still limited by achievable magnetic field strengths, synchrotrons need to be very large at very high energies since the radius of a high-energy particle’s orbit is very large. To further probe the nucleus, physicists need accelerators of greater energy and detectors of shorter wavelength. To do so requires not only greater funding but greater ingenuity as well. Colliding beams used at both the Fermi National Accelerator Laboratory (Fermilab; see Figure 23.11) near Chicago and the LHC in Switzerland are designed to reduce energy loss in particle collisions. Typical stationary particle detectors lose a large amount of energy to the recoiling target struck by the accelerating particle. By providing head-on collisions between particles moving in opposite directions, colliding beams make it possible to create particles with momenta and kinetic energies near zero. This allows for particles of greater energy and mass to be created. Figure 23.10 is a schematic representation of this effect. In addition to circular accelerators, linear accelerators can be used to reduce energy radiation losses. The Stanford Linear Accelerator Center (now called the SLAC National Accelerator Laboratory) in California is home to the largest such accelerator in the world. Figure 23.9 (a) A synchrotron has a ring of magnets and accelerating tubes. The frequency of the accelerating voltages is increased to cause the beam particles to travel the same distance in a shorter time. The magnetic field should also be increased to keep each beam burst traveling in a fixed-radius path. Limits on magnetic field strength require these machines to be very large in order to accelerate particles to very high energies. (b) A positively charged particle is shown in the gap between accelerating tubes. (c) While the particle passes through the tube, the potentials are reversed so that there is another acceleration at the next gap. The frequency of the reversals needs to be varied as the particle is accelerated to achieve successive accelerations in each gap. Figure 23.10 This schematic shows the two rings of Fermilab’s accelerator and the scheme for colliding protons and antiprotons (not to scale). Figure 23.11 The Fermi National Accelerator Laboratory, near Batavia, Illinois, was a subatomic particle collider that accelerated protons and antiprotons to attain energies up to 1 Tev (a trillion electronvolts). The circular ponds near the rings were built to dissipate waste heat. This accelerator was shut down in September 2011. (credit: Fermilab, Reidar Hahn) Check Your Understanding 1. Which of the four forces is responsible for radioactive decay? a. the electromagnetic force Access for free at openstax.org. 23.2 • Quarks 779 b. c. d. the gravitational force the strong nuclear force the weak nuclear force 2. What force or forces exist between an electron and a proton? a. b. c. d. the strong nuclear force, the electromagnetic force, and gravity the weak nuclear force, the strong nuclear force, and gravity the weak nuclear force, the strong nuclear force, and the electromagnetic force the weak nuclear force, the electromagnetic force, and gravity 3. What is the proposed carrier particle for the gravitational force? a. boson b. graviton c. gluon d. photon 4. What is the relationship between the mass and range of a carrier particle? a. Range of a carrier particle is inversely proportional to its mass. b. Range of a carrier particle is inversely proportional to square of its mass. c. Range of a carrier particle is directly proportional to its mass. d. Range of a carrier particle is directly proportional to square of its mass. 5. What type of particle accelerator uses fixed-frequency oscillating electric fields to accelerate particles? cyclotron synchrotron a. b. c. betatron d. Van de Graaff accelerator 6. How does the expanding radius of the cyclotron provide evidence of particle acceleration? a. A constant magnetic force is exerted on particles at all radii. As the radius increases, the velocity of the particle must increase to maintain this constant force. b. A constant centripetal force is exerted on particles at all radii. As the radius increases, the velocity of the particle must decrease to maintain this constant force. c. A constant magnetic force is exerted on particles at all radii. As the radius increases, the velocity of the particle must decrease to maintain this constant force. d. A constant centripetal force is exerted on particles at all radii. As the radius increases, the velocity of the particle must increase to maintain this constant force. 7. Which of the four forces is responsible for the structure of galaxies? a. electromagnetic force b. gravity c. d. weak nuclear force strong nuclear force 23.2 Quarks Section Learning Objectives By the end of the section, you will be able to do the following: • Describe quarks and their relationship to other particles • Distinguish hadrons from leptons • Distinguish matter from antimatter • Describe the standard model of the atom • Define a Higgs boson and its importance to particle physics 780 Chapter 23 • Particle Physics Section Key Terms annihilation antimatter baryon bottom quark charmed quark color down quark flavor gluon hadron Higgs boson Higgs field lepton meson pair production positron quantum chromodynamics quark Standard Model strange quark top quark up quark Quarks “The first principles of the universe are atoms and empty space. Everything else is merely thought to exist…” “… Further, the atoms are unlimited in size and number, and they are borne along with the whole universe in a vortex, and thereby generate all composite things—fire, water, air, earth. For even these are conglomerations of given atoms. And it because of their solidity that these atoms are impassive and unalterable.” —Diogenes Laertius (summarizing the views of Democritus, circa 460–370 B.C.) The search for fundamental particles is nothing new. Atomists of the Greek and Indian empires, like Democritus of fifth century B.C., openly wondered about the most finite components of our universe. Though dormant for centuries, curiosity about the atomic nature of matter was reinvigorated by Rutherford’s gold foil experiment and the discovery of the nucleus. By the early 1930s, scientists believed they had fully determined the tiniest constituents of matter—in the form of the proton, neutron, and electron. This would be only partially true. At present, scientists know that there are hundreds of particles not unlike our electron and nucleons, all making up what some have termed the particle zoo. While we are confident that the electron remains fundamental, it is surrounded by a plethora of similar sounding terms, like leptons, hadrons, baryons, and mesons. Even though not every particle is considered fundamental, they all play a vital role in understanding the intricate structure of our universe. A fundamental particle is defined as a particle with no substructure and no finite size. According to the Standard Model, there are three types of fundamental particles: leptons, quarks, and carrier particles. As you may recall, carrier particles are responsible |
for transmitting fundamental forces between their interacting masses. Leptons are a group of six particles not bound by the strong nuclear force, of which the electron is one. As for quarks, they are the fundamental building blocks of a group of particles called hadrons, a group that includes both the proton and the neutron. Now for a brief history of quarks. Quarks were first proposed independently by American physicists Murray Gell-Mann and George Zweig in 1963. Originally, three quark types—or flavors—were proposed with the names up (u), down (d), and strange (s). At first, physicists expected that, with sufficient energy, we should be able to free quarks and observe them directly. However, this has not proved possible, as the current understanding is that the force holding quarks together is incredibly great and, much like a spring, increases in magnitude as the quarks are separated. As a result, when large energies are put into collisions, other particles are created—but no quarks emerge. With that in mind, there is compelling evidence for the existence of quarks. By 1967, experiments at the SLAC National Accelerator Laboratory scattering 20-GeV electrons from protons produced results like Rutherford had obtained for the nucleus nearly 60 years earlier. The SLAC scattering experiments showed unambiguously that there were three point-like (meaning they had sizes considerably smaller than the probe’s wavelength) charges inside the proton as seen in Figure 23.12. This evidence made all but the most skeptical admit that there was validity to the quark substructure of hadrons. Access for free at openstax.org. 23.2 • Quarks 781 Figure 23.12 Scattering of high-energy electrons from protons at facilities like SLAC produces evidence of three point-like charges consistent with proposed quark properties. This experiment is analogous to Rutherford’s discovery of the small size of the nucleus by scattering α particles. High-energy electrons are used so that the probe wavelength is small enough to see details smaller than the proton. The inclusion of the strange quark with Zweig and Gell-Mann’s model concerned physicists. While the up and down quarks demonstrated fairly clear symmetry and were present in common fundamental particles like protons and neutrons, the strange quark did not have a counterpart of its own. This thought, coupled with the four known leptons at the time, caused scientists to predict that a fourth quark, yet to be found, also existed. In 1974, two groups of physicists independently discovered a particle with this new quark, labeled charmed. This completed the second exoticquark pair, strange (s) and charmed (c). A final pair of quarks was proposed when a third pair of leptons was discovered in 1975. The existence of the bottom (b) quark and the top (t) quark was verified through experimentation in 1976 and 1995, respectively. While it may seem odd that so much time would elapse between the original quark discovery in 1967 and the verification of the top quark in 1995, keep in mind that each quark discovered had a progressively larger mass. As a result, each new quark has required more energy to discover. TIPS FOR SUCCESS Note that a very important tenet of science occurred throughout the period of quark discovery. The charmed, bottom, and top quarks were all speculated on, and then were discovered some time later. Each of their discoveries helped to verify and strengthen the quark model. This process of speculation and verification continues to take place today and is part of what drives physicists to search for evidence of the graviton and Grand Unified Theory. One of the most confounding traits of quarks is their electric charge. Long assumed to be discrete, and specifically a multiple of the elementary charge of the electron, the electric charge of an individual quark is fractional and thus seems to violate a presumed tenet of particle physics. The fractional charge of quarks, which are , are the only structures and found in nature with a nonintegral number of charge . However, note that despite this odd construction, the fractional value of the quark does not violate the quantum nature of the charge. After all, free quarks cannot be found in nature, and all quarks are bound into arrangements in which an integer number of charge is constructed. Table 23.3 shows the six known quarks, in addition to their antiquark components, as will be discussed later in this section. Flavor Symbol Antiparticle Charge[1][2] Up Down Strange Charmed [1]The lower of the [2]There are further qualities that differentiate between quarks. However, they are beyond the discussion in this text. symbols are the values for antiquarks. Table 23.3 Quarks and Antiquarks 782 Chapter 23 • Particle Physics Flavor Symbol Antiparticle Charge[1][2] Bottom Top [1]The lower of the [2]There are further qualities that differentiate between quarks. However, they are beyond the discussion in this text. symbols are the values for antiquarks. Table 23.3 Quarks and Antiquarks While the term flavoris used to differentiate between types of quarks, the concept of color is more analogous to the electric charge in that it is primarily responsible for the force interactions between quarks. Note—Take a moment to think about the electrostatic force. It is the electric charge that causes attraction and repulsion. It is the same case here but with a colorcharge. The three colors available to a quark are red, green, and blue, with antiquarks having colors of anti-red (or cyan), anti-green (or magenta), and anti-blue (or yellow). Why use colors when discussing quarks? After all, the quarks are not actually colored with visible light. The reason colors are used is because the properties of a quark are analogous to the three primary and secondary colors mentioned above. Just as different colors of light can be combined to create white, different colorsof quark may be combined to construct a particle like a proton or neutron. In fact, for each hadron, the quarks must combine such that their color sums to white! Recall that two up quarks and one down quark construct a proton, as seen in Figure 23.12. The sum of the three quarks’ colors—red, green, and blue—yields the color white. This theory of color interaction within particles is called quantum chromodynamics, or QCD. As part of QCD, the strong nuclear force can be explained using color. In fact, some scientists refer to the color force, not the strong force, as one of the four fundamental forces. Figure 23.13 is a Feynman diagram showing the interaction between two quarks by using the transmission of a colored gluon. Note that the gluon is also considered the charge carrier for the strong nuclear force. Figure 23.13 The exchange of gluons between quarks carries the strong force and may change the color of the interacting quarks. While the colors of the individual quarks change, their flavors do not. Note that quark flavor may have any color. For instance, in Figure 23.13, the down quark has a red color and a green color. In other words, colors are not specific to a particle quark flavor. Hadrons and Leptons Particles can be revealingly grouped according to what forces they feel between them. All particles (even those that are massless) are affected by gravity since gravity affects the space and time in which particles exist. All charged particles are affected by the electromagnetic force, as are neutral particles that have an internal distribution of charge (such as the neutron with its magnetic moment). Special names are given to particles that feel the strong and weak nuclear forces. Hadrons are particles that feel the strong nuclear force, whereas leptons are particles that do not. All particles feel the weak nuclear force. This means that hadrons are distinguished by being able to feel both the strong and weak nuclear forces. Leptons and hadrons are distinguished in other ways as well. Leptons are fundamental particles that have no measurable size, while hadrons are composed of quarks and have a diameter on the order of 10 to 15 m. Six particles, including the electron and neutrino, make up the list of known leptons. There are hundreds of complex particles in the hadron class, a few of which (including the proton and neutron) are listed in Table 23.4. Access for free at openstax.org. Category Leptons Hadrons – Mesons[2] Hadrons – Baryons[3] Particle Name Electron Neutrino (e) Muon Neutrino (μ) Tau Neutrino (τ) Pion Kaon Eta Proton Neutron Lambda Omega 23.2 • Quarks 783 Symbol Antiparticle Rest Mass Mean Lifetime (s) [1] [1] Stable Stable Stable [1] Stable Stable 882 0.511 105.7 1,777 139.6 135.0 493.7 497.6 547.9 938.3 939.6 1,115.7 1,672.5 Self Self p n [1]Neutrino masses may be zero. Experimental upper limits are given in parentheses. [2]Many other mesons known [3]Many other baryons known Table 23.4 List of Leptons and Hadrons. There are many more leptons, mesons, and baryons yet to be discovered and measured. The purpose of trying to uncover the smallest indivisible things in existence is to explain the world around us through forces and the interactions between particles, galaxies and objects. This is why a handful of scientists devote their life’s work to smashing together small particles. What internal structure makes a proton so different from an electron? The proton, like all hadrons, is made up of quarks. A few examples of hadron quark composition can be seen in Figure 23.14. As shown, each hadron is constructed of multiple quarks. As mentioned previously, the fractional quark charge in all four hadrons sums to the particle’s integral value. Also, notice that the color composition for each of the four particles adds to white. Each of the particles shown is constructed of up, down, and their antiquarks. This is not surprising, as the quarks strange, charmed, top, and bottom are found in only our most exotic particles. 784 Chapter 23 • Particle Physics Figure 23.14 All baryons, such as the |
proton and neutron shown here, are composed of three quarks. All mesons, such as the pions shown here, are composed of a quark–antiquark pair. Arrows represent the spins of the quarks. The colors are such that they need to add to white for any possible combination of quarks. You may have noticed that while the proton and neutron in Figure 23.14 are composed of three quarks, both pions are comprised of only two quarks. This refers to a final delineation in particle structure. Particles with three quarks are called baryons. These are heavy particles that can decay into another baryon. Particles with only two quarks—a-quark–anti-quark pair—are called mesons. These are particles of moderate mass that cannot decay into the more massive baryons. Before continuing, take a moment to view Figure 23.15. In this figure, you can see the strong force reimagined as a color force. The particles interacting in this figure are the proton and neutron, just as they were in Figure 23.6. This reenvisioning of the strong force as an interaction between colored quarks is the critical concept behind quantum chromodynamics. Figure 23.15 This Feynman diagram shows the interaction between a proton and a neutron, corresponding to the interaction shown in Figure 23.6. This diagram, however, shows the quark and gluon details of the strong nuclear force interaction. Matter and Antimatter Antimatter was first discovered in the form of the positron, the positively charged electron. In 1932, American physicist Carl Anderson discovered the positron in cosmic ray studies. Through a cloud chamber modified to curve the trajectories of cosmic Access for free at openstax.org. rays, Anderson noticed that the curves of some particles followed that of a negative charge, while others curved like a positive charge. However, the positive curve showed not the mass of a proton but the mass of an electron. This outcome is shown in Figure 23.16 and suggests the existence of a positively charged version of the electron, created by the destruction of solar photons. 23.2 • Quarks 785 Figure 23.16 The image above is from the Fermilab 15 foot bubble chamberand shows the production of an electron and positron (or antielectron) from an incident photon. This event is titled pair production and provides evidence of antimatter, as the two repel each other. Antimatter is considered the opposite of matter. For most antiparticles, this means that they share the same properties as their original particles with the exception of their charge. This is why the positron can be considered a positive electron while the antiproton is considered a negative proton. The idea of an opposite charge for neutral particles (like the neutron) can be confusing, but it makes sense when considered from the quark perspective. Just as the neutron is composed of one up quark and two down quarks (of charge , respectively), the antineutron is composed of one anti–up quark and two anti–down and quarks (of charge and , respectively). While the overall charge of the neutron remains the same, its constituent particles do not! A word about antiparticles: Like regular particles, antiparticles could function just fine on their own. In fact, a universe made up of antimatter may operate just as our own matter-based universe does. However, we do not know fully whether this is the case. The reason for this is annihilation. Annihilation is the process of destruction that occurs when a particle and its antiparticle interact. As soon as two particles (like a positron and an electron) coincide, they convert their masses to energy through the equation makes it very difficult for scientists to study antimatter. That said, scientists have had success creating antimatter through highenergy particle collisions. Both antineutrons and antiprotons were created through accelerator experiments in 1956, and an anti–hydrogen atom was even created at CERN in 1995! As referenced in , the annihilation of antiparticles is currently used in medical studies to determine the location of radioisotopes. . This mass-to-energy conversion, which typically results in photon release, happens instantaneously and Completing the Standard Model of the Atom The Standard Model of the atom refers to the current scientific view of the fundamental components and interacting forces of matter. The Standard Model (Figure 23.17) shows the six quarks that bind to form all hadrons, the six lepton particles already considered fundamental, the four carrier particles (or gauge bosons) that transmit forces between the leptons and quarks, and the recently added Higgs boson (which will be discussed shortly). This totals 17 fundamental particles, combinations of which are responsible for all known matter in our entire universe! When adding the antiquarks and antileptons, 31 components make up the Standard Model. 786 Chapter 23 • Particle Physics Figure 23.17 The Standard Model of elementary particles shows an organized view of all fundamental particles, as currently known: six quarks, six leptons, and four gauge bosons (or carrier particles). The Higgs boson, first observed in 2012, is a new addition to the Standard Model. Figure 23.17 shows all particles within the Standard Model of the atom. Not only does this chart divide all known particles by color-coded group, but it also provides information on particle stability. Note that the color-coding system in this chart is separate from the red, green, and blue color labeling system of quarks. The first three columns represent the three familiesof matter. The first column, considered Family 1, represents particles that make up normal matter, constructing the protons, neutrons, and electrons that make up the common world. Family 2, represented from the charm quark to the muon neutrino, is comprised of particles that are more massive. The leptons in this group are less stable and more likely to decay. Family 3, represented by the third column, are more massive still and decay more quickly. The order of these families also conveniently represents the order in which these particles were discovered. TIPS FOR SUCCESS Look for trends that exist within the Standard Model. Compare the charge of each particle. Compare the spin. How does mass relate to the model structure? Recognizing each of these trends and asking questions will yield more insight into the organization of particles and the forces that dictate particle relationships. Our understanding of the Standard Model is still young, and the questions you may have in analyzing the Standard Model may be some of the same questions that particle physicists are searching for answers to today! The Standard Model also summarizes the fundamental forces that exist as particles interact. A closer look at the Standard Model, as shown in Figure 23.18, reveals that the arrangement of carrier particles describes these interactions. Access for free at openstax.org. 23.2 • Quarks 787 Figure 23.18 The revised Standard Model shows the interaction between gauge bosons and other fundamental particles. These interactions are responsible for the fundamental forces, three of which are described through the chart’s shaded areas. Each of the shaded areas represents a fundamental force and its constituent particles. The red shaded area shows all particles involved in the strong nuclear force, which we now know is due to quantum chromodynamics. The blue shaded area corresponds to the electromagnetic force, while the green shaded area corresponds to the weak nuclear force, which affects all quarks and leptons. The electromagnetic force and weak nuclear force are considered united by the electroweak force within the Standard Model. Also, because definitive evidence of the graviton is yet to be found, it is not included in the Standard Model. The Higgs Boson One interesting feature of the Standard Model shown in Figure 23.18 is that, while the gluon and photon have no mass, the Z and W bosons are very massive. What supplies these quickly moving particles with mass and not the gluons and photons? Furthermore, what causes some quarks to have more mass than others? In the 1960s, British physicist Peter Higgs and others speculated that the W and Z bosons were actually just as massless as the gluon and photon. However, as the W and Z bosons traveled from one particle to another, they were slowed down by the presence of a Higgs field, much like a fish swimming through water. The thinking was that the existence of the Higgs field would slow down the bosons, causing them to decrease in energy and thereby transfer this energy to mass. Under this theory, all particles pass through the Higgs field, which exists throughout the universe. The gluon and photon travel through this field as well but are able to do so unaffected. The presence of a force from the Higgs field suggests the existence of its own carrier particle, the Higgs boson. This theorized boson interacts with all particles but gluons and photons, transferring force from the Higgs field. Particles with large mass (like the top quark) are more likely to receive force from the Higgs boson. While it is difficult to examine a field, it is somewhat simpler to find evidence of its carrier. On July 4, 2012, two groups of scientists at the LHC independently confirmed the existence of a Higgs-like particle. By examining trillions of proton–proton collisions at energies of 7 to 8 TeV, LHC scientists were able to determine the constituent particles that created the protons. In this data, scientists found a particle with similar mass, spin, parity, and interactions with other particles that matched the Higgs boson predicted decades prior. On March 13, 2013, the existence of the Higgs boson was tentatively confirmed by CERN. Peter Higgs and Francois Englert received the Nobel Prize in 2013 for the “theoretical discovery of a mechanism that contributes to our understanding of the origin and mass of subatomic particles.” WORK IN PHYSICS Particle Physicist If |
you have an innate desire to unravel life’s great mysteries and further understand the nature of the physical world, a career in particle physics may be for you! Particle physicists have played a critical role in much of society’s technological progress. From lasers to computers, televisions to space missions, splitting the atom to understanding the DNA molecule to MRIs and PET scans, much of our modern society is based on the work done by particle physicists. While many particle physicists focus on specialized tasks in the fields of astronomy and medicine, the main goal of particle physics is to further scientists’ understanding of the Standard Model. This may mean work in government, industry, or 788 Chapter 23 • Particle Physics academics. Within the government, jobs in particle physics can be found within the National Institute for Standards and Technology, Department of Energy, NASA, and Department of Defense. Both the electronics and computer industries rely on the expertise of particle physicists. College teaching and research positions can also be potential career opportunities for particle physicists, though they often require some postgraduate work as a prerequisite. In addition, many particle physicists are employed to work on high-energy colliders. Domestic collider labs include the Brookhaven National Laboratory in New York, the Fermi National Accelerator Laboratory near Chicago, and the SLAC National Accelerator Laboratory operated by Stanford University. For those who like to travel, work at international collider labs can be found at the CERN facility in Switzerland in addition to institutes like the Budker Institute of Nuclear Physics in Russia, DESY in Germany, and KEK in Japan. Shirley Jackson became the first African American woman to earn a Ph.D. from MIT back in 1973, and she went on to lead a highly successful career in the field of particle physics. Like Dr. Jackson, successful students of particle physics grow up with a strong curiosity in the world around them and a drive to continually learn more. If you are interested in exploring a career in particle physics, work to achieve good grades and SAT scores, and find time to read popular books on physics topics that interest you. While some math may be challenging, recognize that this is only a tool of physics and should not be considered prohibitive to the field. High-level work in particle physics often requires a Ph.D.; however, it is possible to find work with a master’s degree. Additionally, jobs in industry and teaching can be achieved with solely an undergraduate degree. GRASP CHECK What is the primary goal of all work in particle physics? a. The primary goal is to further our understanding of the Standard Model. b. The primary goal is to further our understanding of Rutherford’s model. c. The primary goal is to further our understanding of Bohr’s model. d. The primary goal is to further our understanding of Thomson’s model. Check Your Understanding 8. In what particle were quarks originally discovered? a. b. c. d. the electron the neutron the proton the photon 9. Why was the existence of the charm quark speculated, even though no direct evidence of it existed? a. The existence of the charm quark was symmetrical with up and down quarks. Additionally, there were two known leptons at the time and only two quarks. b. The strange particle lacked the symmetry that existed with the up and down quarks. Additionally, there were four known leptons at the time and only three quarks. c. The bottom particle lacked the symmetry that existed with the up and down quarks. Additionally, there were two known leptons at the time and only two quarks. d. The existence of charm quarks was symmetrical with up and down quarks. Additionally, there were four known leptons at the time and only three quarks. 10. What type of particle is the electron? a. The electron is a lepton. b. The electron is a hadron. c. The electron is a baryon. d. The electron is an antibaryon. 11. How do the number of fundamental particles differ between hadrons and leptons? a. Hadrons are constructed of at least three fundamental quark particles, while leptons are fundamental particles. b. Hadrons are constructed of at least three fundamental quark particles, while leptons are constructed of two fundamental particles. c. Hadrons are constructed of at least two fundamental quark particles, while leptons are constructed of three Access for free at openstax.org. 23.2 • Quarks 789 fundamental particles. d. Hadrons are constructed of at least two fundamental quark particles, while leptons are fundamental particles. 12. Does antimatter exist? a. no b. yes 13. How does the deconstruction of a photon into an electron and a positron uphold the principles of mass and charge conservation? a. The sum of the masses of an electron and a positron is equal to the mass of the photon before pair production. The sum of the charges on an electron and a positron is equal to the zero charge of the photon. b. The sum of the masses of an electron and a positron is equal to the mass of the photon before pair production. The sum of the same charges on an electron and a positron is equal to the charge on a photon. c. During the particle production the total energy of the photon is converted to the mass of an electron and a positron. The sum of the opposite charges on the electron and positron is equal to the zero charge of the photon. d. During particle production, the total energy of the photon is converted to the mass of an electron and a positron. The sum of the same charges on an electron and a positron is equal to the charge on a photon. 14. How many fundamental particles exist in the Standard Model, including the Higgs boson and the graviton (not yet observed)? 12 a. 15 b. 13 c. 19 d. 15. Why do gluons interact only with particles in the first two rows of the Standard Model? a. The leptons in the third and fourth rows do not have mass, but the gluons can interact between the quarks through gravity only. b. The leptons in the third and fourth rows do not have color, but the gluons can interact between quarks through color interactions only. c. The leptons in the third and fourth rows do not have spin, but the gluons can interact between quarks through spin interactions only. d. The leptons in the third and fourth rows do not have charge, but the gluons can interact between quarks through charge interactions only. 16. What fundamental property is provided by particle interaction with the Higgs boson? charge a. b. mass spin c. color d. 17. Considering the Higgs field, what differentiates more massive particles from less massive particles? a. More massive particles interact more with the Higgs field than the less massive particles. b. More massive particles interact less with the Higgs field than the less massive particles. 18. What particles were launched into the proton during the original discovery of the quark? a. bosons b. electrons c. neutrons d. photons 790 Chapter 23 • Particle Physics 23.3 The Unification of Forces Section Learning Objectives By the end of the section, you will be able to do the following: • Define a grand unified theory and its importance • Explain the evolution of the four fundamental forces from the Big Bang onward • Explain how grand unification theories can be tested Section Key Terms Big Bang Inflationary Epoch Electroweak Epoch electroweak theory Grand Unification Epoch Grand Unified Theory Planck Epoch Quark Era superforce Theory of Everything Understanding the Grand Unified Theory Present quests to show that the four basic forces are different manifestations of a single unified force that follow a long tradition. In the nineteenth century, the distinct electric and magnetic forces were shown to be intimately connected and are now collectively called the electromagnetic force. More recently, the weak nuclear force was united with the electromagnetic force. As shown in Figure 23.19, carrier particles transmit three of the four fundamental forces in very similar ways. With these considerations in mind, it is natural to suggest that a theory may be constructed in which the strong nuclear, weak nuclear, and electromagnetic forces are all unified. The search for a correct theory linking the forces, called the Grand Unified Theory (GUT), is explored in this section. In the 1960s, the electroweak theory was developed by Steven Weinberg, Sheldon Glashow, and Abdus Salam. This theory proposed that the electromagnetic and weak nuclear forces are identical at sufficiently high energies. At lower energies, like those in our present-day universe, the two forces remain united but manifest themselves in different ways. One of the main consequences of the electroweak theory was the prediction of three short-range carrier particles, now known as the and and that of the predicted characteristics, including masses having those predicted values as given in . boson was predicted to be 90 GeV/c2. In 1983, these carrier particles were observed at CERN with the bosons. Not only were three particles predicted, but the mass of each boson was predicted to be 81 GeV/c2, and How can forces be unified? They are definitely distinct under most circumstances. For example, they are carried by different particles and have greatly different strengths. But experiments show that at extremely short distances and at extremely high energies, the strengths of the forces begin to become more similar, as seen in Figure 23.20. Figure 23.19 The exchange of a virtual particle (boson) carries the weak nuclear force between an electron and a neutrino in this Feynman diagram. This diagram is similar to the diagrams in Figure 23.6 and for the electromagnetic and strong nuclear forces. As discussed earlier, the short ranges and large masses of the weak carrier bosons require correspondingly high energies to create them. Thus, the energy scale on the horizontal axis of Figure 23.20 also corresponds to shorter and shorter di |
stances Access for free at openstax.org. 23.3 • The Unification of Forces 791 (going from left to right), with 100 GeV corresponding to approximately 10−18 m, for example. At that distance, the strengths of the electromagnetic and weak nuclear forces are the same. To test this, energies of about 100 GeV are put into the system. When this occurs, the , and carrier particles become less and less relevant, and the further energy is added, the similar to photons and gluons. carrier particles are created and released. At those and higher energies, the masses of the boson in particular resembles the massless, chargeless photon. As particles are further transformed into massless carrier particles even more , and , , Figure 23.20 The relative strengths of the four basic forces vary with distance, and, hence, energy is needed to probe small distances. At ordinary energies (a few eV or less), the forces differ greatly. However, at energies available in accelerators, the weak nuclear and electromagnetic (EM) forces become unified. Unfortunately, the energies at which the strong nuclear and electroweak forces become the same are unreachable in any conceivable accelerator. The universe may provide a laboratory, and nature may show effects at ordinary energies that give us clues about the validity of this graph. The extremely short distances and high energies at which the electroweak force becomes identical with the strong nuclear force are not reachable with any conceivable human-built accelerator. At energies of about 1014 GeV (16,000 J per particle), distances of about 10 to 30 m can be probed. Such energies are needed to test the theory directly, but these are about 1010 times higher than the maximum energy associated with the LHC, and the distances are about 10 to 12 smaller than any structure we have direct knowledge of. This would be the realm of various GUTs, of which there are many, since there is no constraining evidence at these energies and distances. Past experience has shown that anytime you probe so many orders of magnitude further, you find the unexpected. While direct evidence of a GUT is not presently possible, that does not rule out the ability to assess a GUT through an indirect process. Current GUTs require various other events as a consequence of their theory. Some GUTs require the existence of magnetic monopoles, very massive individual north- and south-pole particles, which have not yet been proven to exist, while others require the use of extra dimensions. However, not all theories result in the same consequences. For example, disproving the existence of magnetic monopoles will not disprove all GUTs. Much of the science we accept in our everyday lives is based on different models, each with their own strengths and limitations. Although a particular model may have drawbacks, that does not necessarily mean that it should be discounted completely. One consequence of GUTs that can theoretically be assessed is proton decay. Multiple current GUTs hypothesize that the stable proton should actually decay at a lifetime of 1031 years. While this time is incredibly large (keep in mind that the age of the universe is less than 14 billion years), scientists at the Super-Kamiokande in Japan have used a 50,000-ton tank of water to search for its existence. The decay of a single proton in the Super-Kamiokande tank would be observed by a detector, thereby providing support for the predicting GUT model. However, as of 2014, 17 years into the experiment, decay is yet to be found. This time span equates to a minimum limit on proton life of grand unifying theories, an acceptable model may still exist. years. While this result certainly does not support many TIPS FOR SUCCESS The Super-Kamiokande experiment is a clever use of proportional reasoning. Because it is not feasible to test for 1031 years in order for a single proton to decay, scientists chose instead to manipulate the proton–time ratio. If one proton decays in 1031 792 Chapter 23 • Particle Physics years, then in one year 10−31 protons will decay. With this in mind, if scientists wanted to test the proton decay theory in one year, they would need 1031 protons. While this is also unfeasible, the use of a 50,000-ton tank of water helps to bring both the wait time and proton number to within reason. The Standard Model and the Big Bang Nature is full of examples where the macroscopic and microscopic worlds intertwine. Newton realized that the nature of gravity on Earth that pulls an apple to the ground could explain the motion of the moon and planets so much farther away. Decays of tiny nuclei explain the hot interior of the Earth. Fusion of nuclei likewise explains the energy of stars. Today, the patterns in particle physics seem to be explaining the evolution and character of the universe. And the nature of the universe has implications for unexplored regions of particle physics. In 1929, Edwin Hubble observed that all but the closest galaxies surrounding our own had a red shift in their hydrogen spectra that was proportional to their distance from us. Applying the Doppler Effect, Hubble recognized that this meant that all galaxies were receding from our own, with those farther away receding even faster. Knowing that our place in the universe was no more unique than any other, the implication was clear: The space within the universe itself was expanding. Just like pen marks on an expanding balloon, everything in the universe was accelerating away from everything else. Figure 23.21 shows how the recession of galaxies looks like the remnants of a gigantic explosion, the famous Big Bang. Extrapolating backward in time, the Big Bang would have occurred between 13 and 15 billion years ago, when all matter would have been at a single point. From this, questions instantly arise. What caused the explosion? What happened before the Big Bang? Was there a before, or did time start then? For our purposes, the biggest question relating to the Big Bang is this: How does the Big Bang relate to the unification of the fundamental forces? Figure 23.21 Galaxies are flying apart from one another, with the more distant ones moving faster, as if a primordial explosion expelled the matter from which they formed. The most distant known galaxies move nearly at the speed of light relative to us. To fully understand the conditions of the very early universe, recognize that as the universe contracts to the size of the Big Bang, changes will occur. The density and temperature of the universe will increase dramatically. As particles become closer together, they will become too close to exist as we know them. The high energies will create other, more unusual particles to exist in greater abundance. Knowing this, let’s move forward from the start of the universe, beginning with the Big Bang, as illustrated in Figure 23.22. Access for free at openstax.org. 23.3 • The Unification of Forces 793 Figure 23.22 The evolution of the universe from the Big Bang onward (from left to right) is intimately tied to the laws of physics, especially those of particle physics at the earliest stages. Theories of the unification of forces at high energies may be verified by their shaping of the universe and its evolution. The Planck Epoch —Though scientists are unable to model the conditions of the Planck Epoch in the laboratory, GeV necessary to unify gravity speculation is that at this time compressed energy was great enough to reach the immense with all other forces. As a result, modern cosmology suggests that all four forces would have existed as one force, a hypothetical superforce as suggested by the Theory of Everything. The Grand Unification Epoch —As the universe expands, the temperatures necessary to maintain the superforce decrease. As a result, gravity separates, leaving the electroweak and strong nuclear forces together. At this time, the electromagnetic, weak, and strong forces are identical, matching the conditions requested in the Grand Unification Theory. The Inflationary Epoch —The separation of the strong nuclear force from the electroweak force during this time is thought to have been responsible for the massive inflation of the universe. Corresponding to the steep diagonal line on the left side of Figure 23.22, the universe may have expanded by a factor of great during this time that it actually occurred faster than the speed of light! Unfortunately, there is little hope that we may be GeV, vastly greater than the limits of modern able to test the inflationary scenario directly since it occurs at energies near accelerators. or more in size. In fact, the expansion was so The Electroweak Epoch —Now separated from both gravity and the strong nuclear force, the electroweak force exists as a singular force during this time period. As stated earlier, scientists are able to create the energies at this stage in the universe’s expansion, needing only 100 GeV, as shown in Figure 23.20. W and Z bosons, as well as the Higgs boson, are released during this time. The Quark Era —During the Quark Era, the universe has expanded and temperatures have decreased to the 794 Chapter 23 • Particle Physics point at which all four fundamental forces have separated. Additionally, quarks began to take form as energies decreased. As the universe expanded, further eras took place, allowing for the existence of hadrons, leptons, and photons, the fundamental particles of the standard model. Eventually, in nucleosynthesis, nuclei would be able to form, and the basic building blocks of atomic matter could take place. Using particle accelerators, we are very much working backwards in an attempt to understand the universe. It is encouraging to see that the macroscopic conditions of the Big Bang align nicely with our submicroscopic particle theory. Check Your Understanding 19. Is there one grand unified theory or multiple grand unifying theories? a. one grand unifying theory b. multiple grand unifying theories 20. In what manner |
is considered a precursor to the Grand Unified Theory? a. The grand unified theory seeks relate the electroweak and strong nuclear forces to one another just as related energy and mass. b. The grand unified theory seeks to relate the electroweak force and mass to one another just as related energy and mass. c. The grand unified theory seeks to relate the mass and strong nuclear forces to one another just as related energy and mass. d. The grand unified theory seeks to relate gravity and strong nuclear force to one another, just as related energy and mass. 21. List the following eras in order of occurrence from the Big Bang: Electroweak Epoch, Grand Unification Epoch, Inflationary Epoch, Planck Epoch, Quark Era. a. Quark Era, Grand Unification Epoch, Inflationary Epoch, Electroweak Epoch, Planck Epoch b. Planck Epoch, Inflationary Epoch, Grand Unification Epoch, Electroweak Epoch, Quark Era c. Planck Epoch, Electroweak Epoch, Grand Unification Epoch, Inflationary Epoch, Quark Era d. Planck Epoch, Grand Unification Epoch, Inflationary Epoch, Electroweak Epoch, Quark Era 22. How did the temperature of the universe change as it expanded? a. The temperature of the universe increased. b. The temperature of the universe decreased. c. The temperature of the universe first decreased and then increased. d. The temperature of the universe first increased and then decreased. 23. Under current conditions, is it possible for scientists to use particle accelerators to verify the Grand Unified Theory? a. No, there is not enough energy. b. Yes, there is enough energy. 24. Why are particles and antiparticles made to collide as shown in this image? a. Particles and antiparticles have the same mass. b. Particles and antiparticles have different mass. c. Particles and antiparticles have the same charge. d. Particles and antiparticles have opposite charges. 25. The existence of what particles were predicted as a consequence of the electroweak theory? a. fermions b. Higgs bosons Access for free at openstax.org. leptons c. d. W+, W-, and Z0 bosons 23.3 • The Unification of Forces 795 796 Chapter 23 • Key Terms KEY TERMS boson positive carrier particle of the weak nuclear force carrier particles are identical under certain circumstances boson negative carrier particle of the weak nuclear Higgs field the field through which all fundamental force boson neutral carrier particle of the weak nuclear force annihilation the process of destruction that occurs when a particle and antiparticle interact antimatter matter constructed of antiparticles; antimatter shares most of the same properties of regular matter, with charge being the only difference between many particles and their antiparticle analogues baryon hadrons that always decay to another baryon Big Bang a gigantic explosion that threw out matter a few billion years ago bottom quark a quark flavor carrier particle a virtual particle exchanged in the transmission of a fundamental force particles travel that provides them varying mass through the transport of the Higgs boson Inflationary Epoch the rapid expansion of the universe by an incredible factor of 10−50 for the brief time from 10−35 to about 10−32 seconds lepton fundamental particles that do not feel the nuclear strong force meson hadrons that can decay to leptons and leave no hadrons pair production the creation of a particle and antiparticle, commonly an electron and positron, due to the annihilation of a photon particle physics the study of and the quest for those truly fundamental particles having no substructure charmed quark a quark flavor, which is the counterpart of pion particle exchanged between nucleons, transmitting the strange quark colliding beam head-on collisions between particles the strong nuclear force between them Planck Epoch the earliest era of the universe, before 10–43 moving in opposite directions seconds after the Big Bang color a property of quarks the relates to their interactions positron a particle of antimatter that has the properties of through the strong force a positively charged electron cyclotron accelerator that uses fixed-frequency alternating electric fields and fixed magnets to accelerate particles in a circular spiral path down quark the second lightest of all quarks Electroweak Epoch the stage before 10−11 back to 10−34 seconds after the Big Bang quantum chromodynamics the theory of color interaction between quarks that leads to understanding of the nuclear strong force quantum electrodynamics on the particle scale the theory of electromagnetism quark an elementary particle and fundamental constituent electroweak theory theory showing connections between of matter that is a substructure of hadrons EM and weak forces Feynman diagram a graph of time versus position that describes the exchange of virtual particles between subatomic particles flavor quark type gluons exchange particles of the nuclear strong force Grand Unification Epoch the time period from 10−43 to 10−34 seconds after the Big Bang, when Grand Unification Theory, in which all forces except gravity are identical, governed the universe Grand Unified Theory theory that shows unification of the strong and electroweak forces graviton hypothesized particle exchanged between two particles of mass, transmitting the gravitational force between them hadron particles composed of quarks that feel the strong and weak nuclear force Quark Era the time period from 10–11 to 10–6 seconds at which all four fundamental forces are separated and quarks begin to exit Standard Model an organization of fundamental particles and forces that is a result of quantum chromodynamics and electroweak theory strange quark the third lightest of all quarks superforce the unification of all four fundamental forces into one force synchrotron a version of a cyclotron in which the frequency of the alternating voltage and the magnetic field strength are increased as the beam particles are accelerated Theory of Everything the theory that shows unification of all four fundamental forces top quark a quark flavor up quark the lightest of all quarks weak nuclear force fundamental force responsible for Higgs boson a massive particle that provides mass to the weak bosons and provides validity to the theory that particle decay Access for free at openstax.org. SECTION SUMMARY 23.1 The Four Fundamental Forces • The four fundamental forces are gravity, the electromagnetic force, the weak nuclear force, and the strong nuclear force. • A variety of particle accelerators have been used to explore the nature of subatomic particles and to test predictions of particle theories. 23.2 Quarks • There are three types of fundamental particles—leptons, quarks, and carrier particles. • Quarks come in six flavors and three colors and occur only in combinations that produce white. • Hadrons are thought to be composed of quarks, with baryons having three quarks and mesons having a quark and an antiquark. • Known particles can be divided into three major groups—leptons, hadrons, and carrier particles (gauge bosons). • All particles of matter have an antimatter counterpart that has the opposite charge and certain other quantum CHAPTER REVIEW Concept Items 23.1 The Four Fundamental Forces 1. What forces does the inverse square law describe? the electromagnetic and weak nuclear force the electromagnetic force and strong nuclear force the electromagnetic force and gravity the strong nuclear force and gravity a. b. c. d. 2. Do the carrier particles explain the loss of mass in nuclear decay? a. no b. yes 3. What happens to the rate of voltage oscillation within a synchrotron each time the particle completes a loop? a. The rate of voltage oscillation increases as the particle travels faster and faster on each loop. b. The rate of voltage oscillation decreases as the particle travels faster and faster on each loop. c. The rate of voltage oscillation remains the same each time the particle completes a loop. d. The rate of voltage oscillation first increases and then remains constant each time the particle completes a loop. 4. Which of the four forces is responsible for ionic bonding? a. electromagnetic force Chapter 23 • Section Summary 797 numbers. These matter–antimatter pairs are otherwise very similar but will annihilate when brought together. • The strong force is carried by eight proposed particles called gluons, which are intimately connected to a quantum number called color—their governing theory is thus called quantum chromodynamics (QCD). Taken together, QCD and the electroweak theory are widely accepted as the Standard Model of particle physics. 23.3 The Unification of Forces • Attempts to show unification of the four forces are called Grand Unified Theories (GUTs) and have been partially successful, with connections proven between EM and weak forces in electroweak theory. • Unification of the strong force is expected at such high energies that it cannot be directly tested, but it may have observable consequences in the as-yet-unobserved decay of the proton. Although unification of forces is generally anticipated, much remains to be done to prove its validity. b. gravity c. strong force d. weak nuclear force 5. What type of particle accelerator uses oscillating electric fields to accelerate particles around a fixed radius track? a. LINAC b. c. SLAC d. Van de Graaff accelerator synchrotron 23.2 Quarks 6. How does the charge of an individual quark determine hadron structure? a. Since the hadron must have an integral value, the individual quarks must be combined such that the average of their charges results in the value of a quark. b. Since the hadron must have an integral value, the individual atoms must be combined such that the sum of their charges is less than zero. c. The individual quarks must be combined such that the product of their charges is equal to the total charge of the hadron structure. d. Since the hadron must have an integral value of charge, the individual quarks must be |
combined such that the sum of their charges results in an 798 Chapter 23 • Chapter Review integral value. 7. Why do leptons not feel the strong nuclear force? a. Gluons are the carriers of the strong nuclear force that interacts between quarks through color interactions, but leptons are constructed of quarks that do not have gluons. b. Gluons are the carriers of the strong nuclear force that interacts between quarks through mass interactions, but leptons are not constructed of quarks and are not massive. c. Gluons are the carriers of the strong nuclear force that interacts between quarks through mass interactions, but leptons are constructed of the quarks that are not massive. d. Gluons are the carriers of the strong nuclear force that interacts between quarks through color interactions, but leptons are not constructed of quarks, nor do they have color constituents. 8. What property commonly distinguishes antimatter from its matter analogue? a. mass b. charge c. energy speed d. 9. Can the Standard Model change as new information is gathered? a. yes b. no account for the remaining excess mass in protons compared to electrons. b. The highly energetic photons connecting the quarks account for the remaining excess mass in protons compared to electrons. c. The antiparallel orientation of the quarks present in a proton accounts for the remaining excess mass in protons compared to electrons. d. The parallel orientation of the quarks present in a proton accounts for the remaining excess mass in protons compared to electrons. 23.3 The Unification of Forces 13. Why is the unification of fundamental forces important? a. The unification of forces will help us understand fundamental structures of the universe. b. The unification of forces will help in the proof of the graviton. c. The unification of forces will help in achieving a speed greater than the speed of light. d. The unification of forces will help in studying antimatter particles. 14. Why are scientists unable to model the conditions of the universe at time periods shortly after the Big Bang? a. The amount of energy necessary to replicate the Planck Epoch is too high. b. The amount of energy necessary to replicate the Planck Epoch is too low. c. The volume of setup necessary to replicate the 10. What is the relationship between the Higgs field and the Planck Epoch is too high. Higgs boson? a. The Higgs boson is the carrier that transfers force d. The volume of setup necessary to replicate the Planck Epoch is too low. for the Higgs field. b. The Higgs field is the time duration over which the Higgs particles transfer force to the other particles. c. The Higgs field is the magnitude of momentum transferred by the Higgs particles to the other particles. 15. What role does proton decay have in the search for GUTs? a. Proton decay is a premise of a number of GUTs. b. Proton decay negates the validity of a number of GUTs. d. The Higgs field is the magnitude of torque transfers 16. What is the name for the theory of unification of all four by the Higgs particles on the other particles. 11. What were the original three flavors of quarks discovered? a. up, down, and charm b. up, down, and bottom c. up, down, and strange d. up, down, and top 12. Protons are more massive than electrons. The three quarks in the proton account for only a small amount of this mass difference. What accounts for the remaining excess mass in protons compared to electrons? a. The highly energetic gluons connecting the quarks fundamental forces? a. b. c. d. the theory of everything the theory of energy-to-mass conversion the theory of relativity the theory of the Big Bang 17. Is it easier for scientists to find evidence for the Grand Unified Theory or the Theory of Everything? Explain. a. Theory of Everything, because it requires of energy b. Theory of Everything, because it requires of energy c. Grand Unified Theory, because it requires Access for free at openstax.org. Chapter 23 • Chapter Review 799 of energy d. Grand Unified Theory, because it requires of energy Critical Thinking Items 23.1 The Four Fundamental Forces 18. The gravitational force is considered a very weak force. Yet, it is strong enough to hold Earth in orbit around the Sun. Explain this apparent disparity. a. At the level of the Earth-to-Sun distance, gravity is the strongest acting force because neither the strong nor the weak nuclear force exists at this distance. b. At the level of the Earth-to-Sun distance, gravity is the strongest acting force because both the strong and the weak nuclear force is minimal at this distance. 19. True or False—Given that their carrier particles are massless, some may argue that the electromagnetic and gravitational forces should maintain the same value at all distances from their source. However, both forces decrease with distance at a rate of a. b. false true 20. Why is a stationary target considered inefficient in a particle accelerator? a. The stationary target recoils upon particle strike, thereby transferring much of the particle’s energy into its motion. As a result, a greater amount of energy goes into breaking the particle into its constituent components. b. The stationary target contains zero kinetic energy, so it requires more energy to break the particle into its constituent components. c. The stationary target contains zero potential energy, so it requires more energy to break the particle into its constituent components. d. The stationary target recoils upon particle strike, transferring much of the particle’s energy into its motion. As a result, a lesser amount of energy goes into breaking the particle into its constituent components. 21. Compare the total strong nuclear force in a lithium atom . to the total strong nuclear force in a lithium ion a. The total strong nuclear force in a lithium atom is thrice the total strong nuclear force in a lithium ion. b. The total strong nuclear force in a lithium atom is twice the total strong nuclear force in a lithium ion. c. The total strong nuclear force in a lithium atom is the same as the total strong nuclear force in a lithium ion. d. The total strong nuclear force in a lithium atom is half the total strong nuclear force in a lithium ion. 23.2 Quarks 22. Explain why it is not possible to find a particle composed of just two quarks. a. A particle composed of two quarks will have an integral charge and a white color. Hence, it cannot exist. b. A particle composed of two quarks will have an integral charge and a color that is not white. Hence, it cannot exist. c. A particle composed of two quarks will have a fractional charge and a white color. Hence, it cannot exist. d. A particle composed of two quarks will have a fractional charge and a color that is not white. Hence, it cannot exist. 23. Why are mesons considered unstable? a. Mesons are composites of two antiparticles that quickly annihilate each other. b. Mesons are composites of two particles that quickly annihilate each other. c. Mesons are composites of a particle and antiparticle that quickly annihilate each other. d. Mesons are composites of two particles and one antiparticle that quickly annihilate each other. 24. Does antimatter have a negative mass? a. No, antimatter does not have a negative mass. b. Yes, antimatter does have a negative mass. 25. What similarities exist between the Standard Model and the periodic table of elements? a. During their invention, both the Standard Model and the periodic table organized material by mass. b. At the times of their invention, both the Standard Model and the periodic table organized material by charge. c. At the times of their invention, both the Standard Model and the periodic table organized material by interaction with other available particles. d. At the times of their invention, both the Standard Model and the periodic table organized material by size. 26. How were particle collisions used to provide evidence of the Higgs boson? 800 Chapter 23 • Chapter Review a. Because some particles do not contain the Higgs boson, the collisions of such particles will cause their destruction. b. Because only the charged particles contain the Higgs boson, the collisions of such particles will cause their destruction and will expel the Higgs boson. c. Because all particles with mass contain the Higgs boson, the collisions of such particles will cause their destruction and will absorb the Higgs boson. d. Because all particles with mass contain the Higgs boson, the collisions of such particles will cause their destruction and will expel the Higgs boson. 27. Explain how the combination of a quark and antiquark can result in the creation of a hadron. a. The combination of a quark and antiquark can result in a particle with an integer charge and color of white, therefore satisfying the properties for a hadron. b. The combination of a quark and antiquark must result in a particle with a negative charge and color of white, therefore satisfying the properties for a hadron. c. The combination of a quark and antiquark can result in a particle with an integer charge and color that is not white, therefore satisfying the properties for a hadron. d. The combination of a quark and antiquark can result in particle with a fractional charge and color that is not white, therefore satisfying the properties for a hadron. 23.3 The Unification of Forces 28. Why does the strength of the strong force diminish under high-energy conditions? a. Under high-energy conditions, particles interacting under the strong force will be compressed closer together. As a result, the force between them will decrease. b. Under high-energy conditions, particles interacting under the strong force will start oscillating. As a result, the force between them will increase. c. Under high-energy conditions, particles interacting under the strong force will have high velocity. As a result, the force between them will decrease. d. Under high-energy conditions, particles interacting under the stro |
ng force will start moving randomly. As a result, the force between them will decrease. 29. If some unknown cause of the red shift, such as light becoming tiredfrom traveling long distances through empty space, is discovered, what effect would there be on cosmology? a. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are moving toward one another. b. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that the galaxies are moving away from one another. c. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are neither moving away from nor moving toward one another. d. The effect would be substantial, as the Big Bang is based on the idea that the red shift is evidence that galaxies are sometimes moving away from and sometimes moving toward one another. 30. How many molecules of water are necessary if scientists -yr estimate of proton decay wanted to check the within the course of one calendar year? a. b. c. d. 31. As energy of interacting particles increases toward the theory of everything, the gravitational force between them increases. Why does this occur? a. As energy increases, the masses of the interacting particles will increase. b. As energy increases, the masses of the interacting particles will decrease. c. As energy increases, the masses of the interacting particles will remain constant. d. As energy increases, the masses of the interacting particles starts changing (increasing or decreasing). As a result, the gravitational force between the particles will increase. Access for free at openstax.org. Performance Task 23.3 The Unification of Forces 32. Communication is an often overlooked and useful skill for a scientist, especially in a competitive field where financial resources are limited. Scientists are often required to explain their findings or the relevance of their work to agencies within the government in order to maintain funding to continue their research. Let’s say you are an ambitious young particle physicist, heading an expensive project, and you need to justify its existence to the appropriate funding agency. Write a brief paper (about one page) explaining why molecularlevel structure is important in the functioning of designed materials in a specific industry. TEST PREP Multiple Choice 23.1 The Four Fundamental Forces 33. Which of the following is not one of the four fundamental forces? a. gravity friction b. strong nuclear c. d. electromagnetic 34. What type of carrier particle has not yet been found? a. gravitons b. bosons c. bosons d. pions 35. What effect does an increase in electric potential have It increases accelerating capacity. It decreases accelerating capacity. on the accelerating capacity of a Van de Graaff generator? a. b. c. The accelerating capacity of a Van de Graaff generator is constant regardless of electric potential. Chapter 23 • Test Prep 801 • First, think of an industry where molecular-level structure is important. • Research what materials are used in that industry as well as what are the desired properties of the materials. • What molecular-level characteristics lead to what properties? One example would be explaining how flexible but durable materials are made up of long-chained molecules and how this is useful for finding more environmentally friendly alternatives to plastics. Another example is explaining why electrically conductive materials are often made of metal and how this is useful for developing better batteries. and the electromagnetic force 23.2 Quarks 37. To what color must quarks combine for a particle to be constructed? a. black b. green red c. d. white 38. What type of hadron is always constructed partially of an antiquark? a. baryon b. lepton c. meson d. photno 39. What particle is typically released when two particles annihilate? a. graviton b. antimatter c. pion d. photno d. Van de Graaff generators do not have the capacity 40. Which of the following categories is not one of the three to accelerate particles. 36. What force or forces exist between a proton and a second proton? a. The weak electrostatic force and strong magnetic force main categories of the Standard Model? a. gauge bosons b. hadrons leptons c. d. quarks b. The weak electrostatic and strong gravitational 41. Analysis of what particles began the search for the Higgs force c. The weak frictional force and strong gravitational force d. The weak nuclear force, the strong nuclear force, boson? a. W and Z bosons b. up and down quarks c. mesons and baryons 802 Chapter 23 • Test Prep d. neutrinos and photons 44. After the Big Bang, what was the first force to separate 42. What similarities exist between the discovery of the quark and the discovery of the neutron? a. Both the quark and the neutron were discovered by launching charged particles through an unknown structure and observing the particle recoil. b. Both the quark and the neutron were discovered by launching electrically neutral particles through an unknown structure and observing the particle recoil. c. Both quarks and neutrons were discovered by studying their deflection under an electric field. 23.3 The Unification of Forces 43. Which two forces were first combined, signifying the eventual desire for a Grand Unified Theory? a. electric force and magnetic forces b. electric force and weak nuclear force c. gravitational force and the weak nuclear force d. electroweak force and strong nuclear force Short Answer 23.1 The Four Fundamental Forces 47. Why do people tend to be more aware of the gravitational and electromagnetic forces than the strong and weak nuclear forces? a. The gravitational and electromagnetic forces act at short ranges, while strong and weak nuclear forces act at comparatively long range. b. The strong and weak nuclear forces act at short ranges, while gravitational and electromagnetic forces act at comparatively long range. c. The strong and weak nuclear forces act between all objects, while gravitational and electromagnetic forces act between smaller objects. d. The strong and weak nuclear forces exist in outer space, while gravitational and electromagnetic forces exist everywhere. 48. What fundamental force is responsible for the force of friction? a. b. c. the electromagnetic force the strong nuclear force the weak nuclear force 49. How do carrier particles relate to the concept of a force field? a. Carrier particles carry mass from one location to another within a force field. b. Carrier particles carry force from one location to another within a force field. Access for free at openstax.org. from the others? a. electromagnetic force b. gravity c. d. weak nuclear force strong nuclear force 45. What is the name of the device used by scientists to check for proton decay? the cyclotron a. the Large Hadron Collider b. the Super-Kamiokande c. the synchrotron d. 46. How do Feynman diagrams suggest the Grand Unified Theory? a. The electromagnetic, weak, and strong nuclear forces all have similar Feynman diagrams. b. The electromagnetic, weak, and gravitational forces all have similar Feynman diagrams. c. The electromagnetic, weak, and strong forces all have different Feynman diagrams. c. Carrier particles carry charge from one location to another within a force field. d. Carrier particles carry volume from one location to another within a force field. 50. Which carrier particle is transmitted solely between nucleons? a. graviton b. photon c. pion d. W and Z bosons 51. Two particles of the same mass are traveling at the same speed but in opposite directions when they collide headon. What is the final kinetic energy of this two-particle system? a. b. c. zero d. infinite the sum of the kinetic energies of the two particles the product of the kinetic energies of the two particles 52. Why do colliding beams result in the location of smaller particles? a. Colliding beams create energy, allowing more energy to be used to separate the colliding particles. b. Colliding beams lower the energy of the system, so it requires less energy to separate the colliding particles. c. Colliding beams reduce energy loss, so less energy is required to separate colliding particles. c. The four fundamental forces are represented by d. Colliding beams reduce energy loss, allowing more their carrier particles, the leptons. energy to be used to separate the colliding particles. d. The four fundamental forces are represented by their carrier particles, the quarks. Chapter 23 • Test Prep 803 23.2 Quarks 53. What two features of quarks determine the structure of a particle? a. b. c. d. the color and charge of individual quarks the color and size of individual quarks the charge and size of individual quarks the charge and mass of individual quarks 54. What fundamental force does quantum chromodynamics describe? the weak nuclear force a. the strong nuclear force b. the electromagnetic force c. the gravitational force d. 55. Is it possible for a baryon to be constructed of two quarks and an antiquark? a. Yes, the color of the three particles would be able to sum to white. b. No, the color of the three particles would not be able to sum to white. 56. Can baryons be more massive than mesons? a. no b. yes 57. If antimatter exists, why is it so difficult to find? a. There is a smaller amount of antimatter than matter in the universe; antimatter is quickly annihilated by its matter analogue. b. There is a smaller amount of matter than antimatter in the universe; matter is annihilated by its antimatter analogue. c. There is a smaller amount of antimatter than matter in universe; antimatter and its matter analogue coexist. d. There is a smaller amount of matter than antimatter in the universe; matter and its antimatter analogue coexist. 58. Does a neutron have an antimatter counterpart? a. No, the antineutron does not exist. b. Yes, the antineutron does exist. 59. How are the |
four fundamental forces incorporated into the Standard Model of the atom? a. The four fundamental forces are represented by their carrier particles, the electrons. 60. Which particles in the Standard Model account for the majority of matter with which we are familiar? a. particles in fourth column of the Standard Model b. particles in third column of the Standard Model c. particles in the second column of the Standard Model d. particles in the first column of the Standard Model 61. How can a particle gain mass by traveling through the Higgs field? a. The Higgs field slows down passing particles; the decrease in kinetic energy is transferred to the particle’s mass. b. The Higgs field accelerates passing particles; the decrease in kinetic energy is transferred to the particle’s mass. c. The Higgs field slows down passing particles; the increase in kinetic energy is transferred to the particle’s mass. d. The Higgs field accelerates passing particles; the increase in kinetic energy is transferred to the particle’s mass. 62. How does mass-energy conservation relate to the Higgs field? a. The increase in a particle’s energy when traveling through the Higgs field is countered by its increase in mass. b. The decrease in a particle’s kinetic energy when traveling through the Higgs field is countered by its increase in mass. c. The decrease in a particle’s energy when traveling through the Higgs field is countered by its decrease in mass. d. The increase in a particle’s energy when traveling through the Higgs field is countered by its decrease in mass. 23.3 The Unification of Forces 63. Why do scientists believe that the strong nuclear force and the electroweak force will combine under high energies? a. The electroweak force will have greater strength. b. The strong nuclear force and electroweak force will achieve the same strength. c. The strong nuclear force will have greater strength. b. The four fundamental forces are represented by 64. At what energy will the strong nuclear force their carrier particles, the gauge bosons. theoretically unite with the electroweak force? 804 Chapter 23 • Test Prep a. b. c. d. 65. While we can demonstrate the unification of certain forces within the laboratory, for how long were the four forces naturally unified within the universe? a. b. c. d. 66. How does the search for the Grand Unified Theory help test the standard cosmological model? a. Scientists are increasing energy in the lab that models the energy in earlier, denser stages of the universe. b. Scientists are increasing energy in the lab that models the energy in earlier, less dense stages of the universe. c. Scientists are decreasing energy in the lab that models the energy in earlier, denser stages of the universe. d. Scientists are decreasing energy in the lab that Extended Response 23.1 The Four Fundamental Forces 69. If the strong attractive force is the greatest of the four fundamental forces, are all masses fated to combine together at some point in the future? Explain. a. No, the strong attractive force acts only at incredibly small distances. As a result, only masses close enough to be within its range will combine. b. No, the strong attractive force acts only at large distances. As a result, only masses far enough apart will combine. c. Yes, the strong attractive force acts at any distance. As a result, all masses are fated to combine together at some point in the future. d. Yes, the strong attractive force acts at large distances. As a result, all masses are fated tocombine together at some point in the future. 70. How does the discussion of carrier particles relate to the concept of relativity? a. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel more slowly than the speed of sound. b. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel at or near the speed of light. Access for free at openstax.org. models the energy in earlier, less dense stages of the universe. 67. Why does finding proof that protons do not decay not disprove all GUTs? a. Proton decay is not a premise of all GUTs, and current GUTs can be amended in response to new findings. b. Proton decay is a premise of all GUTs, but current GUTs can be amended in response to new findings. 68. When accelerating elementary particles in a particle accelerator, they quickly achieve a speed approaching the speed of light. However, as time continues, the particles maintain this speed yet continue to increase their kinetic energy. How is this possible? a. The speed remains the same, but the masses of the particles increase. b. The speed remains the same, but the masses of the particles decrease. c. The speed remains the same, and the masses of the particles remain the same. d. The speed and masses will remain the same, but temperature will increase. c. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel at or near the speed of sound. d. Calculations of mass and energy during their transfer are relativistic, because carrier particles travel faster than the speed of light. 71. Why are synchrotrons constructed to be very large? a. By using a large radius, high particle velocities can be achieved using a large centripetal force created by large electromagnets. b. By using a large radius, high particle velocities can be achieved without a large centripetal force created by large electromagnets. c. By using a large radius, the velocities of particles can be reduced without a large centripetal force created by large electromagnets. d. By using a large radius, the acceleration of particles can be decreased without a large centripetal force created by large electromagnets. 23.2 Quarks 72. In this image, how does the emission of the gluon cause the down quark to change from a red color to a green color? Chapter 23 • Test Prep 805 a. b. c. d. If the velocity of the neutrino is known, then the upper limit on mass of the neutrino can be set. If only the kinetic energy of the neutrino is known, then the upper limit on mass of the neutrino can be set. If either the velocity or the kinetic energy is known, then the upper limit on the mass of the neutrino can be set. If both the kinetic energy and the velocity of the neutrino are known, then the upper limit on the mass of the neutrino can be set. 76. The term force carrier particleis shorthand for the scientific term vector gauge boson. From that perspective, can the Higgs boson truly be considered a force carrier particle? a. No, the mass quality provided by the Higgs boson is a scalar quantity. b. Yes, the mass quality provided by the Higgs boson results in a change of particle’s direction. 23.3 The Unification of Forces 77. If a Grand Unified Theory is proven and the four forces are unified, it will still be correct to say that the orbit of the Moon is determined by the gravitational force. Explain why. a. Gravity will not be a property of the unified force. b. Gravity will be one property of the unified force. c. Apart from gravity, no other force depends on the mass of the object. d. Apart from gravity, no other force can make an a. The emitted red gluon is made up of a green and a red color. As a result, the down quark changes from a red color to a green color. b. The emitted red gluon is made up of an anti-green and an anti-red color. As a result, the down quark changes from a red color to a green color. c. The emitted red gluon is made up of a green and an anti-red color. As a result, the down quark changes from a red color to a green color. d. The emitted red gluon is made up of an anti-green and a red color. As a result, the down quark changes from a red color to a green color. 73. Neutrinos are much more difficult for scientists to find when compared to other hadrons and leptons. Why is this? a. Neutrinos are hadrons, and they lack charge. b. Neutrinos are not hadrons, and they lack charge. c. Neutrinos are hadrons, and they have positive charge. d. Neutrinos are not hadrons, and they have a positive object move in a fixed orbit. charge. 74. What happens to the masses of a particle and its antiparticle when the two annihilate at low energies? a. The masses of the particle and antiparticle are transformed into energy in the form of photons. b. The masses of the particle and antiparticle are converted into kinetic energy of the particle and antiparticle respectively. c. The mass of the antiparticle is converted into kinetic energy of the particle. d. The mass of the particle is converted into radiation energy of the antiparticle. 75. When a star erupts in a supernova explosion, huge numbers of electron neutrinos are formed in nuclear reactions. Such neutrinos from the 1987A supernova in the relatively nearby Magellanic Cloud were observed within hours of the initial brightening, indicating that they traveled to earth at approximately the speed of light. Explain how this data can be used to set an upper limit on the mass of the neutrino. 78. As the universe expanded and temperatures dropped, the strong nuclear force separated from the electroweak force. Is it likely that under cooler conditions, the force of electricity will separate from the force of magnetism? a. No, the electric force relies on the magnetic force and vice versa. b. Yes, the electric and magnetic forces can be separated from each other. 79. Two pool balls collide head-on and stop. Their original kinetic energy is converted to heat and sound. Given that this is not possible for particles, what happens to their converted energy? a. The kinetic energy is converted into relativistic potential energy, governed by the equation . b. The kinetic energy is converted into relativistic mass, governed by the equation . c. The kinetic energy is converted into relativistic potential energy, governed by the equation . 806 Chapter 23 • Test Prep d. Their kinetic energy is converted into relativistic mass, governed |
by the equation . Access for free at openstax.org. Appendix A • Reference Tables 807 APPENDIX A Reference Tables Figure A1 Periodic Table of Elements Prefix Symbol Value Prefix Symbol Value tera giga T G mega M kilo hecto k h deka da 1012 deci d c centi milli m micro nano pico µ n p 109 106 103 102 101 10–1 10–2 10–3 10–6 10–9 10–12 Table A1 Metric Prefixes for Powers of Ten and Their Symbols 808 Appendix A • Reference Tables Prefix Symbol Value Prefix Symbol Value ___ 100 femto f 10–15 Table A1 Metric Prefixes for Powers of Ten and Their Symbols Entity Abbreviation Name Fundamental units Length Mass Time Current Supplementary unit Angle Derived units Force m kg s A rad Energy Power Pressure Frequency Electronic potential Capacitance Charge Resistance Magnetic field meter kilogram second ampere radian newton joule watt pascal hertz volt farad coulomb ohm tesla Nuclear decay rate becquerel Table A2 SI Units Length 1 inch (in.) = 2.54 cm (exactly) 1 foot (ft) = 0.3048 m 1 mile (mi) = 1.609 km Table A3 Selected British Units Access for free at openstax.org. Appendix A • Reference Tables 809 Force 1 pound (lb) = 4.448 N Energy 1 British thermal unit (Btu) = 1.055 × 103 J Power 1 horsepower (hp) = 746 W Pressure 1 lb/in2 = 6.895 × 103 Pa Table A3 Selected British Units Length 1 light year (ly) = 9.46 × 1015 m 1 astronomical unit (au) = 1.50 × 1011 m 1 nautical mile = 1.852 km 1 angstrom(Å) = 10-10 m Area 1 acre (ac) = 4.05 × 103 m2 1 square foot (ft2) 9.29 × 10-2 m3 1 barn (b) = 10-28 m2 Volume 1 liter (L) = 10-3 m3 1 U.S. gallon (gal) = 3.785 × 10-3 m3 Mass 1 solar mass = 1.99 × 1030 kg 1 metric ton = 103 kg 1 atomic mass unit (u) = 1.6605 × 10-27 kg Time 1 year (y) = 3.16 × 107 s 1 day (d) = 86,400 s Speed 1 mile per hour (mph) = 1.609 km / h 1 nautical mile per hour (naut) = 1.852 km / h Angle 1 degree (°) = 1.745x10-2 rad 1 minute of arc (') = 1 / 60 degree 1 second of arc ('') = 1 / 60 minute of arc 1 grad = 1.571 × 10-2 rad Table A4 Other Units 810 Appendix A • Reference Tables Energy 1 kiloton TNT (kT) = 4.2 × 1012 J 1 kilowatt hour (kW h) = 3.60 × 106J 1 food calorie (kcal) = 4186 J 1 calorie (cal) = 4.186 J 1 electron volt (cV) = 1.60 × 10-19 J Pressure 1 atmosphere (atm) = 1.013 × 105 Pa 1 millimeter of mercury (mm Hg) = 133.3 Pa 1 torricelli (torr) = 1 mm Hg = 133.3 Pa Nuclear decay rate 1 curie (Ci) = 3.70 × 1010 Bq Table A4 Other Units Circumference of a circle with radius ror diameter d Area of a circle with radius ror diameter d Area of a sphere with radius r Volume of a sphere with radius r Table A5 Useful formulae Symbol Meaning Best Value Approximate Value c G NA k R σ k Speed of light in vacuum Gravitational constant Avogadro’s number Boltzmann’s constant Gas constant Stefan-Boltzmann Constant Coulomb force constant Table A6 Important Constants Access for free at openstax.org. Symbol Meaning Best Value Approximate Value Appendix A • Reference Tables 811 qe ε0 µ0 h Charge on electron Permittivity of free space Permeability of free space Planck’s constant Table A6 Important Constants Alpha Beta Gamma Delta Epsilon Zeta Eta Theta Iota Kappa Lambda Mu Nu Xi Omicron Pi Rho Sigma Tau Table A7 The Greek Alphabet 812 Appendix A • Reference Tables Upsilon Phi Chi Psi Omega Table A7 The Greek Alphabet Sun mass average radius 1.99 × 1030 kg 6.96 × 108 m Earth-sun distance (average) 1.496 × 1011 m Earth mass 5.9736 × 1024 kg average radius orbital period Moon mass average radius orbital period (average) 6.376 × 106 m 3.16 × 107 s 7.35 × 1022 kg 1.74 × 106 s 2.36 × 106 s Earth-moon distance (average) 3.84 × 108 m Table A8 Solar System Data Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 n 1H 1.008 665 β– 1.007 825 99.985% 2H or D 2.014 102 0.015% 10.37 min 3H or T 3.016 050 β– 12.33 y 3He 3.016 030 1.38 × 10−4 % 0 1 2 neutron Hydrogen Deuterium Tritium Helium 1 1 2 3 3 Table A9 Atomic Masses and Decay Access for free at openstax.org. Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A • Reference Tables 813 3 4 5 6 7 8 9 10 11 Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium 4 6 7 7 9 10 11 11 12 13 14 13 14 15 15 16 18 18 19 20 22 22 23 24 Table A9 Atomic Masses and Decay 4He 4.002 603 ≈100% 6Li 7Li 7Be 9Be 10B 11B 11C 12C 13C 14C 12N 13N 14N 15O 16O 18O 18F 19F 20Ne 22Ne 22Na 23Na 24Na 6.015 121 7.5% 7.016 003 92.5% 7.016 928 EC 53.29 d 9.012 182 100% 10.012 937 19.9% 11.009 305 80.1% 11.011 432 EC, β+ 12.000 000 98.90% 13.003 355 1.10% 14.003 241 13.005 738 β– β+ 14.003 074 99.63% 15.000 108 0.37% 5730 y 9.96 min 15.003 065 EC, β+ 122 s 15.994 915 99.76% 17.999 160 0.200% 18.000 937 EC, β+ 1.83 h 18.998 403 100% 19.992 435 90.51% 21.991 383 9.22% 21.994 434 β+ 22.989 767 100% 23.990 961 β– 2.602 y 14.96 h 814 Appendix A • Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 24Mg 23.985 042 78.99% 27Al 28Si 31Si 31P 32P 32S 35S 35Cl 37Cl 40Ar 39K 40K 40Ca 45Sc 48Ti 51V 52Cr 55Mn 56Fe 59Co 60Co 58Ni 26.981 539 100% 27.976 927 92.23% 2.62h 30.975 362 β– 30.973 762 100% 31.973 907 β– 31.972 070 95.02% 34.969 031 β– 34.968 852 75.77% 36.965 903 24.23% 39.962 384 99.60% 38.963 707 93.26% 14.28 d 87.4 d 39.963 999 0.0117%, EC, β– 1.28 × 109 y 39.962 591 96.94% 44.955 910 100% 47.947 947 73.8% 50.943 962 99.75% 51.940 509 83.79% 54.938 047 100% 55.934 939 91.72% 58.933 198 100% 59.933 819 β– 5.271 y 57.935 346 68.27% 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Magnesium Aluminum Silicon Phosphorus Sulfur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel 24 27 28 31 31 32 32 35 35 37 40 39 40 40 45 48 51 52 55 56 59 60 58 Table A9 Atomic Masses and Decay Access for free at openstax.org. Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A • Reference Tables 815 60 63 64 66 69 72 74 75 80 79 84 85 86 88 90 89 90 90 93 98 98 Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium 60Ni 63Cu 65Cu 64Zn 66Zn 69Ga 72Ge 74Ge 75As 80Se 79Br 84Kr 85Rb 86Sr 88Sr 90Sr 89Y 90Y 90Zr 93Nb 98Mo 59.930 788 26.10% 62.939 598 69.17% 64.927 793 30.83% 63.929 145 48.6% 65.926 034 27.9% 68.925 580 60.1% 71.922 079 27.4% 73.921 177 36.5% 74.921 594 100% 79.916 520 49.7% 78.918 336 50.69% 83.911 507 57.0% 84.911 794 72.17% 85.909 267 9.86% 87.905 619 82.58% 89.907 738 β– 88.905 849 100% 89.907 152 β– 89.904 703 51.45% 92.906 377 100% 97.905 406 24.13% 98Tc 97.907 215 β– Ruthenium 102 102Ru 101.904 348 31.6% 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 Table A9 Atomic Masses and Decay 28.8 y 64.1 h 4.2 × 106 y 816 Appendix A • Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon Cesium Barium Lanthanum Cerium 103 106 107 109 114 115 120 121 130 127 131 132 136 133 134 137 138 139 140 Praseodymium 141 Neodymium Promethium Samarium 142 145 152 Table A9 Atomic Masses and Decay Access for free at openstax.org. 103Rh 106Pd 107Ag 109Ag 114Cd 102.905 500 100% 105.903 478 27.33% 106.905 092 51.84% 108.904 757 48.16% 113.903 357 28.73% 115In 114.903 880 95.7%, β– 120Sn 121Sb 119.902 200 32.59% 120.903 821 57.3% 130Te 129.906 229 33.8%, β– 4.4 × 1014 y 2.5 × 1021 y 127I 131I 132Xe 136Xe 133Cs 134Cs 137Ba 138Ba 139La 140Ce 141Pr 142Nd 145Pm 152Sm 126.904 473 100% 130.906 114 β– 8.040 d 131.904 144 26.9% 135.907 214 8.9% 132.905 429 100% 133.906 696 EC, β– 2.06 y 136.905 812 11.23% 137.905 232 71.70% 138.906 346 99.91% 139.905 433 88.48% 140.907 647 100% 141.907 719 27.13% 144.912 743 EC, α 17.7 y 151.919 729 26.7% Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A • Reference Tables 817 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutecium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium 153 158 159 164 165 166 169 174 175 180 181 184 187 191 192 191 193 195 197 198 199 202 205 Table A9 Atomic Masses and Decay 153Eu 158Gd 159Tb 164Dy 165Ho 166Ho 152.921 225 52.2% 157.924 099 24.84% 158.925 342 100% 163.929 171 28.2% 164.930 319 100% 165.930 290 33.6% 169Tm 168.934 212 100% 174Yb 175Lu 180Hf 181Ta 184W 173.938 859 31.8% 174.940 770 97.41% 179.946 545 35.10% 180.947 992 99.98% 183.950 928 30.67% 187Re 186.955 744 62.6%, β– 190.960 920 β– 191.961 467 41.0% 190.960 584 37.3% 192.962 917 62.7% 194.964 766 33.8% 196.966 543 100% 191Os 192Os 191Ir 193Ir 195Pt 197Au 198Au 199Hg 202Hg 205Tl 4.6 × 1010y 15.4 d 197.968 217 β– 2.696 d 198.968 253 16.87% 201.970 617 29.86% 204.974 401 70.48% 818 Appendix A • Reference Tables Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 82 Lead 206 207 208 210 211 212 209 211 210 218 222 223 226 227 228 232 206Pb 207Pb 208Pb 205.974 440 24.1% 206.975 872 22.1% 207.976 627 52.4% 210Pb 209.984 163 α, β– 211Pb 212Pb 209Bi 211Bi 210Po 218At 210.988 735 211.991 871 β– β– 208.980 374 100% 210.987 255 α, β– 209.982 848 α 218.008 684 α, β– 222Rn 222.017 570 α 223Fr 223.019 733 α, β– 226Ra 227Ac 228Th 226.025 402 α 227.027 750 α, β– 228.028 715 α 232Th 232.038 054 100%, α Bismuth Polonium Astatine Radon Francium 2 Radium Actinium Thorium Protactinium 231 231Pa 231.035 880 Uranium 233 233U 233.039 628 α α 235 236 235U 235.043 924 0.720%, α 236U 236.045 562 α 83 84 85 86 87 88 89 90 91 92 Table A9 Atomic Masses and Decay Access for free at openstax.org. 22.3 y 36.1 min 10.64 h 2.14 min 138.38 d 1.6 s 3.82 d 21.8 min 1.60 × 103 y 21.8 y 1.91 y 1.41 × 1 |
010 y 3.28 × 104 y 1.59 × 103 y 7.04 × 108 y 2.34 × 107 y Atomic number, Z Name Atomic Mass Number, A Symbol Atomic Mass (u) Percent Abundance or Decay Mode Halflife, t1/2 Appendix A • Reference Tables 819 β– β– α α α 238 239 239 Neptunium 238.050 784 99.2745%, α 238U 239U 239.054 289 239Np 239.052 933 Plutonium 239 239Pu 239.052 157 α Americium 243 243Am 243.061 375 α, fission Curium 245 245Cm 245.065 483 Berkelium 245 Californium 249 Einsteinium Fermium 254 253 Mendelevium 255 Nobelium Lawrencium 255 257 247Bk 249Cf 254Es 253Fm 255Md 255No 257Lr 247.070 300 249.074 844 254.088 019 α, β– 253.085 173 EC, α 255.091 081 EC, α 255.093 260 EC, α 257.099 480 EC, α Rutherfordium 261 261Rf 261.108 690 α Dubnium Seaborgium Bohrium Hassium Meitnerium 262 263 262 264 266 262Db 263Sg 262Bh 264Hs 266Mt 262.113 760 α, fission 263.11 86 α, fission 262.123 1 264.128 5 266.137 8 α α α 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 108 Table A9 Atomic Masses and Decay 4.47 × 109 y 23.5 min 2.355 d 2.41 × 104 y 7.37 × 103 y 8.50 × 103 y 1.38 × 103 y 351 y 276 d 3.00 d 27 min 3.1 min 0.646 s 1.08 mim 34 s 0.8 s 0.102 s 0.08 ms 3.4 ms 820 Appendix A • Reference Tables Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent 3H 14C 13N 12.33 y 5730 y 9.96 min 22Na 2.602 y 32P 35S 36Ci 40K 43K 45Ca 51Cr 14.28 d 87.4 d 3.00 × 105 y 1.28 × 109 y 22.3 h 165 d 27.70 d 52Mn 5.59d β– β– β+ β+ β– β– β– β– β– β– EC β+ 100% 100% 100% 90% γ 1.27 100% 0.0186 0.156 1.20 1.20 1.71 0.167 0.710 1.31 0.827 100% 100% 100% 89% 87% γs 0.373 0.618 γ 0.320 0.257 100% 3.69 28% γ s 1.33 52Fe 8.27 h β+ 1.80 43% 1.43 0.169 0.378 59Fe 44.6 d β– s 60Co 5.271 y β– 65Zn 67Ga 244.1 d 78.3 h EC EC 0.273 0.466 0.318 45% 55% γ s 1.10 1.29 100% γ s 1.17 1.33 γ 1.12 γ s 0.0933 0.185 0.300 Table A10 Selected Radioactive Isotopes Access for free at openstax.org. 87% 87% 10% 28% 28% 43% 43% 57% 43% 100% 100% 51% 70% 35% 19% Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent Appendix A • Reference Tables 821 75Se 118.5 d EC 86Rb 18.8 d β– s 85Sr 90Sr 90Y 64.8 d 28.8 y 64.1 h 99mTc 6.02 h 113mIn 99.5 min 123I 131I 13.0 h 8.040 d EC β– β– IT IT EC β– s 0.69 1.77 0.546 2.28 0.248 0.607 others others γ s 0.121 0.136 0.265 0.280 others γ 1.08 20% 65% 68% 20% 9% γ 0.514 1 100% 9% 91% 100% 100% γ γ γ 0.142 0.392 0.159 7% 93% γ s 0.364 others 129Cs 32.3 h EC γ s 0.0400 0.372 0.411 others γ 0.662 95% 5% ≈100% γ s 0.030 137Cs 30.17 y β– s 140Ba 12.79 d β– Table A10 Selected Radioactive Isotopes 0.511 1.17 1.035 100% 100% ≈100% 85% 35% 32% 25% 95% 25% 822 Appendix A • Reference Tables Isotope t1/2 Decay Mode Energy(MeV) Percent T-Ray Energy(MeV) Percent 198Au 197Hg 210Po 226Ra 235U 238U 2.696 d 64.1 h 138.38 d 1.60 × 103 y β– EC α αs 7.038 × 108 y α 4.468 × 109 y αs 1.161 ≈100% 100% 5% 95% 5.41 4.68 4.87 4.68 4.22 4.27 0.044 0.537 others 0.412 0.0733 γ γ 65% 24% ≈100% 100% γ 0.186 1 100% ≈100% γ s Numerous <0.400% γ 0.050 23% 23% 77% 237Np 2.14 × 106 y αs numerous γ s numerous <0.250% 239Pu 2.41 × 104 y αs 4.96 (max.) 5.19 5.23 5.24 11% 15% 73% γ s 7.5 × 10-5 0.013 0.052 others 73% 15% 15% 243Am 7.37 × 103 y αs Max. 5.44 γ s 0.075 others 88% 11% 5.37 5.32 others Table A10 Selected Radioactive Isotopes Symbol Meaning Best Value Approximate Value me Electron mass 9.10938291(40) × 10–31 kg 9.11 × 10–31 kg Table A11 Submicroscopic masses Access for free at openstax.org. Appendix A • Reference Tables 823 Symbol Meaning Best Value Approximate Value mp mn u Proton mass 1.672621777(74) × 10–27 kg 1.6726 × 10–27 kg Neutron mass 1.674927351(74) × 10–27 kg 1.6749 × 10–27 kg Atomic mass unit 1.660538921(73) × 10–27 kg 1.6605 × 10–27 kg Table A11 Submicroscopic masses Substance p(kg/m3) Substance p(kg/m3) Air 1.29 Air (at 20°C and Atmospheric pressure) 1.20 Iron Lead 7.86 × 103 11.3 × 103 Aluminum Benzene Brass Copper Ethyl alcohol Fresh water Glycerin Gold Helium gas Hydrogen gas Ice 2.70 × 103 Mercury 13.6 × 103 0.879 × 103 Nitrogen gas 1.25 8.4 × 103 Oak 0.710 × 103 8.92 × 103 Osmium 22.6 × 103 0.806 × 103 Oxygen gas 1.43 1.00 × 103 Pine 0.373 × 103 1.26 × 103 Platinum 21.4 × 103 1.93 × 103 Seawater 1.03 × 103 1.79 × 10–1 Silver 10.5 × 103 8.99 × 10–2 Tin 7.30 × 103 0.917 × 103 Uranium 18.7 × 103 Table A12 Densities of common substances (including water at various temperatures) Substance Specific Heat (J/kg °C) Substance Specific Heat (J/kg °C) Elemental solids Other solids Aluminum Beryllium Cadmium 900 1830 230 Brass Glass 380 837 Ice (–5 °C) 2090 Table A13 Specific heats of common substances 824 Appendix A • Reference Tables Substance Specific Heat (J/kg °C) Substance Specific Heat (J/kg °C) Copper Germanium Gold Iron Lead Silicon Silver 387 322 129 448 128 703 234 860 1700 Marble Wood Liquids Alcohol (ethyl) 2400 Mercury 140 Water (15 °C) 4186 Gas Steam (100 °C) 2010 Note: To convert values to units of cal/g °C, divide by 4186 Table A13 Specific heats of common substances Substance Melting Point (°C) Latent Heat of Fusion (J/kg) Boiling Point (°C) Latent Heat of Vaporization (J/kg) Helium –272.2 Oxygen –218.79 Nitrogen –209.97 Ethyl Alcohol –114 Water 0.00 Sulfur 119 Lead 327.3 Aluminum 660 Silver 960.80 Gold 1063.00 Copper 1083 5.23 × 103 1.38 × 104 2.55 × 104 1.04 × 105 3.33 × 105 3.81 × 104 3.97 × 105 3.97 × 105 8.82 × 104 6.44 × 104 1.34 × 105 –268.93 –182.97 –195.81 78 100.00 444.60 1750 2516 2162 2856 2562 2.09 × 104 2.13 × 105 2.01 × 105 8.54 × 105 2.26 × 106 2.90 × 105 8.70 × 105 1.05 × 107 2.33 × 106 1.58 × 106 5.06 × 106 Table A14 Heats of fusion and vaporization for common substances Access for free at openstax.org. Appendix A • Reference Tables 825 Materials (Solids) Average Linear Expansion Coefficient (a)(°C)–1 Material (Liquids and Gases) Average Volume Expansion Coefficient (B)(°C)–1 Aluminum 24 × 10–6 Acetone 1.5 × 10–4 Brass and Bronze 19 × 10–6 Concrete 12 × 10–6 Copper 17 × 10–6 Glass (ordinary) 9 × 10–6 Glass (Pyrex) 3.2 × 10–6 Invar (Ni-Fe alloy) Lead Steel 1.3 × 10–6 29 × 10–6 13 × 10–6 Alcohol, ethyl 1.12 × 10–4 Benzene Gasoline Glycerin Mercury 1.24 × 10–4 9.6 × 10–4 4.85 × 10–4 1.82 × 10–4 Turpentine 9.0 × 10–4 Air* at 0°C Helium* 3.67 × 10–3 3.665 × 10–3 * The values given here assume the gases undergo expansion at constant pressure. However, the expansion of gases depends on the pressure applied to the gas. Therefore, gases do not have a specific value for the volume expansion coefficient. Table A15 Coefficients of thermal expansion for common substances Medium v(m/s) Medium v(m/s) Medium v(m/s) Gases Liquids at 25°C Hydrogen 1286 Glycerol Helium Air Air Oxygen 972 343 331 317 Seawater Water Mercury Kerosene Methyl Alcohol Carbon tetrachloride 1904 1533 1493 1450 1324 1143 926 Table A16 Speed of sound in various substances Solids* Pyrex glass Iron Aluminum Brass Copper Gold Lucite Lead Rubber 5640 5950 5100 4700 3560 3240 2680 1322 1600 826 Appendix A • Reference Tables Medium v(m/s) Medium v(m/s) Medium v(m/s) *Values given here are for propagation of longitudinal waves in bulk media. However, speeds for longitudinal waves in thin rods are slower, and speeds of transverse waves in bulk are even slower. Table A16 Speed of sound in various substances Source of Sound B(dB) Nearby jet airplane 150 Jackhammer machine gun 130 Siren; rock concert 120 Subway; power lawn mower 100 Busy traffic Vacuum cleaner Normal Conversation Mosquito buzzing whisper Rustling leaves Threshold of hearing 80 70 60 40 30 10 0 Table A17 Conversion of sound intensity to decibel level Wavelength Range (nm) Color Description 400-430 430-485 485-560 560-590 590-625 625-700 Violet Blue Green Yellow Orange Red Table A18 Wavelengths of visible light Access for free at openstax.org. Appendix A • Reference Tables 827 Substance Index of Refraction Substance Index of Refraction Solids at 20°C Cubic zirconia Diamond (C) Flourite (CaF2) 2.15 2.419 1.434 Liquids at 20°C Benzene 1.501 Carbon disulfide 1.628 Carbon tetrachloride 1.461 Fused quartz (SiO2) 1.458 Ethyl alcohol Gallium phosphide 3.50 Glass, crown Glass, flint Ice (H2O) Polystyrene 1.52 1.66 1.309 1.49 Glycerin Water 1.361 1.473 1.333 Gases at 0°C, 1 atm Air 1.000 293 Sodium chloride (NaCl) 1.544 Carbon dioxide 1.000 45 Note: These values assume that light has a wavelength of 589 nm in vacuum. Table A19 Indices of refraction Hoop or thin cylindrical shell Hollow cylinder Solid cylinder or disk Rectangular plane Long, thin rod with rotation axis through center Long, thin rod with rotation axis through end Solid sphere Thin spherical shell Table A20 Moments of inertia for different shapes μs μk Rubber on dry concrete 1.0 0.8 Table A21 Coefficients of friction for common objects on other objects 828 Appendix A • Reference Tables Steel on steel Aluminum on steel Glass on glass Copper on steel Wood on wood Waxed wood on wet snow Waxed wood on dry snow Metal on metal (lubricated) Teflon on Teflon Ice on ice Synovial joints in humans μs μk 0.74 0.61 0.94 0.53 0.25-0.5 0.14 0.1 0.15 0.04 0.1 0.01 0.57 0.47 0.4 0.36 0.2 0.1 0.04 0.06 0.04 0.03 0.003 Note: All values are approximate. In some cases, the coefficient of friction can exceed 1.0. Table A21 Coefficients of friction for common objects on other objects Material Dielectric Constant ĸ Dielectric Strength* (106V/m) Air (dry) Bakelite Fused quartz Mylar Neoprene rubber Nylon Paper 1.000 59 4.9 4.3 3.2 6.7 3.4 3.7 Paraffin-impregnated paper 3.5 Polystyrene Polyvinyl chloride Porcelain 2.56 3.4 6 Table A22 Dielectric constants 3 24 8 7 12 14 16 11 24 40 8 Access for free at openstax.org. Appendix A • Reference Tables 829 Material Dielectric Constant ĸ Dielectric Strength* (106V/m) Pyrex glass Silicone oil Strontium titanate Teflon Vacuum Water 5.6 2.5 233 2.1 1.000 00 80 Table A22 Dielectric constants 14 15 8 60 ∞ 3 830 Appendix A • Reference Tables Access for free at openstax.org. INDEX A aberration 505 absolute zero 329 absorption spectrum 725 Acceleration 73, 93, 93, 94, 104, 116, 122, 128, 205 acceleration due to gravity 104 Accuracy 26 acoustics 8 a |
ctivity 744 air resistance 162 alpha (α) decay 738 alpha particle 738 alternating current 606 ampere 604 amplitude 180, 390, 417, 457 amplitude modulation 461 analytical method 153 Anger camera 758 angle of incidence 478 angle of reflection 478 angle of refraction 491 angle of rotation 198 angular acceleration 212, 261 angular momentum 261 angular velocity 198, 206 annihilation 753, 785 Antimatter 784 antinode 403, 437 Antoine Henri Becquerel 736 aphelion 230 arc length 198 atomic number 735 atoms 8 Average acceleration 94 Average speed 62 Average velocity 63 B baryons 784 beat frequency 436 Beats 436 becquerel 744 beta ( β β ) decay 738 Big Bang 792 binding energy 317, 699 blackbody 692 Boltzmann constant 359 bottom 781 C capacitor 582 carbon-14 dating 745 carrier particles 773 Celsius scale 328 central axis 481 centrifugal force 205 centripetal acceleration 205, 213 centripetal force 207 cesium atomic clock 20 chain reaction 749 change in momentum 254 charmed 781 chromatic aberration 505 circuit diagrams 613 Circular motion 198, 205 classical physics 8 closed system 286 closed-pipe resonator 439 coefficient of friction 119 Colliding beams 778 collision 256 collisions 262 color 782 complex machine 293 component 154 Compton 705 Compton effect 705 concave lens 499 concave mirror 481 Condensation 341 Index 831 conduction 334, 556 conductors 555 conservation 259 Constant acceleration 99 constructive interference 400 convection 334 conventional current 605 converging lens 498 conversion factor 24, 24 convex lens 498 convex mirror 481 Copernican model 232 corner reflector 494 Coulomb 562 coulomb force 772 Coulomb’s constant 563 Coulomb’s law 563 critical angle 493 critical mass 749 Curie temperature 654 cyclical process 373 cyclotron 777 D damping 435 decay constant 744 decibels 424 deformation 178 degree Celsius ( °C ( °C ) 328 degree Fahrenheit ( °F ( °F ) 328 dependent variable 34, 67 derived units 19 destructive interference 401 dielectric 585 differential interference contrast (DIC) 535 diffraction 525 diffraction grating 533 diffused 479 direct current 606 direction 54 832 Index dispersion 491 displacement 57, 73, 94, 100 distance 56 diverging lens 499 domains 653 Doppler effect 430 down 780 dynamics 7, 116 E E.O. Lawrence 777 eccentricity 234 Edwin Hubble 792 efficiency output 293 Einstein 698 elastic collision 262 electric circuits 612 Electric current 604 electric eye 702 electric field 456, 567 Electric motors 665 electric potential 575 Electric potential energy 572 electric power 632 electrical charge 562 Electricity 8 electromagnet 656 electromagnetic force 772 electromagnetic radiation (EMR) 456 electromagnetism 661 electron 551 electron cloud model 15 Electroweak Epoch 793 electroweak theory 790 emf 673 emission spectrum 724 energy 280, 390 energy-level diagram 726 energy-mass curve 751 English units 19 Enrico Fermi 773 entropy 366 equilibrium position 180 equivalent resistor 618 Access for free at openstax.org. Ernest Rutherford 722 ether 306 excited states 726 experiment 14 exponential relationship 37 external force 116 eyepiece 504 F Fahrenheit scale 328 Faraday 567 Fermi National Accelerator Laboratory 778 ferromagnetic 653 Feynman diagram 774 first law of thermodynamics 361 flavors 780 Fletcher 551 focal length 481 focal point 481 Force 116, 254 force field 567 frame of reference 306 frame-dragging effect 242 frames of reference 314 Fraunhofer lines 725 free-body diagram 116 freefall 124 Freezing 341 Frequency 180, 390, 416, 457 frequency modulation 461 friction 118 fulcrum 215, 290 fundamental 438 fundamental physical 18 fusion reactors 756 G Galilei 115 Galileo 55 gamma (γ) rays 457 Gamma decay 739 gauge pressure 417 Geiger tube 742 General relativity 305 generator 666 geodetic effect 242 geometric optics 478 George Zweig 780 glass rods 550 gluon 782 gluons 774 Grand Unification Epoch 793 Grand Unified Theory (GUT) 790 graph 67, 73 graphical 144 gravitational 129 gravitational constant 238 gravitational force 117, 123, 772 Gravitational potential energy 280 graviton 774, 776 Gravity 104, 123 ground state 726 gun-type bomb 754 H hadrons 780 half-life 743 harmonic motion 416 harmonics 438 head 144 head-to-tail method 144 Hearing 427 Heat 328 heat capacity 332 heat engine 372 Heat pumps 374 Heisenberg uncertainty principle 733 hertz 180 Hideki Yukawa 773 Higgs boson 785 Higgs field 787 Hooke’s law 179 Huygens’s principle 525 hydrogen-like atoms 728 hypothesis 14 joule 281 I ideal gas law 358 ideal mechanical advantage 290 illuminance 467 impulse 254 impulse-momentum theorem 254 in parallel 621 in series 618 incident ray 491 inclined plane 291 independent 162 independent variable 34, 67 index of refraction 488 induction 557, 672 inelastic collision 264 Inertia 120 inertial reference frame 307 Inflationary Epoch 793 infrared (IR) radiation 457 infrasound 427 input work 293 instantaneous acceleration 94 instantaneous speed 62 instantaneous velocity 64 insulator 555 interference 463 Internal energy 361 Inverse proportionality 37 inverse relationship 37 inverse-square law 564 Inversely proportional 123 inversion 403 ionizing radiation 759 iridescence 533 isolated system 260 isotope 736 J K kelvin 20, 329 Kelvin scale 329 Kepler 115 Kepler’s laws of planetary motion 230 kinematic equations 99 kinematics 7, 54, 67, 116 kinematics of rotational motion 214 kinetic 316 kinetic energy 262, 280, 572 Kinetic friction 171 L laminated-coil transformer 669 Large Hadron Collider 776 laser 532 Laser Interferometer Gravitational-Wave Observatory 776 latent heat 341 latent heat of fusion 342 latent heat of vaporization 342 law of conservation of charge 554 law of conservation of energy 286 law of conservation of momentum 260, 262 law of inertia 120 law of reflection 478 law of refraction 478 length contraction 314 Leptons 780 lever 290 lever arm 216 light years 466 LightSail-1 708 line graph 34 Index 833 linear acceleration 213 Linear momentum 254 linear relationships 37 liquid drop model 748 log-log plot 38 logarithmic (log) scale 37 longitudinal wave 391 longitudinal waves 416 loudness 427 Louis de Broglie 732 lumens 466 luminous flux 466 lux 467 M magnetic dipoles 652 magnetic field 456, 654 magnetic flux 675 magnetic induction 672 magnetic pole 650 Magnetism 8 magnetized 653 magnitude 58 Manhattan Project 754 Mass 120, 122, 254 mass defect 317 mass number 735 maximum height 166 maximum height of a projectile 168 Maxwell’s equations 458 mechanical advantage 290 mechanical energy 280 mechanical waves 390 medium 390 Melting 341 meson 774 mesons 784 method of adding percents 29 metric system 21 metric units 21 Michelson–Morley experiment 306 microwave radiation 457 Millikan 551 834 Index model 14 modern physics 8 moment of inertia 261 momentum 259, 262, 315 monochromatic 528 monochromators 536 motion 54, 62 Murray Gell-Mann 780 N natural frequency 435 negative acceleration 94 negative charges 550 net external force 122, 128 net force 116, 128, 129 neutrino 739, 782 Newton 115, 123 Newton’s first law of motion 118 Newton’s laws of motion 115, 116 Newton’s second law of motion 16, 122 Newton’s third law of motion 128 Newton’s universal law of gravitation 238 Niels Bohr 725 node 438 nodes 403 nonohmic 611 normal force 117, 119, 129 north pole 650 nuclear fission 748 Nuclear fusion 751 Nuclear physics 8 nuclear strong force 751 nucleons 735, 773 nuclide 736 O objectives 504 observation 14 Access for free at openstax.org. ocular 504 Ohm’s law 609 ohmic 609 oil-drop 552 open-pipe resonator 439 optics 8 order of magnitude 24 oscillate 179 output work 293 overtones 438 P pair production 785 parfocal 504 Particle physics 771 particle-wave duality 708 Pascal’s principle 16 perihelion 230 period 180, 232, 390 Periodic motion 180 periodic wave 390 permanent magnet 653 permittivity of free space 583 Peter Higgs 787 phase change 341 phase diagram 343 photoelectric effect 698 photoelectron 698 photon 774 photon momentum 705 photons 698 photovoltaic cells 702 physical models 15 physical science 161 physics 6 pigment 460 pion 773 pitch 427 pivot point 215 Planck 693 Planck Epoch 793 Planck’s constant 694 planetary model of the atom 723 plasma 341 plum pudding model 722 point charge 567 point masses 265 polarization 557 polarized light 464 position 54, 67, 73 positive charges 550 positron 784 Positron emission tomography 759 postulates 307 potential difference 575 potential energy 280, 316, 572 Power 281, 423 Precision 27 pressure 358 primary and secondary coils 669 principles 16 projectile 162 Projectile motion 162 Proper length 314 proportionality 122 proton decay 791 proton-proton cycle 753 protons 551 Ptolemaic model 232 pulley 292 pulse wave 390 Q quadratic relationship 37 quantized 695 quantum 694 quantum chromodynamics 774, 782 quantum electrodynamics 774 quantum mechanics 8, 694 Quark Era 793 quarks 780 R rad 760 Radar 461 radian 199 radiation 334 radio waves 457 radioactive 738 Radioactive dating 745 radioactive decay 738 Radioactivity 738 radiopharmaceutical 758 Radiotherapy 760 radius of curvature 198 rarefactions 416 rate 62 ray 478 Rayleigh criterion 538 real image 479 reciprocal 283 recoil 263 reference frame 54 reflection 403, 478 refracted ray 491 refraction 404 relative biological effectiveness 761 relative speed 314 Relativistic 314 relativistic energy 316 relativistic factor 313 Relativistic momentum 315 Relativity 305 Resistance 608 resistor 613 resolution 537 resonance 435 resonate 435 rest mass 315 restoring force 179 resultant 145 resultant vector 147 revolution 199 Richard Feynman 774 right-hand rule 656 roentgen equivalent man 761 rotational inertia 261 Rotational motion 198 Rutherford 551 Rutherford scattering 723 Rydberg constant 731 S scalar 58, 62, 254 scientific law 16 scientific method 14 Scientific models 15 Scientific notation 23 scintillators 742 screw 292 second law of thermodynamics 366 semi-log plot 38 shock wave 432 SI units 19 significant figures 30 simple harmonic motion 179 Simple machines 290 |
simple pendulum 181 Simultaneity 310 Single-photon-emission computer tomography 758 slope 36 Snap Lab 54 Snell’s law 491 solenoid 657 sonic boom 432 Sound 416 sound intensity 423 sound intensity level 424 south pole 650 special relativity 305, 310 specific heat 332 specular 479 speed 54, 62, 122, 417 Spin 198 Standard Model 780 standing waves 402 Stanford Linear Accelerator Center 778 static friction 171 statics 7 stationary 54 steady state 616 strange 780 Index 835 strong nuclear force 737, 772 Sublimation 341 Super-Kamiokande 791 superforce 793 superposition 400 supersonic 432 synchrotron 777 system 119 systems 116 T tagged 758 tail 144 tangent 71 tangential acceleration 213 Tangential velocity 200, 213 temperature 328 tension 129 test charge 567 The net external force 116 theory 16 Theory of Everything 793 theory of relativity 9 therapeutic ratio 760 thermal efficiency 376 thermal energy 328 thermal equilibrium 356 thermodynamics 8 thermonuclear bomb 755 thin-film interference 463 Thomson 550 three-dimensional models 15 thrust 129 time 62, 67, 73, 94, 95 Time dilation 313 top 781 torque 216, 261 torsion balance 562 total energy 316 total internal reflection 493 trajectory 162 transformer 669 translational acceleration 261 transmutation 738 transverse wave 391 836 Index transverse waves 416 trend line 36 twin paradox 314 U U.S. customary units 21 ultrasound 427 ultraviolet (UV) radiation 457 ultraviolet catastrophe 693 uncertainty 28 uniform circular motion 205 uniform electric field 580 universal 16 universal laws 116 up 780 V Van de Graaff 558 Van de Graaff generator 777 Vaporization 341 variables 34 vector 58, 63, 123, 254 vector addition 144 vector quantities 95 Vector subtraction 146 Velocity 63, 73, 93, 94, 99, 116, 122, 205, 254 virtual image 479 visible light 456 voltage 575 W W bosons 775 watts 282 wave 390 wave velocity 394 wavefronts 525 wavelength 394, 419, 457 weak nuclear force 772 wedge 291 weight 119, 123, 129 Werner Heisenberg 733 wheel and axle 291 work 280 work–energy theorem 281 X X-rays 457 Z Z bosons 775 zeroth law of thermodynamics 356 Access for free at openstax.org. |
s and prototypes on the basis of established criteria 4 Unit I 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 5 1-1 QuickLab 1-1 QuickLab Match a Motion Problem What types of motions can you generate using ticker tape? Questions 1. Describe the spacing between dots on each ticker Materials clacker power supply ticker tape Procedure 1 Measure and cut 4 or 5 lengths of ticker tape of length 40 cm (approximately). 2 Pull the ticker tape through the clacker at an even speed. 3 Repeat step 2 for a new length of ticker tape. For each new ticker tape, pull with a different kind of motion: quickly, slowly, speeding up, slowing down, etc. tape. Is it even, increasing, or decreasing? 2. Use the spacing between dots to describe the motion of the ticker tape. Is the ticker tape speeding up, slowing down, or constant? 3. What aspect of motion does the spacing between dots represent? 4. Have you covered all possible motions in the runs you did? If not, which ones did you omit? Give reasons why you omitted some types of motion. Think About It 1. What is motion? 2. What types of motion can objects undergo? 3. What are some words used to describe motion? 4. How can you determine how far and how fast an object moves? 5. What does the term “falling” mean? 6. Describe the motion of falling objects. 7. Which falls faster: a heavy object or a light object? 8. How can a ball that’s moving upward still be falling? 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 1 Graphs and equations describe motion in one dimension. 5 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 6 1.1 The Language of Motion When describing motions, such as the ones in Figures 1.1 and 1.2, you can use many different expressions and words. The English language is rich with them. Many sports broadcasters invent words or expressions to convey action and to excite the imagination. Phrases such as “a cannonating drive from the point” or “blistering speed” have become commonplace in our sports lingo. In physics, you must be precise in your language and use clearly defined terms to describe motion. Kinematics is the branch of physics that describes motion. kinematics: a branch of physics that describes motion Figure 1.2 How would you describe the motions shown in these photos? origin: a reference point position: the straight-line distance between the origin and an object’s location; includes magnitude and direction scalar quantity: a measurement that has magnitude only vector quantity: a measurement that has both magnitude and direction 6 Unit I Kinematics M I N D S O N How Do Objects Move? Study the photos in Figure 1.2. Describe the motion of the puck, the wheelchair, and the harpoon with respect to time. Jot down your descriptions and underline key words associated with motion. Compare your key words with those of a partner. Compile a class list on the chalkboard. When describing motion, certain words, such as speed, acceleration, and velocity, are common. These words have slightly different meanings in physics than they do in everyday speech, as you will learn in the following subsection. Physics Terms It’s Saturday night and the Edmonton Oilers are playing the Calgary Flames. In order to locate players on the ice, you need a reference system. In this case, select the centre of the ice as the reference point, or origin. You can then measure the straight-line distances, d, of players from the origin, such as 5.0 m. If you specify a direction from the origin along with the distance, then you define a player’s position, d , for example, 5.0 m [E] (Figure 1.3). The arrow over the variable indicates that the variable is a vector quantity. The number and unit are called the magnitude of the vector. Distance, which has a magnitude but no direction associated with it, is an example of a scalar quantity. Vector quantities have both magnitude and direction. Position is an example of a vector quantity. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 7 W N S E di 5.0 m origin 26.0 m Figure 1.3 The player’s position is 5.0 m [east of the origin] or simply 5.0 m [E]. The player is at a distance of 5.0 m from the origin. 60.0 m If the player, initially 5.0 m [east of the origin], skates to the east end of the rink to the goal area, his position changes. It is now 25.0 m [east of the origin] or 25.0 m [E] (Figure 1.4). You can state that he has travelled a straight-line distance of 20.0 m, and has a displacement of 20.0 m [E] relative to his initial position. W N S E 25.0 m df 26.0 m origin 60.0 m PHYSICS INSIGHT Technically, if you are standing away from the origin, you are displaced a certain distance and in a certain direction. However, the sign is not used with position unless the object you are referring to has moved from the origin to its current position, that is, unless the object has experienced a change in position. Figure 1.4 The player’s position has changed. A change in position is called displacement. Distance travelled is the length of the path taken to move from one position to another, regardless of direction. Displacement, d , is the change in position. The player’s displacement is written as distance: the length of the path taken to move from one position to another d 20.0 m [E] where is the Greek letter delta that means “change in.” Calculate the change in a quantity by subtracting the initial quantity from the final quantity. In algebraic notation, R Rf Ri. You can calculate the displacement of the player in the following manner: d d f d i 25.0 m [E] 5.0 [E] 20.0 m [E] displacement: a straight line between initial and final positions; includes magnitude and direction info BIT Pilots use radar vectors when landing their aircraft. Radar vectors are instructions to fly in a particular direction and usually include altitude and speed restrictions. Chapter 1 Graphs and equations describe motion in one dimension. 7 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 8 Sign Conventions How would you determine your final distance and displacement if you moved from a position 5.0 m [W] to a position 10.0 m [E] (Figure 1.5)? W 10.0 m 8.0 m 6.0 m 4.0 m 2.0 m 2.0 m 4.0 m 6.0 m 8.0 m 0.0 m E 10.0 m Figure 1.5 The person travels a distance of 5.0 m 10.0 m 15.0 m. What is the person’s displacement? What is the person’s final position relative to the bus stop? To calculate the distance travelled in the above scenario, you need only add the magnitudes of the two position vectors. d 5.0 m 10.0 m 15.0 m To find displacement, you need to subtract the initial position, d i, f. Let d from the final position, d 5.0 m [W] and d f 10.0 m [E]. i d d f d i 10.0 m [E] 5.0 m [W] Note that subtracting a vector is the same as adding its opposite, so the negative west direction is the same as the positive east direction. d 10.0 m [E] 5.0 m [W] 10.0 m [E] 5.0 m [E] 15.0 m [E] PHYSICS INSIGHT When doing calculations with measured values, follow the rules on rounding and the number of significant digits. Refer to pages 876–877 in this book. W N up E S down L R Figure 1.6 Let east be positive and west negative. Similarly, north, up, and right are usually designated as positive. Another way of solving for displacement is to designate the east direction as positive and the west direction as negative (Figure 1.6). The two position 10.0 m [E] 10.0 m. vectors become d i Now calculate displacement: 5.0 m [W] 5.0 m and d f d d f d i 10.0 m (5.0 m) 15.0 m Since east is positive, the positive sign indicates that the person has moved 15.0 m east. Practise finding position and displacement in the next Skills Practice and example. info BIT On April 26, 2004, Stephane Gras of France did 445 chin-ups in one hour. If you consider up as positive, then Gras made 445 positive displacements and 445 negative displacements, meaning that his net displacement was zero! 8 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page Finding Position and Displacement 1. Create a scale using the dimensions of the hockey rink (Figure 1.7). Measuring from the centre of the player’s helmet, (a) find each player’s position relative to the north 26.0 m and south sides of the rink. (b) find each player’s position relative to the east and west sides of the rink. (c) If the player moves from position 2 to position 4 on the rink, what is his displacement? 1 2 5 origin 4 3 60.0 m Figure 1.7 Example 1.1 A traveller initially standing 1.5 m to the right of the inukshuk moves so that she is 3.5 m to the left of the inukshuk (Figure 1.8). Determine the traveller’s displacement algebraically (a) using directions (b) using plus and minus signs Given d i d f 1.5 m [right] 3.5 m [left] Required displacement (d ) inukshuk 3.5 m [left] origin 1.5 m [right] Figure 1.8 Analysis and Solution To find displacement, use the equation d (a) d d i d f 3.5 m [left] 1.5 m [right] 3.5 m [left] (1.5 m [left]) 3.5 m [left] 1.5 m [left] 5.0 m [left] d d i. f i d d f d (b) Consider right to be positive. 1.5 m [right] 1.5 m 3.5 m [left] 3.5 m d f 3.5 m (1.5 m) 3.5 m 1.5 m 5.0 m d i The answer is negative, so the direction is left. Paraphrase The traveller’s displacement is 5.0 m [left] of her initial position. Note that the direction of displacement is relative to initial position, whereas the direction of position is relative to the designated origin, in this case, the inukshuk. Practice Problems 1. Sprinting drills include running 40.0 m [N], walking 20.0 m [N], and then sprinting 100.0 m [N]. What is the sprinter’s displacement from the initial position? 2. To perform a give and go, a basketball player fakes out the defence by moving 0.75 m [right] and then 3.50 m [left]. What is the player’s displacement from the starting position? 3. While building a wall, a bricklayer sweeps the cement back and forth. If she swings her hand back and forth, a distance of 1.70 m, four times, calculate the distance and displacemen |
t her hand travels during that time. Answers 1. 160.0 m [N] 2. 2.75 m [left] 3. 6.80 m, 0 m Chapter 1 Graphs and equations describe motion in one dimension. 9 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 10 For all subsequent problems in this book, you will be using plus and minus signs to indicate direction. This method is more flexible for problem solving and easier to use. Like distance and displacement, speed and velocity is another scalar-vector pair. Speed is the rate at which an object moves. It is a scalar quantity, so it has magnitude only; for example, v 50 km/h (Figure 1.9). Velocity is a vector quantity, so it has both magnitude (speed) and direction. If you are travelling south from Fort McMurray to Lethbridge at 50 km/h, your velocity is written as v 50 km/h [S]. If you designate south as negative, then v 50 km/h. Acceleration is a vector quantity that represents the rate of change of velocity. You will study aspects of displacement, velocity, and acceleration, and their interrelationships, in the sections that follow. Figure 1.9 Scalar or vector? 1.1 Check and Reflect 1.1 Check and Reflect Knowledge 1. What two categories of terms are used to describe motion? Give an example of each. 2. Compare and contrast distance and displacement. 3. What is the significance of a reference point? Applications 4. Draw a seating plan using the statements below. (a) Chad is 2.0 m [left] of Dolores. (b) Ed is 4.5 m [right] of Chad. (c) Greg is 7.5 m [left] of Chad. (d) Hannah is 1.0 m [right] of Ed. (e) What is the displacement of a teacher who walks from Greg to Hannah? 5. A person’s displacement is 50.0 km [W]. What is his final position if he started at 5.0 km [E]? 6. Using an autuk (a type of sealskin racquet), two children play catch. Standing 3.0 m apart, the child on the right tosses the ball to the child on the left, and then moves 5.0 m [right] to catch the ball again. Determine the horizontal distance and displacement the ball travels from its initial position (ignore any vertical motion). 3.0 m 5.0 m 7. Below is a seating plan for the head table at a wedding reception. Relative to the bride, describe the positions of the groom, best man, maid of honour, and flower girl. 0.75 m 0.75 m 0.50 m 0.75 m 0.75 m Flower girl Best man Bride Groom Maid of Honour Ring boy e TEST To check your understanding of scalar and vector quantities, follow the eTest links at www.pearsoned.ca/school/physicssource. 10 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 11 1.2 Position-time Graphs and Uniform Motion You are competing to win the Masters Golf Tournament. The hole is 5.0 m away (Figure 1.10). You gently hit the ball with your club and hold your breath. Time seems to stop. Then, 5.0 s later, it rolls into the hole. You have won the tournament! From section 1.1, you know that displacement is the change in an object’s position. If you replay the sequence of motions of your winning putt in 1.0-s intervals, you can measure the displacements of the golf ball from you, the putter, to the hole (Figure 1.11). Figure 1.10 You can represent motion in sports using vectors and graphs. 0.0 m 0.0 s origin 1.0 m 1.0 s 2.0 m 2.0 s 3.0 m 3.0 s 4.0 m 4.0 s 5.0 m 5.0 s Figure 1.11 What is the golf ball’s displacement after each second? Table 1.1 displays the data from Figure 1.11 for the golf ball’s position from you at 1.0-s intervals. By graphing the data, you can visualize the motion of the golf ball more clearly (Figure 1.12). ▼ Table 1.1 Position-time data Time (s) Position (m [right]) 0.0 1.0 2.0 3.0 4.0 5.0 t0 t1 t2 t3 t4 t5 0.0 1.0 2.0 3.0 4.0 5.0 Velocity Position vs. Time 6.0 5.0 4.0 3.0 2.0 1.0 0..0 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.12 A position-time graph of the golf ball Notice that the graph in Figure 1.12 is a straight line. A straight line has a constant slope. What does constant slope tell you about the ball’s motion? To answer this question, calculate the slope and keep track of the units. Designate toward the hole, to the right, as the positive direction. Chapter 1 Graphs and equations describe motion in one dimension. 11 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 12 e SIM Practise calculating average speed and average velocity. Go to www.pearsoned.ca/ school/physicssource. Recall that slope rise run . For position–time graphs, this equation becomes slope change in position change in time A change in position is displacement. So, the equation for slope becomes PHYSICS INSIGHT Speed has magnitude only. Velocity has both magnitude and direction. velocity: rate of change in position slope d t d d i f ti tf 5.0 m 0.0 m 5.0 s 0.0 s 1.0 m/s The answer is positive, so the golf ball moves at a rate of 1.0 m/s [right]. Notice that the units are m/s (read metres per second). These units indicate speed or velocity. Since displacement is a vector quantity, the slope of the position-time graph in Figure 1.12 gives you the velocity, v, of the ball: the change in position per unit time. Because you have calculated velocity over a time interval rather than at an instant in time, it is the average velocity. d v t Speed and Velocity 4 m/s 4 m/s Figure 1.13 Objects with the same speed can have different velocities. Objects travelling at the same speed can have different velocities. For example, a tram carries passengers across a ravine at a constant speed. A passenger going to the observation deck has a velocity of 4 m/s [right] and a passenger leaving the deck has a velocity of 4 m/s [left] (Figure 1.13). Their speeds are the same, but because they are travelling in opposite directions, their velocities are different. 1-2 Decision-Making Analysis 1-2 Decision-Making Analysis Traffic Safety Is Everyone’s Business The Issue In an average year in Alberta, traffic accidents claim six times more lives than homicide, eight times more lives than AIDS, and 100 times more lives than meningitis. Collisions represent one of the greatest threats to public safety. Background Information In the Canadian 2002 Nerves of Steel: Aggressive Driving Study, speeding was identified as one of two common aggressive behaviours that contribute to a significant percentage of all crashes. The Alberta Motor Association’s Alberta Traffic Safety Progress Report has suggested that a province-wide speed management program could significantly improve levels of road safety, decreasing both speed and casualties. One suggested program is the implementation of the vehicle tachograph, a device required in Europe to improve road safety. 12 Unit I Kinematics e WEB To learn more about how speeding is a key contributing factor in casualty collisions in Alberta, follow the links at www.pearsoned.ca/school/ physicssource. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 13 Analysis Your group has been asked to research different traffic safety initiatives. The government will use the results of your research to make the most appropriate decision. 1. Research (a) how state- or province-wide speed management programs have influenced driver behaviour (b) the societal cost of vehicle crashes (c) driver attitudes toward enforcement of and education about traffic safety issues 2. Analyze your research and decide which management program should be used. 3. Once your group has completed a written report recommending a particular program, present the report to the rest of the class, who will act as representatives of the government and the community. So far, you have learned that the slope of a position-time graph represents a rate of change in position, or velocity. If an object moves at constant velocity (constant magnitude and direction), the object is undergoing uniform motion. A position-time graph for an object at rest is a horizontal line (Figure 1.14). An object at rest is still said to be undergoing uniform motion because its change in position remains constant over equal time intervals. Concept Check (a) Describe the position of dots on a ticker tape at rest. What is the slope of the graph in Figure 1.14? (b) Describe the shape of a position-time graph for an object travelling at a constant velocity. List three possibilities. Frame of Reference uniform motion: constant velocity (motion or rest) at rest: not moving; stationary ) ] .0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 Time (min) 4.0 5.0 Figure 1.14 A position-time graph for a stationary object If you were to designate the hole, rather than the putter, as the origin (starting point) in the golf tournament (Figure 1.15(a)), your data table would start at 5.0 m [left] at time 0.0 s, instead of at 0.0 m and 0.0 s (Table 1.2). The values to the left of the hole are positive. ▼ Table 1.2 Position-time data Time (s) Position (m [left]) t0 t1 t2 t3 t4 t5 0.0 1.0 2.0 3.0 4.0 5.0 5.0 4.0 3.0 2.0 1.0 0.0 5.0 m 0.0 s 4.0 m 1.0 s 3.0 m 2.0 s 2.0 m 3.0 s 1.0 m 4.0 s 0.0 m 5.0 s origin Figure 1.15(a) Designating an origin is arbitrary. In this example, the hole is the origin and all positions are measured relative to it. Chapter 1 Graphs and equations describe motion in one dimension. 13 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 14 info BIT On May 31, 2004 in Moscow, Ashrita Furman of the USA walked 1.6 km while continuously hulahooping in 14 min 25 s. He also holds the world record for the fastest time for pushing an orange with his nose. On August 12, 2004, he pushed an orange 1.6 km in 24 min 36 s. What was his speed, in km/h and m/s, for each case? (See Unit Conversions on page 878.) e WEB In November 2004, at an altitude of 33 000 m, the X-43A recorded a speed of Mach 9. Use the Internet or your local library to research the term “Mach” as used to describe the speed of an object. How did this term originate? What is the difference between Mach and ultrasonic? Write a brief summary of your findings. To learn more about Mach, follow the links at www.pearsoned.ca/school/ physicssource. 14 Unit I Kinematics The corresponding position-time graph is shown in Figure 1.15(b). ) ] .0 5.0 4.0 3.0 2.0 |
1.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.15(b) Compare this graph with the graph in Figure 1.12. If you change your reference frame, the position-time graph also changes. From the graph, slope v d t d d i f ti tf 0.0 m (5.0 m) 5.0 s 0.0 s 1.0 m/s The velocity of the golf ball is 1.0 m/s. What does the negative sign mean? It means the ball is travelling opposite to the direction to which the positions of the ball are measured. It does not mean that the golf ball is slowing down. Since positions, now measured to the left of the hole (the new origin) are designated positive, any motion directed to the right is described as being negative. In this case, you can also see that the ball is decreasing its position from the origin with increasing time. The ball travels to the right toward the hole, decreasing its position each second by 1.0 m — it travels at 1.0 m/s to the right or 1.0 m/s [left] as indicated by the downward slope on the graph. Concept Check Determine how the velocity of the golf ball can be positive if the hole is at the origin. Below is a summary of what you have learned: • The slope of a position-time graph represents velocity. • The velocity is the average velocity for the time interval. • Your choice of reference frame affects the direction (sign) of your answer. • A straight line on a position-time graph represents uniform motion. Comparing the Motion of Two or More Objects on a Position-time Graph You can represent the motions of two objects on one graph, as long as the origin is the same for both objects. You can then use the graph to compare their motions, as in the next example. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 15 Example 1.2 At the end of the school day, student A and student B say goodbye and head in opposite directions, walking at constant rates. Student B heads west to the bus stop while student A walks east to her house. After 3.0 min, student A is 300 m [E] and student B is 450 m [W] (Figure 1.16). (a) Graph the position of each student on one graph after 3.0 min. (b) Determine the velocity in m/s of each student algebraically. Given Choose east to be positive. dd A dd B t 3.0 min 300 m [E] 300 m 450 m [W] 450 m Required (a) position–time graph (b) velocity (v B) A and v BUS STOP bus stop school Lakeview School W N S E+ Student B’s position: 450 m [W] origin Student A’s position: 300 m [E] Figure 1.16 Analysis and Solution (a) Since east is the positive direction, plot student A’s position (3.0 min, 300 m) above the time axis and student B’s position (3.0 min, 450 m) below the time axis (Figure 1.17). Position vs. Time 1.0 2.0 3.0 Time (min 600 400 200 0 200 400 600 Figure 1.17 (b) Convert time in minutes to time in seconds. d t to find the velocity Then use the equation v of each student. 60 s 1 min t 3.0 min 180 s v A 300 m 180 s 1.7 m/s The sign is positive, so the direction is east. v B 450 m 180 s 2.5 m/s The sign is negative, so the direction is west. Paraphrase (b) Student A’s velocity is 1.7 m/s [E] and student B’s velocity is 2.5 m/s [W]. Practice Problems 1. A wildlife biologist measures how long it takes four animals to cover a displacement of 200 m [forward]. (a) Graph the data from the table below. (b) Determine each animal’s average velocity. Animal Time taken (s) Elk Coyote Grizzly bear Moose 10.0 10.4 18.0 12.9 Answers 1. (a 250.0 200.0 150.0 100.0 50.0 0.0 Position vs. Time 0 2 4 6 8 10 12 14 16 18 20 Time (s) (b) Elk: 20.0 m/s [forward] Coyote: 19.2 m/s [forward] Grizzly bear: 11.1 m/s [forward] Moose: 15.5 m/s [forward] Chapter 1 Graphs and equations describe motion in one dimension. 15 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 16 So far, you have learned that the slope of a position–time graph represents velocity. By comparing the slopes of two graphs, you can determine which object is moving faster. From the slopes of the graphs in Figure 1.17, which student is moving faster? When you represent the motions of two objects on the same graph, you can also tell whether the objects are approaching or moving apart by checking if the lines are converging or diverging. An important event occurs at the point where the two lines intersect. Both objects have the same position, so the objects meet at this point. Concept Check Describe the shape of a graph showing the motion of two objects approaching each other. In the next example, two objects start at different times and have different speeds. You will graphically find their meeting point. Example 1.3 Two rollerbladers, A and B, are having a race. B gives A a head start of 5.0 s (Figure 1.18). Each rollerblader moves with a constant velocity. Assume that the time taken to reach constant velocity is negligible. If A travels 100.0 m [right] in 20.0 s and B travels 112.5 m [right] in 15.0 s, (a) graph the motions of both rollerbladers on the same graph. (b) find the time, position, and displacement at which B catches up with A. B A Practice Problems 1. The two rollerbladers in Example 1.3 have a second race in which they each travel the original time and distance. In this race, they start at the same time, but B’s initial position is 10.0 m left of A. Take the starting position of A as the reference. (a) Graph the motions of the rollerbladers. (b) Find the time, position, and B’s displacement at which B catches up with A. Answers 1. (b) t 4.0 s 20.0 m [right] d 30.0 m [right] d 16 Unit I Kinematics distance travelled by A in 5.0 s Figure 1.18 Given Choose right to be positive. d A tA d B tB 100.0 m [right] 100.0 m 20.0 s 112.5 m [right] 112.5 m 15.0 s, started 5.0 s later Required (a) position-time graph (b) time (t), position (d catches up with A ), and displacement (d ) when B 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 17 Analysis and Solution (a) Assume that t 0.0 s at the start of A’s motion. Thus, the position-time graph of A’s motion starts at the origin. A’s final position is 100.0 m at 20.0 s. The position-time graph for B’s motion starts at 0.0 m and 5.0 s (because B started 5.0 s after A). B starts moving after 5.0 s for 15.0 s. Thus, at 20.0 s (5.0 s 15.0 s), B’s position is 112.5 m. Each rollerblader travels with a constant velocity, so the lines connecting their initial and final positions are straight (Figure 1.19(a)). ) ] 120.0 100.0 80.0 60.0 40.0 20.0 0.0 Position vs. Time B A 0.0 5.0 10.0 15.0 20.0 25.0 Time (s) Figure 1.19(a) (b) On the graph in Figure 1.19(a), look for a point of intersection. At this point, both rollerbladers have the same final position. From the graph, you can see that this point occurs at t 15.0 s. The corresponding position is 75.0 m (Figure 1.19(b)). ) ] 120.0 100.0 80.0 60.0 40.0 20.0 0.0 Position vs. Time B A 0.0 5.0 10.0 15.0 20.0 25.0 Time (s) Figure 1.19(b) To find B’s displacement, find the change in position: d Both A and B started from the same position, d both have the same final position at the point of intersection, d f d 75.0 m. 75.0 m 0.0 m i d d i. f 0. Since they 75.0 m The answer is positive, so the direction is to the right. Paraphrase (b) B catches up with A 15.0 s after A started. B’s position and displacement are 75.0 m [right] of the origin. Chapter 1 Graphs and equations describe motion in one dimension. 17 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 18 Example 1.4 From the graph in Example 1.3, find the velocities of the two rollerbladers. Given Choose right to be positive. At the point of intersection (Figure 1.19(b)), d A tA d B tB 75.0 m [right] 75.0 m 15.0 s 75.0 m [right] 75.0 m 15.0 s 5.0 s 10.0 s Required velocities of A and B (vv A, v B) Analysis and Solution To find the velocity of each rollerblader, remember that the slope of a position–time graph is velocity. Because the motions are uniform, the slopes will be constant for each rollerblader. vv vvv A d t 75.0 m 0.0 m 15.0 s 0.0 s 5.0 m/s vvv B 75.0 m 0.0 m 15.0 s 5.0 s 75.0 m 0.0 m 10.0 s 7.5 m/s The answers are both positive, so the direction is to the right. You can see that, in order for B to cover the same distance as A, B must move faster because B started later. Paraphrase A’s velocity is 5.0 m/s [right] and B’s velocity is 7.5 m/s [right]. Practice Problems 1. Suppose rollerblader B gives A a head start of 5.0 s and takes 10.0 s to catch up with A at 100.0 m [right]. Determine the velocities of rollerbladers A and B. Answers 1. A: 6.67 m/s [right] B: 10.0 m/s [right] 18 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 19 1-3 Inquiry Lab 1-3 Inquiry Lab Car Activity Question What are the speeds of two different toy cars? If one car is released 3.0 s after the other, where will they meet? Variables Identify the manipulated, responding, and controlled variables. Materials and Equipment two battery-operated toy cars ticker tape carbon disk spark timer (60 Hz) masking tape ruler graph paper Procedure 1 On a flat surface, such as the floor or lab bench, mark the initial starting position of car 1 with masking tape. Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Analysis 1. Draw a line through the first dot on each ticker tape and label it t = 0 s. 2. Depending on the calibration of your ticker timer, count from the starting position, and place a mark after a fixed number of dots, e.g., 6, 12, 18, 24, etc. Label each mark t1, t2, etc. On a 60-Hz timer, every sixth dot represents 0.10 s. 3. Measure the distance from t = 0 to t1, t = 0 to t2, t = 0 to t3, etc. Record the data in a position–time table. 4. Using an appropriate scale, graph each set of data for each toy car, separately. 5. Determine the slope of the line of best fit for each graph. See pages 872–873 for explicit instruction on how to draw a line of best fit. 2 Using masking tape, attach 1.0 m of ticker tape to the 6. What is the speed of each toy car? end of car 1. 3 Thread the ticker tape through the spark timer (Figure 1.20). 4 Turn the car on. 7. How do the speeds of car 1 and car 2 compare? 8. Assuming |
uniform motion, how far would car 1 travel in 15 s? 9. Assuming uniform motion, how long would it take car 2 5 Turn the spark timer on as you release the car from its to travel 30 m? initial position. 6 Observe the path of car 1 until the ticker tape is used up. Label the ticker tape “car 1.” 7 Repeat steps 2–6 for car 2. recording timer recording tape toy car Figure 1.20 10. Imagine that you release the faster car 3.0 s after the slower car. Graphically determine the position where the two cars meet. Assume uniform motion. e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. In summary, you can see how a position-time graph helps you visualize the event you are analyzing. Calculating the slope of a position-time graph provides new information about the motion, namely, the object’s velocity. In the next sections, you will expand on your graphing knowledge by analyzing motion using a velocity-time graph. Chapter 1 Graphs and equations describe motion in one dimension. 19 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 20 1.2 Check and Reflect 1.2 Check and Reflect Knowledge Applications 1. For an object at rest, what quantities of 8. Two children on racing bikes start from motion remain the same over equal time intervals? 2. For an object travelling at a constant velocity, what quantity of motion remains the same over equal time intervals? 3. Match each ticker tape below with the correct position-time graph. (i) (ii) (iii) (iv.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Position vs. Time A B C D 1.0 2.0 3.0 Time (s) 4.0 5.0 4. Two friends start walking on a football field in the same direction. Person A walks twice as fast as person B. However, person B has a head start of 20.0 m. If person A walks at 3.0 m/s, find the distance between the two friends after walking for 20.0 s and determine who is ahead at this time. Sketch a position-time graph for both people. 5. A camper kayaks 16 km [E] from a camping site, stops, and then paddles 23 km [W]. What is the camper’s final position with respect to the campsite? 6. Sketch a position-time graph for a bear starting 1.2 m from a reference point, walking slowly away at constant velocity for 3.0 s, stopping for 5.0 s, backing up at half the speed for 2.0 s, and finally stopping. 7. Sketch a position-time graph for a student (a) walking east to school with a constant velocity (b) stopping at the school, which is 5 km east of home (c) cycling home with a constant velocity 20 Unit I Kinematics the same reference point. Child A travels 5.0 m/s [right] and child B travels 4.5 m/s [right]. How much farther from the point of origin is child A than child B after 5.0 s? 9. Insect A moves 5.0 m/min and insect B moves 9.0 cm/s. Determine which insect is ahead and by how much after 3.0 min. Assume both insects are moving in the same direction. 10. Describe the motion in each lettered stage for the object depicted by the positiontime graph below 20 16 12 8 4 0 Position vs. Time B A C 0 4 8 12 Time (s) 16 20 11. A mosquito flies toward you with a velocity of 2.4 km/h [E]. If a distance of 35.0 m separates you and the mosquito initially, at what point (distance and time) will the mosquito hit your sunglasses if you are travelling toward the mosquito with a speed of 2.0 m/s and the mosquito is travelling in a straight path? 12. Spotting a friend 5.0 m directly in front of you, walking 2.0 m/s [N], you start walking 2.25 m/s [N] to catch up. How long will it take for you to intercept your friend and what will be your displacement? 13. Two vehicles, separated by a distance of 450 m, travel in opposite directions toward a traffic light. When will the vehicles pass one another if vehicle A is travelling 35 km/h and is 300 m [E] of the traffic light while vehicle B is travelling 40 km/h? When will each vehicle pass the traffic light, assuming the light remains green the entire time? e TEST To check your understanding of uniform motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 21 1.3 Velocity-time Graphs: Uniform and Nonuniform Motion Recently installed video screens in aircraft provide passengers with information about the aircraft’s velocity during the flight (Figure 1.21). x km 0.0 h x km 1.0 h x km 2.0 h x km 3.0 h x km 4.0 h x km 5.0 h Figure 1.22 A plane flies at a constant speed, so the distances within each time interval are equal. Break the plane’s motion into a series of snapshots. Record your data in a data table and then graph it. Figure 1.22 shows the data of the plane’s path. Like position-time graphs, velocity-time graphs provide useful information about the motion of an object. The shape of the velocity-time graph reveals whether the object is at rest, moving at constant speed, speeding up, or slowing down. Suppose an airplane has a cruising altitude of 10 600 m and travels at a constant velocity of 900 km/h [E] for 5.0 h. Table 1.3 shows the velocity-time data for the airplane. If you graph the data, you can determine the relationship between the two variables, velocity and time (Figure 1.23). ▼ Table 1.3 Time (h) Velocity (km/h) [E] Figure 1.21 Video screens are an example of an application of velocitytime graphs. 0.0 1.0 2.0 3.0 4.0 5. 900 800 700 600 500 400 300 200 100 0 0.0 900 900 900 900 900 900 Velocity vs. Time for an Airplane 1.0 2.0 3.0 4.0 5.0 Time (h) Figure 1.23 A velocity-time graph for an airflight Chapter 1 Graphs and equations describe motion in one dimension. 21 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 22 Designating east as the positive direction, the slope of the velocity-time graph is: slope rise run v t v v i f ti tf m k k m 900 900 h h 5.0 h 1.0 h 0 km/h2 From the graph in Figure 1.23, there is no change in the plane’s velocity, so the slope of the velocity-time graph is zero. Notice the units of the slope of the velocity-time graph: km/h2. These units are units of acceleration. Because the plane is moving at a constant velocity, its acceleration is zero. In general, you can recognize acceleration values by their units, which are always distance divided by time squared. In physics, the standard units for acceleration are metres per second per second, which is generally abbreviated to m/s2 (read metres per second squared). e TECH Determine the velocity of an object based on the shape Concept Check of its position-time graph. Go to www.pearsoned.ca/ school/physicssource. (a) What does the slope of a position-time graph represent? (b) What does the slope of a velocity-time graph represent? Non-uniform Motion Although objects may experience constant velocity over short time intervals, even a car operating on cruise control has fluctuations in speed or direction (Figure 1.24). How can you describe and illustrate a change of velocity using the concepts of kinematics? Recall from section 1.2 that an object moving at a constant velocity is undergoing uniform motion. But is uniform motion the only type of motion? Perform the next QuickLab to find out. Figure 1.24 Consider the kinds of changes in velocity this car experiences during the trip. 22 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 23 1-4 QuickLab 1-4 QuickLab Match a Graph Problem What type of motion does each graph describe? 5 Print out the graphs from your experiment. For each graph, construct a table of values for position and time. Materials LM 1-1 (provided by your teacher) ruler motion sensor masking tape Procedure 1 Study the different position-time graphs on LM 1-1. With a partner, decide what type of motion each graph illustrates. 2 Set up the motion sensor to plot position vs. time. 3 Label a starting position with masking tape approximately 1 m in front of the motion sensor. Move away from the motion sensor in such a way that the graph of the motion captured approximates the one on the LM. 4 Switch roles with your partner and repeat steps 1–3. Questions 1. Describe your motion when a horizontal line was being produced on the position–time graph. 2. What relationship exists between the type of motion and change in position? 3. Suggest two different ways in which you could classify the motion described by the four graphs. 4. What would the graph look like if you moved away from and then back toward the motion sensor? 5. What happens to the graph when you move away from your initial position and then move back toward and then beyond your initial position? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Concept Check Which ticker tape in Figure 1.25 represents accelerated motion? Explain. Figure 1.25 Consider an object, such as a drag racer (Figure 1.26), starting from rest and reaching a constant velocity over a time interval (Figure 1.27). During this time interval, the vehicle has to change its velocity from a value of zero to a final non-zero value. An object whose velocity changes (increases or decreases) over a time interval is undergoing acceleration, represented by the variable a. Acceleration is a vector quantity. It is also called non-uniform motion because the object’s speed or direction is changing. Figure 1.26 A drag racer accelerates from rest. acceleration: a vector quantity representing the change in velocity (magnitude or direction) per unit time scale 1.0 m 0.0 m 0.0 s 2.0 m 1.0 s 8.0 m 2.0 s 18.0 m 3.0 s Figure 1.27 This sequence illustrates a car undergoing non-uniform motion. Chapter 1 Graphs and equations describe motion in one dimension. 23 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 24 PHYSICS INSIGHT An object is accelerating if it is speeding up, slowing down, or changing direction. The following scenario illustrates acceleration. A drag race is a 402-m (quarter-mile) contest between two vehicles. Starting from rest, the vehicles leave the starting line at the same time, and the first vehicle to cross the finish line is the winner. A fan records the position of her favourite vehicle during the drag race. Her results are recorded |
in Table 1.4. The position-time graph for this data is shown in Figure 1.28. From the graph, note that the object is speeding up because the displacement between data points increases for each successive time interval. Which ticker tape in Figure 1.25 matches the graph in Figure 1.28? ▼ Table 1.4 Time (s) Position (m [forward]) 0.0 1.0 2.0 3.0 4.0 5.0 0.0 2.0 8.0 18.0 32.0 50.0 Position vs. Time for a Dragster ) ] 50.0 40.0 30.0 20.0 10.0 0.0 0.0 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.28 What does the slope of the graph indicate about the speed of the car? Instantaneous Velocity Instantaneous velocity is the moment-to-moment measure of an object’s velocity. Imagine recording the speed of your car once every second while driving north. These data form a series of instantaneous velocities that describe your trip in detail. Earlier in this section, you learned that determining the velocity of an object from a position-time graph requires calculating the slope of the position-time graph. But how can you obtain the slope of a curve? Remember that each point on the curve indicates the position of the object (in this case, the dragster) at an instant in time. To determine the velocity of an object at any instant, physicists use tangents. A tangent is a straight line that touches a curve at only one point (Figure 1.29(a)). Each tangent on a curve has a unique slope, which represents the velocity at that instant. In order for the object to be at that position, at that time, it must have an instantaneous velocity equal to the slope of the tangent at that point. Determining the slopes of the tangents at different points on a position-time curve gives the instantaneous velocities at different times. Consider forward to be the positive direction. PHYSICS INSIGHT When you calculate the slope of a line or curve at a single point, you are finding an instantaneous value. When you calculate the slope between two points, you are finding an average value. tangent: a straight line that touches a curved-line graph at only one point 24 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 25 Position vs. Time for a Dragster Position vs. Time for a Dragster Position vs. Time for a Dragster 50.0 40.0 30.0 20.0 10..0 0.0 d t 1.0 2.0 3.0 Time (s) 4.0 5.0 50.0 40.0 30.0 20.0 10..0 0.0 d t 1.0 2.0 3.0 Time (s) 4.0 5.0 50.0 40.0 30.0 20.0 10..0 0.0 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.29(a) Figure 1.29(b) Figure 1.29(c) The slope of the tangent at 2.0 s is slope d t 14.0 m 0.0 m 3.0 s 1.0 s 14.0 m 2.0 s 7.0 m/s The slope of the tangent at 3.0 s is slope 30.0 m 0.0 m 4.0 s 1.75 s The slope of the tangent at 4.0 s is slope 47.0 m (15.0 m) 5.0 s 3.0 s 30.0 m 2.25 s 13 m/s 32.0 m 2.0 s 16 m/s The sign is positive, so at 2.0 s, the velocity of the dragster is 7.0 m/s [forward]. At 3.0 s, the velocity of the dragster is 13 m/s [forward]. At 4.0 s, the velocity of the dragster is 16 m/s [forward]. Using Slopes of Position-time Graphs to Draw Velocity-time Graphs You can now create a new table using the slopes of the position-time graphs in Figures 1.29(a), (b), and (c). See Table 1.5. Remember that the slope of a position-time graph is velocity. These slope values are actually instantaneous velocities at the given times. You can use these three velocities to draw a velocity-time graph (Figure 1.30). The resulting velocity-time graph is a straight line that goes through the origin when extended. This means that the dragster has started from rest (0 velocity). The graph has a positive slope. To find the meaning of slope, check the units of the slope of a velocity-time graph. They are (m/s)/s, which simplify to m/s2. These units are the units of acceleration. Since the velocity-time graph in this example is a straight line with non-zero slope, the acceleration of the object is constant, so the object must be undergoing uniformly accelerated motion. info BIT When jet fighters come in to land on an aircraft carrier, they stop so quickly that pilots sometimes lose consciousness for a few seconds. The same thing can happen when a pilot ejects from an aircraft, due to enormous acceleration. uniformly accelerated motion: constant change in velocity per unit time Chapter 1 Graphs and equations describe motion in one dimension. 25 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 26 ▼ Table 1.5 Time (s) Velocity (m/s [forward]) 2.0 3.0 4.0 7.0 13 16 ) ] 20 15 10 5 0 Velocity vs. Time 1.0 2.0 3.0 4.0 5.0 Time (s) Figure 1.30 This velocity-time graph represents an object undergoing uniformly accelerated motion. 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.0 Acceleration vs. Time Just as the slope of a position-time graph reveals the rate at which position changes (velocity), the slope of a velocity-time graph reveals the rate at which velocity changes (acceleration). Calculate the slope of the line in Figure 1.30 as follows, designating forward as positive: 1.0 2.0 Time (s) 3.0 4.0 slope rise run Figure 1.31 An accelerationtime graph for an object undergoing uniformly accelerated motion is a straight line with zero slope. PHYSICS INSIGHT If the acceleration-time graph has a non-zero slope, the acceleration is changing (is nonuniform). The slope of an acceleration-time graph is called jerk, with units m/s3. v v i f t t i f 10 m/s (4 m/s) 2.5 s 1.0 s 4 m/s2 The answer is positive, so the car is accelerating at 4 m/s2 [forward]. The resulting acceleration-time graph is shown in Figure 1.31. You know that the velocity-time graph for an object undergoing uniform motion is a horizontal line (with zero slope, as in Figure 1.23). Similarly, a horizontal line on an acceleration-time graph indicates uniform acceleration. Concept Check If the position-time graph for an object undergoing positive acceleration is a parabola, such as the one in Figure 1.28, what is the shape of the position-time graph for an object undergoing negative acceleration? What would a ticker tape of the motion of an object that is slowing down look like? After driving your all-terrain vehicle (ATV, Figure 1.32) through a field, you see a wide river just ahead, so you quickly bring the vehicle to a complete stop. Notice in Figure 1.33 that, as your ATV slows down, the displacement in each time interval decreases. Figure 1.32 ATVs can undergo a wide variety of motions. 26 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 27 scale 1.0 m 0.0 m 0.0 s 13.5 m 1.0 s 24.0 m 2.0 s 31.5 m 3.0 s 36.0 m 4.0 s 37.5 m 5.0 s Figure 1.33 This ATV is undergoing non-uniform motion. It is accelerating, in this case, slowing down. Example 1.5 shows the calculations and resulting velocity-time graph for an object that is slowing down uniformly. Example 1.5 The position-time data for an ATV approaching a river are given in Table 1.6. Using these data, (a) draw a position-time graph (b) draw a velocity-time graph (c) calculate acceleration Analysis and Solution Designate the forward direction as positive. (a) For the position-time graph, ▼ Table 1.6 Time (s) Position (m [forward]) 0.0 1.0 2.0 3.0 4.0 5.0 0.0 13.5 24.0 31.5 36.0 37.5 plot the data in Table 1.6 (Figure 1.34). Position vs. Time 40.0 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0. Practice Problems 1. Draw a position-time graph from the velocity-time graph given below. Velocity vs. Time ) ] 14 12 10 8 6 4 2 0 0 10 20 Time (s) 30 40 2. Calculate the acceleration using the graph below. Velocity vs. Time ) ] 12 10 Time (s) 10 12 1.0 2.0 3.0 4.0 5.0 Time (s) Answers 1. Position vs. Time Figure 1.34 (b) Since the position-time graph is non-linear, find the slope of the tangent at 2.0 s, 3.0 s, and 5.0 s (Figures 1.35(a), (b), and (c)). ( 40.0 30.0 20.0 10.0 0.0 0.0 Position vs. Time d t 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.35(a) slope d t 32.5 m15.5 m 3.0 s1.0 s 17.0 m 2.0 s 8.5 m/s 250 200 150 100 50 ) ] . 1.0 m/s2 [N] 10 20 Time (s) 30 40 Chapter 1 Graphs and equations describe motion in one dimension. 27 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 28 Position vs. Time ( 40.0 30.0 20.0 10.0 0.0 0.0 1.0 3.0 2.0 Time (s) 4.0 5.0 e MATH Figure 1.35(b) For an alternative method to create a velocity-time graph from the position- time data points, visit www.pearsoned.ca/school/ physicssource ( 40.0 30.0 20.0 10.0 0.0 0.0 Position vs. Time 1.0 2.0 3.0 Time (s) 4.0 5.0 Figure 1.35(c) slope d t 37.0 m(26.0 m) 4.0 s2.0 s 1 1 m .0 s .0 2 5.5 m/s This tangent is a horizontal line, so its slope is zero. The slopes of the tangents give the instantaneous velocities (Table 1.7). Positive signs mean that the direction is forward. Plot the data on a velocity-time graph (Figure 1.36). ▼ Table 1.7 Time (s) Velocity (m/s [forward]) 2.0 3.0 5.0 8.5 5. 15.0 14.0 13.0 12.0 11.0 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 0.0 Velocity vs. Time 1.0 2.0 3.0 Time (s) 4.0 5.0 6.0 (c) Find acceleration by calculating the slope of the Figure 1.36 velocity-time graph. v a t 0.0 m/s (8.5 m/s) 5.0 s 2.0 s 2.8 m/s2 The acceleration of the ATV is 2.8 m/s2. Because the forward direction was designated as positive, the negative sign means that the direction of acceleration is backward. Negative Acceleration Does Not Necessarily Mean Slowing Down In Example 1.5, the value for acceleration is negative. What is the meaning of negative acceleration? When interpreting the sign of acceleration, you need to compare it to the sign of velocity. For example, for the drag racer that is speeding up, the direction of its velocity is the same as the direction of its acceleration (see the calculation of the slope of the velocitytime graph for Figure 1.30). When the directions (signs) of velocity and acceleration are the same (positive or negative), the object is speeding up. PHYSICS INSIGHT v a v a v aa v a speeding up speeding up slowing down slowing down 28 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 29 For the ATV in Example 1.5, the direction of its velocity is opposite to the direction of its acceleration, so it is slowing down. When velocity and acceleration have opposite |
directions (signs), the object slows down. Concept Check (a) Think of two more examples of objects not mentioned in this text that are speeding up and slowing down. In each case, indicate the signs or directions of velocity and acceleration. (b) Under what circumstances can an object have a negative acceleration and be speeding up? (c) You are given a position-time graph that is a curve. How can you use the slope of the tangent to determine whether the object represented in the graph is speeding up or slowing down? (Hint: How does the slope of the tangent change as you move along the position-time curve?) THEN, NOW, AND FUTURE Biomechanics and the Role of the Crash Test Dummy Understanding how biological systems move is a branch of physics known as biomechanics. For automobile manufacturers, understanding how the human body moves during a car accident is very important. To study, collect, and analyze data on how the human body moves during a vehicular collision requires a test subject. Human cadavers were the first test subjects used. While live human testing was valuable, it was limited in its scope due to the physical discomfort required and injury potential for some of the tests. Despite the production of reliable applicable data, most automobile manufacturers discontinued live animal testing in 1993 for moral and ethical reasons. Clearly, a different type of test subject needed to be designed and built. It came in the form of the now recognizable crash test dummy. Sam W. Alderson created “Sierra Sam” in 1949 to test aircraft ejection seats and pilot restraint harnesses. Then came the VIP-50 series and Sierra Stan in the 1950s. Engineers combined the best features of these different models and debuted Hybrid I in 1971. Hybrid I was known as the “50th percentile male” dummy (meaning approximately 50% of men are larger and 50% of men are smaller), with a height of 168 cm and a mass of 77 kg. A year later, Hybrid II, with improved shoulder, spine, and knee responses, was produced to test lap and shoulder belts. Still crude, their use was limited, leading to the advent of the Hybrid III family of crash test dummies that include a 50th percentile male, a 50th percentile female, and two child dummies. This family of crash test dummies is designed to measure spine and rib acceleration, and demonstrate neck movement in rearend collisions. Equipped with a more humanlike spine and pelvis, THOR (Figure 1.37) is the successor of Hybrid III. Its face contains a number of sensors for facial impact analysis. Since front and side air bags have reduced upper body injury, lower extremity injury has become more prevalent. Therefore, THOR is built with an Achilles tendon to better mimic the side-to-side, up-and-down, and rotational movements of the ankle. Even with sophisticated crash test dummies, plastic and steel can only approximate how the human body will move. The study of soft tissue injury can only be accomplished with real-life subjects. Therefore, the future of crash testing will be in cre- ating detailed computer models of human systems. Even though it is slow and cumbersome for full body simulations, the computer has the advantage of repeatability and lower cost. The programmer has the ability to control every variable and repeat each and any event. 1. Why are crash test dummies used? 2. What are some of the advantages of THOR over his previous prototypes? 3. Will crash test dummies become obsolete? Explain. Figure 1.37 THOR Chapter 1 Graphs and equations describe motion in one dimension. 29 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 30 1.3 Check and Reflect 1.3 Check and Reflect Applications 3. Match each velocity-time graph below 1. A sprinter in a championship race accelerates to his top speed in a short time. The velocity-time data for part of the race are given in the table below. Use the data to find the (a) average acceleration from 0.00 s to 0.50 s (b) average acceleration from 0.50 s to 3.00 s (c) average acceleration from 5.00 s to 6.00 s (d) Describe what was happening to the acceleration and velocity over 6.00 s. Time (s) Velocity (m/s [forward]) 0.00 0.12 0.14 0.50 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 9.83 9.93 0.00 0.00 0.00 2.80 5.00 8.00 9.80 10.80 11.30 11.60 11.70 11.80 11.90 11.95 11.97 2. Describe the motion of the object as illustrated in the graph below. Position vs. Time Time (s) 2.0 4.0 6.0 8.0 10. 20.0 0.0 20.0 40.0 60.0 80.0 100.0 30 Unit I Kinematics with the correct statement. (i) negative acceleration (ii) positive acceleration (iii) moving with zero acceleration (iv) stationary object Extensions 4. In your notebook, complete the velocity- time data table for the graph below. Position vs. Time ) ] 36 33 30 27 24 21 18 15 12 9 6 3 0 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) Time (s) Velocity (m/s [forward]) 2.0 4.0 6.0 8.0 e TEST To check your understanding of uniformly accelerated motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 31 1.4 Analyzing Velocity-time Graphs When a plane flies across Alberta with constant speed and direction, it is said to be undergoing uniform motion (Figure 1.38(a)). If you were sitting in the plane, you would experience a smooth ride. An all-terrain vehicle (ATV) bouncing and careening along a rough trail is constantly changing speed and direction in order to stay on the road. A ride in the ATV illustrates non-uniform motion, or acceleration (Figure 1.39(a)). You can distinguish between uniform and non-uniform motion by simple observation and gathering data from your observations (see Figures 1.38(b) and 1.39(b)). There are several ways to interpret the data. One way is to analyze graphs by determining their slopes to obtain further information about an object’s motion, as you did in section 1.3. In this section, you will develop this method further and learn another method of graphical analysis: how to find the area under a graph. First review the information you can obtain from the slopes of position-time and velocity-time graphs. Figure 1.38(a) Uniform motion Figure 1.39(a) Non-uniform motion Velocity vs. Time Velocity vs. Time ) ] Time (h) Figure 1.38(b) A graph representing uniform motion ) ] Time (s) Figure 1.39(b) A graph representing non-uniform motion Chapter 1 Graphs and equations describe motion in one dimension. 31 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 32 Slopes of Graphs Reveal How Objects Move Consider the three photos and velocity-time graphs in Figure 1.40. You can interpret each graph by reading values from it. To gain new information, you must analyze the graph by calculating its slope. The slope describes the object’s motion. Velocity vs. Time Velocity vs. Time Velocity vs. Time ) ] Time (s) Time (s) Time (s) Figure 1.40(a) A velocity-time graph for an object at rest Figure 1.40(b) A velocity-time graph for an object undergoing uniform motion Figure 1.40(c) A velocity-time graph for an object undergoing uniformly accelerated motion Concept Check 1. Sketch position-time graphs for the following: (a) two possibilities for stopped motion (b) two possibilities for uniform motion (c) four possibilities for uniformly accelerated motion Describe each graph in terms of direction of travel and whether the object is speeding up or slowing down. 2. Sketch the corresponding velocity-time graph for each position-time graph in question 1. For each graph, describe how you determined the graph’s shape. By analyzing the units for the slope of a velocity-time graph, m/s2, you know from section 1.3 that the slope of a velocity-time graph represents the acceleration of the object. Concept Check Sketch all the types of acceleration-time graphs you have encountered thus far. Describe the kind of motion that each graph represents. 32 Unit I Kinematics 01-PearsonPhys20-Chap01 7/25/08 7:50 AM Page 33 1-5 Design a Lab 1-5 Design a Lab Tortoise or Hare? The Question In your class, who has the fastest acceleration and the fastest average speed in the 50-m dash? Design and Conduct Your Investigation Make a list of variables that you think are likely to influence the acceleration of each participant. For each variable on your list, write a hypothesis that predicts how changes in that variable will affect the participants’ acceleration. Write a procedure for an investigation that will test the effect of one of these variables on acceleration. Clearly outline all the steps that you will follow to complete your investigation. Identify the responding and manipulated variables. List all the materials and equipment you will need, as well as all safety precautions. Compare your experimental design and procedure with those of your classmates. Identify any strengths and weaknesses. With your teacher’s approval, conduct your investigation. State any problems or questions that you found during your investigation or analysis that would need additional investigation to answer. The Area Under a Velocity-time Graph Represents Displacement Occasionally, due to a medical or other emergency, a pilot must turn the aircraft and land at the same or alternate airport. Consider the graph for the uniform motion of a plane travelling east at 300 km/h for 2.0 h only to turn back west for 0.5 h to make an emergency landing (Figure 1.41). What is the plane’s displacement for this time interval? Unit analysis indicates that the area under a velocity-time graph equals displacement. To calculate displacement using a velocity-time graph, look at the units on the axes. To end up with a unit of displacement (km) from the units km/h and h, you need to multiply: km h h km The shapes in Figure 1.41 are rectangles, so the area under the velocitytime graph is l w (length times width). In this case, find the sum of the areas above and below the time axis. Consider east to be positive. For eastward displacement, the area is above the time axis, so it is positive. For westward displacement, the area is below the time axis, so it |
is negative. For eastward displacement (above the time axis), Velocity vs. Time Time (h) 1.0 2.0 3. 400 200 0 200 400 Figure 1.41 To calculate net displacement, add the areas above and below the time axis. area dd 300 vt km h 600 km (2.0 h) Chapter 1 Graphs and equations describe motion in one dimension. 33 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 34 For westward displacement (below the time axis), dd vt 300 km h 150 km (0.5 h) PHYSICS INSIGHT Note that the answer has one significant digit because there is only one significant digit in 0.5 h. To find the plane’s net displacement, add the two areas. area ddd 600 km (150 km) 600 km 150 km 450 km 5 102 km [E] Because the net area is positive, the plane’s displacement is 5 102 km [E]. Unlike position-time graphs, where you can only calculate the slope to determine velocity, you can use velocity-time graphs to determine both acceleration and displacement, as in the next example. Example 1.6 From the graph in Figure 1.42, calculate (a) displacement (b) acceleration Velocity vs. Time Practice Problems 1. Calculate the displacement and acceleration from the graph. Velocity vs. Time ) ] .4 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 4.0 3.0 2.0 1..0 0.0 2.0 4.0 6.0 8.0 10.0 Time (min) Figure 1.42 0 2 4 6 Time (s) 8 10 Analysis and Solution (a) For displacement, find the sum of the areas under the velocity-time graph (Figure 1.43). Designate east (above the time axis) as the positive direction. Convert minutes to seconds. 2. Use the graph below to determine the displacement of the object. Velocity vs. Time Time (s) 2 4 6 8 10 12 ) ] 10 5 0 5 10 15 20 25 Answers 1. 22 m [N], 0 m/s2 [N] 2. 1.0 102 m [E] or 1.0 102 m [W] 34 Unit I Kinematics 4.0 3.0 2.0 1. Velocity vs. Time A B C 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Time (min) Figure 1.43 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 35 d Region A: vt 3.0 A 4.0 min m s 60 s 1 min (240 s) 3.0 m s 720 m Region B: d B (5.0 min 4.0 min) 1 2 60 s 1 min 3.0 m s 1.0 1.0 m s m s (5.0 min 4.0 min) 60 s 1 min (1.0 min) 60 s 1 min 2.0 1.0 m s m s (1.0 min) 60 s 1 min 1 2 1 2 (120 m) 60 m PHYSICS INSIGHT When calculating total displacement from a velocity-time graph, remember to keep track of whether the area is positive or negative. PHYSICS INSIGHT Check your answer by looking at the units. Do the units reflect the answer that you are asked to find? 120 m Region C: d C 1.0 m s 1.0 m s 300 m d A d B d d C (10.0 min 5.0 min) 60 s 1 min (5.0 min) 60 s 1 min 720 m 120 m 300 m 1140 m 1.1 103 m The answer is positive, so the direction of the displacement is east. (b) For acceleration, find the slope of each section of the graph. In region A: v a t m m 3.0 3.0 s s 240 s 0.0 m/s2 This answer makes sense because the velocity-time graph is a horizontal line. In region B: v a t m m 3.0 1.0 s s 60 s 0.033 m/s2 or 0.033 m/s2[W] Chapter 1 Graphs and equations describe motion in one dimension. 35 01-PearsonPhys20-Chap01 7/25/08 7:54 AM Page 36 Since the third part of the graph is also a horizontal line, its slope is also zero. Paraphrase (a) The displacement is 1.1 103 m [E]. (b) The acceleration is zero in regions A and C and 0.033 m/s2 [W] in region B. Concept Check For the velocity-time graph of a ball thrown up in the air (Figure 1.44), what is the net displacement of the ball? 15.00 10.00 5.00 0.00 5.00 Velocity vs. Time Time (s) 0.50 1.00 1.50 2.00 2.50 ) ] 10.00 15.00 Figure 1.44 Average Velocity from Velocity-time Graphs Objects rarely travel at constant velocity. Think of your journey to school today. Whether you travelled by car or bus, rode a bike, or walked, stop signs, traffic lights, corners, and obstacles caused a variation in your velocity, or rate of travel. If you describe your motion to a friend, you can use a series of instantaneous velocities. The more time instances you use to record your motion, the more details about your trip you can communicate to your friend (Figure 1.45(a)). However, if you were to use the equation v and substitute your total displacement d t for d and your total time of travel for t, you would lose most of the details of your journey. You would obtain a value for your average velocity, v ave (Figure 1.45(b)). Velocity vs. Time Velocity vs. Time ) ] Time (s) v ave Time (s Figure 1.45(a) By using a series of instantaneous velocities at the given times, you can precisely describe your journey. Figure 1.45(b) average velocity of the journey. It describes your journey but the detail of the motions is lost. The straight line represents the If you need to obtain an average velocity value from a velocity-time graph, recall that displacement, d , is the area under the graph. 36 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 37 To find average velocity, determine the area under the velocity-time graph and divide it by the total time. To calculate average velocity when given different displacements over different time intervals, simply add the total displacement and divide by the total time, as shown in the next example. Example 1.7 Find the average velocity of a student who jogs 750 m [E] in 5.0 min, does static stretches for 10.0 min, and then runs another 3.0 km [E] in 30.0 min. Given Choose east to be positive. Convert kilometres to metres. d 1 t1 d 2 t2 d 3 750 m [E] 750 m 5.0 min 0 m 10.0 min 3.0 km [E] 3.0 km 1000 m 1 km 3000 m t3 30.0 min Required average velocity (v ave) Analysis and Solution First add the displacement values. d d 2 d 1 750 m 0 m 3000 m 3750 m d 3 total Then add the time intervals and convert to seconds. The total time elapsed is ttotal t3 t2 t1 5.0 min 10.0 min 30.0 min (45.0 min)60 2700 s s min Average velocity equals total displacement divided by total time elapsed: v ave d t 3750 m 2700 s 1.4 m/s Since the answer is positive, the direction is east. Paraphrase The student’s average velocity is, therefore, 1.4 m/s [E]. Practice Problems 1. A person runs 10.0 m [E] in 2.0 s, then 5.0 m [E] in 1.5 s, and finally 30.0 m [W] in 5.0 s. Find the person’s average velocity. 2. Person A runs the 100-m dash in 9.84 s and then tags person B, who runs 200 m in 19.32 s. Person B then tags an out-of-shape person C, who runs 400 m in 1.90 min. Find the average velocity for the trio. Compare it to each individual’s average velocity. Assume they are all running in a straight line. Answers 1. 1.8 m/s [W] 2. 4.89 m/s [forward] A: 10.2 m/s [forward]; faster than the average velocity for the trio B: 10.4 m/s [forward]; faster than the average velocity for the trio C: 3.51 m/s [forward]; slower than the average velocity for the trio Chapter 1 Graphs and equations describe motion in one dimension. 37 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 38 As you have seen, velocity-time graphs are very useful. They provide the following information: – Reading the velocity-time graph gives you instantaneous velocity values. – Finding the slope of a velocity-time graph gives you an object’s acceleration. – The area under a velocity-time graph gives you the object’s displacement. – You can also determine the average velocity of an object over a time interval from a velocity-time graph. Example 1.8 shows you how to obtain information about an object’s velocity and acceleration. Example 1.8 A bird starts flying south. Its motion is described in the velocity-time graph in Figure 1.46.0 6.0 4.0 2.0 0.0 Velocity vs. Time F E 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0 12.0 13.0 Time (s) –2.0 A –4.0 –6.0 –8.0 B C D From the graph, determine (a) whether acceleration is positive, negative, or zero for each section (b) the value of the acceleration where it is not zero (c) when the bird changes direction Analysis and Solution Consider north to be the positive direction. (a) Acceleration is the slope of each section of the graph. A: Final velocity is more negative than the initial velocity, as the bird is speeding up in the south direction. So the slope of this line is negative. The bird’s acceleration is negative. 1 2 3 4 5 6 7 8 10 11 9 12 Time (s) Figure 1.46 Practice Problems 1. Position vs. Time ) ] 10 1 –2 –3 –4 –5 –6 –7 –8 –9 –10 –11 –12 (a) Describe the motion of the object from the graph above. (b) Draw the corresponding velocity-time graph. (c) Determine the object’s displacement. (d) When is the object stopped? 38 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 39 B: Acceleration is zero because the slope is zero (the graph Answers 1. (a) 3 m/s for 2 s, rest for 3 s, 3 m/s for 4 s, 6 m/s for 1 s (b) Velocity vs. Time ) ] 2 –4 –6 –8 5 10 15 Time (s) (c) 12 m (d) 2–5 s is a horizontal line). C: Acceleration is negative because the slope of the line is negative (as in section A). D: Acceleration is zero because the slope of the line is zero (as in section B). E: Final velocity is positive because the bird is now flying north. So the slope of this line is positive. The bird’s acceleration is positive. F: Acceleration is zero because the slope of the line is zero. (b) Acceleration is not zero for sections A, C, and E. v A.0 4.0 s s 1.0 s 0.0 s 2.0 m/s2 2.0 m/s2 [S] v C.0 6.0 s s 4.0 s 3.0 s 2.0 m/s2 2.0 m/s2 [S] v E.0 6.0 s s 10.0 s 7.0 s m 12.0 s 3.0 s 4.0 m/s2 4.0 m/s2 [N] (c) The bird changes direction at 8.5 s — it crosses the time axis at this instant. Chapter 1 Graphs and equations describe motion in one dimension. 39 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 40 The next example shows you how to use areas to find the displacement of the bird and its average velocity from a velocity-time graph. Example 1.9 From the graph in Figure 1.46, determine (a) the displacement for each section (b) the displacement for the entire flight (c) the average velocity of the flight Analysis and Solution (a) Displacement is the area between the graph and the time axis. Practice Problems 1. In your notebook, redraw the graph in Example 1.8 Practice Problem 1, but label the vertical axis “velocity (m/s [N])”. (a) Find the displacements for 0–2 s, 2–5 s, 5–7 s, 7–9 s, and 9–10 s. (b) Find the object |
’s total displacement. (c) Find the average velocity of the object. Answers 1. (a) 6 m [N], 18 m [N], 6 m [N], 6 m [N], 9 m [N] (b) 15 m [N] (c) 1.5 m/s [N] A: A l w bh 1 2 d (2.0 m s 1 )(1.0 s) (1.0 s)(2.0 2 m s ) 3.0 m B: A l w d (4.0 m s 8.0 m )(3.0 s 1.0 s) C: A l w bh 1 2 d (4.0 m s 1 )(1.0 s) (1.0 s)(2.0 2 m s ) 5.0 m D: A l w d (6.0 m s 18 m )(7.0 s 4.0 s) E: A 1 2 1 bh bh 2 (1.5 s) 6.0 d 1 2 m s (1.5 s) 6.0 1 2 m s 0.0 m F: A l w (6.0 d m s )(3.0 s) 18 m (b) Add all the displacements calculated in (a). The displacement over the entire flight is –16 m. Since north is positive, the displacement is 16 m [S]. (c) v ave d T t m 6 1 . s 0 13 1.2 m/s 40 Unit I Kinematics North is positive, so the average velocity for the flight is 1.2 m/s [S]. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 41 Drawing Position-time and Acceleration-time Graphs from Velocity-time Graphs In this section, you have learned how to use a velocity-time graph to calculate displacement. It is also useful to know how to draw position-time and acceleration-time graphs when given a velocity-time graph. Consider the following trip. A family travelling from Calgary to go camping in Banff National Park moves at 18.0 m/s [forward] in a camper van. The van accelerates for 4.0 s until it reaches a velocity of 30.0 m/s [forward]. It continues to travel at this velocity for 25.0 s. When approaching a check stop, the driver brakes, bringing the vehicle to a complete stop in 15.0 s. The velocity-time graph for the trip is given in Figure 1.47. Velocity vs. Time I II III Figure 1.47 The complete graph of the van’s motion 35.0 30.0 25.0 20.0 15.0 10.0 5..0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 The next example shows you how to create an acceleration-time graph from a velocity-time graph. Example 1.10 Use the velocity-time graph in Figure 1.47 to draw the corresponding acceleration-time graph. Analysis and Solution To find acceleration, calculate the slope for each section of the graph. The velocity-time graph has three distinct sections. The slope in each section is constant. Consider forward to be positive. Velocity vs. Time Section l Time: 0.0 s to 4.0 s ti vi tf vf 0.0 s 18.0 m/s 4.0 s 30.0 m/s 35.0 30.0 25.0 20.0 15.0 10.0 5..0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 Figure 1.48(a) Chapter 1 Graphs and equations describe motion in one dimension. 41 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 42 Practice Problems 1. For each velocity-time graph below, draw the corresponding acceleration-time graph. Velocity vs. Time (a 25 20 15 10 5 0 0 2.0 6.0 10.0 14.0 vf Time (s) (b) Velocity vs. Time ) ] 25 20 15 10 5 0 0 2.0 6.0 Time (s) 10.0 14.0 Answers 1. (a) Acceleration vs. Time ( 15 10 5 0 0 2.0 6.0 Time (s) 10.0 14.0 Acceleration vs. Time 10 5 0 5 Time (s) 2.0 6.0 10.0 14.0 (b) slope a v v i f ti tf 30.0 m/s (18.0 m/s) 4.0 s 0.0 s 3.0 m/s2 Velocity vs. Time Section ll Time: 4.0 s to 29.0 s ti vi tf 4.0 s 30.0 m/s 4.0 s 25.0 s 29.0 s 30.0 m/s v v i f ti tf slope 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 30.0 m/s (30.0 m/s) 29.0 s 4.0 s 0.0 m/s2 Figure 1.48(b) Velocity vs. Time Section lll Time: 29.0 s to 44.0 s ti vi tf 29.0 s 30.0 m/s 29.0 s 15.0 s 44.0 s 0.0 m/s vf slope a v v i f ti tf ) ] 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s) 35.0 40.0 45.0 Figure 1.48(c) 0.0 m/s (30.0 m/s) 44.0 s 29.0 s 2.0 m/s2 Now plot the values on the acceleration-time graph (Figure 1.49). Each section of the graph is a horizontal line because acceleration is constant (uniform). Acceleration vs. Time ) ] 35.0 30.0 25.0 20.0 15.0 10.0 5.0 0.0 5.0 10.0 30.0 35.0 40.0 45.0 5.0 10.0 15.0 20.0 25.0 Time (s) Figure 1.49 The next example shows you how to use a velocity-time graph to generate a position-time graph. 42 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 43 Example 1.11 Sketch a position-time graph from the velocity-time graph in Figure 1.47. Analysis and Solution To sketch the position-time graph, find the area under the velocitytime graph. Consider forward to be positive. In the first part of the velocity-time graph (0.0–4.0 s), area (displacement) is a rectangle and a triangle. The displacement is positive because the areas are above the time-axis. A l w bh 1 2 d 1 (18.0 m/s)(4.0 s) (4.0 s)(30.0 m/s 18.0 m/s) 2 96 m Since the velocity-time graph in this section has a positive slope, the car has positive acceleration, so the corresponding positiontime graph is a parabola that curves upward. On the positiontime graph, sketch a curve from the origin to the point t 4.0 s 96 m (Figure 1.50(a)). and d In the second part of the velocity-time graph (4.0–29.0 s), displacement is a rectangle. It is positive since the area is above the time-axis. A l w (30.0 m/s)(29.0 s 4.0 s) d 750 m Since the velocity-time graph has zero slope in this section, the car moves with constant velocity and the position-time graph is a straight line with a positive slope that extends from t 4.0 s and d t 29.0 s and d 96 m to 96 m 750 100 80 60 40 20 0 0 Position vs. Time 4 Time (s) Figure 1.50(a) Position vs. Time ) ] 1000 900 800 700 600 500 400 300 200 100 0 0 84 24 28 32 12 16 20 Time (s) Figure 1.50(b) Practice Problems 1. Velocity vs. Time ) ] 2 –4 –6 3 6 9 12 15 18 Time (s) (a) Describe the motion of the object illustrated above. Calculate its total displacement. (b) Draw the corresponding position-time graph. Answers 1. (a) travels with uniform motion, changes direction at 10 s, and travels with uniform motion; total displacement: 10 m [forward] (b) Position vs. Time ) ] 60 50 40 30 20 10 0 0 5 15 20 10 Time (s) 846 m (See Figure 1.50(b).) In the third part of the velocity-time graph (29.0– 44.0 s), displacement is a triangle. It is positive since the area is above the time-axis. 1 A bh 2 d (44.0 s 29.0 s)(30.0 m/s) 1 2 225 m Chapter 1 Graphs and equations describe motion in one dimension. 43 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 44 ) ] 1200 1100 1000 900 800 700 600 0 Position vs. Time 0 24 28 32 36 40 44 48 Time (s) Figure 1.50(c) Since the velocity-time graph has a negative slope, the car undergoes negative acceleration, so the slopes of the tangents of the position-time graph decrease (approach zero). The position-time graph is a parabola that curves down, from t 29.0 s and 846 m to d t 44.0 s and 846 m 225 m d 1071 m Position vs. Time (See Figure 1.50(c).) The resulting position-time graph is shown in Figure 1.51 1200 1100 1000 900 800 700 600 500 400 300 200 100 0 0 84 12 16 24 28 32 20 Time (s) 36 40 44 48 Figure 1.51 Concept Check If north is positive, sketch position-time, velocity-time, and acceleration-time graphs for an object (a) speeding up and going north (b) slowing down and going north (c) speeding up and going south (d) slowing down and going south 1.4 Check and Reflect 1.4 Check and Reflect Knowledge 1. On a ticker tape, how can you distinguish between uniform and uniformly accelerated motion? 2. Use the terms “displacement” and “velocity” to describe how uniformly accelerated motion differs from uniform motion. 3. What is the relationship between the slope of a position-time graph and velocity? 4. Compare the shape of a position-time graph for uniform motion with a positiontime graph representing uniformly accelerated motion. 5. What is the relationship between the slope of a velocity-time graph and acceleration? 6. If a velocity-time graph is a straight line with a non-zero slope, what kind of motion is the object undergoing? 44 Unit I Kinematics 7. Determine the displacement of the object whose motion is described by the following graph. Velocity vs. Time ) ] 20.0 15.0 10.0 5.0 0.0 0.0 2.0 6.0 4.0 Time (s) 8.0 10.0 8. Calculate displacement from the velocity-time graph below.0 4.0 3.0 2.0 1.0 0.0 Velocity vs. Time 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 45 9. Describe the velocity-time graph for an object undergoing negative acceleration. 10. What quantity of motion can be determined from the area under a velocity-time graph? (b) Create a graph for the question and check your answer using graphing techniques. 17. Determine acceleration from the velocity- time graph given below. Velocity vs. Time ) ] 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0 10.0 20.0 Time (s) 30.0 40.0 18. A truck travelling forward at 14.0 m/s accelerates at 1.85 m/s2 for 6.00 s. It then travels at the new speed for 35.0 s, when a construction zone forces the driver to push on the brakes, providing an acceleration of 2.65 m/s2 for 7.0 s. Draw the resulting velocity-time and position-time graphs for this motion. Extension 19. Describe the motion of the object illustrated in the graph below. Velocity vs. Time 10.0 5.0 0.0 5.0 5.0 10.0 15.0 20.0 25.0 30.0 Time (s 10.0 e TEST To check your understanding of velocity-time graphs follow the eTest links at www.pearsoned.ca/school/physicssource. 11. Compare and contrast the shape of a velocity-time graph for an object experiencing uniform motion with one experiencing uniformly accelerated motion. 12. Describe the acceleration-time graph of a car travelling forward and applying its brakes. 13. Calculate the acceleration of an object using the velocity-time graph below. Velocity vs. Time 2.0 4.0 6.0 8.0 10.0 Time (s 100 90 80 70 60 50 40 30 20 10 0 10 20 30 40 50 14. Construct an acceleration-time graph using the graph given below. Velocity vs. Time ) ] 12 10 8 6 4 2 0 0 5 10 15 Time (s) Applications 15. A motorbike increases its velocity from 20.0 m/s [W] to 30.0 m/s [W] over a distance of 200 m. Find the acceleration and the time it takes to travel this distance. 16. (a) While driving north from Lake Louise to Jasper, you travel 75 min at a velocity of 70 km/h [N] and another 96 min at 90 km/h [N]. Calculate your average velocity. Chapter 1 Graphs and equations describe motion in one dimension. 45 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 46 1.5 The Kinematics Equations A |
cylindrical piston the length of a football field controls the launch of a fighter plane from the deck of a carrier ship (Figure 1.52). Too much pressure and the nose gear is ripped off; too little pressure and the plane crashes into the ocean. This propulsion system accelerates a 20 000-kg plane from rest to 74 m/s (266 km/h) in just 2.0 s! e TECH Study the physics of jet takeoffs by visiting www.pearsoned.ca/school/ physicssource and viewing the simulation. Figure 1.52 Analyzing complex motions requires many calculations, some of which involve using the kinematics equations you will study in this section. To determine the crucial values required for launching a plane, such as flight deck length, final velocity, and acceleration, physicists and engineers use kinematics equations similar to the ones you will know by the end of this section. In this section, you will practise your analytical skills by learning how to derive the kinematics equations from your current knowledge of graphs and then apply these equations to analyze complex motions such as airplane launches. Velocity vs. Time Concept Check Create a summary chart for the information you can gather by analyzing position-time, velocity-time, and acceleration-time graphs. Use the headings “Graph Type”, “Reading the Graph”, “Slope”, and “Area”. Consider the airplane taking off from a moving aircraft carrier (Figure 1.52). The plane must reach its takeoff speed before it comes to the end of the carrier’s runway. If the plane starts from rest, the velocitytime graph representing the plane’s motion is shown in Figure 1.53. Notice that the slope of the graph is constant. By checking the units on the graph, you know that the slope represents acceleration: rise run Therefore, the velocity-time graph is a straight line. m/s2. In this case, the acceleration is constant (uniform). m/s s 80 70 60 50 40 30 20 10 ) ] .0 1.0 Time (s) 2.0 Figure 1.53 The slope of this velocity-time graph represents the plane’s acceleration. 46 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 47 From Figure 1.53, you can derive the first kinematics equation This equation can also be written as v a t The next example shows you how to apply this equation to solve a problem. Example 1.12 A hybrid car with an initial velocity of 10.0 m/s [E] accelerates at 3.0 m/s2 [E]. How long will it take the car to acquire a final velocity of 25.0 m/s [E]? Given Designate east as the positive direction. v 10.0 m/s [E] 10.0 m/s v 25.0 m/s [E] 25.0 m/s a 3.0 m/s2 [E] 3.0 m/s2 i f Required time (t) Analysis and Solution Use the equation v v v a i t t f Since you are dividing by a vector, and initial and final velocities and acceleration are in the same direction, use the scalar form of the equation. Isolate t and solve. t vi vf a 25 m/s 10 m/s 3.0 m/s2 m 15 s m 3.0 s2 5.0 s Practice Problems 1. A motorcycle with an initial velocity of 6.0 m/s [E] accelerates at 4.0 m/s2 [E]. How long will it take the motorcycle to reach a final velocity of 36.0 m/s [E]? 2. An elk moving at a velocity of 20 km/h [N] accelerates at 1.5 m/s2 [N] for 9.3 s until it reaches its maximum velocity. Calculate its maximum velocity, in km/h. Answers 1. 7.5 s 2. 70 km/h [N] PHYSICS INSIGHT The mathematics of multiplying vectors is beyond this text and division of vectors is not defined. So, when multiplying and dividing vectors, use the scalar versions of the kinematics equations. Paraphrase It will take the car 5.0 s to reach a velocity of 25.0 m/s [E]. As you know from section 1.4, you can calculate the area under a velocitytime graph. By checking the units, you can verify that the area represents displacement: l w m s s m Chapter 1 Graphs and equations describe motion in one dimension. 47 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 48 PHYSICS INSIGHT The area of a trapezoid 1 2 is given by l2)w. (l1 To calculate the displacement (area) from the velocity-time graph in Figure 1.54, you can use the formula for the area of a trapezoid, 1 A (l1 2 l2)w, which is simply the average of the parallel sides multiplied by the base. Velocity vs. Time 1 Area d (vf vi ) t 2 l2 l1 w vf vi ) ] ti t Time (s) tf Figure 1.54 graph to derive the equation d v t. ave Use the area under the velocity-time The second kinematics equation is v i (v d f)t 1 2 v where l1 to apply this equation. v i, l2 f, and w t. The next example shows you how Example 1.13 A cattle train travelling west at 16.0 m/s is brought to rest in 8.0 s. Find the displacement of the cattle train while it is coming to a stop. Assume uniform acceleration. Given Designate west as the positive direction. v v t 8.0 s 16.0 m/s [W] 16.0 m/s 0 m/s [W] 0 m/s i f Required displacement (d ) Analysis and Solution Use the equation d 1 2 (v i vv f)t and solve for d . Practice Problems 1. A hound running at a velocity of 16 m/s [S] slows down uniformly to a velocity of 4.0 m/s [S] in 4.0 s. What is the displacement of the hound during this time? 2. A ball moves up a hill with an initial velocity of 3.0 m/s. Four seconds later, it is moving down the hill at 9.0 m/s. Find the displacement of the ball from its initial point of release. Answers 1. 40 m [S] 2. 12 m 48 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 49 d (16.0 m/s 0 m/s)(8.0 s) 1 2 8.0 64 m (8.0 s) m s The sign is positive, so the train’s direction is west. Paraphrase The cattle train travels 64 m [W] before it stops. You can also calculate the area under a velocity-time graph by considering that the total area under the graph is made up of a triangle and a rectangle (Figure 1.55). vf v at vi vi area area of rectangle l w t vi t 1 2 bh 1 t 2 1 t 2 a1 2 v t a )2 t ( Figure 1.55 You can divide the area under the velocity-time graph into a triangle and a rectangle. info BIT The fastest time of covering 1.6 km while flipping tiddly winks — 52 min 10 s — was achieved by E. Wynn and J. Culliongham (UK) on August 31, 2002. Their speed was 1.8 km/h. Using this data and the displacement equation below, verify that they did indeed travel a distance of 1.6 km. In Figure 1.55, the area of a rectangle represents the displacement of an object travelling with a constant velocity, v i. The height of the rectangle is v i and the base is t. Therefore, the area of the rectangle is equal to v t. The area of the triangle represents the additional displacement resulting from the change in velocity. The height of the triangle is v and the base is t. The area of the triangle is equal to v v i f i 1 2 t(v). But vat. Therefore, the area of the triangle is equal to 1 2 (t)(at) a(t)2. Add both displacements to obtain 1 2 PHYSICS INSIGHT 2 ti (t2) tf ti)2. (t)2 (tf 2, whereas d v 1 t a(t)2 2 i The next example shows you how to apply the third kinematics equation. Chapter 1 Graphs and equations describe motion in one dimension. 49 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 50 Example 1.14 A golf ball that is initially travelling at 25 m/s hits a sand trap and slows down with an acceleration of 20 m/s2. Find its displacement after 2.0 s. Given Assign a positive direction for forward and a negative direction for backward. 25 m/s vi a 20 m/s2 t 2.0 s Practice Problems 1. A skier is moving down a uniform slope at 3.0 m/s. If the acceleration down the hill is 4.0 m/s2, find the skier’s displacement after 5.0 s. 2. A motorcycle travelling at 100 km/h on a flat road applies the brakes at 0.80 m/s2 for 1.0 min. How far did the motorcycle travel during this time? Answers 1. 65 m [down] 2. 2.3 102 m Required displacement (d ) Analysis and Solution Use the equation d v i 25 d m s (2.0 s) 1 2 50 m (40 m) 10 m t a(t)2 to solve for d . 1 2 20 m s2 (2.0 s)2 The sign is positive, so the direction is forward. Paraphrase The displacement of the golf ball is 10 m [forward]. To obtain the fourth kinematics equation, derive the value of a required variable in one equation, substitute the derived value into the second equation, and simplify. v i t f Start with a v Isolate v i. at v v Then substitute v (v v d f i f i 1 2 at for v i into the equation f)t. The equation becomes d 1 2 (v f at v f)t. This equation simplifies to v d 1 t a(t)2 2 f Apply this equation in the next example. info BIT The fastest lava flow ever recorded was 60 km/h in Nyiragongo (Democratic Republic of Congo) on January 10, 1977. At this speed, how far would the lava travel in 2 h 30 min? 50 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 51 Example 1.15 Figure 1.56 A speedboat slows down at a rate of 5.0 m/s2 and comes to a stop (Figure 1.56). If the process took 15 s, find the displacement of the boat. f 0.0 m/s (because the boat comes to rest) Given Let forward be the positive direction. v t 15 s a 5.0 m/s2 (Acceleration is negative because the boat is slowing down, so its sign must be opposite to that of velocity (positive).) Practice Problems 1. If the arresting device on an aircraft carrier stops a plane in 150 m with an acceleration of 15 m/s2, find the time the plane takes to stop. 2. The 1968 Corvette took 6.2 s to accelerate to 160 km/h [N]. If it travelled 220 m [N], find its acceleration. Answers 1. 4.5 s 2. 2.9 m/s2 [N] Required displacement (d ) Analysis and Solution Use the equation d vv f t a(t)2 to solve for d 1 2 . d (0.0 m/s)(15 s) 1 2 (5.0 m/s2)(15 s)2 562.5 m 5.6 102 m The sign is positive, so the direction of displacement is forward. Paraphrase The displacement of the speedboat is 5.6 102 m [forward]. Deriving the fifth and last kinematics equation involves using the difference of squares, another math technique. Isolate t in the equation a v f v i. Remember that, when multiplying t or dividing vectors, use the scalar form of the equation: t vi vf a PHYSICS INSIGHT Recall that the difference of squares is (a b)(a b) a2 b2 Then substitute the expression for t into d (vi d (vi vi vf a 1 2 1 2 vf)t: vf) 2 vi 2 vf a 1 d 2 Chapter 1 Graphs and equations describe motion in one dimension. 51 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 52 info BIT A tort |
oise covered 5.48 m in 43.7 s at the National Tortoise Championships in Tickhill, UK, to set a world record on July 2, 1977. It was moving at 0.45 km/h. The more standard form of the fifth kinematics equation is vf 2 vi 2 2ad This equation is applied in the next example. Example 1.16 Practice Problems 1. A jetliner lands on a runway at 70 m/s, reverses its engines to provide braking, and comes to a halt 29 s later. (a) What is the jet’s acceleration? (b) What length of runway did the jet require to come safely to a complete stop? 2. On-ramps are designed so that motorists can move seamlessly into highway traffic. If a car needs to increase its speed from 50 km/h to 100 km/h and the engine can provide a maximum acceleration of magnitude 3.8 m/s2, find the minimum length of the on-ramp. Answers 1. (a) 2.4 m/s2 [forward] (b) 1.0 km 2. 76 m A bullet accelerates the length of the barrel of a gun (0.750 m) with a magnitude of 5.35 105 m/s2. With what speed does the bullet exit the barrel? Given a 5.35 105 m/s2 d 0.750 m Required final speed (vf) Analysis and Solution Use the equation vf from rest, vi 0 m/s. 2 vi 2 2ad. Since the bullet starts vf vf 2 (0 m/s)2 2(5.35 105 m/s2)(0.750 m) 802 500 m2/s2 802 500 m2/s2 896 m/s Paraphrase The bullet leaves the barrel of the gun with a speed of 896 m/s. It is important to note that the velocity-time graph used to derive the kinematics equations has a constant slope (see Figure 1.54), so the equations derived from it are for objects undergoing uniformly accelerated motion (constant acceleration). General Method of Solving Kinematics Problems Now that you know five kinematics equations, how do you know which one to use to solve a problem? To answer this question, notice that each of the five kinematics equations has four variables. Each kinematics problem will provide you with three of these variables, as given values. The fourth variable represents the unknown value. When choosing your equation, make sure that all three known variables and the one unknown variable are represented in the equation (see Table 1.8). You may need to rearrange the equation to solve for the unknown variable. 52 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 53 ▼ Table 1.8 The Variables in the Five Kinematics Equations Equation v i v f)t v d 1 t a(t)2 2 i v d 1 t a(t)2 2 f 2 vi vf 2 2ad PHYSICS INSIGHT Remember these implied given values. – If the object starts from 0. – If the object comes to rest, v i f a stop, v 0. – If the object experiences uniform motion, a 0. 1.5 Check and Reflect 1.5 Check and Reflect Applications 1. How far will a humanoid robot travel in 3.0 s, accelerating at 1.0 cm/s2 [forward], if its initial velocity is 5.0 cm/s [forward]? 2. What is the displacement of a logging truck accelerating from 10 m/s [right] to 20 m/s [right] in 5.0 s? 3. How far will a car travel if it starts from rest and experiences an acceleration of magnitude 3.75 m/s2 [forward] for 5.65 s? 4. Determine the acceleration of a bullet starting from rest and leaving the muzzle 2.75 103 s later with a velocity of 460 m/s [forward]. 5. Some aircraft are capable of accelerations of magnitude 42.5 m/s2. If an aircraft starts from rest, how long will it take the aircraft to travel down the 2.6-km runway? 6. If a cyclist travelling at 14.0 m/s skids to a stop in 5.60 s, determine the skidding distance. Assume uniform acceleration. 7. Approaching a flashing pedestrianactivated traffic light, a driver must slow down to a speed of 30 km/h. If the crosswalk is 150 m away and the vehicle’s initial speed is 50 km/h, what must be the magnitude of the car’s acceleration to reach this speed limit? 8. A train’s stopping distance, even when full emergency brakes are engaged, is 1.3 km. If the train was travelling at an initial velocity of 90 km/h [forward], determine its acceleration under full emergency braking. 9. A rocket starts from rest and accelerates uniformly for 2.00 s over a displacement of 150 m [W]. Determine the rocket’s acceleration. 10. A jet starting from rest reaches a speed of 241 km/h on 96.0 m of runway. Determine the magnitude of the jet’s acceleration. 11. What is a motorcycle’s acceleration if it starts from rest and travels 350.0 m [S] in 14.1 s? 12. Determine the magnitude of a car’s acceleration if its stopping distance is 39.0 m for an initial speed of 97.0 km/h. 13. A typical person can tolerate an acceleration of about 49 m/s2 [forward]. If you are in a car travelling at 110 km/h and have a collision with a solid immovable object, over what minimum distance must you stop so as to not exceed this acceleration? 14. Determine a submarine’s acceleration if its initial velocity is 9.0 m/s [N] and it travels 1.54 km [N] in 2.0 min. e TEST To check your understanding of the kinematics equations, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 53 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 54 1.6 Acceleration due to Gravity Figure 1.57 Amusement park rides are an application of physics. projectile motion: motion in a vertical plane projectile: an object released or thrown into the air Many amusement parks and midways showcase a ride based solely on acceleration due to gravity. The ride transports thrill seekers up to a dizzying height, allows them to come to rest, and then, without warning, releases them downward before coming to a controlled stop (Figure 1.57). In the previous sections, you learned about objects that move in a horizontal plane. Many objects move in a vertical plane. Flipping a coin, punting a football, and a free throw in basketball are all examples of objects experiencing motion in a vertical plane (Figure 1.58). This type of motion is called projectile motion. A projectile is any object thrown into the air. Projectiles include objects dropped from rest; objects thrown downward or vertically upward, such as a tennis ball for service; and objects moving upward at an angle, such as a punted football. First let’s consider projectile motion of an object moving straight up or down. What is the relationship between an object’s mass and the speed of its fall? Do the next QuickLab to find out. x (horizontal) y (vertical) Figure 1.58 A plane has two dimensions, x and y. 1-6 QuickLab 1-6 QuickLab The Bigger They Are . . . Problem Does mass affect how quickly an object falls? Materials two objects of similar size and shape but different mass, such as a marble and a ball bearing, a die and a sugar cube, a golf ball and a table tennis ball two pans chair Procedure 1 Place a pan on either side of the chair. 2 Standing on the chair, release each pair of objects from the same height at the same time. 3 Listen for the objects hitting the pans. 4 Repeat steps 2 and 3 for other pairs of objects. 5 Repeat steps 2 and 3 from a higher and a lower height. Questions 1. Did the pair of objects land at the same time? 2. How did a change in height affect how long it took the objects to drop? 3. How did a change in the objects’ shape affect how long it took each pair to drop? In the 16th century, Galileo conducted experiments that clearly demonstrated that objects falling near Earth’s surface have a constant acceleration, neglecting air resistance, called the acceleration due to gravity. You can determine the value of the acceleration due to Earth’s gravity by performing the following experiment. info BIT In 1971, astronaut David Scott tested Galileo’s theory with a feather and a hammer. With no air on the Moon, both objects hit the ground at the same time. 54 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 55 1-7 Inquiry Lab 1-7 Inquiry Lab Determining the Magnitude of the Acceleration due to Gravity Required Skills Initiating and Planning Performing and Recording Analyzing and Interpreting Communication and Teamwork Question How can position-time and velocity-time graphs be used to determine the acceleration due to gravity? 7 Using a ruler, measure the position of the object at each time interval and record it in your data table. 8 Plot your collected data on a position-time graph. Materials and Equipment 60-Hz spark timer ticker tape carbon disk power supply small mass metre-stick or ruler Procedure masking tape C-clamp retort stand graph paper cushion 9 With a sweeping motion, practise connecting the dots in a smooth curve that best fits the data. 10 Construct a data table in your notebook for recording instantaneous velocity and time. 11 Draw three tangents on the position-time graph. 12 Calculate the instantaneous velocities at these points by determining the slopes of the tangents. Record the data in your table. 1 Construct a data table in your notebook for recording 13 Plot a velocity-time graph of your collected data. time and position. 14 Draw a line of best fit. 2 Set up materials as shown in Figure 1.59(a), ensuring that the timer is 1.5 m above the floor. 15 Calculate the acceleration experienced by the object, in m/s2, by finding the slope of the velocity-time graph. recording timer Analysis ticker tape mass cushion 1. Determine the experimental value of the magnitude of acceleration due to gravity by averaging your group’s results. 2. Determine the percent error for your experimental value. Assume the theoretical magnitude of a is 9.81 m/s2. 3. Describe the shape of the position-time graph you drew in step 9. Figure 1.59(a) 4. From your graph, describe the relationship between time and displacement for an accelerating object. 3 Attach a 1.5-m strip of ticker tape to the mass and thread the ticker tape through the spark timer. 4 Turn on the spark timer just before your partner releases the mass. 5 Repeat steps 3 and 4 for each person in your group. 6 Analyze the ticker tape by drawing a line through the first distinct dot on the tape. Label it “start”. (On a 60-Hz timer, every sixth dot represents 0.10 s.) Continue labelling your ticker tape as shown in Figure 1.59(b). t 0.10 s start |
t 0.20 s t 0.30 s d 1 d 2 Figure 1.59(b) e LAB d 3 For a probeware activity, go to www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 55 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 56 e TECH Use graphical analysis to determine acceleration due to Earth’s gravity. Go to www.pearsoned.ca/school/ physicssource. vi 0 a 9.81 m/s2 [down] dy Figure 1.60 The time it takes the golf ball to hit the ground depends on the height from which it drops and on the acceleration due to gravity. PHYSICS INSIGHT The equations of parabolas are quadratic equations because they include a power of two, for example, y x2. The equation for the displacement of a vertical projectile is d v 1 t a(t)2. 2 i Gravity Causes Objects to Accelerate Downward Recall the kinematics equations for accelerated motion from section 1.5(t)2 2 1 t a(t)2 2 vf 2 vi 2 2ad You can also apply these equations to motion in a vertical plane. is the acceleration due Because vertical acceleration is due to gravity, a to gravity, or 9.81 m/s2 [down]. If you drop a golf ball from a height of 1.25 m, how long will it take for the ball to reach the ground (Figure 1.60)? Because the ball is moving in only one direction, down, choose down to be positive for simplicity. Since the golf ball is accelerating due to gravity starting from rest, v i 0 and a 9.81 m/s2 [down] 9.81 m/s2 The ball’s displacement can be expressed as 1.25 m [down], or 1.25 m. The equation that includes all the given variables and the unknown variable is d v i t a(t)2. The displacement and acceleration vectors 1 2 are both in the same direction, so use the scalar form of the equation to solve for time. Since vi 0, 1 d at2 2 2d t a 2(1.25 m) m 9.81 s2 0.505 s info BIT Without a parachute, Vesna Vulovic, a flight attendant, survived a fall of 10 160 m when the DC-9 airplane she was travelling in exploded. The golf ball takes 0.505 s to reach the ground when released from a rest height of 1.25 m. Note that the time it takes for an object to fall is directly proportional to the square root of the height it is dropped from: t . If there is 2d a no air resistance, the time it takes for a falling object to reach the ground depends only on the height from which it was dropped. The time does not depend on any other property of the object. 56 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 57 1-8 QuickLab 1-8 QuickLab Could You Be a Goalie for the NHL? Problem What is your reaction time? Materials and Equipment long ruler (30 cm or more) flat surface Procedure 1 Rest your arm on a flat surface with your wrist at the edge. 2 Ask your partner to hold the ruler vertically so that the ruler’s end is just above your hand. 3 Curl your fingers so that the space between your thumb and index finger is large enough for the ruler to pass through easily. 4 Without watching your partner, ask your partner to let go of the ruler without warning. 5 Try to close your thumb and index finger on the ruler as quickly as possible. 6 Record where your hand is on the ruler. 7 Repeat steps 1–6 several times. Questions 1. Determine the average distance the ruler falls in each of your trials. 2. Using the average distance, calculate the time. 3. An average NHL goalie has a reaction time of 0.15 s. How does your reaction time compare with your partner’s? 4. Certain drugs impair reaction time. What would you expect your results in this lab to be if your reaction time were increased? Instead of dropping an object such as a golf ball, what if you threw an object down? By throwing an object straight down, you give the object an initial vertical velocity downward. What effect does an initial velocity have on the motion of the object? The next example will show you. Example 1.17 While cliff diving in Mexico, a diver throws a smooth, round rock straight down with an initial speed of 4.00 m/s. If the rock takes 2.50 s to land in the water, how high is the cliff? Given For convenience, choose down to be positive because down is the only direction of the ball’s motion. v 4.00 m/s [down] 4.00 m/s t 2.50 s a 9.81 m/s2 [down] 9.81 m/s2 i Required height of cliff (d) Analysis and Solution The initial velocity and acceleration vectors are both in the same direction, so use the scalar form of the equation d v 1 t a(t)2. 2 i Chapter 1 Graphs and equations describe motion in one dimension. 57 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 58 d (4.00 m/s)(2.50 s) (9.81 m/s2)(2.50 s)2 1 2 10.0 m 30.7 m 40.7 m Paraphrase The cliff is 40.7 m high. Practice Problems 1. If a rock takes 0.750 s to hit the ground after being thrown down from a height of 4.80 m, determine the rock’s initial velocity. 2. Having scored a touchdown, a football player spikes the ball in the end zone. If the ball was thrown down with an initial velocity of 2.0 m/s from a height of 1.75 m, determine how long it is in the air. 3. An elevator moving downward at 4.00 m/s experiences an upward acceleration of 2.00 m/s2 for 1.80 s. What is its velocity at the end of the acceleration and how far has it travelled? Answers 1. 2.72 m/s [down] 2. 0.43 s 3. 0.400 m/s [down], 3.96 m What Goes Up Must Come Down Circus clowns are often accomplished jugglers (Figure 1.61). If a juggler throws a ball upward, giving it an initial velocity, what happens to the ball (Figure 1.62)? vi a dy Figure 1.61 Juggling is an example of projectile motion. Figure 1.62 The ball’s motion is called vertical projectile motion. When you throw an object up, its height (displacement) increases while its velocity decreases. The decrease in velocity occurs because the object experiences acceleration downward due to gravity (Figure 1.63(a)). The ball reaches its maximum height when its vertical velocity equals zero. In other words, it stops for an instant at the top of its path (Figure 1.63(b)). When the object falls back toward the ground, it speeds up because of the acceleration due to gravity (Figure 1.63(c)). 58 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 59 a vi v 0 a vf a Figure 1.63(a) Stage 1: Velocity and acceleration are in opposite directions, so the ball slows down. Figure 1.63(b) Stage 2: The ball has momentarily stopped, but its acceleration is still 9.81 m/s2 [down], which causes the ball to change direction. Figure 1.63(c) Stage 3: Velocity and acceleration are in the same direction, so the ball speeds up. The next two examples analyze different stages of the same object’s motion. Example 1.18 analyzes the upward part of the motion of an object thrown upward, whereas Example 1.19 analyzes the same object’s downward motion. Example 1.18 A clown throws a ball upward at 10.00 m/s. Find (a) the maximum height the ball reaches above its launch height (b) the time it takes to do so Given Consider up to be positive. 10.00 m/s [up] 10.00 m/s vi a 9.81 m/s2 [down] 9.81 m/s2 Required (a) maximum height above launch height (d) (b) time taken to reach maximum height (t) Analysis and Solution (a) When you throw an object up, as its height increases, its speed decreases because the object is accelerating downward due to gravity. The ball, travelling upward away from its initial launch height, reaches its maximum height when its vertical velocity is zero. In other words, the object stops for an instant at the top of its path up, so vf neglecting air friction, use the equation vf substitute scalar quantities. 0.00 m/s. To find the object’s maximum height, 2 2ad and 2 vi Practice Problem 1. The Slingshot drops riders 27 m from rest before slowing them down to a stop. How fast are they moving before they start slowing down? Answer 1. 23 m/s e WEB Can you shoot an object fast enough so that it does not return to Earth? Research escape velocity. Is it the same regardless of the size of an object? How do you calculate it? Write a brief summary of your findings. To learn more about escape velocity, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 59 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 60 d 2 vi 2 vf 2a m m 2 2 10.00 0.00 s s m 29.81 s2 5.10 m v , where a is the (b) To find the time taken, use the equation a t acceleration due to gravity. Substitute scalar quantities because you are dividing vectors. t vi vf a m m 10.00 0.00 s s m 9.81 s2 1.02 s Paraphrase (a) The ball’s maximum height is 5.10 m above its launch height. (b) It takes the ball 1.02 s to reach maximum height. The next example is a continuation of the previous example: It analyzes the same ball’s motion as it falls back down from its maximum height. Example 1.19 A clown throws a ball upward at 10.00 m/s. Find (a) the time it takes the ball to return to the clown’s hand from maximum height (b) the ball’s final velocity Given Consider up to be positive. v 10.00 m/s [up] 10.00 m/s a 9.81 m/s2 [down] 9.81 m/s2 i Practice Problems 1. A pebble falls from a ledge 20.0 m high. (a) Find the velocity with which it hits the ground. (b) Find the time it takes to hit the ground. Answers 1. (a) 19.8 m/s [down] (b) 2.02 s 60 Unit I Kinematics Required (a) time taken to land (t) (b) final velocity (v f) Analysis and Solution (a) For an object starting from rest at maximum height and accelerating downward due to gravity, its motion is described by the equation d v i t a(t)2, where 1 2 i 0 (at maximum height). For downward motion, the v ball’s displacement and acceleration are in the same direction, so use the scalar form of the equation. For d, substitute 5.10 m (from Example 1.18(a)). Rearrange this equation and substitute the values. 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 61 t 2d a 2(5.10 m) m 9.81 s2 1.02 s Compare this time to the time taken to reach maximum height (Example 1.18(b)). (b) The ball’s final velocity (starting from maximum height) when it info BIT At terminal velocity, parachuters no longer accelerate but fall at a constant speed. Humans have a terminal velocity of about 321 km/h [down] when c |
urled up and about 201 km/h [down] with arms and legs fully extended to catch the wind. v lands on the ground is v a i t at v f f i v 0.00 m/s (9.81 m/s2)(1.02 s) 10.0 m/s The negative sign means that the direction is downward. Paraphrase (a) It takes the ball 1.02 s to return to the clown’s hand. (b) The final velocity at the height of landing is 10.0 m/s [down]. Concept Check (a) Why does it make sense that the time taken to travel up to the maximum height is equal to the time to fall back down to the starting height? (b) What variables determine how long a projectile is in the air? Does the answer surprise you? Why or why not? Position vs. Time ) ] .00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 0.50 1.00 2.00 2.50 1.50 Time (s) You can use the data calculated in Examples 1.18 and 1.19 to plot a position-time graph of the ball’s motion. Because the ball experiences uniformly accelerated motion, the graph is a parabola (Figure 1.64). Figure 1.64 The positiontime graph of a ball thrown vertically upward is a parabola. A Graphical Representation of a Vertical Projectile You can now represent the motion of the juggler’s ball on a position-time graph. Remember that the ball’s motion can be divided into three different stages: Its speed decreases, becomes zero, and then increases. However, the velocity is uniformly decreasing. The graphs that correspond to these three stages of motion are shown in Figure 1.65. Position vs. Time Position vs. Time Position vs. Time ) ] Time (s) Time (s) Time (s) Figure 1.65(a) Consider up to be positive. The ball rises until it stops. Figure 1.65(b) momentarily at maximum height. The ball stops Figure 1.65(c) back down to its launch height. The ball falls Chapter 1 Graphs and equations describe motion in one dimension. 61 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 62 Position vs. Time ) ] Time (s) Figure 1.66 A ball thrown straight up in the air illustrates uniformly accelerated motion. Now put these three graphs together to generate the complete positiontime graph of the ball’s motion. Remember that the ball is actually moving straight up and down, and not in a parabolic path (Figure 1.66). Why is the graph of its motion a parabola rather than a straight vertical line? To generate a corresponding velocity-time graph from the positiontime graph in Figure 1.66, draw a series of tangents at specific time instances. Choosing strategic points will make your task easier. The best points to choose are those that begin and end a stage of motion because they define that stage (Figure 1.67(a)). Position vs. Time (a) 6.00 5.00 4.00 3.00 2.00 1.00 ) ] .00 0.00 0.50 1.00 1.50 2.00 2.50 Time (s) 15.00 10.00 5.00 0.00 5.00 Velocity vs. Time Time (s) 0.50 1.00 1.50 2.00 2.50 (b 10.00 15.00 Figure 1.67 To generate the velocity-time graph in (b) corresponding to the positiontime graph in (a), draw tangents at strategic points. info BIT In 1883, the Krakatoa volcano in Indonesia hurled rocks 55 km into the air. This volcanic eruption was 10 000 times more powerful than the Hiroshima bomb! Can you find the time it took for the rocks to reach maximum height? In Figure 1.67(a), notice that the initial slope of the tangent on the positiontime graph is positive, corresponding to an initial positive (upward) velocity on the velocity-time graph below (Figure 1.67(b)). The last tangent has a negative slope, corresponding to a final negative velocity on the velocity-time graph. The middle tangent is a horizontal line (slope equals zero), which means that the ball stopped momentarily. Remember that the slope of a velocity-time graph represents acceleration. Concept Check What should be the value of the slope of the velocity-time graph for vertical projectile motion? 62 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 63 1.6 Check and Reflect 1.6 Check and Reflect Knowledge 1. Define a projectile. 2. What determines how long it will take an object to reach the ground when released with an initial velocity of zero? Applications 11. A penny is dropped from a cliff of height 190 m. Determine the time it takes for the penny to hit the bottom of the cliff. 12. A coin tossed straight up into the air takes 2.75 s to go up and down from its initial release point 1.30 m above the ground. What is its maximum height? 3. A student drops a bran muffin from the roof of the school. From what height is the muffin dropped if it hits the ground 3.838 s later? 13. If a diver starts from rest, determine the amount of time he takes to reach the water’s surface from the 10-m platform. 4. During a babysitting assignment, a babysitter is constantly picking up toys dropped from the infant’s highchair. If the toys drop from rest and hit the floor 0.56 s later, from what height are they being dropped? 5. A rock takes 1.575 s to drop 2.00 m down toward the surface of the Moon. Determine the acceleration due to gravity on the Moon. 6. At the beginning of a game, a referee throws a basketball vertically upward with an initial speed of 5.0 m/s. Determine the maximum height above the floor reached by the basketball if it starts from a height of 1.50 m. 7. A student rides West Edmonton Mall’s Drop of Doom. If the student starts from rest and falls due to gravity for 2.6 s, what will be his final velocity and how far will he have fallen? 8. If the acceleration due to gravity on Jupiter is 24.8 m/s2 [down], determine the time it takes for a tennis ball to fall 1.75 m from rest. 9. If a baseball popped straight up into the air has a hang time (length of time in the air) of 6.25 s, determine the distance from the point of contact to the baseball’s maximum height. 10. Jumping straight up, how long will a red kangaroo remain in the air if it jumps through a height of 3.0 m? 14. A person in an apartment building is 5.0 m above a person walking below. She plans to drop some keys to him. He is currently walking directly toward a point below her at 2.75 m/s. How far away is he if he catches the keys 1.25 m above the ground? Extensions 15. A rocket launched vertically upward accelerates uniformly for 50 s until it reaches a velocity of 200 m/s [up]. At that instant, its fuel runs out. (a) Calculate the rocket’s acceleration. (b) Calculate the height of the rocket when its fuel runs out. (c) Explain why the rocket continues to gain height for 20 s after its fuel runs out. (d) Calculate the maximum height of the rocket. 16. A ball is dropped from a height of 60.0 m. A second ball is thrown down 0.850 s later. If both balls reach the ground at the same time, what was the initial velocity of the second ball? e TEST To check your understanding of projectiles and acceleration due to gravity, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 63 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 64 CHAPTER 1 SUMMARY Key Terms and Concepts distance displacement velocity uniform motion at rest acceleration non-uniform motion instantaneous velocity tangent uniformly accelerated motion projectile motion projectile acceleration due to gravity kinematics origin position scalar quantity vector quantity Key Equations v f v i)t d vv 1 t a(t)2 2 i d v 1 t a(t)2 2 f 2 vi vf 2 2ad Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. kinematics uniform motion at rest constant velocity accelerated motion uniform displacement : zero dt graph— vt graph— at graph— equations: none displacement : equal changes over equal time intervals dt graph— vt graph— horizontal line (/) at graph— displacement : changes dt graph— curve vt graph— at graph— horizontal line (/) equations: equations: 1 d (vf vi )t 2 vf 2 vi 2 2ad 64 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 65 CHAPTER 1 REVIEW Knowledge 1. (1.1) State two ways in which a vector quantity differs from a scalar quantity. Give an example of each. 2. (1.2) Complete a position-time data table for the motion described by the ticker tape given below. 3. (1.3) Determine the velocity of each object whose motion is represented by the graphs below. (a) (b) (c 12 10 8 6 4 2 0 Position vs. Time 0 2 4 6 8 10 12 Time (s) Position vs. Time Time (min) 2 4 6 8 10 12 Position vs. Time Time (s) 5 10 15 20 ) ] 10 5 0 5 10 15 20 25 ) ] 10 5 0 5 10 15 20 25 30 4. (1.5) What is a vehicle’s displacement if it travels at a velocity of 30.0 m/s [W] for 15.0 min? 5. (1.5) How long will it take a cross-country skier, travelling 5.0 km/h, to cover a distance of 3.50 km? 6. (1.2) Determine the average speed, average velocity, and net displacement from the position-time graph below. Position vs. Time ) ] 30.0 25.0 20.0 15.0 10.0 5.0 0.0 5.0 10.0 15.0 20.0 25.0 30.0 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 18.0 20.0 Time (s) 7. (1.2) Explain how a person standing still could have the same average velocity but different average speed than a person running around a circular track. 8. (1.6) If an object thrown directly upwards remains in the air for 5.6 s before it returns to its original position, how long did it take to reach its maximum height? 9. (1.6) If an object thrown directly upwards reaches its maximum height in 3.5 s, how long will the object be in the air before it returns to its original position? Assume there is no air resistance. 10. (1.6) What is the initial vertical velocity for an object that is dropped from a height, h? Applications 11. A scuba diver swims at a constant speed of 0.77 m/s. How long will it take the diver to travel 150 m at this speed? 12. In 1980, during the Marathon of Hope, Terry Fox ran 42 km [W] a day. Assuming he ran for 8.0 h a day, what was his average velocity in m/s? 13. Explain how the point of intersection of two functions on a position-time graph differs from the point of intersection of two functions on a velocity-time graph. 14. A thief snatches a handbag and runs north at 5.0 m/s. A police officer, 20 |
m to the south, sees the event and gives chase. If the officer is a good sprinter, going 7.5 m/s, how far will she have to run to catch the thief? Chapter 1 Graphs and equations describe motion in one dimension. 65 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 66 Maintaining that acceleration, how long will it take the police car to catch up with the speeding motorist? At what speed would the police car be moving? Explain whether or not this scenario is likely to happen. 25. Two cars pass one another while travelling on a city street. Using the velocity-time graph below, draw the corresponding position-time graph and determine when and where the two cars pass one another. Velocity vs. Time ) ] 20.0 18.0 16.0 14.0 12.0 10.0 8.0 6.0 4.0 2.0 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) 26. Calculate displacement and acceleration from the graph below. Velocity vs. Time Time (h) 1.0 2.0 3.0 4.0 5..0 0.0 1.0 2.0 3.0 4.0 5.0 27. Built in Ontario, the world’s fastest fire truck, the Hawaiian Eagle, can accelerate at 9.85 m/s2 [forward]. Starting from rest, how long will it take the Hawaiian Eagle to travel a displacement of 402 m [forward]? 28. A vehicle is travelling at 25.0 m/s. Its brakes provide an acceleration of 3.75 m/s2 [forward]. What is the driver’s maximum reaction time if she is to avoid hitting an obstacle 95.0 m away? 29. Off-ramps are designed for motorists to decrease their vehicles’ velocities to move seamlessly into city traffic. If the off-ramp is 1.10 km long, calculate the magnitude of a vehicle’s acceleration if it reduces its speed from 110.0 km/h to 60.0 km/h. 15. Calculate the magnitude of a bullet’s acceleration if it travels at a speed of 1200 m/s and stops within a bulletproof vest that is 1.0 cm thick. 16. From the velocity-time graph below, determine how far an elk will travel in 30 min. Velocity vs. Time for an Elk ) ] 80 70 60 50 40 30 20 10 0 0.0 0.5 1.0 Time (h) 1.5 2.0 17. The world record for a speedboat is 829 km/h. Heading south, how far will the boat travel in 2.50 min? 18. How much faster is an airliner than a stagecoach if the stagecoach takes 24 h to travel 300 km and the airliner takes 20 min? 19. A car’s odometer reads 22 647 km at the start of a trip and 23 209 km at the end. If the trip took 5.0 h, what was the car’s average speed in km/h and m/s? 20. A motorcycle coasts downhill from rest with a constant acceleration. If the motorcycle moves 90.0 m in 8.00 s, find its acceleration and velocity after 8.00 s. 21. A cyclist crosses a 30.0-m bridge in 4.0 s. If her initial velocity was 5.0 m/s [N], find her acceleration and velocity at the other end of the bridge. 22. An object with an initial velocity of 10.0 m/s [S] moves 720 m in 45.0 s along a straight line with constant acceleration. For the 45.0-s interval, find its average velocity, final velocity, and acceleration. 23. During qualifying heats for the Molson Indy, a car must complete a 2.88-km lap in 65 s. If the car goes 60 m/s for the first half of the lap, what must be its minimum speed for the second half to still qualify? 24. A car travelling 19.4 m/s passes a police car at rest. As it passes, the police car starts up, accelerating with a magnitude of 3.2 m/s2. 66 Unit I Kinematics 01-PearsonPhys20-Chap01 7/23/08 11:43 AM Page 67 30. Calgary’s CTrain leaves the 10 St. S.W. station at 4:45 p.m. and arrives at 3 St. S.E. at 4:53 p.m. If the distance between the two stops is 3.2 km, determine the CTrain’s average velocity for the trip. 37. A contractor drops a bolt from the top of a roof located 8.52 m above the ground. How long does it take the bolt to reach the ground, assuming there is no air resistance? McKnight – Westwinds (2010) Whitehorn Rundle Marlborough 38. An improperly installed weathervane falls from the roof of a barn and lands on the ground 1.76 s later. From what height did the weathervane fall and how fast was it travelling just before impact? Dalhousie Brentwood N CTrain Map University Banff Trail Lions Park Sunnyside BOWRIVER SAIT/ ACAD/ Jubllee 7 St. S W PRINCE’S ISLAND PARK 4 St. S W 1 St. S W Bridgeland/ M e m orial Zoo Franklin Barlo w/ M ax Bell Olym pic Plaza 3 St. SE City Hall 10 St. S W 8 St. S W 6 St. S W 3 St. S W Centre St Erlton/Stampede Victoria Park/Stampede B O W RIVER 201 Somerset/Bridlewood/Dalhousie 202 10 St./Whitehorn 7 Avenue Free Fare Zone 201 & 202 Transfer Points Chinook Southland Canyon Meadows NEW Shawnessy 39 Avenue Heritage Anderson Fish Creek – Lacombe Somerset/Bridlewood NEW 31. Describe the motion of the truck from the velocity-time graph below. When is the truck at rest? travelling with a uniform velocity? When is its acceleration the greatest? Velocity vs. Time ) ] 20.0 15.0 10.0 5.0 0.0 0.0 2.0 4.0 6.0 Time (s) 8.0 10.0 32. A racecar accelerates uniformly from 17.5 m/s [W] to 45.2 m/s [W] in 2.47 s. Determine the acceleration of the racecar. 33. How long will it take a vehicle travelling 80 km/h [W] to stop if the average stopping distance for that velocity is 76.0 m? 34. The Slingshot, an amusement park ride, propels its riders upward from rest with an acceleration of 39.24 m/s2. How long does it take to reach a height of 27.0 m? Assume uniform acceleration. 35. Starting from rest, a platform diver hits the water with a speed of 55 km/h. From what height did she start her descent into the pool? 36. A circus performer can land safely on the ground at speeds up to 13.5 m/s. What is the greatest height from which the performer can fall? 39. Attempting to beat the record for tallest Lego structure, a student drops a piece from a height of 24.91 m. How fast will the piece be travelling when it is 5.0 m above the ground and how long will it take to get there? Extension 40. Weave zones are areas on roads where vehicles are changing their velocities to merge onto and off of busy expressways. Suggest criteria a design engineer must consider in developing a weave zone. Consolidate Your Understanding Create your own summary of kinematics by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869–871. Use the Key Terms and Concepts listed on page 64 and the Learning Outcomes on page 4. 1. Create a flowchart to describe the changes in position, velocity, and acceleration for both uniform and accelerated motion. 2. Write a paragraph explaining the two main functions of graphing in kinematics. Share your report with another classmate. Think About It Review your answers to the Think About It questions on page 5. How would you answer each question now? e TEST To check your understanding of motion in one dimension, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 1 Graphs and equations describe motion in one dimension. 67 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 68 Vector components describe motion in two dimensions. Figure 2.1 The motion of Canada’s Snowbird precision flight squad can be described using vectors. Imagine being a pilot for the Canadian Snowbirds (Figure 2.1). This precision flight team, composed of highly trained military personnel, performs at air shows across the country. Unlike the flight crew in the cockpit of a commercial airliner, these pilots execute aerobatic manoeuvres that require motion in both horizontal and vertical directions, while being acutely aware of their positions relative to the ground and to each other. In this chapter, you will study motion in one and two dimensions by building on the concepts you learned in Chapter 1. You will use vectors to define position, velocity, and acceleration, and their interrelationships. The vector methods you will learn will allow you to study more complex motions. C H A P T E R 2 Key Concepts In this chapter, you will learn about: two-dimensional motion vector methods Learning Outcomes When you have completed this chapter, you will be able to: Knowledge explain two-dimensional motion in a horizontal or vertical plane interpret the motion of one object relative to another Science, Technology, and Society explain that scientific knowledge is subject to change as new evidence comes to light and as laws and theories are tested, restricted, revised, or reinforced 68 Unit I 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 69 2-1 QuickLab 2-1 QuickLab Taking a One-dimensional Vector Walk Problem How can you add vectors to determine displacement? Materials 30-m tape measure field marker (tape or flag) Procedure 1 Starting at the centre of a football field (or gymnasium), work out a path sequence using six forward and backward displacements to move from your starting position to the goal line. At least two of your six displacements must be oriented in the direction opposite to the direction of the goal line. 2 On the field, mark your starting point with a flag or tape. Define direction axes. 3 Ask your partner to walk the displacements of the path sequence chosen in step 1 while holding the end of the measuring tape. Mark your partner’s endpoint after each displacement (Figure 2.2). 4 Continue the journey, using the measuring tape, until you have walked all the displacements. 5 Mark the final endpoint. 6 Using the measuring tape, determine the displacement from your starting point. 7 Repeat steps 2–6 using two different sequences of the six displacements you originally chose. 5 m [forward] Figure 2.2 Questions 1. What was the total distance you travelled? 2. What was the total displacement? 3. What conclusion can you draw about the order of adding vectors? Think About It 1. How does the order of a series of displacements affect the final position of an object? 2. In order to cross a river in the shortest possible time, is it better to aim yourself upstream so that you end up swimming straight across or to aim straight across and swim at an angle downstream? 3. Why does it take longer to fly across Canada from east to west rather than west to east in the same airplane? 4. How does the angle of a throw affect the time a ba |
ll spends in the air? 5. Two objects start from the same height at the same time. One is dropped while the other is given an initial horizontal velocity. Which one hits the ground first? 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 to your ideas. Chapter 2 Vector components describe motion in two dimensions. 69 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 70 2.1 Vector Methods in One Dimension One of the fastest-growing sports in the world today is snowshoeing (Figure 2.3). The equipment required is minimal and the sport is easy to learn — you need to move forward in a straight line. Despite its simplicity, snowshoeing has great cardiovascular benefits: You can burn up to 1000 calories per hour, which makes it the ultimate cross-training program for athletes. It also allows athletes to explore different terrains and gain a greater appreciation of the outdoors, as well as to test their limits, especially by participating in endurance races! The motions in a showshoe race can be broken up into one-dimensional vector segments. In this section, you will study motion in one dimension using vectors. Figure 2.3 Snowshoeing is an excellent way of enjoying the great outdoors in winter while improving your health. Vector Diagrams In Chapter 1, you used variables and graphs to represent vector quantities. You can also represent vector quantities using vector diagrams. In a diagram, a line segment with an arrowhead represents a vector quantity. Its point of origin is called the tail, and its terminal point (arrowhead) is the tip (Figure 2.4). If the magnitude of a vector is given, you can draw the vector to scale. The length of the line segment depends on the vector’s magnitude. The arrowhead indicates direction. Drawing vector diagrams to represent motion helps you to visualize the motion of an object. Properly drawn, vector diagrams enable you to accurately add vectors and to determine an object’s position. tail tip Figure 2.4 A vector has a tail and a tip. 70 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 71 Choosing Reference Coordinates When describing the motion of an object, there are many ways to describe its direction. You could use adjectives such as forward or backward, up or down, into or out of, and left or right. You can also use compass directions, such as north [N], south [S], east [E], and west [W]. When drawing a vector diagram, it is important to choose which directions are positive and to include these directions on every vector diagram. As you learned in section 1.1 (Figure 1.6), in this unit, forward, up, right, north, and east are usually designated as positive, whereas their opposites are usually considered negative. You may choose your own designation of positive and negative when solving problems, but make sure your reference direction is consistent within each problem and clearly communicated at the beginning of your solution. Practise drawing vectors in the next Skills Practice exercise Representing a Vector Using an appropriate scale and direction convention, draw each of the following vectors. (a) 5 m [forward] (b) 20 m [down] (c) 30 km [north] (d) 150 km [left] Adding Vectors in One Dimension Motion in one dimension involves vectors that are collinear. Collinear vectors lie along the same straight line. They may point in the same or in opposite directions (Figure 2.5). collinear: along the same straight line, either in the same or in opposite directions (a) (b) Figure 2.5 (a) Collinear vectors in the same direction (b) Collinear vectors in opposite directions When more than one vector describes motion, you need to add the vectors. You can add and subtract vectors graphically as well as algebraically, provided they represent the same quantity or measurement. As in mathematics, in which only like terms can be added, you can only add vectors representing the same types of quantities. For example, you can add two or more position vectors, but not a position vector, 5 m [E], to a velocity vector, 5 m/s [E]. In addition, the unit of measurement must be the same. For example, before adding the position vectors 5 m [E] and 10 km [E], you must convert the units of one of the vectors so that both vectors have the same units. In the next example, determine the sum of all the vector displacements graphically by adding them tip to tail. The sum of a series of vectors is called the resultant vector. resultant vector: a vector drawn from the tail of the first vector to the tip of the last vector Chapter 2 Vector components describe motion in two dimensions. 71 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 72 Example 2.1 Contestants in a snowshoe race must move forward 10.0 m, untie a series of knots, move forward 5.0 m, solve a puzzle, and finally move forward 25.0 m to the finish line (Figure 2.6). Determine the resultant vector by adding the vectors graphically. Figure 2.6 Analysis and Solution 1. Choose an appropriate scale and reference direction. 1.0 cm : 5.0 m, forward is positive. 2. Draw the first vector and label its magnitude and direction (Figure 2.7). 10.0 m [forward] Figure 2.7 3. Place the tail of the second vector at the tip of the first vector. Continue to place all the remaining vectors in order, tip to tail (Figure 2.8). 10.0 m [forward] 5.0 m [forward] 25.0 m [forward] Figure 2.8 4. Connect the tail of the first vector to the tip of the last vector. This new vector, which points toward the tip of the last vector, is the resultant vector, R (the purple arrow in Figure 2.9). 72 Unit I Kinematics 02-PearsonPhys20-Chap02 7/25/08 7:57 AM Page 73 Tail (origin) of first vector Tip (terminal point) of last vector Figure 2.9 Find the magnitude of the resultant vector by measuring with a ruler, then convert the measured value using the scale. Remember to include the direction. d 8.0 cm [forward] 5.0 m 1.0 cm 40 m [forward] Practice Problems 1. The coach of the high-school rugby team made the team members run a series of sprints: 5.0 m [forward], 10 m [backward], 10 m [forward], 10 m [backward], 20 m [forward], 10 m [backward], 40 m [forward], and 10 m [backward]. (a) What is their total distance? (b) What is their displacement? Answers 1. (a) 115 m (b) 35 m [forward] In summary, you can see that adding vectors involves connecting them tip to tail. The plus sign in a vector equation tells you to connect the vectors tip to tail in the vector diagram. 2 d d 1: Add the negative of d To subtract collinear vectors graphically (Figure 2.10(a)), you may use one of two methods. For the first method, find d using the equation d 2, tip to tail, as you did in Example 2.1. The negative of a vector creates a new vector that points in the opposite direction of the original vector (Figure 2.10(b)). For the second method, connect the vectors tail to tail. This time, d starts at the tip of d 1 and ends at the tip of d 2 (Figure 2.10(c)). 1 to d (a) (b) (c) d2 d1 d d2 d1 d2 d d1 Figure 2.10 d (a) To subtract two collinear vectors, d 2 or 1 to d 1, graphically, (b) add the negative of d 2 (c) connect the vectors tail to tail and draw the resultant connecting the tip of d 1 to the tip of d 2. Chapter 2 Vector components describe motion in two dimensions. 73 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 74 Recall that the definition of displacement is final position minus initial position, or d i. The next example reviews the algebraic subtraction of vectors, which you learned in Chapter 1, Example 1.1, and also shows you how to subtract vectors graphically. d d f Example 2.2 A sailboat that is initially 15 m to the right of a buoy sails 35 m to the left of the buoy (Figure 2.11). Determine the sailboat’s displacement (a) algebraically and (b) graphically. Figure 2.11 35 m [left] origin 15 m [right] Practice Problems 1. Sprinting drills include running 40.0 m [N], walking 20.0 m [N], and then sprinting 100.0 m [N]. Using vector diagrams, determine the sprinter’s displacement from his initial position. 2. To perform a give and go, a basketball player fakes out the defence by moving 0.75 m [right] and then 3.50 m [left]. Using vector diagrams, determine the player’s displacement from the starting position. 3. While building a wall, a bricklayer sweeps the cement back and forth. If she swings her hand back and forth, a distance of 1.70 m, four times, use vector diagrams to calculate the distance and displacement her hand travels during that time. Check your answers against those in Example 1.1 Practice Problems 1-3. Answers 1. 160.0 m [N] 2. 2.75 m [left] 3. 6.80 m, 0 m 74 Unit I Kinematics Given Consider right to be positive. 15 m [right] 15 m d 35 m [left] 35 m d i f Required displacement (d ) Analysis and Solution (a) To find displacement algebraically, use the equation dd d d f i 35 m (15 m) 35 m 15 m 50 m The sign is negative, so the direction is to the left. (b) To find displacement graphically, subtract the two position vectors. Draw the vectors tail to tail and draw the resultant from the tip of the initial position vector to the tip of the final position vector (Figure 2.12). scale: 1.0 cm : 10 m df di d Figure 2.12 d 5.0 cm [left] 10 m 1.0 cm 50 m [left] Paraphrase The sailboat’s displacement is 50 m [left]. 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 75 For collinear vectors, find displacement by subtracting initial position from final position. Subtract vectors graphically by connecting them tail to tail or by reversing the direction of the initial position vector. Recall from Chapter 1 that direction for displacement is given with respect to initial position. 2.1 Check and Reflect 2.1 Check and Reflect Knowledge 1. Describe the similarities and differences between the two vectors drawn below. 5 m [E] 5 m [W] 2. Using the same scale and reference coordinates, compare the vectors 5 m [N] and 10 m [S]. 3. If the scale vector diagram of 5.0 m [S] is 6.0 cm long, |
what is the length of the scale vector diagram of 20 m [S]? 4. What scale is being used if 5.0 cm represents 100 km? Applications 5. The scale on a National Geographic world map is 1.0 cm : 520 km. On the map, 4.0 cm separates Alberta’s north and south provincial boundaries. What is the separation in kilometres? 6. During a tough drill on a field of length 100 yards, players run to each 10-yard line and back to the starting position until they reach the other end of the field. (a) Write a vector equation that includes all the legs of the run. (b) What is the players’ final displacement? (c) How far did they run? 7. A car drives north 500 km. It then drives three sequential displacements south, each of which is 50 km longer than the previous displacement. If the final position of the car is 50 km [N], find the three displacements algebraically. 8. Are vectors A and B equal? Why or why not? y A B x 9. A bouncy ball dropped from a height of 10.0 m bounces back 8.0 m, then drops and rebounds 4.0 m and finally 2.0 m. Find the distance the ball travels and its displacement from the drop point. e TEST To check your understanding of vectors in one dimension, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 2 Vector components describe motion in two dimensions. 75 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 76 2.2 Motion in Two Dimensions From the boot, the ball flies across the grass into the net and the crowd roars. The enormously successful FIFA Under-19 Women’s World Championship, held in 2002, raised the profile of women’s soccer in Canada and drew crowds totalling almost 200 000 to venues in Edmonton, Vancouver, and Victoria (Figure 2.13). Stars like Charmaine Hooper, Brittany Timko, and Christine Sinclair continue to amaze. From World Championship team members to the Under-6s on the local soccer pitch, performance depends on understanding and coordinating the movement of players and ball across the surface of the field. Figure 2.13 Motion in sports such as soccer can be described by vectors in two dimensions. Playbooks are available for fast-paced games such as hockey and soccer to allow coaches and players to plan the strategies that they hope will lead to success. Sometimes a team can charge straight up the rink or field, but, more often, a series of angled movements is needed to advance the puck or ball (Figure 2.14). For everyone to understand the play, a system is needed to understand motion in two dimensions. Components of Vectors Imagine that you are at one corner of a soccer field and you have to get to the far opposite corner. The shortest path from one corner to the other is a straight diagonal line. Figure 2.15 shows this path to be 150 m. Another way to describe this motion is to imagine an x-axis and a y-axis placed onto the soccer field, with you standing at the point (0, 0). You could move along the length of the field 120 m and then across the field 90 m and end up at the same spot (Figure 2.15). 4 3 2 1 Figure 2.14 This page is taken from a soccer playbook. How many players are involved in this wall pass-in-succession manoeuvre? 76 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 77 Figure 2.15 The diagonal distance from one corner to the opposite corner of a soccer field is 150 m. 150 m 90 m 120 m In this example, the sideline of the soccer field could be considered the x-axis, and the goal line could be the y-axis. The diagonal motion vector can then be separated, or resolved, into two perpendicular parts, or components: the x component and the y component. The diagonal vector is the resultant vector. If you walked along the sideline or x-axis, you would move through a distance of 120 m. This distance is the x component of the diagonal vector. The second part of the walk along the goal line, parallel to the y-axis, is the y component of the diagonal vector. This motion has a distance of 90 m. Figure 2.16 shows the x and y components of the diagonal motion across the soccer field. y components: perpendicular parts into which a vector can be separated Figure 2.16 The resultant vector representing the diagonal walk across the soccer field can be resolved into x and y components. 150 m 90 m y component (width) x (0, 0) 120 m x component (length) Vector Directions Recall that a vector must have a magnitude and a direction. You have just studied how to resolve a vector into its components. Before going further, you need to know how to indicate the direction of vectors in two dimensions. There are two methods commonly used to show direction for vector quantities in two dimensions: the polar coordinates method and the navigator method. Both methods are considered valid ways to describe the direction of a vector. Chapter 2 Vector components describe motion in two dimensions. 77 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 78 info BIT A third method for measuring direction is the bearing method, in which angles are measured clockwise from north, 0° to 360°, so east is 90°, south is 180°, and west is 270°. 90° y II 180° 240° 5 m [240°] III 270° x 0° (360°) I IV N S W 60° 5 m [60° S of W] E info BIT Sailors can now create their sailing plans with a click of a mouse. Digitized maps and global positioning satellites have been combined to allow sailors to create a plan by using the mouse to place vectors on the desired path on screen. The computer calculates the total distance, identifies directions, and estimates the time required for the trip. Figure 2.17 The polar coordinates method for stating vector direction Figure 2.18 The navigator method for stating vector direction Polar Coordinates Method With the polar coordinates method, the positive x-axis is at 0° and angles are measured by moving counterclockwise about the origin, or pole. One complete rotation is 360° — a complete circle. In Figure 2.17, the displacement vector, 5 m [240°], is located in quadrant III in the Cartesian plane. This vector is rotated 240° counterclockwise starting from the positive x-axis. Navigator Method Another method for indicating vector direction is the navigator method. This method uses the compass bearings north [N], south [S], east [E], and west [W] to identify vector directions. In Figure 2.18, the displacement vector 5 m [60° S of W] is between the west and south compass bearings. To draw this vector, start with the second compass bearing you are given in square brackets, west, then move 60° in the direction of the first compass bearing you are given, south. The type of problem will determine the method you use for stating vector directions. Often, it will be clear from the context of the problem which method is preferred. For example, if the question is about a boat sailing [30° N of W], then use the navigator method. If a plane has a heading of 135°, then use the polar coordinates method. In the problems below, you can practise identifying and drawing vectors using the two methods Directions 2. For each vector in question 1, state the direction using the alternative method. Then draw each vector using an appropriate scale and reference coordinates. 1. For each of the following vectors, identify the method used for indicating direction. Then draw each vector in your notebook, using an appropriate scale and reference coordinates. (a) 3 m [0°] (b) 17 m/s [245°] (c) 7 m [65°] (d) 8 m/s [35° W of N] (e) 2 m [98°] (f) 12 m/s [30° S of E] 78 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 79 Concept Check Write the direction [60° S of W] another way using a different starting axis but keeping the angle less than 90°. 2-2 QuickLab 2-2 QuickLab Vector Walk Problem How can you add vectors to determine displacement? Materials 30-m measuring tape large chalkboard protractor field marker (tape or flag) Procedure 1 Using a tree diagram, determine the number of pathways you could take to walk the series of displacement vectors below. Assume that you will start on the centre line of a football field and mark it 0°. (a) 5 m [0°] (b) 12 m [270°] (c) 15 m [90°] (d) 3 m [180°] 7 Use the protractor to estimate the angle of displacement. 8 Repeat steps 3–7 for all the pathways you determined in step 1. 5 m [0°] Figure 2.19 NOTE: Use the same method for determining direction throughout the lab. 2 On a football or large school field, mark your starting point with a flag or tape. Define direction axes. 3 Ask your partner to walk the displacement given in (a) while you hold the end of the measuring tape. Mark your partner’s endpoint (Figure 2.19). Questions 1. What was the total distance you travelled? 2. What was the total displacement? 3. What conclusion can you draw about the order of adding vectors? 4 Continue the journey, using the protractor and measuring tape, until you have walked all the vectors. 5 Mark the final endpoint. 6 Using the measuring tape, determine the displacement from your starting point. Chapter 2 Vector components describe motion in two dimensions. 79 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 80 Adding Two-dimensional Vectors Graphically To sink the eight ball in the front side pocket of a billiard table, you must cause the ball to travel down and across the table (Figure 2.20). The ball’s motion occurs in a plane, or two dimensions, even though its path is linear (Figure 2.21). Figure 2.20 Playing billiards involves two-dimensional motion. Figure 2.21 The path of the billiard ball is linear, but it occurs in two dimensions. Recall from section 2.1 that the plus sign () in a vector equation indicates that you need to connect the vectors tip to tail. Up to this point, you have added collinear vectors only. In this section, you will learn how to add non-collinear vectors. The plus sign still indicates you need to connect the vectors tip to tail while keeping track of their directions. Adding Non-collinear Vectors In section 2.1, you learned that vectors that lie along the same straight line are collinear. Vectors that are not along the same straight line a |
re non-collinear (Figure 2.22). To determine the magnitude and direction of the sum of two or more non-collinear vectors graphically, use an accurately drawn scale vector diagram. Imagine you are walking north a distance of 40 m. Your initial 1. You stop, head west a distance position from your starting point is d 2. To find your disof 30 m, and stop again. Your final position is d placement, you cannot simply subtract your initial position from your final position because the vectors are not collinear. To find your displacement in two dimensions, you need to add the two position vectors: d d d 1 2. From Figure 2.23, you can see that the answer is not 70 m. You would obtain the answer 70 m if you walked in the same direction for both parts of your walk. Because you did not, you cannot directly substitute values into the displacement equation d 2. Instead, you must draw the vectors to scale, connect them tip to tail (because of the plus sign), and measure the magnitude of the resultant. Since d is a vector quantity, you must also indicate its direction. You can find the direction of the resultant using a protractor (Figure 2.24). d d 1 non-collinear: not along a straight line 45° Figure 2.22 Non-collinear vectors lie along different lines. 80 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 81 30 m d 2 W N S E d 30 m 2 W N S E d 50 m 40 m d 1 d 50 m d 40 m 1 scale 10 m 37º scale 10 m d 50 m [37º W of N] Figure 2.23 What is the sum of 2? 1 and d d Figure 2.24 When adding non-collinear vectors graphically, use a protractor to find the direction of the resultant. Eight Steps for Adding Non-collinear Vectors Graphically To find the resultant vector in a non-collinear vector addition statement using the graphical method, follow these eight steps (see Figure 2.25): 1. Create an appropriate scale. 2. Choose a set of reference coordinates. 3. Draw vector 1 to scale. Measure its direction from the tail. 4. Draw vector 2 to scale. Draw its tail at the tip (arrowhead) of vector 1. 5. Draw the resultant vector by connecting the tail of vector 1 to the tail tip of vector 2. W N S E tip resultant vector 2 vector 1 tail tip 6. Measure the magnitude (length) of the resultant. Measure the direction Figure 2.25 Adding vectors (angle) of the resultant from its tail. 7. Use your scale to convert the magnitude of the resultant to its original units. 8. State the resultant vector. Remember to include both magnitude and direction. This method also works for more than two vectors. You can add the vectors in any order. The next example shows you how to add more than two non-collinear vectors graphically. Example 2.3 A camper left her tent to go to the lake. She walked 0.80 km [S], then 1.20 km [E] and 0.30 km [N]. Find her resultant displacement. 1 Given d d d 2 3 0.80 km [S] 1.20 km [E] 0.30 km [N] Required resultant displacement (d R) Chapter 2 Vector components describe motion in two dimensions. 81 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 82 Practice Problems 1. For Example 2.3, add the vectors in two different orders and obtain the resultant for each case. 2. A student runs through a field 100 m [E], then 200 m [S], and finally 50 m [45° S of E]. Find her final position relative to her starting point. Answers 1. 1.30 km [67° E of S] 2. 272 m [60° S of E] Analysis and Solution The three vectors are non-collinear, so add them tip to tail to find the resultant (Figure 2.26). tent 67° d1 0.80 km dR d1 d2 d3 Figure 2.26 scale 0.20 km W N S E dR 1.30 km d3 0.30 km d2 1.20 km Paraphrase The camper’s resultant displacement is 1.30 km [67° E of S]. Distance, Displacement, and Position Figure 2.27 shows the distances a bicycle courier travelled in going along the path from A to D, passing through B and C on the way. Use the information in the diagram, a ruler calibrated in mm, and a protractor to complete the distance, displacement, and position information required in Table 2.1. Assume the bicycle courier’s reference point is A. Complete Table 2.1, then draw and label the displacement vectors AB, BC, and CD, and the position vectors AB, AC, and AD. ▼ Table 2.1 Distance, Displacement, and Position scale 80 m W N S E d 600 m D C Displacement Δd (m) [direction] Distance Δd (m) d Final position (m) [direction] reference point AB BC CD AC AD 82 Unit I Kinematics d 630 m B d 560 m Figure 2.27 d 280 m A 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 83 Determining Components drawn in a Cartesian plane and its two Figure 2.28 shows a vector R components, Rx and Ry, in the x and y directions. The Greek letter theta, , denotes the angle between R and its x and y components form a right triangle. and the x-axis. Vector R y 6.0 4.0 2.0 R 10 km/h 6.0 km/h Ry θ 37° 2.0 x 6.0 8.0 4.0 8.0 km/h Rx Figure 2.28 Drawing components Because the triangle is a right triangle, you can determine components algebraically by using the trigonometric functions sine, cosine, and tangent. You can define each of the trigonometric functions in terms of the sides of a right triangle, like the one in Figure 2.29. Knowing these definitions, you can use the trigonometric functions to help you solve for the components of a vector. To calculate Rx, use the cosine function: Rx R adjacent hypotenuse R cos or Rx cos In Figure 2.28, the x component is: Rx (10 km/h)(cos 37°) 8.0 km/h To calculate Ry, use the sine function: sin opposite hypotenuse Ry R or Ry R sin hypotenuse opposite In Figure 2.28, the y component is: Ry (10 km/h)(sin 37°) 6.0 km/h θ adjacent Figure 2.29 Labelled sides of a right triangle Example 2.4 shows the steps for finding the velocity components of a car travelling in a northeasterly direction using trigonometry. This example uses the navigator method to indicate the direction of the velocity vector. Note that the east direction [E] is the same as the positive x direction in the Cartesian plane, and north [N] is the same as the positive y direction. So, for any vector R , the x component is the same as the east component, and the y component is the same as the north component. PHYSICS INSIGHT a c θ b For a right triangle with sides a and b forming the right angle, c is the hypotenuse. The Pythagorean theorem states that a2 b2 c2. You can find the angle, , in one of three ways: sin a c opposite hypotenuse cos b c adjacent hypotenuse tan a b opposite adjacent Chapter 2 Vector components describe motion in two dimensions. 83 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 84 Example 2.4 Determine the north and east velocity components of a car travelling at 100 km/h [25° N of E]. y 1 0 0 k m /h 25° vx vy x Figure 2.30 Given v 100 km/h [25° N of E] Required velocity component north (y component, vy) velocity component east (x component, vx) Practice Problems 1. A hiker’s displacement is 15 km [40° E of N]. What is the north component of his displacement? 2. A cyclist’s velocity is 10 m/s [245°]. Determine the x and y components of her velocity. 3. A snowmobile travels 65 km [37° E of S]. How far east does it travel? Answers 1. 11 km [N] 4.2 m/s, vy 2. vx 3. 39 km [E] 9.1 m/s Analysis and Solution The vector lies between the north and east directions, so the x and y components are both positive. Since the north direction is parallel to the y-axis, use the sine function, R sin , to find the north component. Since the east Ry direction lies along the x-axis, use the cosine function, Rx R cos , to find the east component. Ry vy Rx vx R sin (100 km/h)(sin 25°) 42.3 km/h R cos (100 km/h)(cos 25°) 90.6 km/h Paraphrase The north component of the car’s velocity is 42.3 km/h and the east component is 90.6 km/h. Concept Check For a vector R in quadrant I (Cartesian method), are Rx and Ry always positive? Determine whether Rx and Ry are positive or negative for vectors in quadrants II, III, and IV. Display your answers in a chart. 84 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 85 Adding Vectors Using Components You can write the magnitude of any two-dimensional vector as the sum of its x and y components. Note that x and y components are perpendicular. Because motion along the x direction is perpendicular to motion along the y direction, a change in one component does not affect the other component. Whether it is movement across a soccer field or any other type of two-dimensional motion, you can describe the motion in terms of x and y components. In Example 2.4, you learned how to determine the x and y com- ponents, Rx and Ry, for a general vector R . In some situations, you already know Rx and Ry, and you must find the magnitude and direc- tion of the resultant vector R . For example, a toy moves 9.0 m right and then 12.0 m across a classroom floor (Figure 2.31). What is the toy’s displacement? Solving this problem algebraically requires two steps: Step 1: Find the magnitude of R . To find the magnitude of the resultant vector, use the Pythagorean theorem. You can use this theorem because the two components, Rx and Ry, form a right triangle with the resultant vector. You are given that Rx R2 Rx R (9.0 m)2 (12.0m)2 9.0 m and Ry 2 Ry 12.0 m. y 2 15 m Step 2: Find the angle of R To find the angle of R , use the tangent function: . R tan opposite adjacent 12.0 m 9.0 m 1.33 tan1(1.33) 53.1° 12.0 9.0 6.0 3.0 12.0 m Ry θ 3.0 Rx 6.0 9.0 m x 9.0 Figure 2.31 Vector components of the movement of a toy across a classroom floor Using the polar coordinates method, the resultant vector direction is [53.1°]. Using the navigator method, the direction is [53.1° N of E]. Using Components 1. Find Rx and Ry for the following vectors: (a) A boat travelling at 15 km/h [45° N of W] (b) A plane flying at 200 km/h [25° E of S] (c) A mountain bike travelling at 10 km/h [N] 2. Find R (a) Rx (b) Rx (c) Rx 7 m and for the following Rx and Ry values: 12 m, Ry 40 km/h, Ry 30 cm, Ry 55 km/h 10 cm Chapter 2 Vector components describe motion in two dimensions. 85 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 86 In general, most vector motion involves adding non-collinear ve |
ctors. Consider the following scenario. During a lacrosse game, players pass the ball from one person to another (Figure 2.32). The ball can then be redirected for a shot on goal. Each of the displacements could involve different angles. In order to find the net displacement, you would use the following sequence of calculations. Four Steps for Adding Non-collinear Vectors Algebraically 1. Determine the x and y components of each vector. 2. Add all components in the x direction. Add all components in the y direction. The sums of the x and y components are the two (perpendicular) components of the resultant vector. 3. To find the magnitude of the resultant vector, use the Pythagorean theorem. 4. To find the angle of the resultant vector, use trigonometric ratios. (See Physics Insight on page 83.) The following example illustrates how to apply these steps. In a lacrosse game (Figure 2.33(a)), player A passes the ball 12.0 m to player B at an angle of 30°. Player B relays the ball to player C, 9.0 m away, at an angle of 155°. Find the ball’s resultant displacement. y B d2 9.0 m 155° 12.0 m d1 30° x C A Figure 2.33(a) on the lacrosse field The path of the ball Figure 2.33(b) as vectors The path of the ball Figure 2.33(b) shows the path of the lacrosse ball as vectors. This problem is different from previous examples because the two vectors are not at right angles to each other. Even with this difference, you can follow the same general steps to solve the problem. Step 1: Determine the x and y components of each vector. Since you are solving for displacement, resolve each displacement vector into its components (Figure 2.34). Table 2.2 shows how to calculate the x and y components. In this case, designate up and right as positive directions. Figure 2.32 The movement of the players and the ball in a lacrosse game could be tracked using vectors. PHYSICS INSIGHT To simplify calculations for finding components, use acute ( 90°) angles. To determine the acute angle when given an obtuse ( 90°) angle, subtract the obtuse angle from 180°. 25° 155° 180° 155° 25° For an angle greater than 180°, subtract 180° from the angle. For example, 240° 180° 60° 86 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 87 ▼ Table 2.2 Resolution of Components in Figure 2.34 x direction y direction d1x d2x (12.0 m)(cos 30°) 10.39 m (9.0 m)(cos 25°) 8.16 m d1y d2y (12.0 m)(sin 30°) 6.00 m (9.0 m)(sin 25°) 3.80 m (Note that d2x is negative because it points to the left, and up and right were designated as positive.) y d2y 30° d2 d2x d1 d1x 25° 155° d1y x Figure 2.34 The path of the lacrosse ball Step 2: Add the x components and the y components separately. Add all the x components together, then add all the y components (see Table 2.3 and Figure 2.35). ▼ Table 2.3 Adding x and y Components in Figure 2.35 x direction y direction dx d2x d1x 10.39 m (8.16 m) 10.39 m 8.16 m 2.23 m dy d2y d1y 6.00 m 3.80 m 9.80 m dx d2x d1x d2y d1y dy Figure 2.35 Add the x and y components separately first to obtain two perpendicular vectors. Step 3: Find the magnitude of the resultant, d To find the magnitude of the resultant, use the Pythagorean theorem (Figure 2.36). . d2 (dx)2 (dy)2 d (dx)2 (dy)2 (2.23m)2 (9.80 m)2 10 m y d dy θ dx x Figure 2.36 The component method allows you to convert non-perpendicular vectors into perpendicular vectors that you can then combine using the Pythagorean theorem. Chapter 2 Vector components describe motion in two dimensions. 87 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 88 y d 77° d2 d1 Figure 2.37 d resultant displacement of the ball. is the Step 4: Find the angle of d. Use the tangent function to find the angle (Figure 2.36). tan opposite adjacent 9.80 m 2.23 m 4.39 x tan1(4.39) 77° The ball’s displacement is, therefore, 10 m [77°], as shown in Figure 2.37. The following example illustrates another situation where the dis- placement vectors are not at right angles. Example 2.5 Practice Problems 1. Find the displacement of a farmer who walked 80.0 m [0°] and then 60.0 m [335°]. 2. Find the displacement of a soccer player who runs 15 m [15° N of E] and then 13 m [5° W of N]. 3. While tracking a polar bear, a wildlife biologist travels 300 m [S] and then 550 m [75° N of E]. What is her displacement? Answers 1. 137 m [349°] 2. 21 m [52° N of E] 3. 272 m [58° N of E] Use components to determine the displacement of a crosscountry skier who travelled 15.0 m [220°] and then 25.0 m [335°] (Figure 2.38). y x 15.0 m [220°] d1 25.0 m [335°] d2 Figure 2.38 Given d d 1 2 15.0 m [220°] 25.0 m [335°] Required displacement (d ) Analysis and Solution Step 1: Use Rx its x and y components. Designate up and to the right as positive. Work with acute angles (Figure 2.39). R sin to resolve each vector into R cos and Ry y y 220° 40° x 335° 25° x d2y d2 d2x d1y d1 d1x Figure 2.39 88 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 89 x direction: d1x (15.0 m)(cos 40°) 11.49 m d2x (25.0 m)(cos 25°) 22.66 m y direction: d1y (15.0 m)(sin 40°) 9.642 m d2y (25.0 m)(sin 25°) 10.57 m Step 2: Add the x and y components. dx dy d1x + d2x 11.49 m 22.66 m 11.17 m d2y d1y 9.642 m (10.57 m) 20.21 m 11.17 m θ 20.21 m d y θ x d d1 d2 Figure 2.40 Figure 2.41 Step 3: To find the magnitude of the resultant, calculate d using the Pythagorean theorem (Figure 2.40). d2 (dx)2 (dy)2 (11.17 m)2 (20.21 m)2 d (11.17 m)2 (20.21 m)2 23.09 m Figure 2.41 shows that the resultant lies in quadrant IV. Step 4: To find the angle, use the tangent function (see Figure 2.40). tan opposite adjacent 20.21 m 11.17 m 1.810 tan1(1.810) 61° From Figure 2.41, note that the angle, , lies below the positive x-axis. Using the polar coordinates method, the angle is 299°. Paraphrase The cross-country skier’s displacement is 23.1 m [299°]. e SIM Practise the numerical addition of two or more vectors. Follow the eSim links at www.pearsoned.ca/school/ physicssource. Chapter 2 Vector components describe motion in two dimensions. 89 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 90 In summary, in order to solve a two-dimensional motion problem, you need to split the motion into two one-dimensional problems by using the vectors’ x and y components. Then add the x and y components separately. To find the magnitude of the resultant, use the Pythagorean theorem. To find the angle of the resultant, use the tangent function. 2.2 Check and Reflect 2.2 Check and Reflect Knowledge 1. What trigonometric functions can be used to determine the x or horizontal component of a vector? Draw diagrams to illustrate your answers. 2. Are the following statements true or false? Justify your answer. (a) The order in which vectors are added is important. (b) Displacement and distance are always equal. 3. Describe when you would use the navigator method to indicate the direction of a vector. Applications 4. A student has created a short computer program that calculates components of vectors drawn with a computer mouse. To demonstrate his program, he drags the mouse to create a vector at 55 cm [30° W of S]. What are the components of the vector? 5. Determine the distance travelled and the displacement for each of the following. (a) Blading through Fish Creek Park in Calgary takes you 5.0 km [W], 3.0 km [N], 2.0 km [E], and 1.5 km [S]. (b) A swimmer travels in a northerly direction across a 500-m-wide lake. Once across, the swimmer notices that she is 150 m east of her original starting position. (c) After leaving her cabin, a camper snowshoes 750 m [90°] and then 2.20 km [270°]. 90 Unit I Kinematics 6. A boat sails 5.0 km [45° W of N]. It then changes direction and sails 7.0 km [45° S of E]. Where does the boat end up with reference to its starting point? 7. A pellet gun fires a pellet with a velocity of 355 m/s [30°]. What is the magnitude of the vertical component of the velocity at the moment the pellet is fired? 8. Tourists on a jet ski move 1.20 km [55° N of E] and then 3.15 km [70° S of E]. Determine the jet ski’s displacement. 9. A jogger runs with a velocity of 6.0 km/h [25° N of W] for 35 min and then changes direction, jogging for 20 min at 4.5 km/h [65° E of N]. Using a vector diagram, determine the jogger’s total displacement and his average velocity for the workout. 10. Given that a baseball diamond is a square, assume that the first-base line is the horizontal axis. On second base, a baseball player’s displacement from home plate is 38 m [45°]. (a) What are the components of the player’s displacement from home plate? (b) Has the runner standing on second base travelled a distance of 38 m? Why or why not? 11. Determine the resultant displacement of a skateboarder who rolls 45.0 m [310°] and 35.0 m [135°]. e TEST To check your understanding of two-dimensional motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 91 2.3 Relative Motion Conveyor belts in the oil sands mines of Northern Alberta are 50 km long (Figure 2.42). Every hour, they move 25 200 t of oil sand from the mine to the extraction plant. What is the speed of the oil sand with respect to the conveyor belt? How fast is the oil sand moving with respect to the ground? How fast is it moving relative to a 21 240-t mechanical drive truck going in the opposite direction? Figure 2.42 A conveyor belt represents an example of uniform and relative motion. Sometimes objects move within a medium that is itself moving. Wind (moving air) affects the motion of objects such as kites, sailboats, and airplanes. Current (moving water) affects the motion of watercraft, wildlife, and swimmers. An Olympic kayak competitor who can paddle with a speed of 5.0 m/s in still water may appear to be going faster to an observer on shore if she is paddling in the same direction as the current. In this case, the velocity of the moving object depends on the location of the observer: whether the observer is on the moving object or observing the moving object from a stationary position. Relative mot |
ion is motion measured with respect to an observer. Concept Check An observer is on a train moving at a velocity of 25 m/s [forward]. A ball rolls at 25 m/s [forward] with respect to the floor of the moving train. What is the velocity of the ball relative to the observer on the train? What is the velocity of the ball relative to an observer standing on the ground? What happens if the ball moves 25 m/s [backward]? How does a moving medium affect the motion of a table tennis ball? relative motion: motion measured with respect to an observer Chapter 2 Vector components describe motion in two dimensions. 91 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 92 2-3 QuickLab 2-3 QuickLab Table Tennis in the Wind Problem How does air movement affect the motion of a table tennis ball? Materials large upright fan table tennis table paddles table tennis ball Procedure 1 With a partner, practise hitting the table tennis ball to each other (Figure 2.43). Figure 2.43 Questions 1. When did the ball move the fastest? the slowest? 2. When the air movement was perpendicular to the ball’s path, did it change the ball’s speed? Did it change the ball’s velocity? Explain. 2 Set up the fan on one side of the table tennis table. 3. Given your results, speculate as to why golfers 3 Hit the table tennis ball straight across the length of the table (a) against the wind (b) with the wind release a tuft of grass into the wind before driving the ball. 4. Describe how wind direction might influence a beach volleyball player’s serve. (c) perpendicular to the wind’s direction e LAB 4 Record how the moving air influences the motion of the ball in each case. For a probeware activity, go to www.pearsoned.ca/school/physicssource. Relative Motion in the Air ground velocity: velocity relative to an observer on the ground air velocity: an object’s velocity relative to still air wind velocity: velocity of the wind relative to the ground A flight from Edmonton to Toronto takes about 3.5 h. The return flight on the same aircraft takes 4.0 h. If the plane’s air speed is the same in both directions, why does the trip east take less time? The reason is that, when travelling eastward from Edmonton, a tailwind (a wind that blows from the rear of the plane, in the same direction as the plane’s motion) increases the airplane’s ground velocity (velocity relative to an observer on the ground), hence reducing the time of travel and, therefore, fuel consumption and cost. scale 100 km/h A Canadian regional jet travels with an air velocity (the plane’s velocity in still air) of 789 km/h [E]. The jet encounters a wind velocity (the wind’s velocity with respect to the ground) of 56.3 km/h [E] (Figure 2.44). (This wind is a west wind, blowing eastward from the west.) What is the velocity of the airplane relative to an observer on the ground? The resultant velocity of the airplane, or ground velocity, is the vector sum of the plane’s air velocity and the wind velocity (Figure 2.45). Let the positive direction be east. vair 789 km/h [E] vwind 56.3 km/h [E] Figure 2.44 The air velocity and wind velocity are in the same direction. 92 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 93 v ground air wind v v 789 km/h 56.3 km/h 845 km/h The sign is positive, so the ground velocity is 845 km/h [E]. scale 100 km/h vair 789 km/h [W] vwind 56.3 km/h [E] scale 100 km/h vwind Figure 2.45 vair vground scale 100 km/h vair vground vwind Figure 2.47 Figure 2.46 The air velocity and wind velocity are in opposite directions. If the jet heads west, from Toronto to Edmonton (Figure 2.46), its v ground resultant velocity becomes v 789 km/h 56.3 km/h 733 km/h v wind air (See Figure 2.47.) The sign is negative, so the ground velocity is 733 km/h [W]. The plane’s speed decreases due to the headwind (wind that approaches from the front). Non-collinear Relative Motion Suppose the jet travelling west from Toronto encounters a crosswind of 56.3 km [N] (Figure 2.48). W N S E vwind= 56.3 km/h vair= 789 km/h vwind= 56.3 km/h vground= 791 km/h θ vair= 789 km/h W N S E Figure 2.48 A plane flies in a crosswind. Figure 2.49 A plane that flies in a crosswind needs to adjust its direction of motion. air ground v v In this case, the velocity of the plane is not aligned with the wind’s velocity. The defining equation for this case is still the same as for the collinear case: v wind. From section 2.2, recall that the plus sign in a two-dimensional vector equation tells you to connect the vectors tip to tail. The resultant vector is the ground velocity, v ground. The ground velocity indicates the actual path of the plane (Figure 2.49). To solve for the ground velocity, notice that the triangle formed is a right triangle, meaning that you can use the Pythagorean theorem to solve for the magnitude of the ground velocity. Chapter 2 Vector components describe motion in two dimensions. 93 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 94 (vground)2 (vair)2 (vwind)2 vground (vair)2 (vw ind)2 (789 km/h)2 (56.3 km/h)2 791 km/h Using the tangent function, the direction of the ground velocity is tan opposite adjacent 56.3 km/h 789 km/h 0.07136 tan1(0.07136) 4.1° From Figure 2.49, the wind blows the airplane off its westerly course in the northerly direction. Hence, the airplane’s ground velocity is 791 km/h [4.1° N of W]. The pilot must take into consideration the effect of the wind blowing the airplane off course to ensure that the plane reaches its destination. What path would the pilot have to take to arrive at a point due west of the point of departure? Remember that, if the vectors are not perpendicular, resolve them into components first before adding them algebraically. Example 2.6 A plane flies west from Toronto to Edmonton with an air speed of 789 km/h. (a) Find the direction the plane would have to fly to compensate for a wind velocity of 56.3 km/h [N]. (b) Find the plane’s speed relative to the ground. W N S E vground vwind Figure 2.50 vair PHYSICS INSIGHT To determine the angle, substitute the magnitudes of the relative velocities into the tangent function. To determine the direction, refer to the vector diagram for the problem. 94 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 95 wind 56.3 km/h [N] Given v v direction of ground velocity is west 789 km/h air Required (a) the plane’s direction (direction of air velocity) (b) ground speed (vground) Analysis and Solution (a) First construct a diagram based on the defining equation, v v v ground wind air vground θ vair vwind Figure 2.51 air and v wind tip to tail. The rules of vector addition tell you to connect the vectors v To find the direction required in order to compensate for the wind velocity, find the angle, . Because the connection of the vectors forms a right triangle, and you know the magnitude of the opposite side (v wind) and the hypotenuse (v air), you can use the sine function to find the angle (Figure 2.51). te i s o p sin t se u n e o hy p p o Practice Problems 1. A swimmer can swim at a speed of 1.8 m/s. The river is 200 m wide and has a current of 1.2 m/s [W]. If the swimmer points herself north, directly across the river, find (a) her velocity relative to the ground. (b) the time it takes her to cross. 2. For a river flowing west with a current of 1.2 m/s, a swimmer decides she wants to swim directly across. If she can swim with a speed of 1.8 m/s, find (a) the angle at which she must direct herself. (b) the time it takes her to cross if the river is 200 m wide. Answers 1. (a) 2.2 m/s [34° W of N] (b) 1.1 102 s 2. (a) [42° E of N] (b) 1.5 102 s o p p p hy sin1 sin1 4.1 se te From Figure 2.51, the angle is [4.1° S of W]. (b) To find the magnitude of the ground velocity, use the Pythagorean theorem. From Figure 2.51, note that the hypotenuse in this case is the air velocity, v air. (vair)2 (vwind)2 (vground)2 (vground)2 (vair)2 (vwind)2 (789 km/h)2 (56.3 km/h)2 6.1935 105 (km/h)2 787 km/h vground Notice that there is a small change in the magnitude of the ground velocity from the previous example of the plane heading west. As the magnitude of the wind velocity increases, the magnitude of the ground velocity and the compensating angle will significantly change. Paraphrase (a) The plane’s heading must be [4.1° S of W]. (b) The plane’s ground speed is 787 km/h. Chapter 2 Vector components describe motion in two dimensions. 95 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 96 PHYSICS INSIGHT For simplicity, Edmonton was assumed to be directly west of Toronto, which, of course, it is not! However, the calculation is still valid because this problem involves straightline motion. To calculate the time it takes to fly from Toronto to Edmonton, use the d equation v . The distance between Edmonton and Toronto is about t 2335 km, but you must decide which value for velocity to use: air velocity or ground velocity. The displacement and velocity vectors in this equation must be aligned. Since you are assuming that displacement is in the west direction, the appropriate velocity to use is the one that is in the westerly direction. In this example, it is the ground velocity, v ground (Figure 2.52). d vground θ vair vwind Figure 2.52 For calculating time, choose the velocity vector that matches the direction of the displacement vector. Consider west to be positive. Since both vectors are in the same direction (west), use the scalar form of the equation to solve for time. t d v 2335 km m 787 k h 2.97 h It takes 2.97 h to fly from Toronto to Edmonton. In the following example, the three velocity vectors do not form a right triangle. In order to solve the problem, you will need to use components. PHYSICS INSIGHT When substituting values into a vector equation, make sure that the values have the same direction (are collinear). If the vectors are not collinear, you need to use graphical or algebraic methods to find the answer. Example 2.7 As a pilot of a small plane, you need to transport three people to an airstrip 350.0 km due west in |
2.25 h. If the wind is blowing at 40.0 km/h [65° N of W], what should be the plane’s air velocity in order to reach the airstrip on time? d = 350.0 km [W] vground vwind = 40.0 km/h [65º N of W] Figure 2.53 96 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 97 40.0 km/h [65 N of W] wind Given v d t 2.25 h 350.0 km [W] Required plane’s air velocity (v air) Analysis and Solution First draw a vector diagram of the problem (Figure 2.54). W N S E vwind Figure 2.54 vground vair Designate north and west as the positive directions. Then calculate the ground velocity from the given displacement and time. If the plane must fly 350.0 km [W] in 2.25 h, its ground velocity is d t ground v Practice Problems 1. An airplane can fly with a maximum air velocity of 750 km/h [N]. If the wind velocity is 60 km/h [15 E of N], what must be the plane’s ground velocity if it is to remain on a course going straight north? 2. What is the air velocity of a jetliner if its ground velocity is 856 km/h [25.0 W of S] and the wind velocity is 65.0 km/h [S]? 3. How long will it take a plane to travel 100 km [N] if its ground velocity is 795 km/h [25 W of N]? Answers 1. 8.1 102 km/h [1 W of N] 2. 798 km/h [27.0 W of S] 3. 0.139 h 350.0 km [W] 2.25 h 155.6 km/h [W] Now find the components of the wind velocity (Figure 2.55). x direction: vwindx (40.0 km/h)(cos 65) 16.9 km/h y direction: vwindy (40.0 km/h)(sin 65) 36.25 km/h The ground velocity is directed west, so its x component is 155.6 km/h and its y component is zero. v Since v v v v wind, rearrange this equation to solve for v v air. ground air air ground wind vwindy vwind 65° vwindx Figure 2.55 Use this form of the equation to solve for the components of the air velocity. Add the x (west) components: vairx vgroundx vwindx 155.6 km/h 16.9 km/h 138.7 km/h Add the y (north) components: vairy vgroundy vwindy 0 36.25 km/h 36.25 km/h Use the Pythagorean theorem to find the magnitude of the air velocity. vair vairx 2 vairy2 (138.7 km/h)2 (36.25 km/h)2 143 km/h Chapter 2 Vector components describe motion in two dimensions. 97 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 98 To find the direction of air velocity, use the tangent function (Figure 2.56). tan opposite adjacent 36.25 km/h 138.7 km/h 0.261 tan1(0.261) 15 138.7 km/h θ 36.25 km/h vair Figure 2.56 The x component is positive, so its direction is west. Since the y component is negative, its direction is to the south. Thus, the direction of the air velocity is [15 S of W]. Paraphrase The airplane’s air velocity is 143 km/h [15 S of W]. Relative Motion in the Water Whereas wind velocity affects the speed and direction of flying objects, watercraft and swimmers experience currents. As with flying objects, an object in the water can move with the current (ground velocity increases), against the current (ground velocity decreases), or at an angle (ground velocity increases or decreases). When the object moves at an angle to the current that is not 90, both the object’s speed and direction change. The following example illustrates how to use components to find velocity. Example 2.8 The Edmonton Queen paddleboat travels north on the Saskatchewan River at a speed of 5.00 knots or 9.26 km/h. If the Queen’s ground velocity is 10.1 km/h [23° E of N], what is the velocity of the Saskatchewan River? vcurrent W N S E vboat vground 23° vcurrent W N S E vboat= 9.26 km/h vground = 10.1 km/h 23º Figure 2.57(a) Figure 2.57(b) 98 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 99 boat ground 9.26 km/h [N] 10.1 km/h [23 E of N] Given v v (Note that the angle is given with respect to the vertical (y) axis (Figure 2.57(a).) vgroundx vgroundy vground Required velocity of current (v current) 23° Analysis and Solution Let north and east be positive. Calculate the current’s velocity using components. First find the components of the ground velocity (Figure 2.58(a)). Figure 2.58(a) x direction: vgroundx vgroundsin (10.1 km/h)(sin 23) 3.946 km/h y direction: vgroundy vgroundcos (10.1 km/h)(cos 23) 9.297 km/h Practice Problems 1. Determine a Sea Doo’s ground velocity if it travels with a constant velocity of 4.50 m/s [W] and encounters a current of 2.0 m/s [20 W of N]. 2. A jogger runs with a velocity of 3.75 m/s [20 N of E] on an Alaskan cruise ship heading north at 13 m/s. What is the jogger’s ground velocity? 3. A ship travelling 55 [W of N] is 65.0 km farther north after 3.0 h. What is the ship’s velocity? Answers 1. 5.5 m/s [20 N of W] 2. 15 m/s [76 N of E] 3. 38 km/h [55 W of N] Since the boat’s velocity is directed north, its y component is 9.26 km/h and its x component is zero. v You are asked to find v v v v v v current ground boat ground boat current current, so rearrange the vector equation accordingly: Use this form of the equation to solve for the components of the current’s velocity. vcurrentx vboatx vcurrenty vgroundx 3.946 km/h 0 3.946 km/h vboaty vgroundy 9.297 km/h 9.26 km/h 0.037 km/h To find the magnitude of the current’s velocity, use the Pythagorean theorem. vcurrent nty)2 ntx)2 (vcurre (vcurre (3.946 km/h)2 (0.037 km/h)2 3.946 km/h To find the direction of the current’s velocity, use the tangent function (Figure 2.58(b)). opposite adjacent tan 0.03700 km/h vcurrent θ tan1 0.037 km/h 3.946 km/h 0.5 3.946 km/h Figure 2.58(b) Since both the x and y components are positive, the directions are east and north, respectively. Therefore, the current’s direction is [0.5 N of E]. Paraphrase The current’s velocity is 3.95 km/h [0.5 N of E]. Chapter 2 Vector components describe motion in two dimensions. 99 02-PearsonPhys20-Chap02 7/25/08 8:05 AM Page 100 In order to find the time required to cross the river, you need to use the velocity value that corresponds to the direction of the object’s displacement, as you will see in the next example. Example 2.9 From Example 2.8, if the river is 200 m wide and the banks run from east to west, how much time, in seconds, does it take for the Edmonton Queen to travel from the south bank to the north bank? Practice Problems 1. A river flows east to west at 3.0 m/s and is 80 m wide. A boat, capable of moving at 4.0 m/s, crosses in two different ways. (a) Find the time to cross if the boat is pointed directly north and moves at an angle downstream. (b) Find the time to cross if the boat is pointed at an angle upstream and moves directly north. Answers 1. (a) 20 s (b) 30 s Given v width of river 200 m 0.200 km 10.1 km/h [23° E of N] ground Required time of travel (t) Analysis and Solution Determine the distance, d, the boat travels in the direction of the boat’s ground velocity, 23° E of N. From Figure 2.59, d 0.200 km cos 23° 0.2173 km N 0.200 km d 23° Figure 2.59 The boat’s ground velocity is 10.1 km/h [23° E of N]. vground t 10.1 km/h d vground 0.2173 km 10.1 km/h 0.02151 h 60 min 1 h 60 s 1 min 77.4 s Paraphrase It takes the Edmonton Queen 77.4 s to cross the river. Relative motion problems describe the motion of an object travelling in a medium that is also moving. Both wind and current can affect the magnitude and direction of velocity. To solve relative motion problems in two dimensions, resolve the vectors into components and then add them using trigonometry. 100 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:17 AM Page 101 2.3 Check and Reflect 2.3 Check and Reflect Knowledge 1. Describe a situation where a wind or current will increase an object’s ground speed. 2. Describe a situation where a wind or current will change an object’s direction relative to the ground, but not its speed in the original direction (i.e., the velocity component in the original, intended direction of motion). 3. Describe a situation when a wind or current will cause zero displacement as seen from the ground. 4. Provide an example other than in the text that illustrates that perpendicular components of motion are independent of one another. Applications 5. A swimmer needs to cross a river as quickly as possible. The swimmer’s speed in still water is 1.35 m/s. (a) If the river’s current speed is 0.60 m/s and the river is 106.68 m wide, how long will it take the swimmer to cross the river if he swims so that his body is angled slightly upstream while crossing, and he ends up on the far bank directly across from where he started? (b) If he points his body directly across the river and is therefore carried downstream, how long will it take to get across the river and how far downstream from his starting point will he end up? 6. A small plane can travel with a speed of 265 km/h with respect to the air. If the plane heads north, determine its resultant velocity if it encounters (a) a 32.0-km/h headwind (b) a 32.0-km/h tailwind (c) a 32.0-km/h [W] crosswind 7. The current in a river has a speed of 1.0 m/s. A woman swims 300 m downstream and then back to her starting point without stopping. If she can swim 1.5 m/s in still water, find the time of her round trip. 8. What is the ground velocity of an airplane if its air velocity is 800 km/h [E] and the wind velocity is 60 km/h [42 E of N]? 9. A radio-controlled plane has a measured air velocity of 3.0 m/s [E]. If the plane drifts off course due to a light wind with velocity 1.75 m/s [25° W of S], find the velocity of the plane relative to the ground. If the distance travelled by the plane was 3.2 km, find the time it took the plane to travel that distance. 10. An airplane is observed to be flying at a speed of 600 km/h. The plane’s nose points west. The wind’s velocity is 40 km/h [45 W of S]. Find the plane’s velocity relative to the ground. 11. A canoe can move at a speed of 4.0 m/s [N] in still water. If the velocity of the current is 2.5 m/s [W] and the river is 0.80 km wide, find (a) the velocity of the canoe relative to the ground (b) the time it takes to cross the river e TEST To check your understanding of relative motion, follow the eTest links at www.pearsoned.ca/school/physicssource. Chapter 2 Vector components describe motion in two dimensions. 101 02-Pe |
arsonPhys20-Chap02 7/24/08 10:17 AM Page 102 2.4 Projectile Motion Sports are really science experiments in action. Consider golf balls, footballs, and tennis balls. All of these objects are projectiles (Figure 2.60). You know from personal experience that there is a relationship between the distance you can throw a ball and the angle of loft. In this section, you will learn the theory behind projectile motion and how to calculate the values you need to throw the fastball or hit the target dead on. Try the next QuickLab and discover what factors affect the trajectory of a projectile. Figure 2.60 Sports is projectile motion in action. 102 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 103 2-4 QuickLab 2-4 QuickLab Projectiles Problem What factors affect the trajectory of a marble? Materials wooden board (1 m 1 m) hammer paint marble newspaper white paper to cover the board two nails elastic band spoon brick masking tape gloves Procedure 1 Spread enough newspaper on the floor so that it covers a larger workspace than the wooden board. 2 Hammer two nails, 7.0 cm apart, at the bottom left corner of the board. Stretch the elastic between them. 3 Cover the board with white paper and affix the 6 Pull the elastic band back at an angle and rest the marble in it. 7 Release the elastic band and marble. Label the marble’s trajectory on the paper track 1. 8 Repeat steps 5–7 for different launch angles and extensions of the elastic band. in Figure 2.61 Questions 1. What is the shape of the marble’s trajectory, paper to the board using masking tape. regardless of speed and angle? 4 Prop the board up on the brick (Figure 2.61). 2. How did a change in the elastic band’s extension 5 Wearing gloves, roll the marble in a spoonful of paint. affect the marble’s path? 3. How did a change in launch angle affect the marble’s path? e LAB For a probeware activity, go to www.pearsoned.ca/school/physicssource. Galileo studied projectiles and found that they moved in two directions at the same time. He determined that the motion of a projectile, neglecting air resistance, follows the curved path of a parabola. The parabolic path of a projectile is called its trajectory (Figure 2.62). The shape of a projectile’s trajectory depends on its initial velocity — both its initial speed and direction — and on the acceleration due to gravity. To understand and analyze projectile motion, you need to consider the horizontal (x direction) and vertical (y direction) components of the object’s motion separately. viy vix dy trajectory: the parabolic motion of a projectile vi Figure 2.62 A projectile has a parabolic trajectory. dx Chapter 2 Vector components describe motion in two dimensions. 103 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 104 2-5 QuickLab 2-5 QuickLab Which Lands First? Problem What is the relationship between horizontal and vertical motion of objects on a ramp? Materials Galileo apparatus (Figure 2.63) steel balls Procedure 1 Set up the Galileo apparatus at the edge of a lab bench. 2 Place a steel ball at the top of each ramp. 3 Release the balls at the same time. 4 Listen for when each ball hits the ground. 5 Using a different ramp, repeat steps 1–4. Questions 1. Which ball landed first? 2. Did the balls’ initial velocity affect the result? If so, how? 3. What inference can you make about the relationship between horizontal and vertical motion? Figure 2.63 From section 1.6, you know that gravity influences the vertical motion of a projectile by accelerating it downward. From Figure 2.64, note that gravity has no effect on an object’s horizontal motion. So, the two components of a projectile’s motion can be considered independently. As a result, a projectile experiences both uniform motion and uniformly accelerated motion at the same time! The horizontal motion of a projectile is an example of uniform motion; the projectile’s horizontal velocity component is constant. The vertical motion of a projectile is an example of uniformly accelerated motion. The object’s acceleration is the constant acceleration due to gravity or 9.81 m/s2 [down] (neglecting friction). Concept Check In a table, classify the horizontal and vertical components of position, velocity, and acceleration of a horizontally launched projectile as uniform or non-uniform motion. Objects Launched Horizontally Suppose you made a new game based on a combination of shuffleboard and darts. The goal is to flick a penny off a flat, horizontal surface, such as a tabletop, and make it land on a target similar to a dartboard beyond the table. The closer your penny lands to the bull’s eye, the more points you score (Figure 2.65). Figure 2.64 Gravity does not affect the horizontal motion of a projectile because perpendicular components of motion are independent. PHYSICS INSIGHT When a projectile is launched, for a fraction of a second, it accelerates from rest to a velocity that has x and y components. 104 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 105 v table horizontal (x) component dy trajectory padded dartboard dx Figure 2.65 An object launched horizontally experiences uniform horizontal motion and uniformly accelerated vertical motion. In the game, once the penny leaves the tabletop, it becomes a projectile and travels in a parabolic path toward the ground. In section 1.6, you studied motion that was caused by acceleration due to gravity. The velocity of an object falling straight down has no horizontal velocity component. In this game, the penny moves both horizontally and vertically, like the ball on the right in Figure 2.64. In this type of projectile motion, the object’s initial vertical velocity is zero. Because the projectile has a horizontal velocity component, it travels a horizontal distance along the ground from its initial launch point. This distance is called the projectile’s range (Figure 2.66). The velocity component in the y direction increases because of the acceleration due to gravity while the x component remains the same. The combined horizontal and vertical motions produce the parabolic path of the projectile. Concept Check (a) What factors affecting projectile motion in the horizontal direction are being neglected? (b) What causes the projectile to finally stop? (c) If the projectile’s initial velocity had a vertical component, would the projectile’s path still be parabolic? Give reasons for your answer. PHYSICS INSIGHT For a projectile to have a non-zero velocity component in the vertical direction, the object must be thrown up, down, or at an angle relative to the horizontal, rather than sideways. range: the distance a projectile travels horizontally over level ground y maximum height range x Solving Projectile Motion Problems In this chapter, you have been working with components, so you know how to solve motion problems by breaking the motion down into its horizontal (x) and vertical (y) components. Figure 2.66 The range of a projectile is its horizontal distance travelled. Chapter 2 Vector components describe motion in two dimensions. 105 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 106 Before you solve a projectile motion problem, review what you already know (Figure 2.67). vi vx vi vy 0 Figure 2.67(a) The projectile is given an initial horizontal velocity. viy vi sin θ vi vix vi cos θ vi cos θ vix vi sin θ viy vi Figure 2.67(b) The projectile is given an initial horizontal velocity and an upward vertical velocity. Figure 2.67(c) The projectile is given an initial horizontal velocity and a downward vertical velocity. x direction – There is no acceleration in this direction, so ax 0. In this text, ax will always be zero. The projectile undergoes uniform motion in the x direction. – The general equation for the initial x component of the velocity can be determined using trigonometry, e.g., vix vi cos . – The range is dx. – Because the projectile is moving in both the horizontal and vertical directions at the same time, t is a common variable. y direction – If up is positive, the acceleration due to gravity is down or negative, so ay 9.81 m/s2. – The y component of the initial velocity can be determined using trigonometry, e.g., viy vi sin . – The displacement in the y direction is dy. – Time (t) is the same in both the x and y directions. ▼ Table 2.4 Projectile Problem Setup x direction y direction ax vix 0 vi cos dx vx t 9.81 m/s2 ay vi sin viy can be positive or negative viy depending on the direction of v i. dy t 1 ay(t)2 2 viy If you check the variables, you can see that they are vi, t, d, and a, 1 a(t)2. In the t all of which are present in the equation d 2 v i horizontal direction, the acceleration is zero, so this equation simplifies t. The next example shows you how to apply these equations. to d v i e MATH To explore and graph the relationship between the velocity and position of an object thrown vertically into the air, visit www.pearsoned.ca/school/ physicssource. 106 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 107 Example 2.10 Head-Smashed-In Buffalo Jump, near Fort Macleod, Alberta, is a UNESCO heritage site (Figure 2.68). Over 6000 years ago, the Blackfoot people of the Plains hunted the North American bison by gathering herds and directing them over cliffs 20.0 m tall. Assuming the plain was flat so that the bison ran horizontally off the cliff, and the bison were moving at their maximum speed of 18.0 m/s at the time of the fall, determine how far from the base of the cliff the bison landed. Practice Problems 1. A coin rolls off a table with an initial horizontal speed of 30 cm/s. How far will the coin land from the base of the table if the table’s height is 1.25 m? 2. An arrow is fired horizontally with a speed of 25.0 m/s from the top of a 150.0-m-tall cliff. Assuming no air resistance, determine the distance the arrow will drop in 2.50 s. 3. What is the horizontal speed of an object if it lands 40.0 m away from the base of a 100-m-tall cliff? Answers 1. 15 cm 2. 30.7 m 3. 8.86 |
m/s Figure 2.68 y d y x d x Figure 2.69 Given For convenience, choose forward and down to be positive because the motion is forward and down (Figure 2.69). x direction vix 18.0 m/s y direction ay dy 9.81 m/s2 [down] 9.81 m/s2 20.0 m Required distance from the base of the cliff (dx) Analysis and Solution Since there is no vertical component to the initial velocity of the 0 m/s. Therefore, the bison experience uniformly bison, viy accelerated motion due to gravity in the vertical direction but uniform motion in the horizontal direction resulting from the run. From the given values, note that, in the y direction, you have all the variables except for time. So, you can solve for time in the y direction, which is the time taken to fall. PHYSICS INSIGHT For projectile motion in two dimensions, the time taken to travel horizontally equals the time taken to travel vertically. e SIM Analyze balls undergoing projectile motion. Follow the eSim links at www.pearsoned.ca/school/ physicssource. y direction: dy t 1 viy ay(t)2 2 0 1 ay 2 t2 Chapter 2 Vector components describe motion in two dimensions. 107 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 108 t 2dy ay 2(20.0 m) 9.81 m/s2 2.019 s x direction: The time taken for the bison to fall vertically equals the time they travel horizontally. Substitute the value for time you found in the y direction to find the range. Since the bison had a uniform horizontal speed of 18.0 m/s, use the equation dx dx vix t. (18.0 m/s)(2.019 s) 36.3 m Paraphrase The bison would land 36.3 m from the base of the cliff. Figure 2.70 Baseball is all about projectile motion. Objects Launched at an Angle Baseball is a projectile game (Figure 2.70). The pitcher throws a ball at the batter, who hits it to an open area in the field. The outfielder catches the ball and throws it to second base. The runner is out. All aspects of this sequence involve projectile motion. Each sequence requires a different angle on the throw and a different speed. If the player miscalculates one of these variables, the action fails: Pitchers throw wild pitches, batters strike out, and outfielders overthrow the bases. Winning the game depends on accurately predicting the components of the initial velocity! vy v sin θ v θ vx v cos θ For objects launched at an angle, such as a baseball, the velocity of the object has both a horizontal and a vertical component. Any vector quantity can be resolved into x and y components using the trigonometric ratios R sin , when is measured relative to Rx the x-axis. To determine the horizontal and vertical compo v cos and nents of velocity, this relationship becomes vx vy R cos and Ry v sin , as shown in Figure 2.71. Solving problems involving objects launched at an angle is similar to solving problems involving objects launched horizontally. The object experiences uniform motion in the horizontal direction, so use the equation dx t. In the vertical direction, the object experiences uniformly accelerated motion. The 1 ay(t)2 still applies, but in this case, viy t general equation dy 2 is not zero. The next example shows you how to apply these equations to objects launched at an angle. vix viy Figure 2.71 The horizontal and vertical components of velocity 108 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 109 Example 2.11 Baseball players often practise their swing in a batting cage, in which a pitching machine delivers the ball (Figure 2.72). If the baseball is launched with an initial velocity of 22.0 m/s [30.0°] and the player hits it at the same height from which it was launched, for how long is the baseball in the air on its way to the batter? Figure 2.72 Given v i 22.0 m/s [30.0°] Practice Problems 1. A ball thrown horizontally at 10.0 m/s travels for 3.0 s before it strikes the ground. Find (a) the distance it travels horizontally. (b) the height from which it was thrown. 2. A ball is thrown with a velocity of 20.0 m/s [30] and travels for 3.0 s before it strikes the ground. Find (a) the distance it travels horizontally. (b) the height from which it was thrown. (c) the maximum height of the ball. Answers 1. (a) 30 m (b) 44 m 2. (a) 52 m (b) 14 m (c) 19 m Required time (t) Analysis and Solution Choose forward and up to be positive (Figure 2.73). First find the components of the baseball’s initial velocity. y viy vi 30.0° vix x Figure 2.73 x direction vix vi cos (22.0 m/s)(cos 30.0) 19.05 m/s y direction viy vi sin (22.0 m/s)(sin 30.0) 11.00 m/s Since the ball returns to the same height from which it was launched, dy 0. With this extra known quantity, you now have enough information in the y direction to find the time the ball spent in the air. PHYSICS INSIGHT Be careful to follow the sign convention you chose. If you chose up as positive, ay becomes 9.81 m/s2. info BIT The world’s fastest bird is the peregrine falcon, with a top vertical speed of 321 km/h and a top horizontal speed of 96 km/h. e WEB The fastest speed for a projectile in any ball game is approximately 302 km/h in jai-alai. To learn more about jai-alai, follow the links at www.pearsoned.ca/school/ physicssource. Chapter 2 Vector components describe motion in two dimensions. 109 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 110 PHYSICS INSIGHT Since the vertical velocity of the ball at maximum height is zero, you can also calculate the time taken to go up and multiply the answer by two. If down is positive, t vi vf a 0 m/s (11.00 m/s) 9.81 m/s2 11.00 m s m 9.81 s2 1.121 s The total time the baseball is in the air is 2 1.121 s 2.24 s. info BIT The longest speedboat jump was 36.5 m in the 1973 James Bond movie Live and Let Die. The boat practically flew over a road. 110 Unit I Kinematics dy viy 1 t ay(t)2 2 0 (11.00 m/s)t (9.81 m/s2)(t)2 1 2 Isolate t and solve. (4.905 m/s2)(t)2 (11.00 m/s)(t) t m 11.00 s m 4.905 s2 2.24 s Paraphrase The baseball is in the air for 2.24 s. How far would the baseball in Example 2.11 travel horizontally if the batter missed and the baseball landed at the same height from which it was launched? Since horizontal velocity is constant, dx t vix (19.05 m/s)(2.24 s) 42.7 m The baseball would travel a horizontal distance of 42.7 m. In the next example, you are given the time and are asked to solve for one of the other variables. However, the style of solving the problem remains the same. In any problem that you will be asked to solve in this course, you will always be able to solve for one quantity in either the x or y direction, and then you can substitute your answer to solve for the remaining variable(s). Example 2.12 A paintball directed at a target is shot at an angle of 25.0. If paint splats on its intended target at the same height from which it was launched, 3.00 s later, find the distance from the shooter to the target. Given Choose down and right to be positive. a ay 25.0 t 3.00 s 9.81 m/s2 [down] 9.81 m/s2 y vi 25.0° Figure 2.74 x 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 111 Required range (dx) Analysis and Solution Use the equation dy viy t ay(t)2. Since the height 1 2 of landing is the same as the launch height, dy y direction: 0. dy viy 1 t ay(t)2 2 0 viy 1 t ay(t)2 2 viy 1 t ay(t)2 2 viy t 1 ay 2 1 9.81 2 m s2 (3.00 s) 14.7 m/s Since down is positive, the negative sign means that the direction of the vertical component of initial velocity is up. x direction: Find the initial horizontal speed using the tangent function. Because there is no acceleration in the x direction, the ball’s horizontal speed remains the same during its flight: ax 0. PHYSICS INSIGHT Alternatively, the time taken to reach maximum height is the same time taken to fall back down to the same height. So, the paintball is at its maximum height at 1.50 s. The speed at maximum height is zero. If up is positive at 0 m/s at (9.81 m/s2)(t) (9.81 m/s2)(1.50 s) 14.7 m/s v i f The sign is positive, so the direction is up. tan opposite adjacent adjacent opposite tan 14.7 m/s tan 25.0° 31.56 m/s vi vix 25.0° Figure 2.75 14.7 m/s From Figure 2.75, the adjacent side is vix and it points to the right, so vix Now find the horizontal distance travelled. dx 31.56 m/s. t vix (31.56 m/s)(3.00 s) 94.7 m Practice Problems 1. Determine the height reached by a baseball if it is released with a velocity of 17.0 m/s [20]. 2. A German U2 rocket from the Second World War had a range of 300 km, reaching a maximum height of 100 km. Determine the rocket’s maximum initial velocity. Paraphrase The distance that separates the target from the shooter is 94.7 m. Answers 1. 1.72 m 2. 1.75 103 m/s [53.1] Chapter 2 Vector components describe motion in two dimensions. 111 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 112 The points below summarize what you have learned in this section. – To solve problems involving projectiles, first resolve the motion into its components using the trigonometric functions, then apply the kinematics equations. – Perpendicular components of motion are independent of one another. – Horizontal motion is considered uniform and is described by the equation d vt, whereas vertical motion is a special case of uniformly accelerated motion, where the acceleration is the acceleration due to gravity or 9.81 m/s2 [down]. – A projectile’s path is a parabola. – In the vertical direction, a projectile’s velocity is greatest at the instant of launch and just before impact, whereas at maximum height, vertical velocity is zero. 2.4 Check and Reflect 2.4 Check and Reflect Knowledge 1. Platform divers receive lower marks if they enter the water a distance away from the platform, whereas speed swimmers dive as far out into the pool as they can. Compare and contrast the horizontal and vertical components of each type of athlete’s motion. 2. For a fixed speed, how does the range depend on the angle, ? 3. (a) For a projectile, is there a location on its trajectory where the acceleration and velocity vectors are perpendicular? Explain. (b) For a projectile, is there a location on its trajectory where the acceleration and velocity |
vectors are parallel? Explain. 4. Water safety instructors tell novice swimmers to put their toes over the edge and jump out into the pool. Explain why, using concepts from kinematics and projectile motion. Applications 5. Participants in a road race take water from a refreshment station and throw their empty cups away farther down the course. If a runner has a forward speed of 6.20 m/s, how far in advance of a garbage pail should he release his water cup if the vertical distance between the lid of the garbage can and the runner’s point of release is 0.50 m? 6. A baseball is thrown with a velocity of 27.0 m/s [35]. What are the components of the ball’s initial velocity? How high and how far will it travel? 112 Unit I Kinematics 7. A football is thrown to a moving receiver. The football leaves the quarterback’s hands 1.75 m above the ground with a velocity of 17.0 m/s [25]. If the receiver starts 12.0 m away from the quarterback along the line of flight of the ball when it is thrown, what constant velocity must she have to get to the ball at the instant it is 1.75 m above the ground? 8. At the 2004 Olympic Games in Athens, Dwight Phillips won the gold medal in men’s long jump with a jump of 8.59 m. If the angle of his jump was 23, what was his takeoff speed? 9. A projectile is fired with an initial speed of 120 m/s at an angle of 55.0 above the horizontal from the top of a cliff 50.0 m high. Find (a) the time taken to reach maximum height (b) the maximum height with respect to the ground next to the cliff (c) the total time in the air (d) the range (e) the components of the final velocity just before the projectile hits the ground Extension 10. Design a spreadsheet to determine the maximum height and range of a projectile with a launch angle that increases from 0 to 90 and whose initial speed is 20.0 m/s. e TEST To check your understanding of projectile motion, follow the eTest links at www.pearsoned.ca/school/physicssource. 02-PearsonPhys20-Chap02 7/25/08 8:15 AM Page 113 CHAPTER 2 SUMMARY Key Terms and Concepts collinear resultant vector components polar coordinates method navigator method non-collinear relative motion ground velocity air velocity wind velocity trajectory range Key Equations v a t d v 1 t a(t)2 2 i d vt Conceptual Overview The concept map below summarizes many of the concepts and equations in this chapter. Copy and complete the map to have a full summary of the chapter. projectile motion Examples of Two-dimensional Motion relative motion Split given values into horizontal (x) and vertical components (y) Set up problem in x and y : x ax 0 vix vi ___ θ y ay 9.81 m/s2 vi ____ θ viy Time is common to both x and y directions. Use the equation d vi Solve in the x direction, then in the y direction, OR solve in the y direction, then in the x direction. t (t)2 a 1 2 State given values: vground _____ [ ] vair _____ [ ] vwind _____ [ ] Split given values into components. Use the equation v_____ vwind vair Use the algebraic vector method to solve for the unknown value. Two-dimensional motion Addition of vectors Graphical method Algebraic method Draw vectors to scale, connect them _________. Split given values into x and y components. Draw a new vector from starting point of diagram to end point. Use Rx R cosθ and _________. Measure length of vector and convert using scale. Solve separate equations in x and y. Measure angle at start of vector. Combine component answers using the Pythagorean theorem: Determine the angle using __________. R x2 y2 √ The answer is called the ___________. It has both magnitude and _________. Chapter 2 Vector components describe motion in two dimensions. 113 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 114 CHAPTER 2 REVIEW Knowledge 1. (2.2) During the Terry Fox Run, a participant 8. (2.4) For an object thrown vertically upward, what is the object’s initial horizontal velocity? travelled from A to D, passing through B and C. Copy and complete the table using the information in the diagram, a ruler calibrated in millimetres, and a protractor. In your notebook, draw and label the displacement vectors AB, BC, and CD and the position vectors AB, AC, and AD. Assume the participant’s reference point is A. d = 5.0 km [E] C 2.0 km scale D d = 3.0 km [E] d = 2.0 km [N] A B Distance Δd (km) Displacement Δd (km) [direction] Final position d (km) [direction] reference point AB BC CD AC AD 2. (2.2) Determine the x and y components of the displacement vector 55 m [222]. 3. (2.4) What is the vertical component for velocity at the maximum height of a projectile’s trajectory? 4. (2.4) During a field goal kick, as the football rises, what is the effect on the vertical component of its velocity? 5. (2.1) Fort McMurray is approximately 500 km [N] of Edmonton. Using a scale of 1.0 cm : 50.0 km, draw a displacement vector representing this distance. 6. (2.1) Give one reason why vector diagrams must be drawn to scale. 7. (2.2) Using an appropriate scale and reference coordinates, graphically solve each of the following: (a) 5.0 m [S] and 10.0 m [N] (b) 65.0 cm [E] and 75.0 cm [E] (c) 1.0 km [forward] and 3.5 km [backward] (d) 35.0 km [right] 45.0 km [left] 114 Unit I Kinematics Applications 9. The air medivac, King Air 200, flying at 250 knots (1 knot 1.853 km/h), makes the trip between Edmonton and Grande Prairie in 50 min. What distance does the plane travel during this time? 10. A golf ball is hit with an initial velocity of 30.0 m/s [55]. What are the ball’s range and maximum height? 11. Off the tee box, a professional golfer can drive a ball with a velocity of 80.0 m/s [10]. How far will the ball travel horizontally before it hits the ground and for how long is the ball airborne? 12. A canoeist capable of paddling north at a speed of 4.0 m/s in still water wishes to cross a river 120 m wide. The river is flowing at 5.0 m/s [E]. Find (a) her velocity relative to the ground (b) the time it takes her to cross 13. An object is thrown horizontally off a cliff with an initial speed of 7.50 m/s. The object strikes the ground 3.0 s later. Find (a) the object’s vertical velocity component when it reaches the ground (b) the distance between the base of the cliff and the object when it strikes the ground (c) the horizontal velocity of the object 1.50 s after its release 14. If a high jumper reaches her maximum height as she travels across the bar, determine the initial velocity she must have to clear a bar set at 2.0 m if her range during the jump is 2.0 m. What assumptions did you make to complete the calculations? 15. An alligator wishes to swim north, directly across a channel 500 m wide. There is a current of 2.0 m/s flowing east. The alligator is capable of swimming at 4.0 m/s. Find (a) the angle at which the alligator must point its body in order to swim directly across the channel (b) its velocity relative to the ground (c) the time it takes to cross the channel 16. A baseball player throws a ball horizontally at 45.0 m/s. How far will the ball drop before reaching first base 27.4 m away? 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 115 17. How much time can you save travelling diagonally instead of walking 750 m [N] and then 350 m [E] if your walking speed is 7.0 m/s? 18. How long will an arrow be in flight if it is shot at an angle of 25 and hits a target 50.0 m away, at the same elevation? 19. A pilot of a small plane wishes to fly west. The plane has an airspeed of 100 km/h. If there is a 30-km/h wind blowing north, find 27. An airplane is approaching a runway for landing. The plane’s air velocity is 645 km/h [forward], moving through a headwind of 32.2 km/h. The altimeter indicates that the plane is dropping at a constant velocity of 3.0 m/s [down]. If the plane is at a height of 914.4 m and the range from the plane to the start of the runway is 45.0 km, does the pilot need to make any adjustments to her descent in order to land the plane at the start of the runway? Consolidate Your Understanding Create your own summary of kinematics by answering the questions below. If you want to use a graphic organizer, refer to Student References 4: Using Graphic Organizers on pp. 869–871. Use the Key Terms and Concepts listed on page 113 and the Learning Outcomes on page 68. 1. Create a flowchart to describe the different components required to analyze motion in a horizontal plane and in a vertical plane. 2. Write a paragraph describing the similarities and differences between motion in a horizontal plane and motion in a vertical plane. Share your thoughts with another classmate. Think About It Review your answers to the Think About It questions on page 69. How would you answer each question now? e TEST To check your understanding of two-dimensional motion, follow the eTest links at www.pearsoned.ca/school/physicssource. (a) the plane’s heading (b) the plane’s ground speed 20. At what angle was an object thrown if its initial launch speed is 15.7 m/s, it remains airborne for 2.15 s, and travels 25.0 m horizontally? 21. A coin rolls off a 25.0° incline on top of a 2.5-m-high bookcase with a speed of 30 m/s. How far from the base of the bookcase will the coin land? 22. Starting from the left end of the hockey rink, the goal line is 3.96 m to the right of the boards, the blue line is 18.29 m to the right of the goal line, the next blue line is 16.46 m to the right of the first blue line, the goal line is 18.29 m right, and the right board is 3.96 m right of the goal line. How long is a standard NHL hockey rink? 23. A plane with a ground speed of 151 km/h is moving 11 south of east. There is a wind blowing at 40 km/h, 45 south of east. Find (a) the plane’s airspeed (b) the plane’s heading, to the nearest degree 24. How long will a soccer ball remain in flight if it is kicked with an initial velocity of 25.0 m/s [35.0]? How far down the field will the ball travel before it hits the ground and what will be its maximum height? 25. At what angle is an object launched if its initial vertical speed is 3.75 m/s |
and its initial horizontal speed is 4.50 m/s? Extensions 26. During the Apollo 14 mission, Alan Shepard was the first person to hit a golf ball on the Moon. If a golf ball was launched from the Moon’s surface with a velocity of 50 m/s [35] and the acceleration due to gravity on the Moon is 1.61 m/s2, (a) how long was the golf ball in the air? (b) what was the golf ball’s range? Chapter 2 Vector components describe motion in two dimensions. 115 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 116 UNIT I PROJECT Are Amber Traffic Lights Timed Correctly? Scenario The Traffic Safety Act allows law enforcement agencies in Alberta to issue fines for violations using evidence provided by red light cameras at intersections. The cameras photograph vehicles that enter an intersection after the traffic lights have turned red. They record the time, date, location, violation number, and time elapsed since the light turned red. The use of red light cameras and other technology reduces the amount of speeding, running of red lights, and collisions at some intersections. The length of time a traffic light must remain amber depends on three factors: perception time, reaction time, and braking time. The sum of perception time and reaction time is the time elapsed between the driver seeing the amber light and applying the brakes. The Ministry of Infrastructure and Transportation’s (MIT) Basic Licence Driver’s Handbook allows for a perception time of 0.75 s and a reaction time of 0.75 s. The braking time is the time it takes the vehicle to come to a full stop once the brakes are applied. Braking time depends on the vehicle’s initial speed and negative acceleration. The MIT’s predicted braking times are based on the assumption that vehicles travel at the posted speed limit and have a uniform acceleration of 3.0 m/s2. Other factors that affect acceleration are road conditions, vehicle and tire performance, weather conditions, and whether the vehicle was travelling up or down hill. If drivers decide to go through an intersection safely (go distance) after a light has turned amber, they must be able to travel not only to the intersection but across it before the light turns red. The go distance depends on the speed of the vehicle, the length of the intersection, and the amount of time the light remains amber. If the driver decides to stop (stop distance), the vehicle can safely do so only if the distance from the intersection is farther than the distance travelled during perception time, reaction time, and braking time. As part of a committee reporting to the Ministry of Infrastructure and Transportation, you must respond to concerns that drivers are being improperly fined for red light violations because of improper amber light timing. You are to decide how well the amber light time matches the posted speed limit and intersection length. Assume throughout your analysis that drivers travel at the posted speed limits. Planning Research or derive equations to determine Assessing Results After completing the project, assess its success based on a rubric* designed in class that considers research strategies experiment techniques clarity and thoroughness of the written report effectiveness of the team’s presentation (a) a car’s displacement during reaction time (b) stop distance (c) go distance (d) amber light time (e) displacement after brakes are applied (f) amount of time elapsed after the brakes are applied Materials • measuring tape, stopwatch Procedure 1 Design a survey to measure the amber light times at 10 different intersections near your school. For each intersection, record its length. Use caution around intersections due to traffic! You may wish to estimate the length of the intersection by counting the number of steps it takes you to cross and measuring the length of your stride. 2 Apply suitable equations to determine appropriate amber light times for the 10 different intersections. 3 Calculate stop distances and go distances for a range (10 km/h) of posted speed limits for each intersection and plot graphs of stop distance and go distance against posted speed. Thinking Further 1. Research the effectiveness of red light cameras in reducing accidents, speeding, and red light violations. Using your research, recommend a course of action to increase vehicle-rail safety at light-controlled railway crossings. 2. Based on your surveys and investigation, recommend whether existing amber light times should be increased, decreased, or left alone. Consider posted speeds against actual speeds and wet against dry surface conditions. 3. Prepare a presentation to the other members of your committee. Include graphs and diagrams. *Note: Your instructor will assess the project using a similar assessment rubric. 116 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 117 UNIT I SUMMARY Unit Concepts and Skills: Quick Reference Concepts Chapter 1 Summary Resources and Skill Building Graphs and equations describe motion in one dimension. Scalar and vector quantities 1.1 The Language of Motion A scalar quantity consists of a number and a unit. Distance, position, and displacement A vector quantity consists of a number, a unit, and a direction. Distance is the length of the path taken to travel from one position to another. Position is the straight-line distance from the origin to the object’s location. Displacement is the change in position. Slope of a position-time graph Slope of a velocity-time graph Position-time, velocity-time, acceleration–time graphs representing accelerated motion 1.2 Position-time Graphs and Uniform Motion A position-time graph for an object at rest is a straight line with zero slope. A position-time graph for an object moving at a constant velocity is a straight line with non-zero slope. The greater the slope of a position-time graph, the faster the object is moving. 1.3 Velocity-time Graphs: Uniform and Non-uniform Motion A velocity-time graph for an object experiencing uniform motion is a horizontal line. The slope of a velocity-time graph represents acceleration. The position-time graph for an object undergoing uniformly accelerated motion is a curve. The corresponding velocity-time graph is a straight line with non-zero slope. The corresponding acceleration-time graph is a horizontal line. Section 1.1 Section 1.1 Figure 1.5 Figures 1.3, 1.4, 1.5, Example 1.1 Figures 1.3, 1.4, 1.5, Example 1.1 Figure 1.14 Figures 1.12, 1.15(b), 1-3 Inquiry Lab Examples 1.2, 1.3, 1-3 Inquiry Lab Figure 1.24 Figures 1.24, 1.30, Example 1.5 Figures 1.28–1.31, Example 1.5 Instantaneous velocity The slope of the tangent on a position-time curve gives instantaneous velocity. Figure 1.29, Example 1.5 Area under and slope of a velocity-time graph Average velocity Velocity-time graphs 1.4 Analyzing Velocity-time Graphs The area under a velocity-time graph represents displacement; slope represents acceleration. Average velocity represents total displacement divided by time elapsed. You can draw acceleration-time and position-time graphs by calculating and plotting slope and area, respectively, of a velocity-time graph. Figure 1.41, Examples 1.6, 1.8, 1.9 Figure 1.45, Examples 1.7, 1.9 Examples 1.10, 1.11 Kinematics equations 1.5 The Kinematics Equations When solving problems in kinematics, choose the equation that contains all the given variables in the problem as well as the unknown variable. Figures 1.53–1.55 Examples 1.12–1.16 Projectile motion straight up and down Maximum height 1.6 Acceleration Due to Gravity Gravity causes objects to accelerate downward. At maximum height, a projectile’s vertical velocity is zero. The time taken to reach maximum height equals the time taken to fall back down to the original height. 1-6 QuickLab, 1-7 Inquiry Lab, 1-8 QuickLab, Examples 1.17–1.19 Figures 1.63–1.65 Examples 1.18, 1.19 Chapter 2 Vector components describe motion in two dimensions. Adding and subtracting vectors 2.1 Vector Methods in One Dimension Add vectors by connecting them tip to tail. Subtract vectors by connecting them tail to tail. Examples 2.1, 2.2 Components Relative motion Projectile motion in two dimensions 2.2 Motion in Two Dimensions To add vectors in two dimensions, draw a scale diagram, or resolve them into their components and use trigonometry to find the resultant. Examples 2.3–2.5, Figures 2.31, 2.33–2.37 2.3 Relative Motion To solve relative motion problems, use trigonometry, with ground velocity as the resultant. If the vectors are not perpendicular, resolve them into their components first. Examples 2.6–2.9 2.4 Projectile Motion The shape of a projectile’s trajectory is a parabola. Horizontal and vertical components of projectile motion are independent. To solve projectile problems in two dimensions, resolve them into their horizontal and vertical components. Then use the kinematics equations. The time taken to travel horizontally equals the time taken to travel vertically. 2-4 QuickLab, Figure 2.62 2-5 QuickLab, Figure 2.64 Examples 2.10–2.12, Figures 2.67, 2.71 Unit I Kinematics 117 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 118 UNIT I REVIEW Vocabulary 1. Using your own words, define these terms: acceleration acceleration due to gravity air velocity at rest collinear components displacement distance ground velocity instantaneous velocity kinematics navigator method non-collinear non-uniform motion origin polar coordinates method position projectile projectile motion range relative motion resultant vector scalar quantity tangent trajectory uniform motion uniformly accelerated motion vector quantity velocity wind velocity Knowledge CHAPTER 1 2. Describe how scalar quantities differ from vector quantities. CHAPTER 2 5. Using a scale of 1.0 cm : 3.5 km, determine the magnitude and direction of the vector below. W N S Applications E R 6. A wildlife biologist records a moose’s position as it swims away from her. Using the graph below, determine the moose’s velocity. Position vs. Time ) ] 30.0 25.0 20.0 15.0 10.0 |
5.0 0 0 2.0 4.0 8.0 6.0 Time (s) 10.0 12.0 7. Sketch a position-time graph for each statement below. Assume that right is positive. (a) object accelerating to the right (b) object accelerating to the left (c) object travelling at a constant velocity left (d) object at rest (e) object travelling with constant velocity right 8. Hockey pucks can be shot at speeds of 107 km/h. If a puck is shot at an angle of 30, determine how long the puck is in the air, how far it will travel, and how high it will be at the peak of its trajectory. 3. Resolve the following vectors into their 9. Sketch two different position-time graphs for objects with a negative velocity. 10. Sketch two different velocity-time graphs for objects with a negative acceleration. components: (a) 5.0 m [90°] (b) 16.0 m/s [20 S of W] 4. Using an appropriate scale and reference coordinates, draw the following vectors: (a) 5.0 m/s [0] (b) 25.0 m/s2 [60 N of E] (c) 1.50 km [120] 118 Unit I Kinematics 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 119 11. From the position-time graph below, determine 16. (a) What is the change in velocity in 10.0 s, which object has the greatest velocity. Time (s) 0.0 2.0 4.0 6.0 8.0 10.0 Time (s 12. Solve each of the following equations for initial velocity, v i, algebraically. v v (a) a i t f (b) d v 1 t at2 2 i (c) d 1 2 (v i v f)t 13. The longest kickoff in CFL history is 83.2 m. If the ball remains in the air for 5.0 s, determine its initial speed. 14. Determine the speed of a raven that travels 48 km in 90 min. 15. Describe the motion of the object illustrated in the graph below. Velocity vs. Time 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1..0 4.0 8.0 6.0 Time (s) 10.0 12.0 as illustrated in the acceleration-time graph below? (b) If the object had an initial velocity of 10 m/s [90], what is its final velocity after 10.0 s? Acceleration vs. Time ) ] ˚ .0 3.0 2.0 1.0 0.0 17. How far will a crow fly at 13.4 m/s for 15.0 min? 18. How long will it take a car to travel from Valleyview to Grande Prairie if its speed is 100 km/h? The map’s scale is 1 cm : 118 km. Wood Buffalo National Park 58 High Level Fort Chipewyan Fort Vermilion 35 ALBERTA Fort McMurray 88 Peace River 63 Grande Prairie Slave Lake 2 40 Valleyview 43 Edmonton 19. A baseball player hits a baseball with a velocity of 30 m/s [25]. If an outfielder is 85.0 m from the ball when it is hit, how fast will she have to run to catch the ball before it hits the ground? 20. Determine the magnitude of the acceleration of a Jeep Grand Cherokee if its stopping distance is 51.51 m when travelling at 113 km/h. 21. What is the velocity of an aircraft with respect to the ground if its air velocity is 785 km/h [S] and the wind is blowing 55 km/h [22 S of W]? 22. An object undergoing uniformly accelerated motion has an initial speed of 11.0 m/s and travels 350 m in 3.00 s. Determine the magnitude of its acceleration. 23. Improperly installed air conditioners can occasionally fall from apartment windows down onto the road below. How long does a pedestrian have to get out of the way of an air conditioner falling eight stories (24 m)? Unit I Kinematics 119 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 120 24. An object is launched from the top of a building with an initial velocity of 15 m/s [32]. If the building is 65.0 m high, how far from the base of the building will the object land? 25. Two friends walk at the same speed of 4.0 km/h. One friend steps onto a travelator moving at 3.0 km/h. If he maintains the same initial walking speed, (a) how long will it take him to reach the end of the 100-m-long travelator? (b) what must be the magnitude of the acceleration of the other friend to arrive at the end of the travelator at the same time? 26. How far will a vehicle travel if it accelerates uniformly at 2.00 m/s2 [forward] from 2.50 m/s to 7.75 m/s? 27. An object is thrown into the air with a speed of 25.0 m/s at an angle of 42°. Determine how far it will travel horizontally before hitting the ground. 28. Determine the average velocity of a truck that travels west from Lloydminster to Edmonton at 110 km/h for 1.0 h and 20 min and then 90 km/h for 100 min. 29. What distance will a vehicle travel if it accelerates uniformly from 15.0 m/s [S] to 35.0 m/s [S] in 6.0 s? 30. From the graph below, determine the instantaneous (b) velocity of the object at 5.0 s, 10.0 s, and 15.0 s. Position vs. Time 70.0 60.0 50.0 40.0 30.0 20.0 10.0 0. 33. Determine the displacement of the blue jay from the velocity-time graph below. Velocity vs. Time for a Blue Jay ) ] 32.0 28.0 24.0 20.0 16.0 12.0 8.0 4.0 0.0 0.0 10.0 20.0 30.0 40.0 50.0 60.0 Time (s) 34. Sketch a position-time graph for an object that travels at a constant velocity of 5.0 m/s for 10 s, stops for 10 s, then travels with a velocity of 2.0 m/s for 20 s. 35. Determine the height reached by a projectile if it is released with a velocity of 18.0 m/s [20]. 36. The bishop is a chess piece that moves diagonally along one colour of square. Assuming the first move is toward the left of the board, determine (a) the minimum number of squares the bishop covers in getting to the top right square. the bishop's displacement from the start if the side length of each square is taken as 1 unit and each move is from the centre of a square to the centre of another square. End 0.0 5.0 10.0 Time (s) 15.0 20.0 31. A speedboat’s engine can move the boat at a velocity of 215 km/h [N]. What is the velocity of the current if the boat’s displacement is 877 km [25 E of N] 3.5 h later? 32. An object starts from rest and travels 50.0 m along a frictionless, level surface in 2.75 s. What is the magnitude of its acceleration? 120 Unit I Kinematics Start 37. A wildlife biologist notes that she is 350 m [N] from the park ranger station at 8:15 a.m. when she spots a polar bear. At 8:30 a.m., she is 1.75 km [N] of the ranger station. Determine the biologist’s average velocity. 38. A bus travels 500 m [N], 200 m [E], and then 750 m [S]. Determine its displacement from its initial position. 02-PearsonPhys20-Chap02 7/24/08 10:18 AM Page 121 39. Match the motion with the correct position-time graph given below. Identify the motion as at rest, uniform motion, or uniformly accelerated motion. (a) an airplane taking off (b) an airplane landing (c) passing a car on the highway 43. A motorcycle stunt rider wants to jump a 20.0-mwide row of cars. The launch ramp is angled at 30 and is 9.0 m high. The landing ramp is also angled at 30 and is 6.0 m high. Find the minimum launch velocity required for the stunt rider to reach the landing ramp. (d) waiting at the red line at Canada Customs (e) standing watching a parade Skills Practice 44. Draw a Venn diagram to compare and contrast (f) travelling along the highway on cruise control vector and scalar quantities. Position vs. Time 45. Draw a Venn diagram to illustrate the concepts 100.0 90.0 80.0 70.0 60.0 50.0 40.0 30.0 20.0 10. ii iii iv 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Time (s) 40. Determine the magnitude of the acceleration of a Jeep Grand Cherokee that can reach 26.9 m/s from rest in 4.50 s. Extensions 41. A penny is released from the top of a wishing well and hits the water’s surface 1.47 s later. Calculate (a) the velocity of the penny just before it hits the water’s surface (b) the distance from the top of the well to the water’s surface 42. A balloonist drops a sandbag from a balloon that is rising at a constant velocity of 3.25 m/s [up]. It takes 8.75 s for the sandbag to reach the ground. Determine of graphical analysis. 46. A swimmer wants to cross from the east to the west bank of the Athabasca River in Fort McMurray. The swimmer’s speed in still water is 3.0 m/s and the current’s velocity is 4.05 m/s [N]. He heads west and ends up downstream on the west bank. Draw a vector diagram for this problem. 47. For an experiment to measure the velocity of an object, you have a radar gun, probeware, and motion sensors. Explain to a classmate how you would decide which instrument to use. 48. Design an experiment to determine the acceleration of an object rolling down an inclined plane. 49. Construct a concept map for solving a twodimensional motion problem involving a projectile thrown at an angle. 50. Explain how you can use velocity-time graphs to describe the motion of an object. Self-assessment 51. Describe to a classmate which kinematics concepts and laws you found most interesting when studying this unit. Give reasons for your choices. 52. Identify one issue pertaining to motion studied in this unit that you would like to investigate in greater detail. 53. What concept in this unit did you find most difficult? What steps could you take to improve your understanding? 54. As a future voter, what legislation would you support to improve vehicular and road safety? (a) the height of the balloon when the sandbag 55. Assess how well you are able to graph the is dropped (b) the height of the balloon when the sandbag reaches the ground (c) the velocity with which the sandbag hits the ground motion of an object. Explain how you determine a reference point. e TEST To check your understanding of kinematics, follow the eTest links at www.pearsoned.ca/school/physicssource. Unit I Kinematics 121 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 122 U N I T II Dynamics Dynamics The design of equipment used in many activities, such as ice climbing, involves understanding the cause of motion. How does gravity affect the climber and the icy cliff? How can understanding the cause of motion help you predict motion? e WEB Explore the physics principles that apply to ice and mountain climbing. Write a summary of your findings. Begin your search at www.pearsoned.ca/school/physicssource. 122 Unit II 03-Phys20-Chap03.qxd 7/24/08 10:37 AM Page 123 Unit at a Glance C H A P T E R 3 Forces can change velocity. 3.1 The Nature of Force 3.2 Newton’s First Law 3.3 Newton’s Second Law 3.4 Newton’s Third Law 3.5 Friction Affects Motion C H A P T E R 4 Gravity extends throughout the u |
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