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### ch29

Course: PHYSICS 201, Fall 2012
School: Rutgers
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Word Count: 8136

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29 CHAPTER PARTICLES AND WAVES ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (a) At the higher temperature, the intensity per unit wavelength is greater, and the aximum occurs at a shorter wavelength (see Section 29.2). 2. (b) An X-ray photon has a much greater frequency than does a microwave photon (see Section 24.2). The X-ray photon also has a greater energy E, because E = hf (Equation 29.2), where h is Plancks...

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29 CHAPTER PARTICLES AND WAVES ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (a) At the higher temperature, the intensity per unit wavelength is greater, and the aximum occurs at a shorter wavelength (see Section 29.2). 2. (b) An X-ray photon has a much greater frequency than does a microwave photon (see Section 24.2). The X-ray photon also has a greater energy E, because E = hf (Equation 29.2), where h is Plancks constant and f is the frequency. The wavelength and frequency of a photon are related by = c/f (Equation 16.1), where c is the speed of light in a vacuum. Since the microwave photon has the smaller frequency, it has the greater wavelength. 3. (c) A green photon has a greater frequency than does a red photon (see Section 24.2). Therefore, the green photon possesses a greater energy E, because E = hf (Equation 29.2), where h is Plancks constant and f is the frequency. 4. (e) As the wavelength of a photon becomes smaller, its frequency and its energy become larger (see Section 29.3). Since the energy of an incident photon is equal to the maximum kinetic energy of the photoelectron plus the work function of the metal, increasing the incident energy increases the maximum kinetic energy of the photoelectron. The work function is a characteristic of the metal, and does not depend on the photon. 5. (d) The maximum kinetic energy of the ejected photoelectrons depends on the energy of the incident photons. Doubling the number of photons per second that strikes the surface does not change the energy of the incident photons, which depends only on the frequency of the photons. Thus, the maximum kinetic energy of the ejected photoelectrons does not change. However, doubling the number of photons per second that strikes the surface means that twice as many photoelectrons per second are ejected from the surface. 6. (b) Whether or not electrons are ejected from the surface of the metal depends on the energy of the incident photons and the work function of the metal The work function depends on the type of metal (e.g., aluminum or copper) from which the plate is made. (See Section 29.3). 7. W0 = 3.6 10 19 J 8. KEmax = 2.7 10 20 J 9. (d) According to the discussion in Section 29.4, a photon has a momentum whose magnitude p is related to its wavelength by p = h/. 231 PARTICLES AND WAVES 10. (b) In the Compton effect, an X-ray photon strikes an electron and, like two billiard balls (particles) colliding on a pool table, the X-ray photon scatters in one direction and the electron recoils in another direction after the collision. 11. (a) In the Compton effect, some of the energy of the incident photon is given to the recoil electron. Therefore, the energy of the scattered photon is less than that of the incident photon. Since the energy of a photon depends inversely on its wavelength (see Section 29.3), the wavelength of the scattered photon is greater than that of the incident photon. 12. = 0.35 nm 13. (c) The de Broglie wavelength depends inversely on the magnitude p of the momentum; = h/p (Equation 29.8), where h is Plancks constant. Particle A, having the smaller charge, has the smaller electric potential energy (see Section 19.2). Consequently, after accelerating through the potential difference, particle A has the smaller kinetic energy, and hence, the smaller momentum. Thus, particle A has the longer de Broglie wavelength. 14. (e) The de Broglie wavelength depends inversely on the magnitude p of the momentum; = h/p (Equation 29.8), where h is Plancks constant. Therefore, as the momentum decreases, the wavelength increases, and vice versa. In A, the proton is moving opposite to the direction of the electric field, so the proton is slowing down, and its momentum is decreasing. In B, the proton is accelerating, and its momentum is increasing. In C, the proton moves parallel to the magnetic field. According to the discussion in Section 21.2, the proton does not experience a force, so its momentum remains constant. In D the proton is moving perpendicular to the direction of the magnetic field. According to the discussion in Section 21.3, such a situation does not change the magnitude of the protons momentum. 15. = 2.1 10 14 m 16. (b) According to the Heisenberg uncertainty principle, the uncertainty y in a particles position is related to the uncertainty py in its momentum by (see Equation 29.10) y h / ( 4 p y ) . If p = 0 kg m/s, then y becomes infinitely large. y Chapter 29 Problems 232 CHAPTER 29 PARTICLES AND WAVES PROBLEMS 1. REASONING AND SOLUTION The energy of a single photon is 34 E = hf = (6.63 10 6 26 J s)(98.1 10 Hz) = 6.50 10 J The number of photons emitted per second is Power radiated 5.0 10 4 W = = Energy per photon 6.50 10 26 J 2. 7 .7 10 29 photons / s REASONING The energy of a photon of frequency f is, according to Equation 29.2, E = hf , where h is Planck's constant. Since the frequency and wavelength are related by f = c / (see Equation 16.1), the energy of a photon can be written in terms of the wavelength as E = hc / . These expressions can be solved for both the wavelength and the frequency. SOLUTION a. The wavelength of the photon is = hc (6.63 10 34 J s)(3.00 10 8 m / s) = = 1.63 10 7 m 18 E 1.22 10 J b. Using the answer from part (a), we find that the frequency of the photon is f= c 3.00 10 8 m / s = = 1.84 10 15 Hz 1.63 10 7 m Alternatively, we could use Equation 29.2 directly to obtain the frequency: f= E 1.22 10 18 J = = 1.84 10 15 Hz h 6.63 10 34 J s c. The wavelength and frequency values shown in Figure 24.9 indicate that this photon corresponds to electromagnetic radiation in the ultraviolet region of the electromagnetic spectrum. 233 PARTICLES AND WAVES 3. REASONING According to Equation 29.3, the work function W0 is related to the photon energy hf and the maximum kinetic energy KEmax by W0 = hf KE max . This expression can be used to find the work function of the metal. SOLUTION KE max is 6.1 eV. The photon energy (in eV) is, according to Equation 29.2, c c h hf = 6.63 10 34 J s 3.00 10 15 Hz F eV J G h 1.601 10 J I = 12.4 eV H K 19 The work function is, therefore, W0 = hf KE max = 12.4 eV 6.1 ev = 6.3 eV 4. REASONING According to Equation 29.3, the relation between the photon energy, the maximum kinetic energy of an ejected electron, and the work function of a metal surface is hf K { = 1 E max 42 4 3 Photon energy Maximum kinetic energy of ejected electron + W0 { Work function Equation 16.1 relates the frequency f of a photon to its wavelength via f = c/, where c is the speed of light in a vacuum. The maximum kinetic energy KE max is related to the mass m 1 and maximum speed vmax of the ejected electron by KEmax = 2 mv 2 (Equation 6.2). With these substitutions, Equation 29.3 becomes hf = KE max + W0 or hc 1 2 = mv + W0 2 max SOLUTION Solving Equation (1) for the wavelength gives = hc 1 mv 2 max 2 + W0 ( 6.63 1034 J s ) ( 3.00 108 m/s ) A = = 2 1 9.11 10 31 kg 7.30 105 m/s + 4.80 1019 J )( ) 2( 2.75 107 m (1) Chapter 29 Problems ( 6.63 1034 J s ) ( 3.00 108 m/s ) B = = 2 1 9.11 1031 kg 5.00 105 m/s + 4.80 10 19 J )( ) 2( 5. 234 3.35 107 m REASONING The energy of the photon is related to its frequency by Equation 29.2, E = hf . Equation 16.1, v = f , relates the frequency and the wavelength for any wave. SOLUTION Combining Equations 29.2 and 16.1, and noting that the speed of a photon is c, the speed of light in a vacuum, we have = 6. c c hc (6.63 10 34 J s)(3.0 10 8 m / s) = = = = 3.1 10 7 m = 19 f ( E / h) E 6.4 10 J 310 nm REASONING The photons of this wave must carry at least enough energy to equal the work function. Then the electrons are ejected with zero kinetic energy. Since the energy of a photon is E = hf according to Equation 29.2, where f is the frequency of the wave, we have that W0 = hf. Equation 16.1 relates the frequency to the wavelength according to f = c/, where c is the speed of light. Thus, it follows that W0 = hc/. SOLUTION Using Equations 29.2 and 16.1, we find that c c h h 6.63 10 34 J s 3.00 10 8 m / s hc W0 = = = 4 .10 10 19 J 485 10 9 m Since 1 eV = 1.60 1019 J, it follows that c W0 = 4 .10 10 19 J 7. REASONING AND SOLUTION KE max = hf Wo = 6 c.63 10 = F eV J G h1.601 10 J I = H K c h J s 3.00 10 8 m / s 9 215 10 m 2 .56 eV Equation 29.3 gives hc Wo 34 19 h b.68 eV g1.60 10 J I = 3.36 10 F 3 G 1 eV J H K 19 19 J 235 PARTICLES AND WAVES Converting to electron volts c KE max = 3.36 10 19 J 8. F h1.601eV J I = G 10 J H K 19 2 .10 eV REASONING The intensity S = 680 W/m2 of the photons is equal to the total amount of energy delivered by the photons per second per square meter of surface area. Therefore, the intensity of the photons is equal to the energy E of one photon multiplied by the number N of photons that reach the surface of the earth per second per square meter: S = NE (1) The energy E of each photon is given by E = hf (Equation 29.2), where h = 6.631034 Js c is Plancks constant and f is the frequency of the photon. We will use f = (Equation 16.1) to determine the frequency f of the photons from their wavelength and the speed c of light in a vacuum. SOLUTION Solving Equation (1) for N, we obtain N = S/E. Substituting E = hf (Equation 29.2) into this result yields N= S S = E hf (2) c (Equation 16.1) into Equation (2), we find that 680 W/m 2 730 10 9 m S S S N= = = = hf c hc 6.63 1034 J 3.00 108 m/s s h Substituting f = ( ( )( )( ) ) = 2.5 1021 photons/ ( s 2 ) m 9. REASONING AND SOLUTION The number of photons per second, N, entering the owl's eye is N = SA / E , where S is the intensity of the beam, A is the area of the owl's pupil, and E is the energy of a single photon. Assuming that the owl's pupil is circular, A = r2 = cd h, where d is the diameter of the owl's pupil. 1 2 2 Combining Equations 29.2 and 16.1, we have E = hf = hc / . Therefore, c h 2 1 (5.0 10 13 W / m 2 ) 2 8.5 10 3 m (510 10 9 m) SA N= = = hc (6.63 10 34 J s)(3.0 10 8 m / s) 73 photons / s Chapter 29 Problems 236 10. REASONING The wavelength of the light shining on the surface is related to the maximum kinetic energy KEmax of the electrons ejected from the surface by c hf = KE max + W0 (Equation 29.3), where h is Plancks constant, f = (Equation 16.1) is the frequency of the light, and W0 is the work function of the surface. Substituting Equation 16.1 into Equation 29.3, we see that hf = hc = KE max + W0 (1) The work function W0 is a property of the metal surface itself, so it remains the same for any wavelength of incident light. When the wavelength of the incident light is 1 = 221 nm, the maximum kinetic energy of the ejected electrons is KE max,1 = 3.281019 J, and when the wavelength is 2, the maximum kinetic energy is twice as great: KE max,2 = 2KEmax,1. We will use this information, with Equation (1), to determine the unknown wavelength 2. SOLUTION Solving Equation (1) for the work function W0 yields W0 = KE max + hc (2) Because the work function W0 isnt affected by changing the wavelength of the incident light from 1 to 2, we have that W0 = KE max, 2 + hc hc = KE max,1 + 2 1 KE max, 2 KE max, 1 + or hc hc = 1 2 (3) Solving Equation (3) for 2 and substituting KEmax,2 = 2 KEmax,1, we obtain 2 = hc hc KE max, 2 KE max, 1 + 1 = hc hc 2KE max, 1 KE max, 1 + 1 = 1 KE max, 1 hc + 1 1 (4) In the last step of Equation (4), we have divided both the numerator and the denominator by the product hc. Substituting the given values the into Equation (4), we find that 237 PARTICLES AND WAVES 2 = 1 ( 3.28 10 J ) 1 + 34 8 ( 6.63 10 J s ) ( 3.00 10 m/s ) 221 109 m 19 = 1.62 107 m = 162 nm 11. REASONING The wavelength of the photon is related to its frequency f by = c/f (Equation 16.1), where c is the speed of light. The frequency of the photon is proportional to its energy E via f = E/h (Equation 29.2), where h is Plancks constant. Thus, = ch/E. The photon energy is equal to the sum of the maximum kinetic energy KE max of the ejected electron and the work function W0 of the metal; E = KE max + W0 (Equation 29.3). Substituting this expression for E into = ch/E gives = ch KE max + W0 (1) 2 The maximum kinetic energy is related to the maximum speed vmax by KE max = 1 mvmax 2 (Equation 6.2), where m is the mass of the electron. 2 SOLUTION Substituting KE max = 1 mvmax into Equation (1), and converting the work 2 function from electron-volts to joules, gives = ch 1 mv 2 max 2 + W0 ( 3.00 108 = 1 2 ( 9.11 1031 m/s ) ( 6.63 1034 J s ) = 1.9 107 m 2 1.60 1019 J kg ) ( 1.2 106 m/s ) + ( 2.3 eV ) 1 eV 12. REASONING The total energy Q delivered by N photons is NE, where E is the energy carried by one photon, so that N = Q/E. Equation 29.2 indicates that the photon energy is E = hf, where h is Plancks constant and f is the frequency. Thus, the number N of photons can be written as QQ N= = (1) E hf Equation 16.1 relates the frequency f of a photon to its wavelength according to f = c/, where c is the speed of light in a vacuum. Therefore, Equation (1) can be expressed as Chapter 29 Problems N= Q Q = hf hc 238 (2) According to Equation 12.4, the heat Q required to raise the temperature of a substance by an amount T is Q = cspecific heatm T, where cspecific heat is the specific heat capacity of the substance and m is its mass. SOLUTION Combining Equation 12.4 with Equation (2), the number of photons required to raise the temperature by an amount T is N= Q hc or N= cspecific heat m T hc Applying this result to each type of photon, we obtain ( 6.0 105 m ) 840 J/ ( kg C ) ( 0.50 kg ) ( 2.0 C ) N infrared = 34 ( 6.63 10 J s ) ( 3.00 108 m/s ) N blue = ( 4.7 107 m ) 840 J/ ( kg C ) ( 0.50 kg ) ( 2.0 C) ( 6.63 1034 J s ) ( 3.00 108 m/s ) = 2.5 1023 = 2.0 1021 13. SSM WWW REASONING AND SOLUTION a. According to Equation 24.5b, the electric field can be found from E = S / ( 0 c ) . The intensity S of the beam is Energy per unit time Nh f Nh c S= = = A A A = ( 1.30 1018 photons/s ) ( 6.63 1034 J s ) 3.00 108 m/s 2 514.5 109 m ( 1.00 103 m ) = 1.60 105 W/m 2 where N is the number of photons per second emitted. Then, E = S / ( 0c ) = 7760 N / C 239 PARTICLES AND WAVES b. According to Equation 24.3, the average magnetic field is B = E/c = 2.59 10 5 T 14. REASONING The heat required to melt the ice is given by Q = mL f , where m is the mass of the ice and L f is the latent heat of fusion for water (see Section 12.8). Since, according to Equation 29.2, each photon carries an energy of E = hf , the energy content of N photons is E Total = Nhf . According to Equation 16.1, f = c / , so we have E Total = Nhc If we assume that all of the photon energy is used to melt the ice, then, E Total = Q , so that Nhc = mL f : E 4 Q Total This expression may be solved for N to determine the required number of photons. SOLUTION a. We find that N= mL f hc = (2.0 kg)(33.5 10 4 J / kg)(620 10 9 m) = 2.1 10 24 photons (6.63 10 34 J s)(3.00 10 8 m / s) b. The number N of molecules in 2.0-kg of water is N = (2.0 kg) F 1 mol I F.022 10 molecules I = 6.7 10 6 J G 10 kg J G 1 mol K 18 H KH 23 3 25 molecules Therefore, on average, the number of water molecules that one photon converts from the ice phase to the liquid phase is N 6.7 10 25 molecules = = 32 molecules / photon N 2.1 10 24 photons 15. SSM REASONING The angle through which the X-rays are scattered is related to the difference between the wavelength of the scattered X-rays and the wavelength of the incident X-rays by Equation 29.7 as Chapter 29 Problems = 240 h( 1 cos ) mc where h is Plancks constant, m is the mass of the electron, and c is the speed of light in a vacuum. We can use this relation directly to find the angle, since all the other variables are known. SOLUTION Solving Equation 29.7 for the angle , we obtain cos = 1 mc ( ) h ( 9.111031 kg ) ( 3.00 108 m/s ) ( 0.2703 109 m 0.2685 109 m ) = 0.26 = 1 6.63 1034 J s = cos 1 ( 0.26 ) = 75 16. REASONING The momentum of the photon is related to its wavelength and Plancks constant h. The momentum (nonrelativistic) of the ball depends on its mass m and speed v. We can set the two momenta equal and solve directly for the speed. SOLUTION The momentum pphoton of the photon and the momentum pball of the ball are pphoton = h (29.6) and pball = mv (7.2) Since pphoton = pball, we have h = mv 4 123 4 14 2 4 3 Momentum of photon Momentum of ball or v= h 6.63 1034 J s = = 4.2 1025 m/s 9 3 m 720 10 m 2.2 10 kg ( )( ) 17. REASONING The frequency f of a photon is related to its energy E by f = E/h (Equation 29.2), where h is Plancks constant. As discussed in Section 29.4, the energy E is related to the magnitude p of the photons momentum by E = pc, where c is the speed of light in a vacuum. By combining these two relations, we see that the frequency can be expressed in terms of p as f = pc/h. SOLUTION 241 PARTICLES AND WAVES a. Substituting values for p, c, and h, into the relation f = pc/h gives f= pc ( 2.3 1029 kg m/s ) ( 3.00 108 m/s ) = = 1.0 1013 Hz 34 h 6.63 10 J s b. An inspection of Figure 24.9 shows that this frequency lies in the infrared region of the electromagnetic spectrum. 18. REASONING Before the scattering, the electron is at rest and has no momentum. Thus, the total initial momentum consists only of the photons momentum, which points along the + x axis. The total initial momentum has no y component. Since the total momentum is conserved, the total momentum after the scattering must be the same as it was before and, therefore, has no y component. The total momentum after the scattering is the sum of the momentum of the scattered photon and that of the scattered electron, and it only has an x component. But the scattered photon is moving along the y axis, so its momentum has no x component. Therefore, the momentum of the electron must have an x component. The total momentum after the scattering is the sum of the momentum of the scattered photon and that of the scattered electron, and it has no y component. But the scattered photon is moving along the y axis, so its momentum points along the y axis. Therefore, this contribution to the total final momentum must be cancelled by part of the momentum of the scattered electron, which must have a component along the +y axis. SOLUTION Since the total momentum is conserved and since the scattered photon has no momentum in the x direction, the momentum of the scattered electron must have an x component that equals the momentum of the incident photon. According to Equation 29.6, the magnitude p of the momentum of the incident photon is p = h/, where h is Plancks constant and is the wavelength. Therefore, the momentum of the scattered electron has a component in the +x direction that is px = h 6.63 10 34 J s = = 7 .37 10 23 kg m / s 9.00 10 12 m The momentum of the scattered electron has a component along the + y direction. This component cancels the momentum of the scattered photon that points along the y direction. To find the momentum of the scattered photon, we first need to determine its wavelength , which we can do using Equation 29.7: Chapter 29 Problems = + 242 h 1 cos 90.0 3 mc 14 2 4 =0 = 9.00 1012 m + 6.63 1034 J s ( 9.11 10 31 )( 8 kg 3.00 10 m/s ) = 1.14 1011 m Again using Equation 29.6, we find that the momentum of the scattered electron has a component in the +y direction that is py = h 6.63 10 34 J s = = 5.82 10 23 kg m / s 1.14 10 11 m 19. REASONING There are no external forces that act on the system, so the conservation of linear momentum applies. Since the photon is scattered at = 180 , the collision is "head-on," and all motion occurs along the horizontal direction, which we take as the x axis. The incident photon is assumed to be moving along the + x axis. For an initially stationary electron, the conservation of linear momentum states that: p 123 Momentum of incident photon = p + pelectron 14 2 43 14 2 4 3 Momentum of scattered photon Momentum of recoil electron where the momentum of the scattered photon is negative since is moves along the x direction (the scattering angle is 180). Using the relation p = h/ (Equation 29.6), where h is Plancks constant and is the wavelength of the photon, we can write the expression for the momentum of the electron as pelectron = p + p = hh 1 1 + = h + SOLUTION Substituting numerical values into the equation above, we have 1 1 pelectron = (6.626 10 34 J s) + = 4.755 10 24 kg m/s 9 9 0.2750 10 m 0.2825 10 m 243 PARTICLES AND WAVES 20. REASONING The wavelength of the incident X-rays is related to the wavelength of h the scattered X-rays by = ( 1 cos ) (Equation 29.7), where h = 6.6261034 Js is mc 8 Plancks constant, c = 2.99810 m/s is the speed of light in a vacuum, m = 9.1091031 kg is the mass of an electron, and = 122.0 is the angle at which the X-rays are scattered. The wavelength of the scattered photon is found from the magnitude p of its h momentum via p = (Equation 29.6). SOLUTION Solving = h ( 1 cos ) (Equation 29.7) for , we obtain mc = h ( 1 cos ) mc (1) h h (Equation 29.6) for yields = , which, on substitution into p Equation (1), gives Solving p = = h h h ( 1 cos ) = ( 1 cos ) mc p mc ( 6.626 1034 J s ) ( 1 cos122.0o ) = 1.856 1024 kg m/s ( 9.109 1031 kg ) ( 2.998 108 m/s ) 6.626 1034 J s = 3.533 1010 m = 0.3533 nm 21. REASONING The change in wavelength that occurs during Compton scattering is given by Equation 29.7: = h ( 1 cos ) mc or ( ) max = h 2h ( 1 cos 180) = mc mc ( ) max is the maximum change in the wavelength, and to calculate it we need a value for the mass m of a nitrogen molecule. This value can be obtained from the mass per mole M of nitrogen ( N 2 ) and Avogadro's number N A , according to m = M / N A (see Section 14.1). SOLUTION Using a value of M = 0.0280 kg/mol , we obtain the following result for the maximum change in the wavelength: Chapter 29 Problems ( ) max ( 244 ) 2 6.63 10 34 J s 2h 2h = = = mc M 0.0280 kg/mol 3.00 108 m/s N c 23 1 A 6.02 10 mol ( ) = 9.50 10 17 m 22. REASONING Energy is conserved during the collision. This means that the energy E of the incident photon must equal the kinetic energy KE of the recoil electron plus the energy E the scattered photon: E = KE + E (1) The energy E of a photon is related to its frequency f by E = hf (Equation 29.2), where h is Plancks constant. The frequency, in turn, is related to the wavelength by f = c/ (Equation 16.1), where c is the speed of light in a vacuum. Substituting f = c/ into E = hf gives (2) E = hc/ The wavelength of the scattered photon depends on the wavelength of the incident photon according to Equation 29.7, so that we have = + h ( 1 cos ) mc Since the photon is scattered straight backward, = 180, and = + h ( 1 cos180 ) = + 2h mc mc (3) SOLUTION The kinetic energy of the recoil electron is given by KE = 1 mv 2 2 (Equation 6.2), where m is its mass and v is its speed. Substituting this expression into Equation (1), we have E = 1 mv 2 + E . Solving for the speed v of the electron gives 2 v= 2 ( E E) m (4) From Equation (2), we also know that E = hc/ and E = hc / . Substituting this expression into Equation (4), we find that the speed of the electron can be written as v= 2h c 1 1 m 245 PARTICLES AND WAVES Since = + v= = 2h [Equation (3)], the speed of the electron is mc 2h c 1 1 2h m + mc 2 ( 6.63 1034 J s ) ( 3.00 108 m/s ) 9.11 1031 kg 1 1 9 34 2 ( 6.63 10 J s ) 0.45000 10 m 0.45000 109 m + 31 8 ( 9.11 10 kg ) ( 3.00 10 m/s ) = 3.22 106 m/s 23. REASONING AND SOLUTION a. We have = ' (h/mc)(1 cos 163) = 0.1819 nm b. For the incident photon E = hf = hc/ = 1.092 10 15 J c. For the scattered photon E ' = hf ' = hc/' = 1.064 10 15 J d. The kinetic energy of the recoil electron is, therefore, KE = E E ' = 2.8 10 17 J 24. REASONING a. Consider one square meter of the sails surface. Each of the N photons that strike this square meter in a one-second interval ( t = 1.0 s) has an initial momentum p that is h determined by its wavelength , according to p = (Equation 29.6), where 34 h = 6.6310 Js is Plancks constant. Each photon is fully reflected, so the final momentum of a photon is equal to p. The magnitude of the change p in a photons Chapter 29 Problems 246 momentum, then, is equal to p = |p (p)| = 2p, and the magnitude P of the net momentum change undergone by all N photons is given by P = N p = 2Np. In order to cause this momentum change, the sail exerts an impulse of magnitude F t = P (Equation 7.4) on the photons, where F is the magnitude of the force exerted on the photons per square meter of the sail. By Newtons Third Law, that force magnitude is equal to the magnitude of the force exerted by the N photons on one square meter of the sail. From Equation 7.4, we then, have that F t = P = 2 Np (1) Lastly, the magnitude F = F of the net force necessary for the sail to attain the desired acceleration of a = 9.8106 m/s2 is given by Newtons Second Law, F = ma (Equation 4.1), where m = 3.0103 kg is the mass of one square meter of the sail. As instructed, we have ignored all other forces acting on the sail. b. The intensity S of the laser beam depends on the total energy delivered to the sail by Total energy S= (Equation 24.4), where A is the area of the sail and t is the time t A interval. The total energy is equal to the number N of photons that strike the area in one second times the energy E of a single photon. Therefore, the intensity of the laser beam is S= Total energy NE = t A t A (2) c 8 (Equation 16.1) is the photons frequency and c = 3.0010 m/s is the speed of light in a vacuum. The energy E of a single photon is given by E = hf (Equation 29.2), where f = SOLUTION a. Solving Equation (1) for N, we obtain N= Substituting p = F t 2p h (Equation 29.6) and F = ma (Equation 4.1) into Equation (3) yields (3) 247 PARTICLES AND WAVES N= F t mat mat = = 2p 2h h 2 ( 3.0 103 kg ) ( 9.8 106 m/s2 ) ( 1.00 s ) ( 225 109 m ) = 5.0 1018 = 2 ( 6.63 1034 J s ) hc c (Equation 16.1) into E = hf (Equation 29.2) gives E = hf = . Substituting this result into Equation (2), we obtain b. Substituting f = S= NE Nhc = t A t A (4) Equation (4) applies to the intensity reaching an area A = 1.0 m2 of the sail in a time t = 1.0 s. Therefore, in order for the sail to accelerate at the desired rate, the intensity of the laser must be S= Nhc ( 5.0 1018 ) ( 6.63 1034 J s ) ( 3.00 108 m/s ) = = 4.4 W/m 2 2) ( 9 t A ( 1.00 s ) ( 1.00 m 225 10 m ) 25. REASONING AND SOLUTION The de Broglie wavelength is given by Equation 29.8 as = h / p , where p is the magnitude of the momentum of the particle. The magnitude of the momentum is p = mv , where m is the mass and v is the speed of the particle. Using this expression in Equation 29.8, we find that = h / ( mv ) , or v= 6.63 10 34 J s h = = 3.05 10 7 m / s 27 14 m 1.67 10 kg 1.30 10 m c h c h The kinetic energy of the proton is 1 1 KE = 2 mv 2 = 2 ( 1.67 10 27 kg)(3.05 10 7 m / s) 2 = 7.77 10 13 J 26. REASONING According to Equation 27.1, the angle that locates the first-order bright fringes (m = 1) is specified by sin = /d, where is the wavelength and d is the separation between the slits. The wavelength of the electron is the de Broglie wavelength, which is given by = h/p (Equation 29.8), where h is Plancks constant and p is the magnitude of the momentum of the electron. Chapter 29 Problems 248 SOLUTION Combining Equations 27.1 and 29.8, we find that the angle locating the firstorder bright fringes is specified by sin = h = d pd Dividing this result for case A by that for case B, we find sin A h / ( pA d ) pB = = sin B h / ( pBd ) pA or pB = ( 1.2 1022 kg m/s ) sin ( 1.6 104 degrees ) = pB = sin ( 4.0 104 degrees ) pA sin A sin B 4.8 1023 kg m/s 27. REASONING AND SOLUTION The de Broglie wavelength is given by Equation 29.8 as = h/p, where p is the magnitude of the momentum of the particle. The magnitude of the momentum is p = mv, where m is the mass and v is the speed of the particle. Using this expression in Equation 29.8, we find that = h mv v= or 6.63 10 34 J s h = = 1.41 10 3 m / s 27 9 m 1.67 10 kg 0.282 10 m c h c h 28. REASONING AND SOLUTION We know that = h/mv. Solving for the mass yields m= 6.63 10 34 J s h = = v 8.4 10 14 m 1.2 10 6 m / s c h c h 6.6 10 27 kg 29. REASONING AND SOLUTION The average kinetic energy of a helium atom is 23 KE = (3/2)kT = (3/2)(1.38 10 21 J/K)(293 K) = 6.07 10 The speed of the atom corresponding to the average kinetic energy is v= 6 b g= 2c.07 10 J h= 1.35 10 2 KE m The de Broglie wavelength is 21 6.65 10 27 kg 3 m/s J 249 PARTICLES AND WAVES = 6.63 10 34 J s h = = mv 6.65 10 27 kg 1.35 10 3 m / s c c h h 7 .38 10 11 m 30. REASONING The de Broglie wavelength of a particle is inversely proportional to the h magnitude p of its momentum, as we see from = (Equation 29.8), where p h = 6.631034 Js is Plancks constant. The electron is moving at a speed v = 0.88c, which is close to the speed of light in a vacuum ( c = 3.00108 m/s). Therefore, we will use mv p= v 2 (Equation 28.3) to determine the magnitude of the electrons relativistic 1 2 c momentum, where m = 9.111031 kg is the electrons mass. SOLUTION Substituting Equation 28.3 into Equation 9.8, we find that = = h h = p mv 2 1 v c2 = h v2 1 2 mv c 2 6.63 1034 J s ( 9.11 1031 kg ) ( 0.88) ( 3.00 108 m/s ) 0.88 c 12 1 = 1.3 10 m c 31. SSM REASONING The de Broglie wavelength is related to Plancks constant h and the magnitude p of the particles momentum. The magnitude of the momentum can be related to the particles kinetic energy. Thus, using the given wavelength and the fact that the kinetic energy doubles, we will be able to obtain the new wavelength. SOLUTION The de Broglie wavelength is = h p (29.8) The kinetic energy and the magnitude of the momentum are 1 2 KE = mv 2 (6.2) p = mv (7.2) Chapter 29 Problems 250 where m and v are the mass and speed of the particle. Substituting Equation 7.2 into Equation 6.2, we can relate the kinetic energy and momentum as follows: 1 2 KE = mv 2 = m2 v 2 p 2 = 2m 2m p = 2m ( KE ) or Substituting this result for p into Equation 29.8 gives = h h = p 2m ( KE ) Applying this expression for the final and initial wavelengths f and i, we obtain f = h 2m ( KE ) f h i = and 2m ( KE ) i Dividing the two equations and rearranging reveals that h f = i 2m ( KE ) f = h 2m ( KE ) i ( KE ) i ( KE ) f f = i or ( KE ) i ( KE ) f Using the given value for i and the fact that KE f = 2 ( KEi ) , we find f = i ( KE ) i ( KE ) f ( = 2.7 10 10 m ) ( KE i 2 KEi ) = 1.9 1010 m h 34 (Equation 29.8), where p is the magnitude of her momentum and h = 6.6310 Js is Plancks constant. We will use p = mv (Equation 7.2) to determine the magnitude p of the womans momentum from her mass m and her speed v at the instant she strikes the water. 32. REASONING The de Broglie wavelength of the woman is found from p = Once the woman jumps from the cliff, she is in free fall with an initial speed of v0 = 0 m/s, and an acceleration a = 9.8 m/s2. Since we have taken upward to be the positive direction, her displacement during the fall is H = 9.5 m. Her final speed v, then, is given by 251 PARTICLES AND WAVES 2 2 v = v0 + 2aH (2.9) h h (Equation 29.8) for yields = . Substituting p = mv p (Equation 7.2) into this result, we obtain SOLUTION Solving p = = h h = p mv (1) Substituting v0 = 0 m/s into Equation (2.9) and taking the square root of both sides, we find that v 2 = ( 0 m/s ) + 2aH 2 or v 2 = 2aH or v = 2aH (2) Substituting Equation (2) into Equation (1), we find the de Broglie wavelength of the woman at the instant she strikes the water to be: h h 6.63 1034 J s = = = = 1.2 1036 m 2 mv m 2aH ( 41 kg ) 2 ( 9.8 m/s ) ( 9.5 m ) 33. REASONING When the electron is at rest, it has electric potential energy, but no kinetic energy. The electric potential energy EPE is given by EPE = eV (Equation 19.3), where e is the magnitude of the charge on the electron and V is the potential difference. When the electron reaches its maximum speed, it has no potential energy, but its kinetic energy is 1 mv 2 . The conservation of energy states that the final total energy of the electron equals the 2 initial total energy: 1 mv 2 = eV 12 12 2 4 4 3 Initial 3 total Final total energy energy Solving this equation for the potential difference gives V = mv 2 / ( 2e ) . The speed of the electron can be expressed in terms of the magnitude p of its momentum by v = p/m (Equation 7.2). The magnitude of the electrons momentum is related to its de Broglie wavelength by p = h/ (Equation 29.8), where h is Plancks constant. Thus, the speed can be written as v = h/(m). Substituting this expression for v into V = mv 2 / ( 2e ) gives V = h 2 / ( 2me 2 ) . Chapter 29 Problems 252 SOLUTION The potential difference that accelerates the electron is ( 6.63 1034 J s ) h2 V= = = 1.86 104 V 2me 2 2 ( 9.11 1031 kg ) ( 1.60 1019 C ) ( 0.900 1011 m ) 2 2 34. REASONING The linear momentum p of a particle is given by p = mv (Equation 7.2), where m and v are its mass and velocity. Since particle A is initially at rest, its momentum is zero. The initial momentum of particle B is p0B = mBv0B. This is also the total initial linear momentum of the two-particle system. After the collision the combined mass of the two particles is mA+ mB, and the common velocity is vf .Thus, the total linear momentum of the system after the collision is pf = (mA+ mB)vf . From Section 7.2, we know that the total linear momentum of an isolated system is conserved. An isolated system is one in which the vector sum of the external forces acting on the system is zero. Since there are no external forces acting on the particles, the two-particle system is an isolated system. Thus, the total linear momentum of the system after the collision equals the total linear momentum before the collision. The de Broglie wavelength is inversely related to the magnitude p of a particles momentum by = h/p (Equation 29.8), where h is Plancks constant. SOLUTION The de Broglie wavelength f of the object that moves off after the collision is given by f = h/pf (Equation 29.8). Since momentum is mass times velocity, the magnitude of the momentum that the object has after the collision is pf = (mA + mB)vf, where vf is the common speed of the two particles. We can evaluate this momentum by using the law of conservation of momentum, which indicates that the total momentum after the collision is the same as it is before the collision. Before the collision only particle B is moving, so that the magnitude of the total momentum at that time has a value of mBv0B, where v0B is the initial speed of particle B. Assuming the particles travel along the + x axis, we write the conservation of linear momentum as follows: + ( mA + mB ) vf = +mBv0B 14243 1 4 4 2 4 43 Total momentum after collision Total momentum before collision Using this result, we find that the desired de Broglie wavelength is f = h h h = = pf ( mA + mB ) vf mBv0B 253 PARTICLES AND WAVES But the term on the far right is just the given de Broglie wavelength of the incident particle B. Therefore, we conclude that f = 2.0 1034 m . 35. SSM REASONING The de Broglie wavelength of the electron is related to the magnitude p of its momentum by = h/p (Equation 29.8), where h is Plancks constant. If the speed of the electron is much less than the speed of light, the magnitude of the electrons momentum is given by p = mv (Equation 7.2). Thus, the de Broglie wavelength can be written as = h/(mv). When the electron is at rest, it has electric potential energy, but no kinetic energy. The electric potential energy EPE is given by EPE = eV (Equation 19.3), where e is the magnitude of the charge on the electron and V is the potential difference. When the electron reaches its maximum speed, it has no potential energy, but its kinetic energy is 1 mv 2 . The 2 conservation of energy states that the final total energy of the electron equals the initial total energy: 1 mv 2 = eV 12 12 2 4 4 3 Initial 3 total Final total energy energy Solving this equation for the final speed gives v = 2eV / m . Substituting this expression for v into = h/(mv) gives = h / 2meV . SOLUTION After accelerating through the potential difference, the electron has a de Broglie wavelength of = h 2meV = 6.63 1034 J s 2 ( 9.11 1031 kg ) ( 1.60 1019 C ) ( 418 V ) = 6.01 1011 m 36. REASONING AND SOLUTION The energy of the photon is E = hf = hc/photon , while the kinetic energy of the particle is KE = (1/2)mv2 = h2/(2m 2). Equating the two energies and rearranging the result gives photon / = (2mc/h). Now the speed of the particle is v = 0.050c, so = h/(0.050 mc), and photon / = 2/0.050 = 4.0 10 1 37. REASONING AND SOLUTION According to the uncertainty principle, the minimum uncertainty in the momentum can be determined from p y y = h / ( 4 ) . Since p y = mv y , it Chapter 29 Problems 254 follows that p y = mv y . Thus, the minimum uncertainty in the velocity of the oxygen molecule is given by h 6.63 10 34 J s v y = = = 8.3 10 6 m/s 26 3 4 m y 4 5.3 10 kg 0.12 10 m ( )( ) 38. REASONING When particles pass through the slit the great majority fall on the screen between the first dark fringes on either side of the central bright fringe. The first dark fringes are located by the angles (below the midpoint) and + (above the midpoint), so this is the minimum range of angles over which the particles spread out. As Figure 29.15 shows, p y 1 the angle is found from = tan p (Equation 1.4), where py is the uncertainty in x the y component py of a particles momentum, and px is the x component of a particles momentum. The given de Broglie wavelength = 0.200 mm of the particles is the wavelength they possess before passing through the slit, when their momentum has only the x component px. Therefore, the x component of each particles momentum is given by px = h (Equation 29.8), where h = 6.631034 Js is Plancks constant. We are given that the uncertainty in the position of each particle along the y direction is equal to one-half the width of the slit: y = 1 W . The minimum uncertainty in the y 2 component py of a particles momentum, then, is found from the Heisenberg uncertainty principle: h p y ( y ) = (29.10) 4 ( ) SOLUTION Solving Equation 29.10 for py and substituting y = 1 W , we obtain 2 p y = Substituting Equation (1) and px = we find that h h h = = 1W 4 ( y ) 4 2 W 2 ( ) p y h 1 (Equation 29.8) into = tan p x (1) (Equation 1.6), 255 PARTICLES AND WAVES p y = tan 1 p x h 1 2 W = tan h = tan 1 2 W 633 109 m = tan 1 3 2 ( 0.200 10 m ) = 0.0289o Therefore, the particles spread out over the range 0.0289 to +0.0289 . 39. SSM WWW REASONING The uncertainty in the electrons position is y = 3.0 15 10 m. The minimum uncertainty py in the y component of the electrons momentum is given by the Heisenberg uncertainty principle as p y = h / ( 4 y ) (Equation 29.10). 15 SOLUTION Setting y = 3.0 10 m in the relation p y = h / ( 4 y ) gives h 6.63 1034 J s p y = = = 1.8 1020 kg m/s 15 4 y 4 ( 3.0 10 m) 40. REASONING Suppose the object is moving along the + y axis. The uncertainty in the objects position is y = 2.5 m. The minimum uncertainty py in the objects momentum is specified by the Heisenberg uncertainty principle (Equation 29.10) in the form ( py) ( y) = h/(4). Since momentum is mass m times velocity v, the uncertainty in the velocity v is related to the uncertainty in the momentum by v = ( py)/m. SOLUTION a. Using the uncertainty principle, we find the minimum uncertainty in the momentum as follows: h p y ( y ) = 4 ( ) py = h 6.63 1034 J s = = 2.11035 kg m/s 4 y 4 ( 2.5 m ) b. For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocity that is given by p y 2.1 1035 kg m/s v y = = = 4.7 1034 m/s m 0.045 kg Chapter 29 Problems 256 c. For an electron this uncertainty in momentum corresponds to an uncertainty in velocity that is given by p y 2.11035 kg m/s v y = = = 2.3 105 m/s m 9.111031 kg 41. SOLUTION The minimum uncertainty y in the position of the particle is related to the minimum uncertainty p y in the momentum via the Heisenberg uncertainty principle. To cast this relationship into a form that gives us the desired percentage for the minimum uncertainty in the speed, we note that the minimum uncertainty in the position is specified as the de Broglie wavelength . We can then express the de Broglie wavelength in terms of Plancks constant h and the magnitude py of the particles momentum. The magnitude of the momentum is related to the mass m and the speed vy of the particle. SOLUTION The percentage minimum uncertainty in the speed is Percentage = v y vy 100% (1) According to the Heisenberg uncertainty principle, the minimum uncertainty p y in the momentum and the minimum uncertainty y in the position of the particle are related according to h p y ( y ) = (29.10) 4 ( ) We know that y is equal to the de Broglie wavelength = h / p y (Equation 29.8), where the magnitude of the momentum is p y = mv y (Equation 7.2). Thus, we have y = = h h = p y mv y Substituting this result for y into Equation 29.10, we obtain h ( p y ) ( y ) = ( p y ) mv y h = 4 (2) The last step in our transformation of the uncertainty principle is to realize that p y = mv y = mv y , since the mass is constant. Substituting this expression for p y ( ) into Equation (2) shows that 257 PARTICLES AND WAVES ( h py mv y ) = mv y ( h mv y ) h = 4 v y or vy = 1 4 Using this result in Equation (1), we find that Percentage = v y vy 100% = 1 100% = 8.0% 4 42. REASONING The mass m of the particle is related to its rest energy E0 by E0 = mc 2 (Equation 28.5). Therefore, if there is a minimum uncertainty E0 in measuring the rest energy of the particle, there will be a corresponding uncertainty m in measuring its mass: E0 = ( m ) c 2 m = or E0 (1) c2 The minimum uncertainty E0 in the particles rest energy is related to the length of time t the particle exists in a state by the Heisenberg uncertainty principle: ( E0 ) ( t ) = 4h (Equation 29.11), where h = 6.631034 Js is Plancks constant. SOLUTION Solving Equation 29.11 for the minimum uncertainty E0, we obtain E0 = h 4 ( t ) (2) Substituting Equation (2) into Equation (1), we find that m = E0 c 2 h 6.63 1034 J s = = 2 4 ( t ) c 4 7.4 1020 s 3.00 108 m/s ( )( ) 2 = 7.9 1033 kg Chapter 29 Problems 258 43. REASONING In order for the person to diffract to the same extent as the sound wave, the de Broglie wavelength of the person must be equal to the wavelength of the sound wave. SOLUTION a. Since the wavelengths are equal, we have that sound = person sound = h mperson v person Solving for v person , and using the relation sound = v sound / f sound (Equation 16.1), we have v person = = h mperson ( v sound / f sound ) = h f sound mperson v sound ( 6.63 10 34 J s)(128 Hz ) = ( 55.0 kg)( 343 m / s) 4 .50 10 36 m / s b. At the speed calculated in part (a), the time required for the person to move a distance of one meter is t= x 1.0 m = v 4.50 10 36 m / s F1.0 h I F1 day I F 1 year I = G s J G h J G days J H KH KH K 3600 24.0 365.25 7.05 10 27 years Factors to convert seconds to years 44. REASONING The energy of a photon is related to its frequency and Plancks constant. The frequency, in turn, is related to the speed and wavelength of the light. Thus, we can relate the energy to the wavelength. The given relationship between the wavelengths will then allow us to determine the unknown energy. SOLUTION The energy E of a photon with frequency f is E = hf (29.2) where h is Plancks constant. The frequency is related to the speed c and wavelength of the light according to c f= (16.1) Substituting this expression for f into Equation 29.2 gives 259 PARTICLES AND WAVES E = hf = h c Applying this result to both sources, we have EB = h c B EA = h and c A Dividing the two expressions gives c EB B A = = EA h c B A h Using the given value for EA and the fact that B = 3A in this result shows that EB = EA A = 2.1 1018 J A = 7.0 10 19 J 3 B A ( ) 45. REASONING The de Broglie wavelength is related to Plancks constant h and the magnitude p of the particles momentum. The magnitude of the momentum is related to the mass m and the speed v at which the bacterium is moving. Since the mass and the speed are given, we can calculate the wavelength directly. SOLUTION The de Broglie wavelength is h p = (29.8) The magnitude of the momentum is p = mv (Equation 7.2), which we can substitute into Equation 29.8 to show that the de Broglie wavelength of the bacterium is = h h 6.63 1034 J s = = = 11018 m 15 p mv 2 10 kg ( 0.33 m/s ) ( ) Chapter 29 Problems 260 46. REASONING AND SOLUTION a. We know E = hc/ for a photon. The energy of the photon is F 10 J I = 8.0 10 1.60 E = 5.0 eV G H 1 eV J K 19 19 J The wavelength is c h c 6.63 10 34 J s 3.00 10 8 m / s hc = = E 8.0 10 19 J h= 2 .5 10 7 m b. The speed of the 5.0-eV electron is v= 2E = m c 2 8.0 10 19 J 9.11 10 31 h= 1.3 10 kg 6 m/s The de Broglie wavelength is = 6.63 10 34 J s h = = 5.6 10 10 m 31 6 mv 9.11 10 kg 1.3 10 m / s c c h h 47. REASONING AND SOLUTION In the first case, the energy of the incident photon is given by Equation 29.3 as hf = KE max + W0 = 0.68 eV + 2.75 eV = 3.43 eV In the second case, a rearrangement of Equation 29.3 yields KE max = hf W0 = 3.43 eV 2.17 eV = 1.26 eV 48. REASONING The speed v of a particle is related to the magnitude p of its momentum by v = p/m (Equation 7.2). The magnitude of the momentum is related to the particles de Broglie wavelength by p = h/ (Equation 29.8), where h is Plancks constant. Thus, the speed of a particle can be expressed as v = h/(m) . We will use this relation to find the speed of the proton. SOLUTION The speeds of the proton and electron are vproton = h mproton proton and velectron = h melectron electron 261 PARTICLES AND WAVES Dividing the first equation by the second equation, and noting that electron= proton, we obtain vproton m m = electron electron = electron velectron mproton proton mproton Using values for melectron and mproton taken from the inside of the front cover, we find that the speed of the proton is m vproton = velectron electron mproton 9.111031 kg ( 6 3 = 2.45 10 m/s = 4.50 10 m/s ) 27 kg 1.67 10 49. REASONING The width of the central bright fringe in the diffraction patterns will be identical when the electrons have the same de Broglie wavelength as the wavelength of the photons in the red light. The de Broglie wavelength of one electron in the beam is given by Equation 29.8, electron = h / p , where p = mv . SOLUTION Following the reasoning described above, we find red light = electron red light = h melectron v electron Solving for the speed of the electron, we have v electron = h 6.63 10 34 J s = = 1.10 10 3 m / s melectron red light (9.11 10 31 kg)(661 10 9 m) 50. REASONING We will first calculate the potential energy of the system at each of the two separations, and then find the energy difference for the two configurations. Since the electric potential energy lost by the system is carried off by a photon that is emitted during the process, the energy difference must be equal to the energy of the photon. The wavelength of the photon can then by found using Equation 29.2 with Equation 16.1: E = hc / . Chapter 29 Problems 262 SOLUTION The initial potential energy of the system is (see Equations 19.3 and 19.6) FI kq GJ r HK L 10 (8.99 = ( 1.6 10 C) M N EPE 1 = eV1 = e 1 19 9 O P Q N m 2 / C 2 )( 8.30 10 6 C) = 2 .84 10 14 J 0.420 m The final potential energy is EPE 2 = eV2 = ( 1.6 10 19 C) L 10 (8.99 M N O P Q N m 2 / C 2 )( 8.30 10 6 C) = 7 .56 10 15 J 1.58 m 9 The energy difference, and therefore the energy of the emitted photon, is E = EPE 1 EPE 2 = 2 .84 10 14 J 7 .56 10 15 J = 2 .08 10 14 J The wavelength of this photon is = hc (6.63 10 34 J s)(3.00 10 8 m / s) = = 9.56 10 12 m E 2.08 10 14 J 51. REASONING Since the net external force acting on the system (the photon and the electron) is zero, the conservation of linear momentum applies. In addition, there are no nonconservative forces, so the conservation of total energy applies as well. Since the photon scatters at an angle of = 180.0 in Figure 29.10, the collision is "head-on." Thus, the motion takes place entirely along the horizontal direction, which we will take as the x axis, with the right as being the positive direction. The conservation of linear momentum gives rise to Equation 29.7, which relates the difference between the scattered and incident X-ray photon wavelengths to the scattering angle of the electron as = b g b g h h 2h 1 cos = 1 cos 180.0 = mc mc mc (1) The conservation of total energy is written as hc 4 Energy of incident photon + 0= { Initial kinetic energy of electron hc 4 Energy of scattered photon + 1 mv 2 2 Final kinetic energy of electron (2) 263 PARTICLES AND WAVES Equations (1) and (2) will permit us to find the wavelength of the incident X-ray photon. SOLUTION Solving Equation (1) for and substituting the result into Equation (2) gives hc = hc + 2h + mc 1 2 mv 2 Algebraically rearranging this result, we obtain a quadratic equation for : 2 + 2 Fh I GJ H K mc 4 .85 10 12 m 2h 2 c h=0 m 1 mv 2 2 9 . 70 10 20 m 2 34 31 8 where we have used h = 6.63 10 J s, m = 9.11 10 kg, c = 3.00 10 m/s, and v = 4.67 106 m/s. Solving this quadratic equation for , we obtain = 3.09 10 10 m
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Rutgers - PHYSICS - 201
CHAPTER 30 THE NATURE OF THE ATOMANSWERS TO FOCUS ON CONCEPTS QUESTIONS1. (d) The Bohr model deals with a single negatively charged electron in orbit about a positivelycharged nucleus. Since only one electron is assumed to be present, the model does no
Rutgers - PHYSICS - 201
CHAPTER 31 NUCLEAR PHYSICSAND RADIOACTIVITYANSWERS TO FOCUS ON CONCEPTS QUESTIONS1.A(e) In the notation Z In, the symbol Z represents the number of protons, and the symbol Arepresents the number of protons plus the number of neutrons.2.(c) Because
Rutgers - PHYSICS - 201
CHAPTER 32 IONIZING RADIATION,NUCLEAR ENERGY, ANDELEMENTARY PARTICLESANSWERS TO FOCUS ON CONCEPTS QUESTIONS1. (c) The biologically equivalent dose (in rems) is the absorbed dose (in rads) times the relativebiological effectiveness. (See Equation 32.4
Istanbul Technical University - ENG - 102
Same sex marriage has been one of the most controversial issue for many years. Weknow that same sex marriage is marriage between two people who have the same sex. Lots ofpeople who are homosexuals(gays, lesbians) are aware that they are not taken seriou
East West University - FINANCE - 302
CHAPTER 3INTEGRATIVE PROBLEM3-33 ASSUME THAT YOU ARE NEARING GRADUATION AND THAT YOU HAVEAPPLIED FOR A JOB WITH A LOCAL BANK. AS PART OF THE BANK'SEVALUATION PROCESS, YOU HAVE BEEN ASKED TO TAKE ANEXAMINATION THAT COVERS SEVERAL FINANCIAL ANALYSISTE
East West University - FINANCE - 302
Case title: Greentree Investment Corporation (GIC)Assignment Questions:1. Perform stakeholder analysis: Where does the relationship exist? Betweenclient and advisor, or client and firm? What is critical in the relationship?The evolution of the sophist
Waterloo - MATH - 118
Monday, March 19 Lecture 30 : Examples on determining the function whichgenerates a series. (Refers to Section 8.8 in your text)After having practiced the problems associated to the concepts of this lecture the student should beable to: Find the functi
Waterloo - MATH - 118
Tuesday, March 20 Lecture 31 : Applications : Approximating the value of a series or afunction. (Refers to Section 8.9 in your text)After having practiced the problems associated to the concepts of this lecture the student should be able to: Use aTaylo
Waterloo - MATH - 118
Wednesday, March 21 Lecture 32 : Applications : Evaluation of definite integrals andlimits. (Refers to Section 8.9 in your text)After having practiced the problems associated to the concepts of this lecture the student should be able to:Use a Taylor se
Waterloo - MATH - 118
Monday, March 26 Lecture 33 : Parametric Equations, curves and tangents (Refersto Section 9.1, 9.2)After having practiced the problems associated to the concepts of this lecture the studentshould be able to : Plot simple parametric equations in the pla
Waterloo - MATH - 118
Tuesday, March 27 Lecture 34 : Polar Coordinates and polar equations (Refers toSection 9.3)After having practiced the problems associated to the concepts of this lecture the student shouldbe able to : Express polar coordinates in Cartesian coordinates
Waterloo - MATH - 118
Wednesday, March 28 Lecture 35 : Graphing polar equations - Tangents to polarcurves (Refers to Section 9.3)After having practiced the problems associated to the concepts of this lecture the student should be able to :Graph a polar equation r = f( ) by
Waterloo - MATH - 118
Monday, April 2 Lecture 36 : Lengths of polar curves and areas of polar regions.(Refers to Section 9.3, 9.4)After having practiced the problems associated to the concepts of this lecture the student shouldbe able to : Find the length of a polar curve w
Waterloo - MATH - 118
Tuesday, January 3 Lecture 1: Integration by substitution (Refers to 6.1 in yourtext)After having practiced using the concepts of this lecture the student should be ableto: define the differential of a function, integrate indefinite integrals by making
Waterloo - MATH - 118
Monday, January 9 Lecture 3 : Trigonometric integrals (Refers to Section 6.3 inyour text)After having practiced the problems associated to the concepts of this lecture the studentshould be able to: Apply some of the common trigonometric identities to t
Waterloo - MATH - 118
Tuesday, January 10 Lecture 4 : More Integration Methods : TrigonometricSubstitution (Refers to Section 6.4 in your text)After having practiced the problems associated to the concepts of this lecture the studentshould be able to: Solve integrals of fun
Waterloo - MATH - 118
Wednesday, January 10 Lecture 5 : Still More Integration Methods : Completingthe square and rational functions. (Not explicitly in your text)After having practiced the problems associated to the concepts of this lecture the studentshould be able to: So
Waterloo - MATH - 118
Wednesday, January 11 Lecture 6 : Rational functions : Partial fractions (Refersto Section 6.5 of your text)After having practiced the problems associated to the concepts of this lecture the studentshould be able to: Solve integrals containing rational
Waterloo - MATH - 118
Monday, January 16 Lecture 7 : Improper integrals. (Refers to section 6.6 in yourtext.)After having practiced the problems associated to the concepts of this lecture the studentshould be able to: Recognize the two different types of improper integrals,
Waterloo - MATH - 118
Wednesday, January 18 Lecture 9 : Error estimation for numerical integration.(Refers to section 5.3)After having practiced the problems associated to the concepts of this lecture the student shouldbe able to: Find error's bounds when applying either th
Waterloo - MATH - 118
Wednesday, January 25 Lecture 12 : Second-order Differential equations. (Notexplicitly discussed in your text)After having practiced the problems associated to the concepts of this lecture the student shouldbe able to: Solve second-order differential e
Waterloo - MATH - 118
Thursday, February 10 Lecture 20 : Alternating series (Refers to Section 8.2 inyour text)After having practiced the problems associated to the concepts of this lecture the student shouldbe able to: Define an alternating series, state the alternating se
Waterloo - MATH - 118
Monday, February 27 Lecture 21 : Absolute convergence and conditionalconvergence. (Refers to Section 8.5 in your text)After having practiced the problems associated to the concepts of this lecture the studentshould be able to: Define &quot;converges absolut
Waterloo - MATH - 118
Tuesday February 28 Lecture 22 : Ratio test and Root test. (Refers to Section 8.5 inyour text)After having practiced the problems associated to the concepts of this lecture the student should beable to: Apply the ratio test and the root test to determi
Waterloo - MATH - 118
Wednesday, February 29 Lecture 23 : Error bounds for series approximations.(Refers to Section 8.2 and 8.3 in your text)After having practiced the problems associated to the concepts of this lecture the student should beable to: Approximate the value of
Waterloo - MATH - 118
Monday, March 5 Lecture 24 : Power Series and their interval of convergence.(Refers to Section 8.6 in your text)After having practiced the problems associated to the concepts of this lecture the studentshould be able to: Define a power series centered
Waterloo - MATH - 118
Tuesday, March 6 Lecture 25 : Expressing a power series as a function andoperations on power series. (Refers to Section 8.7 )After having practiced the problems associated to the concepts of this lecture the student should beable to: Determine the sum,
Waterloo - MATH - 118
Wednesday, March 7 Lecture 26 : Expressing functions as a power series:Derivatives and Integrals of power series. (Refers to Sections 8.6 and 8.7 )After having practiced the problems associated to the concepts of this lecture the student should beable
Waterloo - MATH - 118
Monday, March 12 Lecture 27 : Taylor series and Maclaurin series generated by afunction f(x). (Refers to Section 8.8 your text)After having practiced the problems associated to the concepts of this lecture the student should be able to:Define a Taylor
Waterloo - MATH - 118
Tuesday, March 13 Lecture 28 : Taylor's remainder theorem: Convergence of aseries to its generator (Refers to Section 8.8 in your text)After having practiced the problems associated to the concepts of this lecture the student should be able to:Define t
Waterloo - MATH - 118
Wednesday, March 14 Lecture 29 : Binomial series (Refers to Section 8.8 in yourtext)After having practiced the problems associated to the concepts of this lecture the student should be able to:State and apply the Binomial series theorem.29.1 Introduct
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Waterloo - PHYS - 125
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Waterloo - PHYS - 125
Waterloo - PHYS - 125
Waterloo - PHYS - 125
Waterloo - PHYS - 125
Waterloo - PHYS - 125
Waterloo - PHYS - 125
Waterloo - PHYS - 125
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Waterloo - PHYS - 125
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Waterloo - PHYS - 125
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Waterloo - PHYS - 125
Waterloo - PHYS - 125
UDiversity of WaterlooFinal ExaminationUnJ~tyoiWaterloo,Winter TERM2002(TemI)(Year)iNSTRUcnONS:PART A: 6 multiple-choice questionsword1S InarkBeachI OUESTION GRADEa) ANSWER CARDS Wll.L BE COLLBCI'ED AT 11:30 a.m.,b) Mark. ourcardin darkpenci
Waterloo - PHYS - 125
University of WaterlooFinal ExaminationUniversity ofWaterloo,Sprine TERM 2004(Tenn)(Year)INSTRUCnONS:PART A: 6 multiple-choicequestionsworth 5 markseacha) ANSWER CARDS WILL BE COLLECTED AT 11:30 a.m.b) Mark your card in dark pencil. Avoid erasu
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UNlVERSffY OF WATERLOOK-SECfIONDEPARTMENT OF PHYSICSNAME (print)Physics 125(SectionsIBI, IBK, IBM, IBQ)I.D.#- Physicsfor Engineers-SIGNATUREFINAL EXAMINA nONWinter Tenn 1999April 12. 19999:00 am- 12:00pmINSTRUCTOR (Cross your Prof!)DLeslieCo
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PRIVACY POLICY &amp; MARKED WORKDue to recent changes in the Universitys Privacy Policy, the students name, ID andmark, should never be left together out in public.As a result, students are required to have a cover sheet on their assignment uponsubmission
Waterloo - PHYS - 125
Waterloo - PHYS - 125
Waterloo - PHYS - 125