Solution Manual for Modern Physics for Scientists and Engineers, 4th Edition

28 Chapter 3 The Experimental Basis of Quantum Theory Chapter 3 1. The required field is homogeneous within the desired region and decreases in magnitude to zero as rapidly as possible outside that region. The magnitude of the field is B  E / v0 . The best design is an electromagnet with flat, parallel pole faces that are large compared with the distance between them. But no matter what the design, it is impossible to eliminate edge effects. 2. eE  evB B E (2.5 105 V/m)   0.11T v (2.2 106 m/s) 3. Assume the speed is exact. Non-relativistically use the energy obtained by accelerating through a potential difference causes an increase in kinetic energy: eV    12 mv 2  31 7 mv 2  9.1094 10 kg 1.80 10 m/s  V   921.06 V 2e 2 1.6022 1019 C  2 Relativistically eV  K  (  1)mc2 :        511keV/c 2  2 1 V   1    c  923.59 V . 2 e 7      1.80 10 m/s  1      8  2.9979 10 m/s    The results differ by about 2.5 volts, or about 0.27%. Relativity is required only if that level of precision is needed. 4. eE  evB so E  vB   4.0 106 m/s 1.2 102 T   4.8 104 V/m © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 3 The Experimental Basis of Quantum Physics 2 29 2 1 2 1  F    1  eE    eE 2 y  at            2 2  m   v0  2  m   v0  2mv02 1.602 10 C  4.8 10 V/m   0.02 m   1.0552 10 m  10.6 cm  2  9.109 10 kg  4.0 10 m/s  19 2 4 1 31 2 6 5. The forces are shown below. 6. At terminal velocity the net force is zero, so Ff  fvt  mg and vt  mg / f . 7. vt  mg mg  f 6  r 4 m    volume    r 3 3 2  vt 4   g  2g r vt    r 3   Solving for r: r  3  9 2g  3   6 r  1.82 10 kg·m  s 1.3 10 m/s   3.47μm  vt 3 8. (a) r  3 2g  2  9.80 m/s 2  900 kg/m3  5 1 1 3 3 4 4 (b) m  V   r 3    900 kg/m3  3.47 106 m   1.58 1013 kg 3 3 13 2 mg 1.58 10 kg  9.80 m/s    1.19 109 kg/s (c) f  vt 1.3 103 m/s 1 1   1  9. Lyman:    RH 1  2    RH1  1.096776 107 m1   91.2 nm     1 1   1 1  Balmer:    RH  2  2    4 RH1  4 1.096776 107 m1   364.7 nm   2   © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 30 Chapter 3  10. d  400 mm1 The Experimental Basis of Quantum Theory   2.5μm  2500 nm and   d sin  in first order. Also tan   y / x so 1 y  x tan    3.0 m  tan sin 1 ( / d )  .   656.5nm   Red: y   3.0 m  tan sin 1     81.6 cm  2500 nm      486.3nm   Blue-green: y   3.0 m  tan sin 1     59.5 cm  2500 nm      434.2nm   Violet: y   3.0 m  tan sin 1     52.9 cm  2500 nm     11. d  420 mm1   2.381 μm ; 1   656.5nm for red. We know n  d sin  so  n   . For n = 1 (first order) we find  d    656.5 nm   y  x tan    2.8 m  tan sin 1     80.3 cm .  2381 nm    Similarly for n = 2 we find y =185.1 cm, and for n = 3 we find y = 412.2 cm.   sin 1  Therefore the separations are: between n =1 and n = 2, y  185.1cm  80.3cm  104.8cm ; between n = 2 and n = 3, y  412.2cm  185.1cm  227.1cm . 12. Use Equation (3.13) with n = 4 for the Brackett series and n = 5 for the Pfund series. The largest wavelengths occur for the smallest values of k. 1 1   1 1    1 1  Brackett:    RH  2  2    1.096776 107 m-1   2  2   .  4 k    4 k   For k = 5,   4.052 μm ; for k = 6,   2.626 μm ; for k =7,   2.166 μm ; for k = 8,   1.945 μm . 1 1   1 1    1 1  Pfund:    RH  2  2    1.096776 107 m-1   2  2   .  5 k    5 k   For k = 6,   7.460 μm ; for k = 7,   4.654 μm ; for k = 8,   3.741 μm ; for k = 9,   3.297 μm . © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 3 The Experimental Basis of Quantum Physics 31 13. Beginning with Equation (3.10) with n = 1, we have   d sin  with d  0.20mm . d  d cos  . Assuming the spectrum was viewed in the forward direction, Therefore d d     d . An angle of 0.50 minutes of arc corresponds then   0 and cos  1 so d  to 1.45 104 rad so   d    0.20 103 m 1.45 104 rad  29 nm.  1 1   RH  2  2  with RH  1.096776 107 m1 and n = 2 for  n k  1  1 1 the Balmer series. With k = 3 we find  1.096776 107 m1  2  2   656.47nm.  2 3  This wavelength, found with the smallest value of k, is referred to as the hydrogen alpha or H line. Using the equation above with k = 4 we find the hydrogen beta or H  14. (a) Use Equation (3.13) 1 wavelength equals 486.27 nm. Similarly with k = 5 the hydrogen gamma or H wavelength equals 434.17 nm, and finally with k = 6 the hydrogen delta or H wavelength equals 410.29 nm. (b) Because only three wavelengths are observed, the source must moving relative to the detector. The wavelengths have increased, so the source must be moving away from the detector. [See Equation (2.34) for example.] The H line at 656.47 nm is redshifted out of the visible. (c) Because the wavelengths are known and c   f , we will use the reciprocal of Equation (2.33). We select this equation since the wavelengths are larger and thus the  1  source is receding from the observer. observed  . Using algebra to solve the source 1  equation for  and substituting the smallest wavelength gives  observed   453.4nm     1   1  source   410.29nm     0.0995 . 2 2  observed   453.4nm      1  410.29nm   1    source  2 2 Therefore the speed is v  0.10c or v  3.0 107 m/s . Using other pairs of wavelengths gives a similar result. If the object is rotating, then one side would be moving toward and another side away from the detector so a range of wavelengths would be observed. This effect can be used to determine the rotation speeds. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 32 Chapter 3 The Experimental Basis of Quantum Theory  1 1   RH  2  2  with RH  1.096776 107 m1 and n = 3 for the  n k  1 1 1 Paschen series. With k = 4 we find  1.096776 107 m1  2  2 ;   1875.63 nm .  3 4  With k = 5,   1282.17 nm ; with k = 6,   1094.12 nm ; with k = 7,   1005.22 nm; with k = 8,   954.86 nm. 15. (a) Use Equation (3.13) 1 (b) The observed spectral lines have been Doppler shifted. It might appear as if the wavelengths have been blueshifted since the largest observed wavelength is smaller than the largest expected wavelength using the Paschen series. However it is more likely that the wavelengths have been redshifted and the calculated wavelength just below 1000 nm in part (a) corresponds to the observed wavelength of 1046.1 nm. (c) Using the formula from Problem 14 and noting from part (b) that the star is receding 2  observed   1334.5nm     1   1 1282.17nm    source    0.040 or v  1.20 107 m/s . from the detector   2 2  observed   1334.5nm      1  1282.17nm   1    source  2 16. (a) To obtain a charge of +1 with three quarks requires two charges of +2e/3 and one of charge  e / 3 . Three quarks with charge + e/3 would violate the Pauli Exclusion Principle for spin 1/2 particles. (b) To obtain a charge of zero we could have either two e / 3 and one 2e / 3 or one 2e / 3 and two e / 3 . At this point in the text there is no reason to prefer either choice. (The latter turns out to be correct.) 2.898 103 m  K  0.69 mm 17. (a) max  4.2 K 2.898 103 m  K  9.89 μm (b) max  293 K 2.898 103 m  K  1.16 μm (c) max  2500 K 2.898 103 m  K  0.322 μm (d) max  9000 K 2.898 103 m  K  1.932 1011 K 18. (a) T  14 1.50 10 m 2.898 103 m  K  1.932 106 K (b) T  1.50 109 m 2.898 103 m  K  4528 K (c) T  640 109 m © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 3 The Experimental Basis of Quantum Physics 33 2.898 103 m  K  2.898 103 K 1m 2.898 103 m  K  1.42 105 K (e) T  204 m 4 P1  T14 T14  2300 K   19. (a) ; so P1  P0 4    P0  42.7 P0 ; The power increases by a factor P0  T04 T0  900 K  of 42.7. (d) T  (b) To double the power output, the ratio of temperatures to the fourth power must equal 4  T  2.  1   2 . Solving we find T1  1070 K .  900 K  2.898 103 m·K  9.35 μm 310K (b) At this temperature the power per unit area is 20. (a) max  R   T 4   5.67 108 W·m2  K 4   310K   524W/m2 . The total surface area of a 4 cylinder is 2 r  r  h   2  0.13 m 1.75 m  0.13 m   1.54 m2 so the total power is P   524 W/m2 1.54 m2   807 W. (c) The total energy radiated in one day is the power multiplied by the time; E  P  t  807 W  86400 s   6.97 107 J. 2000 kcal   2 106 cal    4.186 J/cal   8.37 106 J . There are several assumptions. First, a cylinder may overestimate the total surface area; second, radiation is minimized by hair covering and clothing. 21. max  2.898 103 m  K  1.447 108 m 200, 273 K 22. We know from Example 3.8 that in the long-wavelength limit, the two expressions are the same. By comparing the expressions, we can also note that the Rayleigh-Jeans spectral distribution will be larger than the Planck expression for a given  and T. [Compare graphs or calculations using Equations (3.22) and (3.23) one can obtain using Excel, Mathcad, or similar program.] So we want   2 c 2 h   2 ckT 1  . Simplifying we find  0.95  4    5   hc / kT  1         e  kT  kT  1 1  x and this simplifies to  0.95 x  1/ x . Let . We  0.95    hc /  kT hc e  1  1  hc  e © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 34 Chapter 3 The Experimental Basis of Quantum Theory can solve this transcendental equation to find x = 9.83 so 9.83   kT hc . Substituting numerical values for the constants we find 9.83 1.986 1025 J  m  9.83hc    2.44 105 m  24.4 μm. 23 1 kT 1.38 10 J  K  5800K  2.898 103 m  K  966 nm which is in the near infrared. 3000 K 24. The graph is a characteristic Planck law curve with a maximum at   966 nm (see Problem 23). (a) Numerical integration of the ( , T ) function shows that approximately 8.1% of the radiated power is between 400 nm and 700 nm. Details of the calculation are: 710 710  4.798  106  5  hc  5 16 2 c 2 h  exp    d   3.74  10 exp    410      d  That is 410   kT  23. max  7 7 7 7  3.71105 W/m2 the power per unit area emitted over visible wavelengths. Over all wavelengths we know the power per unit area is R   T 4  4.59 106 W/m2 . Therefore the fraction 3.7128 105 W/m2  0.081 . emitted in the visible is 4.593 106 W/m2 (400 nm, T )  0.073 ; (966nm, T) (b) Using computed numerical intensity values (700 nm, T )  0.754 . (966nm, T) 2 c 2 h  hc  exp    . The exponential goes    kT  to zero faster than  5 , so the intensity approaches zero in this limit. 25. In this limit exp  hc /  kT   1 , so ( , T )  5 26. (a) At this temperature the power per unit area is R   T 4   5.67 108 W  m2  K 4   293 K   419 W/m2 . For the basketball, a sphere 4 of radius r = 12.5 cm, we obtain P  R  4 r 2    419 W/m2   4  0.125 m   82.3 W . 2 (b) At this temperature the power per unit area is R   T 4   5.67 108 W  m2  K 4   310 K   524 W/m2 . Assume the body is roughly 4 cylindrical with a radius of 13 cm and a height of 1.65 m. The total surface area of a cylinder is 2 r  r  h   2  0.13 m 1.78 m   1.45 m2 so the total power is P   524 W/m2 1.45 m2   760 W. Numerical values will vary depending on estimates of the human body size. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 3 The Experimental Basis of Quantum Physics 35 2.898 103 m  K  9.35 μm 310 K 2  2  2 a n 2 2  n x     n x  28. Taking derivatives ;  sin   a sin     . Substituting these 2 2 2 2 t t L  L  x  L  1  2 a  n x   n2 2  n x   values into the wave equation produces 2 2 sin     a 2 sin    0 c t L  L    L  27. max  n c 2L 2a n2 2c 2   a   2 a where   . Since   for this 2 2 L n t L system and c   f , then   2 f . which simplifies to 29. Let r  nx2  ny2  nz2 be the radius of a three-dimensional number space with the ni the three components of that space. Then let dN be the number of allowed states between r and r  dr . This corresponds to the number of points in a spherical shell of number 1 space, which is dN   4 r 2 dr where we have used the fact that 4 r 2 dr is the 8 “volume” of the shell (area 4 r 2 by thickness dr ), and the 1/8 is due to the fact that only positive numbers ni are allowed, so only 1/8 of the space is available. Also r 2  nx2  ny2  nz2  2Lf 2 L2 4 L2 f 2  2L   or r  . Then from this dr    df . Putting 2 2 2 c  c c  c  1    2 Lf  2 L 4 L3 2 everything together: dN   4 r 2 dr  r 2 dr   df  f df  8 2 2 c  c c3 2 30. From the diagram at right we see that the average x-component of the velocity c of  electromagnetic radiation within the cavity is  /2   ccos 2 r sin  d . c    2 r sin  d 2 0 x /2 2 0 1 Letting u  cos  we have cx  c  udu 0 1  du  c . 2 0 On average only one-half of the photons are traveling to the right. Thus the mean velocity of photons traveling to the right is c/4. c Therefore power =  intensity  area   dU  A . 4 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 36 Chapter 3 The Experimental Basis of Quantum Theory 31. For classical oscillators the Maxwell-Boltzmann distribution gives n( E )  A exp   E / kT   A exp   E  where E  nhf and   1/ kT . The mean energy  is E   En( E )   nhf exp   nhf  n 0   n 0  n 0 n 0  n( E )  exp   nhf  . Notice that E    ln  exp    nhf  . Now  n0 letting x  exp   hf  we see that by Taylor series   exp   nhf   1  x  x  …  1  x  for x  1. 1 2 n 0 E 2 hf exp    hf    1 ln 1  x    ln 1  x     1  exp    hf  hf . Using the result of Problem 29 (along with a factor of 2 for two exp  hf / kT   1 photon polarizations) we can see that 4 2 hf 8 hf 3 / c3 U ( f ,T )  2 3 f  . c exp  hf / kT   1 exp  hf / kT   1 To change from U to requires the factor c/4 (Problem 30), and changing from a frequency distribution requires a factor c /  2 , because with f  c /  we have E df   c /  2 d  . Putting these together ( , T )  2 8 h /  3 1  c  c  2 c h  .  2   5 exp  hf / kT   1    4   exp  hc /  kT   1 32. Energy per photon  hf   6.626 1034 J  s  98.1106 s1   6.50 1026 J 1 photon  7.69 10 photons/s 5.0 10 J/s  6.50 10 J 4 29 26 33. (a) Energy per photon  hf   6.626 1034 J  s 1100 103 s1   7.29 1028 J 180 J/s  1 photon  2.47 1029 photons/s 7.29 1028 J (b) energy per photon  h  3.00 108 m/s  17   6.626 1034 J  s     2.48 10 J 9  8  10 m   c 1 photon  7.26 1018 photons/s 17 2.48 10 J 1 photon 1 MeV  2.811014 photons/s (c) 180 J/s  13 4 MeV 1.60 10 J 180 J/s  © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 3 The Experimental Basis of Quantum Physics  37 2.93 eV hc 1  hc   7.08 1014 Hz : eV0    so V0 =     ; 15 h 4.136 10 eV  s  e  1 1240 eV  nm  V0    2.93 eV   0.333 V e  380 nm  34. ft  35. t   hc   1240 eV  nm  267.2 nm . 4.64 eV If the wavelength is halved (to   133.6 nm), then K hc    36. Notice that 1240 eV  nm  4.64 eV  4.64 eV 133.6 nm hc   1240eV  nm  2.33eV   , so photoelectrons will be produced. 532 nm 19  105 electrons   1 photon   1 eV   1.60 10 C  3 2  10 J/s    1 photon   2.33eV  1.60 1019 J   electron   8.58 nA      hc 1240 eV  nm 37.     4.59 eV ; K  2.0 eV  hf   ; t 270 nm K  2.0 eV  4.59eV f    1.59 1015 Hz 15 h 4.136 10 eV  s 1240 eV  nm  hc  38. E  100    100  203 eV 610 nm   39. eV01  hc / 1   and eV02  hc / 2   . Subtracting these equations and rearranging we find h e V02  V01  1 1 c    2 1   e  2.3 V  1.0 V  3.00 10 m/s   2071nm  2601nm   4.40 1015 eV  s . 8 This is about 6% from the accepted value. For the work function we use the first set of data (the second set should give the same result):  4.40 1015 eV  s 3.00 108 m/s   1.0 eV  4.1 eV hc    eV01  1 260 109 m © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 38 Chapter 3 40. hf    K or hc  The Experimental Basis of Quantum Theory 2    K    12 mvmax ; Using 4.63 eV as the work function for tungsten (see table 3.3) and 9.111031 kg as the mass of the electron, one can solve for  .     hc 1240 eV  nm    2   12 mvmax   4.63eV  12  9.111031 kg 1.4 106 m/s 2 1eV19  1.6 10 J   2  1.2110 nm  121nm 1.24 106 V  m 1.24 106 V  m   0.0413 nm 41. min  V 30 kV hc 1240 eV  nm   2.48 1017 m. A photon produced by bremsstrahlung is still an K 5 1010 eV x ray, even though this falls outside the normal range for x rays. 42.   43.   1240 eV  nm  0.0496 nm 2.5 104 eV 44. Because the electron is accelerated through a potential difference, it has kinetic energy equal to K  eV0 . As described in the text, Equation (3.37) gives the minimum wavelength (assuming the work function is small). So 6  hc   1  1.240 10 V·m min        3.54 1011 m  3.54 102 nm . 4 e V 3.5  10 V   0  The work function for tungsten from Table 3.3 is 4.63 eV. If we include the work function, the energy available for the photon and the wavelength of the photon will change by 4.63 parts out of 35000 or about 0.013% which is very small. 45. From Figure (3.19) we observe that the two characteristic spectral lines occur at wavelengths of 6.4 1011 m and 7.2 1011 m . Rearrange Equation (3.37) to solve for hc 1 1.24 106 V  m   17.2 kV and we have used the larger of e min 7.2 1011 m the two wavelengths in order to determine the minimum potential difference. the potential V0 ; V0  h 1  cos   so at maximum cos  1 and mc 2 1240 eV  nm   2h 2hc     1.01105 . This corresponds to 2   mc mc   511.0 keV  480 nm  46.     4.85 1012 m and therefore is not easily observed. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 3 The Experimental Basis of Quantum Physics 39 47. The maximum change in the photon’s energy is obtained in backscattering (   180 ), so 2h  4.853 1012 m. The photon’s original wavelength was 1  cos  2 and   mc hc 1240 eV  nm    0.0310 nm  3.10 1011 m and the new wavelength is E 40000 eV        3.585 1011 m. The electron’s recoil energy equals the change in the photon’s energy, or K hc   hc 1240 eV  nm 1240 eV  nm    5411 eV  5.41 keV 2   3.10 10 nm 3.585 102 nm 48. Use the Compton scattering formula but with the proton’s mass and   90 : h h hc 1240 eV  nm    1.32fm . 1  cos    2  mc mc mc 938.27 MeV 49. We find the Compton wavelength using c  photon energy is E  hc c requirements are high. 50.    0.004  h hc 1240 eV  nm  2  1.32 fm. The mc mc 938.27 MeV  938 MeV . In principle this could be observed, but the energy c 1  cos   so   250c 1  cos   ; (a)   250  2.43 1012 m 1  cos30   8.14 1011 m (b)   250  2.43 1012 m 1  cos90   6.08 1010 m (c)   250  2.43 1012 m 1  cos170   1.21109 m 51. By conservation of energy we know the electron’s recoil energy equals the energy lost by hc hc hc       hc   the photon: K    .         /   hf hc  hf  hf     Using       we have K  .           1   /   1    h h Conservation of px : pe cos   cos   ;   h h pe cos    cos  . (1)   Conservation of p y : pe sin   h sin   0 ;  © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 40 Chapter 3 pe sin   h sin   The Experimental Basis of Quantum Theory (2) h sin    Dividing equation (2) by equation (1), tan   . h h  cos    h sin  h  1  cos   h mc tan   1  cos  Using      we have: .   h h cos  mc     h 1  cos  mc   h  Multiplying numerator and denominator by    1  cos   , we have: mc    h sin   sin  tan    . h2    h  1  cos   h  1  cos     h cos    mc   mc sin    Using the trig identity:  cot   , we find: 1  cos   2  1 1       tan   cot    cot    cot   . h  2  1 h  2  1  hf  2  mc mc mc 2 hf     Inverting the equation gives cot   1  2  tan   .  mc  2 52.      c 1  cos    hc  c 1  cos   ; E 1240 eV  nm   2.43 103 nm 1  cos110   5.17 pm 3 650 10 eV hc 1240 eV  nm E    2.40 105 eV  240 keV 3   5.17 10 nm   By conservation of energy we find: Ke  E  E  650keV  240keV  410keV which agrees with the K formula in the previous problem. Also from the previous  110  hf       650 keV  cot   1  2  tan    1  tan problem we have:    3.245 so  mc   2   511 keV   2    17.1 . 53. For   90 we know      c  2.00243nm ;  c 2.43 103 nm    1.22 103   2 nm  © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 3 The Experimental Basis of Quantum Physics 41 54. E  2mc2  2  938.3 MeV   1877 MeV . This energy could come from a particle accelerator. 55. a) To find the minimum energy consider the zero-momentum frame. Let Ee be the energy of the electron in that frame, and E0 is the rest energy of the electron. From conservation of energy and momentum: Ee  E0 hf  pe  or hf  Ee2  E02 . c c energy: hf  Ee  3E0 or hf  3E0  Ee . 2 2 momentum: Squaring and subtracting these two equations gives: 0  10E02  6Ee E0 which simplifies 5 to Ee  E0 . This tells us that for the transformation from the lab frame to the zero3 momentum frame,   5 / 3 and v  0.8c . Then from the momentum equation we have in 25E02 4  E02  E0 . 9 3 In the lab the photon’s energy is obtained using a Doppler shift: 1  4 1  0.8 hf lab  hf  E0  4 E0  2.04 MeV 1  3 1  0.8 the zero-momentum frame: hf  b) The proton’s rest energy is Mc 2 . Now as in (a) we let the proton’s energy in the lab frame be Ep and conservation of momentum and energy give momentum: hf  E p2   Mc 2  ; 2 energy: hf  E p  2E0  Mc2 .  Mc   2E  2E Mc . This is very close to Squaring and subtracting, we find E  2 2 p 2 0 2 0 2 E0  Mc 2 E p  Mc 2 . Therefore the zero-momentum and lab frames are equivalent, and we conclude hflab  2E0  1.02 MeV . 56. The maximum energy transfer occurs when   180 so that   (h / mc)(1  cos )  2h / mc . By conservation of energy the kinetic energy of the hc hc hc hc electron is K  E  E     . Multiplying through by  (   ) we        find  (   ) K  hc(   )  hc  hc or  2 K    K  hc  0 . This is a quadratic equation that with numerical values can be solved for  to find hc 1.24 keV  nm  104 keV .   1.20 1011 m . Then E    1.20 102 nm © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 42 Chapter 3 The Experimental Basis of Quantum Theory 57. The power of the light received is intensity times area, or 2 P  IA   4.0 1011 W/m2    4.5 103 m    2.5 1015 W . The energy of each   hc  6.626 10  34 J  s  3.00 108 m/s   3.6 1019 J . Thus, the number 5.50 107 m 1 photon  6900 photons . of photons per second is 2.5 1015 J/s  3.6 1019 J photon is E  58. (a) max   2.898 103 m  K  2.13 106 m 1358 K (b) The peak is well into the infrared part of the spectrum. However, the shape of the “tail” of the distribution implies that more red photons are emitted than from any other part of the visible spectrum, and so we see red. 59. To find the asteroid mass m, note that Earth (matter) would supply an equal mass m to the GM E2 process, so 2mc 2  . Solving for m we find: 2 RE 11 3 1 2 24 GM E2  6.67 10 m  kg  s  5.98 10 kg  m   1.04 1015 kg . 2 3 8 4 RE c 2 4  6378 10 m  3.00 10 m/s  2  3 1.04 1015 kg      4  5000 kg/m3     Evaluating the energy: 1/3 1/3  3m  Then r     4  GM E2  6.67 10 E  2 Re 11  3.68 km which is relatively small. m3  kg 1  s 2  5.98 1024 kg  2  6378 10 m  3 2  1.87 1032 J ; E 1.87 1032 J   8.9 1012 . There is a lot of energy in the nuclear arsenals 5000  4.2 1015 J  annihilation process! 60. For maximum recoil energy the scattering angle is   180 and   0 . Then as usual 2h   . Using the result of Problem 56, mc  /  2h / mc 2hf / mc 2 K hf  hf  hf . 1   /  1  2h / mc 1  2hf / mc 2 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 44 Chapter 3 The Experimental Basis of Quantum Theory (b) If 30th magnitude corresponds to 2 1020 W/m2 and each change in magnitude by one represents a change in brightness of 1001/ 5  2.5119 then we must increase by a factor of 100   (2.5119)  3.98 10 . So 6 magnitude corresponds to 7.96 10 W/m . 1/ 5 24 24 th 9 11 2 If the diameter of the pupil is 6.5 mm, then the power reaching the retina is 2 P  7.96 1011 W/m2   3.25 103 m    2.64 1015 J/s. With each photon carrying    photons  energy 4.09 1019 J , then 2.64 1015 J/s  4.09 1019 J  n  and s   3 n  6.45 10 photons each second. 64. The laser’s power is 25 kW = 2.5 × 104 J/s. The energy of a single photon is 34 8 hc  6.626 10 J  s  3.00 10 m/s  E   1.88 1019 J . Thus,  1.06 106 m 1 photon 2.5 104 J/s  1.3 1023 photons/s . 1.88 1019 J 65. The laser’s power is 25 kW = 2.5 × 104 J/s. The energy of a single photon is 34 8 hc  6.626 10 J  s  3.00 10 m/s  E   1.88 1019 J . Thus, 6  1.06 10 m 1 photon 2.5 104 J/s  1.3 1023 photons/s 19 1.88 10 J 66. From Example 3.5, the sun uses 3.91 × 1026 J/s. At that rate, the sun’s lifetime is 5 106 J/kg 2.0 1030 kg = 2.6 1010 s or about 820 years! This estimate is off, because 3.911026 J/s the sun uses nuclear energy, not chemical. 67. (a) max  2.898 103 m  K  3.01107 m = 301 nm 9600 K This is actually in the ultraviolet part of the spectrum, but the shape of the distribution function guarantees that most of the visible photons will be in the violet-blue region, which explains the star’s blue appearance. (b) Assuming a perfect blackbody,   2 4 P   T 4 A  5.67 108 W/  m2  K 4   9600 K   4 1.6  6.96 108 m     27  7.5110 W This is about 19 times the value for the sun, reported in Example 3.5. © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 3 The Experimental Basis of Quantum Physics 45 68. Kinetic energy is related to momentum (nonrelativistically) by K  p 2 / 2m , so an electron of mass m and kinetic energy K has momentum p  2mK . Thus the magnitude of the electron’s change in momentum is p  2mKf  2mKi  7.5 1024 (SI units) By conservation, this is the momentum of the recoiling nucleus. The nucleus then has  p   9.2 1023 J = 5.8 104 eV This is negligible, compared with kinetic energy K  2 2m the energy (5 keV) of the x ray produced in the process. 69. For the 10-keV gamma ray, the electron absorbs half this amount, so its kinetic energy is 5 keV. That is less than 1 percent of its rest energy, so we’ll use a non-relativistic 1 calculation K  mv 2 so 2 v  2K / m  2  5000 eV  1.60 1019 J/eV  / 9.111031 kg  4.2 107 m/s . Half of the 300-MeV gamma ray’s energy is 150 MeV, which is highly relativistic. In K 150 MeV this case K    1 E0 so   1    294 and   295 . Solving for E0 0.511 MeV 1 speed v,   295  , which yields β = 0.999994 . Thus v = 0.999994 c. 1  2 70. If n gamma rays strike the target, the energy deposited is Q = nE0, where E0 = 100 MeV is the energy of each gamma ray. Converting to SI units, 1.60 1013 J E0  100 MeV   1.60 1011 J . The relationship between thermal energy 1 MeV and temperature is Q  mcV T , where cV is the specific heat. Solving for n: Q  nE0  mcV T ; n  mcV T  2.5 kg  430 J/(kg  K)  0.010 K    6.7 1011 11 E0 1.60 10 J © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.