# Physics exercises_solution: Chapter 38

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## Physics exercises_solution: Chapter 38

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Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 38

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## Nội dung Text: Physics exercises_solution: Chapter 38

1. c 3.00  108 m s 38.1: f   7  5.77  1014 Hz λ 5.20  10 m h 6.63  10 34 J  s p   1.28  10 27 kg  m s λ 5.20  10 7 m E  pc  (1.28  10 27 kg  m s) (3.00  108 m s)  3.84  10 19 J  2.40 eV. 38.2: a) Pt  (0.600 W) (20.0  10 3 s)  0.0120 J  7.49  1016 eV. hc b) hf   3.05  10 19 J  1.90 eV. λ Pt c)  3.94  1016. hf E (2.45  106 eV) (1.60  10 19 J eV) 38.3: a) E  hf  f    5.91  10 20 Hz. h 6.63  10 34 J  s c 3.00  108 m s b) λ    5.08  10 13 m. f 5.91  10 20 Hz c) λ is of the same magnitude as a nuclear radius. dN (dE dt ) P Pλ (12.0 W ) (2.48  10 7 m) 38.4:     dt (dE dN ) hf hc hc  1.50  1019 photons sec. 1 2 38.5: mvmax  hf   2  (6.63  10  34 J  s )  c  (5.1 eV) (1.60  10 19 J eV )    2.35  10 m 7   3.04  10  20 J 2(3.04  10 20 J)  vmax  31  2.58  105 m s . 9.11  10 kg hc  c  38.6: E  hf    hf   h 1.45  1015 Hz  7  λ0  2.72  10 m   2.301019 J  1.44 eV.
2. 38.7: a) f  c λ  5.0  1014 Hz b) Each photon has energy E  hf  3.31  10 19 J. Source emits 75 J s so (75 J s) (3.31  10 19 photons/s)  2.3  10 20 photons s c) No, they are different. The frequency depends on the energy of each photon and the number of photons per second depends on the power output of the source. 38.8: For red light λ  700 nm hc   hf  λ (6.626  10 34 J  s) (3.00  108 m s)  (700  10 9 m)  1eV   2.84  10 19 J   1.6  10 19   J   1.77 eV 38.9: a) For a particle with mass, K  p 2 2m . p2  2 p1 means K 2  4 K1 . b) For a photon, E  pc. p2  2 p1 means E2  2 E1. 38.10: K max  hf   Use the information given for λ  400 nm to find  : (6.626  10 34 J  s) (2.998  108 m s)   hf  K max  9  (1.10 eV) (1.602  10 19 J eV) 400  10 m  3.204  10 19 J Now calculate K max for λ  300 nm : (6.626  10 34 J  s) (2.998  108 m s) K max  hf     3.204  10 19 J 300  10 9 m  3.418  10 19 J  2.13 eV
3. hc 38.11: a) The work function   hf  eV0   eV0 λ (6.63  10 34 J  s) (3.00  108 m s)   7  (1.60  10 19 C) (0.181 V) 2.54  10 m 19  7.53  10 J. hc hc The threshold frequency implies   hf th   λ th  λ th  (6.63  10 34 J  s) (3.00  108 m s)  λ th  19  2.64  10 7 m. 7.53  10 J b)   7.53  10 19 J  4.70 eV, as found in part (a), and this is the value from Table 38.1. 38.12: a) From Eq. (38.4), 15  (4.136  10 eV  s) (3.00  10 m s) 8 1  hc V      2.3 V  2.7 V. e λ  (2.50  10 7 m) b) The stopping potential, multiplied by the electron charge, is the maximum kinetic energy, 2.7 eV. 2K 2 eV 2(1.60  10 19 C) (2.7 V) c) v    31  9.7  105 m s. m m (9.11 10 kg) 38.13: a) E  pc  (8.24  10 28 kg  m s) (3.00  108 m s) 2.47  10 19 J  2.47  10 19 J   1.54 eV 1.60  10 19 J eV h h (6.63  10 34 J  s) b) p  λ   8.05  10 7 m. λ p (8.24  10 28 kg  m s) This is infrared radiation. 38.14: a) The threshold frequency is found by setting V = 0 in Eq. (40.4), f 0   h . hc hc b)   hf 0   7  5.35  10 19  3.34 eV. λ 3.72  10 m
4. hc (6.63  10 34 J  s) (3.00  108 m s) 38.15: a) E   7  2.31  10 19 J  1.44 eV. λ 8.60  10 m So the internal energy of the atom increases by 1.44 eV to E  6.52 eV  1.44 eV   5.08 eV. hc (6.63  10 34 J  s) (3.00  108 m s) b) E   7  4.74  10 19 J  2.96 eV. λ 4.20  10 m So the final internal energy of the atom decreases to E  2.68 eV  2.96 eV  5.64 eV. 38.16: a)  E1  20 eV. b) The system starts in the n = 4 state. If we look at all paths to n = 1 we find the 4-3, 4-2, 4-1, 3-2, 3-1, and 2-1 transitions are possible (the last three are possible in combination with the others), with energies 3 eV, 8 eV, 18 eV, 5 eV, 15 eV, and 10 eV, respectively. c) There is no energy level 8 eV above the ground state energy, so the photon will not be absorbed. d) The work function must be more than 3 eV, but not larger than 5 eV. 1  1 1 38.17: a)  R  2  2  (Balmer series implies final state is n = 2) λ 2 n  1  1 1  21 H  n  5 :  R    R λ  4 25  100 100 100 λ  m  4.33  10 7 m  433 nm 21R 21(1.10  10 7 ) c 3.00  108 m s b) f   7  6.93  1014 Hz λ 4.33  10 m c) E  hf  2.87 eV. (4 3) (4 3) 38.18: Lyman: largest is n  2, λ    122 nm, in the ultraviolet. R (1.097  10 7 m 1 ) 1 Smallest is n  , λ   91.2 nm, also ultraviolet. Paschen: largest is R (144 7) 9 n  4, λ   1875 nm, in the infrared. Smallest is n  , λ   820 nm, also R R infrared.
5. hc (6.626  10 34 J  s) (3.000  108 m s) 38.19: E 3  g   2 λ1 5.890  10 7 m  3.375  10 19 J  2.109 eV. hc (6.626  10 34 J  s) (3.000  108 m s) E 1  g   2 λ2 5.896  10 7 m  3.371  10 19 J  2.107 eV. E 3  1  2.00  10 3 eV. 2 2 38.20: a) Equating initial kinetic energy and final potential energy and solving for the separation radius r, 1 (92e) (2e) r 4 ε0 K 1 (184) (1.60  10 19 C)   5.54  10 14 m. 4 ε0 (4.78  10 J C) 6 b) The above result may be substituted into Coulomb’s law, or, the relation between the magnitude of the force and the magnitude of the potential energy in a Coulombic field is K (4.78  10 6 eV) (1.6  10 19 J ev) F   13.8 N. r (5.54  10 14 m) q1q2 (2e) (82e) 164 (1.60  10 19 C) 2 38.21: a) U    4 ε0 r 4 ε0 r 4 ε0 (6.50  10 14 m)  U  5.81  10 13 J  3.63  10 6 eV  3.63 MeV. b) K 1  U 1  K 2  U 2  K 1  U 2  5.81  10 13 J  3.63 MeV. 1 2 2K 2(5.81 10 13 J) c) K  mv  v    27  1.32  10 7 m s . 2 m 6.64  10 kg 1  1  c 38.22:  R 1   (4) 2 , so λ  97.0 nm and f  λ  3.09  10 Hz.  15 λ  
6. 38.23: a) Following the derivation for the hydrogen atom we see that for Be 3 all we need do is replace e 2 by 4e 2 . Then 1 m(4e 2 ) 2   13.60 eV  En (Be3 )   2 2 2  16 En (H)  En (Be3 )  16  . ε0 8n h  n2  So for the ground state, E1 (Be3 )  218 eV. b) The ionization energy is the energy difference between the n   and n  1 levels. So it is just 218 eV for Be 3 , which is 16 times that of hydrogen. 1 m ( 4e 2 ) 2  1 1   1 1  c)   2  2   (1.74  108 m 1 )  2  2  . n  n  λ 8 0 h c  1 n2  2 3  1 n2  1  1 So for n  2 to n  1,  (1.74  108 m 1 ) 1    1.31  108 m λ  4 9  λ  7.63  10 m. This is 16 times shorter than that from the hydrogen atom. ε n2h2 1 d) rn  (Be3 )  0  rn (H).  m ( 4e ) 4 2 h 38.24: a), b) For either atom, the magnitude of the angular momentum is  2 1.05  10 34 kg  m 2 s. 38.25: En   (13.6 eV) n 2 , so this state has n  13.6 1.51  3. In the Bohr model. L  n so for this state L  3  3.16  10 34 kg  m 2 s . 38.26: a) We can find the photon’s energy from Eq. 38.8  1 1  1 1 E  hcR 2  2   (6.63  10 34 J  s) (3.00  108 m s) (1.097  10 7 m 1 )  2  2  2 n  2 5   4.58  10 19 J. The corresponding wavelength is λ  hc  434 nm. E b) In the Bohr model, the angular momentum of an electron with principal quantum number n is given by Eq. 38.10 h Ln . 2 Thus, when an electron makes a transition from n = 5 to n = 2 orbital, there is the following loss in angular momentum (which we would assume is transferred to the photon): h 3(6.63  10 34 J  s) L  (2  5)   3.17  10 34 J  s. 2 2π However, this prediction of the Bohr model is wrong (as shown in Chapter 41).
7. 1 e2 (1.60  10 19 C) 2 38.27: a) v n  : n  1  v1   2.18  10 6 m/s  0 2nh  0 2 (6.63  10 34 J  s) v h  2  v2  1  1.09  106 m s 2 v h  3  v3  1  7.27  105 m s . 3 2rn 2 0 n 2 h 2 me 2 4ε0 n 3 h 3 2 b) Orbital period    vn 1  0  e 2 2nh me 4 4 0 (6.63  10 34 J  s) 3 2 n  1  T1  31 19  1.53  10 16 s (9.11  10 kg) (1.60  10 C) 4 n  2 : T2  T1 (2) 3  1.22  10 15 s n  3 : T3  T1 (3) 3  4.13  10 15 s. 1.0  10 8 s c) number of orbits  15  8.2  10 6. 1.22  10 s 38.28: a) Using the values from Appendix F, keeping eight significant figures, gives R  1.0973731  10 7 m 1 . (Note: On some standard calculators, intermediate values in the calculation may have exponents that exceed 100 in magnitude. If this is the case, the numbers must be manipulated in a different order.) hc b) Using the eight-figure value for R gives E   hcR  2.1798741  10 18 λ J  13.605670 eV. c) Using the value for the proton mass as given in Appendix F gives R  1.0967758  10 7 m 1 , so R  5970 m 1.
8. n5 s ( E E ) kT 38.29:  e  5s 3 p n3 p But E 5 s  20.66 eV  3.306  10 18 J E3 p  18.70 eV  2.992  10 18 J  E  3.14  10 19 J n5 s 19 J ) (1.38  10 23 J K 300 K ) a)  e ( 3.14  10  1.15  10 33. n3 p 19 J ) (1.38  10 23 J K 600 K ) b) e  (3.14  10  3.39  10 17. 19 23 c) e  (3.14  10 J ) (1.38  10 J K1200 K )  5.82  10 9. d) The 5s state is not highly populated compared to the 3 p state, so very few atoms are able to make the required energy jump to produce the 632.8 nm light. n2 p3 2  ( E2 p 3 2  E2 p1 2 ) KT 38.30: e . n2 p1 / 2 hc (6.626  10 34 J)(3.000  108 m s) From the diagram E3 / 2 g    3.375  10 19 J. λ1 5.890  10 7 m hc (6.626  10 34 J )(3.000  108 m s) E1 2  g    3.371  10 19 J. λ2 5.896  10 7 m so E3 / 2 1/ 2  3.375  10 19 J  3.371  10 19 J  4.00  10 22 J. n2 p3 / 2  22 J ) (1.38  10  23 J / K500 K).  e ( 4.00  10  0.944. So more atoms are in the 2 p1 2 state. n2 p1 / 2 hc (6.63  10 34 J  s)(3.00  108 m s) 38.31: E    1.88  10 20 J. λ 1.06  10 5 m Total energy in 1 second from the laser E  Pt  7.50  10 3 J. So the number of E 7.50  10 3 J photons emitted per second is   4.00  1017. E 1.88  10 20 J c hc 38.32: 20.66 eV  18.70 eV  1.96 eV  3.14  10 19 J, and λ    632 nm, in good f E agreement.
9. hc 38.33: eVAC  hf max  λ min hc (6.63  10 34 J  s)(3.00  108 m s)  λ min   19  3.11  10 10 m eVAC (1.60  10 C)(4000 V) This is the same answer as would be obtained if electrons of this energy were used. Electron beams are much more easily produced and accelerated than proton beams. hc 38.34: a)  8.29  10 3 V  8.29 kV. b) The shortest wavelength would correspond eλ hc to the maximum electron energy, eV , and so λ   0.0414 nm. eV 38.35: An electron’s energy after being accelerated by a voltage V is just E  eV . The hc hc most energetic photon able to be produced by the electron is just: λ   E eV (6.63  10 34 J  s)(3.00  108 m s) λ 19  8.29  10 11 m  0.0829 nm. (1.60  10 C)(1.5  10 V) 4 λ 38.36: a) From Eq. (38.23), cos   1  , and so a) (h mc) λ  0.0542 nm  0.0500 nm, 0.0042 nm cos   1   0.731, and   137 . 0.002426 nm 0.0021 nm b) λ  0.0521 nm  0.0500 nm. cos   1   0.134.   82.3. 0.002426 nm c) λ  0, the photon is undeflected, cos   1 and   0. h 38.37: λ  λ  (1  cos  ) mc h  λmax  λ  (1  (1))(  180 is chosen to maximize λ.) mc 2h λ mc λ max  6.65  10 11 m  2(2.426  10 12 m)  7.14  10 11 m.
10. hc 38.38: a) λ   0.0691 nm. b) λ  λ  (h mc)(1  cos  )  (2.426  10 12 m)  eV (1  cos 45.0)  7.11  10 13 m, so λ  0.0698 nm. hc c) E   17.8 keV. λ h 38.39: λ  λ  (1  cos  ) mc 2h λ  λ  λ,   180 ; λ  mc Δλ 2h   9.70  10 6 λ mcλ 38.40: The change in wavelength of the scattered photon is given by Eq. 38.23 Δλ h h  (1  cos  )  λ  (1  cos  ). λ mcλ mc Δλ  λ Thus, (6.63  10 34 J  s) λ  27 (1  1)  2.65  10 14 m. (1.67  10 kg)(3.00  10 m/s)(0.100) 8 38.41: The derivation of Eq. (38.23) is explicitly shown in Equations (38.24) through (38.27) with the final substitution of h p  h λ and p  h λ yielding λ  λ  (1  cos  ). mc 2.90  10 3 m  K 2.90  10 3 m  K 38.42: T   9  7.25  103 K. λm 400  10 m 2.90  10 3 m  K 2.90  10 3 m  K 38.43: λ m    1.06  10 3 m  1.06 mm. This is in T 2.728 K the microwave part of the electromagnetic spectrum. 38.44: From Eq. (38.30), 2.898  10 3 m  K c a) λ m   0.966 mm, and f   3.10  1011 Hz. 3.00 K λm Note that a more precise value of the Wien displacement law constant has been used. b) A factor of 100 increase in the temperature lowers λm by a factor of 100 to 9.66 m and raises the frequency by the same factor, to 3.10  1013 Hz. c) Similarly, λ m  966 nm and f  3.10  1014 Hz.
11. 38.45: a) H  Aeσ T 4 ; A  r 2l 14 14  H   100 W  T     (0.20  10 3 m) 2 (0.30 m)(0.26)(5.671  10 8 W m 2  K 4 )    Aeσ    T  2.06  10 K 4 b) λ mT  2.90  10 3 m  K; λ m  141 nm Much of the emitted radiation is in the ultraviolet. 38.46: (a) Wien’s law: λ m  k T 2.90  10 3 K  m λm   9.7  10 8 m  97 nm 30,000 K This peak is in the ultraviolet region, which is not visible. The star is blue because the largest part of the visible light radiated is in the blue violet part of the visible spectrum (b) P  σAT 4 (Stefan-Boltzmann law)  W  (100, 000)(3.86  10 26 W)   5.67  10 8 2 4  (4R 2 )(30,000 K )  m K  R  8.2  10 m 9 8.2  109 m Rstar Rsun   12 6.96  108 m (c) The visual luminosity is proportional to the power radiated at visible wavelengths. Much of the power is radiated nonvisible wavelengths, which does not contribute to the visible luminosity. 2hc 2 x2 38.47: Eq. (38.32): I ( λ)  but e x  1  x   . . .  1  x for λ 5 (e hc λkT  1) 2 2hc 2 2ckT x  1  I ( λ)  5   Eq. (38.31), which is Rayleigh’s distribution. λ (hc λkT ) λ4
12. 38.48: a) As in Example 38.10, using four-place values for the physical constants, hc  95.80, from which λkT I (λ)λ 4  6.44  10 38. σT b) With T  2000 K and the same values for λ and λ, hc  14.37 λkT and so I ( λ)λ 4  7.54  10 6. σT hc I (λ)λ c) With T  6000 K,  4.790 and 4  1.36  10 3. λkT σT d) For these temperatures, the intensity varies strongly with temperature, although for even higher temperatures the intensity in this wavelength interval would decrease. From the Wien displacement law, the temperature that has the peak of the corresponding distribution in this wavelength interval is 5800 K (see Example 38.10), close to that used in part (c). 38.49: a) To find the maximum in the Planck distribution: dI d  2hc 2  (2hc 2 ) 2hc 2 (  λ 2 )   5    0  5 5    5 λ dλ dλ  λ (e  1)    λ (e  1) λ (e  1) 2   5(e λ  1)λ     5e λ  5   λ  hc  Solve 5  x  5e x where x   . λ λkT α hc Its root is 4.965, so  4.965  λ  . λ (4.965)kT hc (6.63  10 34 J  s)(3.00  108 m s) b) λ mT    2.90  10 3 m  K. (4.965)k (4.965)(1.38  10 23 J K) 38.50: Combining Equations (38.28) and (38.30), 2.90  10 3 m  K λm  ( I σ)1/4 (2.90  10 3 m  K)  (6.94  10 6 W/m 2 5.67  10 8 W m 2  K 4 )1/4  8.72  10 7 m  872 nm.
13. E (mole) 1.00  10 5 J 38.51: a) Energy to dissociate an AgBr molecule is just E    1 mole 6.02  10 23 1.66  10 19 J  1.04 eV. (1.60  10 19 J eV) hc (6.63  10 34 J  s)(3.00  108 m s) b) λ   19  1.20  10 6 m. E 1.66  10 J c 3.00  108 m/s c) f   6  2.51  1014 Hz. λ 1.20  10 m 34 6.63  10 26 J d) E  hf  (6.63  10 J  s)(1.00  10 Hz)  8 19  4.14  10 7 eV. 1.60  10 J eV e) Even though a 50-kW radio station emits huge numbers of photons, each individual photon has insufficient energy to dissociate the AgBr molecule. However, the individual photons in faint visible light do have enough energy. h 38.52: a) Assume a non-relativistic velocity and conserve momentum  mv   λ h v . mλ 2 1 1  h  h2 b) K  mv 2  m   . 2 2  mλ  2mλ 2 K h2 λ h c)  2   . Recoil becomes an important concern for small m and E 2mλ hc 2mcλ small  since this ratio becomes large in those limits. d) hc (6.63  10 34 J  s)(3.00  108 m s) E  10.2 eV  λ    1.22  10 7 m  122 nm. E (10.2 eV)(1.60  10 19 J eV) (6.63  10 34 J  s) 2 K  8.84  10 27 J  5.53  10 8 eV. 2(1.67  10  27 kg)(1.22  10 7 m) 2 K 5.53  10 8 eV   5.42  10 9. This is quite small so recoil can be neglected. E 10.2 eV 38.53: Given a source of spontaneous emission photons we can imagine we have a uniform source of photons over a long period of time (any one direction as likely as any other for emission). If a certain number of photons pass out though an area A, a distance D from the source, then at a distance 2D, those photons are spread out over an area (2) 2  4 times the original area A (because of how surface areas of spheres increase). Thus the number of photons per unit area DECREASES as the inverse square of the distance from the source.
14. hc 38.54: a) λ 0  , and the wavelengths are: cesium: 590 nm, copper: 264 nm, E potassium: 539 nm, zinc: 288 nm. b) The wavelenghts of copper and zinc are in the ultraviolet, and visible light is not energetic enough to overcome the threshold energy of these metals. 38.55: a) Plot: Below is the graph of frequency versus stopping potential. Threshold frequency is when the stopping potential is zero. 1.89  f th  Hz  4.60  1014 Hz 4.11  10 15 c (3.00  108 m s) b) Threshold wavelength is λ th    6.52  10 7 m. f th 4.60  10 Hz 14 c) The work function is just hf th  (6.63  10 34 J  s)(4.60  1014 Hz)  3.05  10 19 J  1.91 eV. V h d) The slope of the graph is m  0   h  me f e  (4.11  10 15 V Hz)(1.60  10 19 C)  6.58  10 34 J  s dN (dE dt ) P (200 W )(0.10) 38.56: a) See Problem 38.4:     dt (dE dN ) hf h(5.00  1014 Hz) 6.03  1019 photons sec. (dN dt ) b) Demand  1.00  1011 photons sec cm 2 . So r  4r 2 1/ 2  6.03  1019 photons sec    4 (1.00  1011 photons sec  cm 2 )    6930 cm  69.3 m.  
15. hc 38.57: a) Recall eV0   λ hc hc hc  1 1  e(V02  V01 )    V0     λ 2 λ1 λ e  2 λ1   b) (6.63  10 34 J  s)(3.00  108 m s)  1 1  V0  11   2.65  10 7 m  2.95  10 7 m   0.477 V.  (1.60  10 C)   So the change in the stopping potential is an increase of 0.739 V. 38.58: From Eq. (38.13), the speed in the ground state is v1  Z (2.19  10 6 m s). Setting c v1  gives Z = 13.7, or 14 as an integer. b) The ionization energy is E  Z 2 (13.6 10 mc 2 eV), and the rest mass energy of an electron is 0.511 MeV, and setting E  gives Z 100 = 19.4, or 19 as an integer. h 38.59: λ  λ  (1  cos  ) mc 2h   180 so λ  λ   0.09485 m mc p  h λ  6.99  10  24 kg  m s b) E  E   Ee ; hc λ  hc λ  Ee 1 1  λ  λ Ee  hc    (hc)  1.129  10 16 J  705 eV  λ λ  λλ  38.60: a) The change in wavelength of the scattered photon is given by Eq. 38.23 h h λ  λ  (1  cos  )  λ  λ  (1  cos  )  mc mc (6.63  10 34 J  s) (0.0830  10 9 m)  (1  1)  0.0781 nm. (9.11  10 31 kg)(3.00  108 m s) b) Since the collision is one-dimensional, the magnitude of the electron’s momentum must be equal to the magnitude of the change in the photon’s momentum. Thus,  1  1 34  1 1  1 pe  h    (6.63  10 J  s)  9  (10 m )  λ λ   0.0781 0.0830   23  23  1.65  10 kg  m s  2  10 kg  m s . c) Since the electron is non relativistic (   0.06), pe2 Ke   1.49  10 16 J  10 16 J. 2m
16. m1m2 207me m p 38.61: a) mr    1.69  10 28 kg. m1  m2 207me  m p b) The new energy levels are given by Eq. (38.18) with me replaced by mr . 1 mr e 4  mr    13.60 eV  En       02 8n 2 h 2  me     n2   1.69  10 28 kg    13.60 eV    9.11  10 31 kg       n2   2.53  103 eV   E1  2.53 keV. n2 1 1 (2.53  103 e V 4  (2.53  103 eV)) (1.60  10 19 J eV)) c)  ( E1  E2 )  λ hc (6.63  10 34 J  s) (3.00  108 m s)  1.53  10 9 m 1  λ  6.55  10 10 m. 38.62: a) The levels are E 4  1.0 eV, E3  5.0 eV, E 2  8.0 eV, and E1  10.0 eV. b) We can go from 4-3(4 eV), 4-2(7 eV), and 4-1(9 eV) directly, but also 3-2(3 eV), 3-1(5 eV), and 2-1(2 eV) after starting from 4. 38.63: a) The maximum energy available to be deposited in the atom is hc (6.63  10 34 J  s) (3.00  108 m s) E   8  2.33  10 18 J  14.5 eV λ 8.55  10 m but the inoization energy of hydrogen is 13.6 eV, so the maximum kinetic energy is 14.5 eV  13.6 eV  0.900 eV.   13.6 eV  b) If some of the electrons were in the n  2 state  E 2   3.4 eV  , then we  4  would expect a maximum kinetic energy of 14.5 eV  3.4 eV  11.1 eV , which is exactly 10.2 eV above that measured in part (a), explaining the anomoly.
17. 38.64: a) In terms of the satellite’s mass M , orbital radius R and orbital period T , 2 2  2  4 MR 2 2 n L MR 2     h h T  hT Using the given numerical values, n  1.08  10 46  b) The angular momentum of the satellite in terms of its orbital speed V , mass, and radius is L  MVR, so V2 L2 V 2  ( L MR) 2 , and its centripetal acceleration is  2 3 R M R Newton’s law of gravitation can then be expressed as GM earth ML2 L2  2 3 , or R   R2 M R GM earth M If L  nh 2 ,  h2  R  n2  2   4 GM M   kn  2  earth  c) R  2knn, and for the next orbit, n  1, and R  n(h 2 4 2 GM earth M ). Insertion of numerical values from Appendix F and using n from part (a) gives R  1.5  10 39 m, which is (d) not observable. (e) The quantum and classical orbit do correspond, either would be correct, but only the classical calculation is useful. h nh 38.65: a) Quantization of angular momentum implies L  mvr  n v  2 2mr But 14 mv 2 mv 2 n2h2  n2h2  F   Dr   r2    rn   2  4 mD    r D 4 2 mr 2 D   1 1 b) The energy E  K  U  mv 2  Dr 2 , since F   Dr is completely analogous 2 2 to 1 2 1 D  1 F  kx  U  kx . So E  m   r 2   Dr 2  Dr 2 2 2 m  2 nh D nh  En  D    . 2 mD m 2 D h D h c) Photon energies E  Ei  E f  (ni  n f )   E  n where n  m 2 m 2 integers  0. d) This could describe a charged mass attached to a spring, being spun in a circle.
18. md 38.66: a) mr,d  me  me (1  (me md )) 1 , and insertion of the numerical me  md (4 3) (4 3) values gives mr,d  0.999728 me . b) Let λ 0  , λ  , where R is the R R Rydberg constant evaluated with m  mr,p  0.999456, so mr,p (4 3) m R  R, and λ  , R  r,d R  me R me Then 4 1R R m m  λ      λ 0  e  e   0.033 nm. m  3 R  R R   r,p mr,d  38.67: a) The H  line is emitted by an electron in the n  3 energy level,  13.60 eV E3   1.51 eV. The ground state energy is E1  13.60 V, so one must (3) 2 add at least 13.60 eV  1.51 eV=12.09 eV if the H  line is to be emitted. b) The possible emitted photons are 3  2, 3  1, and 2  1, with the wavelengths 1 1  13.60 eV  1   2  12  . This yields the wavelengths of given by   ( Ei  E f )  λ hc hc n   f ni  658 nm, 103 nm, and 122 nm for the respective photons above.
19. n2  ( E  E ) kT  ( Eex  Eg ) 38.68:  e ex g T  . n1  n2  kln  n   1  13.6 eV Eex  E 2   3.4 eV. 4 E g  13.6 eV. E ex  E g  10.2 eV  1.63  10 18 J. n2 a)  10 12. n1  (1.63  10 18 J) T  4275 K. (1.38  10  23 J K ) ln(10 12 ) n2 b)  10 8. n1  (1.63  10 18 J) T  6412 K. (1.38  10 23 J K ) ln(10 8 ) n2 c)  10  4. n1  (1.63  10 18 J) T  12824 K. (1.38  10 23 J K ) ln(10 4 ) d) For absorption to take place in the Balmer series, hydrogen must start in the n  2 state. From part (a), colder stars have fewer atoms in this state leading to weaker absorption lines.
20. 38.69: The transition energy equals the sum of the recoiling atom’s kinetic energy and the photon’s energy. Etr  Ek  E  hc hc  E    Etr  Ek  λ  . λ Etr  Ek hc If the recoil is neglected λ  Etr  1 1  hc  1   λ  hc  E  E  E   E  1  E E  1     tr k tr  tr  k tr  hc      1  Ek      1     Etr   Etr   E  Ek E   λ  hc k   hc  E2  2   k λ 2  tr  (hc λ)  hc  Conservation of momentum, assuming atom initially at rest, yields: h Pk2 h2 P   Pk  Ek   λ 2m 2mλ 2  h2  2 h  λ    2mhcλ 2  λ  2mc    (6.63  10 34 J  s) b) For the hydrogen atom: λ   27  6.6  10 16 m and 2(1.67  10 kg)(3.00  10 m s)8 it doesn’t depend on n. 2h 38.70: a)   180 so (1  cos )  2  λ   0.0049 nm, so λ  0.1849 nm. mc 1 1  b) E  hc    2.93  10 17 J  183 eV.  λ λ  This will be the kinetic energy of the electron. c) The kinetic energy is far less than the rest mass energy, so a non-relativistic calculation is adequate; v  2 K m  8.02  10 6 m s .