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Problems & Exercises

1.

(a) 0.070 eV

(b) 14

3.

(a) 2.21×1034 J2.21×1034 J

(b) 2.26×10342.26×1034

(c) No

4.

263 nm

6.

3.69 eV

8.

0.483 eV

10.

2.25 eV

12.

(a) 264 nm

(b) Ultraviolet

14.

1.95 × 10 6 m/s 1.95 × 10 6 m/s

16.

(a) 4.02×1015/s4.02×1015/s

(b) 0.256 mW

18.

(a) –1.90 eV–1.90 eV

(b) Negative kinetic energy

(c) That the electrons would be knocked free.

20.

6.34 × 10 9 eV 6.34 × 10 9 eV , 1.01 × 10 27 J 1.01 × 10 27 J

22.

2 . 42 × 10 20 Hz 2 . 42 × 10 20 Hz

24.
hc = 6.62607 × 10 34 J s 2.99792 × 10 8 m/s 10 9 nm 1 m 1.00000 eV 1.60218 × 10 19 J = 1239.84 eV nm 1240 eV nm hc = 6.62607 × 10 34 J s 2.99792 × 10 8 m/s 10 9 nm 1 m 1.00000 eV 1.60218 × 10 19 J = 1239.84 eV nm 1240 eV nm
26.

(a) 0.0829 eV

(b) 121

(c) 1.24 MeV

(d) 1.24×1051.24×105

28.

(a) 25.0 × 103 eV 25.0 × 103 eV

(b) 6.04 × 1018 Hz 6.04 × 1018 Hz

30.

(a) 2.69

(b) 0.371

32.

(a) 1.25 × 1013 photons/s 1.25 × 1013 photons/s

(b) 997 km

34.

8.33 × 10 13 photons/s 8.33 × 10 13 photons/s

36.

181 km

38.

(a) 1.66 × 10 32 kg m/s 1.66 × 10 32 kg m/s

(b) The wavelength of microwave photons is large, so the momentum they carry is very small.

40.

(a) 13.3 μm

(b) 9.38×10-29.38×10-2 eV

42.

(a) 2.65×1028kgm/s2.65×1028kgm/s

(b) 291 m/s

(c) electron 3.86×1026 J3.86×1026 J, photon 7.96×1020 J7.96×1020 J, ratio 2.06×1062.06×106

44.

(a) 1.32×1013 m1.32×1013 m

(b) 9.39 MeV

(c) 4.70×102 MeV4.70×102 MeV

46.

E=γmc2mc2E=γmc2mc2 and P=γmuP=γmu, so

EP = γmc2 γmu = c2 u . EP = γmc2 γmu = c2 u .

As the mass of particle approaches zero, its velocity uu will approach cc, so that the ratio of energy to momentum in this limit is

limm→0 E P = c2 c = c limm→0 E P = c2 c = c

which is consistent with the equation for photon energy.

48.

(a) 3 . 00 × 10 6 W 3 . 00 × 10 6 W

(b) Headlights are way too bright.

(c) Force is too large.

49.

7.28 × 10 –4 m 7.28 × 10 –4 m

51.

6.62 × 10 7 m/s 6.62 × 10 7 m/s

53.

1.32 × 10 –13 m 1.32 × 10 –13 m

55.

(a) 6.62×107 m/s6.62×107 m/s

(b) 22.9 MeV22.9 MeV

57.
15.1 keV
59.

(a) 5.29 fm

(b) 4.70×1012 J4.70×1012 J

(c) 29.4 MV

61.

(a) 7.28×1012 m/s7.28×1012 m/s

(b) This is thousands of times the speed of light (an impossibility).

(c) The assumption that the electron is non-relativistic is unreasonable at this wavelength.

62.

(a) 57.9 m/s

(b) 9.55×109 eV9.55×109 eV

(c) From Table 29.1, we see that typical molecular binding energies range from about 1eV to 10 eV, therefore the result in part (b) is approximately 9 orders of magnitude smaller than typical molecular binding energies.

64.

29 nm,

290 times greater

66.

1 . 10 × 10 13 eV 1 . 10 × 10 13 eV

68.

3 . 3 × 10 22 s 3 . 3 × 10 22 s

70.

2.66 × 10 46 kg 2.66 × 10 46 kg

72.

0.395 nm

74.

(a) 1.3 × 10 19 J 1.3 × 10 19 J

(b) 2 . 1 × 10 23 2 . 1 × 10 23

(c) 1 . 4 × 10 2 s 1 . 4 × 10 2 s

76.

(a) 3.35×105 J3.35×105 J

(b) 1.12×10–3 kgm/s1.12×10–3 kgm/s

(c) 1.12×10–3 m/s1.12×10–3 m/s

(d) 6.23×10–7 J6.23×10–7 J

78.

(a) 1.06×1031.06×103

(b) 5.33×1016kgm/s5.33×1016kgm/s

(c) 1.24×1018m1.24×1018m

80.

(a) 1 . 62 × 10 3 m/s 1 . 62 × 10 3 m/s

(b) 4 . 42 × 10 19 J 4 . 42 × 10 19 J for photon, 1 . 19 × 10 24 J 1 . 19 × 10 24 J for electron, photon energy is 3 . 71 × 10 5 3 . 71 × 10 5 times greater

(c) The light is easier to make because 450-nm light is blue light and therefore easy to make. Creating electrons with 7.43 μeV 7.43 μeV of energy would not be difficult, but would require a vacuum.

81.

(a) 2 . 30 × 10 6 m 2 . 30 × 10 6 m

(b) 3 . 20 × 10 12 m 3 . 20 × 10 12 m

83.

3 . 69 × 10 4 ºC 3 . 69 × 10 4 ºC

85.

(a) 2.00 kJ

(b) 1.33×105kgm/s1.33×105kgm/s

(c) 1.33×105 N1.33×105 N

(d) yes

87.

(a) p= h λ = 6.63× 10 34 550× 10 9 kg m/s=1.21 kg× 10 27  m/s p= h λ = 6.63× 10 34 550× 10 9 kg m/s=1.21 kg× 10 27  m/s

(b) p= h λ = 6.63× 10 34 650× 10 9 kg m/s=1.02× 10 27 kg m/s p= h λ = 6.63× 10 34 650× 10 9 kg m/s=1.02× 10 27 kg m/s

(c) Yes, conservation of momentum applies.

(d) The photon with the longer wavelength has less momentum than the one with the shorter wavelength.

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