College Physics 2e

# Chapter 29

### 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) $2.65×10−28kg⋅m/s2.65×10−28kg⋅m/s$

(b) 291 m/s

(c) electron $3.86×10−26 J3.86×10−26 J$, photon $7.96×10−20 J7.96×10−20 J$, ratio $2.06×1062.06×106$

44.

(a) $1.32×10−13 m1.32×10−13 m$

(b) 9.39 MeV

(c) $4.70×10−2 MeV4.70×10−2 MeV$

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) $4.70×10−12 J4.70×10−12 J$

(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) $9.55×10−9 eV9.55×10−9 eV$

(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) $1.12×10–3 kg⋅m/s1.12×10–3 kg⋅m/s$

(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) $5.33×10−16kg⋅m/s5.33×10−16kg⋅m/s$

(c) $1.24×10−18m1.24×10−18m$

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) $1.33×10−5kg⋅m/s1.33×10−5kg⋅m/s$

(c) $1.33×10−5 N1.33×10−5 N$

(d) yes

87.

(a)

(b)

(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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