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Chapter 38 - 38.1 Hz 10 5.77 m 10 5.20 s m 10 3.00 14 7 8...

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Unformatted text preview: 38.1: Hz 10 5.77 m 10 5.20 s m 10 3.00 λ 14 7 8 × = × × = =- c f eV. 2.40 J 10 3.84 ) s m 10 (3.00 s) m kg 10 (1.28 s m kg 10 1.28 m 10 5.20 s J 10 6.63 λ 19 8 27 27 7 34 = × = × ⋅ × = = ⋅ × = × ⋅ × = =----- pc E h p 38.2: a) eV. 10 7.49 J 0.0120 s) 10 (20.0 W) (0.600 16 3 × = = × =- Pt b) eV. 1.90 J 10 3.05 λ 19 = × = =- hc hf c) . 10 94 . 3 16 × = hf Pt 38.3: a) Hz. 10 5.91 s J 10 6.63 eV) J 10 (1.60 eV) 10 (2.45 20 34 19 6 × = ⋅ × × × = = ⇒ =-- h E f hf E b) m. 10 5.08 Hz 10 5.91 s m 10 3.00 λ 13 20 8- × = × × = = f c c) λ is of the same magnitude as a nuclear radius. 38.4: hc W hc P hf P dN dE dt dE dt dN m) 10 (2.48 ) (12.0 λ ) ( ) ( 7- × = = = = sec. photons 10 1.50 19 × = 38.5: φ- = hf mv 2 max 2 1 J 10 3.04 eV J 10 (1.60 eV) (5.1 m 10 2.35 s J 10 (6.63 20 19 7 34 ) )---- × × × ⋅ × =- = c . s m 10 2.58 kg 10 9.11 J) 10 2(3.04 5 31 20 max × = × × = ⇒-- v 38.6: ×- × =- =- =- m 10 2.72 Hz 10 1.45 λ 7 15 c h hc hf hf E φ eV. 1.44 J 10 2.30 19 = × =- 38.7: a) Hz 10 5.0 λ 14 × = = c f b) Each photon has energy J. 10 .31 3 19- × = = hf E Source emits s photons 10 2.3 photons/s) 10 (3.31 ) s J (75 so s J 75 20 19 × = ×- 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 nm 700 λ = eV 1.77 J 10 1.6 1eV J 10 2.84 m) 10 (700 ) s m 10 (3.00 s) J 10 (6.626 λ 19 19 9 8 34 = × × = × × ⋅ × = = =---- hc hf φ 38.9: a) For a particle with mass, . 4 means 2 . 2 1 2 1 2 2 K K p p m p K = = = b) For a photon, . 2 means 2 . 1 2 1 2 E E p p pc E = = = 38.10: φ- = hf K max Use the information given for : find to nm 400 λ φ = J 10 3.204 eV) J 10 (1.602 eV) (1.10 m 10 400 ) s m 10 (2.998 s) J 10 (6.626 19 19 9 8 34 max---- × = ×- × × ⋅ × =- = K hf φ Now calculate : nm 300 λ for max = K eV 2.13 J 10 3.418 J 10 3.204 m 10 300 ) s m 10 (2.998 s) J 10 (6.626 19 19 9 8 34 max = × = ×- × × ⋅ × =- =---- φ hf K 38.11: a) The work function eV λ eV- =- = hc hf φ J. 10 7.53 V) (0.181 C) 10 (1.60 m 10 2.54 s) m 10 (3.00 s) J 10 (6.63 19 19 7 8 34---- × = ×- × × ⋅ × = ⇒ φ The threshold frequency implies φ φ hc hc hf th th = ⇒ ⇒ = th λ λ m. 10 2.64 J 10 7.53 ) s m 10 (3.00 s) J 10 (6.63 λ 7 19 8 34 th--- × = × × ⋅ × = ⇒ b) eV, 4.70 J 10 7.53 19 = × =- φ as found in part (a), and this is the value from Table 38.1. 38.12: a) From Eq. (38.4), V. 2.7 V 2.3 m) 10 (2.50 s) m 10 (3.00 s) eV 10 (4.136 λ 1 7 8 15 =- × × ⋅ × = - =-- φ hc e V b) The stopping potential, multiplied by the electron charge, is the maximum kinetic energy, 2.7 eV....
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Chapter 38 - 38.1 Hz 10 5.77 m 10 5.20 s m 10 3.00 14 7 8...

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