chap38 - 38.1 f p E c h pc 3.00 108 m s 5.77 1014 Hz 5.20...

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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 0 c h hc hf hf E φ eV. 1.44 J 10 2.30 19 = × = -
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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 0 0 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.
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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. c) s. m 10 9.7 kg) 10 (9.11 V) (2.7 C) 10 2(1.60 2 2 5 31 19 × = × × = = = - - m eV m K v 38.13: a) ) s m 10 (3.00 ) s m kg 10 (8.24 8 28 × × = = - pc E eV 1.54 eV J 10 1.60 J 10 2.47 J 10 2.47 19 19 19 = × × = × = - - - b) m. 10 8.05 s) m kg 10 (8.24 s) J 10 (6.63 λ λ 7 28 34 - - - × = × × = = = p h h p This is infrared radiation. 38.14: a) The threshold frequency is found by setting V = 0 in Eq. (40.4), . 0 h f φ = b) eV. 3.34 10 5.35 m 10 72 . 3 λ 19 7 0 = × = × = = = - - hc hc hf φ 38.15: a) . eV 1.44 J 10 2.31 m 10 8.60 ) s m 10 (3.00 s) J 10 (6.63 λ 19 7 8 34 = × = × × × = = - - - hc E γ So the internal energy of the atom increases by = + - = eV 1.44 eV 6.52 to eV 44 . 1 E eV. 5.08 - b) eV. 2.96 J 10 4.74 m 10 4.20 s) m 10 (3.00 s) J 10 (6.63 λ 19 7 8 34 = × = × × × = = - - - hc E γ So the final internal energy of the atom decreases to eV. 5.64 eV 2.96 eV 2.68 - = - - = E 38.16: a) eV. 20 1 = - E 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.
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