1086_PartUniversity Physics Solution

1086_PartUniversity Physics Solution - The Wave Nature of...

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The Wave Nature of Particles 39-11 (b) 34 7 27 35 distance 0.80 m 7.3 10 s(1 y/3.156 10 s) 2.3 10 y velocity 1.1 10 m/s t == = × × = × × Since you walk through doorways much more quickly than this, you will not experience diffraction effects. EVALUATE: A 1 kg object moving at 1 m/s has a de Broglie wavelength 34 6.6 10 m, λ which is exceedingly small. An object like you has a very, very small at ordinary speeds and does not exhibit wavelike properties. 39.56. (a) 19 7 2.58 eV 4.13 10 J, with a wavelength of 4.82 10 m 482 nm hc E E −− × = = × = (b) 34 28 9 7 (6.63 10 J s) 6.43 10 J 4.02 10 eV. 22 ( 1 . 6 4 1 0 s ) h E π t π ×⋅ Δ= = = × = × Δ× (c) , so ( ) 0, and Eh c E E EE λλ + Δ = Δ = Δ , so 28 71 6 7 19 6.43 10 J ( 4 . 8 21 0 m ) 7 . 5 01 0 m 7 . 5 0 nm . 4.13 10 J ⎛⎞ × Δ= Δ = × = × = × ⎜⎟ × ⎝⎠ 39.57. IDENTIFY: The electrons behave as waves whose wavelength is equal to the de Broglie wavelength. SET UP: The de Broglie wavelength is = h/mv , and the energy of a photon is E = hf = hc/ . EXECUTE: (a) Use the de Broglie wavelength to find the speed of the electron. () ( ) 34 31 9 6.626 10 J s 9.11 10 kg 1.00 10 m h v m ×× = 7.27 × 10 5 m/s which is much less than the speed of light, so it is nonrelativistic. (b) Energy conservation gives eV = ½ mv 2 . V = mv 2 /2 e = (9.11 × 10 –31 kg)(7.27 × 10 5 m/s) 2 /[2(1.60 × 10 –19 C)] = 1.51 V (c) K = eV = e (1.51 V) = 1.51 eV, which is about ¼ the potential energy of the NaCl crystal, so the electron would not be too damaging. (d) E = hc/ = (4.136 × 10 –15 eV s)(3.00 × 10 8 m/s)/(1.00 × 10 –9 m) = 1240 eV which would certainly destroy the molecules under study. EVALUATE: As we have seen in Problems 39.10 and 39.43, when a particle and a photon have the same wavelength, the photon has much more energy. 39.58. sin sin , and ( ) ( 2 ), and so θθ hp h mE ′′ = arcsin sin 2 h mE ′ = . 34 11 31 3 19 (6.63 10 J s)sin 35.8 arcsin 20.9 (3.00 10 m) 2(9.11 10 kg)(4.50 10 )(1.60 10 J eV) θ + ° ° × × 39.59. (a) The maxima occur when 2s i n d θ m = as described in Section 38.7. (b) 2 hh p mE . 34 10 37 19 (6.63 10 J s) 1.46 10 m 0.146 nm 2(9.11 10 kg)(71.0 eV) 1.60 10 J/eV × = . 1 sin (Note: This 2 m θ m d = is the order of the maximum, not the mass.) 10 1 11 (1)(1.46 10 m) sin 53.3 . 2(9.10 10 m) × × (c) The work function of the metal acts like an attractive potential increasing the kinetic energy of incoming electrons by . e φ An increase in kinetic energy is an increase in momentum that leads to a smaller wavelength. A smaller wavelength gives a smaller angle (see part (b)).
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1086_PartUniversity Physics Solution - The Wave Nature of...

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