hw23 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 23...

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 23. 38.40, 38.60, 38.62, 38.73; Q39.02. Q39.04 38.40. a) p h = λ . To get p , use m p m v m mv K 2 2 2 2 2 2 2 1 = = = , so mK p 2 = . Therefore, J/eV) 10 eV)(1.6 10 kg)(1.00 10 11 . 9 ( 2 s J 10 63 . 6 2 19 3 31 34 - - - × × × × = = mK h λ λ = 3.88 ×10 -11 m = 38.8 pm b) For the photon, c E p = , so J/eV) 10 eV)(1.6 10 (1.00 s) / m 10 0 . 3 s)( J 10 63 . 6 ( / 19 3 34 34 - - - × × × × = = = = E hc c E h p h λ λ = 1.24 ×10 -9 m = 1.24 nm c) J/eV) 10 eV)(1.6 10 kg)(1.00 10 68 . 1 ( 2 s J 10 63 . 6 2 19 3 27 34 - - - × × × × = = mK h λ λ = 9.04 ×10 -13 m = 904 fm 38.60. a) m 10 0 . 0 1 m/s) 10 s)(3.0 eV 10 (4.14 12 8 15 - - × × × = = λ hc E E = 1.24 ×10 5 eV = 124 keV b) This is a Compton collision. In a head-on collision, the photon bounces back in the opposite direction, so φ = 180°. Calculate the wavelength change of the photon, and from that find the energy lost by the photon. The energy lost by the photon is gained by the electron. m 10 85 . 4 ) 180 cos 1 ( m/s) 10 kg)(3.0 10 (9.11 s J 10 6.63 ) cos 1 ( 12 8 31 34 - - - × = ° - × × × = - = φ λ c m h e Then, λ λ λ hc hc E photon - + = , so ) ( ) ( λ λ λ λ λ λ λ λ λ + - + = + - = hc hc hc E electron , and
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