# hw23 - PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS...

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Unformatted text preview: PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 23. 38.40, 38.60, 38.62; Q39.02. Q39.04 38.40. a) p h = . 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 = 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 . 3 s)( J 10 63 . 6 ( / 19 3 34 34--- = = = = E hc c E h p h = 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 = 9.04 10-13 m = 904 fm 38.60. a) m 10 . 1 m/s) 10 s)(3.0 eV 10 (4.14 12 8 15-- = = 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 = 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 from that find the energy lost by the photon....
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## This note was uploaded on 04/07/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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hw23 - PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS...

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