52. (a) We use the value 1240nm eVhc=⋅: Ehcphotonnm eVnmkeV==⋅=λ1240100124...(b) For the electron, we have Kpmhmhcmceee==⋅FHGIKJ=22222212 05111240150//.λλbgbgbgMeVeV nmeV. (c) In this case, we find 9photon61240nm eV1.24 10 eV 1.24GeV.1.00 10 nmE−⋅×=×(d) For the electron (recognizing that 1240 eV·nm = 1240 MeV·fm) Kpcmcmchcmc=+−=+−=⋅FHGIKJ+−×22222222212400511chbgbg/.λMeV fmfmMeVMeV= 1.24 10 MeV = 1.24GeV.
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.