ch38-p052 - 52(a We use the value hc = 1240nm eV E photon =...

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52. (a) We use the value 1240nm eV hc =⋅ : E hc photon nm eV nm keV == = λ 1240 100 124 . .. (b) For the electron, we have K p m h m hc mc ee e = = F H G I K J = 2 22 2 2 2 1 2 0511 1240 150 // . λλ b g b g b g MeV eV nm eV. (c) In this case, we find 9 photon 6 1240nm eV 1.24 10 eV 1.24GeV. 1.00 10 nm E × = × (d) For the electron (recognizing that 1240 eV·nm = 1240 MeV·fm) K p c m cm c h c m c =+ = + = F H G I K J +− × 2 2 2 2 2 2 2 2 2 1240 0511 ch bg b g / . λ MeV fm fm MeV MeV = 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.

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