ch37-p081 - 81. (a) For a proton (using Table 37-3), we...

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81. (a) For a proton (using Table 37-3), we have 2 2 938MeV 6.65GeV 1 (0.990) p Em c γ == = which gives 2 6.65GeV 938MeV 5.71GeV p KEm c =− = = . (b) From part (a), 6.65GeV E = . (c) Similarly, we have 2 2 (938MeV)(0.990)/ () / 6 . 5 8 G e V / 1 (0.990) pp c p mv mc c c γγ β = = . (d) For an electron, we have 2 2 0.511MeV
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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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