p38_033 - 33. (a) Using K = me c2 ( 1) (Eq. 38-49) and me...

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33. (a) Using K = m e c 2 ( γ 1) (Eq. 38-49) and m e c 2 = 511 keV = 0 . 511 MeV (Table 38-3), we obtain γ = K m e c 2 +1= 1 . 00 keV 511 keV +1=1 . 00196 . Therefore, the speed parameter is β = r 1 1 γ 2 = r 1 1 1 . 00196 2 =0 . 0625 . (b) We could Frst Fnd β and then Fnd γ , as illustrated here: With K =1 . 00 MeV, we Fnd β = s 1 µ 1 . 00 MeV 0 . 511 MeV +1 2 =0 . 941 and γ =1 / p 1 β 2 =2 . 96. (c) ±inally, K = 1000 MeV, so
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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