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48. We interpret the given 10 GeV = 10000 MeV to be the kinetic energy of the proton. Using Table 383
and Eq. 3849, we Fnd
γ
=
K
m
p
c
2
+1=
10000 MeV
938 MeV
+1=11
.
66
,
and (from Eq. 388)
β
=
r
1
−
1
γ
2
=0
.
9963
.
Therefore, using the equation introduced in problem 46, we obtain
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 Fall '08
 SPRUNGER
 Physics, Energy, Kinetic Energy

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