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49. We interpret the given 2
.
50 MeV = 2500 keV to be the kinetic energy of the electron. Using Table 383
and Eq. 3849, we Fnd
γ
=
K
m
e
c
2
+1=
2500 keV
511 keV
+1=5
.
892
,
and (from Eq. 388)
β
=
r
1
−
1
γ
2
=0
.
9855
.
Therefore, using the equation introduced in problem 46 (with “
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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.
 Fall '08
 SPRUNGER
 Physics, Energy, Kinetic Energy

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