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34. From Eq. 3849,
γ
=(
K/mc
2
) + 1, and from Eq. 388, the speed parameter is
β
=
p
1
−
(1
/γ
)
2
.
(a) Table 383 gives
m
e
c
2
= 511 keV = 0
.
511 MeV, so the Lorentz factor is
γ
=
10
.
0MeV
0
.
511 MeV
+1=20
.
57
,
and the speed parameter is
β
=
s
1
−
1
(20
.
57)
2
=0
.
9988
.
(b) Table 383 gives
m
p
c
2
= 938 MeV, so the Lorentz factor is
γ
=1+10
.
0MeV
/
938 MeV = 1
.
01, and
the speed parameter is
β
=
r
1
−
1
1
.
01
2
=0
.
145
.
(c) If we refer to the data shown in problem 36, we ±nd
m
α
=4
.
0026 u, which (using Eq. 3843) implies
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 Fall '08
 SPRUNGER
 Physics

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