7. From Eq. 3845, the Lorentz factor would be
γ
=
E
mc
2
=
1
.
5
×
10
6
eV
20 eV
= 75000
.
Solving Eq. 388 for the speed, we ±nd
γ
=
1
p
1
−
(
v/c
)
2
=
⇒
v
=
c
r
1
−
1
γ
2
which implies that the di²erence between
v
and
c
is
c
−
v
=
c
µ
1
−
r
1
−
1
γ
2
¶
≈
c
µ
1
−
µ
1
−
1
2
γ
2
+
···
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 Fall '08
 SPRUNGER
 Physics

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