49. Let the distance in question be r. The initial kinetic energy of the electron is K
i
=
1
2
m
e
v
2
i
, where
v
i
= 3.2
× 10
5
m/s. As the speed doubles, K becomes 4K
i
. Thus
∆U =
−e
2
4πε
0
r
=
−∆K = −(4K
i
− K
i
) =
−3K
i
=
−
3
2
m
e
v
2
i
,
or
r
=
2e
2
3(4πε
0
)m
e
v
2
i
=
2(1.6
× 10
−19
C)
2
8.99
× 10
9 N
·m
2
C
2
3(9.11
× 10
−19
kg)(3.2
× 10
5
m/s)
2
=
1.6
× 10
−9
m .