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 .