An alpha nucleus of energy `(1)/(2)m nu^(2)` bombards a heavy nucleus of charge `Ze` . Then the distance of closed approach for the alpha nucleus will be proportional to

Options

A. `(1)/(ze)`
B. `V^(2)`
C. `(1)/(m)`
D. `(1)/(v^(4))`

Answer

C

Explanation

`(1)/(2)mv^(2)=((ze)(2e))/(4piin_(0)d_(min))"then"d_(min) prop(1)/(m)`