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/m
B. `1//v^(4)`
C. 1/Ze
D. `v^(2)`

Answer

A

Explanation

Distance of closest approach KE = PE
`rArr (1)/(2) mv^(2) = (Ze^(2))/(4pi epsi_(0) r_(0))`
`rArr r_(0)= (Ze^(2))/(2mv^(2) pi epsi_(0)) rArr r_(0) prop (1)/(m)`