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

`(1)/(2)mv^(2)=(1)/(4 pi epsilon_(0)).(Ze.2e)/(r )`
`r=(Ze^(2))/(pi epsilon_(0)mv^(2))`
`r prop (1)/(m)`