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

From energy conservation, kinetic energy is converted into potential energy
`1/2mv^(2) =(k(q_(1).q_(2))/r_(0))`
`1/2mv^(2)=(k xx (2e)(Ze))/r_(0)`
`r_(0)=(4kZe^(2))/(mv^(2))`
`r_(0) propto 1/v^(2)` and `r_(0) propto 1/m`