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))`

Explanation

At the distance of closest approach d,
Kinetic energy = Potential energy
`(1)/(2)mv^(2)=(1)/(4piepsilon_(0))((2e)(Ze))/(d)`
where, Ze = charge of target nucleus,
2e = charge of alpha nucleus
`(1)/(2)mv^(2)=` kinetic energy of alpha nucleus of mass m moving with velocity v
or `d=(2Ze^(2))/(4piepsilon_(0)((1)/(2)mv^(2))) therefore dprop (1)/(m)`