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

An `alpha`-particle of mass `m` possesses initial velocity `v`, when it is at a large distance form the nucleus of an atom having atomic `Z`. At the distance of closest approach, the kinetic energy of `alpha`-particle is completely convered into potential energy. Mathematical,
`(1)/(2)mv^(2) = (1)/(4pi epsilon_(0)) ((2e)(Ze))/(r_(0))`
`r_(0) = (1)/(4pi epsilon_(0)) (2Ze^(2))/((1)/(2)mv^(2))`