When an `alpha-`particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze' its distance of closest approach from the nucleus depends on `m` as :

Options

A. `(1)/(m)`
B. `(1)/(sqrtm)`
C. `(1)/(m^(2))`
D. m

Answer

A

Explanation

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy
`rArr (1)/(2) mv^(2)= (KQq)/(d) rArr d prop (1)/(m)`