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

Answer

B

Explanation

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