When an a-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 v as:

Options

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

Answer

C

Explanation

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy Kinetic energy
Kinetic energy K.E `1/2 mv^2`
potential energy `P.E = (KQq )/( r ) `
`1/2 mv^2 = (KQ q )/( r ) implies r prop (1)/(v^2)`