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

Answer

D

Explanation

Assuming that there is no loss of energy, at the distance of closest approach form the nucleus.
K.E. of `alpha` particle =P.E. of `alpha` particle
`1/2mv^(2)=(Ze)/(4pi in_(0)r_(0)) (2e)`
`r_(0) = (2Ze^(2))/(4pi in_(0)mv^(2)//2)=(Ze)/(pi in_(0)mv^(2))`
Clearly, `r_(0) prop1/m`