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

Options

A. `(Qq)/(4pi epsi_(0)m^(2))`
B. `(Qq)/(2pi epsi_(0)mv^(2))`
C. `(Qq mv^(2))/(2)`
D. `(Qq)/(mv^(2))`

Answer

B

Explanation

According to the question,
For the distance of closest approach kinetic energy will totally be converted to potential energy
Hence, `" " (1)/(2) mv^(2) =(1)/(4piepsi_(0)) (Qq)/(r_(0)) rArr r_(0)=(Qq)/(2piepsi_(0)mv^(2))`