If and `alpha`-paricle of mass m, charged q and velocity v is incident on a nucleus charge Q and mass m, then the distance of closest approach is

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))`