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)/(4piepsi_(0)m^(2)v^(2))`
B. `(Qq)/(2piepsi_(0)mv^(2))`
C. `(Qqmv^(2))/(2)`
D. `(Qq)/(mv^(2))`

Answer

B

Explanation

Distance of closest approach,
`r_(0)=(Zke^(2))/(KE)=(qQ)/(4piepsi_(0)1//2mv^(2))=(qQ)/(2piepsi_(0)mv^(2))`