An `alpha`-particle accelerated through V volt is fired towards a nucleus . Its distance of closest approach is r. If a proton acceleration throught the same potential is fired towards the same nucleus, the distance of closest approach of proton will be xr. Find value of x.

Options

A. r
B. 2r
C. `(r)/(2)`
D. `(r)/(4)`

Answer

A

Explanation

Decreases in kinetic energy = increases in potential energy
`:. " " qV=(1)/(4piepsi_(0))*(Q*q)/(r)`
`:. " " r=(Q)/((4piepsi_(0))V)`