The distance of closest approach of an `alpha`-particle fired at nucleus with momentum p is d. The distence of closest approach when the `alpha`-particle is fired at same nucleus with momentum 3p will be

Options

A. `(r_(0))/(2)`
B. `4r_(0)`
C. `(r_(0))/(4)`
D. `2r_(0)`

Answer

A

Explanation

For closest approach,
`k= (1)/(4pi epsi_(0)) (e^(2))/(r_(0)) rArr r_(0) prop (1)/(k)`
`rArr (r_(0))/(r._(0)) = (k.)/(k)= (2k)/(k) rArr r._(0) = (r_(0))/(2)`