Question

The distance of the closest approach of an $\alpha -$particle fired towards a nucleus with momentum $p$ is $r$. If the momentum of the $\alpha -$particle is $2p$, the corresponding distance of the closest approach is -

• A. $r/2$
• B. $2r$
• C. $4r$
• D. $r/4$

Answer 1

The correct option is D $r/4$

From law of conservation of mechanical energy,

$K{E}_i+P{E}_i=K{E}_f+P{E}_f$

$\Rightarrow KE+0=0+\frac{kq\left(2e\right)}{r}$

$\Rightarrow r\propto \frac{1}{KE}$

$\Rightarrow r\propto \frac{1}{p^2}$

Therefore, if momentum is doubled, the distance of the closest approach will become one-fourth.

Hence, option $\left(D\right)$ is the correct answer.