The distance of closest approach of an $α -$ particle projected towards a nucleus with momentum $p$ is $r$ . What will be the distance of closest approach, when the momentum of projected $α -$ particle is $2 p$ ?

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$2 r$

$4 r$

$\frac{r}{2}$

$\frac{r}{4}$

answer is D.

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Detailed Solution

On doubling the momentum K.E. will become 4 times

$K . E . = \frac{K_{q_{1} q_{2}}}{r} r \propto \frac{1}{K E} s o r_{i} {(K E)}_{i} = r_{f} {(K E)}_{f} \frac{r_{f}}{r_{i}} = \frac{{(K E)}_{i}}{{(K E)}_{f}} = \frac{1}{4} r_{f} = \frac{r}{4}$

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