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Offline Centres

The closest distance of approach of an $α - p a r t i c l e$ travelling with a velocity V towards a stationary nucleus is 'd'. For the closest distance to become d/4 towards a stationary nucleus of the same charge the velocity of projection of the $α - p a r t i c l e$ has to be

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$2 V$

$\frac{3 V}{2}$

$\frac{V}{56}$

$6 V$

answer is B.

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Detailed Solution

$\frac{1}{2} m v^{2} = \frac{1}{4 {πε}_{0}} \frac{q_{1} q_{2}}{d} o r v \propto \frac{1}{\sqrt{d}} \frac{v_{2}}{v_{1}} = \sqrt{\frac{d_{1}}{d_{2}}} = \sqrt{\frac{d_{1}}{\frac{d_{1}}{4}}} = 2$

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