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Q.

The closest distance of approach of an $α$ - particle travelling with a velocity V towards a stationary nucleus is ‘d’. For the closest distance to become d/3 towards a stationary nucleus of double the charge the velocity of projection of the $α$ - particle has to be

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a

3V/2

b

6V

c

V/56

d

$\sqrt{6} V$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} m_{α} V_{α}^{2} = \frac{1}{4 {πε}_{0}} . \frac{q_{α} \times q}{d} \\ V_{α}^{2} \propto \frac{q}{d} \Rightarrow V_{2}^{2} = (\frac{q_{2}}{q_{1}}) (\frac{d_{1}}{d_{2}}) V_{1}^{2} \Rightarrow V_{2} = \sqrt{6} V_{1} \end{array}$

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