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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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6 V

$\frac{3 V}{2}$

$\frac{V}{56}$

$\sqrt{6} V$

answer is B.

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Detailed Solution

The kinetic energy of the $α$ -particle will be converted into the potential energy of the system,

$\begin{array}{l} \frac{1}{2} m_{α} V_{α}^{2} = \frac{1}{4 {πε}_{0}} \cdot \frac{q_{α} \times q}{d} \\ V_{α}^{2} \propto \frac{q}{d} \end{array}$

$\Rightarrow V_{2}^{2} = (\frac{r_{2}}{q_{1}}) (\frac{d_{1}}{d_{2}}) V_{1}^{2} \Rightarrow V_{2} = \sqrt{6} V$

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