The ratio of momenta of an electron and an $α$ -particle which are accelerated from rest by a potential difference of 100 volt is

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$\sqrt{\frac{2 m_{e}}{m_{α}}}$

$\sqrt{\frac{m_{e}}{m_{α}}}$

$\sqrt{\frac{m_{e}}{2 m_{α}}}$

answer is D.

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Detailed Solution

$Momentum p = \sqrt{2 mK}; where K = kinetic energy = Q . V \Rightarrow p = \sqrt{2 mQV} \Rightarrow p \propto \sqrt{mQ} \Rightarrow \frac{p_{e}}{p_{α}} = \sqrt{\frac{m_{e} Q_{e}}{m_{α} Q_{α}}} = \sqrt{\frac{m_{e}}{2 m_{α}}}$

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