The ratio of momentum of electron and an alpha particle which are accelerated from rest through a potential difference of 100 volt is

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\sqrt{(2 m e / m)}

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\sqrt{(m e / m)}

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\sqrt{(m e / 2 m)}

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Solution

The correct option is D $\sqrt{(m e / 2 m)}$
Momentum of the particle $P = \sqrt{2 m K}$
where kinetic energy $K = q V$
$⟹ P = \sqrt{2 m q V}$
Or $P \propto \sqrt{m q}$ $(∵ V = c o n s t a n t)$
$⟹ \frac{P_{e}}{P_{α}} = \sqrt{\frac{m_{e} q_{e}}{m_{α} q_{α}}}$
Given : $m_{α} = m$ $q_{α} = 4 q_{e}$
$⟹$ $\frac{P_{e}}{P_{α}} = \sqrt{\frac{m_{e} q_{e}}{m 4 q_{e}}} = \sqrt{\frac{m_{e}}{2 m}}$

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

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