"An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:"

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"

We know,

Momentum $(p) = \sqrt {2m{E_k}}$

and ${E_k} = q{V_{acc}}$

$\therefore$ $p = \sqrt {2mq\,{V_{acc}}}$

Both $\alpha$ particle and proton are passed through same potential difference.

$\therefore$ ${\left( {{V_{acc}}} \right)_\alpha } = {\left( {{V_{acc}}} \right)_p} = v$

$\therefore$ ${p_\alpha } = \sqrt {2{m_\alpha }{q_\alpha }v}$

${p_p} = \sqrt {2{m_p}{q_p}v}$

$\therefore$ $${{{p_\alpha }} \over {{p_p}}} = \sqrt {{{{m_\alpha }{q_\alpha }} \over {{m_p}{q_p}}}} $$

$= \sqrt {{{4{m_p} \times 2e} \over {{m_p} \times e}}}$

$= \sqrt {{8 \over 1}}$

$= {{2\sqrt 2 } \over 1}$

"

An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

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An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

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