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.

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

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

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

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

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

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

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