An α particle and a proton are accelerated from rest through the same potential difference. The ratio…

For $\alpha$ particle, $m_{\alpha }=4$ a.m.u and charge $q_{\alpha }=2e$.

For proton, $m_p=1$ a.m.u and charge $q_p=e$.

As the charged particle are moving through same potential, the kinetic energy acquired by the particle will be equal to electric potential energy i.e., $\frac{1}{2}m{v}^2=qV$, where, $V$ is potential.

Using the relation between momentum $\left(p\right)$ and kinetic energy, we have 

$p=\sqrt{2mE}=\sqrt{2mqV}$

Then, the ratio of linear momenta of two particles is $\frac{p_{\alpha }}{p_p}=\sqrt{\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}}=\sqrt{\frac{4}{1}\times \frac{2}{1}}$

$=\frac{2\sqrt{2}}{1}=2\sqrt{2}:1$