An α particle and a proton are accelerated from rest through the...

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., $\dfrac{1}{2}m{v}^{2} = qV$, 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 $\dfrac{{p}_{\alpha}}{{p}_{p}} = \sqrt{\dfrac{{m}_{\alpha}{q}_{\alpha}}{{m}_{p}{q}_{p}}} = \sqrt{\dfrac{4}{1} \times \dfrac{2}{1}}$

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