The ratio of wavelengths of proton and deuteron accelerated by...

Kinetic energy gained by a charged particle accelerated by a potential $V$ is $qV$

$KE = qV$

Using relation between momentum and kinetic energy

$⇒ \dfrac{{p}^{2}}{2m} = qV ⇒ p = \sqrt{2mqV}$

We know that momentum of charged particle is $p = \dfrac{h}{\lambda}$.

Thus, $\lambda = \dfrac{h}{\sqrt{2mqV}}$

Now, $\dfrac{{\lambda}_{p}}{{\lambda}_{d}} = \sqrt{\dfrac{{m}_{d}{V}_{d}}{{m}_{p}{V}_{p}}}$

$⇒ \dfrac{1}{\sqrt{2}} = \sqrt{\dfrac{2{V}_{d}}{{V}_{p}}} ⇒ \dfrac{{V}_{p}}{{V}_{d}} = 4$