The ratio of wavelengths of proton and deuteron accelerated by potential $V_p$ and $V_d$ is $1:\sqrt{2}$. Then, the ratio of $V_p$ to $V_d$ will be

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

$KE=qV$

Using relation between momentum and kinetic energy

$\Rightarrow \frac{p^2}{2 \mathrm{m}}=qV\Rightarrow p=\sqrt{2\mathrm{mqV}}$

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

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

Now, $\frac{{\lambda }_p}{{\lambda }_d}=\sqrt{\frac{m_d{V}_d}{m_p{V}_p}}$

$$\begin{gathered} \Rightarrow \frac{1}{\sqrt{2}}=\sqrt{\frac{2{\mathrm{V}}_{\mathrm{d}}}{V_p}} \\ \Rightarrow \frac{V_p}{V_d}=4 \end{gathered}$$