1. Use de Broglie wavelength formula

For a particle accelerated through a potential difference $V$,

$\lambda = \frac{h}{p}$

and the kinetic energy gained is

$qV = \frac{p^2}{2m}$

So,

$p = \sqrt{2mqV}$

Hence,

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

Thus,

$\lambda \propto \frac{1}{\sqrt{mV}}$

for particles with the same charge magnitude.

2. Apply to proton and deuteron

Both proton and deuteron are singly charged, so $q=e$ for both.

Let

• proton mass $= m_p$
• deuteron mass $= 2m_p$

Then,

$\lambda_p \propto \frac{1}{\sqrt{m_p V_p}}$

$\lambda_d \propto \frac{1}{\sqrt{(2m_p) V_d}}$

Therefore,

$\frac{\lambda_p}{\lambda_d} = \sqrt{\frac{2m_p V_d}{m_p V_p}} = \sqrt{\frac{2V_d}{V_p}}$

3. Use the given ratio

Given,

$\lambda_p : \lambda_d = 1 : \sqrt{2}$

So,

$\frac{\lambda_p}{\lambda_d} = \frac{1}{\sqrt{2}}$

Now,

$\sqrt{\frac{2V_d}{V_p}} = \frac{1}{\sqrt{2}}$

Squaring both sides,

$\frac{2V_d}{V_p} = \frac{1}{2}$

$4V_d = V_p$

Thus,

$V_p : V_d = 4 : 1$

4. Check options

• A: $1:1$ ❌
• B: $\sqrt{2}:1$ ❌
• C: $2:1$ ❌
• D: $4:1$ ✅

So the correct option is D.