1. Use de-Broglie relation

For a particle accelerated from rest through a potential difference $V$:

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

So,

$p = \sqrt{2MqV}$

The de-Broglie wavelength is

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2MqV}}$

Thus,

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

for the same accelerating potential $V$.

2. For proton

For a proton:

• mass $= m_p$
• charge $= e$

Hence,

$\lambda_p = \frac{h}{\sqrt{2m_p eV}}$

3. For $\alpha$-particle

For an $\alpha$-particle:

• mass $= 4m_p$
• charge $= 2e$

Hence,

$\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)V}}$ $\lambda_\alpha = \frac{h}{\sqrt{16m_p eV}}$

4. Take the ratio

$\frac{\lambda_\alpha}{\lambda_p} = \frac{h/\sqrt{16m_p eV}}{h/\sqrt{2m_p eV}}$

Or more directly using $\lambda \propto 1/\sqrt{Mq}$:

$\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p\,e}{(4m_p)(2e)}}$

$\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{1}{8}} = \frac{1}{\sqrt{8}}$

Given,

$\frac{\lambda_\alpha}{\lambda_p} = \frac{1}{\sqrt{m}}$

So,

$m = 8$

5. Check options

• A: $2$ ❌
• B: $16$ ❌
• C: $8$ ✅
• D: $4$ ❌

Therefore, the correct answer is Option C.