1. Use energy gained through a potential difference

When a charged particle is accelerated from rest through a potential difference $V$, the gained kinetic energy is

$K = qV$

Also,

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

So,

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

which gives

$p = \sqrt{2mqV}$

Thus, momentum is proportional to

$p \propto \sqrt{mq}$

for the same potential difference $V$.

2. Write mass and charge of each particle

• For a proton: $m_p = m, \quad q_p = e$

• For an $\alpha$-particle: $m_\alpha = 4m, \quad q_\alpha = 2e$

3. Find the ratio of momenta

$\frac{p_\alpha}{p_p} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$

Substitute values:

$\frac{p_\alpha}{p_p} = \sqrt{\frac{(4m)(2e)}{(m)(e)}}$

$\frac{p_\alpha}{p_p} = \sqrt{8} = 2\sqrt{2}$

Hence,

$p_\alpha : p_p = 2\sqrt{2} : 1$

4. Match with the options

This corresponds to Option B.

5. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

So, the derived answer agrees with the stored answer.