1. Use de Broglie relation

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

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

So momentum is

$p = \sqrt{2mqV}$

Hence de Broglie wavelength,

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

Therefore,

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

since $h, V$ are same for both particles.

2. Write values for proton and alpha particle

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

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

3. Find the ratio

$\frac{\lambda_p}{\lambda_\alpha} = \frac{h/\sqrt{2m_peV}}{h/\sqrt{2m_\alpha q_\alpha V}} = \sqrt{\frac{m_\alpha q_\alpha}{m_p q_p}}$

Substitute values:

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

$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{8} = 2\sqrt{2} \approx 2.8$

4. Match with option

So the correct option is:

$\boxed{2.8}$

which is Option B.