A proton and an alpha particle accelerated through different potential differences have the same de-Broglie wavelength. The ratio of their respective accelerating potentials is

8 : 1

The correct answer is Option (1) → 8 : 1

de-Broglie wavelength: $\lambda = \frac{h}{p}$

Momentum of a particle accelerated through potential $V$: $p = \sqrt{2 m q V}$

Given that proton ($m_p, q_p = e$) and alpha particle ($m_\alpha = 4 m_p, q_\alpha = 2e$) have the same $\lambda$, then:

$\lambda_p = \lambda_\alpha \text{ implies }\frac{h}{\sqrt{2 m_p e V_p}} = \frac{h}{\sqrt{2 (4 m_p) (2e) V_\alpha}}$

Simplify: