Q 12-11-028NEETJEE MainTop questionMedium
A proton and an alpha particle are accelerated through the same potential difference. The ratio of their de Broglie wavelengths $\lambda_p : \lambda_\alpha$ is
Answer: (D) $2\sqrt{2} : 1$
$\lambda = \dfrac{h}{\sqrt{2mqV}}$. The alpha particle has $4$ times the mass and $2$ times the charge:
$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m \times 2e}{m \times e}} = \sqrt{8} = 2\sqrt{2}$$
Solution by Sreeraj P, M.Sc Physics