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) $2\sqrt{2} : 1$
Concept. $\lambda = \dfrac{h}{\sqrt{2mqV}}$.
Step 1. Ratio. Alpha: $4m$, $2e$:
$$\frac{\lambda_p}{\lambda_\alpha} = \sqrt{\frac{4m \times 2e}{m \times e}} = 2\sqrt2$$
Answer. $2\sqrt2 : 1$ (Option D).
Solution by Sreeraj P, M.Sc Physics