Q 12-11-093JEE MainJEE Main 2023 (12 Apr, Shift 1)Easy
A proton and an $\alpha$-particle are accelerated from rest by $2\ \text{V}$ and $4\ \text{V}$ potentials, respectively. The ratio of their de-Broglie wavelength is
Answer: (C) $4:1$
$\lambda=\dfrac h{\sqrt{2mqV}}$. Proton: $mq V=m\cdot e\cdot2$; alpha: $4m\cdot2e\cdot4=32me$.
$$\frac{\lambda_p}{\lambda_\alpha}=\sqrt{\frac{32}{2}}=4$$
Solution by Sreeraj P, M.Sc Physics