Q 12-11-079JEE MainJEE Main 2023 (29 Jan, Shift 2)Easy
The ratio of de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\dfrac{1}{\sqrt m}$. The value of $m$ is
Answer: (C) 8
$\lambda=\dfrac{h}{\sqrt{2mqV}}$, so for the same $V$:
$$\frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_pq_p}{m_\alpha q_\alpha}}=\sqrt{\frac{1}{4\times2}}=\frac{1}{\sqrt8}$$
So $m=8$.
Solution by Sreeraj P, M.Sc Physics