Q 12-11-128JEE MainJEE Main 2021 (22 Jul, Shift 1)Easy
An electron of mass $m_e$ and a proton of mass $m_p$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is
Answer: (C) $\sqrt{\frac{m_p}{m_e}}$
$\lambda = \dfrac{h}{\sqrt{2mqV}}$ with equal $q$ and $V$, so $\dfrac{\lambda_e}{\lambda_p} = \sqrt{\dfrac{m_p}{m_e}}$.
Solution by Sreeraj P, M.Sc Physics