Q 12-11-143JEE MainJEE Main 2020 (6 Sep, Shift 1)Easy
An electron, a doubly ionized helium ion ($\text{He}^{++}$) and a proton are having the same kinetic energy. The relation between their respective de Broglie wavelengths $\lambda_e$, $\lambda_{\text{He}^{++}}$ and $\lambda_p$ is:
Answer: (C) $\lambda_e > \lambda_p > \lambda_{\text{He}^{++}}$
$\lambda = \dfrac{h}{\sqrt{2mK}}$, so for equal kinetic energy $\lambda \propto \dfrac{1}{\sqrt m}$.
$m_e < m_p < m_{\text{He}}$, hence $\lambda_e > \lambda_p > \lambda_{\text{He}^{++}}$.
Solution by Sreeraj P, M.Sc Physics