Q 12-11-072JEE MainJEE Main 2024 (9 Apr, Shift 1)Easy
A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:
Answer: (A) $\lambda_\alpha < \lambda_p < \lambda_e$
$\lambda = \dfrac{h}{\sqrt{2mK}}$, so for equal kinetic energy $\lambda \propto \dfrac{1}{\sqrt m}$. Since $m_\alpha > m_p > m_e$:
$$\lambda_\alpha < \lambda_p < \lambda_e$$
Solution by Sreeraj P, M.Sc Physics