Q 12-11-105JEE MainJEE Main 2022 (25 Jun, Shift 2)Medium
A proton, a neutron, an electron and an $\alpha$-particle have the same energy. If $\lambda_p$, $\lambda_n$, $\lambda_e$ and $\lambda_\alpha$ are the de Broglie wavelengths of proton, neutron, electron and $\alpha$-particle respectively, then choose the correct relation from the following
Answer: (B) $\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$
For the same kinetic energy $K$, $\lambda = \dfrac{h}{\sqrt{2mK}}\propto\dfrac1{\sqrt m}$.
Masses: $m_\alpha > m_n > m_p > m_e$ (the neutron is slightly heavier than the proton), so $\lambda_\alpha < \lambda_n < \lambda_p < \lambda_e$.
Solution by Sreeraj P, M.Sc Physics