Q 12-11-095JEE MainJEE Main 2023 (6 Apr, Shift 1)Medium
The kinetic energy of an electron, $\alpha$-particle and a proton are given as $4K$, $2K$ and $K$ respectively. The de-Broglie wavelength associated with electron ($\lambda_e$), $\alpha$-particle ($\lambda_\alpha$) and the proton ($\lambda_p$) are as follows:
Answer: (B) $\lambda_\alpha<\lambda_p<\lambda_e$
$\lambda=\dfrac h{\sqrt{2mK}}$. Compare $mK$: electron $4m_eK$ (tiny), proton $m_pK$, alpha $4m_p\times2K=8m_pK$.
Larger $mK$ means shorter $\lambda$: $\lambda_\alpha<\lambda_p<\lambda_e$.
Solution by Sreeraj P, M.Sc Physics