Q. An electron, a proton and an alpha particle have KE of 16$E$, $4E$ and $E$ respectively. What is the qualitative order of their de-Broglie wavelengths?
A) $\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$
B) $\lambda_{p} = \lambda_{\alpha} > \lambda_{e}$
C) $\lambda_{p} < \lambda_{e} < \lambda_{\alpha}$
D) $\lambda_{\alpha} < \lambda_{e} \approx \lambda_{p}$
Explanation
(a)
$\lambda = \frac{h}{mv},ie,\lambda \propto \frac{1}{\sqrt{mE}}$ and $m > \ > \ > E$
Thus, correct order is $\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.