Q. If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
A) Zero
B) Less than that of a proton
C) More than that of a proton
D) Equal to that of a proton
Explanation
(c) Given, $\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mKE}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ KE = \frac{1}{2m\lambda^{2}}$
As $\lambda$ is same for both electron and photon
So, $KE \propto \frac{1}{m}$
Hence, kinetic energy will be maximum for particle with lesser mass, $ie,$ electron.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.