Q. The rest energy of an electron is $0.511\ MeV$. The electron is accelerated from rest to a velocity $0.5\ c$. The change in its energy will be
A) $0.026\ MeV$
B) $0.051\ MeV$
C) $0.079\ MeV$
D) $0.105\ MeV$
Explanation
(c) $\mathrm{\Delta}E = mc^{2} - m_{0}c^{2} = \frac{m_{0}c^{2}}{\sqrt{1 - \left( v^{2}/c^{2} \right)}} - m_{0}c^{2}$
$= m_{0}c^{2}\left( \frac{1}{\sqrt{1 - \left( v^{2}/c^{2} \right)}} - 1 \right) = 0.511\left( \frac{1}{\sqrt{0.75}} - 1 \right)$
$= 0.079\ MeV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.