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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
Chhattisgarh PMTChhattisgarh PMT 2005
Solution:
The relation between the rest energy and moving energy is
$E=\frac{E_{0}}{\sqrt{1-v^{2} / c^{2}}}$
$\because v=0.5\,c$
$\therefore E=\frac{E_{0}}{\sqrt{1-0.25}}$
$=\frac{E_{0}}{\sqrt{0.75}}$
Change in energy $\Delta E=E-E_{0}$
$=\frac{E_{0}}{\sqrt{0.75}}-E_{0}$
$=E_{0}\left(\frac{1}{\sqrt{0.75}}-1\right)$
$=0.511\left(\frac{1}{\sqrt{0.75}}-1\right)$
$=0.079\, MeV$