Question H3.4: An electron of rest energy 0.511 MeV is accelerated through ...
An electron of rest energy 0.511 MeV is accelerated through a potential difference of 5.0 MV in a lab.
(a) What is its total energy with respect to the lab?
(b) What is its speed with respect to the lab?
(a) The total energy of the electron will increase by
q V=1 e \times 5.0 \times 10^{6} volt =5.0 MeV
and so the total energy is
\begin{aligned}E &=m_{0} c^{2}+q V=0.511 MeV +5.0 MeV \\&=5.511 MeV\end{aligned}
(b) We know that
E=\gamma m_{0} c^{2}5.511=\gamma \times 0.511
\begin{aligned}\gamma &=\frac{5.511}{0.511} \\&=10.785\end{aligned}
Since \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} it follows that
10.785=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\sqrt{1-\frac{v^{2}}{c^{2}}}=\frac{1}{10.875}(=0.0927)
1-\frac{v^{2}}{c^{2}}=0.008597
v = 0.996 c
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