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Question

# An electron is accelerated so that its kinetic energy is greater than its rest energy $m_{0} c^{2}$ by a factor of 999. What is the speed of the electron in each case?

Solution

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From the condition given in the task we get:

\begin{align*} (\gamma -1 )mc^2&=999mc^2\rightarrow \gamma=1000 \end{align*}

Solving for $v$ we get:

\begin{align*} \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}&=6\\ v&=c\sqrt{1-\big(\frac{1}{1000}\big)^2}\\ &\approx c\big[ 1-\frac{1}{2}\big(\frac{1}{1000} \big)^2 \big]\\ &=\boxed{0.9999995c} \end{align*}

We used the binomial expansion from the appendix A-5.

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