Question

At what speed, as a fraction of $c$, must an electron move so that its total energy is$10\%$ more than its rest mass energy?

Short answer

$v=1.24\times {10}^{8}m/s$

Step-by-step solution

Given Information

We have given that, $c=3\times {10}^{8}m/s$

Total energy is$10\%$more.

Simplify

Given values are: $c=3\times {10}^{8}m/s$and total energy is $10\%$more.

To find the speed $v$, we use the below given formula.

role="math" localid="1649826808978" $\frac{m{c}^2}{\sqrt{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}}=m{c}^2+10\%\text{of}m{c}^2$

Now, put the given values in the above formula to find the speed.

$$\begin{gathered} \frac{m{c}^2}{\sqrt{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}}=m{c}^2+0.1\times m{c}^2 \\ \frac{m{c}^2}{\sqrt{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}}=1.1m{c}^2 \\ \frac{1}{\sqrt{\left(1-{\displaystyle \frac{v^2}{c^2}}\right)}}=1.1 \\ \sqrt{\left(1-\frac{v^2}{c^2}\right)}=\frac{1}{1.1} \\ 1-\frac{v^2}{c^2}=\frac{1}{1.21} \\ \frac{v^2}{c^2}=\frac{0.21}{1.21} \\ {v}^2=c^2·\frac{21}{121} \\ v=c·\sqrt{\frac{21}{121}} \\ v=1.24\times {10}^{8}m/s \end{gathered}$$