Question

At what speed, as a fraction of $c$, is a particle’s total energy twice its rest energy?

Short answer

$v=\sqrt{\frac{3}{2}}c$

Step-by-step solution

Given Information

We have to find the speed, as a fraction of $c$, is a particle’s total energy twice its rest energy?

Simplify

Total energy formula is

$E_o=\frac{m{c}^2}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}$

The rest of energy formula is

$E=m{c}^2$

Now we will find the total energy twice its rest energy.

$$\begin{gathered} E_o=2E \\ \frac{m{c}^2}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}=2m{c}^2 \\ m{c}^2=\sqrt{1-\frac{v^2}{c^2}}·2m{c}^2 \\ \frac{c^2}{2c^2}=\sqrt{1-\frac{v^2}{c^2}} \\ \frac{1}{2}=\sqrt{1-\frac{v^2}{c^2}} \\ \frac{1}{4}=1-\frac{v^2}{c^2} \\ 4v^2=3c^2 \\ {v}^2=\frac{3}{4}{c}^2 \\ v=\sqrt{\frac{3}{2}}c \end{gathered}$$