Question

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

Answer 1

The rest energy of a particle is
$E_0=m_0 c^2$
where $m_0$ is the rest mass of the particle and c is the speed of light.

The total energy of a relativistic particle is
$E=mc^2 = \frac{m_0 c^2}{ \sqrt{1- \frac{v^2}{c^2} } }$
where v is the speed of the particle.

We want the total energy of the particle to be twice its rest energy, so that
$E=2E_0$
which means:
$\frac{m_0c^2}{ \sqrt{1- \frac{v^2}{c^2} } }=2m_0 c^2$
$\frac{1}{ \sqrt{1- \frac{v^2}{c^2} } }=2$
From which we find the ratio between the speed of the particle v and the speed of light c:
$\frac{v}{c}= \sqrt{1- (\frac{1}{2})^2 } =0.87$
So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.