Question

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

Short answer

The speed of particle is 0.866c.

Step-by-step solution

Given information

We have given that, particle’s total energy twice its rest energy.

Step 1.

We know that the formula of total energy is :

$E={{\rm Y}}_{p}m{c}^2$

Here, m = mass of particle

v = velocity of particle

${{\rm Y}}_p$= Lorentz factor

We have given that total energy is twice of rest mass energy.

$E=2E_0$

Here $E$= total energy

$E_0$= rest mass energy

Step 2.

Substituting $E=2E_0$in the formula of total energy :

$2m{c}^2={{\rm Y}}_{p}m{c}^2$

${{\rm Y}}_p=2$

We know that ${{\rm Y}}_p=\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$, substituting this value of ${{\rm Y}}_p$:

$\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}=2$

$\frac{v^2}{c^2}=\frac{3}{4}$

$v=0.866c$