Question

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

Short answer

The speed of particle is 0.943c.

Step-by-step solution

Given information

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

Step 1. 

The formula of total energy is $E={{\rm Y}}_{p}m{c}^2$.

Here, m = mass of particle

c = speed of light

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

We know that total energy is the sum of rest mass energy and kinetic energy.

$E=E_0+K$

Here, $E_0$=rest mass energy

K = kinetic energy

Substituting both values of total energy in formula :

${{\rm Y}}_{p}m{c}^2=E_0+K$

Step 2.

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

Substituting value in formula :

${{\rm Y}}_{p}m{c}^2=E_0+2E_0$

We know that $E_0=m{c}^2$

Substituting the value of $E_0$in formula :

${{\rm Y}}_{p}m{c}^2=3m{c}^2$

${{\rm Y}}_p=3m$

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

Substituting the value of ${{\rm Y}}_p$in formula :

$\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}=3$

$1-\frac{v^2}{c^2}=\frac{1}{9}$

$v=\frac{\sqrt{8}}{3}c$

$v=0.943c$