Question

Question: What must be the momentum of a particle with mass m so that the total energy of the particle is 3.00 times its rest energy?

Short answer

Answer

The momentum of the particle is 2.83 mc.

Step-by-step solution

Relativistic momentum of a particle.

The relativistic momentum of particle is defined as

$p=\gamma mv$

Where,$\gamma$ is the Lorentz factor, mis the rest mass and v is the speed of the particle.

Determine Lorentz factor from the given relation.

The given relation is

$\mathbf{E}\mathbf{=}\mathbf{3}{\mathbf{E}}_{\mathbf{o}}$

Here, E is the total energy and E₀ is the rest mass energy.

$$\begin{gathered} \gamma m{c}^2=3m{c}^2 \\ \gamma =3 \end{gathered}$$

Determine speed of the particle from the Lorentz factor 

.

The Lorentz factor is defined in terms of speed is

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-}{V}^2/C^2} \\ \frac{v}{c}=\sqrt{1-\frac{1}{{\gamma }^2}} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} \frac{v}{c}=\sqrt{1-\frac{1}{{\left(3\right)}^2}} \\ v=0.9428c \end{gathered}$$

Substitute Lorentz factor and speed of the particle in momentum expression and solve as:

$$\begin{gathered} p=\left(3\right)m\left(0.9428c\right) \\ =2.83mc \end{gathered}$$

Hence, the momentum of the particle is 2.83 mc .