Chapter 37: Q51P (page 1148)

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?

Expert verified

Answer

The momentum of the particle is 2.83 mc.

Step by step solution

The relativistic momentum of particle is defined as

$p = γ m v$

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

The given relation is

$E = 3 E_{o}$

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

$γ m c^{2} = 3 m c^{2} γ = 3$

The Lorentz factor is defined in terms of speed is

$γ = \frac{1}{\sqrt{1 -} V^{2} / C^{2}} \frac{v}{c} = \sqrt{1 - \frac{1}{γ^{2}}}$

Substitute the values and solve as:

$\frac{v}{c} = \sqrt{1 - \frac{1}{{(3)}^{2}}} v = 0.9428 c$

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

$p = (3) m (0.9428 c) = 2.83 m c$

Hence, the momentum of the particle is 2.83 mc .

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