Question

What is the magnitude of the relativistic momentum of a proton with a relativistic total energy of 3.8 × 10-10 J?

Answer 1

he relativistic momentum of a proton is determined by the relativistic equation:

p = γmu

where γ = 1/√(1 - u ²/c ²), m is the proton rest mass, u is the speed of the proton and c is the speed of light.

The total energy is given by

E = γmc ²

where mc ² is the proton rest mass energy. (The total energy is the kinetic energy plus the rest mass energy which is captured in the given equation via gamma)

So γ = E/mc ²

For the proton, mc ² = 938.27201 MeV is the rest mass energy.

1 eV = 1.60218 * 10^(-19) J

Given: E = 3.8 × 10^(-10) J

So γ = 3.8 × 10^(-10)J/(938.27201 * 10^6 * 1.60218 * 10^(-19)J) : we converted MeV to J

γ = 2.52......(i)

From γ = 1/√(1 - u ²/c ²) we get

γ ² = 1/(1 - u ²/c ²)

γ ² (1 - u ²/c ²) = 1

γ ² - γ ²u ²/c ² = 1

u ²/c ² = (γ ² -1)/γ ²

Substituting for γ = 2.52, we get

u/c = 0.921

u = 0.921c ...(ii)

Now we are ready to find the momentum:

p = γmu

p = 2.52 * 1.67262 * 10^(-27) kg * 0..921 * 2.99792 * 10^8 m/s

where we substituted for the mass of the proton = 1.67262 * 10^(-27) kg and the speed of light c = 2.99792 * 10^8 m/s

p = 11.6* 10^(-19) kg.m/s