Question

What is the magnitude of the relativistic momentum of a proton with a relativistic total energy of 3.0 × 10-10 j?

Answer 1

The relativistic total energy of a particle is given by:

`E^2= (pc)^2+(m_0 c^2)^2`
where
p is the particle momentum
c is the speed of light

`m_0` is the rest mass of the particle

If we re-arrange the equation, we find

`p= (1)/(c) √((E^2-(m_0c^2)^2)`
and by using

`c=3 \cdot 10^8 m/s`

`m_0 = 1.67 \cdot 10^(-27) kg` (proton mass)

we find the momentum of the proton:

`p= (1)/(3\cdot 10^8 m/s) \sqrt{(3.0 \cdot 10^(-10)J)^2-(1.67\cdot 10^(-27)kg (3\cdot 10^8 m/s)^2)^2} =`

`=8.65 \cdot 10^(-19) kg m/s`