Consider two photons one with energy 1 2 MeV traveling to t

Consider two photons, one with energy 1 = 2 MeV traveling to the right, and the other with energy 2 = 3 MeV moving to the left. The two photons collide head-on and produce a positron-electron pair. Suppose the electron and the positron move along the same axis as the photons.

What are the final energies (Ee and Ee+ ) and velocities (ve and ve+ ) of the electron and the positron?

Solution

two photons, one with energy 1 = 2 MeV travelling to the right,
and the other with energy 2 = 3 MeV moving to the left

electron, with rest mass m_e = 0.511MeV/c^2
total energy of two protons ,E 5 MeV

now the electron- positron pair formed will have equal energies , let each of theirs energy be E\", then
E\" = 1/2 * E = 2.5 MeV

The velocity can be found by recalling the expression for total energy (E_e = gamma*m_e * c^2):
gamma = E_e/(m_e *c^2) = 2.5 MeV/0.511 MeV = 4.892
This corresponds to a velocity
v = c sqrt[1 -(1/gamma^2)] = c sqrt[1 -(1/4.892^2)] = c * sqrt (0.958) = 0.9789 c

as rest omass of positron is same as that of electron and both have same energies , their velocities will be also same. v = 0.9789c