Question

Suppose that an electron and a positron collide head-on. Both have kinetic energy of 1.20 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Ephoton of one of these photons

Answer 1

Answer:

E = 1.711 MeV

Explanation:

From the law of the conservation of energy:

$K.E_{e}+K.E_p + E_{e}+E_{p} = 2 E$

where,

$K.E_e=K.Ep=$ the kinetic energy of positron and electron = 1.2 MeV

$E_e=E_p =$ Rest energy of the electron and the positron = 0.511 MeV

E = Energy of Photon = ?

Therefore,

$1.2\ MeV + 1.2\ MeV + 0.511\ MeV + 0.511\ MeV = 2E\\\\E = \frac{3.422\ MeV}{2}$

E = 1.711 MeV