Question

An electron (rest mass me) of energy E makes a head-on collision with a positron (positron is electron’s antiparticle, it has the same mass as electron, but opposite charge) In collision the two particles annihilate each other and are replaces by two photons (γ rays) of equal energy, each traveling at equal angles θ with electron’s direction of motion. Find 1. The energy of each photon. 2. The momentum p of each photon. 3. The angle θ.

Problem 3. Electron-positron annihilation An electron (rest mass me) of energy E makes a head-on collision with a positron (positron is electron's antiparticle, it has the same mass as electron, but opposite charge) In collision the two particles annihilate each other and are replaces by two photons (7 rays) of equal energy, each traveling at equal angles with electron's direction of motion. Find 1. The energy e of each photon. 2. The momentum p of each photon. 3. The angle . 1 - Exam 2

Answer 1

a) By conservation of energy, the total energy of the electron and positron is equal to the total energy of the 2 photons.

$E+m_ec^2 = 2\epsilon$

Therefore the energy of each phton is

$\epsilon=\frac{E+m_ec^2 }{2}$

b) The momentum of photon is given by

p= €/c = E + mec 2c

c)

The mass energy of the electron is given by the relation

2.4 E? - pc = mec

or

Hence

$p_e=\frac{\sqrt{E^2-m_e^2 c^4}}{c}$

Consrrvation of momentum in X direction gives

$2p\cos\theta =p_e=\frac{\sqrt{E^2-m_e^2 c^4}}{c}$

or

$\frac{2(E+m_ec^2)}{2c}\cos\theta=\frac{\sqrt{E^2-m_e^2c^4}}{c}$

or

$\cos\theta=\frac{\sqrt{E^2-m_e^2c^4}}{E+m_ec^2}$

or

$\theta=\cos^{-1}\left(\frac{\sqrt{E^2-m_e^2c^4}}{E+m_ec^2} \right )$