# Annihilation of Electron by Positron to Produce at least 2 Gamma Rays

by tomnomnom
Tags: annihilation, conservation laws, electrton, gamma, positron
 P: 5 1. The problem statement, all variables and given/known data The problem is listed as follows: Show that conservation of energy and momentum require at least two gamma rays to e emitted in the annihilation of an electron by a positron. 2. Relevant equations p(initial) = p(final) E(initial) = E(final) Total rest mass = 1.0218 MeV/c^2 p=hf/c E=hf 3. The attempt at a solution I'm checking to see whether my answer to this question is sufficient. The book (Wong) introduces the idea of a proton and anti-proton annihilation at rest, so I assumed that this process could occur at rest as well. Is that ok? I first tried the production of a single gamma ray. However, since initial p=0, final p=0 as well. However, since initial E=/=0, and initial E = final E, then final E =/=0. These two are in contradiction to each other, since p=hf/c and E=hf. Combining, this gives p=E/c. Since c=/= 0, this equation is impossible. I then explained that if the rays were travelling in opposite directions, this would allow for final p=0, fulfilling initial p=final p. Do you guys believe that this is a sufficient answer? It is based upon the presumption that such an interaction can occur at rest. Let me know, thanks. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
 P: 5 Just came up with a new thought, I could create a situation such that the electron and positron are approaching each other in such a manner that initial p=0. This would allow the dismissal of the "at rest" presumption. Thoughts?
 Sci Advisor HW Helper P: 11,863 This is the simplest exercise in quantum field theory, showing that the vertex of QED can't have all 3 particles on the mass shell. You only need the energy-momentum equation describing the mass shell. $$p^2 = m^2 \, \mbox{for the electron/positron and} \, p^2=0 \, \mbox{for the photon}$$

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