Question

23. A positron and an electron can annihilate each other and form 3

gamma rays. Two gammas are detected. One has energy of 225 keV, the other has 357 keV. What is the energy of the 3rd gamma ray?

Answer 1

m = 9.11E-31 mass of electron

M = 1.822E-13 mass of electron and anti-electron

E = M c^2 = 1.822E-31 * (3.00E8)^2 = 1.64E-13 Joules

E = 1.64E-13 J / 1.60E-19 J/EV = 1025 KEV energy released

(225 + 357) kEV = 582 kEV

E(3d ray) = 1025 - 582 = 443 kEV energy of 3'd ray