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Energy of resulting Gamma Rays after particle annihilation

  1. Mar 28, 2010 #1
    1. The problem statement, all variables and given/known data
    An electron and a positron are moving side by side in the +x direction at 0.50c when they annihilate each other creating two gamma rays. What is the energy of each photon?


    2. Relevant equations
    (mc2 + K+) + (mc2 + K-) = E1 + E2


    3. The attempt at a solution

    I solved out that K = mc2 = 0.511MeV ( [tex]\gamma[/tex] - 1 ) where [tex]\gamma[/tex] = 1.155. Therefore K should equal 79.1keV. Plug this into the equation up in that "relevant equations" section and I get E1 + E2 = 1.18MeV.

    I know the answer is E1 = .885 MeV in the +x direction and E2 = .295 MeV in the -x direction as my prof gave us the answers. These values verify that the calculations I performed earlier are correct.

    However, I don't know how to obtain the answers. It seems like it should be stupidly simple, but I just can't figure it out. I tried using hc/[tex]\lambda[/tex] = E to solve for wavelength and then tried plugging stuff in again to get energy but that doesn't work. And even if it did, how would I know direction? If anyone can just give me that slight push in the right direction I'd be very grateful!
     
  2. jcsd
  3. Mar 28, 2010 #2
    You have one equation and two unknowns (E_1, E_2). So you might want to use conservation of momentum to help solve for both unknowns.
     
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