1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Energy of resulting Gamma Rays after particle annihilation
  1. Particle annihilation (Replies: 3)

  2. Gamma ray emission (Replies: 15)

  3. Counting gamma rays (Replies: 1)