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Homework Help: Pion decaying into 2 photons

  1. Apr 30, 2009 #1
    1. The problem statement, all variables and given/known data
    The neutral pion is an unstable particle that decays very quickly after its creation into two photons (“particles” of light: v = c, mo = 0). The pion has a rest-mass energy of 135 MeV. Consider a pion that has a kinetic energy of 90 MeV
    1) Determine the v of this pion
    2) Determine the momentum of the pion
    3) Determine the sum of the energies of the photons

    2. Relevant equations
    1) moc2[tex]\gamma[/tex] = KE + Rest E
    2) mov[tex]\gamma[/tex] = p
    3) Ephotons= pc

    3. The attempt at a solution
    I'm pretty sure I got 1 and 2. For 1 I added the KE + RE, converted to proper units, and finally got v = .80c
    For 2 I got p = 9.6 x 10-20 kg m/s
    Now I'm not so sure about #3. I think Ephotons = pc because momentum is conserved, and all of the pion's momentum gets turned into the photons' momentum. However, what about adding up the original rest mass energy + KE of the pion? If I do that I get a bigger energy then if I do pc. Thanks for all the help!
  2. jcsd
  3. Apr 30, 2009 #2


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    Just using pure energy conservation, how much energy is there before and after the decay?
  4. Apr 30, 2009 #3
    According to my calculations, E before = (90 + 135) MeV = 225 MeV
    After E = pc = 181.25 MeV
  5. May 1, 2009 #4
    I calculated p a different way this time, p = sqrt(2mKE) where m is the relativistic mass. When I do this then multiply p by c to get E, I get E = 201 MeV. Oh boy...

    edit: actually I'm not sure if that equation works for speeds close to c...
    Last edited: May 1, 2009
  6. May 1, 2009 #5


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    Hrm, you're letting the calculations get bogged down too much. The change in energy is zero, right? You know initial energy, so final energy must be.. the same.
  7. May 1, 2009 #6
    I get what you're saying, but I don't understand why the equations don't work here. It seems to me that momentum isn't conserved.
    Last edited: May 1, 2009
  8. May 1, 2009 #7
    and there is one more part to this problem which I thought I could get myself, but I can't. Here is a pic of the question/diagram

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