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Components of Compton Spectrum

  1. Dec 19, 2009 #1
    Hey can someone explain the significants of the different components of the Compton spectrum? I know that the Compton edge comes from the incident angle of the photon approaching 180 degrees and that its the maximum energy that can be transferred from the photon to the electron without reaching the photoelectric peak.

    But what I'm not sure of is the significants of the single and double escape peaks ? Also at the photo electric peak is the photon absorbed and then remitted or just all of its momentum transferred to the electron?

    Anyways I have an exam on this on monday and missed the lecture on it i guess.


  2. jcsd
  3. Dec 19, 2009 #2


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    Is one asking about gamma spectroscopy in general.

    The gamma spectrum depends on the energy levels of the initial gamma ray. When a gamma source produces a gamma ray with energy > 1.022 MeV, then it can produce electron-positron (e-e+) pairs. When a e-e+ is produced, the e+ eventually looses energy and is annihilated with an electron to produce to gamma rays, each of 0.511 MeV. Now on or both gamma rays can escape the scintillation detector. So in addition to a photopeak at E, there are peaks at E-0.511 and E-2(0.511) = E - 1.022 MeV.

    There is also the Compton edge which is at E - Emax, where Emax is the maximum energy a gamma ray can loose by scattering 180°.
  4. Dec 19, 2009 #3
    so the single escape peak would just correspond to the energy of a photon that was produced by an electron-positron annihilation that has already happened? leftovers from the initial collision so to speak? and the double would just be the same thing happening again?
  5. Dec 19, 2009 #4
    ok i think i kinda got it now.

    the single escape peak (E-.511) is just a photon that was created by the annihilation bouncing off another e- in the system right?

    what does the double escape peak (E-(2*.511))correspond to though?
  6. Dec 19, 2009 #5


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    The single escape peak comes from the loss of the 0.511 MeV gamma (single gamma) that has escaped from the system, but the other one is caught. So the energy counted is total E minus the 0.511 MeV that escapes.

    Two 0.511 MeV gammas have escaped, i.e. some of the energy is detected by the energy deposited by the electron-positron pair, but when the positron annihilates, it is possible that both 0.511 MeV gammas escape from the system.
  7. Dec 19, 2009 #6

    so simple now:rolleyes:

    thanks very much you probably just added 10% to my final exam mark !
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