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Scintillator Compton Edge

  1. Nov 11, 2006 #1
    I am trying to calculate the energy at which the Compton edge occurs for a sample of Cs137 using a NaI(Tl) scintillator. I know the original energy of the gamma ray was 661.6keV. and that for the Compton edge the angle between the electron and the line of incident must be 0degrees (for max energy of photon). but this just give the value of the photopeak (incident gamma ray). How do i calculate the energy of the emitted photons from Compton scattering at the Compton edge??

    [i have calculated the beginning of the Compton continuum using an angle of 180 degrees between the electron and line of incident (minimum energy of the photon) and this gives a value which is about the same as my graph]
    i have looked at a few sites but am fairly confused as to why the Compton edge is at a lower energy to the initial gamma ray photon.

    Many thanks
  2. jcsd
  3. Nov 11, 2006 #2


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    Last edited by a moderator: Apr 22, 2017
  4. Nov 11, 2006 #3
    According to these equations the Compton edge should be the same energy as the photo-peak, as when theta=0 the equation is just 662keV/1.
    i don't understand how it is possible to calculate the Compton edge, with these equations.
  5. Nov 11, 2006 #4


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    I think you are misinterpreting theta.

    Last edited by a moderator: Apr 22, 2017
  6. Nov 11, 2006 #5
    ok so plugging in theta=180 and E=2MeV to http://nucleus.wpi.edu/Reactor/pics/rscineq2.jpg gives E' as .23MeV as is shows in the diagram.
    But then theta=0 (ie gamma ray has most energy and electron is not scattered) E'=2MeV. this is not what is shown in the diagram.

    how do you work out the 1.77MeV Compton edge?!
    Last edited by a moderator: Apr 22, 2017
  7. Nov 12, 2006 #6
    ok sorry i was being really stupid the 1.77MeV comes from 2MeV-0.23MeV.

    many thanks for your help.
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