Co60, Gamma Spec, Different Counts in Peaks

  1. jcsd
  2. To explain my uncertainty more.

    For each Gamma with 1.1MeV (appearing>99%) theire will be a Gamma with 1.3MeV(appearing>99%) but in the Spectrum the 1.1MeV Gamma ist counted about 10% often.

    How could that be explained?
  3. jtbell

    Staff: Mentor

    What kind of detector produced the spectrum? The detector might be less efficient at the higher energy.
  4. Thank You

    A NaI(Tl) Scintillation Detector.

    They have a Detection Efficiency and its depending on the Energy and more.
  5. mfb

    Staff: Mentor

    Where does that number 10% come from?
    The background is clearly different for the two energies so this has to be subtracted. Did you make a fit to the spectrum? The resolution can be different, so peak height and total number of events in the peak don't have to be proportional.

    I would expect the lower efficiency as main reason. Maybe pair production outside the detector can be an issue as well.
  6. I think that it is due to the decreasing slope of efficiency vs energy as was stated.
    Also, the QED cross section drops with energy so if efficiency was flat I would expect a (small) decrease in counts as energy increases.
  7. jtbell

    Staff: Mentor

    The two peaks at the far right end of the spectrum are due to the photoelectric effect. The interaction cross-section σ for the photoelectric effect decreases rapidly with increasing photon energy. See here for example:

    Scroll down near the bottom of the page to the section "Photon interactions" and see the graph of σpe.
  8. It is not necessarily *only* photoelectric effect that takes place there.
    It could be successive compton scattering with all the energy absorbed within the detector, taken that the detector is thick enough.
    A 10% effect is an order of magnitude difference in counts and thus in the cross section. This argument is not supported by the plot you pointed us to if you look at the range 1.17 - 1.3 MeV!
    Thus one may conclude that the effect is mainly detector efficiency and the decreasing cross section is a smaller additional effect.

    If RotBlau had the efficiency curve of the detector in question, it would be most enlightening.
    Take this for example from the web, although I don't know how relevant it might be
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted