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Scintillation detector's output

  1. Feb 13, 2015 #1
    I'm looking at scintillation detectors and I'm quite confused to the output of the crystal. My notes just say emission output has intensity proportional to input energy. How though? I mean surely for the Photoelectric effect there is a huge spectrum of output energies depending upon shell jumped. If you consider 1 of these energies you would need multiple ionisations to occur to increase intensity output. If multiple are occurring, why? Additionally how can you get mostly PE absorptions at high energy, like ~140kev. Is it say a 140kev photon is absorbed, releasing a 138kev electron? That causes 69 further ionisations. That would make some sense but I'm still confused to what exactly is going on. Plus even with 70 ionisations taking place, if you consider 1 energy output there is only a chance of this occurring. It could be k-alpha, k-beta ect. Are all considered?
     
  2. jcsd
  3. Feb 13, 2015 #2

    Vanadium 50

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    Which you do.

    Why not? Why should an electron suddenly stop ionizing after it's done so once?

    And yes, the energy of ionization varies. But you're taking an average, and the Central Limit Theorem helps you here. Also, remember that all the energy has to end up somewhere. Double that energy, and twice as much has to end up somewhere as well.
     
  4. Feb 25, 2015 #3
    I just found it odd that it was possible for a photon to undergo ~50 photoelectric absorption's all relatively instantaneously. Well with any large degree of occurrence. Turns out that's what the detector relies upon, thanks.
     
  5. Feb 25, 2015 #4

    mfb

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    No, the photons hit electrons, those move through the material and lead to the excitations, losing a bit of energy each time.
     
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