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In gamma spectrometry, energy is deposited in the crystal and the electron from the valence band gets excited to the conduction band. Both bands have degenerate energy levels. More the energy deposition, more is the no. of electron-hole pair. There is no difference in the energy of the electron or hole. Then how the energy info is reconstructed in the gamma spectrum? Assume 662 keV and 1460 keV gamma is incident simultaneously of equal activity of 1 nCi and counting by HPGe detector.

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  • $\begingroup$ In a semiconductor sensor what you measure is the current pulse. More current means more e-h pairs means more energy in the gamma. In a scintillator you measure light emission, and more light means more energy. $\endgroup$
    – Jon Custer
    Commented Mar 15, 2020 at 15:46
  • $\begingroup$ Sir, You are correct. But my question is when simultaneously 2 difference energy gammas are incident on the crystal, with more energy one produces more e-h pair and the e-h pair have same energy as with low energy gamma. Suppose1 nci 137Cs and 1 nCi 40K is incident. Being more energy, 40K will make more e-h pair than 137Cs. But in both cases the e-h pair will have same energy as the conduction band has degenerate energy level. Then how in MCA 2 different gamma shall appear? Why the energy info is not lost? $\endgroup$
    – ggs
    Commented Mar 18, 2020 at 14:30
  • $\begingroup$ Because the count rate has to be kept low enough to prevent the ‘pile-up’ from happening too often. Otherwise you can get what appear to be higher energy counts as the total current of the two come out close enough together in time. A well known nuclear data acquisition problem to those versed in the art. There are various pile-up models around to try and remove the background. $\endgroup$
    – Jon Custer
    Commented Mar 18, 2020 at 14:34
  • $\begingroup$ Sir, I understand your concern. But pl ignore pile up problem. Assume the dead time is less than 1%. I am not getting any explanation of my doubt anywhere. Pl feel free to ask more Q, so that I can let you know the exact doubt, if not clear from my Q and comments. $\endgroup$
    – ggs
    Commented Mar 18, 2020 at 14:47
  • $\begingroup$ Can I pl get the answer now? $\endgroup$
    – ggs
    Commented Jun 30, 2020 at 4:35

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