In the HPGe detector, any gamma energy (> e-h pair production E) shall excite the electron of the valence band and send it to the conduction band. Due to reverse bias, the conduction band e shall be collected and used for gamma counting in the HPGe detector. Isn't the energy information will be lost? Suppose I have 60Co source with 1173 and 1332 keV gamma energy. Does the e-h pair produced from this gamma shall have the memory of the incident gamma energy? As in the end, we are getting gamma spectrometry containing precise gamma energy vs Intensity in MCA.
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1$\begingroup$ Many e-h pairs are generated. The energy is proportional to the current pulse from all pairs (note that both electrons and holes contribute to the current). $\endgroup$– Jon CusterCommented Oct 11, 2018 at 12:37
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$\begingroup$ Dear Sir, It's not clear to me yet. The current pulse is proportional to the intensity. Can you pl elaborate pl. @Jon Custer $\endgroup$– ggsCommented Oct 11, 2018 at 13:28
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A gamma ray incident on the HPGe detector will deposit energy in the detector producing a number of ion pairs that is proportional to the amount of energy deposited. When you collect the electrons, you get a current pulse where the pulse height is proportional to the energy deposited. Accumulate all these pulses to get your energy spectrum where you have pulse height on the x-axis and the number of pulses with that height on the y-axis.
Use your HPGe detector with some radioactive sources that emit gamma rays at known energies to calibrate current pulse height to energy.
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1$\begingroup$ 'Ion pairs' is incorrect, that should be 'electron-hole' pairs. And the hole current counts as well in a diode. $\endgroup$ Commented Oct 15, 2018 at 21:22
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$\begingroup$ Oh yes, you're right. I'm usually working with ionization chambers so Ion pairs is what I think of first $\endgroup$– imabugCommented Oct 15, 2018 at 23:43