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Below I have two plots from a gamma spectrum which I've been analyzing. The first plot is between a low energy range, the second between a significantly higher energy range. It is clear that the FWHMs (Full Width Half Maxima) of the peaks in the spectrum increase with energy. This is apparently on each of the spectra I've seen and seems to be an integral property of all of them.

200-300 KeV

1.7-1.8 MeV

Generally speaking, the peaks/Gamma lines in a Gamma spectra are wider at higher energies.

What is the exact reason for this? I do assume intuitively that there would be more inherent variance in the energy of peaks at higher energies. Whether this is a correct assumption is part of my question. Is the increase in width to do with error in the measurement of a particular Gamma? Do the Gammas at higher energies genuinely have a larger fluctuation in their energy? Is it a combination of both?

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  • $\begingroup$ Is this from a germanium detector? BTW, gamma isn't capitalized. $\endgroup$
    – user4552
    Jun 3, 2018 at 2:33
  • $\begingroup$ Long time ago for me now, but I this was indeed a HPGe detector. $\endgroup$
    – Matt
    Jun 5, 2018 at 10:59

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As Arthur Desrosiers mentions, there are 3 main interactions which govern the deposition of energy into the detector. These, however, are not the reason for the effect you are describing. The increase in FWHM is due to the statistical fluctuation of electron-hole pairs created and their contribution to a small current created in the detector, converted to a voltage pulse.

Based on your spectra, I'm guessing the data is from a Ge(Li) or HPG detector. In those detectors, the photon will produce several hundred to several thousand electron-hole pairs (EHPs) in the semiconductor latice of germanium. The high-voltage bias on the detector (it's a diode) will sweep these electrons and holes across the crystal, creating a current pulse in the bias. The preamplifier on the detector converts this to a voltage pulse which is then processed.

There is a statistical variation in the number of EHPs created with a standard deviation of approximately $s=\sqrt{N}$, where $N$ is the average number of EHPs created for a given energy. Say it requires $\epsilon$ electron volts to create one EHP in the crystal (0.7 eV for germanium), then a full-energy deposition by a 100 keV photon will produce $N=\frac{100,000}{\epsilon}$ EHPs (about 140000 in germanium). The standard deviation will be about 315 so the roughly Gaussian shaped peak will have a standard deviation of about 0.2 keV due strictly to the EHP statistics. Other contributions will come from the processing electronics. If the energy of the photon goes up, the standard deviation will increase because $N$ increases.

Pair production and Compton scattering will produce effects which are many keV away from the full energy peak.

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  • $\begingroup$ This sounds basically right, although for a germanium detector, which is what I've worked with, IIRC Poisson statistics is only one contribution to the width. For example, neutron damage will worsen the resolution of a detector, and I don't think that's because of a difference in $N$. $\endgroup$
    – user4552
    Jun 3, 2018 at 2:37
  • $\begingroup$ Thanks, @BenCrowell I believe that the damage to the crystal lattice in the semiconductor detectors like HPG and Ge(Li) affects the EHP collection process near damage sites. That would definitely be another statistical factor. $\endgroup$
    – Bill N
    Jun 3, 2018 at 12:53
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This is something I never really understood, but Glen Knoll offers the following in pp. 116 of his book "Radiation Detection and Measurement":

The energy resolution of the detector is conventionally defined as the FWHM divided by the location of the peak centroid $H_0$. The energy resolution $R$ is thus a dimensionless fraction conventionally expressed as a percentage. Semiconductor diode detectors used in alpha spectroscopy can have an energy resolution less than 1%, whereas scintillation detectors used in gamma-ray spectroscopy normally show an energy resolution in the range of 3-10%.

$R={\rm FWHM\over H_0}$

So, from a purely definitional point of view, if we hold the resolution constant, and increase the energy, the FWHM must also increase.

I think really, the direction of inference in this question is the wrong way. You have an observation (FWHM increases as energy does). Now, the task is to come up with a theory as to why this is true (and test of course).

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  • $\begingroup$ This doesn't answer the question. $\endgroup$
    – user4552
    Jun 3, 2018 at 2:35
  • $\begingroup$ Well, yes but the point wasn't to answer the question, but rather to take a step in a direction towards answering it. $\endgroup$ Jul 2, 2018 at 21:28
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The answer to the question is due to the fact that the number of charge carriers increased.

Glen Knoll offers the following in pp. 117 of his book "Radiation Detection and Measurement": The width parameter σ determines the FWHM of any Gaussian through the relation FWHM = 2.35σ. The standard deviation σ of the peak in the pulse height spectrum is then σ = K*sqrt(N). Here the K is a proportionality constant, and N is the charge carriers, particularly in a scintillator detector.

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The gamma photon energy is a function of the energy levels in the nucleus. There is some uncertainty in these levels. However, the data that you present in the chart is the amount of energy deposited in a detector, where the uncertainty is determined by the uncertainty in the interaction of the photon with the material of the detector and the processing of that physical interaction. At low energies, the photoelectric effect (absorption) dominates; at mid energies, the Compton effect (scattering) dominates.

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  • $\begingroup$ None of this has anything to do with the FWHM of the photopeaks, which has to do with the physics of the specific type of detector (here probably a germanium). $\endgroup$
    – user4552
    Jun 3, 2018 at 2:34

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