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The following plot shows data collected from a Co-60 coincidence experiment. The detectors used were NaI(T) scintillation detectors.

One detector was gated around the 1.33 MeV peak and the second detector collected the data shown below. I have been trying to figure out what the two peaks are around 200keV.

enter image description here

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  • $\begingroup$ How did you calibrate your detector? Are you confident it is accurate over the full energy range? How are you fitting the peaks? $\endgroup$
    – Chris
    Sep 6, 2020 at 8:00
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    $\begingroup$ I am fairly confident in my calibrations. I used a number of sources (Cs-137, Co-60, Na-22, Mn-54) to calibrate by fitting Gaussian functions to the photopeaks using Scipy curve fit. My other calibrated spectra all have features at the expected values. $\endgroup$
    – Anna
    Sep 6, 2020 at 8:24
  • $\begingroup$ Cobalt 60 has peaks at 1.17 and 1.33. I don't know where you get your 347 and 826 from. And the two peaks are usually about the same size - your 1.3 peak looks very small $\endgroup$ Sep 6, 2020 at 8:49
  • $\begingroup$ @RogerJBarlow Those are the single escape and double escape peaks of the 1.33MeV peak. $\endgroup$
    – user273992
    Sep 6, 2020 at 9:22
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    $\begingroup$ As described in the question, this is a coincidence experiment so this detector collected data with gating controlled by another detector (180 degrees away) that was centred on the 1.33MeV peak. So whenever the gating detector recorded a 1.33MeV gamma, a corresponding coincident gamma was recorded on this spectrum which is why the 1.17MeV peak is dominant in this spectrum. $\endgroup$
    – Anna
    Sep 6, 2020 at 10:53

1 Answer 1

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You have already identified the peak at about 195 keV as backscatter peak.

Apparently, you have used a NaI(Tl) detector. Photoelectric absorption by iodine of NaI results in a characteristic x-ray with 28 keV. If this x-ray exits the detector crystal, it results in a secondary peak 28 keV below the corresponding photopeak.

Since 195 keV minus 28 keV is 167 keV, we may conclude that your peak at 167 keV is the iodine x-ray escape peak corresponding to the backscatter peak.

You cannot see the iodine x-ray escape peak corresponding to the photopeak at 1.17323 MeV because it is hidden in the spread of the photopeak.

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  • $\begingroup$ Yes, but it should be the backscatter of the 1.17 MeV peak which would be at 210 keV and its iodine K-alpha escape peak (surprisingly strong). Calibration may be a bit off. $\endgroup$
    – user137289
    Sep 6, 2020 at 10:33

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