Why in 22Na spectrum is only one backscatter peak evident when we have two of different energy (511 keV and 1275 keV) gamma rays in the spectrum?
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$\begingroup$ Welcome to Physics SE. Coud you tell - what detector do you use (HPGe?), can you post the picture? Basically, there is one $\gamma$ and $\epsilon\beta+$ that you expect to generate 511 keV somewhere. Also - the backscatter is not clear enough to see what you mean. $\endgroup$– jaromraxCommented Apr 6, 2017 at 7:17
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$\begingroup$ It is a sodium iodide crystal activated with thallium NaI(Tl) This is the procedure I followed: 1. Place the 22Na source in front of the detector and run the MCA in the non-gated mode, until a smooth spectrum is obtained. 2. Identify and estimate the energy of the annihilation quanta (511 keV) and the high energy gamma ray from neon de-excitation (1275 keV). 3. Also note the Compton edges for each absorption peak and the backscatter peak. 4. Place the 22Na source in the middle in between the NaI detector and the organic scintillator detector. $\endgroup$– Rafiul NakibCommented Apr 6, 2017 at 7:35
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$\begingroup$ 6. Run the MCA in the gated (coincidence) mode, until a smooth spectrum is obtained. 5. Compare this spectrum with the previous one and explain the differences. $\endgroup$– Rafiul NakibCommented Apr 6, 2017 at 7:37
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$\begingroup$ Ok. So you have two detectors setup. Until 5, you dont care about the other one. i NaI detector has quite a poor resolution and also depends on its state. You probably see one large bump, that can cover all the effects. ii check the voltage on NaI. iii check the signal iv check the MCA setting, (if you acquire the peak top in case of analog). You may post a picture, although - it wouldnt help much in case of problem in detector setting $\endgroup$– jaromraxCommented Apr 6, 2017 at 10:41
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$\begingroup$ Calculate the maximum kinetic energy of the electron from a Compton event for each energy of the photons? These will be totally deposited in the detector. Consider whether a NaI(Tl) detector would have resolution to show these as distinct peaks. $\endgroup$– Bill NCommented Apr 6, 2017 at 20:54
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1 Answer
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This is how the typical spectrum should look. 
However, your crystal can be quite bad and you see like one large bump. I remember old crystals having 700 keV resolution. Nothing to do.
Or
the tension is too low
the MCA is not properly set and maximum of the signal is not recorded (gate)
if you have a digital acquisition, you may need lot of work and understanding to tune correctly
Edit: 1. the picture is an illustrative one taken from https://i.sstatic.net/n0BzJ.jpg. Purpose is to give a hint how such a spectrum can look. The author of the question did not see anything reasonable up to now, nor he posted a picture.
The labeling is not mine, Compton edge label should be also at 1000 keV near the photopeak.
I modified labeling a picture a little bit.
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$\begingroup$ There is some mislabling in the figure. The thing labeled "Compton edge" is the single escape peak; the edge is the steep slope near 325 keV. In this data set the two run together due to finite detector resolution, but they are quite distinct in a high resolution spectrum. And what is up with "KeV" label for the graph, anyway?!? The illustrative figure in the Wikipedia page for "Compton edge" is better labeled, but because it displays germanium detector data won't match what the OP should expect to see. $\endgroup$ Commented Apr 6, 2017 at 14:21
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$\begingroup$ Secondly, though the statistics in the figure are not sufficient to do anything with them (only maybe enough to positively identify them) there are hints of both the Compton edge for the photopeak (near 1050 keV) and the annihilation+photo coincidence near 1800 keV. $\endgroup$ Commented Apr 6, 2017 at 14:23
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$\begingroup$ @dmckee - I think you made a mistake - you will have difficulties to get single/double escape peaks in this picture. As I edited the answer: the picture and labels are just copied from imgur, KeV should be keV, but it is not a big deal... and just a matter of opinion, 20kcnts in a peak is good enough. $\endgroup$– jaromraxCommented Apr 6, 2017 at 15:44
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1$\begingroup$ I really mean it about the Compton edge, but it appears I mispoke about the peak aroun 175 keV. Not sure right now what that is. The Compton edge for the annihilation peak is at ... ::punches calculator:: ... 340 keV, and it is an edge. The structure labeled as Compton edge is around 175 keV, and it is a peak. Wrong place. Wrong shape. $\endgroup$ Commented Apr 6, 2017 at 16:03
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$\begingroup$ I modified the arrows a bit. Although I think the answer does not help much, as @rafiul-nakib is silent now $\endgroup$– jaromraxCommented Apr 6, 2017 at 16:15