Energy Reconstruction in Gamma Spectroscopy

In summary: No, you don't forget the first time you calibrate detectors, but you might forget what α looked like when you last calibrated them.
  • #1
Cmertin
58
0
I have a question about energy reconstruction. I'm writing up a paper, and I'm trying to understand why this is the reason, before putting it in my paper. I know from experimental evidence that it is the case, though I would like to know if there is a reason.

Let's say you have a sufficiently long crystal-scintillator detector, on the order of 50 cm or so. On each end of the crystal is one photo tube. Now, the energy that each photo tube receives can be written as the following formula (I know why, but I'd rather not explain if I don't have to)

[itex]E_{1}=\widetilde{E}_{T} \cdot e^{-x \cdot \alpha}[/itex]
[itex]E_{2}=\widetilde{E}_{T} \cdot e^{-\alpha \cdot (\ell-x)}[/itex]

Where [itex]E_{1,2}[/itex] are the corresponding energies for each detector 1 and 2, [itex]\alpha[/itex] is the light attenuation factor for said detector, [itex]\ell[/itex] is the total length of the detector, [itex]x[/itex] is the position of the incident photon before it interacts with the crystal, and [itex]\widetilde{E}_{T}[/itex] is the theoretical value for the incident gamma - ie the total energy of the gamma.

Now, my question is, why when recreating the energy based off of the energy received from each detector is the equation

[itex]E_{T}^{\prime} = \sqrt{E_{1} \cdot E_{2}}[/itex]

instead of

[itex]E_{T}^{\prime} = E_{1} + E_{2}[/itex]

where [itex]E_{T}^{\prime}[/itex] is the reconstructed energy based on the two PMT's​

Can anyone help me understand this?

Thanks in advanced.
 
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  • #2
Just calculate both to see it. A multiplication of the energies corresponds to an addition of the exponents in the exponential. The first one gives the correct result, the second one does not.
 
  • #3
mfb said:
Just calculate both to see it. A multiplication of the energies corresponds to an addition of the exponents in the exponential. The first one gives the correct result, the second one does not.

I've done that and got the following result, but I'm not sure how it reads as the total energy. What I get is

[itex]E_{T}^{\prime}=\widetilde{E}_{T} \cdot e^{-\alpha \cdot \frac{\ell}{2}}[/itex]​

However, how I don't see how the relative positions to each photo tube take effect, since they cancel out when you work it out. Or, is this simply saying that the total energy is equal to when the source is at the center of the detector? But I can't follow that logic because it's been reconstructed at different positions.

Edit: I just had some brain food, a doughnut, and I think I came to the realization what this is saying. It's saying that the total energy is independent of the position of the incident gammas relative to the PMTs. Therefore, by using this relation, one always winds up with the same value that one were to have gotten if the source was in the center and the energy that each PMT "read" was equal. Yes? Sorry for the previous bit, I'm a bit tired.
 
Last edited:
  • #4
It's saying that the total energy is independent of the position of the incident gammas relative to the PMTs. Therefore, by using this relation, one always winds up with the same value that one were to have gotten if the source was in the center and the energy that each PMT "read" was equal.
Right. The effect of x is cancelled, which is exactly what you want. α and l have to be known to reconstruct the energy afterwards.
 
  • #5
mfb said:
Right. The effect of x is cancelled, which is exactly what you want. α and l have to be known to reconstruct the energy afterwards.


Awesome. Thank you. I kind of feel dumb for asking such a simple question now. And I understand that α and l have to be known to reconstruct the energy, you never forget the first time you calibrate detectors, do you?
 
  • #6
α could change with irradiation, so it might be better to repeat the calibration from time to time. But that depends on the setup.
 

1. What is energy reconstruction in gamma spectroscopy?

Energy reconstruction in gamma spectroscopy is the process of determining the energy of gamma rays emitted by radioactive sources. This is done by measuring the energy of the gamma rays using a detector and then using mathematical algorithms to reconstruct the original energy of the gamma ray.

2. Why is energy reconstruction important in gamma spectroscopy?

Energy reconstruction is important in gamma spectroscopy because it allows for accurate identification and quantification of radioactive materials. It also helps in determining the energy levels of the gamma rays, which can provide valuable information about the source of the radiation.

3. How is energy reconstruction performed in gamma spectroscopy?

Energy reconstruction is performed by using a detector, such as a scintillation detector or a high-purity germanium detector, to measure the energy of the gamma rays. The detector converts the energy of the gamma rays into electrical signals, which are then analyzed using computer software to reconstruct the original energy of the gamma ray.

4. What are the factors that can affect energy reconstruction in gamma spectroscopy?

There are several factors that can affect energy reconstruction in gamma spectroscopy, including the type of detector used, the calibration of the detector, and the resolution of the detector. Other factors such as dead time, background radiation, and detector efficiency can also impact the accuracy of energy reconstruction.

5. Are there any limitations to energy reconstruction in gamma spectroscopy?

Yes, there are limitations to energy reconstruction in gamma spectroscopy. These limitations include the energy range that the detector is capable of measuring, the resolution of the detector, and the accuracy of the calibration. Additionally, the presence of multiple gamma rays with similar energies can make it difficult to accurately reconstruct the individual energies of each ray.

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