# Energy Reconstruction in Gamma Spectroscopy

1. Mar 11, 2013

### Cmertin

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)

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

Where $E_{1,2}$ are the corresponding energies for each detector 1 and 2, $\alpha$ is the light attenuation factor for said detector, $\ell$ is the total length of the detector, $x$ is the position of the incident photon before it interacts with the crystal, and $\widetilde{E}_{T}$ 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

$E_{T}^{\prime} = \sqrt{E_{1} \cdot E_{2}}$

$E_{T}^{\prime} = E_{1} + E_{2}$

where $E_{T}^{\prime}$ is the reconstructed energy based on the two PMT's​

Can anyone help me understand this?

2. Mar 11, 2013

### Staff: Mentor

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. Mar 12, 2013

### Cmertin

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

$E_{T}^{\prime}=\widetilde{E}_{T} \cdot e^{-\alpha \cdot \frac{\ell}{2}}$​

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: Mar 12, 2013
4. Mar 12, 2013

### Staff: Mentor

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. Mar 12, 2013

### Cmertin

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. Mar 12, 2013

### Staff: Mentor

α could change with irradiation, so it might be better to repeat the calibration from time to time. But that depends on the setup.