LB Thesis Thin Window Analysis

This will be a detailed explanation of how to trace the isotope of interest's activity back to its original value. The sample of interest here is the oven ash sample with inventory number 170063. The measurement was made on Detector A at the IAC and it was a split run using a Co-60 flag to denote the changing of samples between a pure selenium sample and a mixture of selenium and sage ash. The target information can be found below:

Nickel Foil: 0.2783g

Outer Se Pellets: 0.0971g

Sage Ash: 0.5111g

Inner Se Pellet: 0.0523g

Half Life of Se-81: 57.28 Minutes

Decay Constant: [math] 2.016838863 \times 10^{-4} \frac{1}{seconds} [/math]

Thin Window Histograms

Each histogram has its number of counts weighted by the mass of the sample.

Since the resolution of Detector A at the IAC is about 1 keV, the window in which the signal is viewed should have a width of 2 (whether you plot channel number or energy along the horizontal axis). The signals were fit with a function of the form

[math] f(x) = C + Ae^{\frac{(x-\overline{x})^2}{2 \sigma^2}} [/math]

Here, the constant value is to assess the value for the background. The background must be removed from the signal in order to have a more accurate analysis as it is not a part of the signal that we are interested in. Below are sample plots of the original signal and the signal viewed in a thin 2 channel window:

This signal corresponds to the 103 keV line of Se-81. I plotted channel number instead of energy to maintain integer bins and avoid a picket fencing effect in the signal.

In looking at the thin window, the total number of counts within the window is given by the "Integral" value in the statistics box, which I will denote as I.

[math] Stats Box Integral = I [/math]

Since this is a counting experiment, the error in the signal will be given by

[math] \sigma_{I} = \sqrt{I} [/math]

Dealing With Background

Now that I have discussed how to get the number of counts along with the uncertainty in the number of counts, I will now discuss how to remove the background from a signal.

In the statistics box, there is a value called "background" which corresponds to the constant value in the fit function. To removed the background from the number of counts in the statistics box, we must integrate the "background" value across the region of interest. In other words

[math] Background = B = \int_{113}^{115} C d(Channel) = C(115-113) [/math]

Since the "background" value in the statistics box has an error with it and by integrating we are essentially just multiplying by a constant, so in this case

[math] \sigma_{B} = \int_{113}^{115}\sigma_{C} d(Channel) = \sigma_{C}(115-113) [/math]

Now that we have a value for the background along with its uncertainty, we can find the true signal of interest within the window. To find this use the relations

[math] True Signal = N = I - B [/math]

[math] \sigma_{N} = \sqrt{\sigma_{I}^2 + \sigma_{B}^2} [/math]

Now we have values for the number of counts without background along with the uncertainty in that value.

Here it is worth noting that there is some decay during the measurement period.

This is the general method for finding the number of counts for a background subtracted signal within a region of interest. This will be needed for creating plots to find the original activity of a sample.