LB Se-79 Soil Analysis

Below is a description of how the activity of the Se-79 in the May 2017 soil/Se was determined.

Activity of Se-79m

First, find the activity of the Se-79m state which has a half life of 3.92 [math] \pm [/math] 0.01 minutes. The line of interest here is 95.73 [math] \pm [/math] 0.03 keV

Histogram analysis

The most active sample was the 10% Se/soil mixture's pure selenium witness pellet, so let's analyze that first. The histogram for the sample from 615<t<1115 s is shown below.

The line in the stats box labeled "integral" represents the total number of counts in the window, which is

I = 7787 [math] \pm [/math] 88.24 Counts

and the background is

B = 5053.2 [math] \pm [/math] 60.6 Counts

Now subtract the background from the total number of counts in the window

N = 2733.8 [math] \pm [/math] 107.05

Now divide by the runtime to get the activity

[math] A_{Measure} [/math] = [math] \frac{2733.8 \pm 107.05}{500} = 5.5 \pm 0.2 [/math] Hz

Note this is the measurement some time after the beam was turned off, so we can correct this back using the known half life of Se-79m ([math] \lambda [/math] = 0.0029 [math] \pm [/math] 0.00008 [math] s^{-1} [/math])

The efficiency for this run is [math] 0.29% \pm 0.0087% [/math]

So the efficiency corrected activity for the pure selenium pellet's 95 keV line is

[math] A_{\epsilon} = \frac{A_{Beam Off}}{\epsilon} [/math]

[math] A_{Beam Off} = A_{Measure} e^{\lambda t} = 431.10 Hz \pm 54.19 [/math] Hz

The efficiency was measured by placing a Ba-133 source at the same position as the pure Se sample and taking the ratio of the measured counts of the 81 keV line to the theoretical number of counts that the sample should have given off, which is computed by

[math] (Branching Ratio) \times (A_{Source}) \times (3.7 \times 10^{10} \frac{Hz}{Ci}) [/math]

More details can be found at https://wiki.iac.isu.edu/index.php/LB_May_2017_Det_A_Efficiency

Now divide by the efficiency to get the corrected activity

[math] A_{\epsilon} = \frac{A_{Beam Off}}{\epsilon} = 148655.17 \pm 19211.01 [/math] Hz

Now since [math] A = \lambda N [/math], we can find the total number of activated nuclei to be [math] 5.13 \times 10^7 \pm 6.8 \times 10^6 [/math] nuclei.

Now using [math] A = \lambda N [/math] with Se-79's decay constant, we can find the activity assuming a 100% branching ratio from the metastable state to the excited state of Se-79, which is ([math] \lambda = 6.74 \times 10^{-14} \pm 5.79 \times 10^{-20} s^{-1}) [/math]