LB MDA/Se Mass Calculations

MDA and Se mass Calculations

To find the mass of the selenium in the irradiated horse feed sample, we need some masses and volumes of the horse feed. A vial containing 20mL of regular horse feed was massed. The mass of the vial was 13.6406g and the total mass of the vial and the horse feed was 31.8504g. This means the mass of the horse feed is [math]31.8504g - 13.6406g = 18.2098g[/math]. Since this mass is in a 20mL = 0.02L container, the density of the horse feed (assuming the mass difference between the non-irradiated and the irradiated horse feed is negligible) is

[math]\frac{18.2098g}{0.02L} = 910.49 \frac{g}{L}[/math].

Now for the irradiated horse feed the mass of the container is 25.0259g and the total mass of the container and the sample is 43.7529g. The mass of the sample then is 43.7529g - 25.0259g = 18.727g. Now using the density found before we can find the volume of the irradiated horse feed within the container.

[math]\rho = 910.49\frac{g}{L} = \frac{18.727}{V},  V = \frac{18.727g}{910.49 \frac {g}{L}} = 0.0206L.[/math]

The EMSL report says that there are [math]0.56\frac{mg}{L}[/math] in the horse feed, so the mass of the selenium in the horse feed sample is 

[math]0.56\frac{mg}{L}\times 0.0206L = 0.011536mg[/math].

Each of the calibration runs were used at position C in detector D. Below is the runlist for the isotopes used. The intervals used were (117:127) and (272:282) respectively.

Source	Serial #	Reference Date	Activity	Start	Stop	Live
Co-57	129735	07/01/08	1.074 micro Ci	16:17:37	16:27	601.10
Ba-133	129790	07/01/08	1.188 micro Ci	16:29:14	16:32	208.259

For the Co-57, the activity as of 09/07/16 is 0.0005 microCi and the intensity of the 122.3 keV line is 85.6%

= [math]\left (0.856 \right )\left ( 0.0005 \times 10^{-6} \mbox{Ci} \right) \left (\frac{ (3.7 \times 10^{10} \mbox{Hz}}{\mbox{Ci}} \right)= 15.84 Hz [/math] for the 122.3 line

For the Ba-133, the activity as of 09/07/16 is 0.69 microCi and the intensity of the 277.05 keV line is 7.164%

= [math]\left (0.07164 \right )\left ( 0.69 \times 10^{-6} \mbox{Ci} \right) \left (\frac{ (3.7 \times 10^{10} \mbox{Hz}}{\mbox{Ci}} \right)= 1828.97 Hz [/math] for the 277.05 line

Run	Source	Energy (keV)	Expected Rate (Hz)	HpGe Rate (Hz)	HpGe Det D Efficiency (%)
Eff_C_Co_57.root	Co-57	122.3	15.84	0.4526 - 0.2144	1.5
Eff_C_Ba_133.root	Ba-133	277.05	1828.97	20.58 - 0.07641	1.12

Now that the efficiency has been found, we must find the background rate to find the MDA. To find the background rate I used ROOT and plotted the energy spectra for Se_B_005 and HorseFeed_NoIrr. The windows of interest are the same as above. Each run's time was cut down to 1 hour.

First I will find the MDA for the window from (117:127). In Se_B_005, the activity was found to be 0.8433 HZ and in the HorseFeed_NoIrr the activity was found to be 0.2133 Hz, which gives a background rate of 0.63 Hz -> 37.6 cpm.

Now we can compute the MDA using this reference https://wiki.iac.isu.edu/index.php/File:Nwsltr-43re.pdf

I found the MDA to be [(2.71 + 4.65*(37.6 * 60)^1/2]/(60*0.015) = 249.1 dpm

Now I will find the MDA for the window from (272:282). In Se_B_005, the activity was found to be 0.7408 Hz and in HorseFeed_NoIrr the activity was found to be 0.08278. This means that the background activity is 0.66 Hz -> 39.48 cpm.

Now we can compute the MDA = [(2.71 + 4.65*(39.48 * 60)^1/2]/(60*0.012) = 318.1 dpm