LB Se-79 Soil Analysis - Revision history

Brenleyt: /* Histogram analysis */

2018-10-16T20:45:15Z

Histogram analysis

2018-10-16T19:58:55Z

2018-10-15T20:56:05Z

Histogram analysis

2018-10-15T20:51:48Z

Histogram analysis

2018-10-15T20:35:05Z

Histogram analysis

2018-10-15T20:32:42Z

Histogram analysis

2018-10-15T20:30:26Z

Histogram analysis

2018-10-15T20:25:39Z

Histogram analysis

2018-10-15T20:19:53Z

Histogram analysis

2018-10-15T20:18:30Z

Histogram analysis

← Older revision		Revision as of 20:45, 16 October 2018
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	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 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.

		+	[[File:LB Se79HistoAnalysis.png\|200px]]

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

← Older revision		Revision as of 19:58, 16 October 2018
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	<math> A = \lambda_{Se-79} N = 3.43 \times 10^{-6} \pm 4.55 \times 10^{-7} </math> Hz		<math> A = \lambda_{Se-79} N = 3.43 \times 10^{-6} \pm 4.55 \times 10^{-7} </math> Hz

−	Which, converted into curies is <math> 9.27 \times 10^{-17} \pm 1.23 \times 10^{-7} </math> Ci. This is well below the 16 pCi limit	+	Which, converted into curies is <math> 9.27 \times 10^{-17} \pm 1.23 \times 10^{-7} </math> Ci. This is well below the 16 pCi limit.

← Older revision		Revision as of 20:56, 15 October 2018
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	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>		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>
		+
		+	<math> A = \lambda_{Se-79} N = 3.43 \times 10^{-6} \pm 4.55 \times 10^{-7} </math> Hz
		+
		+	Which, converted into curies is <math> 9.27 \times 10^{-17} \pm 1.23 \times 10^{-7} </math> Ci. This is well below the 16 pCi limit

← Older revision		Revision as of 20:51, 15 October 2018
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	<math> A_{\epsilon} = \frac{A_{Beam Off}}{\epsilon} = 148655.17 \pm 19211.01 </math> Hz		<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>

← Older revision		Revision as of 20:35, 15 October 2018
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	More details can be found at https://wiki.iac.isu.edu/index.php/LB_May_2017_Det_A_Efficiency		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

← Older revision		Revision as of 20:32, 15 October 2018
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	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>)		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		<math> A_{Beam Off} = A_{Measure} e^{\lambda t} = 431.10 Hz \pm 54.19 </math> Hz

← Older revision		Revision as of 20:30, 15 October 2018
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	<math> A_{Beam Off} = A_{Measure} e^{\lambda t} = 431.10 Hz \pm 54.19 </math> Hz		<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

@@ Line 22: / Line 22: @@
 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>
+<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>)
-<math> A_{Beam Off} = A_{Measure} e^{\lambda t} </math>
+<math> A_{Beam Off} = A_{Measure} e^{\lambda t} = 431.10 Hz \pm 54.19 </math> Hz

← Older revision		Revision as of 20:18, 15 October 2018
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	==Histogram analysis==		==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 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
		+
		+	A = <math> \frac{2733.8 \pm 107.05}{500} = 5.5 \pm 0.2 </math>
		+
		+	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>