Difference between revisions of "Se170063 Thin Window Analysis"

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Line 12: Line 12:
 
||Original Window ||[[File:170063 PureSe 400 640Sec OGWindow.png|100px]] || [[File:170063 PureSe 1100 1360 Seconds OGWindow.png|100px]] || [[File:170063 PureSeSpec 1875 2150Sec OGWindow.png|100px]] || [[File:170063 PureSeSpec 2650 2930Sec OGWindow.png|100px]] || [[File:170063 PureSeSpec 3400 3690Sec OGWindow.png|100px]] || [[File:170063 PureSeSpec 4120 4400Sec OGWindow.png|100px]] || [[File:170063 PureSeSpec 4840 5130 OGWindow.png|100px]]  
 
||Original Window ||[[File:170063 PureSe 400 640Sec OGWindow.png|100px]] || [[File:170063 PureSe 1100 1360 Seconds OGWindow.png|100px]] || [[File:170063 PureSeSpec 1875 2150Sec OGWindow.png|100px]] || [[File:170063 PureSeSpec 2650 2930Sec OGWindow.png|100px]] || [[File:170063 PureSeSpec 3400 3690Sec OGWindow.png|100px]] || [[File:170063 PureSeSpec 4120 4400Sec OGWindow.png|100px]] || [[File:170063 PureSeSpec 4840 5130 OGWindow.png|100px]]  
 
|-
 
|-
−
||Expanded Window || [[File:170063 PureSe Spect 400 640Sec ExpWindow.png|100px]] || [[File:170063 PureSeSpectrum 1100 1360Sec ExpWindow.png|100px]] || [[File:170063 PureSeSpec 1875 2150 ExpWindow.png|100px]] || [[File:170063 PureSeSpec 2650 2930Sec ExpWindow.png|100px]] ||[[File:170063 PureSeSpec 3400 3690Sec ExpWindow.png|100px]] || [[File:170063 PureSeSpec 4120 4400Sec ExpWindow.png|100px]] || [[File:170063 PureSeSpec 4840 5130Sec ExpWindow.png|100px]]
+
||Expanded Window || [[File:170063 PureSe Spect 400 640Sec ExpWindow.png|100px]] || [[File:170063 PureSeSpec 1100 1360 ExpWindow 1 11.png|100px]] || [[File:170063 PureSeSpec 1875 2150 ExpWidnow 1 11.png|100px]] || [[File:170063 PureSeSpec 2650 2930 ExpWindow 1 11.png|100px]] ||[[File:170063 PureSeSpec 3400 3940 ExpWindow 1 11.png|100px]] || [[File:170063 PureSeSpec 4120 4400 ExpWindow 1 11.png|100px]] || [[File:170063 PureSeSpec 4840 5130 ExpWindow 1 11.png|100px]]
 
|-
 
|-
 
||Maximum of Histogram|| 260980 || 241756 || 220348 || 191476 || 165549 || 139528 || 130930
 
||Maximum of Histogram|| 260980 || 241756 || 220348 || 191476 || 165549 || 139528 || 130930
Line 54: Line 54:
 
||Original Window || [[File:170063 MixSpec 0 300 OGWindow.png|100px]] || [[File:170063 MixSpec 730 1020Sec OGWindow.png|100px]] || [[File:170063 MixSpec 1480 1775Sec OGWindow.png|100px]] || [[File:170063 MixSpec 2250 2550Sec OGWindow.png|100px]] || [[File:170063 MixSpec 3050 3300Sec OGWindow.png|100px]] || [[File:170063 MixSpec 3775 4050Sec OGWindow.png|100px]] || [[File:170063 MixSpec 4480 4770Sec OGWindow.png|100px]]
 
||Original Window || [[File:170063 MixSpec 0 300 OGWindow.png|100px]] || [[File:170063 MixSpec 730 1020Sec OGWindow.png|100px]] || [[File:170063 MixSpec 1480 1775Sec OGWindow.png|100px]] || [[File:170063 MixSpec 2250 2550Sec OGWindow.png|100px]] || [[File:170063 MixSpec 3050 3300Sec OGWindow.png|100px]] || [[File:170063 MixSpec 3775 4050Sec OGWindow.png|100px]] || [[File:170063 MixSpec 4480 4770Sec OGWindow.png|100px]]
 
|-
 
|-
−
||Expanded Window || [[File:170063 MixSpec 0 300Sec ExpWindow.png|100px]] || [[File:170063 MixSpec 730 1020Sec ExpWindow.png|100px]] || [[File:170063 MixSpec 1480 1775Sec ExpWindow.png|100px]] || [[File:170063 MixSpec 2250 2550Sec ExpWindow.png|100px]] || [[File:170063 MixSpec 3050 3300Sec ExpWindow.png|100px]] || [[File:170063 MixSpec 3775 4050Sec ExpWindow.png|100px]] ||[[File:170063 MixSpec 4480 4770Sec ExpWindow.png|100px]]
+
||Expanded Window || [[File:170063 MixSpec 0 300 ExpWindow 1 11.png|100px]] || [[File:170063 MixSpec 730 1020 ExpWindow 1 11.png|100px]] || [[File:170063 MixSpec 1480 1775 ExpWindow 1 11.png|100px]] || [[File:170063 MixSpec 2250 2550 ExpWindow 1 11.png|100px]] || [[File:170063 MixSpec 3050 3300 ExpWindow 1 11.png|100px]] || [[File:170063 MixSpec 3775 4050 ExpWindow 1 11.png|100px]] ||[[File:170063 MixSpec 4480 4770 ExpWindow 1 11.png|100px]]
 
|-
 
|-
 
||Maximum of Histogram|| 173289 || 155770 || 135740 || 124656 || 88582 || 80993 || 77781
 
||Maximum of Histogram|| 173289 || 155770 || 135740 || 124656 || 88582 || 80993 || 77781
Line 439: Line 439:
 
|| || 400 <t< 640 sec || 1100 < t < 1360 sec || 1875 < t < 2150 || 2650 < t < 2930 sec || 3400 < t < 3690 sec || 4120 < t < 4400 sec || 4840 < t < 5130 sec  
 
|| || 400 <t< 640 sec || 1100 < t < 1360 sec || 1875 < t < 2150 || 2650 < t < 2930 sec || 3400 < t < 3690 sec || 4120 < t < 4400 sec || 4840 < t < 5130 sec  
 
|-  
 
|-  
−
||.dat file entry ||7.58 +/- 0.003 || 7.45 +/- 0.003 || 7.32 +/- 0/003 || 7.15 +/- 0.004 || 6.94 +/- 0.004 || 6.82 +/- 0.004 || 6.68 +/- 0.004
+
||.dat file entry ||7.58 +/- 0.11 || 7.45 +/- 0.10 || 7.32 +/- 0.10 || 7.15 +/- 0.10 || 6.94 +/- 0.13 || 6.82 +/- 0.11 || 6.68 +/- 0.11
 +
|-
 +
|}
 +
 
 +
Below is a table for the self comparison of the sample runs
 +
 
 +
{| border="3"  cellpadding="5" cellspacing="0"
 +
|| || Run 1 || Run 2 || Run 3 || Run 4 || Run 5 || Run 6 ||
 +
|-
 +
|| Original Measurement (Hz) || 1961.89 +/- 216.27 || 1713.22 +/- 175.05 || 1510.32 +/- 151.17 || 1274.03 +/- 130.1 || 1036.14 +/- 131.17 || 913.17 +/- 105.31
 +
|-
 +
||n+1 Measurement (Hz) || 1713.22 +/- 175.05 || 1510.32 +/- 151.17 || 1274.03 +/- 130.1 || 1036.14 +/- 131.17 || 913.17 +/- 105.31 || 793.67 +/- 90.19
 +
|-
 +
||Time Correction Factor (Sec) ||720 || 790 || 780 || 760 || 710 ||730
 +
|-
 +
||n+1 Measurement Time Corrected (Hz) ||1980.97 +/- 202.41 || 1771.19 +/- 177.28 || 1491.08 +/- 152.26 || 1207.78 +/- 152.90 || 1053.76 +/- 121.52 || 919.56 +/- 104.50
 +
|-
 +
||n Measurement/n+1 Corrected Measurement || 0.99 +/- 0.15 || 0.97 +/- 0.14 || 1.01 +/- 0.14 || 1.05 +/- 0.17 || 0.98 +/- 0.17 || 0.99 +/- 0.16
 
|-
 
|-
 
|}
 
|}
Line 582: Line 599:
 
From this we can see that  
 
From this we can see that  
  
−
<math> I \pm \sigma_I =  238600 \pm 488.47 Counts </math>  
+
<math> I \pm \sigma_I =  238600 \pm 39100 Counts </math>  
  
 
and the integrated background is given by  
 
and the integrated background is given by  
Line 590: Line 607:
 
Now we can find the background subtracted signal  
 
Now we can find the background subtracted signal  
  
−
<math> N \pm \sigma_N = 209680 \pm 502.27 Counts</math>  
+
<math> N \pm \sigma_N = 209680 \pm 39100 Counts</math>  
  
 
Now convert this into an activity by dividing by the runtime  
 
Now convert this into an activity by dividing by the runtime  
  
−
<math> A_{Measure} \pm \sigma_{A_{Measure}} = \frac{209680}{300} \pm \frac{502.27}{300} = 698.93 \pm 1.67 Hz </math>
+
<math> A_{Measure} \pm \sigma_{A_{Measure}} = \frac{209680}{300} \pm \frac{39100}{300} = 698.93 \pm 130.33 Hz </math>
  
 
Now we can find the true value using the integrated measurement
 
Now we can find the true value using the integrated measurement
Line 601: Line 618:
  
  
−
<math> \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.72 Hz </math>
+
<math> \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 134.31 Hz </math>
  
 
So now the true value for the activity is  
 
So now the true value for the activity is  
  
−
<math> A_{True} \pm \sigma_{A_{True}} = 720.29 \pm 1.72 Hz </math>
+
<math> A_{True} \pm \sigma_{A_{True}} = 720.29 \pm 134.31 Hz </math>
  
 
Now we must correct for the dead time. The 1st run had an average count rate of 1401.98 Hz, which corresponds to a dead time of 2.53 +/- 0.31
 
Now we must correct for the dead time. The 1st run had an average count rate of 1401.98 Hz, which corresponds to a dead time of 2.53 +/- 0.31
Line 617: Line 634:
 
So the real activity for the 2nd measurement of the Mixed sample is  
 
So the real activity for the 2nd measurement of the Mixed sample is  
  
−
<math> A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 738.99  \pm 2.94 </math>
+
<math> A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 738.99  \pm 137.82 </math>
  
 
==Activity of the Fifth Measurement==
 
==Activity of the Fifth Measurement==
Line 627: Line 644:
 
From this we can see that  
 
From this we can see that  
  
−
<math> I \pm \sigma_I =  170100 \pm 412.43 Counts </math>  
+
<math> I \pm \sigma_I =  170100 \pm 28500 Counts </math>  
  
 
and the integrated background is given by  
 
and the integrated background is given by  
Line 635: Line 652:
 
Now we can find the background subtracted signal  
 
Now we can find the background subtracted signal  
  
−
<math> N \pm \sigma_N = 148200 \pm 424.69 Counts</math>  
+
<math> N \pm \sigma_N = 148200 \pm 28500 Counts</math>  
  
 
Now convert this into an activity by dividing by the runtime  
 
Now convert this into an activity by dividing by the runtime  
  
−
<math> A_{Measure} \pm \sigma_{A_{Measure}} = \frac{148200}{250} \pm \frac{424.69}{250} = 592.8 \pm 1.70 Hz </math>
+
<math> A_{Measure} \pm \sigma_{A_{Measure}} = \frac{148200}{250} \pm \frac{28500}{250} = 592.8 \pm 114 Hz </math>
  
 
Now we can find the true value using the integrated measurement
 
Now we can find the true value using the integrated measurement
Line 646: Line 663:
  
  
−
<math> \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.74 Hz </math>
+
<math> \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 116.90 Hz </math>
  
 
So now the true value for the activity is  
 
So now the true value for the activity is  
  
−
<math> A_{True} \pm \sigma_{A_{True}} = 607.87 \pm 1.74 Hz </math>
+
<math> A_{True} \pm \sigma_{A_{True}} = 607.87 \pm 116.90 Hz </math>
  
 
Now we must correct for the dead time. The 1st run had an average count rate of 1025.13 Hz, which corresponds to a dead time of 1.92 +/- 0.26
 
Now we must correct for the dead time. The 1st run had an average count rate of 1025.13 Hz, which corresponds to a dead time of 1.92 +/- 0.26
Line 662: Line 679:
 
So the real activity for the 2nd measurement of the Mixed sample is  
 
So the real activity for the 2nd measurement of the Mixed sample is  
  
−
<math> A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 619.70  \pm 2.42 </math>
+
<math> A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 619.70  \pm 119.2 </math>
  
 
==Sixth Mix Sample Measurement==
 
==Sixth Mix Sample Measurement==
Line 672: Line 689:
 
From this we can see that  
 
From this we can see that  
  
−
<math> I \pm \sigma_I =  153200 \pm 391.41 Counts </math>  
+
<math> I \pm \sigma_I =  153200 \pm 28000 Counts </math>  
  
 
and the integrated background is given by  
 
and the integrated background is given by  
Line 680: Line 697:
 
Now we can find the background subtracted signal  
 
Now we can find the background subtracted signal  
  
−
<math> N \pm \sigma_N = 132140 \pm 403.93 Counts</math>  
+
<math> N \pm \sigma_N = 132140 \pm 28000 Counts</math>  
  
 
Now convert this into an activity by dividing by the runtime  
 
Now convert this into an activity by dividing by the runtime  
  
−
<math> A_{Measure} \pm \sigma_{A_{Measure}} = \frac{132140}{275} \pm \frac{403.93}{275} = 480.51 \pm 1.47 Hz </math>
+
<math> A_{Measure} \pm \sigma_{A_{Measure}} = \frac{132140}{275} \pm \frac{28000}{275} = 480.51 \pm 101.81 Hz </math>
  
 
Now we can find the true value using the integrated measurement
 
Now we can find the true value using the integrated measurement
Line 691: Line 708:
  
  
−
<math> \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.51 Hz </math>
+
<math> \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 104.66 Hz </math>
  
 
So now the true value for the activity is  
 
So now the true value for the activity is  
  
−
<math> A_{True} \pm \sigma_{A_{True}} = 493.96 \pm 1.51 Hz </math>
+
<math> A_{True} \pm \sigma_{A_{True}} = 493.96 \pm 104.66 Hz </math>
  
 
Now we must correct for the dead time. The 1st run had an average count rate of 898.19 Hz, which corresponds to a dead time of 1.65 +/- 0.34
 
Now we must correct for the dead time. The 1st run had an average count rate of 898.19 Hz, which corresponds to a dead time of 1.65 +/- 0.34
Line 707: Line 724:
 
So the real activity for the 2nd measurement of the Mixed sample is  
 
So the real activity for the 2nd measurement of the Mixed sample is  
  
−
<math> A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 502.25  \pm 2.32 </math>
+
<math> A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 502.25  \pm 106.43 </math>
  
 
==Activity of the 7th Measurement==
 
==Activity of the 7th Measurement==
Line 717: Line 734:
 
From this we can see that  
 
From this we can see that  
  
−
<math> I \pm \sigma_I =  147900 \pm 384.58 Counts </math>  
+
<math> I \pm \sigma_I =  147900 \pm 26400 Counts </math>  
  
 
and the integrated background is given by  
 
and the integrated background is given by  
Line 725: Line 742:
 
Now we can find the background subtracted signal  
 
Now we can find the background subtracted signal  
  
−
<math> N \pm \sigma_N = 128282 \pm 396.48 Counts</math>  
+
<math> N \pm \sigma_N = 128282 \pm 26400 Counts</math>  
  
 
Now convert this into an activity by dividing by the runtime  
 
Now convert this into an activity by dividing by the runtime  
  
−
<math> A_{Measure} \pm \sigma_{A_{Measure}} = \frac{128282}{290} \pm \frac{396.48}{290} = 442.35 \pm 1.38 Hz </math>
+
<math> A_{Measure} \pm \sigma_{A_{Measure}} = \frac{128282}{290} \pm \frac{26400}{290} = 442.35 \pm 91.03 Hz </math>
  
 
Now we can find the true value using the integrated measurement
 
Now we can find the true value using the integrated measurement
Line 736: Line 753:
  
  
−
<math> \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.42 Hz </math>
+
<math> \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 93.72 Hz </math>
  
 
So now the true value for the activity is  
 
So now the true value for the activity is  
  
−
<math> A_{True} \pm \sigma_{A_{True}} = 455.41 \pm 1.42 Hz </math>
+
<math> A_{True} \pm \sigma_{A_{True}} = 455.41 \pm 93.72 Hz </math>
  
 
Now we must correct for the dead time. The 1st run had an average count rate of 846.40 Hz, which corresponds to a dead time of 1.65 +/- 0.34
 
Now we must correct for the dead time. The 1st run had an average count rate of 846.40 Hz, which corresponds to a dead time of 1.65 +/- 0.34
Line 750: Line 767:
 
So the real activity for the 2nd measurement of the Mixed sample is  
 
So the real activity for the 2nd measurement of the Mixed sample is  
  
−
<math> A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} =  463.05 \pm 2.16 </math>
+
<math> A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} =  463.05 \pm 95.31 </math>
  
 
{| border="3"  cellpadding="5" cellspacing="0"
 
{| border="3"  cellpadding="5" cellspacing="0"
 
|| || 0<t<300 || 730 <t< 1020 || 1480<t<1775 || 2250<t<2550 || 3050<t<3300 || 3775<t<4050 || 4480<t<4770
 
|| || 0<t<300 || 730 <t< 1020 || 1480<t<1775 || 2250<t<2550 || 3050<t<3300 || 3775<t<4050 || 4480<t<4770
 
|-
 
|-
−
||Data File Entry || 6.90 +/- 0.004 || 6.85 +/- 0.004 ||6.73 +/- 0.004 || 6.61 +/- 0.004 || 6.43 +/- 0.004 || 6.23 +/- 0.005 || 6.14 +/- 0.005
+
||Data File Entry || 6.90 +/- 0.30 || 6.85 +/- 0.22 ||6.73 +/- 0.19 || 6.61 +/- 0.19 || 6.43 +/- 0.19 || 6.23 +/- 0.21 || 6.14 +/- 0.21
 +
|-
 +
|}
 +
 
 +
 
 +
Below is a table to check the activities against later measurements by tracing the activity back in time and taking a ratio
 +
 
 +
{| border="3"  cellpadding="5" cellspacing="0"
 +
|| || Run 1 || Run 2 || Run 3 || Run 4 || Run 5 || Run 6
 +
|-
 +
|| Original Measurement (Hz) || 994.73 +/- 304.327 || 948.20 +/- 211.33 || 839.55 +/- 163.38 || 738.99 +/- 137.82 || 619.70 +/- 119.20 || 502.25 +/- 106.43
 +
|-
 +
||n+1 Measurement (Hz) || 948.20 +/- 211.33 || 839.55 +/- 163.38 ||738.99 +/- 137.82 || 619.70 +/- 119.20 ||502.25 +/- 106.43 || 463.05 +/- 95.31
 +
|-
 +
||Time Correction Factor (Sec) ||720 || 755 || 775 || 750 || 750 || 720
 +
|-
 +
||n+1 Measurement Time Corrected (Hz) ||1096.39 +/- 244.36 || 977.63 +/- 190.25 || 864.01 +/- 161.14 || 720.90 +/- 138.67 || 584.27 +/- 123.81 || 535.42 +/- 110.21
 +
|-
 +
||n Measurement/n+1 Corrected Measurement || 0.91 +/- 0.33 || 0.97 +/- 0.29 || 0.97 +/- 0.26|| 1.02 +/- 0.27 || 1.06 +/- 0.3 || 0.94 +/- 0.28
 
|-
 
|-
 
|}
 
|}
  
 
==Plots==
 
==Plots==
 +
 +
Below is a table for the plots of the cross sample ratios
 +
 +
{| border="3"  cellpadding="5" cellspacing="0"
 +
|| || Run 1 || Run 2 || Run 3 || Run 4 || Run 5 || Run 6 || Run 7
 +
|-
 +
|| Mix Measurement (Hz) || 994.73 +/- 304.327 || 948.20 +/- 211.33 || 839.55 +/- 163.38 || 738.99 +/- 137.82 || 619.70 +/- 119.2 || 502.25 +/- 106.43 || 463.05 +/- 95.31
 +
|-
 +
||Original Pure Se Measurement (Hz) || 1961.89 +/- 216.27 || 1713.22 +/- 175.05 || 1510.32 +/- 151.17 || 1274.03 +/- 130.1 || 1036.14 +/- 131.17 || 913.17 +/- 105.31 || 793.67 +/- 90.19
 +
|-
 +
||Time Correction Factor (Sec) ||340 || 340 ||375 || 380 || 390 || 350 || 360
 +
|-
 +
||Time Corrected Mix Measurement (Hz) ||1065.33 +/- 325.93 || 1015.50 +/- 226.33 || 905.51 +/- 176.22 || 797.85 +/- 148.80 || 670.41 +/- 128.95 || 538.98 +/- 114.214  || 497.92 +/-  102.49
 +
|-
 +
|| Time Corrected Mix Measurement/Pure Se Measurement || 0.54 +/- 0.18 || 0.59 +/- 0.14 || 0.60 +/- 0.13 || 0.63 +/- 0.13 || 0.65 +/- 0.15  || 0.59 +/- 0.14 || 0.63 +/- 0.15
 +
|-
 +
|}
  
 
Below are the time vs. frequency plots
 
Below are the time vs. frequency plots
  
−
[[File:170063 MixHL DTCorrected.png|200px]]
+
[[File:170063 MixHLPlot 1 11 18.png|200px]]
−
[[File:170063 PureSeHL DTCorrected.png|200px]]
+
[[File:170063 PureSeHLPlot 1 11 18.png|200px]]
 +
 
 +
By exponentiating we find that the initial activity of the Mixture is 1056.48 +/- 2.91 Hz, while the pure sample has an initial activity of 2162.90 +/- 5.16 Hz which gives a ratio of
 +
 
 +
<math> 0.488 \pm 0.002 </math>
  
 
To check consistency I would time correct measurements and take a ratio to compare, for example measurement 1 was taken and a 2nd measurement was taken some time later. So correct the rate of the second measurement back to the first measurement by using the radioactive decay equation, then take a ratio of these 2 numbers.
 
To check consistency I would time correct measurements and take a ratio to compare, for example measurement 1 was taken and a 2nd measurement was taken some time later. So correct the rate of the second measurement back to the first measurement by using the radioactive decay equation, then take a ratio of these 2 numbers.
  
−
[[File:170063 MixRateRatios.png|200px]]
+
[[File:170063 Mix RateRatios 1 11 18.png|200px]]
−
[[File:170063 PureSeRateRatios.png|200px]]
+
[[File:170063 PureSe RateRatio 1 11 18.png|200px]]
  
 
Another check to do is to take the ratio of the mixture and the pure selenium sample (time corrected to the same time for each measurement) and see if the ratio here is relatively constant
 
Another check to do is to take the ratio of the mixture and the pure selenium sample (time corrected to the same time for each measurement) and see if the ratio here is relatively constant
  
−
[[File:170063 CrossSampleRatios.png|200px]]
+
[[File:170063 CrossSample RateRatios 1 11 18.png|200px]]
 
   
 
   
−
By exponentiating we find that the initial activity of the Mixture is 1056.48 +/- 2.91 Hz, while the pure sample has an initial activity of 2162.90 +/- 5.16 Hz which gives a ratio of  
+
By exponentiating we find that the initial activity of the Mixture is 1082.15 +/- 174.89 Hz, while the pure sample has an initial activity of 2169.13 +/- 175.01 Hz which gives a ratio of  
  
−
<math> 0.488 \pm 0.002 </math>  
+
Thin Window Mean + 2 Sigma:<math> 0.50 \pm 0.09 </math> ,Original Window + 2 Sigma or original window
  
 
We can also check the ratio of the nickel with the selenium mixture. The rate of the nickel foil was <math> 1.127931 \times 10^6 \pm 7.258 \times 10^3 Hz </math>  
 
We can also check the ratio of the nickel with the selenium mixture. The rate of the nickel foil was <math> 1.127931 \times 10^6 \pm 7.258 \times 10^3 Hz </math>  
Line 791: Line 847:
  
 
308985.7143 +/- 882.6876 Hz for the pure selenium sample
 
308985.7143 +/- 882.6876 Hz for the pure selenium sample
−
 
−
We can also look at the ratios between the nickel and the selenium ash mixture
 
−
 
−
[[File:170063 NickelMix RateRatios.png|200px]]
 

Latest revision as of 19:19, 11 January 2018

In order to try to pin down the ratio of pure selenium to the selenium in the soil, try using a 2 channel window to find the signal. The uncertainty in the number of counts in the signal will be determined by subtracting the counts from the 2 channel window from the counts from a window expanded by 1 standard deviation, then divided by the original number of counts in the 2 channel window. Since the max value is fixed, it does not appear in the stats box, but I have included it in the table. It would seem that widening the gaussian function by half of its standard deviation yields results that agree up to 3 digits after the decimal with the result of widening the gaussian by its standard deviation. This would imply not much has changed between these methods.

Changes Made to table:

12/14: Error in signal changed to [math] \sigma_I = I_{Expanded} - I [/math] (try to not underestimate error) This includes changing the expanded window to a window expanded off of the 2 channel window

400 <t< 640 sec 1100 < t < 1360 sec 1875 < t < 2150 2650 < t < 2930 sec 3400 < t < 3690 sec 4120 < t < 4400 sec 4840 < t < 5130 sec
Thin Window 170063 PureSe ThinWindow 400 640Sec.png 170063 PureSe 1100 1360Sec ThinWindow.png 170063 PureSeSpec 1875 2150Sec ThinWindow.png 170063 PureSeSpec 2650 2930Sec ThinWindow.png 170063 PureSeSpec 3400 3690Sec ThinWindow.png 170063 PureSeSpec 4120 4400Sec ThinWindow.png 170063 PureSeSpec 4840 5130Sec ThinWindow.png
Original Window 170063 PureSe 400 640Sec OGWindow.png 170063 PureSe 1100 1360 Seconds OGWindow.png 170063 PureSeSpec 1875 2150Sec OGWindow.png 170063 PureSeSpec 2650 2930Sec OGWindow.png 170063 PureSeSpec 3400 3690Sec OGWindow.png 170063 PureSeSpec 4120 4400Sec OGWindow.png 170063 PureSeSpec 4840 5130 OGWindow.png
Expanded Window 170063 PureSe Spect 400 640Sec ExpWindow.png 170063 PureSeSpec 1100 1360 ExpWindow 1 11.png 170063 PureSeSpec 1875 2150 ExpWidnow 1 11.png 170063 PureSeSpec 2650 2930 ExpWindow 1 11.png 170063 PureSeSpec 3400 3940 ExpWindow 1 11.png 170063 PureSeSpec 4120 4400 ExpWindow 1 11.png 170063 PureSeSpec 4840 5130 ExpWindow 1 11.png
Maximum of Histogram 260980 241756 220348 191476 165549 139528 130930
Original Background 10560 +/- 51.58 10560 +/- 51.41 9496 +/- 48.9 8898 +/- 47.1 9418 +/- 48.2 7485 +/- 43 7032 +/- 31.1
Expanded Background 10750 +/- 42.42 10630 +/- 42.09 9674 +/- 40.2 9029 +/- 38.7 9248 +/- 39.1 7634 +/- 35.5 7161 +/- 34.2
Signal Background (Take Larger Error) 10560 +/- 51.58 10560 +/- 51.41 9496 +/- 48.9 8898 +/- 47.1 9418 +/- 48.2 7485 +/- 43 7161 +/- 34.2
Integrated Background 21120 +/- 103.16 21120 +/- 102.82 18992 +/- 97.8 17796 +/- 94.2 18836 +/- 96.4 14970 +/- 86 14322 +/- 68.4
Signal in Thin Window 466700 444100 412800 356800 304100 258700 233500
Signal In Expanded Window 515800 487300 452200 391400 340200 286800 258400
Error in Signal 49100 43200 39400 34600 36100 28100 24900
Signal In Thin Window w/ Error 466700 +/- 49100 444100 +/- 43200 412800 +/- 39400 356800 +/- 34600 304100 +/- 36100 258700 +/- 28100 233500 +/- 24900
Background Subtracted Signal 445580 +/- 49100 422980 +/- 43200 393808 +/- 39400 339004 +/- 34600 285264 +/- 36100 243730 +/- 28100 219178 +/- 24900
Time Correction Factor (Sec) 400 370 395 400 350 345 360
Corrected Counts 483015.53 +/- 109095.5248 455750 +/- 115595.0622 426463.63 +/- 114048.1804 367485.52 +/- 97638.49218 306127.92 +/- 75691.81615 261292.55 +/- 67639.40782 235683.31 +/- 56257.91145
.dat file entry 7.607165162 +/- 0.2258633895 7.469018062 +/- 0.2536369988 7.346511269 +/- 0.2676838585 7.179649592 +/- 0.2656934406 6.96187741 +/- 0.2472555138 6.838606337 +/- 0.2588646627 6.700363353 +/- 0.2387012956

Below is the table for the Se-Ash Mixture


0 <t< 300 sec 730 < t < 1020 sec 1480 < t < 1775 2250 < t < 2550 sec 3050< t < 3300 sec 3775 < t < 4050 sec 4480 < t < 4770 sec
Thin Window 170063 MixSpec 0 300Sec ThinWindow.png 170063 MixSpec 730 1020Sec ThinWindow.png 170063 MixSpec 1480 1775Sec ThinWindow.png 170063 MixSpec 2250 2550Sec ThinWindow.png 170063 MixSpec 3050 3300Sec ThinWindow.png 170063 MixSpec 3775 4050Sec ThinWindow.png 170063 MixSpec 4480 4770Sec ThinWindow.png
Original Window 170063 MixSpec 0 300 OGWindow.png 170063 MixSpec 730 1020Sec OGWindow.png 170063 MixSpec 1480 1775Sec OGWindow.png 170063 MixSpec 2250 2550Sec OGWindow.png 170063 MixSpec 3050 3300Sec OGWindow.png 170063 MixSpec 3775 4050Sec OGWindow.png 170063 MixSpec 4480 4770Sec OGWindow.png
Expanded Window 170063 MixSpec 0 300 ExpWindow 1 11.png 170063 MixSpec 730 1020 ExpWindow 1 11.png 170063 MixSpec 1480 1775 ExpWindow 1 11.png 170063 MixSpec 2250 2550 ExpWindow 1 11.png 170063 MixSpec 3050 3300 ExpWindow 1 11.png 170063 MixSpec 3775 4050 ExpWindow 1 11.png 170063 MixSpec 4480 4770 ExpWindow 1 11.png
Maximum of Histogram 173289 155770 135740 124656 88582 80993 77781
Original Background 29440 +/- 84.93 20580 +/- 70.31 16420 +/- 62.52 14460 +/- 58.47 10950 +/- 50.65 10530 +/- 49.90 9809 +/- 48.2
Expanded Background 28960 +/- 69.12 20580 +/- 57.79 16500 +/- 51.60 14500 +/- 48.2 10630 +/- 41.29 10460 +/- 41.01 9814 +/- 39.7
Signal Background (Take Larger Error) 29440 +/- 84.93 20580 +/- 70.31 16420 +/- 62.52 14460 +/- 58.47 10950 +/- 50.65 10530 +/- 49.90 9809 +/- 48.2
Integrated Background 58880 +/- 169.86 41160 +/- 140.62 32840 +/- 131.04 28920 +/- 116.94 21900 +/- 101.3 21060 +/- 99.8 19618 +/- 96.4
Signal in Thin Window 342200 300100 267200 238600 170100 153200 147900
Signal In Expanded Window 423600 357800 312800 277700 198600 181200 174300
Error In Signal 81400 57700 45600 39100 28500 28000 26400
Signal with Error 342200 +/- 81400 300100 +/- 57700 267200 +/- 45600 238600 +/- 39100 170100 +/- 28500 153200 +/- 28000 147900 +/- 26400
Background Subtracted Signal 283320 +/- 81400 258940 +/- 57700 234360 +/- 45600 209680 +/- 39100 148200 +/- 28500 132140 +/- 28000 128282 +/- 26400
.dat file entry 6.850549805 +/- 0.3278271919 6.794470731 +/- 0.3027342241 6.677638317 +/- 0.3181179382 6.549555363 +/- 0.3026659672 6.384857074 +/- 0.3184246458 6.174846148 +/- 0.3010453307 6.092105322 +/- 0.2951787468


Below is a plot of the ratio of the activities where the pure selenium sample has been time corrected back to match the run time of the mixture.

170063 MixVsPureRatio ThinWindowMethod.png

Note that this does give the expected result that the ratio between the two samples is statistically the same as 0.5, which is the ratio of the masses. The only problem here is that the error is quite large, The average of the uncertainty divided by the mean is 39.6%.

Below are the half life plots for the Pure Selenium sample and the ash mixture


170063 MixHLPlot ThinWindowMethod.png

170063 PureSeHLPlot ThinWindowMethod.png


The slope of the mixture plot gives a half life of 63.48 +/- 26.85 minutes. This does overlap, but the error is extremely large.

The slope of the pure selenium sample plot gives a half life of 56.08 +/- 16.33 minutes. Again this does overlap, but the error is still quite large.

The ratio of the initial activities is 0.49 +/- 0.13. Now this does fall within 1 standard deviation of 0.5, which is the value we want.


Dead Time Corrected

I have shown that by tracing the activity back in time for the pure selenium sample that the activities are the same here LB Thesis Thin Window Analysis, but now I must make sure that this will actually give the ratio that is expected.

Begin with the third measurement that was taken for the pure selenium sample

Pure Se Analysis

This will be using a new method as of 12/14 to find the error in the signal. To do this, find the intrinsic error by

[math] \frac{I_{Expanded}-I}{I} [/math]

Where the expanded integral value is the integral from the stats box after the 2 channel window has been increased by 2 sigma. This number will then be multiplied by the background subtracted thin window signal. So the error in the signal is

[math] \sigma_I = I_{Expanded} - I [/math]

First Pure Se Measurement

Below is the histogram of interest

170063 PureSe ThinWindow 400 640Sec.png

From this we can see that

[math] I \pm \sigma_I = 466700 \pm 49100 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 21120 \pm 103.16 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 445580 \pm 49100 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{445580}{240} \pm \frac{49100}{240} = 1856.58 \pm 204.58 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 240} \times 1856.58 \times 240 \times \lambda}{e^{\lambda \times 240}-1} = 1901.86 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 209.57 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 1901.86 \pm 209.57 Hz [/math]

Now we must correct for the dead time. The third run had an average count rate of 1387.06 Hz, which corresponds to a dead time of 3.06 +/- 0.30

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1901.86 \times \frac{1}{0.9694} = 1961.89 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.0306 [/math]

[math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the third measurement of the Pure Se sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1961.89 \pm 216.27 [/math]

Second Pure Se Measurement

Below is the histogram of interest

170063 PureSe 1100 1360Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 444100 \pm 43200 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 21120 \pm 102.82 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 422980 \pm 43200 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{422980}{260} \pm \frac{43200}{260} = 1626.85 \pm 166.15 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 260} \times 1626.85 \times 260 \times \lambda}{e^{\lambda \times 260}-1} = 1669.88 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 170.54 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 1669.88 \pm 170.54 Hz [/math]

Now we must correct for the dead time. The third run had an average count rate of 1387.06 Hz, which corresponds to a dead time of 2.53 +/- 0.31

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1669.88 \times \frac{1}{0.9747} = 1713.22 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.025 [/math]

[math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the third measurement of the Pure Se sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1713.22 \pm 175.05 [/math]

Third Pure Se Measurement

Below is the histogram of interest

170063 PureSeSpec 1875 2150Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 412800 \pm 39400 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 18992 \pm 97.8 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 393808 \pm 39400 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{393808}{275} \pm \frac{39400}{275} = 1432.03 \pm 143.27 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 275} \times 1432.03 \times 275 \times \lambda}{e^{\lambda \times 275}-1} = 1472.11 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 147.28 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 1472.11 \pm 147.28 Hz [/math]

Now we must correct for the dead time. The third run had an average count rate of 1387.06 Hz, which corresponds to a dead time of 2.53 +/- 0.3

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1472.11 \times \frac{1}{0.9747} = 1510.32 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003 [/math]

[math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the third measurement of the Pure Se sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1510.32 \pm 151.17 [/math]

Below is a table representing the numbers that will be put into a data file to be linearly fit (Also using the values from the thesis portion)

Fourth Pure Se Measurement

Now do the analysis for the 4th measurement taken on the pure selenium sample.

Below is the histogram of interest

170063 PureSeSpec 2650 2930Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 356800 \pm 34600 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 17796 \pm 94.2 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 339004 \pm 34600 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{339004}{280} \pm \frac{34600}{280} = 1210.73 \pm 123.57 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 280} \times 1210.73 \times 280 \times \lambda}{e^{\lambda \times 280}-1} =1245.24 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 127.09 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 1245.24 \pm 127.09 Hz [/math]

Now we must correct for the dead time. The 4th run had an average count rate of 1274.32 Hz, which corr5esponds to a dead time of 2.26 +/- 0.33

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1245.24 \times \frac{1}{0.9774} = 1274.03 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.004 [/math]

[math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the third measurement of the Pure Se sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1274.03 \pm 130.1 [/math]

Fifth Pure Se Measurement

Now do the analysis for the 5th measurement taken on the pure selenium sample.

Below is the histogram of interest

170063 PureSeSpec 3400 3690Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 304100 \pm 36100 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 18836 \pm 96.4 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 285264 \pm 36100 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{285264}{290} \pm \frac{36100}{290} = 983.67 \pm 124.48 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 290} \times 983.67 \times 290 \times \lambda}{e^{\lambda \times 290}-1} = 1012.72 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 128.16 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 1012.72 \pm 128.16 Hz [/math]

Now we must correct for the dead time. The 5th run had an average count rate of 1168.06 Hz, which corresponds to a dead time of 2.26 +/- 0.33

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1012.72 \times \frac{1}{0.9774} = 1036.14 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003 [/math] [math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the third measurement of the Pure Se sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1036.14 \pm 131.17 [/math]

Sixth Pure Se Measurement

Now do the analysis for the 6th measurement taken on the pure selenium sample.

Below is the histogram of interest

170063 PureSeSpec 4120 4400Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 258700 \pm 28100 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 14970 \pm 86 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 243730 \pm 28100 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{243730}{280} \pm \frac{28100}{280} = 870.46 \pm 100.36 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 280} \times 870.46 \times 280 \times \lambda}{e^{\lambda \times 280}-1} = 895.27 Hz [/math]

[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 103.22 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 895.27 \pm 103.22 Hz [/math]

Now we must correct for the dead time. The 6th run had an average count rate of 1168.06 Hz, which corresponds to a dead time of 1.96 +/- 0.26

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 895.27 \times \frac{1}{0.9804} = 913.17 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003 [/math] [math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the 6th measurement of the Pure Se sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 913.17 \pm 105.31 [/math]

Seventh Pure Se Measurement

Now do the analysis for the 7th measurement taken on the pure selenium sample.

Below is the histogram of interest

170063 PureSeSpec 4840 5130Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 233500 \pm 24900 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 14322 \pm 68.4 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 219178 \pm 24900 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{219178}{290} \pm \frac{24900}{290} = 755.79 \pm 85.86 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 290} \times 755.79 \times 290 \times \lambda}{e^{\lambda \times 290}-1} = 778.11 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 88.40 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 778.11 \pm 88.40 Hz [/math]

Now we must correct for the dead time. The 7th run had an average count rate of 1012 Hz, which corresponds to a dead time of 1.96 +/- 0.26

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 778.11 \times \frac{1}{0.9804} = 793.67 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003 [/math] [math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the 7th measurement of the Pure Se sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 793.67 \pm 90.19 [/math]

400 <t< 640 sec 1100 < t < 1360 sec 1875 < t < 2150 2650 < t < 2930 sec 3400 < t < 3690 sec 4120 < t < 4400 sec 4840 < t < 5130 sec
.dat file entry 7.58 +/- 0.11 7.45 +/- 0.10 7.32 +/- 0.10 7.15 +/- 0.10 6.94 +/- 0.13 6.82 +/- 0.11 6.68 +/- 0.11

Below is a table for the self comparison of the sample runs

Run 1 Run 2 Run 3 Run 4 Run 5 Run 6
Original Measurement (Hz) 1961.89 +/- 216.27 1713.22 +/- 175.05 1510.32 +/- 151.17 1274.03 +/- 130.1 1036.14 +/- 131.17 913.17 +/- 105.31
n+1 Measurement (Hz) 1713.22 +/- 175.05 1510.32 +/- 151.17 1274.03 +/- 130.1 1036.14 +/- 131.17 913.17 +/- 105.31 793.67 +/- 90.19
Time Correction Factor (Sec) 720 790 780 760 710 730
n+1 Measurement Time Corrected (Hz) 1980.97 +/- 202.41 1771.19 +/- 177.28 1491.08 +/- 152.26 1207.78 +/- 152.90 1053.76 +/- 121.52 919.56 +/- 104.50
n Measurement/n+1 Corrected Measurement 0.99 +/- 0.15 0.97 +/- 0.14 1.01 +/- 0.14 1.05 +/- 0.17 0.98 +/- 0.17 0.99 +/- 0.16

Se-Sage Mix Analysis

Activity of the First Measurement

Below is the histogram of interest

170063 MixSpec 0 300Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 342200 \pm 81400 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 58880 \pm 169.86 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 283320 \pm 81400 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{283320}{300} \pm \frac{84100}{300} = 944.4 \pm 280.33 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 300} \times 944.4 \times 300 \times \lambda}{e^{\lambda \times 300}-1} = 973.26 Hz [/math]

[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 288.90 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} =973.26 \pm 288.90 Hz [/math]

Now we must correct for the dead time. The 1st run had an average count rate of 2434.44 Hz, which corresponds to a dead time of 5.06 +/- 0.39

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 944.4 \times \frac{1}{0.9494} = 994.73 Hz [/math]

[math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the 1st measurement of the Mixed sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 994.73 \pm 304.327 [/math]

Activity of the Second Measurement

Below is the histogram of interest

170063 MixSpec 730 1020Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 300100 \pm 57700 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 41160 \pm 140.62 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 258940 \pm 57700 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{258940}{290} \pm \frac{57700}{290} = 892.90 \pm 198.97 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 290} \times 892.90 \times 290 \times \lambda}{e^{\lambda \times 300}-1} = 919.27 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 204.84 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 919.27 \pm 204.84 Hz [/math]

Now we must correct for the dead time. The 1st run had an average count rate of 1756.30 Hz, which corresponds to a dead time of 3.06 +/- 0.30

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 919.27 \times \frac{1}{0.9694} = 948.29 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.0.003 [/math] [math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the 2nd measurement of the Mixed sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 948.20 \pm 211.33 [/math]

Activity of the Third Measurement

Below is the histogram of interest

170063 MixSpec 1480 1775Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 267200 \pm 45600 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 32840 \pm 131.04 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 234360 \pm 45600 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{234360}{295} \pm \frac{45600}{295} = 794.44 \pm 154.58 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 295} \times 794.44 \times 295 \times \lambda}{e^{\lambda \times 295}-1} = 818.31 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 159.22 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 818.31 \pm 159.22 Hz [/math]

Now we must correct for the dead time. The 1st run had an average count rate of 1401.98 Hz, which corresponds to a dead time of 2.53 +/- 0.31

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 818.31 \times \frac{1}{0.9747} = 839.55 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003 [/math] [math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the 2nd measurement of the Mixed sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 839.55 \pm 163.38 [/math]

Fourth Mix Sample Measurement

Below is the histogram of interest

170063 MixSpec 2250 2550Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 238600 \pm 39100 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 28920 \pm 116.94 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 209680 \pm 39100 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{209680}{300} \pm \frac{39100}{300} = 698.93 \pm 130.33 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 300} \times 698.93 \times 300 \times \lambda}{e^{\lambda \times 300}-1} = 720.29 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 134.31 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 720.29 \pm 134.31 Hz [/math]

Now we must correct for the dead time. The 1st run had an average count rate of 1401.98 Hz, which corresponds to a dead time of 2.53 +/- 0.31

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 720.29 \times \frac{1}{0.9747} = 738.99 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003 [/math]

[math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]>

So the real activity for the 2nd measurement of the Mixed sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 738.99 \pm 137.82 [/math]

Activity of the Fifth Measurement

Below is the histogram of interest

170063 MixSpec 3050 3300Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 170100 \pm 28500 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 21900 \pm 101.3 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 148200 \pm 28500 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{148200}{250} \pm \frac{28500}{250} = 592.8 \pm 114 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 250} \times 592.8 \times 250 \times \lambda}{e^{\lambda \times 250}-1} = 607.87 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 116.90 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 607.87 \pm 116.90 Hz [/math]

Now we must correct for the dead time. The 1st run had an average count rate of 1025.13 Hz, which corresponds to a dead time of 1.92 +/- 0.26

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 607.8 \times \frac{1}{0.9808} = 619.70 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003 [/math]

[math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the 2nd measurement of the Mixed sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 619.70 \pm 119.2 [/math]

Sixth Mix Sample Measurement

Below is the histogram of interest

170063 MixSpec 3775 4050Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 153200 \pm 28000 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 21060 \pm 99.8 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 132140 \pm 28000 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{132140}{275} \pm \frac{28000}{275} = 480.51 \pm 101.81 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 275} \times 480.51 \times 275 \times \lambda}{e^{\lambda \times 275}-1} = 493.96 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 104.66 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 493.96 \pm 104.66 Hz [/math]

Now we must correct for the dead time. The 1st run had an average count rate of 898.19 Hz, which corresponds to a dead time of 1.65 +/- 0.34

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 493.96 \times \frac{1}{0.9835} = 502.25 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.004 [/math]

[math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the 2nd measurement of the Mixed sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 502.25 \pm 106.43 [/math]

Activity of the 7th Measurement

Below is the histogram of interest

170063 MixSpec 4480 4770Sec ThinWindow.png

From this we can see that

[math] I \pm \sigma_I = 147900 \pm 26400 Counts [/math]

and the integrated background is given by

[math] B \pm \sigma_B = 19618 \pm 96.4 Counts[/math]

Now we can find the background subtracted signal

[math] N \pm \sigma_N = 128282 \pm 26400 Counts[/math]

Now convert this into an activity by dividing by the runtime

[math] A_{Measure} \pm \sigma_{A_{Measure}} = \frac{128282}{290} \pm \frac{26400}{290} = 442.35 \pm 91.03 Hz [/math]

Now we can find the true value using the integrated measurement

[math] A_{True} = \frac{e^{\lambda \times 290} \times 442.35 \times 290 \times \lambda}{e^{\lambda \times 290}-1} = 455.41 Hz [/math]


[math] \sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 93.72 Hz [/math]

So now the true value for the activity is

[math] A_{True} \pm \sigma_{A_{True}} = 455.41 \pm 93.72 Hz [/math]

Now we must correct for the dead time. The 1st run had an average count rate of 846.40 Hz, which corresponds to a dead time of 1.65 +/- 0.34

[math] A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 455.41 \times \frac{1}{0.9835} = 463.05 Hz [/math]

[math] \sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.004 [/math] [math] \sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}} [/math]

So the real activity for the 2nd measurement of the Mixed sample is

[math] A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 463.05 \pm 95.31 [/math]

0<t<300 730 <t< 1020 1480<t<1775 2250<t<2550 3050<t<3300 3775<t<4050 4480<t<4770
Data File Entry 6.90 +/- 0.30 6.85 +/- 0.22 6.73 +/- 0.19 6.61 +/- 0.19 6.43 +/- 0.19 6.23 +/- 0.21 6.14 +/- 0.21


Below is a table to check the activities against later measurements by tracing the activity back in time and taking a ratio

Run 1 Run 2 Run 3 Run 4 Run 5 Run 6
Original Measurement (Hz) 994.73 +/- 304.327 948.20 +/- 211.33 839.55 +/- 163.38 738.99 +/- 137.82 619.70 +/- 119.20 502.25 +/- 106.43
n+1 Measurement (Hz) 948.20 +/- 211.33 839.55 +/- 163.38 738.99 +/- 137.82 619.70 +/- 119.20 502.25 +/- 106.43 463.05 +/- 95.31
Time Correction Factor (Sec) 720 755 775 750 750 720
n+1 Measurement Time Corrected (Hz) 1096.39 +/- 244.36 977.63 +/- 190.25 864.01 +/- 161.14 720.90 +/- 138.67 584.27 +/- 123.81 535.42 +/- 110.21
n Measurement/n+1 Corrected Measurement 0.91 +/- 0.33 0.97 +/- 0.29 0.97 +/- 0.26 1.02 +/- 0.27 1.06 +/- 0.3 0.94 +/- 0.28

Plots

Below is a table for the plots of the cross sample ratios

Run 1 Run 2 Run 3 Run 4 Run 5 Run 6 Run 7
Mix Measurement (Hz) 994.73 +/- 304.327 948.20 +/- 211.33 839.55 +/- 163.38 738.99 +/- 137.82 619.70 +/- 119.2 502.25 +/- 106.43 463.05 +/- 95.31
Original Pure Se Measurement (Hz) 1961.89 +/- 216.27 1713.22 +/- 175.05 1510.32 +/- 151.17 1274.03 +/- 130.1 1036.14 +/- 131.17 913.17 +/- 105.31 793.67 +/- 90.19
Time Correction Factor (Sec) 340 340 375 380 390 350 360
Time Corrected Mix Measurement (Hz) 1065.33 +/- 325.93 1015.50 +/- 226.33 905.51 +/- 176.22 797.85 +/- 148.80 670.41 +/- 128.95 538.98 +/- 114.214 497.92 +/- 102.49
Time Corrected Mix Measurement/Pure Se Measurement 0.54 +/- 0.18 0.59 +/- 0.14 0.60 +/- 0.13 0.63 +/- 0.13 0.65 +/- 0.15 0.59 +/- 0.14 0.63 +/- 0.15

Below are the time vs. frequency plots

170063 MixHLPlot 1 11 18.png 170063 PureSeHLPlot 1 11 18.png

By exponentiating we find that the initial activity of the Mixture is 1056.48 +/- 2.91 Hz, while the pure sample has an initial activity of 2162.90 +/- 5.16 Hz which gives a ratio of

[math] 0.488 \pm 0.002 [/math]

To check consistency I would time correct measurements and take a ratio to compare, for example measurement 1 was taken and a 2nd measurement was taken some time later. So correct the rate of the second measurement back to the first measurement by using the radioactive decay equation, then take a ratio of these 2 numbers.

170063 Mix RateRatios 1 11 18.png 170063 PureSe RateRatio 1 11 18.png

Another check to do is to take the ratio of the mixture and the pure selenium sample (time corrected to the same time for each measurement) and see if the ratio here is relatively constant

170063 CrossSample RateRatios 1 11 18.png

By exponentiating we find that the initial activity of the Mixture is 1082.15 +/- 174.89 Hz, while the pure sample has an initial activity of 2169.13 +/- 175.01 Hz which gives a ratio of

Thin Window Mean + 2 Sigma:[math] 0.50 \pm 0.09 [/math] ,Original Window + 2 Sigma or original window

We can also check the ratio of the nickel with the selenium mixture. The rate of the nickel foil was [math] 1.127931 \times 10^6 \pm 7.258 \times 10^3 Hz [/math]

So using the efficiency corrected rate for the soil, we find that the ratio is

[math] 0.134 \pm 0.001 [/math]


The measurement was made 20cm away from the detector face, which has an efficiency of 0.70 +/- 0.0011, so the measurements with the efficiency taken into account are

150925.7143 +/- 392.5923 Hz for the mixture and

308985.7143 +/- 882.6876 Hz for the pure selenium sample

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