Se170063 Thin Window Analysis

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 $\sigma_I = I_{Expanded} - I$ (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
Original Window
Expanded Window
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
Original Window
Expanded Window
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.

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

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

$\frac{I_{Expanded}-I}{I}$

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

$\sigma_I = I_{Expanded} - I$

First Pure Se Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 466700 \pm 49100 Counts$

and the integrated background is given by

$B \pm \sigma_B = 21120 \pm 103.16 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 445580 \pm 49100 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{445580}{240} \pm \frac{49100}{240} = 1856.58 \pm 204.58 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 240} \times 1856.58 \times 240 \times \lambda}{e^{\lambda \times 240}-1} = 1901.86 Hz$

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

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 1901.86 \pm 209.57 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1901.86 \times \frac{1}{0.9694} = 1961.89 Hz$

$\sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.0306$

$\sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}}$

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1961.89 \pm 216.27$

Second Pure Se Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 444100 \pm 43200 Counts$

and the integrated background is given by

$B \pm \sigma_B = 21120 \pm 102.82 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 422980 \pm 43200 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{422980}{260} \pm \frac{43200}{260} = 1626.85 \pm 166.15 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 260} \times 1626.85 \times 260 \times \lambda}{e^{\lambda \times 260}-1} = 1669.88 Hz$

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

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 1669.88 \pm 170.54 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1669.88 \times \frac{1}{0.9747} = 1713.22 Hz$

$\sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.025$

$\sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}}$

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1713.22 \pm 175.05$

Third Pure Se Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 412800 \pm 39400 Counts$

and the integrated background is given by

$B \pm \sigma_B = 18992 \pm 97.8 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 393808 \pm 39400 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{393808}{275} \pm \frac{39400}{275} = 1432.03 \pm 143.27 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 275} \times 1432.03 \times 275 \times \lambda}{e^{\lambda \times 275}-1} = 1472.11 Hz$

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

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 1472.11 \pm 147.28 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1472.11 \times \frac{1}{0.9747} = 1510.32 Hz$

$\sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003$

$\sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}}$

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1510.32 \pm 151.17$

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

From this we can see that

$I \pm \sigma_I = 356800 \pm 34600 Counts$

and the integrated background is given by

$B \pm \sigma_B = 17796 \pm 94.2 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 339004 \pm 34600 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{339004}{280} \pm \frac{34600}{280} = 1210.73 \pm 123.57 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 280} \times 1210.73 \times 280 \times \lambda}{e^{\lambda \times 280}-1} =1245.24 Hz$

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

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 1245.24 \pm 127.09 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1245.24 \times \frac{1}{0.9774} = 1274.03 Hz$

$\sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.004$

$\sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}}$

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1274.03 \pm 130.1$

Fifth Pure Se Measurement

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

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 304100 \pm 551.45 Counts$

and the integrated background is given by

$B \pm \sigma_B = 18836 \pm 96.4 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 285264 \pm 559.81 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{285264}{290} \pm \frac{559.81}{290} = 983.67 \pm 1.93 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 290} \times 983.67 \times 290 \times \lambda}{e^{\lambda \times 290}-1} = 1012.72 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.99 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 1012.72 \pm 1.99 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 1012.72 \times \frac{1}{0.9774} = 1036.14 Hz$

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

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 1036.14 \pm 4.05$

Sixth Pure Se Measurement

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

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 258700 \pm 508.63 Counts$

and the integrated background is given by

$B \pm \sigma_B = 14970 \pm 86 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 243730 \pm 515.85 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{243730}{280} \pm \frac{515.85}{280} = 870.46 \pm 1.84 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 280} \times 870.46 \times 280 \times \lambda}{e^{\lambda \times 280}-1} = 895.27 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.89 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 895.27 \pm 1.89 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 895.27 \times \frac{1}{0.9804} = 913.17 Hz$

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

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 913.17 \pm 3.09$

Seventh Pure Se Measurement

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

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 233500 \pm 483.22 Counts$

and the integrated background is given by

$B \pm \sigma_B = 14322 \pm 68.4 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 219178 \pm 488.04 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{219178}{290} \pm \frac{488.04}{290} = 755.79 \pm 1.68 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 290} \times 755.79 \times 290 \times \lambda}{e^{\lambda \times 290}-1} = 778.11 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.73 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 778.11 \pm 1.73 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 778.11 \times \frac{1}{0.9804} = 793.67 Hz$

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

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 793.67 \pm 2.75$

	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

Se-Sage Mix Analysis

Activity of the First Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 342200 \pm 584.98 Counts$

and the integrated background is given by

$B \pm \sigma_B = 58880 \pm 169.86 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 283320 \pm 609.14 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{283320}{300} \pm \frac{609.14}{300} = 944.4 \pm 2.03 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 300} \times 944.4 \times 300 \times \lambda}{e^{\lambda \times 300}-1} = 973.26 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 2.09 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} =973.26 \pm 2.09 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 944.4 \times \frac{1}{0.9494} = 994.73 Hz$

$\sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}}$

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 994.73 \pm 4.75$

Activity of the Second Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 300100 \pm 547.81 Counts$

and the integrated background is given by

$B \pm \sigma_B = 41160 \pm 140.62 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 258940 \pm 565.57 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{258940}{290} \pm \frac{547.81}{290} = 892.90 \pm 1.89 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 290} \times 892.90 \times 290 \times \lambda}{e^{\lambda \times 300}-1} = 919.27 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.95 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 919.27 \pm 1.95 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 919.27 \times \frac{1}{0.9694} = 948.29 Hz$

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

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 948.20 \pm 3.56$

Activity of the Third Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 267200 \pm 516.91 Counts$

and the integrated background is given by

$B \pm \sigma_B = 32840 \pm 131.04 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 234360 \pm 533.26 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{234360}{295} \pm \frac{533.26}{295} = 794.44 \pm 1.81 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 295} \times 794.44 \times 295 \times \lambda}{e^{\lambda \times 295}-1} = 818.31 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.86 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 818.31 \pm 1.86 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 818.31 \times \frac{1}{0.9747} = 839.55 Hz$

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

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 839.55 \pm 3.28$

Fourth Mix Sample Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 238600 \pm 488.47 Counts$

and the integrated background is given by

$B \pm \sigma_B = 28920 \pm 116.94 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 209680 \pm 502.27 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{209680}{300} \pm \frac{502.27}{300} = 698.93 \pm 1.67 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 300} \times 698.93 \times 300 \times \lambda}{e^{\lambda \times 300}-1} = 720.29 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.72 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 720.29 \pm 1.72 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 720.29 \times \frac{1}{0.9747} = 738.99 Hz$

$\sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003$

$\sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}}$>

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 738.99 \pm 2.94$

Activity of the Fifth Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 170100 \pm 412.43 Counts$

and the integrated background is given by

$B \pm \sigma_B = 21900 \pm 101.3 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 148200 \pm 424.69 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{148200}{250} \pm \frac{424.69}{250} = 592.8 \pm 1.70 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 250} \times 592.8 \times 250 \times \lambda}{e^{\lambda \times 250}-1} = 607.87 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.74 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 607.87 \pm 1.74 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 607.8 \times \frac{1}{0.9808} = 619.70 Hz$

$\sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.003$

$\sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}}$

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 619.70 \pm 2.42$

Sixth Mix Sample Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 153200 \pm 391.41 Counts$

and the integrated background is given by

$B \pm \sigma_B = 21060 \pm 99.8 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 132140 \pm 403.93 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{132140}{275} \pm \frac{403.93}{275} = 480.51 \pm 1.47 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 275} \times 480.51 \times 275 \times \lambda}{e^{\lambda \times 275}-1} = 493.96 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.51 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 493.96 \pm 1.51 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 493.96 \times \frac{1}{0.9835} = 502.25 Hz$

$\sigma_{DTC} = \frac{\sigma_{DT}}{1-DT^2} = 0.004$

$\sigma_{A_{True}^{DT}} = \sqrt{\frac{\sigma_{A_{True}^{DT}}^2}{(1-DT)^2} + \frac{A_{True}^2 \sigma_{DT}^2}{(1-DT)^4}}$

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 502.25 \pm 2.32$

Activity of the 7th Measurement

Below is the histogram of interest

From this we can see that

$I \pm \sigma_I = 147900 \pm 384.58 Counts$

and the integrated background is given by

$B \pm \sigma_B = 19618 \pm 96.4 Counts$

Now we can find the background subtracted signal

$N \pm \sigma_N = 128282 \pm 396.48 Counts$

Now convert this into an activity by dividing by the runtime

$A_{Measure} \pm \sigma_{A_{Measure}} = \frac{128282}{290} \pm \frac{396.48}{290} = 442.35 \pm 1.38 Hz$

Now we can find the true value using the integrated measurement

$A_{True} = \frac{e^{\lambda \times 290} \times 442.35 \times 290 \times \lambda}{e^{\lambda \times 290}-1} = 455.41 Hz$

$\sigma_{A_{True}} = \sqrt{(\frac{t\times e^{\lambda t} \times \lambda \times \sigma_{A_{Measure}}}{e^{\lambda t}-1})^2 + (\frac{e^{\lambda t}(A t e^{\lambda t} - \lambda^2 - A t + A t \lambda)}{(e^{\lambda t}-1)^2})^2 \times \sigma_{\lambda}^2} = 1.42 Hz$

So now the true value for the activity is

$A_{True} \pm \sigma_{A_{True}} = 455.41 \pm 1.42 Hz$

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

$A_{True}^{DT} = A_{True} \times \frac{1}{1-DT} = 455.41 \times \frac{1}{0.9835} = 463.05 Hz$

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

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

$A_{True}^{DT} \pm \sigma_{A_{True}^{DT}} = 463.05 \pm 2.16$

	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

Plots

Below are the time vs. frequency plots

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.

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

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

$0.488 \pm 0.002$

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

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

$0.134 \pm 0.001$

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

We can also look at the ratios between the nickel and the selenium ash mixture