Difference between revisions of "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.

12/14: Error in signal changed to (try to not underestimate error) This includes changing the expanded window to a window expanded off of the 2 channel window

 400

Below is the table for the Se-Ash Mixture

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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.

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

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

First Pure Se Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

Second Pure Se Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

Third Pure Se Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

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

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

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

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

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

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

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

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

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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

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

Activity of the Second Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

Activity of the Third Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

Fourth Mix Sample Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

>

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

Activity of the Fifth Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

Sixth Mix Sample Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

Activity of the 7th Measurement

Below is the histogram of interest

From this we can see that

and the integrated background is given by

Now we can find the background subtracted signal

Now convert this into an activity by dividing by the runtime

Now we can find the true value using the integrated measurement

So now the true value for the activity is

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

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

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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

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

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 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: ,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

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

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