# Difference between revisions of "Se170063 Thin Window Analysis"

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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 CrossSample Ratio.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 | 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 |

## Revision as of 18:13, 7 December 2017

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 error in the signal will be found by integrating the fit gaussian function over the window and "widening" it by half its standard deviation. 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.

Below is the table for the Se-Ash Mixture

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

## 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

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

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

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

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