Se170063 Pure Se Wide Gauss Window Expansion

In this example I only weighted the histogram by the mass of the sample, which in the case of the pure selenium sample is 0.0971g.

	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
Original Window Counts	564700	536100	497100	432200	380500	319900	289400
Original Window Background (Integrated)	85711	86719	78500.2	73399.5	76327.6	61106.5	57459.7
Original Window Difference	478989	449381	418599.8	358800.5	304172.4	243572.4	231940.3
Expanded Window Counts	587300	558100	517500	451400	398700	336000	304500
Expanded Window Background	108563	108152	98915.2	92183.5	93245	77423.5	72660.2
Expanded Window Difference	478737	449948	418584.8	359216.5	305455	258576.5	231839.8
Error in counts	252*	567	15*	416	1282.6	15004.1	100.5 *
.dat file entry	7.598793988 +/- 0.0005261081152	7.454944728 +/- 0.0012617356	7.327899514 +/- 0.0000358337	7.1557322 +/- 0.0011594187	6.955469002 +/- 0.0042166876	6.768379905 +/- 0.0616001649	6.684354367 +/- 0.0004333011

Using this method there seemed to be a problem with the background. When finding the error in the counts, sometimes the value was negative. Below is the example of the last timed run where 4840 < t < 5130

Here you can see that the total number of counts in the window is 289400, while the integral of the background, which is found in the terminal is 57459.7. So we have a difference of 231940.3 counts. The expanded window is shown below.

Now here it is shown that the total number of counts within the window is 304500 counts, while on the terminal we can see that the integral of the background is 72660.2 counts, which gives a difference of 231839.8 counts. Subtracting these two numbers to find the error in the number of counts, we find that

[math] Expanded Counts - Original Counts = 231839.2 - 231940.3 = -100.8 counts [/math]

This result is slightly troubling, but it did not occur in the ash + se sample at all. It seems that the macro may be overcompensating for the background.

Below is a plot of the activity vs. time to get the half life and initial activity

The root window gives a slope of -0.000211997 +/- 1.3423e^-7. This gives us a half life of

54.49 +/- 0.03 Minutes

We can find the initial activity by using the constant value given, which is 7.72506 +/- 0.000255985.

[math] A_t = e^{7.72506} = 2264.39 Hz [/math]

[math] \sigma_{A_t} = e^{7.72506}*\sigma_{A_t} = 0.58 Hz [/math]

Now we can correct for the efficiency, which will be the same factor as the mixed (which should cancel)

[math] A_{t'} = \frac{A_t}{\epsilon} = \frac{2264.39}{0.0070} = 323484.28 Hz [/math]

while the error is

[math] \sigma_{A_{t'}} = \frac{A_t * \sigma_{\epsilon}}{\epsilon^2} = 508.33 Hz [/math]

Now we must trace this signal back to t = 0 for the mixture (since that is what it will be compared to). The mixture was measured 400 seconds prior to the pure sample, which gives

[math] A_0 = A_{t'}*e^{\lambda*400} = 352112.00 Hz [/math]

while the error is

[math] \sqrt{e^{2\lambda t}\sigma_{A_t}^2 + A_{t}^2t^2e^{2\lambda t}\sigma_{\lambda}^2 } = 553.64 Hz [/math]