Se170063 Nickel Foil Thin Gauss Window Expansion

This analysis was done by weighting the histogram by the mass of the sample. The window is [1360,1365] and was expanded by one channel on each side to find the error in the counts. Since the window is smaller, a constant fit was no long acceptable as the constant kept becoming negative and would yield a negative integral, which is not a good representation of background that is clearly there. Instead I used a linear function plus a gaussian to fit, then integrated the line across the window.

	5/25/17	5/26/17	5/27/17	5/29/17	5/30/17	5/31/17	6/1/17
Original Window Counts	92120	165000	201400	80830
Original Window Background (Integrated)	12773.2	37500	75000	11215.6
Original Window Difference	79346.8	127500	193900	69614.4
Expanded Window Counts	95730	166000	215500	83820
Expanded Window Background	921.179	9000	2818.01	6681.52
Expanded Window Difference	94808.821	157000	212681.99	77138.48
Error in counts	15462.021	29500	18781.99	7524.08
Position	50 cm	30 cm	20cm	20 cm	10cm	10cm	10cm
Efficiency	0.00017	0.000422	0.000847	0.000847	0.0031	0.0031	0.0031
.dat file entry	14.25308619 +/- 0.1948663462	13.81819743 +/- 0.231372549	13.4567054 +/- 0.0968643115	12.51979457 +/- 0.1080822359

Below is a plot of the activity as a function of time

This gives the constant of 14.5046 +/- 0.00345043. Now find the initial activity

[math] A_t = e^{14.5046} = 1991900.96 Hz [/math]

Since the plot is made with reference to the mixture's measurement time (first point plotted 108 minutes after the mix was measured) means that the constant should be the activity at the time the mixture was being measured.