Se170063 Nickel Foil Thin Gauss Window Expansion

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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

170063 Nickel WideGauss HLPlot.png

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.