A-5 APPENDIX A
Example Calculation:
Determine the detector efficiency of Am-241 in a well detector with an opening of 0.625 inches and the
source located in the well at 1 inch below the surface.
Geometric Efficiency for a Well: I.D. = 0.625 inches
X - 1.00 inch
G.E. = 97.7% (see earlier example)
Detector Efficiency for Am-241: A = 0.095 µCi @ T = 8/15/84
T
1/2
= 432.2
t = 5/1/93 (date counted in well)
Am-241 was programmed for gain of 12 and an ROI of 50 to 69 keV.
100 second counts resulted in the following:
S = 99,403
B = 83
Now calculate the individual parts of the equation for (D.E.
*
G.E.)
Count rate = (S-B)/100 s = 993 cps
Disintegration rate = A µCi
*
37000 dps/µCi = 3515 dpm
Decay Correction = 2
(T-t)T
1/2
=
2
(84.71 y - 93.42 y)/432.2 y
= 0.986
D.E.
*
G.E. = 0.286
*
100% = 28.6%
D.E. = {(D.E.
*
G.E.)/G.E.}
*
100% = (28.6/97.7)
*
100 = 29.3%
We should now check our result for common sense. The photon emission from Am-241 are 59.5 keV at
35.9% intensity and 26.3 keV at 2.4% intensity. The interaction probability for a 60 keV photon is very
high in the Nal well detector (Nal thickness on sides and bottom is about 0.625 inches), the bulk of this
interaction will be photoelectric which puts most of the counts recorded in the photo peak. The ROI is set
to integrate the photo peak of 59.5 keV. Therefore, we would expect a detector efficiency to be a little less
than photon intensity at 59.5 keV which it is.
Example of a Well Compared to a Probe for Counting Efficiency
If we assume the following parameters:
Isotope: Cs-137
D.E. Probe: 9.86%
D.E. Well: 13.47%
G.E. Well: 96.816% for a source placed 2.54 cm deep in the well
G.E. Probe: 0.278% for a source placed 21 cm in front of the probe
Probe: D.E.
*
G.E. = 9.86%
*
0.278% = .02741%
Well: D.E.
*
G.E. = 13.47%
*
96.816% = 13.04%
A.) If the detector counts 54 cpm that is equivalent to the following dpm for the parameters listed above.
Probe: 54 cpm/.02741% = 197,008 dpm
Well: 54 cpm/13.04% = 414 dpm
CONTENTS
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