A-8 APPENDIX A Biodex Medical Systems, Inc. © 2014
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 range 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.
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