Use detector physical area, not effective area

[Steve Frantz <sfrantz@YAHOO.COM>]

I just had a discussion about something that I'd like to pass

along. Excuse me if this was so obvious that everyone else

already knows it.



An alpha probe has a physical area, say 76 cm^2, but a smaller

open (effective) area of, say 50 cm^2. The difference of 26 cm^2

is the part of the probe covered by the metal grid that holds

the mylar window in place.



Some think that in calculations you should use the effective

surface area of the probe, 50 cm^2, since the rest of the probe

is blocked. So if the detector had a calibration efficiency of

25% and you read 100 cpm, you would report 800 dpm/100 cm^2 (100

cpm / 0.25 * 100/50 cm^2).



But, it turns out you should be using 76 cm^2 to get the answer

of 526 dpm/cm^2, about a third less!



The reason is that the difference in areas was taken into

account during the calibration. The calibration source was also

blocked by the metal grid. The explanation is in MARSSIM

(NUREG-1575, Rev. 1) page 6-29. Here's the example I used to

convince myself....



Say my calibration source is exactly 76 cm^2 big and has a

strength of 152 dpm (4?). For simplicity, suppose I put a

detector with an 100% efficient scintillator on it. I'd read 50

cpm, because half of the alphas go down and only 50 out of 76 of

the remaining alphas make it through the grid (effective area).

Thus, I'll report an efficiency of 32.9% (50 cpm /152 dpm).



If I now measure a contaminated surface and it reads 50 cpm, the

contamination should be reported as 200 dpm/100 cm^2  (50 cpm /

0.329 * 100/76 cm^2), using the 76 cm^2; _NOT_ 304 dpm/100 cm^2 

(50 cpm / 0.329 * 100/50 cm^2), using the 50 cm^2.



The reason is obvious; if we measure the calibration source with

this detector, we know we will read 50 cpm, and we know (by

NIST) that the source is 152 dpm/76 cm^2, which equals 200

dpm/100 cm^2.



If the source is smaller than the physical size, the process is

the same because the grid is over the whole surface and always

blocks 26/76 of the incoming radiation. 



The difference between physical and effective area is accounted

for in the detector efficiency.



So, use the physical probe area, not the effective area to

calculate contaminatin levels. Strange, eh?



-Stephen Frantz

Reed College

sfrantz@yahoo.com





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