Re: Use detector physical area, not effective area

[Steve Frantz <sfrantz@YAHOO.COM>]

Ted,



I'm sorry, I don't agree unless the source is an actual point

source.



The difference between physical and effective areas is the metal

grid covering the detector. Anywhere you put a source, some of

it will be covered by some part of the grid. If the grid covers

20% of the detector, then 20% of the alphas are blocked. So you

use the physical size since that included the 20% loss.



If the source is an actual point source, you should measure

different efficiencies depending on whether the grid happens to

cover it. They you could easily measure two radically different

efficiencies.



Bottom line, if 20% of the calibration source is covered by the

grid, so is the actual in-field contamination. The grid is part

of the detector efficiency.



Other opinions?



-Stephen Frantz

-sfrantz@yahoo.com



--- Ted de Castro <tdc@xrayted.com> wrote:

> Ok - if the source is the size of the whole detector and is

> thus

> effectively a flood source of known Ci/unit area.  BUT when

> the source

> is smaller than the effective area only using the effective

> area makes

> any sense.  I'd think most calibrations are done with a source

> smaller

> than the effective area.

> 

>



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