[ RadSafe ] Infinite Thickness of KCl

[Joseph Shonka]

Erik

You can simply invoke conservation of energy and divide by 2.  This is
theoretically true.  For compton scattering (off free electrons), the number
of electrons per gram of matter is largely similar for all matter except for
hydrogen ( for which it is twice as large).  With KCl and 1.46 Mev, you are
in a region where compton dominates.

In an infinite source, every cc has the same energy released, and also
absorbed.  If you convert from energy absorbed (rads) to energy absorbed in
air (R) (about a 10% correction at compton energies for most matter), one
can show that:

R/hr = V * Ebar, where V is the microcuries per gram, and Ebar is the
average energy per decay.  A very simple rule.

This is similar to an old rule used in medical physics for unit density
tissue, I have expanded it for any matter and density where compton
scattering is the dominant interaction.

For your case of K-40, you have 416 pCi/g and 0.105*1.46 = 0.1533 MeV Ebar,
so you have 64 uR/hr from a semi-infinite source.  The source area should be
>10X the detector area in surface when the detector is close to the surface
to satisfy semi-infinite conditions.   As for how thick does it have to be
to get semi-infinite dimensions, at compton energies, 3 mean free paths is
about 15 grams per square centimeter for any material (and especially for
low Z that does not have a lot of H), or about 7.5 centimeters at your
asserted density.

Just remember R/hr + V * E.  I will let my European friends give you the
equation in SI units.  It is too painful for me to state it that way.

Regards,

Joe Shonka
On Tue, Sep 13, 2011 at 1:35 PM, Dixon, John E. (CDC/ONDIEH/NCEH) <
gyf7 at cdc.gov> wrote:

> Erik,
>
> This could be a candidate case for MicroShield. You could set the photon
> energy for K-40 and alter the source thickness manually and then run the
> program. This would take several iterations. For this case, I would not use
> the transmission factor relationship since you are searching for the
> thickness of source material which is going to emit gammas no matter what
> the thickness. You just need to find the thickness where the fluence appears
> not to change.
>
> John Dixon
>
> -----Original Message-----
> From: radsafe-bounces at health.phys.iit.edu [mailto:
> radsafe-bounces at health.phys.iit.edu] On Behalf Of
> Nielsen.Erik at epamail.epa.gov
> Sent: Tuesday, August 30, 2011 12:50 PM
> To: radsafe at health.phys.iit.edu
> Cc: liste de distribution pour les RADIOCHIMISTEs
> Subject: [ RadSafe ] Infinite Thickness of KCl
>
>  I am trying to determine the infinite thickness of the 1460 keV gamma
> emission (10.5% abundance) of a planar source of KCl (416 pCi/g) and a
> density of 1.98 g/cc.
>
> In other words, what is the thickness of a KCl planar source where any
> additional thickness does not increase the surface dose rate?
>
> I have scoured my references but have not located the appropriate formula
> for calculating this value.
>
> I would appreciate any references or suggestions for an appropriate
> calculation.
>
> Erik C. Nielsen
> USEPA, National Air and Radiation Environmental Laboratory
> 540 South Morris Ave.
> Montgomery, AL 36115
> Phone 334-270-3475
> Fax 334-270-3454
>
> "Those who do not read are no better off than those who cannot"
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-- 
Joseph J. Shonka, Ph.D.
Shonka Research Associates, Inc.
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