# [ RadSafe ] Infinite Thickness of KCl

Forsee, Gary Gary.Forsee at illinois.gov
Tue Aug 30 15:38:41 CDT 2011

```Afternoon Erik,

I'm going stick my head out there and take a shot at this:

Assuming you are blocking any dose contribution from alpha and beta, and
only measuring the gamma contribution then I would recommend solving the
equation below:

T = exp( -mu / p * x )

Set the transmittance (T) equal to zero, indicating the point where the
1460 keV gamma is completely absorbed.  -mu/p (email wont support the
Greek mu)is the mass attenuation coefficient, specific to the keV of the
gamma and the density of the material - input as cm2/g.  The website
below has the mass attenuation coefficients for varying Z of materials
across the energy ranges of 10 keV to 1 MeV.

http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html

'x' is the thickness of material (in centimeters) that will be required
to absorb the 1460 keV gamma.

G. Forsee

-----Original Message-----
[mailto:radsafe-bounces at health.phys.iit.edu] On Behalf Of
Nielsen.Erik at epamail.epa.gov
Sent: Tuesday, August 30, 2011 11:50 AM
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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