Re: dose rate from plane source

[Michael McNaughton <mcnaught@lanl.gov>]

At 06:55 PM 9/14/99 -0500, you wrote:
>Regarding the plane source equation:
>How far away from the plane is the determination of dose rate ?

For a plane source, the dose rate is independent of the distance.

>How big is the plane?

infinite

>Need to show your derivation.
Very briefly:
If "a" is the activity per unit volume (Ci/m^3)
h is the half-value-layer thickness (meters)
b is the mean depth [i.e., b=h/ln(2)],  
E is the energy per emission (MeV)
x is in the plane (and disappears in the integral),
y is perpendicular to the plane (and cancels from the final equation), 
x^2+y^2=r^2, (r also disappears in the integral)
and sin(theta)=y/r, 
then using the rule of thumb: dose rate (R/h) = 0.5 C E, 
the dose rate (R/h) =
integral from 0 to infinity of [0.5 E a b sin(theta) 2 pi x dx /r^2]
= E pi a b

>-----Original Message-----
>From: Michael McNaughton <mcnaught@lanl.gov>
>To: Multiple recipients of list <radsafe@romulus.ehs.uiuc.edu>
>Date: Tuesday, September 14, 1999 11:18 AM
>Subject: dose rate from plane source
>
>
>>I have not seen a rule of thumb for the dose rate for a plane source, so I
>>derived the following simple rule of thumb.
>>If E is the average gamma energy per emission in MeV,
>>a is the activity in Ci/m^3,
>>h is the half-value layer in meters,
>>pi = 3.14, and ln(2) = 0.693,
>>then the dose rate in R/h = (pi)(E)(a)(h)/ln(2)
>>
>>I would appreciate any comments.
>>
>>mike

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