Re: Thermalisation optimisation

[azapantis@mgdestmx01.erin.gov.au]

_______________________________________________________________________________
Subject: Thermalisation optimisation
From:    radsafe@romulus.ehs.uiuc.edu at mgdestmx01
Date:    3/2/97  8:06 PM

Hi Ant!

Actually, all you need to find out (from where I don't know) is the Fast
Diffusion length (also called Slowing down length) for polyethylene for 4.5MeV
neutrons. The FDL is the average straight line distance covered by a neutron
before being thermalised. Use that together with neutron shielding data (NCRP38
might be useful) to determine the optimum thickness for your moderator. 

Alternatively, if you are a maths wiz (which I am not) you could try and
calculate it. 

The log of average energy loss (x) of a neutron per collision with a nucleus is
dependant ONLY on the mass of the nucleus;

              x = - ln (E/Eo)
               E = energy of neutron after collision
               Eo = energy of neutron before collision

also,
               x = 1 + [(a.lna)/(1-a)]
               a = [(M-m)/M+m)]^2
               M = mass of nucleus
               m = mass of neutron


If you plug the numbers into the above, you will find that x = 1 for hydrogen
and 0.159 for carbon. Thus E/Eo for a neutron colliding with a hydrogen nucleus
will always be 0.37 and for carbon, 0.85.

Since you know the formula for polyethylene (I don't!) you should be able to
work out an average nucleus mass and determine x and hence E/Eo (on average) for
polyethylene. (is this valid?????) 

Then you have to work out how many collisions (n) are required to reduce the
energy from 4.5MeV to 0.025eV.

      ie   4.5E6.(E/Eo)^n = 0.025
thus             (E/Eo)^n = 0.025/4.5E6
             ln[(E/Eo)^n] = ln(0.025/4.5E6)
               n.ln(E/Eo) = ln(0.025/4.5E6)
therefore               n = [ln(0.025/4.5E6)]/[ln(E/Eo)]

Now is when I get stuck! There are 2 things we need to account for which I do
not know how to do.

1. The cross section for elastic scattering will change (I think?) with energy
as the neutron slows down.

2. The path of the neutron will be tortuous, not a straight line.

If you can get around these problems, then, knowing the number of atoms per cc
in polyethylene, the number of collisions required, the cross section for
elastic collisions as a function of neutron energy and the relationship between
linear path length and actual path length (of the neutron) you will be able to
calculate the thickness of polyethylene needed. (I think)

I've probably made some errors. Any suggestions from the nuclear physicists out
there?

Regards

                       

Alex Zapantis                                  
Environmental Radiation Officer     
Office of the Supervising Scientist		
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Email: azapantis@dest.gov.au
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