
Lesson 20  Uncollided dose from point sourcesWith this lesson, we begin the study of approximate methods of determining the dose distribution in shielding situations. The first step is to determine the dose contribution from that subset of source particles that reach the detector without collision  this contribution to the dose is referred to as the uncollided dose. In addition to being the easiest component of the dose to handle mathematically, we will see in Chapter 7 that there is a fairly accurate method (involving "buildup factors") to estimate the total gamma ray dose from the uncollided gamma ray dose. (Unfortunately, this technique does not work too well for neutrons.)In contrast to the discussion in the book, which builds up the ideas from the simplest (monoenergetic point source in a vacuum) to the general case, I will work backwards from the general case. The general equation for the uncollided dose rate at a point (t for "target") due to sources distributed in a volume (s for "source") is given by: If you can dissect this equation and understand its parts, then you understand this lesson. The only tricky part of this equation is the exponential term at the end. I have invented a variable "s" to denote the distance along a straightline path between the source position, , and the detector position, . (Also, I note that the upper limit of the exponential integral is a little hard to make out: it is supposed to be , the distance between the source and detector.) Note that in this equation
is a variable (source position inside
), whereas is fixed.
Polyenergetic Point SourceAs advertised, I will now work backwards to simpler representations of the source. For a source that is fixed in space (but still emits particles of all energies), the source distribution is a Dirac delta in space:where I have done what the textbook authors did: I divided the source into a strength (in units of particles/sec) and an energy distribution (which integrates to 1 and has units of 1/MeV). The Dirac delta is a distribution as well and has units of 1/cc. (For convenience, I placed the source at the origin.) I you plug this into the previous equation, the Dirac delta has its usual property of "extracting" the value of the function at the point that makes the Dirac delta function go to 0, that is: and the previous equation becomes: Monoenergetic Point SourceI we further simplify by making the source emit particles at a particular energy, , we are replacing the energy distribution, N(E) with a Dirac delta function in energy:Substituting this gives us: Monoenergetic Point Source in a Homogeneous Attenuating MediumOur final simplification is to make the medium homogeneous, which has the effect of making:(since all points the same distance from the source will have the same flux) and (since the cross sections will be the same everywhere). Substitution of these gives us the simplest version we will get: 
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