There are three mechanisms that are important in mass shielding. First, a charged particle excites electrons for many hundreds of angstroms about its trajectory. This excitation extracts kinetic energy at a roughly constant rate for relativistic particles and acts as a braking mechanism. For relativistic protons in low-Z matter this "linear energy transfer" is 2 MeV/g-cm^-2 of matter. If the thickness of the mass shield is great enough a particle of finite kinetic energy is stopped. This is the least effective shielding mechanism in matter for relativistic particles.
The second mechanism is nuclear attenuation. For silicon dioxide the average nuclear cross section is 0.4 barn (10^-24 cm^2). Thus if a charged particle traverses far enough in the shield (composed of silicon dioxide) it collides with a nucleus and loses energy by inelastic collisions with the nuclear matter. The measure of how far a particle must travel to have a substantial chance of nudear collision is the mean free path, which for silicon dioxide is 106 g/cm^2 . This mechanism is an exponential damper of primary beam particles.
Opposing the beam clearing tendency of nuclear attenuation is the creation of energetic secondary particles. For each nuclear collision there is beam loss from nuclear excitation, and beam enhancement (though with overall energy degradation through the increase of entropy) from the secondaries emitted by the excited nuclei. These secondary particles are, of course, attenuated themselves by further nuclear collisions with roughly the same mean free path as the primary particles.
The calculation of a mass shield can only be properly done by Monte Carlo simulation of the various pathways that the interactions can take. Two approximations often used either assume that secondary particles supplement nuclear attenuation well enough that only electron excitation stops the beam (the "ionization" approximation); or that secondary particle creation is negligible, resulting in beam attenuation by both nuclear attenuation and electron excitation ("ionization + exponential" approximation). Figure 5-26 plots the exposure rate versus all three formulations of shielding effectiveness for a cosmic ray spectrum incident on a copper shield. As the shield thickness becomes greater than approximately one mean free path length (for nuclear attenuation) the Monte Carlo result begins to show behavior that parallels the slope of the ionization + exponential approximation. The remaining difference is that the Monte Carlo result is a scale multiple of the ionization + exponential curve. This behavior then provides the following new approximation:
= initial exposure rate of
primary beam,
1 = mean free path for nuclear collision,
x = distance (g/cm^2) of shield traversed.
Since the secondary particle production factor for a high-Z nucleus like copper is greater than for low-Z nuclei this approximation should be conservative using the factor of 5 for secondary production. Because thick shields (>> 1) are being considered, this equation's sensitivity to error is greatest in the value of 1 used, which fortunately can be accurately determined (106 g/cm^2 for silicon dioxide).
To calculate doses behind a shield the result is used that brad in carbon is liberated by 3 X 10^7 cm^-2 minimally ionizing, singly-charged particles. Within a factor of 2 this is valid for all materials. Assuming a quality factor of unity for protons, and an omnidirectional flux of protons at 3/cm^2-sec (highest value during the solar cycle) the dose formula is
x = shield thickness (g/cm^2 ).
This is a conservative formula good to perhaps a factor of 2 in the thick shield regime. A spherical shell shield is assumed with human occupancy at the sphere's center. If the dose to be received is set at 0.25 rem/yr, to be conservative with the factor of 2, a required thickness of Moon dust (silicon dioxide) of 441 g/cm^2 is derived to protect the habitat.
As a final point, note that an actual shield will generally not have spherical symmetry. To handle this case the shield geometry must be subdivided into solid angle sections which, due to slant angles, have differing effective thicknesses. Equation (3) then has to be integrated over all the solid angle sections to calculate the received dose.
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Curator: Al Globus If you find any errors on this page contact Al Globus. |
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