Talk:Radiation shielding

From Marspedia
Jump to: navigation, search

^(M Lamontagne)

I changed the meaning of the magnetic shielding section as it excluded magnetic shielding without sufficient analysis. In particular it didn't refer to minimagnetospheres, and their existence ;on the moon. The mechanism of a minimagnetosphere may be electric rather than simply magnetic, with particles being ionised by the shield and then deflected by the electrical field potential. Cosmic :rays might be deflected by very large shields with low densities.


Some material (the Beer Lamberth law, linear attenuation coefficients, and half value layers) appears to apply only to electromagnetic radiation, so I added a subheading to specify this and allow a comparison of these considerations with particle considerations.  If I am wrong about this please correct me.

Hi Jim, they actually apply to both types of radiation but in different ways. A text I wrote a few years ago on fusion reactors (high energy neutrons act like ions):

Most materials are quite transparent to neutrons and X-rays, but not perfectly so.  If the material is transparent enough and thin enough, most high energy radiation will go right through. The behavior of this at the atomic level is complex, but at the macroscopic level, the simple equation for attenuation of high energy neutrons and X-rays is the Beer-Lambert law:

I = Io*e-(μ/ρen)ρl      (X-rays) (6a)

I = Io*e-(μ/ρ)ρl     (Neutrons) (6b)


I = intensity at depth ‘l’ into the material

Io = Original intensity

μ = attenuation coefficient (m2/kg)

ρ = density (kg/m3)

l = thickness into the material (m)

μ/ρen= mass energy attenuation coefficient (m2/kg)

The attenuation coefficient ‘μ’ is an experimental value that varies from material to material and according to the type of radiation.  The mass energy attenuation coefficient ‘μ/ρen’ is a form used to express the attenuation coefficient when it is modified to take into account the generation of secondary radiation in the material.  This coefficient has been tabulated by the NIST for X-rays. For neutrons, a slightly less precise mass attenuation coefficient is available from various sources (5). These are summarized in the following table for a power level of 15 MeV:

Mass attenuation coefficient Neutron

at 15 MeV


at 15 MeV

μ/ρ (m2/kg) μ/ρen(m2/kg)
Hydrogen 0.06580 0.00225
Lithium 0.01570 0.00114
Beryllium 0.01140 0.00123
Boron 0.00836
Carbon 0.00803 0.001959
Oxygen 0.00451 0.001483
Iron 0.00346 0.002108
Nickel 0.00355 0.002234
Copper 0.00347 0.002174
Tungsten 0.00160 0.003072
Lead 0.00150 0.004972
Bismuth 0.00147 0.003123
Uranium 0.00154 0.003259

Table 3 - Mass and mass energy attenuation coefficients (3)(4)(5)

From Table 3, we can see that X-rays rays are best stopped by dense materials since μ/ρ gets larger with denser materials, while neutrons are best stopped by light materials. In a very general way, X-rays interact with electrons, while neutrons interact with nuclei.  So X-ray opaque materials must have the maximum number of electrons, while the best neutron absorbers must have the most tightly packed nuclei.

X-ray Example

Beryllium is often recommended as a material with good X-ray transparency. Its density is 1848 kg/m3 (ρ). From the NIST tables, at an X-ray energy of 10 Mev, the mass energy attenuation coefficient is 0.014 cm2/g, or 0.0014 m2/kg (μ/ρ).

If we suppose a thickness of 1cm of beryllium, then,from equation (6a):

I/Io = e-(μ/ρen)ρl = e-0.0014*1848*0.01  = 0.98

At this thickness, the beryllium has absorbed 2% of the radiation intensity, allowing the remainder to escape. That’s pretty good, but how much energy is that 2% absorbed?  Well, if the beryllium pipe wall is 2m away from the Z-pinch, for example, the X-ray radiation power (from Table 2) is 0.5 x 27,000 MW per m2.  So 2% of that is about 280 MW per m2.  That’s still a lot of waste energy (heat) so it is very important to minimize the mass in the Z-pinch area.

Neutron Example

The neutron mass attenuation coefficient for high energy neutrons in Beryllium is 0.0157 m2/kg.  That’s almost ten times higher than for X-rays. For the same 1cm thickness and equation (6b):

I/Io = e-(μ/ρ)ρl= e-0.0157*1848*0.01  = 0.75; therefore the beryllium has absorbed 25% of the radiation intensity.  Very intense cooling will probably be required, even for this semi-transparent material.

However, it is important to note that this description doesn’t quite tell the whole story.  In fact, many of the neutrons are not absorbed at all but are re-emited as secondary neutron radiation in a process called neutron scattering.  So the actual energy absorbed is less than what the Beer-Lambert equation calculates. Consult your local nuclear physicist for details!

Perhaps we might add the above table as an example, although the energy levels are a it high.