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Below an answer to a recent question about shielding against radiation on a trip to Mars, there was this comment from dubu (emphasis mine):

[..] For high-energy particles entering dense matter, Bremsstrahlung is joined by pair production, and both together create particle showers. [..] A known problem in radiation shielding (thin shielding might be worse than no shielding at all).

Thus,

How thick must a shield be to be better than no shield at all?

Assuming the shield is used on a trip between planets of our solar system and is intended to protect the human crew.

I'm looking for both theoretical papers / calculations as well as practical studies (if there are any).

uhoh
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Daniel Jour
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  • I'd appreciate help with tagging :) – Daniel Jour Jun 22 '17 at 18:27
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    Actually Bremsstrahlung would make mostly electromagnetic ($e^+, e^-, \gamma$) showers, whereas nuclear collisions would make the $\pi$, $\mu$, neutron, proton, and lighter ions in the showers, although for the highest energies it gets messier. There will be plenty of papers and calculations, but they may disagree with each other, it's a messy problem. @kimholder has dealt with these issues, and I think there are some Q&A here with some links to some papers. If I remember correctly it's the spallation products, neutrons and protons in particular that ruin your day, and your DNA. – uhoh Jun 23 '17 at 00:05
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    In addition to the ionization effects from radiation exposure, fast neutrons and protons can knock a proton (hydrogen atom) right out of your DNA molecules. The situation reminds me of Steve Martin's character in the movie The Jerk where he exclaims "It's the cans! Stay away from the cans!" It's a poor analogy, but there probably is a "worst thickness" of material that will convert cosmic rays into more damaging forms, and you would indeed prefer either less or more thickness. – uhoh Jun 23 '17 at 00:15
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    different but related Is radiation dose from cosmic rays higher behind 50 cm of shielding, or lower? – uhoh Mar 18 '19 at 01:29

1 Answers1

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First of all let's define what the radiation is:

  1. alpha (helium nucleus i.e. two protons and two neutrons)
  2. beta (electrons)
  3. gamma (high energy photons)
  4. neutrons
  5. ions (bare stripped nucleus)
  6. other stuff like visible light, IR, UF. X-RAY is not considered here, X-RAY is gamma.

The only thing that is important to consider is a nuclear cross section which is different for every type and every protection material.

Protection:

  1. alpha is easy, it's a heavy and well-charged particle. And there is not much of them in the space. The energy of the ones from the space is low. 1 mm of a kitchen tin foil will be a good protection.
  2. beta. There are lots of electrons in the space and they can have wild energy, ranging from 10 to 10^20, and this is huge. For example, the large hadron collider at CERN is able to produce just 10^13 tops. There is nothing in the world that can stop such a high energy stuff. However, the good news is that the nuclear cross section of such a high energy particles is relatively small. 3mm of the kitchen foil is enough just as a placebo shield and in order to stop some low energy ones.
  3. gamma. this is simple. the heavier the stuff the better. The earth is protected by 10km of air, which has the same level of protection as 10m of water, or 1.4m of steel, or 0.9m of lead, or 0.1m of depleted uranium. Sounds scary but 0.1m of U-238 can be the best option to protect from this kind of radiation.
  4. neutron. this is the most difficult one. The best shielding is the light stuff - water, hydrogen, oxygen, graphite, etc. Ideally - by other neutrons. However, the nuclear cross section raises as its energy drops! The slower neutron is the more dangerous it is! Also when a neutron is stopped by some materials like cadmium, the neutron absorption is accompanied by a strong emission of gamma which requires more gamma protection. You can easily google for neutron protection articles like this, but overall, either not protect at all or establish an earth-like level of protection. The earth is protected by ~10 km of air which has about 25% of oxygen, hydrogen, and water. I would estimate that 3m of water will be a good earth-like protection from neutrons. Water is better than other material - it is rich in hydrogen.
  5. Ions, UF, IR - the same stuff as alpha

Overall, 3mm of tin foil + 3m of water + 0.1m U-238 and the Martian astronauts will be just fine.

Here is a scan from one work, it's in Russian but you should get the idea, the legend here is:

  • X is the aluminum shield thickness gram/cm^2
  • Y is the radiation absorbed dose mRad/24hours
  • 1: total dose
  • 2: primary protons
  • 3: cascade protons
  • 4: cascade neutrons
  • 5: evaporative neutrons
  • 6: evaporative protons

Dependency between the cosmic radiation power and shield thickness

ilyakharlamov
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  • This is looking like really great answer, thanks! Does the plot represent the initial distribution of radiation in deep space far from the sun? Would it look much different if there were high solar activity? – uhoh Jul 16 '17 at 03:20
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    The plot represents some sort of average amount. During the high solar activity, it's complicated. Some particles stay the same, the intensity of other particles can be as high as 10, 100 and even 1000 times greater than average. – ilyakharlamov Jul 16 '17 at 03:34
  • Is it mostly short-range stuff? few MeV protons and X-rays, I'm guessing that once the thickness is larger than a few centimeters it doesn't matter much? – uhoh Jul 16 '17 at 03:47
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    GOES program studies this kind of solar stuff. For protons, tin foil will protect from everything, however, gamma can go up 100x sometimes which will require log2 of 100 ~7 times thicker shield which is not possible. So the guys will suffer well. – ilyakharlamov Jul 16 '17 at 04:30
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    You are dismissing ions as alpha, stopped by a very thin shield. The issue is stuff coming in at extreme energies where you get secondary particles from the collision. That's not going to be stopped by the tinfoil. – Loren Pechtel Jul 16 '17 at 04:35
  • @LorenPechtel and ilyakharlamov OK I see, thanks! Maybe it's a good idea to add a sentence in the answer, near the plot, just mentioning that the plot applies to some sort of average radiation situation, and would not apply in the case of a large solar event. – uhoh Jul 16 '17 at 04:59
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    @uhoh It's not just a solar event, but the cosmic rays. The issue is that you actually increase the radiation dose by adding an inadequate shield--while you stop the light stuff you turn individual high energy particles into showers. – Loren Pechtel Jul 16 '17 at 05:24
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    @LorenPechtel you misunderstand my comment. I'm familliar with the content of the question, the answer, and issues of radiation shielding in general. However, in my comments I am trying to establish how applicable the plot shown here applies to space environment. Because it is written in Russian, and I don't know the source of this plot, I'm asking ilyakharlamov to add a little more explanation about this plot to the body of the answer so that future readers of the answer will understand that during a major solar even, this plot would not apply. – uhoh Jul 16 '17 at 05:31
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    I'm not really sure why you're suggesting adding a full-thickness gamma-ray shield of depleted uranium to a full-thickness water neutron shield. The mass of water would do double duty, since gamma rays are stopped almost entirely by mass. (Greater density is handy for other reasons, since it is more efficient geometrically to protect a three-dimensional space with a thinner shell, but it has almost no effect on linear photon absorption: a photon going through 1 tonne/m^2 of hydrogen is absorbed just as well as 1 tonne/m^2 of tungsten.) – Nathan Tuggy Jul 16 '17 at 19:21
  • @NathanTuggy I wonder if high atomic weight can actually be a small disadvantage for electromagnetic showers? While things made from elements like water & paraffin have about 0.5 electron per AMU, uranium drops below 0.4 due to neutron excess. Pure hydrogen would be an exceptional case as it has nearly one electron per AMU. For the lower energy photons and electrons in a shower, the high binding energy of inner electrons for heavy elements (tens of keV up to 100 keV) makes them less available for scattering as well. Water & paraffin are also attractive for fast neutron-slowing from hydrogen. – uhoh Jul 17 '17 at 05:31
  • This scan ... is the data from some simulation, or from experiments? If I understand this correctly the total radiation dose drops below the value with no shield somewhere between 100 g/cm^2 and 200 g/cm^2 .... using a density of 2.71 g/cm^3 this would call for an aluminium shield between approx. 37 cm and 74 cm? (Note: To reduce the total radiation instead of increasing it, not to make it "safe" behind that shield) – Daniel Jour Sep 16 '17 at 13:44
  • This answer seems to provide a good explanation of what "good" shielding would be, but it doesn't seem to answer the actual question of where the crossover is between helping/hurting? I don't see any discussion at all of the secondary particles that the question is about. – mbrig Nov 12 '19 at 23:49
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    @mbrig Agreed. If I understand the plot, the total dose (top curve) only drops below its no-shielding asymptote (left side) once the aluminum shield thickness exceeds about 160 g/cm^2 or 60 cm. The answer badly needs to include a discussion of this. – nanoman Jun 04 '21 at 05:59