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).


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).

  • $\begingroup$ I'd appreciate help with tagging :) $\endgroup$ Jun 22, 2017 at 18:27
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    $\begingroup$ 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. $\endgroup$
    – uhoh
    Jun 23, 2017 at 0:05
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    $\begingroup$ 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. $\endgroup$
    – uhoh
    Jun 23, 2017 at 0:15
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    $\begingroup$ different but related Is radiation dose from cosmic rays higher behind 50 cm of shielding, or lower? $\endgroup$
    – uhoh
    Mar 18, 2019 at 1:29

1 Answer 1


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.


  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

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    $\begingroup$ 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. $\endgroup$ Jul 16, 2017 at 3:34
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    $\begingroup$ 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. $\endgroup$ Jul 16, 2017 at 4:30
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    $\begingroup$ 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. $\endgroup$ Jul 16, 2017 at 4:35
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    $\begingroup$ @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. $\endgroup$ Jul 16, 2017 at 5:24
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    $\begingroup$ @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. $\endgroup$
    – uhoh
    Jul 16, 2017 at 5:31

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