I understand that, for example, a thick enough sheet of lead can absorb gamma radiation, but I want to understand what actually happens at the molecular/atomic/subatomic level. Also, can the same logic be applied to cosmic particles? I have tried Googling for an answer, but to no avail. Can someone enlighten me?

  • $\begingroup$ It looks to me that you are asking for a review of any kind of possible interaction between matter and various kind of projectiles. Whole books have been written on this! The very short summary is that the energy is dispersed through (multiple) scattering, being transferred by the recoils to the material structure as phonons and finally dispersed as heat. Damages to molecules, nuclei or even nucleons can of course take place according to the energy and the kind of the projectiles. $\endgroup$ – DarioP Jul 21 '14 at 10:43

Radiation can be several things, but since you specifically mentioned lead shielding, let's look at X-rays - a lot of what you learn applies to other radiation as well.

To stop radiation it needs to interact with "something" that makes it give up its energy and momentum. This is how you get the radiation to stop going in the direction it was going.

Now X-rays typically interact with matter (atoms) in one of three ways:

At low energies you can have the photoelectric effect: the energy of the radiation is completely absorbed by the electrons of the atom - so the photon "disappears" and the electron gets all the energy (minus whatever energy was needed to get it detached from the atom - the bonding energy). Electrons don't travel very far in matter, so the energy is usually absorbed once a photoelectric interaction occurs. The probability of this interaction depends on the energy of the photon and the $Z$ (atomic number) of the atom - higher $Z$ means much higher probability (I have seen $Z^4$ relationships but I'm not sure how well those hold, and over what range.)

As the energy of the photon increases above the K-edge of the atom, you get Compton scatter dominating: this is an elastic collision between the photon and the electrons in the material and it results in a transfer of momentum and energy from the photon to the electron. The famous Compton equation shows the relationship between incident and final energy of the photon as:


Where $m_0$ is the rest mass of the electron and $\theta$ is the angle between the incident photon with energy $E$ and final energy $E'$.

The more electrons there are in your material, the more effective the stopping power in this range (above 80 keV or so). This is why lead, depleted uranium, bismuth, tungsten, and other such materials are good for shielding.

At very high energies, you can get pair production: the photon (with more than 1.022 MeV energy) creates an electron/positron pair "out of thin air", giving up 1.022 MeV of energy (which is turned into mass of the particles created).

So to recap: X-rays shielding works by interaction of electrons with photons. Higher density materials improve the probability of Compton scatter; higher atomic number increase photoelectric interaction cross section. Typically, one talks about the half value thickness: the thickness of material that stops half the radiation. Because shielding is a probabilistic process, there is no such thing as "perfect shielding".

One more point about density of the shielding material:
In some situations, you care about stopping the radiation in the shortest possible distance. This happens for example in a radiation pinhole camera (used in SPECT systems), where you want to have a small opening to let radiation through, but need to stop all radiation outside of that. Such an aperture has to be made of the densest high-Z material you can find. People usually choose gold for this application (http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=949378) - the figure of merit here is the product of density and specific scatter cross section, the linear attenuation coefficient with units of $m^{-1}$. The larger this number, the more efficient the material is at stopping radiation in a short distance. A couple of examples (all values at 100 keV, attenuation data from http://physics.nist.gov/cgi-bin/Xcom/xcom2-t):

symbol   Z  density    sigma   lambda
           (g/cm^3)  (cm^2/g)   (/cm)
  Ir    77   22.5      4.86     109
  Pt    78   21.5      4.99     107
  Au    79   19.3      5.16     100

As you can see, for this particular example the shortest stopping length is obtained for iridium - because although it has a slightly lower Z than gold, it has significantly higher density.

When you are interested in "bulk radiation protection", for example in nuclear reactors, then the question is simply "how do I get a lot of shielding for not a lot of money". Now the size of the shield does not matter very much, and you end up with water - a very cheap and abundant material which is capable of stopping radiation (not just gamma rays, but neutrons as well). This is the material of choice for shielding (spent) reactor fuel. You may have seen the pictures of the blue-glowing fuel rods under water:

enter image description here

source of this image: http://spectrum.ieee.org/image/37182

The glow is Cerenkov radiation - resulting from the fact that particles are traveling "faster than the speed of light". In this case, that's faster than the speed of light in water - which is of course less than the speed of light in vacuum because of the refractive index of water.

My point is - as long as you put "lots of electrons in the way" you will eventually stop gamma rays: if you need to stop them in a short distance, you need dense high-Z material, but that is not always necessary.

In radiation protection, it is recognized that the combination of shielding, distance, and time to exposure all play a role in keeping radiation dose to people as low as possible: ALARA "As Low As Reasonably Achievable".

Finally - a wikipedia link

  • $\begingroup$ Water is neither dense (relative to lead, for example) nor does it contain high atomic number atom. Then why do people consider water as a shielding agent? $\endgroup$ – Icarus Jul 20 '14 at 21:30
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    $\begingroup$ Water provides you with lots and lots of electrons for very little money. That makes it a good shielding option for large scale shielding (where size does not matter) $\endgroup$ – Floris Jul 20 '14 at 21:38
  • $\begingroup$ Based on what I understand from your post, if a material is not dense enough, the probability of a photon hitting one of the electrons becomes much lower. Correct? $\endgroup$ – Icarus Jul 20 '14 at 21:41
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    $\begingroup$ @Icarus The figure of merit is roughly the number of electrons per unit area in traversing the whole of the shielding. Dense materials allow you to have a high figure of merit in a short distance (which is very often the thing you want), but in other applications the best engineering deal may be found from maximizing electrons per unit area per dollar. That is often when water gets used. For instance, the on-site cooling ponds for spend fuel at reactor facilities are not compact, but they are cheap (and they provide passice cooling as well). $\endgroup$ – dmckee --- ex-moderator kitten Jul 21 '14 at 1:15
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    $\begingroup$ Water, and light elements in general, are good for neutrons. $\endgroup$ – DarioP Jul 21 '14 at 10:52

There are several different things labeled "radiation". Gamma rays are electromagnetic radiation, similar to visible light but at a higher frequency. X- rays are also electromagnetic radiation. For electromagnetic radiation, elements with heavy nuclei are good shielding. See this Wikipedia article on protection against electromagnetic radiation.

Also called radiation are high speed protons and atomic nuclei.

And the high speed nuclei can vary a lot in velocity. High speed nuclei from outside the solar system are called Cosmic Rays. These tend to be a lot faster than the high speed ions coming from the solar wind or the Van Allen Belts.

Galactic Cosmic Rays often are moving at close to light speed. When a such a high speed nuclei or proton strikes a massive nuclei (such as a lead nucleus), it is like a cue ball breaking a rack on a pool table. You have particles going every which way forming secondary cosmic rays. To avoid this shower of secondary particles, atoms with small nuclei are desirable. So hydrogen rich compounds may be better shielding against GCRs. Water is often suggested as shield against GCRs.

  • $\begingroup$ That makes sense about GCRs but what about electromagnetic radiation? $\endgroup$ – Icarus Jul 20 '14 at 21:21
  • $\begingroup$ @Icarus good point. I've edited the first paragraph because of this comment. $\endgroup$ – HopDavid Jul 21 '14 at 1:33

The two existing answers have noted that EM radiation (X-rays, gamma) is effectively stopped by electrons. There are at least 4 other types common types of radiation:

  1. Alpha particles (2 protons, 2 neutrons - essentially He4 2+)
  2. Beta particles (single electron)
  3. Neutrons
  4. Ions (other than alpha particles)

The first thee are commonly generated by nuclear decay reactions; the 4th category is relevant as it's part of the cosmic particles mentioned. Shielding for them differs. Alpha's are quite easy to stop, as are ions. Pretty much any thin layer of matter will stop them - a centimeter of air will have already have a measurable effect.

Beta's are electrons and are therefore easily stopped by many materials. Because they're lighter than alpha's, yet have similar energies, they travel faster. As a result, they penetrate further than alpha's.

Neutrons are the odd one out, as they're electrically neutral. No electron cloud is going to stop them; they're stopped by nuclei. But it doesn't take a particularly heavy nucleus for that. Light elements have less electron orbitals and can therefore pack more nuclei in a given volume, which compensates for the fact that each nucleus is smaller.

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    $\begingroup$ There is a second advantage to light nuclei in stopping neutrons. The target-frame, kinetic-energy loss of the incident particle is largest when the projectile and target have the same mass. Neutrons hitting $^1\mathrm{H}$ nuclei dump energy faster than when they hit any other nucleus. $\endgroup$ – dmckee --- ex-moderator kitten Jul 22 '14 at 23:32

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