radiation
I have a plastic container and want to make sure that infrared radiation (specifically, in the 750-850 nanometer range) cannot pass through it. Would wrapping it in aluminum foil do the trick? If not, what household material would be best? The fluence of the radiation is around 22J/cm^2.
Thank you very much!
Lead is used to block radiation because:
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:
$$E'=\frac{E}{1+\frac{E}{m_0c^2}(1+cos\theta)}$$
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:
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
Best Answer
The chart for the reflectance of aluminum tells us that it should do the job for you. But it will likely block anything else within that spectral range, see the plot! reflectivity aluminum