A device emits 0.2 μSv/h of gamma rays. How thick does an aluminum sheet need to be to completely stop radiation from coming out ? What equation is to be used to calculate this ?
[Physics] How to calculate gamma radiation shielding
electromagnetic-radiationradiationradioactivity
Related Solutions
There are three processes by which gamma rays interact with matter: the photoelectric effect, Compton scattering, and pair production.
The photoelectric effect is an interaction between the gamma ray and an electron. It's forbidden by conservation of energy and momentum unless there is some other body present as well, such as an atomic nucleus. For this reason, the probability of the photoelectric effect is proportional not just to the density of electrons but also approximately to $Z^n$, where $Z$ is the atomic number and $n$ is about 4 to 5.
Compton scattering can occur without the presence of anything besides an electron, so it only depends on electron density.
Pair production goes like $Z^2$ at typical gamma-ray energies.
For these reasons, the best shielding against gamma rays is achieved with a substance that has a high density of electrons (which correlates with a high mass density) and also a high $Z$. Lead has these properties. It's also cheap. There are elements with a higher $Z$, such as bismuth, polonium, and uranium, but they aren't cheap, and their atomic numbers are only slightly higher.
Yes, heavy shielding is needed primarily for gamma radiation. Neutron radiation (with energies seen in fission reactors) is easily stopped with boron-10 (isotopically enriched boric acid in water).
While alpha and beta radiation is easier to shield, it is even more dangerous if alpha and beta active particles (dust) is consumed by human, because they will irradiate you for many years, and all their energy would be consumed by your body. So it's obviously important to physically contain high-pressure radioactive material inside reactor.
Regarding shielding of gamma radiation: it is usually done by materials with high atomic mass (lead, depleted uranium e.t.c). It can be done with lighter materials with comparable mass (i.e. water shield must be ~10 times thicker, than lead one). Depending on gamma ray energy, you might need about 1-10cm of lead to consume 50% of gamma radiation. Some more details are here : http://en.wikipedia.org/wiki/Gamma_ray#Shielding
Best Answer
X-rays are attenuated when they pass through any material, and the amount they are attenuated depends on how far they travel in the material and that material's mass attenuation coefficient. The NIST web site has an excellent set of pages on this at http://www.nist.gov/pml/data/xraycoef/index.cfm.
NB the x-ray intensity falls exponentially, so the x-rays will never fall to zero. What the calculation described on the NIST web page will tell you is the thickness you need to reduce the intensity to below some specified limit.