Answer to Question #5362 Submitted to "Ask the Experts"

Q

What are the properties of lead which make it so useful for shielding from x rays and other forms of radiation? Are the x rays absorbed by the lead or are they simply deflected? If they are absorbed, then what happens to the energy which was absorbed?

A

The radiation shielding properties of lead are primarily due to its density. Lead has a density of 11.35 grams/cubic centimeter (g/cm³). This means that its mass is more tightly packed into a smaller volume than most other elements. The denser a material is, the more likely an ionizing photon will interact with an orbital electron and be absorbed. Density generally rises with increasing atomic number. A few other elements have greater densities than lead and consequently would be better for shielding than lead, but they are either very rare and expensive (e.g., gold [19.32 g/cm³] and platinum [21.45 g/cm³]), toxic (e.g., mercury [13.55 g/cm³]) or already radioactive (e.g., uranium [18.95 g/cm³] and plutonium [19.84 g/cm³]). Lead is usually used for shielding because it is relatively cheap, plentiful, and easily formed into shielding bricks or leaded sheets. When the volume of a shield is a design consideration, titanium or even depleted uranium (primarily ²³⁸U) are sometimes substituted for lead. Please refer to the Environmental Chemistry Web site for additional information on the densities of different elements.

The way that most ionizing photons (gamma and x rays) interact with matter is through two primary processes, the photoelectron effect or Compton scattering. In the photoelectric effect, the energy from a photon is entirely absorbed by an inner orbital electron within the target atom. As a result, the electron is kicked out of the atom, turning the photon's energy into kinetic energy by moving the electron and leaving the atom in an ionized (excited) state. The ejected electron's kinetic energy is quickly expended by collisions with surrounding electrons and generally travels only a few millimeters at most. In Compton scattering, the photon is only partially absorbed by the ejection of an outer orbital electron from the atom. The photon itself is scattered at a lower energy than it was originally, depending on its scattering angle. If the photon is scattered in a forward direction, less energy is deposited into the recoil electron and if the photon is scattered in a backwards direction, more of its energy is imparted into the recoil electron. This continues multiple times until all or most of the photon's energy is absorbed by interaction with the orbital electrons within the shield. Eventually, the photon's energy is converted into heat, but usually the increased heat is imperceptible. In fact, extremely large sources of radiation, shielded in leaded containers, will feel slightly warmer to the touch than their surroundings.

However, it should be recognized that no matter how thick or dense a shield is created, some photons will still penetrate the shield. Ionizing photons like x rays or gamma rays can be attenuated, but never totally absorbed by a shield. Properly engineered shields are designed to absorb most of the photons, so only a few percent of the photons actually can penetrate all the way through the shield.

Mike Bohan, RSO