MadSci Network: Physics |
Although I am not a physicist or a radiologist, I work with radioactive isotopes in my biomedical research and found a quick answer to your question in my handy Radiation Safety Training Manual (published by the Radiation Safety Office of the University of Pittsburgh, to give credit where it's due). The following is extracted and paraphrased from that manual: Gamma rays are attenuated exponentially, therefore theoretically it is not possible to stop the radiation completely. However, the exposure can be reduced by various amounts. For example, if a thickness of a particular shielding material reduces the radiation to one-half the initial amount (a "half-value layer", or HVL), then the thickness of three such layers will reduce the dose to one-eighth (1/2 x 1/2 x 1/2) the initial amount. Similarly, three "tenth-value layers" (TVL) will reduce the dose to 1/1000 of the initial amount (1/10 x 1/10 x 1/10). Shielding materials of high density and high atomic number, such as lead, are generally the most effective absorbers or shields for gamma rays. Thus less thickness and weight per square foot is required for such materials. However, steel, brick, concrete, or other materials can provide the same degree of shielding if used in appropriately greater thicknesses. Below are some examples of tenth- value layers for some gamma-emitting radionuclides of various energies: Tenth Value Layer Radionuclide Energy (MeV) lead (cm) concrete(in) ____________________________________________________________ Cs-137 0.662 2.1 6.2 Co-60 1.33 4.0 8.1 Cr-51 0.32 0.7 4.3 I-125 0.035 0.13 1.0 Na-22 1.27 4.0 8.1 As you can see, 10 cm of lead will reduce gamma radiation from these radianuclides to various extents, from almost completely for I-125 to about 1/500 the initial amount for Co-60.
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