|
||||||||||||||||
Answer to Question #2026 Submitted to "Ask the Experts"Category: Radiation Basics — Interaction Coefficients The following question was answered by an expert in the appropriate field: Q
Please help me with the values of the mass attenuation coefficient of lead (Pb) for gamma rays(60Co). The problem is, I find two values at different energies (that is, at 1.17 MeV, I find 0.06 kg/m2 and at 1.33 MeV, I find 0.053 kg/m2). I heard that at a certain energy, only a certain percentage of the mass attenuation coefficient is used. Could you please clarify to me how much percentage of mass attenuation coefficient is used at 1.17 MeV and how much is used at 1.33 MeV?
A
You are correct that each gamma ray has a unique attenuation coefficient. The coefficients—which are directly related to the total interaction cross section in the target material—vary with photon energy. The application of the coefficient is not given, so I will assume that the intended use is the calculation of exposure-rate shielding. As a general rule, it is not a good idea to combine attenuation or absorption coefficients when performing calculations where exposure or dose rates will be the final product. Since exposure per unit fluence also varies with photon energy, it is usually required that calculations be performed at each energy independently and then combined into a final result. This isn't always possible in real life—especially when considering the addition of shielding to an existing (but unknown) arrangement—so best guesses must suffice. Because 60Co has two photons with very similar energies, the use of a single "effective" attenuation coefficient (representing out-of-beam scatter and total absorption) would work well. (Please note that the units of the attenuation coefficients you have quoted are inverted and the values are too large by a factor of 10; I have corrected them here.) Using a weighted sum (by abundance) with two photons of equal abundance (that is, 100% for each), the effective attenuation coefficient calculates to be 0.006 × ½ + 0.0053 × ½ = 0.0056 m2/kg. Again, while this will work well for 60Co, I would not recommend the approach for more complex spectra. Attenuation coefficients represent total interaction probabilities but do not account for in-scattered photons or secondary photon production. At times, some individuals have used mass energy absorption coefficients, rather than total mass attenuation coefficients, in selected shielding calculations, in an attempt to accout for effects of in-scattered photons, but such methods provide only approximate solutions. Calculations, using the total attenuation coefficients, may have to include appropriate terms for the buildup factor that accounts for the increased dose from these scattered photons. Robert L. Coleman Oak Ridge National Laboratory
Answer posted on 30 May 2002. The information and material posted on this Web site is intended as general reference information only. Specific facts and circumstances may alter the concepts and applications of materials and information described herein. The information provided is not a substitute for professional advice and should not be relied upon in the absence of such professional advice specific to whatever facts and circumstances are presented in any given situation. Answers are correct at the time they are posted on the Web site. Be advised that over time, some requirements could change, new data could be made available, or Internet links could change. For answers that have been posted for several months or longer, please check the current status of the posted information prior to using the responses for specific applications.
|
||||||||||||||||
| This page last updated 18 December 2009. Ask Question | Search ATE | Site Map | Privacy Statement | Disclaimer | Security Notice | Webmaster | ||||||||||||||||
