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Category: Medical and Dental Equipment/Shielding — Shielding

The following question was answered by an expert in the appropriate field:

Q
I know that 1 cm of lead has the same shielding ability as 6 cm of concrete. The question I have is that we use iodine-131 in our pharmacy and we have a 15.24 cm poured concrete wall separating our space from the neighbor's space. When we handle the iodine-131 it is handled 2.44 meters away from the poured concrete wall. Is there a mathematical formula to estimate the exposure an individual would receive on the other side of the poured concrete wall? We hang a badge on our side of the wall on a monthly basis to measure exposure and it reads about 0.2 millisievert (mSv) on our side of the wall. What would it read on the other side of the 15.24 cm thick concrete wall?

A

The relative shielding effectiveness of lead and concrete is highly dependent on photon energy. At 100 keV, the relative half-value thicknesses in concrete and lead are 150 to 1. At 1 Mev it is 5.4 to 1 and at 364 keV (the principal energy of iodine-131) it is 14 to 1.

The unshielded exposure rate from 37 megabecquerels (MBq) of iodine-131 is 5.8 E-8 Coulomb/kilogram/hour (C/kg/hr) at 1 meter. To calculate the exposure from iodine-131, multiply 5.8 E-8 by the activity of the source (A) in MBq, divide by the square of the distance (d) in meters, and multiply by the time of exposure (t) in hours.

Exposure = (5.8 E-8) At .

d  2

The effect of 15.24 cm of concrete on 364 keV photons would be to reduce the exposure to about 24 percent of the unshielded exposure. The HVL for 364 keV photons is 2.93 cm (in concrete). The 15.24 cm of concrete reduces the exposure to a factor of  0.027. However, the buildup factor (including scatter from broad beam conditions) would be 8.9, resulting an a net attenuation of 0.24. Of course, iodine-131 is multienergetic and the calculation for shielding for iodine-131 is more complicated. But, the above will give you an order-of-magnitude value. You would estimate that the 0.2 mSv value would be reduced to 0.048 mSv with 15.24 cm of concrete, not considering inverse square effects.

Data for the above was taken from NUREG/CR-5740, "New Gamma-Ray Buildup Factor Data for Point Kernel Calculations: ANS-6.4.3, Standard Reference Data," 1991, Oak Ridge National Laboratory.

David Close, CHP

Answer posted on 24 January 2008. The information posted on this web page is intended as general reference information only. Specific facts and circumstances may affect the applicability of concepts, 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. To the best of our knowledge, answers are correct at the time they are posted. Be advised that over time, requirements could change, new data could be made available, and Internet links could change, affecting the correctness of the answers. Answers are the professional opinions of the expert responding to each question; they do not necessarily represent the position of the Health Physics Society.