Answer to Question #7162 Submitted to "Ask the Experts"
The following question was answered by an expert in the appropriate field:
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 .
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