Answer to Question #7257 Submitted to "Ask the Experts"

Category: Medical and Dental Equipment/Shielding — Shielding

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

Q

I want to make shielding calculations for a Molybdenum-99/Technetium-99m (99Mo/99mTc) production laboratory. I need some information like tenth value layers (TVL) [in cm lead (Pb) and cm concrete] and the specific gamma ray constant (in µSv×m2/MBq×h) of 99Mo.


A

There is no easy answer to your question. 99Mo produces 30 gammas, in energies from 40.5 kiloelectron volts (keV) to 1.05 megaelectron volts (MeV), plus K x-rays of 18-21 keV. The most abundant gammas are 140 keV at 4.52 percent, 181 keV at 5.99 percent, 366 keV at 1.2 percent, 740 keV at 12.1 percent, and 778 keV at 4.26 percent (National Nuclear Data Center, Brookhaven National Laboratories). (I am neglecting the gammas from 99mTc.) Thus, as shielding is added, the radiation gets "harder" and the TVL gets larger. With heavy shielding, the 740 keV and the 778 keV photons become the ones of concern. The TVL for 750 keV photons is 2.03 cm in lead or 13.4 cm in concrete. This does not include buildup. Calculating shielding for these multienergetic sources requires more than a simple evaluation by TVL.

The reference I found for the gamma ray constant for 99Mo is 1.3E-5 (C/kg)×cm2/MBq×hr (Radiological Health Handbook 1970). Of course, that value is for an unshielded source. Please consider contracting with a health physicist in your area to provide support in this project.

David Close, CHP
 

Ask the Experts is posting answers using only SI (the International System of Units) in accordance with international practice. To convert these to traditional units we have prepared a conversion table. You can also view a diagram to help put the radiation information presented in this question and answer in perspective. Explanations of radiation terms can be found here.
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