Answer to Question #7943 Submitted to "Ask the Experts"

Category: Environmental and Background Radiation — Rocks, Minerals, and Mines

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

Q

I am trying to calculate an activity for natural uranium rock samples from a measured dose rate. In total, 35 kg of rock (not all uranium, in fact barely any—but for safety's sake 35 kg), gave a combined mean dose rate of about 6.5 µSv h-1. I need to convert this measurement into bequerels per gram to satisfy health and safety concerns. How do I calculate this? I am also having trouble finding dose rates for alpha radiation.

A

With the information you have provided it is not possible to make a fully reliable evaluation of the amount of uranium present in the rock. Assuming the rocks you describe are not contaminated with man-made radionuclides, the most likely natural sources of gamma radiation exposure are 238U and its progeny, 232Th and its progeny, and 40K, and the relative contributions from each depend on the relative amounts of these species, which are quite dependent on the origin and characteristics of the rocks that you are considering. The measured dose rate is sufficiently high that we can assume that the 40K is likely not a major dose contributor.

The dose rate from the rocks also depends on the geometric arrangement when the measurement was made. We can estimate radioactivity content by making a conservative assumption that all of the measured dose rate resulted from gamma and x radiation from 238U and progeny in the rock and that all progeny were in secular equilibrium with the uranium.

If we assume a density of about 2.5 g cm-3 for the rock, and it were spread out in a cylindrical pattern to a depth of about 15 cm, the diameter of the cylindrical volume would be about 35 cm, neglecting the effects of air spaces among the rocks. Disregarding the effect of air spaces among the rocks, if the measurement was made very close to the rock surface near the center of the cylinder the assumption of energy spatial equilibrium (ESE) would be reasonable since all dimensions from the measurement point to any source would be several photon mean free paths.

The principle we shall actually apply is ½ ESE to account for the fact that the dose point is being irradiated from only one side. If we consider all the significant gamma and x rays from 238U and progeny we would find a total photon energy emission of about 3.0 MeV per disintegration, and the average energy is about 0.5 MeV. Under ESE the energy absorbed per unit mass in the rock is equal to the energy emitted per unit mass, from which we calculate the dose rate in the rock at the rock surface, Drock (in Gy h-1), as

Drock = (1/2)(C)(0.5 MeV)(1.6 x 10-10 Gy/MeV g-1)(3,600 s h-1),

where C is the 238U concentration in Bq g-1. The dose rate in soft tissue, DT, at the same point would be about

DT = Drock(μen/ρ)T/( μen/ρ)rock

where the tissue-to-rock mass energy absorption coefficient ratio would be evaluated at the average photon energy of 0.5 MeV. For lack of other specific information for the rock, I shall use the coefficient values for concrete (to simulate rock) and ICRU-44 soft tissue composition available from NIST. Using the above equations we have

DT =(1/2)(C)(0.5 MeV)(1.6 x 10-10 Gy/MeV g-1)(3,600 s h-1)(0.03269 cm2 g-1/0.03033 cm2 g-1).

If we assume that the tissue dose rate measured in Sv h-1 is approximately equal to the tissue absorbed dose rate in Gy h-1, we may correlate the reading with the concentration and solve the above equation for C. For the measured value of 6.5 x 10-6 Sv h-1, which I assume is a net reading associated only with gamma response (this means that the detector walls must be sufficiently thick to stop high-energy beta radiation from 234m Pa decay that might be present), we then solve for C and obtain

C = 42 Bq g-1.

This activity concentration would translate to a uranium content of 0.34 weight percent, which would imply a rather high-grade uranium ore.

Keep in mind that the above type of calculation is meant to be demonstrative, but the accuracy depends on the validity of the assumptions made regarding the detector response, the possible presence of other contributing radionuclides in the rock, and the geometry conditions under which the measurement was made (including the lack of impact of air spaces between the rocks).

Regarding your mention of alpha radiation, you should have no concern about dose from the alpha radiation unless you are considering internal exposure. Some alpha radiation might be seen with a thin-window detector held close to the ore if uranium is present very close to the ore surface, but the radiation would be attenuated in the dead skin layer of the body and would not be capable of producing dose to live tissue. As noted earlier, it is important to make sure that the detector you used responded only to the gamma and x radiations. Any response to beta or alpha radiation would result in the reading not being equal to either the photon dose rate or the total dose rate.

I hope this is sufficient to get you going in the right direction.

George Chabot, PhD, 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.
Answer posted on 26 January 2009. 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.