Answer to Question #4817 Submitted to "Ask the Experts"

Category: Radiation Basics — Photons

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

Q

What value should be taken as the average gamma energy of 226Ra?

A

Most of the significant gamma radiation from 226Ra decay comes from the radioactive progeny 214Pb and its daughter, 214Bi. These are produced following the decay of 226Ra to 222Rn, which then decays to 214Pb. Since 222Rn is a gas it will escape, to varying degrees, from unsealed sources, and the gamma radiation from the 214Pb and 214Bi may not be significant in such cases. In sealed sources that prevent leakage of 222Rn, the 222Rn, 214Pb, and 214Bi each reach the same activity level as that of the 226Ra within a few weeks of preparation of the source.

There are dozens of different gamma rays of varying energies and yields produced by the decay of 226Ra and its progeny; energies range from less than 50 keV to about 2.5 MeV. The energies and numbers of gamma rays that escape from the source depend on how the source is fabricated, in particular the type of material used for the source encapsulation and its thickness. Generally, for most practical sealed sources of 226Ra, photons with energies less than about 50 keV are not very important from a dose perspective.

If we neglect photons lower in energy than 50 keV and also consider only photons with yields greater than 1%, we obtain an effective gamma energy of about 0.74 MeV (obtained by multiplying each photon energy by its respective fractional yield, summing up all such products, and dividing the result by the sum of all the photon yields). Herman Cember's text (Introduction to Health Physics, 3rd ed.) gives the average energy as about 0.7 MeV, and Report 112 of the National Council on Radiation Protection and Measurements (Calibration of Survey Instruments Used in Radiation Protection for the Assessment of Ionizing Radiation Fields and Radioactive Surface Contamination, 1991) gives an effective gamma energy of 800 keV (p. 54).

Depending on the purpose for which the gamma radiation is being used, you may require better definition of the energy distribution from the source, in which instance you may have to make corrections for attenuation of photons of differing energies within the source materials or possibly make measurements of the gamma ray distribution being emitted from the source. The latter process requires use of a gamma spectrometric system and setup of the source to produce an acceptable photon fluence at the detector used, which would probably be a germanium detector in conjunction with a multichannel gamma analyzer. I hope this answer is sufficient for your needs.

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 27 September 2005. 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.