Answer to Question #7434 Submitted to "Ask the Experts"

Category: Radiation Basics

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


Kindly tell me the difference between branching fraction and adundance of a radionuclide. How can abundance or branching fraction can be mare than 100 percent as in 111In, 11C, and other radionuclides?


In addition to radionuclide abundance and branching fraction that you mention, we should also consider the concept of radiation yield since I believe it is the latter that you are probably confused about.

The abundance of a particular species refers to the atom fraction of that nuclide that is found in nature. The term is used only in reference to stable nuclides and radionuclides that have sufficiently long half-lives that they are still present on earth. For example, the fractional atom abundances of the two important uranium isotopes, 238U (half-life of 4.5 x 109 years) and 235U (half-life of 7.0 x 108 years), are 0.992745 and 0.0072. A third natural isotope of uranium, 234U, has an atom abundance of 0.000055; note that the 234U has a half-life of only 2.5 x 105 years, not long enough to ensure its survival since the beginning of the earth, but it is produced as part of the decay chain of 238U and is continually produced as the 238U decays. The abundance of any atomic species, including naturally occurring radionuclides, cannot exceed 1.0.

The concept of branching fraction, as applied to decay of a radionuclide, means that the radionuclide decays by more than one possible mode. For example, a number of radionuclides that contain an excess number of protons in their nuclei may decay by both electron capture and by positron emission; both decay events result in reduction of the proton number by one. The branching fraction refers to the fraction of the time the decays occurs by the respective route. For example, the radionuclide 58Co decays by both electron capture and positron emission with respective branching fractions of 0.85 and 0.15. The sum of branching fractions cannot exceed 1.0. A value of 1.0 for a branching fraction implies that the nuclide decays by only a single mode.

This brings us to the notion of radiation yield, which is also sometimes referred to as radiation abundance (but not radionuclide abundance). This refers to the fractional yield (number of that type of radiation emitted per disintegration of the nucleus) of a particular radiation emitted during the decay process. The major decay processes are beta minus (or negatron) decay, beta plus (or positron) decay, electron capture decay, and alpha decay. These are associated, respectively with the emission of beta minus particles, positrons, x rays, and alpha particles. As noted above, if branch decay is possible, the sum of the branching fractions for a particular radionuclide cannot exceed 1.0; therefore, the sum of the yields of the major particulate radiations (beta particles or alpha particles) that result from decay of that radionuclide also cannot exceed 1.0. In addition to the major decay radiations, however, gamma radiation is often present as a result of deexcitation of the nucleus following the decay event. It is possible for the sum of the gamma ray yields to exceed 1.0 because deexcitation may occur through sequential emissions from different nuclear excited states. One of the most common examples of this process occurs in the case of 60Co. This radionuclide decays by beta minus emission to an excited nuclear state of 60Ni, which immediately deexcites by emitting a 1.17 MeV gamma ray that leads to a second lower excited state, and this state deexcites with the emission of a second gamma ray of 1.33 MeV. Each of these gamma rays occurs every time a 60Co nucleus decays by beta minus emission, which it does 100 percent of the time; therefore, the total gamma yield (or gamma abundance) is 2.0 (2 gamma rays per disintegration). In the case of the 111In that you mention, the nucleus decays100 percent of the time by electron capture, which leads to an excited state of 111Cd. This excited state deexcites 90.5 percent of the time with the emission of a 0.171 MeV gamma ray (in the other 10.5 percent of the deexcitations a conversion electron is emitted rather than a gamma ray); similarly to the 60Co case, this initial deexcitation leads to a lower excited state that deexcites with the emission of a second gamma ray with a yield of 0.940 and an energy of 0.245 MeV. The sum of the two gamma ray yields for this decay is then 1.845.

In the case of a radionuclide that decays by positron emission, there is another factor that adds more photons to the process; these are the annihilation photons that result when the positron that is emitted slows down and encounters an electron. The two antibodies annihilate, and two annihilation photons, each of 0.511 MeV, are produced. These are often shown when you look up decay scheme data, frequently being presented as if they were gamma rays, although they are actually not since they do not originate in the nucleus. At any rate, since there are two such photons produced per positron emitted, a radionuclide that decays 100 percent of the time by positron emission will exhibit an annihilation photon yield of 2.0. This is more-or-less the case for the11C that you cite, since it decays almost 100 percent (99.8 percent) of the time by positron emission.

I hope this discussion resolves most of your confusion about the concepts of nuclide abundance, branching fraction, and radiation yield.

George Chabot, PhD, CHP

Answer posted on 16 May 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.