Answer to Question #9866 Submitted to "Ask the Experts"

Category: Radiation Basics

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

I have been pondering a comparison between the reduction of a photon field shown by the inverse square law and that of an exponential decay. Since dose-rate formulas and shielding calculations both use attenuation/absorption coefficients (allowing for correct units), the basic formulas are not the same, the curves (when distances are short) are not the same, and knowing the dose-rate/photon field when distances are short draws the question "why use exponential formulas when fields drop off through the inverse-square law in any given medium?"

Why not use exponential decay formulas from distance = 0 all the way instead of the 1/d2 formulas?

Is this seen as an approximation?

The inverse square dependence of photon fluence rate or dose rate on distance from a source applies to the primary photons emitted isotropically from a source whose dimensions are small compared to the distance from the source to the dose point (referred to as a point isotropic source). The process of exponential attenuation of photons applies at any distance in any medium and to any source, but when dealing with transmission media that present little attenuation we often neglect attenuation and consider only the reduction in intensity with distance. This is often the case when dealing with sources in air when the photons are sufficiently energetic that air attenuation is not a significant factor.

The common approximate exposure rate equation, X = 6CEn d2, where X is the exposure rate (R per hour), C is the source activity (Ci), E is the photon energy (MeV), n is the photon yield (number of photons of energy E per disintegration of the radionuclide), and d is the distance (feet) from a point isotropic source, assumes no significant attenuation in the medium between the source and dose point. This is a legitimate assumption when the distance is not excessive and the photon energy is not very low.

For example, at one meter in air from a source that emits 1 MeV photons, the reduction in intensity of primary photons by attenuation in the air is about 1 percent. If the medium were water, however, the loss of primary photons would have been about 99.9 percent, as calculated as 1 minus the fractional transmission—i.e., 1 – e-µd, where µ is the linear attenuation coefficient for 1 MeV photons in water (0.0707 cm-1) and d is 100 cm.

The fact is that the processes of intensity reduction by distance and by attenuation do not operate exclusively in many situations involving exposure to photons. It is necessary to consider both processes in order to account properly for the reduction in primary photon intensity. Note that I have referred several times to primary photons—i.e., those emitted directly by the source. In addition to these, it is often necessary to account for the effects of secondary photons that arise as a consequence of interactions of the primary photons in the media through which the primary photons travel. Compton-scattered photons are among the most important in this regard. The impact that these secondary photons have on increasing the dose rate at the point of interest may be substantial and is often accounted for in calculations by the use of a buildup factor that represents the factor by which the primary photon dose (rate) is increased by these secondary radiations. If you are interested in a more detailed discussion of this process, see the information available at this link on the HPS website.

In summary, we do not generally have the option of choosing either distance or attenuation as an approach to calculating dose rate from a source; we must consider both reduction of intensity by increasing distance from the source and reduction by attenuation in any media between the source and dose point. It is only in cases when one of these reduction mechanisms becomes insignificant compared to the other that it is appropriate to neglect the lesser effect.

I hope this clarifies the situation for you.

George Chabot, PhD, CHP

Answer posted on 2 September 2011. 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.