Answer to Question #8659 Submitted to "Ask the Experts"

Category: Radiation Basics — Interaction Coefficients

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

Q
I measured the linear attenuation coefficient of lead for a 60Co teletherapy machine (1.17 MeV and 1.33 MeV photon energies). My results showed the linear attenuation coefficient depended on the field size of the machine (e.g., 5 x 5, 10 x 10, 15 x 15). If it is okay, then how can we give a mean attenuation coefficient for lead?
A

I cannot answer your question completely without knowing the specific geometries that you used in your measurements, but I shall attempt to provide information that should be helpful.

In order to measure the attenuation coefficient for the primary gamma rays from 60Co, it is necessary to use what is called a good geometry setup. What this means is that the primary beam must be narrow, preferably like a pencil beam, and should be normally incident on the lead. In the usual procedure, lead plates would be added one at a time to yield increasing thickness. The detector should be placed a good distance away from the lead plates (I would estimate at least 10 beam diameters and at least five times the largest lead-plate dimension), and the source-to-detector distance should remain constant through all the measurements. Ideally, the detector should be shielded on its sides to reduce background and so that only photons incident on the front face of the detector are measured. The idea of this geometry is to minimize production of scattered photons and to prevent photons that are scattered in the lead from reaching the detector. The larger the distance the detector is from the lead, the smaller will be the solid angle subtended at the detector by the lead surface; this minimizes any photons scattered at small forward angles from reaching the detector.

When the beam is not highly collimated, as apparently is your situation in which various field sizes were tried, the beam is incident on a significant lead-plate area, providing the opportunity for more scattering of photons in the lead and more scattered photons then reaching the detector. One would expect that the greater the field size on the plates the greater would be the extent of scattered photons reaching the detector. If the detector is simply recording the number of photons, or possibly an intensive quantity such as dose, one would expect to record more photons for the larger field size, yielding a likely smaller value of the apparent attenuation coefficient. This poor geometry setup results then in measuring not only the primary photons but some of the scattered photons as well. The attenuation coefficients you obtain will not be the best representations of the true primary photon attenuation coefficients, and the larger the field size incident on the lead the greater should be the difference between the measured value and the true value. Assuming an acceptable distance of the detector from the lead, as the field size decreases the determined value of the attenuation coefficient should more closely approach the true value.

Thus, you should minimize the field size and attempt to maintain a good geometry arrangement in order to obtain a realistic estimate of the attenuation coefficient for the primary photons. It is not appropriate to use an average value for the three field sizes. Because 60Co emits two slightly different energy photons (1.17 and 1.33 MeV), the shielding experiment is not quite as clean as it is with monoenergetic photons, but the energies are close enough so that you should obtain good results with the attenuation coefficient having a value close to that for 1.25 MeV photons. It is a good idea to plot the data, fractional transmission on a log scale vs. lead thickness, to observe the transmission curve, which should be very nearly a straight line on the semilogarithmic plot. Use a least squares fitting routine to obtain the best fit to the data. The negative of the slope of the semilogarithmic plot represents the value of the attenuation coefficient. Good luck.

George Chabot, PHD, CHP

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