Answer to Question #12285 Submitted to "Ask the Experts"

Category: Radiation Basics — Photons

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

Q

Which material, wood or Lucite, will produce more Compton scatter radiation when irradiated with x rays?

A

While this question appears rather straightforward, the answer may depend on the particular circumstances of the irradiation—in particular, the energies of the photons incident on the material, the relative thicknesses of the two materials of interest, and the specific characteristics of the materials being exposed. As a consequence, my answer will be a bit more extensive than you might have anticipated, and I shall not be able to provide an absolute answer to your question.

We should first note the similarities and the differences between the two materials, wood and Lucite (a tradename for polymethyl methacrylate). Different woods may have different compositions, and these can also be affected by humidity, since many woods will absorb humidity from the air, increasing the possible hydrogen and oxygen content of the woods. Lucite is not particularly hygroscopic, and its composition is not noticeably affected by moisture or humidity. The density of Lucite is relatively constant, being about 1.19 grams per centimeter3 (g cm-3). The densities of woods may vary considerably, varying from about 0.1 g cm-3 for some balsa woods to about 1.3 g cm-3 for some ironwood. Many common woods, such as most of the pines, maples, and oaks, have densities that range from about 0.4 g cm-3 to 0.8 g cm-3.

The probability of a Compton scatter event occurring in a given material depends directly on the scattering cross section and on the electron density of the material (the electrons represent the scattering sites). The magnitude of the cross section is also dependent on the energies of the incident photons. Additionally, there is an angular dependence for the scattering of photons that may be interpreted through application of differential scattering cross sections, which I shall not specifically discuss.

A typical composition by weight of many common woods is about 50% carbon, 42% oxygen, 6% hydrogen, 1% nitrogen, and 1% other elements (e.g., Ca, K, Na, Mg, Fe, Mn). Lucite has the weight composition of 59.98% carbon, 31.96% oxygen, and 8.054% hydrogen. Based on these compositions, we can readily calculate the electron densities of the two materials; in so doing we obtain 3.18 × 1023 electrons per gram of wood (assuming calcium as representative of the "other" components) and 3.24 × 1023 electrons per gram of Lucite. The two values are very similar, differing by only about 1.9%.

The most important parameter affecting the mass interaction cross section is the effective atomic number of the material. For the wood composition given, the effective atomic number can be calculated as 6.69; the effective atomic number for Lucite is calculated as 6.24. This difference is not very large; however, it can become an influencing factor, especially when dealing with very low-energy photons. At low energies, the photoelectric cross section becomes relatively more important, and this cross section varies approximately as the effective atomic number raised to the fifth power. As an example, the photoelectric cross section for Lucite at 20 kiloelectronvolts (keV) photon energy is 0.30 cmg-1, and the Compton scattering cross section is about 0.20 cmg-1. We would then expect the photoelectric cross section for wood of the above composition to be about (0.30 cmg-1)(6.69/6.24)5 = 0.42 cm2 g-1. The effect of this 40% increase in the photoelectric cross section is to make this interaction notably more probable in wood than in Lucite, thus reducing the expected degree of Compton scatter in wood compared to Lucite. A similar effect, though generally less so, might be expected at very high energies, especially greater than about 10 megaelectronvolts (MeV), when the pair production cross section, which varies as about the square of the material's atomic number, will increase more notably in wood than in Lucite, also leading to somewhat reduced Compton scatter. At intermediate photon energies, from about 100 keV to 3 MeV, we would not expect much difference in Compton scatter between the two materials, on an equal mass basis, with the Lucite offering possibly slightly enhanced scatter associated with the slightly greater electron density.

This conclusion, however, should also consider possible effects associated with the respective thicknesses of the materials. As noted above, wood density is typically considerably less than Lucite density. Thus, for a given thickness there will be considerably fewer electron-scattering sites (i.e., electrons) available per unit area of wood irradiated compared to Lucite, thus markedly reducing the extent of backscatter. One could negate this effect by using infinitely thick samples of wood and Lucite; by infinitely thick, from a radiological viewpoint, I mean a thickness equivalent to about five photon mean free paths in the material (i.e., 5/µ, where µ is the photon linear attenuation coefficient in the material). For 1 MeV photons in Lucite, µ is about 0.0818 cm-1, and an infinite thickness would be about 61 cm—a substantial thickness. If wood had a density of about 0.6 g cm-3, an infinite thickness, calculated similarly, would be about 120 cm. Thus, if you were measuring Compton scattered photons coming back from the front surface being irradiated, the large difference in thicknesses would yield a markedly reduced Compton scatter component for wood compared to Lucite, simply because of a geometry effect that causes fewer of the photons scattered at greater depths in the wood to reach the measurement point near the front surface. We could look at other geometry effects for other measurement locations, but I don't think that is necessary.

For the conditions we have so far assumed, involving moderate-energy photons, we might expect the scattered-photon fluence near the surface of photon incidence to be greater for the Lucite than for typical woods. This conclusion would apply whether referring to moderate, finite thicknesses (such as a few cm), or radiologically infinite thicknesses. For very low, or possibly very high-energy photons, relatively fewer Compton scatter events will occur in wood compared to Lucite, thus increasing the differential between Lucite and wood even more. A modifying consideration that would have to be applied, depending on the specific energies of the scattered photons being considered and on the thickness of the scattering material, is the attenuation of the scattered photons reaching the measurement point. For example, backscattered photons from 1 MeV incident photons have energies of about 0.20 MeV. Such photons traversing a thickness of 1 cm in Lucite would be attenuated by about 15%, while the same photons traversing 1 cm in the described wood would be attenuated by about 8%. This effect works in opposition to the electron-density effect and the geometry effect that we mentioned, and it becomes greater as thicknesses increase. For relatively modest and equal thicknesses of the two materials, I would expect the reduced number of electrons in wood, compared to Lucite, to have a greater effect than the attenuation of the scattered photons for moderate-energy incident photons, leading to a higher fluence of backscattered photons from Lucite compared to wood. Other situations would require further analysis.

George Chabot, PhD

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 22 December 2017. 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.