I don’t like the

mass attenuation coefficient; it bugs me and it has weird units. Yet researchers studying the attenuation of

x-rays in materials usually quote the mass attenuation coefficient rather than the

linear attenuation coefficient in their publications.

As x-rays pass through a material, they fall off exponentially as exp(−

*μL*), where

*L* is the distance (m) and

*μ* is the linear attenuation coefficient (m

^{−1}). But often researchers multiply and divide by the

density *ρ*, so the exponential becomes exp(−(

*μ*/

*ρ*)(

*ρL*)), where

*μ*/

*ρ* is the mass attenuation coefficient (m

^{2}/kg) and

*ρL* is the

areal density (kg/m

^{2}).

In Chapter 15 of

*Intermediate Physics of Medicine and Biology*,

Russ Hobbie and I explain some of the advantages of using

*μ*/

*ρ*.

The mass attenuation coefficient has the advantage of being independent of the density of the target material, which is particularly useful if the target is a gas. It has an additional advantage if Compton scattering is the dominant interaction. If *σ*_{tot} = *Zσ*_{C}, then *μ*_{atten}/*ρ* = *Zσ*_{C}*N*_{A}/*A* [*Z* is the atomic number, *A* the mass number, *N*_{A} is Avogadro’s number, and *σ*_{C} is the Compton cross section]. Since *Z*/*A* is nearly 1/2 for all elements except hydrogen, this quantity changes very little throughout the periodic table. This constancy is not true for the photoelectric effect or pair production. Figure 15.10 plots the mass attenuation coefficient vs energy for three substances spanning the periodic table. It is nearly independent of *Z* around 1 MeV where Compton scattering is dominant. The *K* and *L* absorption edges can be seen for lead; for the lighter elements they are below 10 keV. Figure 15.11 shows the contributions to *μ*_{atten}/*ρ* for air from the photoelectric effect, incoherent scattering, and pair production. Tables of mass attenuation coefficients are provided by the National Institute of Standards and Technology (NIST) at http://www.nist.gov/pml/data/xcom/index.cfm.

Let me offer an example where it makes sense to consider the mass attenuation coefficient.

Imagine you have a large box of area

*S*. You measure a mass

*M* of the fluid

pentane and poor it into the box. Then you place a source of x-rays under the box, directed upwards. You measure the intensity of the radiation incident on the underside of the box to be

*I*_{o}, and then move your detector to above the box and measure the intensity of radiation that passes through the pentane to be

*I*. Finally, use your ruler to measure the thickness of the pentane layer,

*L*.

You now have enough data to determine both the linear attenuation coefficient and the mass attenuation coefficient of pentane. For the linear attenuation coefficient, use the relationship

*I* =

*I*_{o} exp(−

*μL*) and solve for

*μ* = ln(

*I*_{o}/

*I*)/

*L*. You can also calculate the density

*ρ* =

*M*/(

*SL*). If you want the mass attenuation coefficient, you can now easily determine it:

*S* ln(

*I*_{o}/

*I*)/

*M*. You can also calculate the areal density:

*M*/

*S*.

Next you perform the same experiment on

neopentane. You use the same box with area

*S* and measure out the same mass

*M* of fluid. You find that

*I*_{o}/

*I* is unchanged, but

*L* is about 6% larger. You conclude the linear attenuation coefficient and the density both decrease by 6%, but the mass attenuation coefficient and the areal density are unchanged.

Why is

*I*_{o}/

*I *the same for both fluids? Pentane and neopentane are

isomers. They have exactly the same chemical formula, C

_{5}H

_{12}, but they have different structures. Pentane is an unbranched

hydrocarbon and neopentane has a central carbon bonded to four other carbon atoms. Because the mass

*M* of both substances is the same, the number of the atoms is the same in each case. The x-ray attenuation only depends on the number of atoms and the type of atoms, but not how those atoms are arranged (the density). This is one advantage of the mass attenuation coefficient: it depends only on the atoms and not their arrangement.

You can calculate the mass attenuation coefficient and the areal density without knowing

*L*. If for some reason

*L* were difficult to measure, you could still determine the mass attenuation coefficient even if you could not calculate the linear attenuation coefficient.

In a gas, the number of molecules is fixed but the density depends on the pressure and temperature. The mass attenuation coefficient does not change with the pressure and temperature. Again, it just depends on the atoms and not their distribution.

Water has a density of 1 g/cm

^{3}. If you express the mass attenuation coefficient in cm

^{2}/g and the linear attenuation coefficient in cm, then the mass attenuation coefficient and the linear attenuation coefficient have the same numerical value. Most tissue has a density close to that of water, so this trick works well for tissue too.

Given these advantages, have I started liking the mass attenuation coefficient? No, I still think it’s weird. But I can tolerate it a little better now.