KAHIBARO
Table of Contents
Meaning and Definition
When radiation passes through matter, its intensity decreases because some photons interact with the material. A useful way to describe this weakening is with the attenuation coefficient. If we divide the linear attenuation coefficient by the density of the material, we obtain the mass attenuation coefficient.
The mass attenuation coefficient is written as
$$
\frac{\mu}{\rho}
$$
where $\mu$ is the linear attenuation coefficient and $\rho$ is the density of the absorber.
This quantity tells us how strongly a material attenuates radiation per unit mass, rather than per unit length. That makes it especially useful when comparing the same substance in different physical forms, or comparing different materials on an equal mass basis.
The mass attenuation coefficient is defined by
$$
\frac{\mu}{\rho}
$$
Its SI unit is typically $ \mathrm{m^2 \, kg^{-1}} $, and a commonly used practical unit is $ \mathrm{cm^2 \, g^{-1}} $.
Why Density Is Removed
The linear attenuation coefficient $\mu$ depends on how much material is packed into a given volume. A dense material usually has more atoms in a given thickness, so it tends to attenuate more strongly per meter. By dividing by density, we remove this packing effect and focus more directly on the intrinsic interaction strength per unit mass.
This is helpful because two samples of the same material can have different densities, for example a compressed solid and a porous form. Their linear attenuation coefficients may differ, but their mass attenuation coefficients are often much closer.
If the exponential attenuation law is written as
$$
I = I_0 e^{-\mu x}
$$
and we use $\mu = \left(\frac{\mu}{\rho}\right)\rho$, then
$$
I = I_0 e^{-\left(\frac{\mu}{\rho}\right)\rho x}
$$
The product $\rho x$ is called the mass thickness, with units such as $\mathrm{g/cm^2}$.
Mass Thickness
A very important idea in this topic is that attenuation can be described using mass per unit area instead of geometric thickness. If a slab has thickness $x$ and density $\rho$, then its mass thickness is
$$
m = \rho x
$$
Using this, the attenuation law becomes
$$
I = I_0 e^{-\left(\frac{\mu}{\rho}\right)m}
$$
This form is very convenient in radiation physics, especially when the absorber is thin, layered, or not uniform in ordinary thickness.
For attenuation written in terms of mass thickness,
$$
I = I_0 e^{-\left(\frac{\mu}{\rho}\right)\rho x}
= I_0 e^{-\left(\frac{\mu}{\rho}\right)m}
$$
where
$$
m = \rho x
$$
is the mass thickness.
Units
Because $\mu$ has units of inverse length and $\rho$ has units of mass per volume, the units of $\mu/\rho$ are area per mass.
$$
\frac{\mu}{\rho} \sim \frac{1/\mathrm{length}}{\mathrm{mass}/\mathrm{volume}} = \frac{\mathrm{area}}{\mathrm{mass}}
$$
The two most common unit systems are shown below.
| Quantity | Common unit |
|---|---|
| Linear attenuation coefficient, $\mu$ | $\mathrm{cm^{-1}}$ or $\mathrm{m^{-1}}$ |
| Density, $\rho$ | $\mathrm{g/cm^3}$ or $\mathrm{kg/m^3}$ |
| Mass attenuation coefficient, $\mu/\rho$ | $\mathrm{cm^2/g}$ or $\mathrm{m^2/kg}$ |
| Mass thickness, $\rho x$ | $\mathrm{g/cm^2}$ or $\mathrm{kg/m^2}$ |
Physical Interpretation
A larger mass attenuation coefficient means that, for the same mass per unit area, the material reduces the radiation intensity more strongly. A smaller value means the radiation passes through more easily.
For photons, the mass attenuation coefficient depends on both the material and the photon energy. Different interaction processes contribute to it, such as photoelectric absorption, Compton scattering, and pair production. The detailed physics of those processes belongs to other chapters, but here it is enough to know that the total mass attenuation coefficient changes with energy because the dominant interaction mechanism changes with energy.
In general, for a given material, $\mu/\rho$ is not constant for all photon energies.
Relation to Microscopic Interactions
The mass attenuation coefficient is connected to the probability of interaction with the atoms in the material. Since it is normalized by density, it is a convenient bridge between bulk behavior and atomic scale properties.
For compounds and mixtures, the mass attenuation coefficient can often be estimated from the weighted contributions of the components. If a material contains several elements with mass fractions $w_i$, then approximately
$$
\left(\frac{\mu}{\rho}\right)_{\text{mixture}} = \sum_i w_i \left(\frac{\mu}{\rho}\right)_i
$$
where $w_i$ is the fraction of the total mass due to element $i$.
For a mixture or compound, an approximate rule is
$$
\left(\frac{\mu}{\rho}\right)_{\text{mixture}} = \sum_i w_i \left(\frac{\mu}{\rho}\right)_i
$$
where the mass fractions satisfy
$$
\sum_i w_i = 1
$$
Comparison with Linear Attenuation Coefficient
The difference between linear and mass attenuation coefficients is often a source of confusion, so it is worth comparing them directly.
| Quantity | Symbol | Meaning | Depends on density directly | Typical unit |
|---|---|---|---|---|
| Linear attenuation coefficient | $\mu$ | Attenuation per unit length | Yes | $\mathrm{cm^{-1}}$ |
| Mass attenuation coefficient | $\mu/\rho$ | Attenuation per unit mass thickness | No, density factor removed | $\mathrm{cm^2/g}$ |
If density is known, you can move from one form to the other using
$$
\mu = \left(\frac{\mu}{\rho}\right)\rho
$$
and
$$
\frac{\mu}{\rho} = \frac{\mu}{\rho}
$$
The second expression is just the definition, but it reminds us that the conversion is simple once density is known.
Example Calculation
Suppose a material has density
$$
\rho = 2.0 \, \mathrm{g/cm^3}
$$
and a mass attenuation coefficient
$$
\frac{\mu}{\rho} = 0.15 \, \mathrm{cm^2/g}
$$
Then the linear attenuation coefficient is
$$
\mu = \left(0.15 \, \mathrm{cm^2/g}\right)\left(2.0 \, \mathrm{g/cm^3}\right)
= 0.30 \, \mathrm{cm^{-1}}
$$
If the slab thickness is
$$
x = 5.0 \, \mathrm{cm}
$$
then
$$
I = I_0 e^{-\mu x} = I_0 e^{-(0.30)(5.0)} = I_0 e^{-1.5}
$$
so
$$
\frac{I}{I_0} \approx 0.223
$$
About $22.3\%$ of the original intensity remains.
Using mass thickness gives the same result. The mass thickness is
$$
m = \rho x = (2.0)(5.0) = 10.0 \, \mathrm{g/cm^2}
$$
Then
$$
I = I_0 e^{-\left(\frac{\mu}{\rho}\right)m}
= I_0 e^{-(0.15)(10.0)}
= I_0 e^{-1.5}
$$
which matches exactly.
Graphical View
A thicker absorber means a larger mass thickness, so the intensity falls exponentially as mass thickness increases.
The curve starts at $I/I_0 = 1$ when no material is present, and then decreases exponentially as the mass thickness increases.
Practical Importance
Mass attenuation coefficients are widely tabulated for many materials and photon energies. These tables are used in radiation shielding, medical imaging, dosimetry, and detector design. Because the quantity is normalized by density, the tables are easier to apply across different situations.
For example, when comparing shielding materials, the linear attenuation coefficient tells how effective a certain thickness is, while the mass attenuation coefficient helps compare materials independent of how tightly packed they are.
Key Formula Summary
Main relations for the mass attenuation coefficient:
$$
\frac{\mu}{\rho} = \text{mass attenuation coefficient}
$$
$$
\mu = \left(\frac{\mu}{\rho}\right)\rho
$$
$$
m = \rho x
$$
$$
I = I_0 e^{-\mu x}
$$
$$
I = I_0 e^{-\left(\frac{\mu}{\rho}\right)m}
$$
Units:
$$
\frac{\mu}{\rho} \rightarrow \mathrm{cm^2/g} \quad \text{or} \quad \mathrm{m^2/kg}
$$
Final Insight
The mass attenuation coefficient is a way to express attenuation strength per unit mass of material rather than per unit length. This makes it a powerful and widely used quantity for comparing materials and analyzing photon attenuation in a form that is less tied to physical density and sample geometry.