Physics · Medical Physics · A2 Level

Attenuation of X-Rays

Every visible feature on an X-ray image — bone standing out sharply against soft tissue, a lung's outline against the ribcage — comes down to one underlying piece of physics: different materials absorb X-rays by different amounts, and that absorption follows a precise exponential law. This lesson works through that law and the graphical skills used to apply it.

9 min read

The exponential attenuation law

As an X-ray beam passes through matter, it loses intensity — photons are absorbed or scattered out of the beam by the atoms they pass. For a beam of initial intensity $I_0$ passing through a thickness $x$ of a uniform material, the transmitted intensity $I$ is given by:

$$I = I_0e^{-\mu x}$$

Here $\mu$ is the linear attenuation coefficient of the material, with units of m$^{-1}$ (or sometimes cm$^{-1}$). It measures how strongly a given material attenuates the beam per unit thickness — a larger $\mu$ means the material is a more effective absorber, and intensity falls off more rapidly with increasing thickness.

Worked exampleApplying the attenuation law

An X-ray beam of initial intensity $I_0 = 2.4\times10^{4}\text{ W m}^{-2}$ passes through a $3.5\text{ cm}$ thickness of tissue with linear attenuation coefficient $\mu = 0.22\text{ cm}^{-1}$. Find the transmitted intensity.

Use $I = I_0e^{-\mu x}$ directly, keeping units of $\mu$ and $x$ consistent (both in cm here).

$I = (2.4\times10^{4})\,e^{-(0.22)(3.5)} = (2.4\times10^{4})\,e^{-0.77}$

The exponent $\mu x$ must be dimensionless, which it is here since both are in cm.

$I = (2.4\times10^{4})(0.463) \approx 1.11\times10^{4}\text{ W m}^{-2}$

So roughly 46% of the original intensity is transmitted through this thickness of tissue.

answer

The mass attenuation coefficient

The linear attenuation coefficient $\mu$ depends on how tightly packed a material's atoms are — that is, on its density $\rho$ — as well as on the material's intrinsic ability to absorb X-rays. This can be a nuisance when comparing materials, since the same substance can exist in different physical states (a gas compressed to different pressures, for instance) with very different densities but the same underlying atomic composition.

The mass attenuation coefficient removes this dependence on physical density:

$$\mu_m = \frac{\mu}{\rho}$$

with units of m$^2$ kg$^{-1}$ (or cm$^2$ g$^{-1}$). Because dividing by density cancels out the effect of how compactly the material's atoms happen to be packed, $\mu_m$ characterises a material's intrinsic attenuating ability — a genuine property of the material's atomic composition, independent of its physical state. This makes $\mu_m$ a more convenient quantity for tabulating and comparing materials' attenuating properties, or for handling the same material at different densities using a single reference value.

Half-value thickness

The half-value thickness, $x_{1/2}$, of a material is the thickness needed to reduce the transmitted intensity of a beam to exactly half its initial value. This is the exact same mathematical idea as a half-life in radioactive decay — an exponential quantity falling to half its initial value over a fixed interval — except here the "interval" is a thickness in space, rather than a duration in time.

Same mathematics as half-life

If half-life has been covered elsewhere as $t_{1/2} = \dfrac{\ln 2}{\lambda}$ for radioactive decay, half-value thickness is the direct spatial analogue: $\mu$ plays exactly the role that the decay constant $\lambda$ plays in the time-based exponential decay law. The derivation below stands on its own either way.

DerivationHalf-value thickness in terms of μ

Start from $I = I_0e^{-\mu x}$. By definition, at $x = x_{1/2}$ the transmitted intensity has fallen to half: $I = \tfrac{1}{2}I_0$.

This is exactly what "half-value thickness" means.

Substituting: $\tfrac{1}{2}I_0 = I_0e^{-\mu x_{1/2}}$, so $\tfrac{1}{2} = e^{-\mu x_{1/2}}$

The initial intensity $I_0$ cancels from both sides.

Take natural logs of both sides: $\ln\left(\tfrac{1}{2}\right) = -\mu x_{1/2}$, so $-\ln 2 = -\mu x_{1/2}$

$\ln(1/2) = -\ln 2$, a standard log identity.

Rearranging gives the result: $x_{1/2} = \dfrac{\ln 2}{\mu}$

A larger $\mu$ (stronger attenuator) gives a smaller half-value thickness, as expected.

QED

For example, if a material has $\mu = 0.15\text{ cm}^{-1}$, its half-value thickness is $x_{1/2} = \dfrac{\ln 2}{0.15} \approx 4.6\text{ cm}$ — every additional 4.6 cm of that material halves the transmitted intensity again.

Differential tissue absorption and image contrast

Different tissues have measurably different linear attenuation coefficients. Dense, compact tissue such as bone has a relatively high $\mu$, while soft tissue has a lower $\mu$, and air-filled spaces (such as the lungs) attenuate the beam hardly at all. When an X-ray beam passes through a real patient, it passes through a mixture of these different tissue types simultaneously along different paths, and the transmitted intensity pattern reaching the detector varies according to exactly what each part of the beam passed through.

Bone, with its high attenuation coefficient, casts a strong "shadow": very little intensity is transmitted through it, so it appears light or white on a typical radiograph.
Soft tissue and air-filled spaces, with lower attenuation coefficients, let much more of the beam through, so they appear darker on the same image.

This difference in transmitted intensity across the detector, arising directly from differential tissue attenuation, is exactly what makes internal anatomical structure visible on an X-ray image at all. Without a meaningful difference in $\mu$ between adjacent structures, there would be no way to distinguish them on the resulting image — which is precisely the problem that contrast media (covered in the previous lesson) are introduced to solve for tissues that would otherwise attenuate too similarly.

Log-graph practice

Because $I = I_0e^{-\mu x}$ is an exponential relationship, plotting raw intensity $I$ against thickness $x$ produces a curve, from which $\mu$ is difficult to read off accurately. Taking logarithms converts this into a straight-line relationship, which is far easier to analyse. Taking natural logs of both sides of the attenuation law:

$$\ln I = \ln I_0 - \mu x$$

This has exactly the form of a straight line, $y = c + mx$, if $y = \ln I$ is plotted against $x$: the graph has a $y$-intercept of $\ln I_0$ and, critically, a gradient of $-\mu$. Plotting $\log_{10}I$ against $x$ instead works the same way, just with a rescaled gradient. This graphical technique — converting an exponential relationship into a straight line by taking logs, and reading a physical quantity from the gradient — is a standard, directly examined analysis skill.

Worked exampleReading μ from a log-graph gradient

A student plots $\ln I$ (vertical axis, no units since $I$ was in W m$^{-2}$) against thickness $x$ (horizontal axis, in cm) for attenuation data through a tissue sample. The resulting straight line passes through the points $(0, 9.8)$ and $(8.0, 7.4)$.

Two points on the straight line are enough to find its gradient.

Gradient $= \dfrac{7.4 - 9.8}{8.0 - 0} = \dfrac{-2.4}{8.0} = -0.30\text{ cm}^{-1}$

The gradient of a $\ln I$ against $x$ graph is $-\mu$, by the linear form derived above.

Since gradient $= -\mu$, we have $\mu = 0.30\text{ cm}^{-1}$.

The negative sign in the gradient simply reflects that intensity decreases with increasing thickness.

answer

Key takeaways

X-ray attenuation follows $I = I_0e^{-\mu x}$, where $\mu$ is the linear attenuation coefficient of the material.
The mass attenuation coefficient $\mu_m = \mu/\rho$ removes the dependence on physical density, giving an intrinsic, density-independent measure of a material's attenuating ability.
Half-value thickness $x_{1/2} = \dfrac{\ln 2}{\mu}$ is the thickness needed to halve the transmitted intensity — mathematically identical to half-life applied to a spatial thickness.
Different tissues have different attenuation coefficients: bone (high $\mu$) casts a strong shadow, while soft tissue and air (lower $\mu$) transmit more of the beam.
This differential absorption between tissue types is the physical basis of all X-ray image contrast.
Taking logs of the attenuation law gives $\ln I = \ln I_0 - \mu x$, a straight-line relationship between $\ln I$ and thickness $x$.
Plotting $\ln I$ (or $\log_{10}I$) against thickness produces a straight line of gradient $-\mu$, letting $\mu$ be read directly from experimental data.

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