X-ray Attenuation Revision Notes | A Level AQA Physics

1. Why X-rays Fade as They Travel Through Matter

Picture a beam of X-rays entering a patient's body. With every millimetre it travels, some photons are absorbed or scattered — and the beam grows weaker. The question is: how does it weaken, and can we predict it mathematically?

The answer turns out to be beautifully systematic. Each thin layer of material removes the same fraction of whatever intensity remains — not the same absolute amount. That pattern is the hallmark of exponential decay, and it is exactly what governs X-ray attenuation.

Three physical processes cause this attenuation: photoelectric absorption, Compton scattering, and pair production. At the diagnostic X-ray energies used in medical imaging (roughly 20–100 keV), photoelectric absorption and Compton scattering dominate. The net effect is that photons are continuously removed from the beam as it penetrates deeper into tissue.

2. Exponential Attenuation and the Key Equation

Think back to radioactive decay — the activity halves in each successive half-life, regardless of how much activity remains. X-ray attenuation follows the same mathematical logic. Each successive layer of equal thickness removes the same proportion of the beam, so the intensity falls exponentially with depth.

The equation that captures this is one you must know precisely for the exam. Here it is, built from first principles:

Notice the negative sign in the exponent: as $x$ increases, $e^{-\mu x}$ decreases, so $I$ falls. A large value of $\mu$ means the material is a strong attenuator — the beam loses intensity rapidly. A small $\mu$ means the material is relatively transparent to X-rays.

The chart above shows the key contrast: bone (high $\mu$) attenuates the beam far more rapidly than soft tissue (low $\mu$). This difference in attenuation is precisely what creates contrast on an X-ray image — bone appears white because very little intensity reaches the detector behind it.

3. The Linear Attenuation Coefficient μ

The linear attenuation coefficient $\mu$ is the material's 'stopping power' per unit length. A high $\mu$ means the material removes a large fraction of the beam per metre of thickness. Its SI unit is m⁻¹, though you will also encounter cm⁻¹ in data tables — always check and convert before substituting into the equation.

Crucially, $\mu$ depends on both the material and the X-ray photon energy. Higher-energy X-rays are harder to stop, so $\mu$ decreases as photon energy increases. This is why radiographers can adjust the 'hardness' of the X-ray beam by changing the tube voltage — a harder beam penetrates more deeply and delivers less dose to superficial tissue.

Typical values give a sense of scale: for soft tissue at diagnostic energies, $\mu \approx 20 \text{ m}^{-1}$; for bone, $\mu \approx 600 \text{ m}^{-1}$; for lead, $\mu$ can exceed $10^4 \text{ m}^{-1}$. These large differences explain why lead aprons are effective radiation shields.

4. The Mass Attenuation Coefficient μ_m

Here is a subtlety worth pausing on. Lead has a very high $\mu$ partly because it is extremely dense — pack more atoms into a given volume and you intercept more photons. But what if you want to compare how well two materials attenuate X-rays per unit mass, independent of how densely they are packed? That is where the mass attenuation coefficient $\mu_m$ comes in.

The mass attenuation coefficient is particularly useful when comparing materials of different densities, or when the same material is present in different physical states. For example, water vapour and liquid water have very different densities but the same $\mu_m$, because the attenuation per kilogram of water molecules is unchanged.

5. Half-Value Thickness

The half-value thickness $x_{\frac{1}{2}}$ is the thickness of a specific material that reduces the X-ray intensity to exactly half its incident value. It plays the same conceptual role as half-life in radioactive decay — and the derivation follows the same algebra.

To derive it, set $I = \frac{1}{2} I_0$ in the attenuation equation and solve for $x$. The $I_0$ cancels, which tells you something important: the half-value thickness is a property of the material and the beam energy alone — it does not depend on the initial intensity.

Worked example: Deriving the Half-Value Thickness Formula

Starting from $I = I_0 e^{-\mu x}$, derive an expression for the half-value thickness $x_{\frac{1}{2}}$.

1. $$\frac{1}{2} I_0 = I_0 \, e^{-\mu x_{\frac{1}{2}}}$$ — We want the thickness at which exactly half the intensity is transmitted.
2. $$\frac{1}{2} = e^{-\mu x_{\frac{1}{2}}}$$ — I₀ appears on both sides and is non-zero, so it cancels cleanly. This confirms that x½ is independent of the initial intensity.
3. $$\ln\left(\frac{1}{2}\right) = -\mu x_{\frac{1}{2}}$$ — The natural log is the inverse of the exponential function. $\ln(e^{-\mu x}) = -\mu x$.
4. $$-\ln 2 = -\mu x_{\frac{1}{2}}$$ — $\ln(\frac{1}{2}) = \ln(2^{-1}) = -\ln 2$. The negative signs on both sides cancel.
5. $$x_{\frac{1}{2}} = \frac{\ln 2}{\mu}$$ — Divide both sides by μ. Since ln 2 ≈ 0.693, a larger μ gives a smaller half-value thickness — strongly attenuating materials have thin half-value thicknesses.

Answer: $$x_{\frac{1}{2}} = \frac{\ln 2}{\mu} \approx \frac{0.693}{\mu}$$

This result is directly analogous to the half-life formula $t_{\frac{1}{2}} = \frac{\ln 2}{\lambda}$ from radioactive decay, where $\lambda$ is the decay constant. If you remember one, you can reconstruct the other. The half-value thickness is used clinically to characterise shielding materials and to specify the quality of an X-ray beam.

6. Worked Calculations

Exam questions on attenuation typically ask you to calculate transmitted intensity, find the thickness needed to achieve a target attenuation, or determine the half-value thickness. The algebra always flows from the same equation — the skill is in rearranging it confidently. Work through both examples below before attempting the retrieval check.

Worked example: Calculating Transmitted Intensity

An X-ray beam of incident intensity $I_0 = 4.0 \times 10^3$ W m⁻² passes through 30 mm of soft tissue. The linear attenuation coefficient of soft tissue for this beam is $\mu = 20$ m⁻¹. Calculate the transmitted intensity. [3 marks]

1. $$x = 30 \text{ mm} = 30 \times 10^{-3} \text{ m} = 0.030 \text{ m}$$ — μ is in m⁻¹, so x must be in metres. Always convert before substituting. [M1]
2. $$\mu x = 20 \text{ m}^{-1} \times 0.030 \text{ m} = 0.60$$ — The units cancel: m⁻¹ × m = dimensionless. The exponent must be dimensionless.
3. $$e^{-0.60} = 0.5488...$$ — Use your calculator. This factor tells you what fraction of the beam is transmitted.
4. $$I = I_0 \, e^{-\mu x} = 4.0 \times 10^3 \times 0.5488 = 2195 \text{ W m}^{-2}$$ — $4.0 \times 10^3 \times 0.5488 = 4000 \times 0.5488 = 2195.2$ W m⁻². [M1]
5. $$I \approx 2200 \text{ W m}^{-2} \quad (2 \text{ s.f.})$$ — The data is given to 2 significant figures, so round accordingly. Include units. [A1]

Answer: $I \approx 2200$ W m⁻² (2 s.f.)

Worked example: Finding the Half-Value Thickness and Comparing Materials

A material has a linear attenuation coefficient of $\mu = 600$ m⁻¹. (a) Calculate the half-value thickness. (b) Calculate the thickness needed to reduce intensity to 12.5% of its incident value. [4 marks]

1. $$x_{\frac{1}{2}} = \frac{\ln 2}{\mu} = \frac{0.6931}{600 \text{ m}^{-1}}$$ — Substitute directly. ln 2 = 0.6931 to 4 significant figures. [M1]
2. $$x_{\frac{1}{2}} = \frac{0.6931}{600} = 1.155 \times 10^{-3} \text{ m} \approx 1.2 \text{ mm}$$ — $0.6931 \div 600 = 0.001155$ m. Converting: $0.001155 \times 1000 = 1.155$ mm. [A1]
3. $$12.5\% = \frac{1}{8} = \left(\frac{1}{2}\right)^3$$ — Reducing to 1/2 requires 1 half-value thickness; to 1/4 requires 2; to 1/8 requires 3. So we need 3 half-value thicknesses. [M1]
4. $$x = 3 \times x_{\frac{1}{2}} = 3 \times 1.155 \times 10^{-3} \text{ m} = 3.47 \times 10^{-3} \text{ m} \approx 3.5 \text{ mm}$$ — $3 \times 1.155 = 3.465$ mm, rounded to 2 s.f. gives 3.5 mm. [A1]

Answer: (a) $x_{\frac{1}{2}} \approx 1.2$ mm; (b) $x \approx 3.5$ mm to reduce intensity to 12.5%

7. Applying Attenuation to Medical Imaging

The reason X-ray imaging works clinically is precisely because different tissues have very different values of $\mu$. Bone contains calcium, a relatively high atomic number element, which gives it a much larger $\mu$ than the hydrogen, carbon, and oxygen that dominate soft tissue. The beam is strongly attenuated by bone and weakly attenuated by muscle — producing the contrast that makes a fracture visible.

The mass attenuation coefficient $\mu_m$ is particularly useful here. Iodine-based contrast agents are injected into blood vessels before imaging. Iodine has a high $\mu_m$ relative to soft tissue, so blood vessels filled with contrast agent appear bright white on the image — even though the density of the contrast solution is not dramatically different from surrounding tissue.

Lead shielding in X-ray rooms exploits the same physics. Lead has an exceptionally high $\mu$, so even a few millimetres of lead reduces the transmitted intensity by many orders of magnitude. Radiographers stand behind lead screens or wear lead aprons to protect tissues that are not being imaged.

8. Exam Technique and Common Mistakes

Attenuation questions are reliable mark-earners if you follow a disciplined method. The most frequent errors are unit mismatches and sign errors in logarithm rearrangements. The guidance below addresses the mistakes that most commonly cost students marks in this topic.

Taken together, the three spec requirements — the exponential attenuation law, the definitions of μ and μ_m, and the half-value thickness — form a tightly connected set. Mastering the algebra of $I = I_0 e^{-\mu x}$ and its rearrangements, alongside the physical interpretation of each coefficient, gives you the tools to answer every standard exam question on this topic.

The full attenuation model required by AQA is I = I0 e^(−μx), where μ is the linear attenuation coefficient and x is thickness. The mass attenuation coefficient is defined by μm = μ/ρ, so it removes the simple effect of density and allows more direct comparison between materials.