Absorption of X-rays (A-level only) — AQA A-Level Physics

This topic covers how X-ray photons are removed from a beam as they pass through matter, focusing on the exponential attenuation law I = I0 e^(-mu x). You will learn the three main interaction mechanisms: the photoelectric effect, Compton scattering, and pair production, and how their relative importance depends on photon energy and the atomic number of the absorber. Understanding these processes is essential for explaining why different materials are used for shielding and why contrast agents such as barium and iodine are used in medical imaging.

The concept of absorption of X-rays is fundamental to medical physics, radiography, and radiation protection. It builds on your knowledge of photon energy (E = hf), the photoelectric effect, and atomic structure. In AQA A-Level Physics, this topic often appears in the Medical Physics option and in questions about attenuation calculations, half-value thickness, and the design of X-ray machines and shields.

• →The intensity of a collimated X-ray beam decreases exponentially with thickness: I = I0 e^(-mu x), where mu is the linear attenuation coefficient (units m^-1 or cm^-1).
• →The linear attenuation coefficient mu depends on the photon energy and the absorber's atomic number and density; it is the sum of contributions from photoelectric absorption, Compton scattering, and pair production.
• →The photoelectric effect dominates at low photon energies and for high-Z materials; it is proportional to Z^3 and E^-3, making lead an excellent shield for diagnostic X-rays.
• →Compton scattering dominates at higher energies (around 0.1-10 MeV) and depends on the electron density of the material, not strongly on Z.
• →Half-value thickness (HVT) is the thickness of material required to reduce the intensity to half its original value; it is related to mu by HVT = ln(2)/mu.

• States that intensity decreases exponentially with absorber thickness because each equal layer removes the same fraction of the photons remaining.
• Recalls and applies I = I₀e^(−μx), identifying I₀ as incident intensity, I as transmitted intensity, μ as attenuation coefficient and x as thickness.
• Uses the linear form ln I = ln I₀ − μx to find μ from the gradient of a graph of ln I against x.
• Calculates half-value thickness from x½ = ln 2 / μ and explains that it is independent of the initial intensity.
• Explains that μ depends on the material and on photon energy, so comparison of absorbers requires like-for-like conditions.
• Interprets a steeper exponential curve or larger gradient as a larger attenuation coefficient and therefore a more strongly attenuating material.

• 💡Write the exponential equation before substituting numbers, then rearrange symbolically so the examiner can follow your method.
• 💡When a graph of ln I against x is given, read the gradient from well-separated points on the line of best fit and quote μ with the unit m⁻¹.
• 💡Check that your answer for transmitted intensity is less than the incident intensity; a value greater than I₀ signals a sign or rearrangement error.
• 💡Link each calculation back to the physical situation, for example stating that a larger μ means more photons are removed per metre.
• 💡Always show your working clearly in attenuation calculations: write the formula, substitute numbers with units, and give your answer to an appropriate number of significant figures (usually 2 or 3).
• 💡When explaining why a material is a good shield, refer to both the photoelectric effect and Compton scattering, and state how each depends on atomic number and photon energy.
• 💡In questions about medical imaging, link the absorption of X-rays to the contrast between bone and soft tissue, and explain how contrast agents increase absorption by providing high-Z atoms.

• Treating attenuation as linear and subtracting a fixed intensity for each centimetre; the correct model is a constant fractional decrease, giving an exponential curve.
• Using logarithms to base ten when the equation is written with e; either use natural logarithms throughout or convert the coefficient consistently.
• Forgetting to convert thickness to metres before dividing by μ, which gives a value of μ in the wrong unit; convert mm or cm to m first.
• Assuming half-value thickness depends on I₀; it is fixed by μ alone, so doubling the incident intensity does not change x½.
• Students often think that X-rays are completely absorbed by a material once they exceed a certain thickness. In reality, absorption is exponential, so the intensity never reaches zero, though it becomes negligible after several half-value thicknesses.
• Many students believe that the linear attenuation coefficient mu is a constant for a given material. It actually depends strongly on the photon energy, so mu changes with the X-ray tube voltage (kVp).
• Some students confuse the linear attenuation coefficient (mu) with the mass attenuation coefficient (mu/rho). The mass attenuation coefficient is independent of density and is often tabulated; to find mu, you must multiply by the density.

1. 1Day 1: Review the exponential attenuation law and learn the definitions of linear attenuation coefficient, half-value thickness, and mass attenuation coefficient. Practice rearranging I = I0 e^(-mu x) to solve for mu, x, or I.
2. 2Day 2: Study the three interaction mechanisms (photoelectric, Compton, pair production) and create a table comparing their dependence on photon energy and atomic number. Use your textbook or reliable online resources.
3. 3Day 3: Work through calculations involving half-value thickness and attenuation, including converting between mu and mu/rho. Attempt past paper questions on these calculations.
4. 4Day 4: Practice explanation questions: why lead is used for shielding, why contrast agents are used, and how the absorption of X-rays produces an image. Write model answers and check against mark schemes.
5. 5Day 5: Complete a timed past paper section on this topic, then review your mistakes and revisit any weak areas. Use active recall to test yourself on key definitions and formulas.

• 📋Calculation of transmitted intensity or thickness using I = I0 e^(-mu x). Advice: always convert units to be consistent and check the exponent is dimensionless.
• 📋Explanation of why a particular material is suitable for shielding or contrast, linking to photoelectric effect and Compton scattering. Advice: mention Z^3 dependence and energy dependence.
• 📋Graph interpretation: given a graph of ln(I) against x, determine mu from the gradient. Advice: remember that ln(I) = ln(I0) - mu x, so gradient = -mu.
• 📋Comparison of the effectiveness of different materials or thicknesses for X-ray attenuation. Advice: use the concept of half-value thickness and consider the energy of the X-rays.