Absorption of X-rays (A-level only): Exponential attenuation.

The idea

When a narrow beam of X-rays passes through matter, photons are removed by absorption and scattering, so the beam intensity falls. Because each thin layer removes a constant fraction of the photons still present, the decrease is exponential rather than linear. The transmitted intensity I after thickness x is I = I₀e^(−μx), where I₀ is the incident intensity and μ is the attenuation coefficient, measured in m⁻¹. A larger μ or a thicker absorber gives a smaller I. Taking natural logarithms gives ln I = ln I₀ − μx, so a graph of ln I against x is a straight line of gradient −μ. The half-value thickness x½, where I = I₀/2, is related to μ by x½ = ln 2 / μ. This explains why radiographers choose photon energies and shielding thicknesses carefully.

What earns the marks

• 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.

Common mistakes and how to put them right

• Treating attenuation as linear and subtracting a fixed intensity for each centimetre.

  Put it right: The correct model is a constant fractional decrease, giving an exponential curve.

• Using logarithms to base ten when the equation is written with e.

  Put it right: 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.

  Put it right: Convert mm or cm to m first.

• Assuming half-value thickness depends on I₀.

  Put it right: It is fixed by μ alone, so doubling the incident intensity does not change x½.

Exam tips

• 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.