Attenuation of X-rays and half-value thickness

Overview

Half-value thickness of a material for X-rays and plotted to turn attenuation data into a straight line.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   State the attenuation equation for a parallel beam of X-rays passing through matter.

   Model answer

   The transmitted intensity is I = I₀e^(−μx), where I₀ is the incident intensity in W m⁻², μ is the linear attenuation coefficient in m⁻¹ and x is the thickness in m.

2. 2

   What does the linear attenuation coefficient of a material represent?

   Model answer

   It is the fraction of the beam intensity removed per unit thickness of the material. Its unit is m⁻¹, or mm⁻¹ where thicknesses are quoted in millimetres.

3. 3

   Define the half-value thickness of a material for X-rays.

   Model answer

   The half-value thickness is the thickness that reduces the transmitted intensity of a beam to one half of its incident value. It is measured in metres or millimetres.

4. 4

   Derive the relationship between the half-value thickness and the linear attenuation coefficient.

   Model answer

   Setting I = I₀/2 in I = I₀e^(−μx) gives e^(−μx½) = 0.5, so μx½ = ln 2. The half-value thickness is therefore x½ = ln 2/μ = 0.693/μ.

5. 5

   A tissue has a linear attenuation coefficient of 0.25 mm⁻¹ at a given photon energy. Calculate its half-value thickness.

   Model answer

   x½ = 0.693/0.25 = 2.8 mm. Working back, e^(−0.25 × 2.77) = e^(−0.693) = 0.500, exactly one half of the incident intensity.

6. 6

   A beam passes through 3.0 mm of a material with μ = 0.20 mm⁻¹. Calculate the fraction of the intensity transmitted.

   Model answer

   I/I₀ = e^(−0.20 × 3.0) = e^(−0.60) = 0.55, so 55% is transmitted. Working back, ln 0.549 = −0.600, which divided by 0.20 mm⁻¹ gives 3.0 mm.

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