recall and use I = I0e–μx for the attenuation of X-rays in matter

Production, Uses and Attenuation of X‑rays – Cambridge 9702 (A‑Level Physics)

1. How X‑rays are produced

• Electron bombardment of a high‑Z metal target

  • Electrons are emitted from a heated cathode (filament) and accelerated through a potential difference \(V\) by a high‑voltage supply.
  • Each electron gains kinetic energy \(E_{\text{k}} = eV\) (where \(e = 1.60\times10^{-19}\,\text{C}\)).

• Conversion of kinetic energy to photons

  • If the whole kinetic energy of one electron is converted into a single photon, the photon has the highest possible energy (shortest wavelength).

    \[

    \lambda_{\min}= \frac{hc}{eV}\approx\frac{1240\;\text{nm·kV}}{V\;(\text{kV})}

    \]

  • Example: For an 80 kV tube

    \[

    \lambda_{\min}= \frac{1240}{80}=0.0155\;\text{nm}\;( \approx 80\;\text{keV})

    \]

• Radiation processes

  • Bremsstrahlung (braking radiation) – a continuous spectrum produced when fast electrons are decelerated in the electric field of the nuclei of the target.
  • Characteristic radiation – discrete lines that appear when an inner‑shell electron is ejected and an outer‑shell electron falls to fill the vacancy.

    \[

    h\nu = E{K}-E{L},\;E{K}-E{M},\dots

    \]

    The energies (and hence wavelengths) depend on the atomic number \(Z\) of the target; high‑\(Z\) materials (e.g. tungsten, \(Z=74\)) give higher‑energy characteristic lines.

2. Typical X‑ray tube – labelled schematic

3. Uses of X‑rays

• Medical imaging

  • Radiography – a single‑projection image; contrast arises from differences in linear attenuation coefficient \(\mu\) of tissues (e.g. bone vs. soft tissue).
  • Computed tomography (CT) – the X‑ray tube rotates around the patient, acquiring many projections. Reconstruction algorithms convert the set of projections into a series of cross‑sectional “slices”.
  • Contrast studies – high‑\(Z\) contrast agents (iodine, barium) increase the local \(\mu\) (photoelectric absorption \(\propto Z^{3}\)), making blood vessels or the gastrointestinal tract appear brighter.

• Radiation therapy – high‑energy X‑rays (MeV range) are used to deposit dose inside tumours while sparing superficial tissue; the depth‑dose curve is governed by attenuation.
• Industrial radiography – non‑destructive testing of welds, castings, and structural components.
• Crystallography & material analysis – X‑ray diffraction determines crystal structures; X‑ray fluorescence identifies elemental composition.

4. Attenuation of X‑rays in matter

When a mono‑energetic X‑ray beam of initial intensity \(I_{0}\) passes through a homogeneous absorber of thickness \(x\), the transmitted intensity \(I\) follows the exponential attenuation law

\[

I = I_{0}\,e^{-\mu x}

\]

• \(\mu\) – linear attenuation coefficient (units cm\(^{-1}\) in the syllabus; 1 cm\(^{-1}\)=100 m\(^{-1}\)).
• \(x\) – thickness of the absorber (cm).
• \(\mu\) is strongly energy‑dependent; tables in the syllabus give \(\mu\) for specific photon energies.

4.1. Half‑value layer (HVL)

The thickness that reduces the intensity to one‑half of its original value:

\[

\text{HVL}= \frac{\ln 2}{\mu}

\]

Example (lead, 30 keV): \(\mu_{\text{Pb}} = 15\;\text{cm}^{-1}\) → \(\text{HVL}= \frac{0.693}{15}=0.046\;\text{cm}\) (≈ 0.5 mm of lead). This is the figure used when designing shielding.

4.2. Mass attenuation coefficient

Because \(\mu\) scales with material density \(\rho\), the mass attenuation coefficient is defined as

\[

\frac{\mu}{\rho}\quad\text{(units cm}^{2}\,\text{g}^{-1}\text{)}

\]

Tables in the specification list \(\mu/\rho\). To obtain the linear coefficient for a particular material, multiply by its density:

\[

\mu = \left(\frac{\mu}{\rho}\right)\rho

\]

4.3. Factors influencing \(\mu\) (qualitative trends)

5. Worked examples

5.1. Aluminium – reduce intensity to 10 %

1. Given: \(\displaystyle\frac{I}{I{0}} = 0.10\), \(\mu{\text{Al}}(50\;\text{keV}) = 1.2\;\text{cm}^{-1}\).
2. Re‑arrange the attenuation law:

   \[

   x = -\frac{1}{\mu}\ln\!\left(\frac{I}{I_{0}}\right)

   \]

3. Substitute:

   \[

   x = -\frac{1}{1.2}\ln(0.10)=\frac{2.3026}{1.2}=1.92\;\text{cm}

   \]

≈ 1.9 cm of aluminium will attenuate a 50 keV beam to 10 % of its original intensity.

5.2. Lead – 50 % intensity reduction (HVL) at 30 keV

1. From the table (see §6) \(\mu_{\text{Pb}}(30\;\text{keV}) = 15\;\text{cm}^{-1}\).
2. Using the HVL formula:

   \[

   \text{HVL}= \frac{\ln 2}{\mu}= \frac{0.693}{15}=0.046\;\text{cm}

   \]

   (≈ 0.5 mm of lead).

3. Thus a sheet of lead only half a millimetre thick halves a 30 keV X‑ray beam.

6. Typical linear attenuation coefficients and densities

All linear coefficients are given in cm\(^{-1}\) to match the Cambridge syllabus. When a question supplies \(\mu\) in m\(^{-1}\) convert by dividing by 100.

7. Summary – key points to remember

• The exponential attenuation law \(I = I_{0}e^{-\mu x}\) governs how X‑ray intensity falls with material thickness.
• Linear attenuation coefficient \(\mu\) depends on photon energy, atomic number and density; tables give \(\mu/\rho\) for specific energies.
• Half‑value layer \(\text{HVL}= \ln2/\mu\) is a convenient way of expressing shielding thickness; e.g. 0.5 mm lead halves a 30 keV beam.
• Mass attenuation coefficient \(\mu/\rho\) allows direct comparison of different elements irrespective of state.
• In medical imaging, low‑energy photons are removed with aluminium filters to reduce patient dose while preserving image contrast.
• Design of protective shielding (lead aprons, walls) and dose calculations both rely on the attenuation law and the appropriate \(\mu\) values.
• Typical exam tasks:

  • Calculate \(\lambda_{\min}\) for a given tube voltage.
  • Use \(I = I_{0}e^{-\mu x}\) to find required thickness or transmitted intensity.
  • Convert between \(\mu\) and \(\mu/\rho\) using material density.
  • Interpret how changes in photon energy, \(Z\) or \(\rho\) affect image contrast and shielding.