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

Production and Use of X‑rays

1. How X‑rays are produced

X‑rays are generated when high‑energy electrons are decelerated rapidly upon striking a metal target. Two

processes contribute to the emitted radiation:

• Bremsstrahlung (braking radiation) – a continuous spectrum produced by the

  deceleration of electrons in the electric field of the nuclei.

• Characteristic radiation – discrete lines resulting from electron transitions

  between inner atomic shells of the target material.

2. Typical X‑ray tube arrangement

A high voltage source accelerates electrons from a heated cathode toward an anode (usually tungsten). The

resulting X‑ray beam exits through a thin window and may be filtered or collimated before use.

3. Uses of X‑rays

1. Medical imaging (radiography, CT scans)
2. Industrial radiography for non‑destructive testing
3. Crystallography and material analysis
4. Radiation therapy for cancer treatment

4. Attenuation of X‑rays in matter

When an X‑ray beam passes through a material of thickness \$x\$, its intensity \$I\$ is reduced according to the

exponential attenuation law:

\$\$

I = I_0 e^{-\mu x}

\$\$

• \$I_0\$ – initial intensity (before the material)
• \$I\$ – transmitted intensity (after the material)
• \$\mu\$ – linear attenuation coefficient (units: \$\text{m}^{-1}\$)
• \$x\$ – thickness of the material (m)

5. Factors affecting the attenuation coefficient \$\mu\$

• Photon energy – higher energy X‑rays are less attenuated.
• Atomic number \$Z\$ of the absorber – high‑\$Z\$ materials have larger \$\mu\$.
• Density of the material – denser materials provide greater attenuation.

6. Example calculation

Determine the thickness of aluminium required to reduce a 50 ke \cdot X‑ray beam to 10 % of its original intensity.

1. Given: \$I/I0 = 0.10\$, \$\mu{\text{Al}}\$ for 50 keV \$\approx 1.2 \times 10^{2}\ \text{m}^{-1}\$.
2. Re‑arrange the attenuation equation:

\$\$

\frac{I}{I0}=e^{-\mu x}\;\Longrightarrow\; x = -\frac{1}{\mu}\ln\!\left(\frac{I}{I0}\right)

\$\$

Substituting the numbers:

\$\$

x = -\frac{1}{1.2\times10^{2}}\ln(0.10)

= \frac{1}{1.2\times10^{2}}\times 2.3026

\approx 1.92\times10^{-2}\ \text{m}

= 1.9\ \text{cm}

\$\$

Thus, about 1.9 cm of aluminium will reduce the beam to 10 % of its original intensity.

7. Typical linear attenuation coefficients

8. Key points to remember

• The attenuation law \$I = I_0 e^{-\mu x}\$ is exponential; a plot of \$\ln I\$ versus \$x\$ is a straight line.
• Half‑value layer (H \cdot L) is the thickness that reduces intensity to \$I_0/2\$ and is given by \$\text{H \cdot L}= \ln 2 / \mu\$.
• In medical imaging, filters (e.g., aluminium) are used to remove low‑energy photons that would increase patient dose without improving image quality.
• Understanding attenuation is essential for designing shielding and for quantitative analysis of X‑ray spectra.