calculate the energy of the gamma-ray photons emitted during the annihilation of an electron-positron pair

Production and Use of X‑rays (Cambridge A‑Level Physics 9702 – Section 24.2)

Learning Objectives

• Calculate the energy of the photons produced in electron‑positron annihilation (γ‑rays).
• Explain how X‑rays are generated by electron bombardment – Bremsstrahlung and characteristic radiation.
• Apply the photon‑energy relation \(E = hf = \dfrac{hc}{\lambda}\) to determine the minimum wavelength of X‑rays.
• Use the exponential attenuation law \(I = I_{0}e^{-\mu x}\) to predict X‑ray absorption in matter.
• Describe the basic principles of radiographic imaging and computed tomography (CT).
• Recall safety measures and shielding requirements for ionising radiation.

1. Photon Energy – a Quick Reminder

For any photon

\[

E = hf = \frac{hc}{\lambda}

\]

where

• \(h = 6.626\times10^{-34}\ \text{J·s}\) (Planck constant)
• \(c = 3.00\times10^{8}\ \text{m·s}^{-1}\) (speed of light)
• \(f\) = frequency, \(\lambda\) = wavelength

Energy is often expressed in electron‑volts (eV) using \(1\ \text{eV}=1.602\times10^{-19}\ \text{J}\).

2. Production of X‑rays by Electron Bombardment

2.1 Bremsstrahlung (braking radiation)

• Fast electrons are decelerated in the electric field of nuclei in a metal target.
• The loss of kinetic energy is emitted as a continuous spectrum of X‑ray photons.
• The most energetic photon corresponds to the complete conversion of the electron’s kinetic energy.

2.2 Characteristic X‑rays

• Occur when an incident electron ejects an inner‑shell electron from a target atom.
• An outer‑shell electron drops into the vacancy, releasing a photon whose energy equals the difference between the two atomic energy levels.
• These appear as sharp lines superimposed on the Bremsstrahlung background.

2.3 Influence of Target Material

2.4 Tube Filtration

• Thin sheets of low‑Z metal (commonly aluminium) are placed in the beam path.
• They preferentially absorb low‑energy photons, “hardening” the spectrum and reducing patient dose.
• Filtration is specified in mm Al; a typical clinical tube uses 2–3 mm Al.

2.5 Minimum Wavelength

When the entire kinetic energy \(eV\) of the electron is converted into a single photon:

\[

\lambda_{\min} = \frac{hc}{eV}

\]

where \(V\) is the accelerating potential (V) and \(e = 1.602\times10^{-19}\ \text{C}\).

Worked Example – Minimum wavelength for a 30 kV X‑ray tube

1. Insert the values:

   \[

   \lambda_{\min}= \frac{(6.626\times10^{-34}\ \text{J·s})(3.00\times10^{8}\ \text{m·s}^{-1})}

   {(1.602\times10^{-19}\ \text{C})(3.0\times10^{4}\ \text{V})}

   \]

2. Calculate:

   \[

   \lambda_{\min}= \frac{1.9878\times10^{-25}}{4.806\times10^{-15}}

   \approx 4.13\times10^{-11}\ \text{m}=0.0413\ \text{nm}

   \]

3. Interpretation: photons with wavelengths shorter than 0.041 nm cannot be produced by a 30 kV tube.

3. Electron–Positron Annihilation – Gamma‑ray Production

3.1 Why annihilation yields γ‑rays

• The rest‑mass energy of an electron (511 keV) is far larger than typical atomic transition energies.
• When an electron (\(e^{-}\)) meets its antiparticle, the positron (\(e^{+}\)), the total rest‑mass energy is converted into electromagnetic radiation in the γ‑ray region.

3.2 Energy released

\[

E{\text{total}} = 2\,m{e}c^{2}

\]

3.3 Energy of each photon

• Conservation of linear momentum requires two photons emitted in opposite directions.
• Each photon carries half the total energy:

  \[

  E{\gamma}=m{e}c^{2}=511\ \text{keV}

  \]

• If the electron and positron have additional kinetic energy, the photons share that extra energy, so each photon’s energy will be \(>511\ \text{keV}\).

Key Physical Constants

Worked Example – Energy in joules

1. Calculate:

   \[

   E_{\gamma}= (9.109\times10^{-31}\ \text{kg})(3.00\times10^{8}\ \text{m·s}^{-1})^{2}

   = 8.187\times10^{-14}\ \text{J}

   \]

2. Convert to electron‑volts:

   \[

   E_{\gamma}= \frac{8.187\times10^{-14}\ \text{J}}{1.602\times10^{-19}\ \text{J eV}^{-1}}

   \approx 5.11\times10^{5}\ \text{eV}=511\ \text{keV}

   \]

3. Result: each photon carries 511 keV, a typical γ‑ray energy.

4. Interaction of X‑rays with Matter – Attenuation

The intensity of a narrow‑beam X‑ray after passing through a material of thickness \(x\) is

\[

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

\]

• \(I_{0}\) – incident intensity
• \(I\) – transmitted intensity
• \(\mu\) – linear attenuation coefficient (depends on photon energy and material)

Worked Example – Bone vs. Muscle (30 keV photons)

1. Typical coefficients: \(\mu{\text{bone}} = 0.50\ \text{cm}^{-1}\), \(\mu{\text{muscle}} = 0.15\ \text{cm}^{-1}\).
2. For a 2 cm path:

   \[

   \frac{I{\text{bone}}}{I{0}} = e^{-0.50\times2}=e^{-1}=0.37

   \]

   \[

   \frac{I{\text{muscle}}}{I{0}} = e^{-0.15\times2}=e^{-0.30}=0.74

   \]

3. Conclusion: bone attenuates X‑rays roughly twice as much as muscle, producing the contrast seen on a radiograph.

5. Imaging Applications

5.1 Radiography (plain X‑ray imaging)

• Object placed between X‑ray source and detector (film or digital sensor).
• Differences in \(\mu\) give varying exposure on the detector → image contrast.
• Key factors for image quality:

  • Tube voltage (controls photon energy and \(\lambda_{\min}\)).
  • Exposure time (controls total number of photons).
  • Object‑detector distance (affects geometric sharpness).

5.2 Computed Tomography (CT)

• Rotating X‑ray tube and a ring of detectors acquire many projection images around the patient.
• Each projection records the line integral \(\displaystyle \int \mu\,\text{d}s\) through the body.
• Reconstruction algorithms (e.g., filtered back‑projection) combine the projections to produce a cross‑sectional image.
• CT provides quantitative information (Hounsfield units) because the reconstructed value is directly related to \(\mu\).

6. Safety and Shielding

• Both X‑rays and γ‑rays are ionising and can damage biological tissue.
• ALARA principle – keep exposure As Low As Reasonably Achievable by:

  • Minimising exposure time.
  • Maximising distance from the source.
  • Using appropriate shielding.

• Lead (Pb) is the most common shielding material. Approximate attenuation for 100 keV photons:

  • 0.5 mm Pb → ≈ 90 % reduction.
  • 2 mm Pb → ≈ 99 % reduction.

• Personal protective equipment (PPE) includes lead aprons, thyroid shields, and lead glasses.

7. Common Pitfalls

• Assuming a single photon is emitted in annihilation – momentum conservation forces two photons of equal energy.
• Confusing X‑ray and γ‑ray terminology: X‑rays arise from electron transitions or Bremsstrahlung; γ‑rays originate from nuclear or particle‑antiparticle processes.
• Neglecting unit conversion (J ↔ eV) when calculating photon energies.
• Overlooking the contribution of kinetic energy: if the electron and positron have extra kinetic energy, each photon’s energy will be \(>511\ \text{keV}\).
• Applying the simple exponential attenuation law to thick, scattered beams – it is valid only for narrow, unscattered beams.

8. Summary

• X‑ray production: high‑energy electrons decelerated in a target give Bremsstrahlung (continuous) and characteristic (line) radiation. The shortest wavelength is \(\lambda_{\min}=hc/eV\). Target material and tube filtration shape the spectrum.
• Electron–positron annihilation: total rest‑mass energy \(2m_{e}c^{2}\) is converted into two γ‑ray photons, each of 511 keV (or higher if kinetic energy is present).
• Attenuation: \(I = I_{0}e^{-\mu x}\); differences in \(\mu\) produce contrast in radiographs.
• Imaging: radiography uses a single projection; CT acquires many projections and reconstructs cross‑sections.
• Safety: ionising radiation demands ALARA, lead shielding, and PPE.