Interaction of Radiation with Matter

3.1 Charged Particle Interactions (Alpha, Beta)

As charged particles pass through matter they interact through the Coulomb force with the orbital electrons and nuclei of atoms in the medium. As a result of these interactions the particle loses energy and eventually stops. The mechanisms by which charged particles interact with matter depend on the type of particle (heavy/light), its energy and the properties of the medium.

3.1.1 Heavy Charged Particles (Alpha Particles, Protons)2 questions

Main interaction mechanisms:

1. Ionisation (the dominant mechanism):

2. Excitation:

Behaviour of alpha particles in matter:

Figure 3.1: Paths of alpha and beta particles in matter and the Bragg curve of the alpha particle (schematic).

3.1.2 Light Charged Particles (Beta Particles – Electrons and Positrons)6 questions

Beta particles (electrons and positrons) are much lighter than alpha particles (mass ~1/7300 of an alpha) and carry a charge of only -1e or +1e.

Main interaction mechanisms:

1. Ionisation and Excitation (Collisional/Electronic Interactions):

2. Bremsstrahlung (Braking Radiation):

3. Elastic Scattering:

4. Annihilation (Positrons Only):

Behaviour of beta particles in matter:

Figure 3.2: Choice of shielding for beta emitters: bremsstrahlung production with lead alone, Perspex and two-layer shielding (schematic).

3.1.3 Basic Concepts in Charged Particle Interactions4 questions

Linear Energy Transfer (LET):

Specific Ionisation:

Stopping Power:

W Value:

3.1.4 The X-ray Tube: Bremsstrahlung and Characteristic X-rays5 questions

The X-rays used in CT and planar radiography also arise from the interaction of electrons with matter. In the X-ray tube, electrons released from the cathode are accelerated by the tube voltage (kV) and made to strike an anode, usually made of tungsten.

Bremsstrahlung (continuous spectrum):

Characteristic X-rays (sharp peaks in the spectrum):

Effect of tube parameters on the spectrum:

Figure 3.3: X-ray spectrum at 80 and 120 kV: continuous bremsstrahlung component and the characteristic peaks of tungsten (schematic).

3.2 Photon Interactions (Photoelectric, Compton, Pair Production)1 questions

Gamma and X-ray photons are uncharged and ionise matter indirectly. Unlike charged particles they do not lose energy continuously along their path; they transfer all or part of their energy in a single interaction. Imaging physics and radiation protection in nuclear medicine are built on these interactions. What distinguishes an X-ray from a gamma ray is not its energy but its origin: gamma rays come from the nucleus, X-rays from electron processes outside the nucleus (characteristic X-rays, bremsstrahlung). Gamma rays usually have higher energies, but an X-ray and a gamma photon of equal energy interact with matter in exactly the same way.

3.2.1 Coherent (Rayleigh) Scattering1 questions

3.2.2 Photoelectric Effect4 questions

The photoelectric effect is the complete absorption of a photon by an inner-shell electron.

Mechanism:

1. The incoming photon interacts with an inner-shell electron (usually K shell).
2. All the photon's energy is absorbed and the photon disappears.
3. The electron is ejected from the atom; it is called a photoelectron.
4. Kinetic energy of the photoelectron: E_k = E_photon − E_binding
5. An electron from an outer shell fills the inner-shell vacancy, emitting a characteristic X-ray or an Auger electron.

Factors determining the probability of the photoelectric effect:

Absorption Edges:

Importance in nuclear medicine:

3.2.3 Compton Scattering7 questions

Compton scattering is the interaction of a photon with a free or loosely bound outer-shell electron.

Mechanism:

1. The incoming photon interacts with an outer-shell electron 2. The photon transfers part of its energy to the electron 3. The electron is ejected from the atom → Compton electron (recoil electron) 4. The photon is scattered with lower energy → scattered photon

5. Energy conservation: Ephoton(incident) = Ephoton(scattered) + Ek(electron) + Ebinding

Compton equation:

E′ = E / [1 + (E/511) × (1 − cos θ)]

Energy distribution by scattering angle:

E'(180°) = E / [1 + 2E/511]

Figure 3.4: Variation of scattered photon energy with scattering angle in Compton scattering (140 and 511 keV).

Factors determining the probability of Compton scattering:

Importance in nuclear medicine:

3.2.4 Pair Production2 questions

Pair production is the conversion of a high-energy photon into an electron–positron pair in the Coulomb field of a nucleus.

Mechanism:

1. The photon disappears as it passes close to the nucleus 2. An electron (e⁻) and a positron (e⁺) are created 3. The positron later annihilates with an electron, producing 2 × 511 keV photons

Threshold energy:

Factors determining the probability of pair production:

Importance in nuclear medicine:

3.2.5 Photodisintegration

Figure 3.5: Photon–matter interactions: photoelectric effect, Compton scattering, pair production and coherent scattering (schematic).

3.2.6 Dominance of Photon Interactions by Energy and Z2 questions

Which type of photon interaction dominates depends on the photon energy and the atomic number (Z) of the medium:

Transition energies for soft tissue (Z ≈ 7.4):

Transition energies for lead (Z = 82):

Conclusion: In the nuclear medicine energy range (70-511 keV) Compton scattering dominates in soft tissue, whereas the photoelectric effect dominates in lead shielding.

Figure 3.6: Regions of dominant interaction by photon energy and atomic number of the medium; transition energies for soft tissue and lead.

3.2.7 Comparison Table of Photon Interactions

3.2.8 Ionising Radiation Threshold and Medical Imaging Windows3 questions

Distinguishing ionising from non-ionising radiation:

The three radiation windows used in medical imaging:

There are three wavelength ranges in which tissue does not absorb radiation completely, thus allowing imaging:

3.3 Attenuation Coefficient and HVL/TVL

3.3.1 Photon Attenuation

As a photon beam passes through matter, photons are removed from the beam by interactions (absorbed or scattered). This process is called "attenuation". The narrow-beam attenuation equation:

I(x) = I₀ × e^−µx

Total linear attenuation coefficient:

µ_total = µ_{photoelectric} + µ_Compton + µ_{pair production} + µ_coherent

3.3.2 Linear Attenuation Coefficient (µ)

3.3.3 Mass Attenuation Coefficient (µ/ρ)

3.3.4 Half-Value Layer (HVL)4 questions

The HVL is the thickness of material that reduces the intensity of a photon beam by half:

x = HVL when I = I₀/2

Relationship between HVL and µ:

HVL = 0.693 / µ = ln 2 / µ

Attenuation by several HVLs:

Important HVL values in nuclear medicine (lead):

3.3.5 Tenth-Value Layer (TVL)1 questions

The TVL is the thickness of material that reduces the intensity of a photon beam to one tenth:

x = TVL when I = I₀/10

Relationship between TVL and µ:

TVL = 2.303 / µ = ln 10 / µ

Relationship between HVL and TVL:

TVL = 3.32 × HVL

Use of the TVL:

Figure 3.7: Exponential attenuation in a narrow beam (HVL and TVL) and HVL values in lead for important radionuclides.

3.3.6 Broad-Beam Geometry and the Build-up Factor1 questions

In narrow-beam geometry scattered photons do not reach the detector. In real clinical conditions (broad beam), however, scattered photons also contribute to the detector:

I(x) = B × I₀ × e^−µx

Figure 3.8: Narrow- and broad-beam geometry: contribution of scattered photons in a broad beam and the build-up factor (schematic).

3.4 Inverse Square Law

3.4.1 Basic Principle

The intensity of radiation from a point source decreases in inverse proportion to the square of the distance from the source:

Formula:

I₁ / I₂ = d₂² / d₁² or I₁ × d₁² = I₂ × d₂² = constant

Conditions for validity:

1. Point source (or a distance large enough relative to the source) 2. Radiation must be emitted isotropically (equally in all directions) 3. Absorption and scattering in the medium must be negligible (vacuum or short distances in air)

3.4.2 Practical Applications4 questions

Example 1: If the dose rate at 1 metre from a Tc-99m source is 4 mR/h, the dose rate at 2 metres is:

I₂ = I₁ × (d₁/d₂)² = 4 × (1/2)² = 4 × 1/4 = 1 mR/h

Example 2: If the dose rate at 0.5 metre is 20 mR/h, at 2 metres it is:

I₂ = 20 × (0.5/2)² = 20 × 1/16 = 1.25 mR/h

Clinical importance:

3.4.3 Limitations of the Inverse Square Law

Figure 3.9: Inverse square law: increase of area with distance for a point source and fall in dose rate according to source geometry.

3.5 Principles of Radiation Protection (Time, Distance, Shielding)

The basic aim of radiation protection is to keep radiation exposure as low as possible, in line with the ALARA (As Low As Reasonably Achievable) principle. There are three basic methods of protection: time, distance and shielding.

3.5.1 Time

Principle: the total dose is reduced by reducing the time spent near the radiation source.

Formula:

Total dose = dose rate × time

D = Ḋ × t

Practical applications:

3.5.2 Distance

Principle: the dose is reduced by moving away from the source (inverse square law).

Practical applications:

Distance is the most effective protection method because it costs nothing and, owing to the inverse square law, even small increases in distance give large dose reductions.

3.5.3 Shielding

Principle: reducing radiation intensity by placing a suitable material between the source and the person.

Choice of shielding material:

For gamma and X-rays:

For beta particles:

For alpha particles:

For neutrons:

Figure 3.10: Choice of shielding material by type of radiation: alpha, beta, gamma/X-ray and neutron (schematic).

3.5.4 The ALARA Principle2 questions

ALARA (As Low As Reasonably Achievable): Radiation exposure should be kept as low as reasonably achievable, taking social and economic factors into account.

Components of ALARA:

1. Time: Minimise exposure time 2. Distance: Stay as far from the source as possible 3. Shielding: Use suitable shielding material 4. Activity: Use the lowest activity possible without compromising diagnostic quality 5. Training: Train staff in radiation safety

ALARA in nuclear medicine practice:

3.5.5 Specific Gamma Ray Constant (Γ)2 questions

The specific gamma ray constant is used to calculate the dose rate produced by a radionuclide at a given distance. Definition: the dose rate at 1 cm from a point source with an activity of 1 mCi (37 MBq).

Formula:

Ḋ = Γ × A / d²

Γ values of important radionuclides (R·cm²/mCi·h):

3.5.6 Shielding Calculations1 questions

Calculating the required lead thickness:

Determine the target attenuation → calculate the number of HVLs required → find the lead thickness.

Example: lead thickness required to reduce the dose from an I-131 source to 12.5% (87.5% reduction):

Example: reducing the dose to 6.25% for Tc-99m:

Key Points for the Board