2. Key Definitions
- Radioactive Tracer: A chemical compound where one or more atoms have been replaced by a radioactive isotope. It is introduced into the body (injection, ingestion, or inhalation) and is absorbed by the specific tissue or organ under investigation.
- Positron ($\beta^{+}$ particle): The antiparticle of the electron. It has the same rest mass ($m_e = 9.11 \times 10^{-31}$ kg) as an electron but an equal and opposite charge ($+1.60 \times 10^{-19}$ C).
- Annihilation: A process that occurs when a particle interacts with its corresponding antiparticle. The two particles disappear, and their total mass-energy is converted into the energy of two gamma-ray photons.
- Line of Response (LOR): The straight path connecting two opposite detectors in a PET ring. When two photons are detected simultaneously, the annihilation event is known to have occurred somewhere along this line.
3. Content
The Function of the Radioactive Tracer
A tracer is designed to target specific physiological processes. The most common tracer is Fluorodeoxyglucose (FDG), which uses the isotope Fluorine-18.
- Absorption: FDG is a glucose analogue. Tissues with high metabolic rates (such as cancer tumours, the brain, or the heart) absorb glucose rapidly. Consequently, the tracer concentrates in these "active" areas.
- Isotope Choice: The isotope must be a positron emitter.
- Half-life Considerations: The tracer must have a short half-life (e.g., F-18 has $t_{1/2} \approx 110$ minutes).
- It must be long enough to allow for synthesis, transport, and the duration of the scan.
- It must be short enough to ensure the patient's radiation exposure is minimised after the procedure.
Positron Emission ($\beta^{+}$ Decay)
Inside the nucleus of the tracer, a proton decays into a neutron, a positron, and an electron neutrino: $$\mathbf{p \to n + e^+ + \nu_e}$$ Once emitted, the positron travels a very short distance (typically $< 1$ mm) through the surrounding tissue. During this travel, it loses kinetic energy through collisions with atoms until it interacts with an electron.
The Physics of Annihilation
When a positron ($e^+$) meets an electron ($e^-$) in the tissue, they annihilate. This event is governed by two fundamental conservation laws:
- Conservation of Mass-Energy: The combined rest mass of the electron and positron is converted into electromagnetic energy.
- Total initial mass = $2m_e$.
- Total energy released $E = \Delta m c^2 = (2m_e)c^2$.
- Conservation of Momentum:
- Before annihilation, the positron has lost most of its kinetic energy, so the total linear momentum of the $e^+/e^-$ pair is approximately zero.
- To ensure the final momentum is also zero, the process must produce two photons of equal energy travelling in exactly opposite directions.
Calculating Gamma-Ray Photon Energy
The energy of the photons is derived from the mass-energy equivalence principle.
The Derivation:
- Total mass annihilated: $m = 2m_e$
- Total energy produced: $E_{total} = (2m_e)c^2$
- Since two identical photons are produced to conserve momentum, the energy of one photon ($E_\gamma$) is: $$\mathbf{E_\gamma = m_e c^2}$$
Standard Values for Calculation:
- Mass of electron ($m_e$): $9.11 \times 10^{-31}$ kg
- Speed of light ($c$): $3.00 \times 10^8$ m s⁻¹
- Elementary charge ($e$): $1.60 \times 10^{-19}$ C
Worked Example 1 — Photon Energy and Frequency
Question: Calculate the frequency of a gamma-ray photon produced during the annihilation of an electron and a positron. Step 1: Calculate the energy of one photon in Joules. $$E = m_e c^2$$ $$E = (9.11 \times 10^{-31} \text{ kg}) \times (3.00 \times 10^8 \text{ m s}^{-1})^2$$ $$E = 8.199 \times 10^{-14} \text{ J}$$ Step 2: Use the photon energy equation to find frequency. $$E = hf \implies f = \frac{E}{h}$$ $$f = \frac{8.199 \times 10^{-14} \text{ J}}{6.63 \times 10^{-34} \text{ J s}}$$ Step 3: Final Answer. $$f = 1.24 \times 10^{20} \text{ Hz}$$
Detection and Image Reconstruction
The PET scanner consists of a ring of gamma-ray detectors (scintillation crystals) surrounding the patient.
- Coincidence Detection: The scanner only records an event if two photons arrive at opposite detectors at almost the same time (within nanoseconds). This is called a "coincidence event."
- The Line of Response (LOR): Each coincidence event defines a straight line (the LOR) passing through the point of annihilation.
- Arrival Time Difference ($\Delta t$):
- If the annihilation occurs exactly in the center of the ring, the photons travel equal distances and arrive simultaneously ($\Delta t = 0$).
- If the event is closer to one detector, that photon arrives slightly earlier.
- The computer uses the difference in arrival times to calculate the exact position of the annihilation along the LOR.
- Image Formation: By processing millions of these LORs and the specific locations determined by $\Delta t$, the computer reconstructs a 3D map of the tracer concentration. High-concentration areas indicate high metabolic activity.
Worked Example 2 — Locating the Annihilation
Question: An annihilation event occurs along a Line of Response. One photon is detected $1.20 \times 10^{-10}$ s before the other. Calculate the distance of the annihilation event from the midpoint of the LOR. Step 1: Understand the relationship. Let $x$ be the distance from the midpoint. One photon travels $(d/2 - x)$ and the other travels $(d/2 + x)$. The path difference is $\Delta d = (d/2 + x) - (d/2 - x) = 2x$. The time difference is $\Delta t = \frac{\Delta d}{c} = \frac{2x}{c}$. Step 2: Rearrange for $x$. $$x = \frac{\Delta t \times c}{2}$$ Step 3: Substitute and solve. $$x = \frac{(1.20 \times 10^{-10} \text{ s}) \times (3.00 \times 10^8 \text{ m s}^{-1})}{2}$$ $$x = \frac{0.036 \text{ m}}{2} = 0.018 \text{ m}$$ Answer: The event occurred 1.80 cm from the midpoint.
4. Key Equations
| Equation | Description | Status |
|---|---|---|
| $E = mc^2$ | Mass-energy equivalence (Total energy) | Data Sheet |
| $E_\gamma = m_e c^2$ | Energy of one annihilation photon | Memorise |
| $p \to n + e^+ + \nu_e$ | Equation for $\beta^+$ decay in the tracer | Memorise |
| $E = hf = \frac{hc}{\lambda}$ | Photon energy related to frequency/wavelength | Data Sheet |
| $\Delta d = c \Delta t$ | Path difference from arrival time delay | Use Logic |
5. Common Mistakes to Avoid
- ❌ Wrong: Forgetting the neutrino ($\nu_e$) in the $\beta^+$ decay equation.
- ✓ Right: Always include the neutrino; it is required for the conservation of lepton number.
- ❌ Wrong: Doubling the mass in $E = mc^2$ but then forgetting to divide by 2 for a single photon.
- ✓ Right: Remember that the energy of one photon is exactly equal to the rest mass energy of one electron ($0.511$ MeV).
- ❌ Wrong: Stating that the tracer emits gamma rays.
- ✓ Right: The tracer emits positrons. The gamma rays are a result of the subsequent annihilation with electrons in the tissue.
- ❌ Wrong: Confusing the tracer with the gamma rays.
- ✓ Right: The tracer is the source of the positrons; the gamma rays are the signal detected by the scanner.
- ❌ Wrong: Using the mass of a proton instead of an electron for annihilation.
- ✓ Right: Annihilation in PET occurs between a positron and an electron. Use $m_e$.
6. Exam Tips
- Conservation Laws: If a question asks "Why are two photons emitted in opposite directions?", you must mention two things:
- The initial momentum of the system is zero.
- To conserve momentum, the two photons must have equal momentum in opposite directions.
- The "Why" of Tracers: When asked why a specific tracer is used, always link the biological molecule (e.g., glucose) to the tissue function (e.g., high metabolism in tumours).
- Significant Figures: Cambridge 9702 is strict. If the constants in the data sheet are given to 3 s.f. (like $c = 3.00 \times 10^8$), provide your final answer to 3 s.f.
- Unit Conversions: Practice converting Joules to eV and MeV.
- To go from J to eV: Divide by $1.60 \times 10^{-19}$.
- To go from eV to J: Multiply by $1.60 \times 10^{-19}$.
- The Role of the Computer: If asked how the image is formed, emphasize the processing of arrival time differences to locate the position of the tracer concentration.