3.10.6.3
PET scanning and effective half-life
Medical Physics | AQA A-Level Physics
- PET (positron emission tomography) uses a positron-emitting tracerA radioactive substance (such as fluorine-18 attached to glucose, known as FDG) that emits positrons ($\beta^+$ particles) as it decays. Areas with high metabolic activity absorb more tracer., typically fluorine-18 (F-18) attached to a glucose molecule (FDG). F-18 has a half-life of about 110 minutes.
- The tracer accumulates in metabolically active areas (e.g. tumours, the brain). When F-18 decays, it emits a positronThe antiparticle of the electron. It has the same mass as an electron but positive charge. When a positron meets an electron, they annihilate. ($\beta^+$). The positron travels a short distance before meeting an electron.
- AnnihilationThe process in which a particle and its antiparticle meet and convert their combined mass into energy. A positron-electron annihilation produces two gamma photons, each with energy 511 keV, travelling in opposite directions (180 degrees apart). occurs: the positron and electron destroy each other, producing two gamma photons each of energy 511 keV, emitted in opposite directions (180 degrees apart).
- A ring of detectors around the patient records both photons simultaneously. The line of responseThe straight line connecting the two detectors that simultaneously detect the pair of 511 keV gamma photons from a single annihilation event. The annihilation occurred somewhere along this line. connecting the two detectors locates the annihilation event. Many such lines are used to build a 3D image.
- Crucially, the effective half-lifeThe time for the activity of a tracer in the body to fall to half, accounting for both radioactive decay (physical half-life $T_P$) and biological removal (biological half-life $T_B$). Always shorter than either $T_P$ or $T_B$ alone. $T_E$ accounts for both physical decay and biological removal: $$\frac{1}{T_E} = \frac{1}{T_P} + \frac{1}{T_B}$$ where $T_P$ is the physical half-life and $T_B$ is the biological half-life. The effective half-life is always the shortest of the three.
Worked Example
A tracer has a physical half-life of 6.0 hours and a biological half-life of 4.0 hours. Calculate the effective half-life.
Show Solution
1
Use the effective half-life equation
$$\frac{1}{T_E} = \frac{1}{T_P} + \frac{1}{T_B} = \frac{1}{6.0} + \frac{1}{4.0} = \frac{2 + 3}{12} = \frac{5}{12}$$
2
Invert to find $T_E$
$$T_E = \frac{12}{5} = 2.4 \text{ hours}$$
Answer
$T_E = 2.4$ hours, which is shorter than both the physical and biological half-lives.