PET scanning
| English | Chinese | Pinyin |
|---|---|---|
| tracer | 示踪剂 | shì zōng jì |
| metabolism | 代谢 | dài xiè |
| positron emission tomography | 正电子发射断层扫描 | zhèng diàn zi fā shè duàn céng sǎo miáo |
| positron | 正电子 | zhèng diàn zi |
| antiparticle | 反粒子 | fǎn lì zi |
| annihilate | 湮灭 | yān miè |
| energy | 能量 | néngliàng |
| momentum | 动量 | dòngliàng |
| electron | 电子 | diàn zi |
Two photons, exactly back to back
- A PET scanner does not watch the tracer. It watches for two gamma photons arriving at opposite sides of the ring at the same instant.
- That pair must have come from a single point somewhere on the straight line joining the two detectors. One coincidence, one line.
- Collect a few million of those lines from every angle and the intersection is a three-dimensional map of where the tracer went.
- The whole method rests on one conservation law forcing the photons to be back to back. This lesson is why.
The tracer
- A tracer 示踪剂 is a substance containing radioactive nuclei that is introduced into the body, usually bound to a molecule such as glucose, and is absorbed by the tissue being studied.
- A tumour has a high metabolism 代谢, so it takes up more glucose-tagged tracer than the tissue around it, and shows up bright.
- For positron emission tomography 正电子发射断层扫描 (PET) the tracer must be a beta-plus emitter, so that it gives out a positron 正电子. Fluorine-18 on a glucose analogue is the standard one.
- The half-life is a compromise: short enough that the activity is high during the scan and the patient's dose afterwards is small, but not so short that it decays before reaching the tissue. Fluorine-18 has $110$ minutes, oxygen-15 only $2$, so oxygen-15 is made on the spot.
A PET tracer is:
It is a β$^{+}$ emitter on a molecule (like glucose) that active tissue absorbs more of.
Why is a tracer's half-life chosen as a compromise? Select all that apply.
Fluorine-18 has 110 minutes; oxygen-15 has only 2 and must be made on the spot. Leaving activity in the patient afterwards is the thing to avoid, not to aim for.
Annihilation
- When a particle meets its antiparticle 反粒子 they annihilate 湮灭: their mass becomes electromagnetic energy. Mass-energy and momentum are both conserved.
- The four-mark explanation: the positron travels a few millimetres and meets an electron 电子 in the tissue; the pair annihilates, mass converting to energy; this appears as two gamma-ray photons; because momentum 动量 must be conserved and the pair had almost none, the photons travel in opposite directions with equal momenta.
- A single photon could not conserve momentum. That sentence is what the fourth mark is for, and it is the reason the whole technique works.
One event, one straight line
PET scan route
Follow positron emission to a ring of detected photons.
When the emitted positron meets an electron, they:
Matter and antimatter annihilate: their mass turns into two photons.
The two annihilation photons travel in opposite directions.
The pair was almost at rest, so to conserve momentum the photons go back-to-back.
Put the four-mark explanation of annihilation in PET in order.
The last step is the one that earns the fourth mark: a single photon could not conserve momentum, which is exactly why the technique works.
The energy of each photon
- Energy conservation shares the pair's rest energy 能量 between two identical photons:
- So each photon carries $m_e c^2 = 8.2\times10^{-14}\ \text{J} = 0.51\ \text{MeV} = 511\ \text{keV}$.
- Every PET photon has this same energy, which is what lets the scanner reject anything that is not $511\ \text{keV}$ as scatter or background.
Each photon from electron–positron annihilation has an energy of (in keV):
Each carries $m_e c^{2} \approx 0.511\ \text{MeV} = 511\ \text{keV}$.
Each annihilation photon carries an energy of ____ keV.
That is m_e c^2, the rest energy of ONE particle. The pair's total is 1.02 MeV. Writing 2 m_e c^2 per photon is the standard error.
Worked example: the annihilation photons
- Find the total energy released when a slow-moving positron and electron annihilate, and the wavelength of each photon.
- Total: $E = 2m_e c^2 = 2(9.11\times10^{-31})(3.00\times10^{8})^2 = 1.64\times10^{-13}\ \text{J} = 1.02\ \text{MeV}$.
- Shared equally: $8.2\times10^{-14}\ \text{J}$ each.
- $\lambda = \dfrac{hc}{E} = \dfrac{(6.63\times10^{-34})(3.00\times10^{8})}{8.2\times10^{-14}} = 2.4\times10^{-12}\ \text{m}$, a gamma ray.
- If the particles had been moving at $4.9\times10^{7}\ \text{m/s}$, each would add only $1.1\times10^{-15}\ \text{J}$, about $1\%$ of its rest energy. The rest energy dominates, which is why every annihilation gives very nearly $511\ \text{keV}$.
An annihilation photon has energy 8.2e-14 J. What is its wavelength, in units of 1e-12 m?
lambda = hc/E = 2.4e-12 m, firmly in the gamma range. Every PET photon has this same energy, which lets the scanner reject anything else as scatter.
Detecting a coincidence
- The two photons leave the body in opposite directions and strike a ring of detectors around the patient.
- Recording two simultaneous arrivals, a coincidence, fixes the line the annihilation lay on. A single detection on its own tells you nothing about position.
- Comparing the two arrival times more precisely narrows the position along that line. This is time-of-flight PET.
- The photons must escape the body to be detected at all, which is why gamma rays are used rather than the positron itself: a positron travels only a few millimetres before it annihilates.
Two photons detected at the same instant — a ____ — fix the line the annihilation happened on.
Many such coincidence lines, from many angles, let a computer build a 3-D map of the tracer.
A single detected gamma photon is enough to locate an annihilation event.
One detection gives no position at all. It takes two simultaneous arrivals on opposite sides, a coincidence, to fix the line the event lay on.
Building the image
- Many coincidences, from many angles, give many lines. The computer finds where the lines concentrate, and that is where the tracer is.
- The result is a three-dimensional map, not of anatomy but of tracer concentration, which means of metabolic activity.
- That is the difference from a CT scan worth stating: CT shows structure, PET shows function. Modern scanners often do both and overlay them.
Match each scan to what it shows.
A coincidence gives a LINE, not a point; the point comes from where many lines cross. Modern scanners often overlay the two images.
Marks that slip away
- The two photons go in opposite directions because momentum must be conserved. Say the conservation law, not just the direction.
- Each photon is $511\ \text{keV}$, which is $m_e c^2$, not $2m_e c^2$. The pair's total is $1.02\ \text{MeV}$.
- The tracer is a beta-plus emitter. A gamma emitter would not annihilate and could not be located by coincidence.
- The half-life is a compromise: short for a low dose, but long enough to reach the tissue.
- A coincidence gives a line, not a point. The point comes from where many lines cross.
You've got it
- a tracer is radioactive nuclei introduced into the body and taken up by the tissue studied; PET needs a beta-plus emitter such as fluorine-18
- a positron annihilates with an electron, and momentum conservation forces two gamma photons in opposite directions
- each photon carries $m_e c^2 = 0.51\ \text{MeV}$, a wavelength of about $2.4\times10^{-12}\ \text{m}$
- a coincidence at the detector ring fixes the line of the event, and many lines from many angles build a 3-D map of tracer concentration