Nuclear Medicine and Radiation Therapy – Radiation and Medical Imaging (Senior 6 Physics) | Ganzaa

Radiation and Medical Imaging

Lessons

Senior 6 · Physics

Radiation and Medical Imaging

5 lessons

Lesson 5 of 5

Nuclear Medicine and Radiation Therapy

25 min

A cow that is milked every morning

Early each week, a busy nuclear-medicine department receives a heavy, lead-shielded container about the size of a small bucket. Staff call it the "cow". Inside, molybdenum-99 (half-life 65.9 h) is fixed to a small column of alumina, decaying into technetium-99m. Each morning a technologist runs salt water through the column and "milks" off the technetium, which has grown back overnight, then attaches it to chemicals that carry it to bone, heart muscle, the kidneys or the thyroid. After a week the molybdenum has decayed to about a sixth of its activity and the cow is replaced. Radiography, CT and ultrasound show the body's structure; nuclear medicine shows what its organs are doing. The same physics, turned up to much higher doses, is also used to destroy cancers.

Radioactive tracers

A radiopharmaceutical is a radionuclide attached to a molecule that the body handles in a known way: a phosphate compound settles in growing bone, a glucose analogue collects in very active cells, iodine is taken up by the thyroid. Injected or swallowed, it goes where the chemistry takes it, and a detector outside the body records where the radiation comes from. The amounts are tiny (Worked example 1), so they do not disturb the processes they reveal.

A good imaging tracer:

emits gamma rays only, or mainly: alpha and beta particles cannot get out of the body to the detector, and would only add dose;
has a gamma energy of about 100 keV to 200 keV: enough to escape the body, yet low enough to be stopped in the detector and absorbed by the lead collimator;
has a half-life comparable with the time of the test (hours): long enough to prepare and image, short enough to leave little activity behind;
is chemically suitable for labelling, and available on site.

Radionuclide	Half-life	Emission used	Typical use
technetium-99m	6.01 h	γ, 140 keV	bone, heart, kidney and thyroid scans
iodine-123	13.2 h	γ, 159 keV	thyroid imaging
fluorine-18	110 min	β⁺ (two 511 keV photons)	PET with FDG, a glucose analogue
iodine-131	8.02 days	β⁻ (and γ, 364 keV)	treating thyroid cancer and an overactive thyroid

Technetium-99m is an excited state ("m" for metastable) of technetium-99; it decays to the ground state by emitting a gamma ray, with no beta particle, which keeps the patient's dose low.

Definition

A radioactive tracer (radiopharmaceutical) is a compound labelled with a radionuclide, given to a patient in a very small amount, so that its path and where it collects in the body can be followed from the radiation it emits. Example: technetium-99m attached to a phosphate compound collects where bone is being repaired or invaded by cancer.

Worked example 1: preparing a dose

A heart scan needs 740 MBq of technetium-99m at 11:00. (a) What activity must be prepared at 08:00? (b) How many technetium-99m nuclei are in the dose at 11:00, and what is their mass? ( $T_{\frac{1}{2}} = 6.01\,\mathrm{h}$ , molar mass $99\,\mathrm{g\,mol^{-1}}$ , $N_A = 6.02 \times 10^{23}\,\mathrm{mol^{-1}}$ .)

Solution

(a) The dose decays for 3.0 h, so $A_0 = A \times 2^{\frac{t}{T}} = 740 \times 2^{\frac{3.0}{6.01}} = 740 \times 1.414 = 1.05 \times 10^{3}\,\mathrm{MBq}$ , about 1.05 GBq.

(b) $\lambda = \frac{\ln 2}{6.01 \times 3600} = 3.20 \times 10^{-5}\,\mathrm{s^{-1}}$ ; $N = \frac{A}{\lambda} = \frac{7.40 \times 10^{8}}{3.204 \times 10^{-5}} = 2.31 \times 10^{13}$ . Mass $= \frac{2.31 \times 10^{13}}{6.02 \times 10^{23}} \times 99 = 3.8 \times 10^{-9}\,\mathrm{g}$ , about 4 nanograms.

The gamma camera

Figure 5.1: A gamma camera (not to scale).

The collimator, a thick lead plate pierced by thousands of parallel holes, lets through only photons travelling straight towards the camera. Without it, photons from every part of the body would reach every part of the crystal and there would be no image. Longer, narrower holes give a sharper image but let fewer photons through, so the scan takes longer.
The scintillation crystal, sodium iodide with a trace of thallium, NaI(Tl), absorbs each photon and gives a flash of light whose brightness is proportional to the photon's energy.
An array of photomultiplier tubes (PMTs) turns the light into electrical pulses. In each tube the light releases electrons from a photocathode (the photoelectric effect), and a chain of electrodes called dynodes multiplies them: each electron striking a dynode knocks out several more.
The position circuit compares the pulses from neighbouring tubes to locate each flash, and the pulse-height analyser accepts only pulses in a narrow energy window around 140 keV. In Lesson 1 you found that a 140 keV photon scattered through 90° keeps only 110 keV: the window rejects such photons, which would appear to come from the wrong place.

Over several minutes the computer counts hundreds of thousands of photons into an image. If the camera rotates around the patient, the computer can reconstruct slices, as in CT: this is SPECT (single-photon emission computed tomography).

Worked example 2: inside the photomultiplier and the energy window

(a) A photomultiplier has 10 dynodes, and each releases 4 electrons for every electron that hits it. Find the number of electrons reaching the anode for each electron released by the photocathode. (b) The pulse-height analyser accepts energies within ±10% of 140 keV. Would a photon Compton-scattered through 30° be accepted? ( $m_ec^2 = 511\,\mathrm{keV}$ .)

Solution

(a) Gain $= 4^{10} = 1.05 \times 10^{6}$ : one electron becomes about a million, a pulse big enough to measure.

(b) Window: 126 keV to 154 keV. $E' = \frac{140}{1 + \frac{140}{511}(1 - \cos 30^\circ)} = \frac{140}{1 + 0.274 \times 0.134} = \frac{140}{1.0367} = 135\,\mathrm{keV}$ . It is accepted. Photons scattered through small angles cannot be rejected, and they blur the image slightly: no energy window is perfect.

Whole-body bone scans from a gamma camera, taken a few hours after injecting a technetium-99m phosphate compound, front (A–V) and back (P–V) views. In 1A, dark spots in the left ribs show cancer that had spread to the bone; in 1B, after treatment, they have gone. Image: Mado and colleagues, CC BY-SA 2.0, via Wikimedia Commons

Positron emission tomography (PET)

PET uses tracers labelled with positron emitters, above all fluorine-18 in FDG (fluorodeoxyglucose), which collects in cells that use a lot of glucose, such as many cancers, the working brain and heart muscle. Each positron travels about a millimetre in tissue and then annihilates with an electron, producing two 511 keV photons, as you saw in Unit 6. Because the pair had almost no momentum, the two photons fly off in opposite directions.

Figure 5.2: Coincidence detection in a PET scanner.

A ring of scintillation detectors surrounds the patient. When two detectors on opposite sides fire within a few nanoseconds of each other, a coincidence circuit records an annihilation somewhere on the straight line joining them, the line of response. No lead collimator is needed, because the pair of photons itself defines the direction, so PET is much more sensitive than a gamma camera. Millions of lines are reconstructed into slices. Modern scanners also measure the tiny difference $\Delta t$ in the arrival times (time of flight), which places the event at a distance $\frac{c\Delta t}{2}$ from the midpoint of the line. Most PET scanners are combined with a CT scanner (PET/CT), so that the activity map can be laid over the anatomy; from Lesson 2, the combined effective dose is high, about 23 mSv for a whole-body scan.

Worked example 3: locating an annihilation by time of flight

In a PET scanner, the two 511 keV photons from one annihilation reach detectors A and B, which face each other across the ring, and the photon at B arrives 0.40 ns after the photon at A. (a) Where on the line did the annihilation happen? (b) Why must the coincidence window be only a few nanoseconds?

Solution

(a) If the event is a distance $x$ from the midpoint, towards A, the photon to B travels $2 x$ further. So $2x = c\Delta t$ and $x = \frac{(3.00 \times 10^{8})(0.40 \times 10^{-9})}{2} = 0.060\,\mathrm{m}$ : 6.0 cm from the midpoint, on A's side.

(b) Light crosses a ring about 80 cm wide in under 3 ns, so two photons from one annihilation always arrive within a few nanoseconds. A longer window would accept more "random" pairs: two photons from different annihilations that happen to arrive close together, which put events on false lines.

Radiation therapy

Imaging tries to give the smallest useful dose. Radiotherapy does the opposite on purpose: it gives a tumour a dose high enough to kill its cells, typically a few tens of grays in total, while keeping the dose to the healthy tissue around it as low as possible. Every treatment plan is built around this balance.

External beam therapy (teletherapy). A medical linear accelerator (linac) accelerates electrons to several MeV with microwaves and fires them at a tungsten target, producing a megavoltage X-ray beam (about 6 MV to 18 MV) by the braking radiation of Unit 7; for tumours near the surface it can use the electron beam directly. Older units use a cobalt-60 source (gamma rays of 1.17 MeV and 1.33 MeV, half-life 5.27 years). The machine rotates around the patient and aims beams from several directions, all crossing at the tumour, and a multileaf collimator of thin tungsten leaves shapes each beam to the tumour's outline (Figure 5.3(a)).
Brachytherapy. Small sealed sources, such as iridium-192 (half-life 73.8 days) or iodine-125 seeds, are placed inside or right next to the tumour, for example in the cervix or prostate. By the inverse-square law the dose falls steeply with distance, so the tumour receives a high dose and organs a few centimetres away much less.
Radionuclide therapy. An unsealed source is chosen to collect in the target. Iodine-131 is swallowed, concentrated by thyroid tissue, and its beta particles, with a range in tissue of at most about 2 mm, destroy that tissue from within.
Proton therapy. A proton loses energy slowly at first and then very fast just before it stops, depositing most of its dose in a sharp Bragg peak, with almost nothing beyond it (Figure 5.3(b)). It is valuable for tumours next to the spinal cord or in children, but the machines are very expensive.

Figure 5.3: (a) Crossfire from several beams. (b) Schematic depth-dose curves for megavoltage X-rays and for protons.

Megavoltage X-rays have two advantages. Compton scattering dominates, so bone does not absorb much more than soft tissue (Lesson 1, Worked example 4). And the fast secondary electrons are thrown forward, so the dose builds up over the first centimetre or two (about 1.5 cm for 6 MV) instead of being largest at the skin: the skin is spared. The total dose is usually given in daily fractions, for example 2 Gy a day, five days a week, because healthy cells repair radiation damage between fractions better than most tumour cells do.

Worked example 4: why several beams?

A tumour lies 10 cm deep. A cobalt-60 beam in tissue has $\mu = 0.063\,\mathrm{cm^{-1}}$ . Ignoring the inverse-square law, scatter and build-up, find the dose where each beam enters the body if the tumour is to receive 2.0 Gy (a) from one beam, (b) from four equal beams from different directions, all 10 cm deep.

Solution

Fraction reaching the tumour: $e^{-0.063 \times 10} = e^{-0.63} = 0.53$ .

(a) Entrance dose $= \frac{2.0}{0.5315} = 3.8\,\mathrm{Gy}$ , nearly twice the tumour dose.

(b) Each beam must deliver 0.50 Gy at the tumour: entrance dose $= \frac{0.50}{0.5315} = 0.94\,\mathrm{Gy}$ .

Conclusion: with one beam the healthy tissue in front of the tumour receives more than the tumour; with four, the tumour receives 2.0 Gy and the tissue along each path at most about 0.94 Gy. Modern machines go further and spread the beam over a continuous arc.

Choosing the right examination

Method	What it shows	Ionising?	Main limitation
radiograph (Unit 7)	bone, lungs, foreign bodies	yes (low dose)	overlapping shadows; poor soft-tissue contrast
CT (Lesson 4)	cross-sections of all tissues, fast	yes (higher dose)	dose; less soft-tissue contrast than MRI
ultrasound (Lesson 3)	soft tissue, fluid, blood flow, the fetus, in real time	no	blocked by bone and gas; depends on the operator
MRI (Lesson 4)	soft tissue, brain, spinal cord, joints	no	slow, costly, noisy; unsafe with some metal implants
endoscopy (Lesson 4)	the inside surface of hollow organs, with biopsy	no	invasive; sees only surfaces
gamma camera, SPECT, PET	function: blood flow, metabolism, uptake	yes (from inside)	coarse detail; patient is radioactive for a while

The best test answers the doctor's question with the least risk: justification and ALARA, from Lesson 2, applied to the choice of machine itself.

Warning

Common mistakes. (1) Saying the gamma camera sends radiation into the patient: the radiation comes out of the patient, from the tracer. (2) Choosing an alpha or beta emitter for imaging: those particles cannot reach a detector outside the body. (3) Saying the collimator focuses gamma rays like a lens: it cannot bend them; it only absorbs those not travelling along its holes. (4) Thinking a radiotherapy patient becomes radioactive: external beams leave no activity behind; only patients given a source (brachytherapy or iodine-131) carry activity for a while.

Tip

For any tracer question check "G-E-H-C": Gamma emitter? Energy about 100 keV to 200 keV? Half-life of hours? Chemistry that takes it to the organ? For therapy, reverse the first: choose a beta (or alpha) emitter, because you want the energy to stay inside.

How does a PET scan work?, by Imperial College London. Watch for how the tracer is made and injected, and how pairs of photons reaching opposite detectors build the image.

✦

Practice

State three properties of technetium-99m that make it suitable for imaging.
A hospital receives a molybdenum-99 generator with an activity of 20 GBq. What is its activity one week (168 h) later? ( $T_{\frac{1}{2}} = 65.9\,\mathrm{h}$ )
Explain the purpose of (a) the collimator, (b) the NaI(Tl) crystal and (c) the pulse-height analyser in a gamma camera.
A 140 keV photon is Compton-scattered through 60°. Find its new energy and say whether a window of 126 keV to 154 keV accepts it.
In a PET scanner the two photons from an annihilation arrive at opposite detectors 0.25 ns apart. How far from the midpoint of the line was the annihilation?
A patient receives iodine-131 (half-life 8.02 days) for an overactive thyroid. Ignoring excretion, what fraction of the activity remains after 24 days? Why may the patient be asked to avoid close contact with small children for a short time?
A brachytherapy source gives a dose rate of 1.2 Gy per hour at 1.0 cm. Estimate the dose rate at 3.0 cm, stating the assumption.
Evaluate this claim: "PET is better than CT, so all cancer patients should have PET instead of CT."

Show answersHide answers

It emits gamma rays only (no beta), at 140 keV (escapes the body and is stopped well by the detector); its half-life of 6.01 h suits a test of a few hours; it can be attached to many compounds; it is available daily from a generator.
$A = 20 \times 2^{-\frac{168}{65.9}} = 20 \times 0.171 = 3.4\,\mathrm{GBq}$ .
(a) Absorbs photons not travelling straight along its holes, so each point of the crystal sees only the part of the body directly beneath it. (b) Converts each gamma photon into a flash of light proportional to its energy. (c) Accepts only pulses near 140 keV, rejecting scattered photons that would put counts in the wrong place.
$E' = \frac{140}{1 + \frac{140}{511}(1 - 0.5)} = \frac{140}{1.137} = 123\,\mathrm{keV}$ : below 126 keV, so it is rejected.
$x = \frac{c\Delta t}{2} = \frac{(3.00 \times 10^{8})(0.25 \times 10^{-9})}{2} = 0.0375\,\mathrm{m} \approx 3.8\,\mathrm{cm}$ , towards the detector that fired first.
24 days is about 3 half-lives: $2^{-\frac{24}{8.02}} = 0.13$ . The patient's thyroid still contains iodine-131, whose gamma rays (364 keV) leave the body; following ALARA (time and distance), close contact with children, who are more sensitive, is limited for a few days.
Inverse-square law: $1.2 \times \left(\frac{1.0}{3.0}\right)^2 = 0.13\,\mathrm{Gy}$ per hour, assuming a point source and neglecting absorption in tissue (which makes the real value a little lower).
Too strong. PET shows metabolic activity and finds active cancer and its spread, but its images are coarse, its dose is higher, it is expensive and scarce, and some cancers take up little FDG. CT shows anatomy in fine detail, quickly and cheaply. They answer different questions, which is why they are often combined (PET/CT); each test must be justified for each patient.

Key points

Tracers: gamma emitters of about 100 keV to 200 keV with half-lives of hours, carried by chemistry to an organ; technetium-99m (140 keV, 6.01 h) from a molybdenum-99 generator is the workhorse.
Gamma camera: lead collimator, NaI(Tl) crystal, photomultipliers (gain $\delta^n$ ), position and energy circuits; SPECT gives slices.
PET: β⁺ annihilation gives two 511 keV photons in opposite directions; coincidence detection and time of flight ( $x = \frac{c\Delta t}{2}$ ) locate the event.
Radiotherapy: maximise tumour dose, minimise healthy-tissue dose: MV beams from many directions, skin sparing, fractionation, brachytherapy, iodine-131, protons (Bragg peak).

1A cow that is milked every morning

2Radioactive tracers

3The gamma camera

4Positron emission tomography (PET)

5Radiation therapy

6Choosing the right examination

A cow that is milked every morning

Radioactive tracers

The gamma camera

Positron emission tomography (PET)

Radiation therapy

Choosing the right examination