Questions on radionuclide imaging and the gamma camera

Question 1

State what is meant by a medical tracer.

Accepted Answer

A radioactive nuclide chemically attached to a biologically active molecule, such as a glucose analogue, which the body takes to wherever that molecule is used (1). The radiation it emits from there is detected to map the tracer's location (1).

Question 2

Explain why a tracer used with a gamma camera should be a gamma emitter, and why its half-life must be chosen with care.

Accepted Answer

Gamma photons are penetrating enough to escape the body and be detected outside (1); alpha particles are absorbed within a fraction of a millimetre of tissue and beta minus particles within a few millimetres, so both deliver dose but no image (1). (A PET tracer is a positron emitter, but there too it is photons that leave the patient: the pair of 511 keV gammas made when the positron annihilates.) The half-life must be long enough for the tracer to be prepared, administered and imaged, but short enough that the patient is not left radioactive for weeks (1).

Question 3

An X-ray image and a tracer image answer different questions about a patient. State the difference.

Accepted Answer

X-ray and CT images show structure: what the anatomy looks like (1). A tracer image shows function: where the body is actually using the labelled molecule, so which tissue is active, inflamed or failing (1).

Question 4

Technetium-99m has a half-life of 6.0 hours, so a dose made at a distant factory would decay before reaching the hospital. Explain how hospitals nevertheless have technetium-99m available every morning.

Accepted Answer

The hospital keeps a generator containing the longer-lived parent nuclide, molybdenum-99 (1). Technetium-99m forms continuously as the molybdenum decays, and is drawn off each day as it is needed (1).

Question 5

In PET scanning for tumours, the positron emitter fluorine-18 is attached to a glucose analogue. Suggest why a glucose analogue is a good choice of carrier molecule.

Accepted Answer

Many tumours are metabolically very active and consume glucose faster than the tissue around them (1), so the tracer accumulates in the tumour and the annihilation photons come mostly from where the disease is (1).

Question 6

Describe the roles of the collimator and the scintillator crystal in a gamma camera.

Accepted Answer

The lead collimator admits only gamma photons travelling nearly parallel to its channels, so each detected photon's origin lies along a known line (1); oblique photons are absorbed by the slats (1). The scintillator converts each admitted gamma into a flash of light whose position is recorded, building a map of the tracer's location one photon at a time (1).

Question 7

Technetium-99m has a half-life of 6.0 hours. Calculate the fraction of a dose's activity remaining 24 hours after preparation.

Accepted Answer

24 hours is four half-lives (1) A/A₀ = (1/2)⁴ = 0.0625 (1) ≈ 1/16, about 6% (1)

Question 8

Describe what happens when a positron emitted by a PET tracer meets an electron, and state the energy of each photon produced.

Accepted Answer

The positron and electron annihilate, so both particles vanish (1) and their rest energy becomes two gamma photons travelling in opposite directions (1). Each photon carries 511 keV, the rest energy of one electron through E = mc² (1).

Question 9

Show that the rest energy of an electron is about 8 × 10⁻¹⁴ J, and convert it to keV.

Accepted Answer

E = mc² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² (1) E = 8.20 × 10⁻¹⁴ J (1) Dividing by 1.60 × 10⁻¹⁹ J eV⁻¹ gives 5.12 × 10⁵ eV ≈ 512 keV (511 keV with more precise constants) (1)

Question 10

Explain why the two annihilation photons travel in opposite directions.

Accepted Answer

The electron-positron pair is very nearly at rest when it annihilates, so its total momentum is close to zero (1). Momentum is conserved, so the two photons must carry equal and opposite momenta: they leave back to back (1).

Question 11

A tracer has a physical half-life of 6.0 hours and is also cleared from the body biologically with a half-life of 12 hours. Calculate the effective half-life of the dose, and explain why it is shorter than either value alone.

Accepted Answer

1/T_E = 1/T_B + 1/T_P = 1/12 + 1/6.0 (1), so T_E = 4.0 hours (1). Radioactive decay and biological clearance both remove activity from the patient at the same time, so together they act faster than either would alone (1).

Question 12

Iodine-131 is given to a patient to destroy overactive thyroid tissue. Explain how this treatment uses the same principle as a medical tracer, and how its aim differs from imaging.

Accepted Answer

As with a tracer, the nuclide rides the body's own chemistry: iodine is taken up and concentrated by the thyroid, so the radiation is delivered where it is wanted (1). The aim differs because the radiation is meant to deposit its energy in the thyroid and destroy the tissue (1). Absorption of radiation inside the patient, useless for imaging, is exactly what the treatment requires (1).

Question 13

A sealed implant containing a beta-emitting nuclide is placed inside a tumour. Explain why a beta emitter suits this use although it is unsuited to imaging with a gamma camera, and state why it is the physical half-life, rather than the effective half-life, that matters for such an implant.

Accepted Answer

Beta particles are absorbed within a few millimetres of tissue, so almost all their energy is deposited in the tumour itself and tissue a centimetre away receives very little: the absorbed dose in the tumour is far greater than in the tissue around it (1). That same short range makes a beta emitter useless with a gamma camera, because the radiation cannot leave the patient to be detected (1). The source is sealed and held in place, so the body cannot excrete it: there is no biological clearance and the activity falls by decay alone, at the physical half-life (1).

Question 14

Explain how a PET scanner uses coincidence detection to locate the tracer.

Accepted Answer

The patient lies inside a ring of gamma detectors (1). When two 511 keV photons arrive at opposite sides of the ring at effectively the same instant, they are taken to come from one annihilation (1), which must lie on the straight line joining the two detectors (1). Recording hundreds of thousands of such coincidence lines, the scanner finds where they intersect: the places where the tracer, and so the biological activity, is concentrated (1).

Question 15

Calculate the frequency of a 511 keV annihilation photon.

Accepted Answer

E = 511 keV = 511 × 10³ × 1.60 × 10⁻¹⁹ = 8.18 × 10⁻¹⁴ J (1) f = E/h = 8.18 × 10⁻¹⁴/(6.63 × 10⁻³⁴) = 1.23 × 10²⁰ Hz (1)

Question 16

Fluorine-18, used in PET, has a half-life of 110 minutes. A hospital needs at least one eighth of the delivered activity remaining at scan time. Calculate the maximum delay between delivery and scanning.

Accepted Answer

One eighth is three half-lives (1) t = 3 × 110 = 330 minutes, that is 5.5 hours (1)

Question 17

A patient asks whether a PET scan means being irradiated by the machine. Explain what the scanner actually does, and where the dose really comes from.

Accepted Answer

The scanner emits nothing: it is purely a detector, listening for gamma photons leaving the body (1). The radiation dose comes from the injected tracer itself, whose positrons annihilate inside the patient (1). Choosing a short half-life and a small activity keeps that dose low while still giving enough coincidences for an image (1).

Question 18

A PET detector ring has a radius of 40 cm. An annihilation occurs 4.5 cm from the centre of the ring, on the line joining two opposite detectors. Calculate the difference in the arrival times of the two photons at the detectors (c = 3.00 × 10⁸ m s⁻¹), and comment on what this means for the scanner's timing electronics.

Accepted Answer

One photon travels 2 × 4.5 = 9.0 cm further than the other; the ring radius cancels (1). Δt = 9.0 × 10⁻²/(3.00 × 10⁸) = 0.30 ns (1). A genuine pair from anywhere in the body must be accepted, and an annihilation near the edge of the ring can put nearly the full 80 cm difference between the paths, up to about 2.7 ns, so the coincidence window is a few nanoseconds wide and this 0.30 ns falls comfortably inside it. Timing much finer than a nanosecond is what a time-of-flight scanner adds, using the difference to locate the annihilation along the line between the detectors rather than to accept the pair (1).

Question 19

Three radionuclides are available for a kidney investigation in which imaging ends 30 minutes after the tracer is administered. Nuclide P emits beta minus particles and has a half-life of 8.0 days. Nuclide Q emits gamma photons and has a half-life of 6.0 hours. Nuclide R emits gamma photons and has a half-life of 90 s. Deduce which nuclide should be chosen, giving a reason for rejecting each of the other two.

Accepted Answer

Q should be chosen: its gamma photons escape the body to be detected (1), and a half-life of hours is long enough to prepare, administer and image, yet short enough that the dose falls away soon after the test (1). P is rejected because beta minus particles are absorbed inside the patient, giving dose but no image, and 8.0 days leaves the patient radioactive long afterwards (1). R is rejected because 30 minutes is 20 half-lives, so its activity would be almost entirely gone before imaging ends (1).

Question 20

In radiotherapy a beam of megavoltage X-rays is rotated about the patient, entering along 12 equally spaced directions in turn and crossing the tumour on every one of them, and delivering an equal dose to the tissue on each entry path. Calculate the dose the tumour receives as a multiple of the dose received by the tissue on one entry path. The attenuation coefficient of soft tissue is 0.25 cm⁻¹ for a kilovoltage beam and 0.030 cm⁻¹ for the megavoltage beam. Calculate the fraction of each beam reaching a tumour 12 cm deep, and hence explain why the megavoltage beam is used.

Accepted Answer

The tumour lies on all 12 paths while each entry path is used once, so the tumour receives 12 times the dose of the tissue on any one entry path (1). Kilovoltage: I/I₀ = e^−0.25 × 12 = 0.050 (1). Megavoltage: I/I₀ = e^−0.030 × 12 = 0.70 (1). About 14 times as large a fraction of the megavoltage beam survives to the tumour, so far more of its energy is delivered at that depth rather than in the tissue in front of it (1). The two features act together: the rotation shares the entry dose among twelve paths while the tumour collects all of it, and the harder beam carries its energy to that depth, so the absorbed dose in the tumour greatly exceeds the dose anywhere in the healthy tissue (1).