OCR A Level Physics A: Gamma camera, PET scanning and ultrasound (H556)

This study guide follows the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. It covers 6.5.2 (Diagnostic methods in medicine), statements (a) to (e), and 6.5.3 (Using ultrasound), statements (a) to (g). X-rays, attenuation and CAT scanning (6.5.1) are taught separately. Next: the revision notes and practice set. See the course hub, the checklist and the free 10-minute diagnostics.

Data used: c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J s; densities and sound speeds are given in each example. Photon energy E = hf is in the photons guide and v = fλ in the wave motion guide.

What this unit covers

Statement	What you must be able to do
6.5.2(a)	Medical tracers: technetium-99m and fluorine-18
6.5.2(b), (c)	Gamma camera components, image formation and diagnosis
6.5.2(d), (e)	PET: annihilation, image formation, diagnosis; issues of an expensive scanner
6.5.3(a), (b)	Ultrasound above 20 kHz; piezoelectric effect; transducer
6.5.3(c)	A-scan and B-scan
6.5.3(d), (e)	Z = ρc; Ir/I0 = (Z2 − Z1)² / (Z2 + Z1)²
6.5.3(f)	Impedance matching; the gel
6.5.3(g)	Doppler shift Δf/f = 2v cos θ / c for blood speed

6.5.2(a) Medical tracers

A medical tracer is a radioactive substance, attached to a molecule the body handles in a known way, that is put into a patient so that its radiation can be detected from outside. The molecule carries it to the organ or process under study.

A good tracer emits gamma rays, which leave the body, and has a half-life long enough to use but short enough to keep the dose small.

Technetium-99m (Tc-99m) is a gamma emitter: its photons (about 140 keV) escape the body, and it gives out no alpha or beta. Its half-life is about 6 hours. The “m” means metastable: an excited nucleus that decays by gamma emission.

Fluorine-18 (F-18) is a positron (β⁺) emitter with a half-life of about 110 minutes, attached to a glucose-like molecule (fluorodeoxyglucose) for PET scanning. It must be made close to where it is used.

Worked example 1. A Tc-99m injection of 600 MBq is used 9.0 h later (half-life 6.0 h): 1.5 half-lives, so A = 600 × (1/2)^1.5 = 212 MBq.

Worked example 2. F-18 of activity 370 MBq is made at a cyclotron and reaches the scanner 3.0 h later (half-life 110 min).

180/110 = 1.636 half-lives; fraction left = (1/2)^1.636 = 0.322
A = 370 × 0.322 = 119 MBq

About two thirds has decayed in transit, so F-18 needs a nearby supply and a tight schedule.

6.5.2(b) and (c) The gamma camera

A gamma camera makes an image of the gamma photons coming from tracer inside the patient. Follow a photon through in order.

1. Collimator. A thick lead plate with many narrow parallel holes. Only photons travelling nearly along the axis of a hole get through; photons travelling at an angle are absorbed by the lead walls (the septa). Without it the image would be a blur.
2. Scintillator. A crystal (sodium iodide with a small thallium impurity is the usual one). A gamma photon is absorbed and its energy is re-emitted as a flash of many visible or near-ultraviolet photons.
3. Photomultiplier tubes. An array of tubes sits behind the crystal. Light hitting a tube’s photocathode releases electrons by the photoelectric effect, and a chain of dynodes multiplies them into a measurable pulse. The tube nearest the flash gives the biggest pulse.
4. Computer and display. The computer compares pulses from neighbouring tubes to locate the flash and adds one count there. Thousands of counts build the image, with brighter areas meaning more tracer.

The collimator absorbs most photons, so few are detected and scans take a long time.

Diagnosis (6.5.2(c)). The gamma camera shows function. The tracer compound is chosen to go where the doctor wants to look. In a bone scan, Tc-99m on a bone-seeking compound collects where bone is active, so a hot spot can indicate a fracture or tumour. A series of images over time shows how fast an organ takes up and clears the tracer, and low uptake (a cold spot) can show a region that is not working.

6.5.2(d) and (e) PET scanning

Positron emission tomography (PET) uses a positron emitter such as F-18. The key physics is annihilation.

A positron is the antiparticle of the electron. After emission it travels about a millimetre through tissue, slows down, and meets an electron. The pair annihilate: their mass is converted to energy in the form of two gamma photons.

• Momentum is conserved: the pair has almost none before, so the photons leave in opposite directions.
• Energy is conserved: each photon carries the rest energy of one electron.

Worked example 3. Each photon carries 0.511 MeV, the rest energy mc² of an electron. Find its frequency and wavelength.

E = 0.511 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.18 × 10⁻¹⁴ J
f = E/h = 1.23 × 10²⁰ Hz;  λ = c/f = 2.43 × 10⁻¹² m

The PET scanner. The patient lies inside a ring of detectors (scintillators with photomultiplier tubes). When two detectors on opposite sides record photons at almost the same instant, the computer takes it as one annihilation lying somewhere along the straight line joining those two detectors. Millions of such lines are recorded. The computer finds where the lines cross most densely and builds a map (a tomographic image) of where the positrons were emitted.

No collimator is needed, because the line comes from the pair of photons, so PET detects a much larger fraction of the emitted radiation than a gamma camera does. Sharpness is limited by the distance the positron travels before annihilating.

Diagnosis (6.5.2(e)). F-18 is attached to fluorodeoxyglucose, a molecule that cells take up like glucose. Tissue that uses a lot of glucose shows up bright: active tumours, and the more active parts of the brain or heart. PET can therefore show a tumour that is working, and whether a treatment has reduced its activity. Compared with a gamma camera, PET is more sensitive but needs the tracer made nearby.

Issues in equipping a hospital with an expensive scanner. You may be asked to weigh these:

• Cost of buying and running the scanner, supplying F-18, and trained staff.
• How many patients would benefit, against other uses of the money.
• Radiation dose to patients and staff, against the value of the diagnosis.
• Travel and waiting times if no scanner is available locally.

6.5.3(a) to (c) Ultrasound basics

Ultrasound is a longitudinal wave with frequency greater than 20 kHz. Medical scanners usually use a few megahertz. Since λ = v/f, a higher frequency gives a shorter wavelength and finer detail, but it is absorbed more quickly and penetrates less deeply.

The piezoelectric effect. In a piezoelectric crystal, a p.d. across it makes it change shape, and squeezing it produces a p.d. One crystal does both jobs. An alternating p.d. at the right frequency makes the crystal vibrate and emit ultrasound pulses. Between pulses the same crystal receives echoes, which make it vibrate and give a small alternating p.d.

A-scan. One transducer sends a short pulse in a fixed direction and records echoes against time. The display is a graph of echo strength against time (and so depth). Each peak marks a boundary. If an echo returns after time t, the boundary is at depth d = ct/2 (the factor 2 is because the pulse goes there and back).

B-scan. An array of transducers (or a moved one) takes many A-scans across the body, and the computer combines them into a two-dimensional image in which brightness shows echo strength.

Worked example 4. An echo from a boundary returns 78 μs after the pulse. The speed of ultrasound in the tissue is 1540 m s⁻¹.

d = ct/2 = 1540 × 78 × 10⁻⁶ / 2 = 0.060 m = 6.0 cm

To image to 18 cm, the next pulse must wait at least 2 × 0.18 / 1540 = 234 μs.

6.5.3(d) and (e) Acoustic impedance and reflection

The acoustic impedance of a medium is Z = ρc, where ρ is the density and c the speed of ultrasound in it. The unit is kg m⁻² s⁻¹.

At a boundary, part of the intensity is reflected and part is transmitted. For a beam at right angles to the boundary:

Ir / I0 = (Z2 − Z1)² / (Z2 + Z1)²

where I0 is the incident intensity, Ir the reflected intensity, and Z1 and Z2 the impedances on either side. The order of the media does not matter. The transmitted fraction is 1 − Ir/I0.

Worked example 5. Fat has density 920 kg m⁻³ and ultrasound speed 1450 m s⁻¹. Muscle has density 1080 kg m⁻³ and speed 1560 m s⁻¹. Bone has density 1900 kg m⁻³ and speed 4080 m s⁻¹.

Z(fat) = 1.334 × 10⁶;  Z(muscle) = 1.685 × 10⁶;  Z(bone) = 7.75 × 10⁶  (kg m⁻² s⁻¹)

fat to muscle:  (1.685 - 1.334)² / (1.685 + 1.334)² = 0.0135
muscle to bone: (7.752 - 1.685)² / (7.752 + 1.685)² = 0.413

A small difference in Z gives a weak echo, so most of the beam goes on past soft-tissue boundaries. A large difference reflects much of the beam, so structures behind bone are hard to image.

6.5.3(f) Impedance matching and gel

Air has a very small impedance (about 408 kg m⁻² s⁻¹, from ρ = 1.2 kg m⁻³ and c = 340 m s⁻¹).

Worked example 6. Take skin as 1.60 × 10⁶ and the gel as 1.52 × 10⁶ kg m⁻² s⁻¹.

air to skin: (1.60×10⁶ - 408)² / (1.60×10⁶ + 408)² = 0.9990
gel to skin: (1.60 - 1.52)² / (1.60 + 1.52)² = 6.6 × 10⁻⁴

Without gel, 99.9% is reflected. The gel is spread between transducer and skin to remove air: its impedance is close to that of skin, so there is little reflection (impedance matching) and nearly all the beam enters.

6.5.3(g) The Doppler effect and blood speed

Ultrasound reflected from moving blood cells has a different frequency from the one emitted. The frequency shift is

Δf / f = 2v cos θ / c

where f is the emitted frequency, v the blood speed, θ the angle between the beam and the direction of blood flow, and c the speed of ultrasound in tissue. The 2 arises because moving blood acts first as a moving receiver, then as a moving source.

Worked example 7. A 5.0 MHz beam is directed at 60° to blood flowing at 0.45 m s⁻¹. Take c = 1570 m s⁻¹.

Δf = 2 × 5.0×10⁶ × 0.45 × cos 60° / 1570 = 1.4 × 10³ Hz

Worked example 8. A 4.0 MHz beam at 50° gives a shift of 2.4 kHz. Find v.

v = Δf c / (2 f cos θ) = 2.4×10³ × 1570 / (2 × 4.0×10⁶ × cos 50°) = 0.73 m s⁻¹

At θ = 90° the shift is zero, so the beam must not be across the vessel.

Common errors

• Saying the gamma camera detects positrons, or that PET needs one photon rather than two detected together.
• Saying the gel is “to stop friction” rather than to match impedance.
• Using d = ct instead of d = ct/2 for echoes.
• Using Δf/f = v/c without the 2 and the cos θ.
• Leaving out the square in the reflection equation.

Official syllabus

OCR Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), for first assessment in 2017, published by OCR: sections 6.5.2 and 6.5.3.