The production and use of X-rays, ultrasound imaging and the acoustic impedance principle, and the use of radioactive tracers in medicine.

What this dot point is asking

CCEA wants you to describe how X-rays are produced and used in imaging, explain ultrasound imaging and the role of acoustic impedance and coupling gel, and describe how radioactive tracers are used in diagnosis. Questions mix description with a short calculation, often the ultrasound depth equation or a half-life decay, so you must be confident with both the physics and the arithmetic.

X-rays

Electrons are released from a heated filament by thermionic emission, then accelerated through a potential difference $V$ before striking the target. Each electron gains kinetic energy $E_k = eV$, so a tube voltage of $V = 80\ \text{kV}$ gives electrons an energy of $eV = 1.6 \times 10^{-19} \times 80\,000 = 1.3 \times 10^{-14}\ \text{J}$. The maximum photon energy equals this electron energy, setting the shortest wavelength in the beam through $E_{\max} = hf_{\max} = eV$.

X-rays are ionising, so the dose to the patient is kept as low as reasonably achievable. Lead shielding and collimators limit the beam to the region of interest, and a contrast medium (such as a barium meal, which strongly absorbs X-rays) can be used to make soft-tissue structures such as the gut visible.

Ultrasound

A coupling gel is smeared on the skin to remove the air gap between the transducer and the body. The impedance of air is roughly $400\ \text{kg m}^{-2}\,\text{s}^{-1}$ while that of skin is around $1.6 \times 10^{6}\ \text{kg m}^{-2}\,\text{s}^{-1}$; substituting these into the reflection equation gives a reflected fraction close to one, so almost all the ultrasound would bounce off. The gel has an impedance close to that of skin, matching the two media so the beam enters the body. The time delay of each reflected pulse gives the depth of each boundary, building up an image as the transducer scans across the surface.

Radioactive tracers

A radioactive tracer is a substance containing a radioisotope that is introduced into the body and concentrates in a target organ (for example technetium-99m taken up by the thyroid or bone). A gamma emitter is used so the radiation passes out of the body to be detected by a gamma camera.

The half-life must be long enough to complete the scan yet short enough to limit exposure. Technetium-99m, with a half-life of about $6\ \text{hours}$, is ideal: its activity has fallen to a small fraction within a day.

Worked example: tracer activity and half-life

Examples in context

Example 1. A chest X-ray. The ribs and spine absorb many photons and appear white, while the air-filled lungs absorb little and appear dark, so a fluid-filled or solid region (such as a tumour or pneumonia) shows as an abnormal bright patch against the dark lung field. The whole image is a map of attenuation built from the absorption contrast described above.

Example 2. A prenatal ultrasound scan. Pulses of ultrasound at a few megahertz are sent into the abdomen through coupling gel. Reflections from the boundaries between amniotic fluid, soft tissue and developing bone return at different times, and the scanner converts each round-trip time into a depth using $2d = ct$, assembling a real-time image. Ultrasound is preferred here because, unlike X-rays, it is non-ionising and safe for the fetus.

Try this

Q1. Explain why a coupling gel is used in ultrasound scanning. [2 marks]

• Cue. It removes the air gap, because the large acoustic impedance difference between air and skin would otherwise reflect almost all the ultrasound.

Q2. State two properties a radioisotope should have to be a good medical tracer. [2 marks]

• Cue. It should be a gamma emitter (to escape the body) and have a short half-life (to limit the dose).

Q3. An ultrasound pulse returns from a boundary $6.0 \times 10^{-5}\ \text{s}$ after emission. The speed of sound in tissue is $1500\ \text{m s}^{-1}$. Find the depth of the boundary. [2 marks]

• Cue. $2d = ct = 1500 \times 6.0 \times 10^{-5} = 0.090\ \text{m}$, so $d = 0.045\ \text{m} = 4.5\ \text{cm}$.