Medical physics

24.1 Production and use of ultrasound

The piezoelectric transducer

A piezo-electric crystal changes shape when a p.d. is applied across it. Conversely, it generates an e.m.f. when its shape changes.

• Generating: an alternating p.d. at the crystal's natural frequency makes it resonate, so it vibrates strongly, producing ultrasound (above 20 kHz; medical uses are in the MHz range).
• Detecting: returning ultrasound makes the crystal vibrate, and this gives an alternating e.m.f.

The same transducer does both, by alternating between short pulses and listening.

Diagnosis by reflection

A pulse of ultrasound is sent into the body. Some of it reflects at each boundary between tissues. The time delay shows the depth of the boundary (distance = speed × time / 2, because the sound goes there and back), and the intensity of the echo shows the kind of boundary.

A gel is put between the transducer and skin. Without it, there would be air, and nearly all of the ultrasound would reflect at the air–skin boundary.

Acoustic impedance

Specific acoustic impedance of a medium is:

$Z = \rho c$

where $\rho$ is its density and $c$ is the speed of sound in it.

The intensity reflection coefficient at a boundary is:

$\frac{I_R}{I_0} = \frac{(Z_1 - Z_2)^2}{(Z_1 + Z_2)^2}$

Attenuation

Intensity falls exponentially with distance $x$ through a material:

$I = I_0 e^{-\mu x}$

where $\mu$ is the attenuation coefficient. Higher-frequency ultrasound gives better detail but is attenuated more.

24.2 Production and use of X-rays

Producing X-rays

Electrons are accelerated through a large p.d. and hit a metal target. Most of the energy becomes heat, so the target is often a metal like tungsten and is rotated or cooled.

The electrons decelerate suddenly, producing a continuous spectrum of X-rays with a sharp minimum wavelength. This happens when one electron gives all of its kinetic energy to one photon:

$eV = \frac{hc}{\lambda_{\text{min}}}$

Imaging and contrast

X-rays passing through the body are attenuated by different amounts: $I = I_0e^{-\mu x}$. Bone has a much larger $\mu$ than soft tissue.

Contrast is the difference in the degree of blackening between structures in the image. Good contrast makes tissues easy to tell apart. Soft tissues have similar $\mu$, so they have poor contrast. A contrast medium such as barium, which absorbs X-rays strongly, can be swallowed to show the gut.

CT scanning

Computed tomography (CT) produces a 3D image:

1. Multiple X-ray images of one section are taken from different angles, by rotating the tube around the patient.
2. A computer combines them into a 2D image of that section.
3. This is repeated along an axis to obtain many sections.
4. The 2D images of the sections are combined into a 3D image, which can be viewed from any angle.

24.3 PET scanning

Tracers and annihilation

A tracer is a substance containing radioactive nuclei that can be introduced into the body and is then absorbed by the tissue being studied.

Positron emission tomography (PET) uses a tracer that decays by β⁺ decay, emitting positrons. A common example is a fluorine-18 compound related to glucose, which gathers in tissue that is using more energy, such as many tumours.

Annihilation occurs when a particle interacts with its antiparticle. Mass–energy and momentum are conserved in the process.

How the scan works

1. A positron emitted from the tracer travels a very short distance, then meets an electron in the tissue, and they annihilate.
2. This gives two gamma-ray photons travelling in opposite directions, in order to conserve momentum, since the total momentum beforehand was close to zero.
3. The photons leave the body and are detected by a ring of detectors around the patient.
4. From the arrival times of the pair, the computer locates where each annihilation happened, and it builds an image of the tracer concentration in the tissue.

Energy of the gamma photons

The rest mass of the electron and the positron turns into the energy of two photons, so each photon has the energy of one electron's mass:

$E = m_ec^2 = 9.11 \times 10^{-31} \times (3.00 \times 10^{8})^2 = 8.2 \times 10^{-14}\text{ J}$

This is about 0.51 MeV for each photon. The wavelength is then $\lambda = hc/E = 2.4 \times 10^{-12}$ m.