AS & A Level Physics · Area 2 · A Level Content

Medical physics

Medical physics uses waves and nuclear decay to see inside the body. This topic covers ultrasound generation and detection by a piezo-electric transducer, reflection at tissue boundaries and acoustic impedance, attenuation, X-ray production and contrast, computed tomography, and PET scanning with its annihilation gamma-ray photons.

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Syllabus coverage

  • Understand that a piezo-electric crystal changes shape when a p.d. is applied across it and generates an e.m.f. when its shape changes, and understand how ultrasound waves are generated and detected by a piezoelectric transducer
  • Understand how the reflection of pulses of ultrasound at boundaries between tissues can be used to obtain diagnostic information about internal structures
  • Define the specific acoustic impedance of a medium as Z = rho c, where c is the speed of sound in the medium, and use IR / I0 = (Z1 - Z2)^2 / (Z1 + Z2)^2 for the intensity reflection coefficient of a boundary between two media
  • Recall and use I = I0 e^(-mu x) for the attenuation of ultrasound in matter
  • Explain that X-rays are produced by electron bombardment of a metal target, and calculate the minimum wavelength of the X-rays produced from the accelerating p.d.
  • Understand the use of X-rays in imaging internal body structures, including an understanding of the term contrast in X-ray imaging, and recall and use I = I0 e^(-mu x) for the attenuation of X-rays in matter
  • Understand that computed tomography (CT) scanning produces a 3D image of an internal structure by first combining multiple X-ray images taken in the same section from different angles to obtain a 2D image of the section, then repeating this along an axis and combining 2D images of multiple sections
  • Understand that a tracer is a substance containing radioactive nuclei that can be introduced into the body and is then absorbed by the tissue being studied, and recall that a tracer that decays by beta-plus decay is used in positron emission tomography (PET scanning)
  • Understand that annihilation occurs when a particle interacts with its antiparticle and that mass-energy and momentum are conserved in the process, and explain that in PET scanning the positrons emitted by the decay of the tracer annihilate when they interact with electrons in the tissue, producing a pair of gamma-ray photons travelling in opposite directions
  • Calculate the energy of the gamma-ray photons emitted during the annihilation of an electron-positron pair, and understand that the gamma-ray photons from an annihilation event travel outside the body and can be detected, so that an image of the tracer concentration in the tissue is created by processing the arrival times of the photons

Part of the AS & A Level Physics syllabus 9702.

The X-ray tube and the ultrasound pulse-echo techniqueX-rays are produced when fast electrons are stopped by a heavy target. Ultrasound images from the time taken for a pulse to travel toa boundary and back.The X-ray tubeevacuated glass tubecathode (heated filament)electron beamaccelerating voltage betweencathode and anode: 100 kVanode (tungsten target)the anode rotates so thatthe target does not meltX-ray beamthe X-rays leave through a thin windowUltrasound: the pulse-echo techniquetransducer (a piezoelectric crystal)skinboundary: the acoustic impedance changesincident pulsereflected pulsepart is transmittedd = c t / 2, with c = 1540 m s-1 in soft tissue= 1540 x 52 x 10^-6 / 2 = 0.040 m = 4.0 cmvoltage (V)020406080100time (microseconds)time delay t = 52 microsecondsX-rays come from fast electrons stopped by the target: a continuous spectrum from deceleration plus sharp characteristic lines.The shortest wavelength has all the electron energy in one photon: h f max = e V, so lambda min = h c / (e V).Ultrasound: a pulse reflects where the acoustic impedance changes, so the depth is d = c t / 2, because the pulse travels there andback.A higher frequency gives better resolution but is absorbed more, so ultrasound of a few MHz is used at 1540 m s-1 in soft tissue.
X-rays and ultrasound used to image the inside of the body: an X-ray tube in which electrons accelerated through a high p.d. strike a metal target to produce a beam of X-rays that passes through the patient to a detector, and an ultrasound transducer that transmits pulses into the body and receives the echoes reflected from boundaries between tissues.

Piezo-electric transducers: generating and detecting ultrasound

A piezo-electric crystal, such as quartz or a lead zirconate titanate ceramic, changes shape when a potential difference is applied across it, and it generates an e.m.f. across it when its shape is changed. The effect works in both directions, and that is exactly why one crystal can both send and receive a pulse of ultrasound. A transducer is that crystal mounted in a housing with electrodes on its opposite faces, backed with a damping material so that it rings for only a few cycles.

To generate ultrasound you apply an alternating p.d. at the resonant frequency of the crystal across the electrodes. The crystal is cut so that its thickness is half a wavelength of sound in the material, so it vibrates strongly at that frequency and radiates a beam of ultrasound into whatever it is pressed against. To detect ultrasound the process is reversed: an arriving wave forces the crystal to vibrate, the vibration changes its shape, and it produces an alternating e.m.f. of the same frequency across its faces. That tiny signal is amplified and displayed. Most scanners use the same crystal alternately as transmitter and receiver, switching between the two thousands of times a second.

The audible limit of hearing is about 20 kHz, so ultrasound is anything above that, and medical scanners work between about 1 MHz and 20 MHz. Frequency decides the trade-off that shapes every scan: a shorter wavelength gives better resolution, because two boundaries closer together than about one wavelength cannot be told apart, but the attenuation coefficient also rises with frequency, so a higher frequency is absorbed sooner and penetrates less deeply. Abdominal and obstetric scanning therefore uses about 2 MHz to 5 MHz, while an eye, which is only a few centimetres across, is scanned at 10 MHz to 20 MHz. Ultrasound is used in preference to X-rays wherever it can do the job because it is not ionising and so carries no radiation risk.

Because the pulse is short, the scanner can time it. The crystal is excited for a few microseconds, then listens for echoes. The time between transmitting the pulse and receiving an echo gives the depth of the boundary, and because pulses are repeated many times a second, the image can be updated fast enough to show a moving fetus or a beating heart.

  • Piezo-electric effect: a p.d. across the crystal makes it change shape, and changing its shape makes it produce a p.d.
  • An alternating p.d. at the resonant frequency (set by the thickness of the crystal) generates the ultrasound.
  • An incoming ultrasound wave vibrates the crystal, which generates an e.m.f. that is amplified and displayed.
  • Medical ultrasound is 1 MHz to 20 MHz: higher frequency gives better resolution but is attenuated more.
  • Ultrasound is non-ionising, which is why it is preferred for scanning a fetus.
One crystal, two jobs: a changing p.d. moves the crystal and sends a pulse; a moving crystal produces a p.d. and detects the echo.

Echoes from boundaries and specific acoustic impedance

A pulse of ultrasound travels through tissue until it reaches a boundary between two media. Part of the pulse is reflected there and part continues, and the reflected pulse - the echo - returns to the transducer. The echo is detected and timed, and because the pulse has travelled down to the boundary and back again, the depth is depth = c t / 2, where c is the speed of sound in the tissue (about 1540 m s-1 for soft tissue) and t is the time between transmitting and receiving. A boundary that reflects strongly gives a large echo; one that reflects weakly gives a small echo, and that difference is what builds up the picture.

How much is reflected depends on the specific acoustic impedance of each medium, defined as Z = rho c, where rho is the density of the medium in kg m-3 and c is the speed of sound in it in m s-1. The unit of Z is therefore kg m-2 s-1. Physically, Z measures how hard it is to make the medium move: a dense, stiff medium such as bone has a large Z and resists being vibrated, while air has a tiny Z because it is light and springy.

The fraction of the incident intensity reflected at a boundary is the intensity reflection coefficient, IR / I0 = (Z1 - Z2)^2 / (Z1 + Z2)^2. For example, at a fat-muscle boundary Z1 = 1.4 x 10^6 kg m-2 s-1 and Z2 = 1.7 x 10^6 kg m-2 s-1, so the difference is 3.0 x 10^5 and the sum is 3.1 x 10^6. Then IR / I0 = (3.0 x 10^5)^2 / (3.1 x 10^6)^2 = 9.0 x 10^10 / 9.6 x 10^12 = 9.4 x 10-3, so only about 1% of the intensity is reflected and 99% carries on. That is why echoes from soft tissue are so weak that the receiver has to be very sensitive.

The equation also explains the coupling gel. The specific acoustic impedance of air is about 4.3 x 10^2 kg m-2 s-1 and that of soft tissue about 1.6 x 10^6 kg m-2 s-1, so (Z1 - Z2)^2 / (Z1 + Z2)^2 is almost 1: a thin layer of air between the transducer and the skin would reflect about 99.9% of the ultrasound straight back and almost nothing would enter the body. A gel whose impedance is close to that of soft tissue removes that boundary and lets the pulse through. The same idea explains why an air-filled lung, a bone surface or a gallstone produces such a strong echo, and why a crack in a metal casting - a metal-air boundary - is so easy to detect.

MediumDensity / kg m-3Speed of sound / m s-1Specific acoustic impedance Z / kg m-2 s-1
Air1.33304.3 x 10^2
Fatabout 95014501.4 x 10^6
Water100015001.5 x 10^6
Soft tissue (average)about 105015401.6 x 10^6
Muscleabout 107015901.7 x 10^6
Boneabout 180033006.0 x 10^6
A big mismatch reflects almost everything and a small mismatch reflects almost nothing. Tissue-to-tissue boundaries reflect only about 1%, so the scanner depends on a very sensitive receiver.

Attenuation, and turning echoes into a picture: A-scans and B-scans

Ultrasound does not travel through tissue for free: it is attenuated by absorption, as energy is turned into internal energy of the tissue, and by scattering at small irregularities. The intensity that survives a path length x is given by I = I0 e^(-mu x), where mu is the linear attenuation coefficient of the tissue in m-1 and x is in metres. Because mu x must be dimensionless, convert every length to metres before you substitute. Take ultrasound of intensity 5.0 x 10-3 W m-2 entering a tissue with mu = 12 m-1 and travelling 8.0 cm: mu x = 12 x 0.080 = 0.96, so I = 5.0 x 10-3 x e^(-0.96) = 5.0 x 10-3 x 0.38 = 1.9 x 10-3 W m-2. Note how the intensity falls by an ever-decreasing amount and never reaches zero, because the attenuation is exponential rather than a fixed loss per centimetre.

An A-scan (amplitude scan) displays the received signal as the height of a voltage spike against time on an oscilloscope-like display. The transmitter pulse starts the time base, and each boundary along the line of the beam produces a spike at a time that gives its depth. An A-scan therefore gives distances along one line - the length of an eye, the position of the midline of the brain, or the depth of a flaw inside a casting - but it is not a picture of a cross-section.

A B-scan (brightness scan) moves the transducer across the body, and instead of plotting the echo height it sets the brightness of a spot from the echo amplitude, using the position of the transducer for the horizontal coordinate and the echo time for the depth. Electronics combines thousands of these lines and builds up a two-dimensional cross-section of the body, which is what most people picture when they think of an ultrasound scan. Scanning fast enough produces a moving image, so the movement of a heart valve or of a fetus can be watched directly.

Attenuation and depth of penetration are inseparable. mu increases roughly in proportion to frequency, so doubling the frequency roughly doubles the attenuation in dB. A 2 MHz beam may penetrate 20 cm of soft tissue, whereas a 15 MHz beam used to measure the thickness of a cornea is useful over only a centimetre or so. Choosing the frequency is a compromise between the resolution you want and the depth you need to reach.

FeatureA-scanB-scan
DisplayEcho amplitude as the height of a spikeEcho amplitude as the brightness of a spot
TransducerHeld in one fixed placeMoved across the surface
OutputDepths of boundaries along one lineA two-dimensional cross-section
Typical useEye length before cataract surgery, flaw detection in metalObstetric and abdominal imaging, echocardiography
Halve the time of flight before you call it a depth. The pulse has made the journey twice.

Producing X-rays: electron bombardment and the minimum wavelength

X-rays are produced when fast-moving electrons are stopped by a metal target. In an X-ray tube, a tungsten filament is heated so that electrons are emitted from it by thermionic emission, and a large potential difference - typically 30 kV to 150 kV - accelerates them across an evacuated tube towards a heavy metal target, usually tungsten. When an electron strikes the target it is decelerated violently, and most of its kinetic energy becomes internal energy in the target while a small fraction, well under 1%, is radiated as X-ray photons. Because so much energy goes into heating, the target is a rotating anode so that the hot spot moves, and it is cooled by oil or water.

The spectrum produced has two parts. The continuous (bremsstrahlung) spectrum comes from electrons that lose different amounts of energy in different numbers of deflections, so every photon energy up to a maximum is possible. Superimposed on it are sharp characteristic peaks at wavelengths fixed by the energy levels of the target atoms. The continuous spectrum has a sharp short-wavelength cut-off, lambda_min, produced when a single electron gives all of its kinetic energy to a single photon: eV = hc / lambda_min, so lambda_min = hc / (eV) - or, if you prefer the frequency form, fmax = eV / h.

A worked value. With an accelerating p.d. of 80 kV, eV = 1.60 x 10-19 x 8.0 x 10^4 = 1.28 x 10-14 J and hc = 6.63 x 10-34 x 3.00 x 10^8 = 1.99 x 10-25 J m, so lambda_min = 1.99 x 10-25 / 1.28 x 10-14 = 1.55 x 10-11 m. Every photon in the beam has a wavelength longer than this, and the cut-off is a hard edge on the spectrum.

The two controls on the tube do different jobs. Raising the accelerating p.d. increases the energy of each electron, which moves the cut-off to shorter wavelength and makes the beam more penetrating, so the image improves; the cut-off depends only on the p.d. and not on the target material. Raising the filament current boils off more electrons per second, which increases the intensity of the beam without changing the cut-off, because the energy each individual electron can hand over is fixed by the p.d. alone. The target material fixes the wavelengths of the characteristic lines and helps determine how efficiently X-rays are produced.

  • Electrons are released by thermionic emission from a heated tungsten filament.
  • They are accelerated through a high p.d. and stopped by a heavy metal target such as tungsten.
  • Almost all of the kinetic energy becomes heat, so the anode rotates and is cooled.
  • The continuous spectrum ends at lambda_min = hc / (eV), where one electron gives all of its energy to one photon.
  • Increasing the p.d. shortens lambda_min and makes the beam more penetrating; increasing the filament current increases the intensity.
lambda_min = hc / (eV) depends only on the accelerating p.d., because it is set by the energy carried by one electron.

X-ray imaging, attenuation and contrast in tissue

A single X-ray image is a shadow picture. X-rays leave a small source, pass through the patient and reach a detector or film on the far side. Where the beam passes through material that absorbs strongly, less intensity reaches the detector, so that part of the film is less exposed and appears white. Bone, which absorbs well, shows white; fat and muscle show grey; air-filled lung shows nearly black because it absorbs almost nothing.

X-rays are removed from the beam by the same exponential law as ultrasound, I = I0 e^(-mu x), but the physics of the removal is quite different. Being ionising electromagnetic radiation, an X-ray photon interacts with the atoms of the material in one of two main ways. In photoelectric absorption the photon is completely absorbed and its energy is used to eject a tightly bound electron from an inner shell; this is a strong interaction with high atomic number atoms and low-energy photons, and it grows roughly as the cube of the atomic number. In Compton scattering the photon collides with a loosely bound outer electron, transfers part of its energy and is deflected out of the beam in a new direction, so it never reaches the detector. Compton scattering depends mainly on the number of electrons present, and therefore on the density of the tissue rather than on the atomic number.

The attenuation of X-rays in a material is described by its linear attenuation coefficient mu in m-1, which depends on the photon energy, the density of the material and the atomic number of its atoms. Because bone contains calcium and phosphorus (higher Z and denser) and soft tissue is mostly water, bone has a much larger mu. Contrast in an X-ray image is the difference in the transmitted intensity between two parts of the image, and it is largest where the two tissues differ most in mu x. Bone against soft tissue gives good contrast; soft tissue against soft tissue, such as a tumour inside a muscle, gives almost none, because mu x is nearly the same. That is why contrast media of high atomic number are used: a suspension of barium sulphate is swallowed to outline the digestive tract, and iodine compounds are injected into blood vessels, because they absorb X-rays much more strongly than the tissue around them.

X-rays are ionising, so every X-ray examination delivers a dose to living tissue and that is the reason for the safety rules. Absorbed radiation can ionise atoms and molecules in cells, breaking chemical bonds and producing free radicals, which can kill cells outright or damage DNA so that a cell mutates or starts to divide out of control and becomes a cancer. The effect of the radiation on matter is therefore ionisation, and its consequences in tissue depend on the dose and on which tissue is exposed. Radiographers use the smallest exposure that gives a usable image, collimate the beam to the region of interest, use lead shielding over the gonads and a lead apron for anyone else in the room, stand behind a lead screen, and keep a record of the dose received by staff. The dose from a diagnostic X-ray must always be justified by the benefit of the diagnosis.

FeaturePhotoelectric absorptionCompton scattering
What happensThe photon is absorbed completely and ejects an inner-shell electronThe photon scatters off an outer electron, losing energy and changing direction
Depends strongly onAtomic number of the atoms and photon energyNumber of electrons present, so on density
Contribution to imageLarge differences between bone and soft tissueA general background of scattered radiation that reduces contrast
Effect on muIncreases sharply with atomic numberRoughly proportional to density
Contrast is a difference in transmitted intensity. Two tissues give contrast only if their mu x values differ, which is why high-atomic-number contrast media are swallowed or injected.
The X-ray tube and the ultrasound pulse-echo techniqueX-rays are produced when fast electrons are stopped by a heavy target. Ultrasound images from the time taken for a pulse to travel toa boundary and back.The X-ray tubeevacuated glass tubecathode (heated filament)electron beamaccelerating voltage betweencathode and anode: 100 kVanode (tungsten target)the anode rotates so thatthe target does not meltX-ray beamthe X-rays leave through a thin windowUltrasound: the pulse-echo techniquetransducer (a piezoelectric crystal)skinboundary: the acoustic impedance changesincident pulsereflected pulsepart is transmittedd = c t / 2, with c = 1540 m s-1 in soft tissue= 1540 x 52 x 10^-6 / 2 = 0.040 m = 4.0 cmvoltage (V)020406080100time (microseconds)time delay t = 52 microsecondsX-rays come from fast electrons stopped by the target: a continuous spectrum from deceleration plus sharp characteristic lines.The shortest wavelength has all the electron energy in one photon: h f max = e V, so lambda min = h c / (e V).Ultrasound: a pulse reflects where the acoustic impedance changes, so the depth is d = c t / 2, because the pulse travels there andback.A higher frequency gives better resolution but is absorbed more, so ultrasound of a few MHz is used at 1540 m s-1 in soft tissue.
X-rays and ultrasound compared as imaging techniques: an X-ray tube in which accelerated electrons strike a metal target to produce a beam of X-rays that passes through the body to a detector, alongside an ultrasound transducer that sends pulses into the body and receives the echoes reflected from boundaries between tissues.

Computed tomography: many projections, one slice

A single X-ray image has a fundamental weakness: the body is three-dimensional but the film is flat, so everything along a beam is superimposed. A tumour lying in front of a rib may be completely hidden by the brighter image of the bone, and no amount of staring at the film will separate them. Computed tomography (CT) solves this by measuring the beam from many directions and letting a computer do the separating.

In a CT scanner, an X-ray tube and a ring of detectors rotate around the patient in one plane. At each of several hundred positions a narrow fan of X-rays crosses the same thin section of the body and the detectors record how much intensity has been transmitted, which is a measure of the total mu x along each of hundreds of different lines. Because each line passes through many different tissues, any single measurement is a sum of contributions, but the whole set of sums, taken at enough different angles, can be solved by computer to give the attenuation of each small volume element (voxel) in the section. The result is a two-dimensional image of that section in which each pixel is labelled with a number related to mu, on the Hounsfield scale: air is about -1000, water is 0, and dense bone is about +1000. Nothing is hidden, because the overlapping has been undone mathematically.

To build a three-dimensional image, the patient is moved a little along an axis of the body and the whole process is repeated for the next section. Stacking the reconstructed 2D sections one behind another gives a volume that can be displayed as any chosen slice or viewed from any angle, which is what makes CT so useful for planning surgery and for locating a tumour precisely. A modern scanner reconstructs slices about 1 mm thick, and a whole scan takes only a few seconds.

The drawback of gathering all this information is the radiation dose. A CT scan uses many exposures rather than one, so the dose to the patient is typically two or three orders of magnitude larger than that from a single plain radiograph of the same region. CT is therefore reserved for the cases where the extra information changes what is done - a suspected bleed in the head, a tumour that must be measured, a complex fracture - and the benefit of the diagnosis has to be weighed against the risk from the ionising radiation.

CT reconstructs the attenuation coefficient of every voxel in a slice from many projections, then stacks slices along the axis to give a 3D image - and delivers a much larger dose than a plain radiograph.

PET scanning: tracers, positron annihilation and the gamma camera

A tracer is a substance containing radioactive nuclei that is introduced into the body and absorbed by the tissue being studied, so that the tissue can be imaged from outside. In positron emission tomography the tracer is one that decays by beta-plus decay. The most common is a glucose analogue labelled with fluorine-18, which has a half-life of 110 minutes and is made in a cyclotron. Cells that use a lot of glucose - cancer cells, and the brain - take up the tracer most, so the concentration of the tracer is a map of metabolic activity rather than of anatomy.

In beta-plus decay a proton-rich nucleus converts a proton into a neutron and emits a positron, the antiparticle of the electron: proton -> neutron + positron. The positron is the positively charged partner of the electron, with the same mass, and in tissue it travels only about a millimetre before it meets an electron. When a particle meets its antiparticle the two annihilate: both masses disappear and their total mass-energy is released as radiation. Two particles of rest mass 9.11 x 10-31 kg each therefore release 2 m c^2 = 2 x 9.11 x 10-31 x (3.00 x 10^8)^2 = 1.6 x 10-13 J, and because two photons are produced, each carries half of it: 8.2 x 10-14 J, which is 5.1 x 10^5 eV (about 511 keV).

Both conservation laws decide what the photons do. Mass-energy is conserved, which fixes the energy of each photon at m e c^2 for an electron at rest. Momentum is also conserved, and the electron and positron are almost at rest when they annihilate, so their total momentum is close to zero; two photons of equal energy therefore have to fly off in exactly opposite directions so that their momenta cancel. That single fact is the basis of the whole technique. It also means the process is impossible with one photon, because a single photon cannot carry away both the energy and the momentum.

The two gamma-ray photons are penetrating enough to leave the body, unlike the positron, which is stopped within about a millimetre. A ring of detectors surrounds the patient, and when two detectors on opposite sides record a photon at the same instant - a coincidence - the annihilation event must have happened somewhere on the straight line joining them. Each coincidence therefore defines a line in space, and after many thousands of events the computer adds the lines together (back-projection) to produce an image of the tracer concentration in each part of the tissue. Recording the arrival times precisely is what makes the coincidence possible, and because the two photons are detected together, absorption in the surrounding tissue affects both of them and largely cancels out, so a PET scan gives a quantitative measure of activity rather than just a picture.

  • The tracer decays by beta-plus decay, emitting a positron: proton -> neutron + positron.
  • The positron travels about 1 mm in tissue and annihilates with an electron.
  • Mass-energy is conserved, so the pair's rest energy 2 m c^2 appears as two photons of 5.1 x 10^5 eV each.
  • Momentum is conserved, so the two photons travel in exactly opposite directions.
  • Detectors in coincidence define a line of response; processing many lines gives an image of tracer concentration.
Two photons, 511 keV each, exactly 180 degrees apart. Mass-energy fixes the energy; momentum fixes the opposite directions.
PET scanning and the gamma cameraA positron-emitting tracer is detected from the two gamma photons it produces. In a gamma camera a collimator makes sure each pointof the image comes from one place.Positron emission tomography (PET)the tracercollects here511 keVgamma photon511 keVgamma photonring of detectorsthe positron travels a few mm, then annihilates with an electron,making two gamma photons of 511 keV that fly off in opposite directions.detectors on opposite sides record one photon each at the same instant,and that coincidence fixes the line along which the tracer lies.The gamma cameraphotomultiplier tubesto the displayscintillator (sodium iodide) crystallight photonscollimator: lead with long narrow holesabsorbed by the leadpatientPET: the tracer emits positrons; each one annihilates and gives two 511 keV gamma photons that leave in opposite directions.Detectors round the patient pick them up in coincidence, which fixes the line through the tracer.Gamma camera: the collimator passes only photons travelling along its holes, and the scintillator turns each one into light.Only about 1 in 10000 photons is detected, but the position of every one is known, so an image of the tracer is built up.
Gamma-ray production and detection in PET scanning: a tracer nucleus decays by beta-plus decay, the emitted positron annihilates with an electron in the tissue to produce two gamma-ray photons travelling in opposite directions, and the arrival of the two photons at detectors on opposite sides of the patient in coincidence fixes the line on which the annihilation took place.

Key terms you must be able to define

Piezo-electric effect
The property of certain crystals that they change shape when a p.d. is applied across them and generate an e.m.f. across opposite faces when their shape is changed, which allows one crystal to act as both the transmitter and the receiver of ultrasound.
Specific acoustic impedance
Z = rho c, the product of the density of a medium and the speed of sound in it, measured in kg m-2 s-1; it measures how strongly a medium resists being made to vibrate, and the size of the mismatch between two media decides how much ultrasound is reflected at their boundary.
Intensity reflection coefficient
The fraction of the intensity of a wave that is reflected at a boundary between two media, given for ultrasound by IR / I0 = (Z1 - Z2)^2 / (Z1 + Z2)^2, where Z1 and Z2 are the specific acoustic impedances of the two media.
Attenuation coefficient
The constant mu in I = I0 e^(-mu x), measured in m-1, describing how quickly the intensity of a beam falls as it travels through a medium by absorption and scattering; it increases with the frequency of ultrasound and depends on photon energy, density and atomic number for X-rays.
A-scan
An ultrasound display in which the amplitude of each returning echo is shown as the height of a spike against a time base, so the depths of boundaries along one fixed line of the beam can be measured.
B-scan
An ultrasound display in which the brightness of a spot is set by the amplitude of the echo and the transducer is moved across the body, so that a two-dimensional cross-sectional image is built up from many lines.
Contrast (X-ray imaging)
The difference in the transmitted X-ray intensity between two parts of an image, which arises when the tissues have different values of mu x; it is large for bone against soft tissue and very small for one soft tissue against another.
Tracer
A substance containing radioactive nuclei that is introduced into the body and absorbed by the tissue being studied, so that the distribution of the tissue can be imaged by detecting the radiation emitted from it.
Positron annihilation
The process in which a positron meets an electron and both masses are converted into energy, producing two gamma-ray photons of equal energy (about 5.1 x 10^5 eV each) travelling in opposite directions, because both mass-energy and momentum are conserved.

Applied to the real world

Cambridge sets 9702 in unfamiliar contexts - a fairground ride, a power station, a hospital scanner, a satellite, an astronomical observatory. Use these as models for how to apply the physics, then write your own.

In medicine

An obstetric B-scan uses a transducer of about 2 MHz to 5 MHz pressed against the mother's abdomen with coupling gel. Pulses are transmitted many times a second and the echoes are displayed as bright spots, giving a real-time cross-section of the fetus that shows its size, its position and the movement of its heart. Ultrasound is chosen rather than X-rays because it is not ionising, so neither the mother nor the fetus receives a radiation dose.

In medicine

Before cataract surgery the length of the eye is measured to within a tenth of a millimetre by an A-scan at 10 MHz to 20 MHz, and the result is used to choose the power of the artificial lens. The same pulse-echo method is used to measure the thickest part of the cornea through a water bath. These are high frequencies because the distances involved are only a couple of centimetres, and the short wavelength is needed to resolve the cornea and the lens from the retina.

In medicine

A plain X-ray of the abdomen shows very little contrast, because most of the gut contents have much the same linear attenuation coefficient as the surrounding soft tissue. A patient swallows a suspension of barium sulphate, which contains barium, an atom of high atomic number, so the barium strongly absorbs X-rays by photoelectric absorption and outlines the digestive tract in bright white on the image. Iodine compounds injected into the blood do the same job for the kidneys and blood vessels.

In medicine

Computed tomography, developed by Godfrey Hounsfield in the early 1970s, reconstructs a cross-section from hundreds of X-ray transmission measurements taken around the same slice. Each pixel is labelled with a Hounsfield number derived from the attenuation coefficient, with water at 0 and dense bone near +1000. A scan of a head takes only a few seconds, but the dose to the patient is hundreds of times larger than that of a single radiograph, because every one of the many slices is a separate exposure.

In medicine

In PET scanning, fluorine-18 is attached to a glucose analogue and injected into the patient. Because fluorine-18 has a half-life of 110 minutes it must be made in a cyclotron close to the scanner, and the tracer must be used within a few hours. Cancer cells take up the labelled glucose most rapidly, and the annihilation photons they emit are detected in coincidence by a ring of detectors, so a tumour shows up as a region of high tracer concentration even when its anatomy looks normal on a CT scan.

In engineering

A weld or a metal casting is tested for internal flaws with an ultrasound A-scan at about 2 MHz to 5 MHz. A crack or a void is a metal-air boundary, so its specific acoustic impedance differs enormously from that of the steel and almost all of the ultrasound is reflected from it, giving a large echo spike at a time that fixes its depth. The test is non-destructive, so the component can go into service afterwards, and the same method is used to check the thickness of pipework from the outside.

In the laboratory

A school pulse-echo experiment uses one transducer coupled with oil to one face of a Perspex or steel block. The oscilloscope time base is triggered by the transmitted pulse, and the echo from the far face appears as a second pulse, so the speed of ultrasound in the block is 2 x thickness divided by the time between the two pulses. The echo from the near face shows that a small fraction of the pulse is reflected even where the two media are fairly well matched.

Flashcards

Click a card to flip it. Mark honestly — the tracker uses this.

Q1What is the piezo-electric effect?
A piezo-electric crystal changes shape when a p.d. is applied across it, and generates an e.m.f. across it when its shape is changed. Because the effect works both ways, one crystal can transmit a pulse and then detect the echo.
Q2Why is coupling gel spread between an ultrasound transducer and the skin?
Air has a specific acoustic impedance of about 4.3 x 10^2 kg m-2 s-1 and soft tissue about 1.6 x 10^6 kg m-2 s-1, so a thin air layer would reflect about 99.9% of the ultrasound. The gel has an impedance close to that of soft tissue, so the mismatch is removed and the pulse enters the body.
Q3Define the specific acoustic impedance of a medium.
Z = rho c, the product of the density of the medium and the speed of sound in it, measured in kg m-2 s-1. It is a measure of how strongly the medium resists being vibrated.
Q4Give the equation for the intensity reflection coefficient of a boundary between two media.
IR / I0 = (Z1 - Z2)^2 / (Z1 + Z2)^2, where Z1 and Z2 are the specific acoustic impedances of the two media.
Q5What is the intensity reflection coefficient at a fat-muscle boundary, and what does it tell you?
With Z = 1.4 x 10^6 and 1.7 x 10^6 kg m-2 s-1, IR / I0 = (3.0 x 10^5)^2 / (3.1 x 10^6)^2 = 9.4 x 10-3, so only about 1% of the intensity is reflected. Echoes from soft-tissue boundaries are very weak, so the receiver must be very sensitive.
Q6State the equation for the attenuation of ultrasound or X-rays in matter, and say what each symbol means.
I = I0 e^(-mu x). I0 is the incident intensity, I is the intensity after travelling a distance x through the medium in metres, and mu is the linear attenuation coefficient in m-1. The product mu x has no unit.
Q7How is the depth of a reflecting boundary found from a pulse-echo measurement?
Depth = c t / 2, where c is the speed of sound in the tissue and t is the time between transmitting the pulse and receiving the echo, because the pulse travels to the boundary and back again.
Q8What is the difference between an A-scan and a B-scan?
An A-scan shows the amplitude of each echo as the height of a spike against time and gives the depths of boundaries along one fixed line. A B-scan shows the amplitude as the brightness of a spot and is taken with the transducer moved across the body, so a two-dimensional cross-section is built up.
Q9How does the attenuation of ultrasound depend on frequency, and what follows from that?
The attenuation coefficient increases with frequency, so a higher frequency penetrates less deeply. It does, however, give a shorter wavelength and therefore better resolution, so the frequency chosen is a compromise between resolution and depth.
Q10How are X-rays produced in an X-ray tube?
Electrons are emitted by thermionic emission from a heated tungsten filament, accelerated through a high p.d. and stopped by a heavy metal target such as tungsten, where their kinetic energy is converted mainly into heat and partly into X-rays. The rotating anode and oil cooling stop the target melting.
Q11Give the expression for the minimum wavelength of the X-rays produced in a tube, and say what it depends on.
lambda_min = hc / (eV), from eV = hc / lambda_min for one electron giving all of its kinetic energy to one photon. It depends only on the accelerating p.d. V, not on the material of the target.
Q12What is meant by contrast in X-ray imaging, and what makes it large?
Contrast is the difference in transmitted X-ray intensity between two parts of the image. It is large when the tissues have very different values of mu x, which happens when they differ in density or in atomic number, so bone gives good contrast against soft tissue and a high-atomic-number contrast medium is needed to show one soft tissue against another.
Q13What is a tracer?
A substance containing radioactive nuclei that is introduced into the body and absorbed by the tissue being studied, so that the tissue can be imaged from outside by detecting the radiation it emits. A PET tracer decays by beta-plus decay.
Q14In PET scanning, what is produced when a positron annihilates with an electron, and why?
Two gamma-ray photons, each of energy about 5.1 x 10^5 eV (511 keV), travelling in exactly opposite directions. Mass-energy conservation fixes the total energy released as 2 m c^2, and momentum conservation requires the two photons to carry equal and opposite momenta, so a single photon is impossible.
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Exam-style questions

9 questions, 31 marks. Write your answer, then reveal the mark scheme and tick each point you actually made. Answering honestly here is what makes the rest of the site useful.

StateAO12 marks
State what happens when a potential difference is applied across a piezo-electric crystal, and what happens when the shape of such a crystal is changed.

Mark scheme — tick each point you actually made

Examiner tip: The effect works both ways, and that is the point of the question. One mark for each direction, so do not give the same idea twice.
CalculateAO24 marks
Ultrasound of frequency 2.0 MHz (2.0 x 10^6 Hz) travels through soft tissue at a speed of 1540 m s-1. (a) Calculate the wavelength of the ultrasound in the tissue. (b) A pulse is transmitted into the tissue and the echo from a boundary between two tissues is received 26 microseconds later. Calculate the depth of the boundary below the surface.

Mark scheme — tick each point you actually made

Examiner tip: The pulse goes down to the boundary and back, so halve the distance the pulse has travelled. Convert MHz to Hz and microseconds to seconds before you substitute, and give every answer with its unit.
ExplainAO23 marks
Explain why a coupling gel is spread between an ultrasound transducer and the patient's skin before a scan is taken.

Mark scheme — tick each point you actually made

Examiner tip: Quote the impedance mismatch and its consequence. Saying only that the gel 'helps the waves get through' without mentioning acoustic impedance will not score full marks.
CalculateAO24 marks
At a boundary between fat and muscle the specific acoustic impedances are 1.4 x 10^6 kg m-2 s-1 and 1.7 x 10^6 kg m-2 s-1. Calculate the intensity reflection coefficient of the boundary, and state the fraction of the incident intensity that is transmitted.

Mark scheme — tick each point you actually made

Examiner tip: Square the difference and square the sum separately, then divide. Only about 1% of the intensity is reflected, which is the reason that echoes from boundaries between soft tissues are so weak.
CalculateAO23 marks
Ultrasound of intensity 5.0 x 10-3 W m-2 enters a tissue whose linear attenuation coefficient is 12 m-1. Calculate the intensity of the ultrasound after it has travelled 8.0 cm through the tissue.

Mark scheme — tick each point you actually made

Examiner tip: mu x must have no unit, so convert 8.0 cm to 0.080 m. Give the intensity with its unit, W m-2.
CalculateAO24 marks
X-rays are produced in a tube when electrons are accelerated through a potential difference of 80 kV and strike a metal target. Calculate the minimum wavelength of the X-rays produced.

Mark scheme — tick each point you actually made

Examiner tip: Use the accelerating p.d. in volts: 80 kV = 8.0 x 10^4 V. Every photon produced has this wavelength or longer, and the cut-off does not depend on the target material.
SuggestAO23 marks
Suggest why a bone shows up much more clearly than a muscle of the same thickness in an X-ray image.

Mark scheme — tick each point you actually made

Examiner tip: Contrast is a difference in the transmitted intensity, not simply a difference in appearance. Link the larger attenuation coefficient of bone to the larger difference in the intensity reaching the detector.
DescribeAO33 marks
A student uses a single ultrasound transducer and an oscilloscope to determine the speed of ultrasound in a block of Perspex. Describe the measurements the student should take and how the speed is then calculated.

Mark scheme — tick each point you actually made

Examiner tip: The pulse passes through the block and back, so the distance is twice the thickness. Name every quantity in the equation and give its unit.
ExplainAO25 marks
A tracer that decays by beta-plus decay is used in positron emission tomography (PET) scanning. Explain how the gamma-ray photons used to form the image are produced and how they are used to show where the tracer has collected in the tissue.

Mark scheme — tick each point you actually made

Examiner tip: Quote the two conservation laws separately: mass-energy conservation for the energy of the photons, and momentum conservation for the 180 degrees between them. State that it is the gamma-ray photons, not the positrons, that leave the body.

Exam technique for this topic

  • In any pulse-echo calculation, remember that the pulse travels to the boundary and back. The time gives twice the depth, so the depth is c t / 2. Candidates who forget the factor of two lose a mark more often than for any other reason on this topic.
  • Convert before you substitute: MHz to Hz (2.0 MHz = 2.0 x 10^6 Hz), microseconds to seconds, cm to m. Then check that mu x has no unit, because getting that wrong always shows up as a unit error in the answer.
  • For the intensity reflection coefficient, square the difference and square the sum, then divide. Do not square the whole fraction, and do quote the answer as a fraction or percentage - the mark scheme expects a comment that about 1% reflection at a tissue-tissue boundary is why the receiver must be very sensitive.
  • In a question on the minimum wavelength of X-rays, start from eV = hc / lambda_min and say that one electron gives all of its kinetic energy to one photon. State that the cut-off depends only on the accelerating p.d., while the target element fixes the wavelengths of the characteristic lines.
  • For PET, quote the two conservation laws separately: mass-energy conservation gives the energy of each photon as m e c^2 (about 5.1 x 10^5 eV), and momentum conservation gives the 180 degrees between them. Say clearly that it is the gamma-ray photons, not the positrons, that leave the body and are detected.
  • Never write that a positron is an electron, or that a beta-plus particle is the same as a beta-minus particle. A positron carries a positive charge; the annihilation, and every calculation that follows from it, depends on that.

Why students lose marks here

A student wroteA student wrote: 'A boundary is 3.0 cm below the surface of the tissue and the speed of ultrasound is 1540 m s-1, so the echo is received 0.030 / 1540 = 19 microseconds after the pulse is transmitted.'
Why it loses marksThe pulse does not stop at the boundary. It travels down to the boundary and back again, so the path length is twice the depth. Using the depth as the path length halves the calculated time.
The answer that scoresThe path length is 2 x 0.030 = 0.060 m, so the time is 0.060 / 1540 = 3.9 x 10-5 s = 39 microseconds. Going the other way, a measured time of 39 microseconds corresponds to a depth of 1540 x 39 x 10-6 / 2 = 0.030 m = 3.0 cm, which agrees. Always write depth = c t / 2 and substitute the time, never the depth.
A student wroteA student wrote: 'In PET scanning the positrons emitted by the tracer travel out of the body and are detected by the ring of detectors, so the position of the tracer can be worked out.'
Why it loses marksA positron is a charged particle, so it interacts strongly with the tissue it passes through and has a range of only about a millimetre. It annihilates long before it reaches the surface, and it is the two gamma-ray photons produced in the annihilation that escape and are detected.
The answer that scoresThe positron travels about 1 mm through the tissue and then annihilates with an electron. The annihilation converts the two rest masses into two gamma-ray photons of about 5.1 x 10^5 eV each, travelling in opposite directions. Those photons are penetrating enough to leave the body, and a coincidence recorded by two detectors on opposite sides of the patient shows that an annihilation took place somewhere on the line between them.
A student wroteA student wrote: 'An X-ray beam loses half of its intensity in the first 2 cm of bone, so after 4 cm of bone the intensity is zero because all of it has been absorbed.'
Why it loses marksAttenuation is exponential, not a fixed loss per centimetre. Each successive layer of material transmits the same fraction of what reaches it, so the intensity falls by an ever smaller amount and never reaches zero.
The answer that scoresUse I = I0 e^(-mu x). With mu = 60 m-1, 2.0 cm gives I / I0 = e^(-1.2) = 0.30, so 30% is transmitted, and 4.0 cm gives I / I0 = e^(-60 x 0.040) = e^(-2.4) = 0.091, so 9.1% is transmitted. The intensity is never zero, because e^(-mu x) only approaches zero as x becomes very large.
The X-ray tube and the ultrasound pulse-echo techniqueX-rays are produced when fast electrons are stopped by a heavy target. Ultrasound images from the time taken for a pulse to travel toa boundary and back.The X-ray tubeevacuated glass tubecathode (heated filament)electron beamaccelerating voltage betweencathode and anode: 100 kVanode (tungsten target)the anode rotates so thatthe target does not meltX-ray beamthe X-rays leave through a thin windowUltrasound: the pulse-echo techniquetransducer (a piezoelectric crystal)skinboundary: the acoustic impedance changesincident pulsereflected pulsepart is transmittedd = c t / 2, with c = 1540 m s-1 in soft tissue= 1540 x 52 x 10^-6 / 2 = 0.040 m = 4.0 cmvoltage (V)020406080100time (microseconds)time delay t = 52 microsecondsX-rays come from fast electrons stopped by the target: a continuous spectrum from deceleration plus sharp characteristic lines.The shortest wavelength has all the electron energy in one photon: h f max = e V, so lambda min = h c / (e V).Ultrasound: a pulse reflects where the acoustic impedance changes, so the depth is d = c t / 2, because the pulse travels there andback.A higher frequency gives better resolution but is absorbed more, so ultrasound of a few MHz is used at 1540 m s-1 in soft tissue.
X-ray imaging compared with ultrasound imaging: X-rays are produced in a tube and pass through the body, where their intensity falls exponentially as I = I0 e^(-mu x) with the path length, while an ultrasound transducer sends pulses that are partly reflected at boundaries between tissues.
A student wroteA student wrote: 'A higher ultrasound frequency gives a clearer image because it is not attenuated as much, so more of it reaches the deeper tissues.'
Why it loses marksThe attenuation coefficient increases with frequency, so a higher-frequency beam loses intensity faster, not more slowly. Its advantage is a shorter wavelength, which gives better resolution and lets two nearby boundaries be seen as separate echoes.
The answer that scoresLarger mu means a smaller I for the same value of x, so higher-frequency ultrasound penetrates less deeply. The benefit of the higher frequency is resolution: two boundaries closer together than about one wavelength merge into a single echo. Choosing a frequency is therefore a compromise between resolution and depth - about 2 MHz to 5 MHz for an abdomen, but 10 MHz to 20 MHz for an eye.