PhysicsCore24 min read

Medical Physics

Ultrasound, X-rays and PET — three ways of seeing inside without cutting

This topic appears in:

A Level Physics (9702) · ch.24
01

Ultrasound: listening to the echoes

Ultrasound is sound above the range of human hearing, typically 1 to 15 MHz in medical use. It is produced and detected by the same device: a piezoelectric transducer, a crystal that changes shape when a voltage is applied and generates a voltage when it is deformed. That reversibility is why one probe both sends and receives.

A pulse travels into the body, and at every boundary between different tissues part of it reflects. Timing those echoes gives depth, since the speed in tissue is known. The strength of each echo depends on how different the two tissues are, measured by their acoustic impedance.

acoustic impedanceZ = ρ creflection coefficient at a boundary:I_r / I_i = (Z₂ − Z₁)² / (Z₂ + Z₁)²depth from timing: d = ct / 2(there and back)the factor of 2 accounts for the round trip
Z
acoustic impedancedensity × speed of sound in that tissue
ρ, c
density and speedproperties of the tissue
d = ct/2
the depthhalved because the pulse travels there and back

Why the gel matters

The impedance of air is thousands of times smaller than that of skin, so almost all the ultrasound would reflect straight off the surface and never enter the body. A coupling gel with an impedance close to skin removes the air gap and lets the pulse in. This is impedance matching, and it is a standard exam question rather than a practical detail.

02

X-rays: attenuation and contrast

X-rays are produced by accelerating electrons through a large potential difference and stopping them abruptly in a metal target. Most of the energy becomes heat; a small fraction becomes X-ray photons, with the maximum photon energy set by the accelerating voltage.

Unlike ultrasound, X-rays are detected in transmission — the image is a shadow. Different tissues absorb different amounts, and the intensity falls exponentially with the thickness of material traversed. Bone attenuates far more than soft tissue, which is why bones show clearly and why soft-tissue contrast is poor without an added contrast medium.

I = I₀ e^(−µx)half-value thickness: x½ = ln2 / µ = 0.693 / µmaximum photon energy: E_max = eVµ depends on both the material and the photon energythe same exponential form as capacitor discharge and radioactive decay
µ
the attenuation coefficientlarger for denser, higher-Z material such as bone
x
thickness traversedin the same units as 1/µ
x½
half-value thicknessthe thickness that halves the intensity
Worked example

A beam of X-rays passes through 3.0 cm of tissue with attenuation coefficient 0.25 cm⁻¹. Find the fraction of the intensity transmitted, and the half-value thickness.

  1. I/I₀ = e^(−µx) = e^(−0.25 × 3.0) = e^(−0.75).Substituting into the attenuation equation; the units of µ and x must match.
  2. = 0.472, so about 47% is transmitted.Slightly under half, so the thickness must be slightly more than one half-value thickness — a useful consistency check.
  3. x½ = ln2/µ = 0.693/0.25.The half-value thickness follows directly from the attenuation coefficient.
  4. = 2.77 cm.Confirming the check: 3.0 cm is a little more than one half-value thickness, so a little under half gets through.

47% transmitted; x½ = 2.77 cm

03

PET: annihilation, and why it locates so precisely

Positron emission tomography works differently from both. A tracer — a molecule such as glucose labelled with a positron-emitting isotope — is injected and accumulates where metabolism is most active. This makes PET a functional scan, showing what tissue is doing rather than what it looks like.

Each emitted positron travels a very short distance before meeting an electron. The two annihilate, and their combined mass becomes energy as two gamma photons of 0.51 MeV each. Momentum conservation forces those photons to travel in almost exactly opposite directions.

That back-to-back emission is what makes the technique work. A ring of detectors around the patient records pairs arriving simultaneously, and the annihilation must have happened somewhere on the line joining them. Many such lines intersect at the source, and a computer reconstructs the distribution of the tracer.

annihilation: e⁻ + e⁺ → 2γenergy of each photon: E = mc²= (9.11 × 10⁻³¹)(3.00 × 10⁸)²= 8.20 × 10⁻¹⁴ J = 0.51 MeVposition along the line from the time difference:Δd = c Δt / 2each photon carries the rest energy of one electron
0.51 MeV
the photon energythe rest energy of an electron or positron
2γ
two photonsone alone could not conserve momentum
Δt
arrival time differencelocates the event along the line of response

The three techniques compared

  1. Ultrasound: reflection, non-ionising, safe in pregnancy, poor through bone or gas.
  2. X-ray: transmission shadow, ionising, excellent for bone, poor soft-tissue contrast.
  3. PET: emission from within, ionising, shows function rather than structure, expensive.
  4. Ultrasound and X-ray image structure; PET images activity.
  5. Only ultrasound is non-ionising, which is the deciding factor for obstetric scanning.
  6. All three exponentials — attenuation, discharge and decay — share the same mathematics.
04

Choosing the right technique

Exam questions in this topic usually describe a clinical situation and ask which method suits it, with reasons. The answer turns on three considerations: whether ionising radiation is acceptable, whether the target is bone or soft tissue, and whether the question is about structure or function.

Ultrasound is the only non-ionising option, which settles obstetric scanning immediately. It is also cheap and portable, but it cannot pass through bone or gas, so it is useless for imaging the brain in an adult or a lung.

X-rays excel at bone precisely because dense, high-atomic-number material attenuates strongly — the contrast that makes a fracture obvious is the same physics that makes soft tissue nearly invisible. PET, being expensive and requiring a radioactive tracer with a short half-life, is reserved for questions about metabolic activity that no structural image could answer.

SituationBest choiceDeciding reason
Scanning a fetusultrasoundnon-ionising, so no risk to the fetus
Suspected fractured wristX-raybone attenuates far more than tissue
Finding an active tumourPETshows metabolic activity, not just shape
Imaging a moving heart valveultrasoundreal-time and safe for repeated use
Imaging the adult brainnot ultrasoundthe skull reflects almost all of it

X-ray attenuation follows the same exponential absorption shown here: equal thicknesses of material remove equal fractions of the beam, which is why the half-value thickness is a constant for a given material and photon energy.

Practice questions

5 questions · 17 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Explain why a coupling gel is used between an ultrasound transducer and the skin.
Model answer

Air has a far lower acoustic impedance than skin, so at an air–skin boundary almost all the ultrasound would be reflected and very little would enter the body. The gel has an impedance close to that of skin, which removes the air gap and allows the pulse to be transmitted.

Examiner tip. Both the large impedance mismatch and the resulting near-total reflection are needed.

SQ2[2 marks]
Explain why two gamma photons, rather than one, are produced when a positron and an electron annihilate.
Model answer

Momentum must be conserved. The electron and positron have almost no momentum before the event, so the total afterwards must be near zero. A single photon would carry momentum in one direction, which is impossible — so two photons are emitted in almost exactly opposite directions.

Examiner tip. The answer is momentum conservation. Energy conservation alone would permit a single photon.

SQ3[2 marks]
State what is meant by the half-value thickness, and explain why it does not depend on the initial intensity.
Model answer

The half-value thickness is the thickness of a material that reduces the transmitted intensity to half its incident value. It is independent of the initial intensity because the attenuation is exponential: equal thicknesses remove equal fractions, not equal amounts.

Examiner tip. The same reasoning applies to radioactive half-life and to capacitor discharge — all three are exponential.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
An ultrasound pulse returns from a boundary after 65 µs. The speed of sound in the tissue is 1540 m s⁻¹. Find the depth of the boundary. Also calculate the acoustic impedance of tissue of density 1060 kg m⁻³.
Full working
  1. Distance travelled = ct = 1540 × 65 × 10⁻⁶ = 0.1001 mThe total path length, there and back.[1]
  2. Depth = 0.1001/2 = 0.050 m = 5.0 cmHalving for the round trip is the step most often forgotten.[1]
  3. Z = ρc = 1060 × 1540Acoustic impedance is density times speed.[1]
  4. Z = 1.63 × 10⁶ kg m⁻² s⁻¹Typical of soft tissue, and far greater than air.[1]

Depth 5.0 cm; Z = 1.63 × 10⁶ kg m⁻² s⁻¹

Exam questions

1 · 7 marks

Multi-part questions with a full mark scheme.

Q1[7 marks]
(a) State one advantage and one disadvantage of ultrasound compared with X-ray imaging.
(b) X-rays of intensity I₀ pass through 4.0 cm of bone with µ = 0.60 cm⁻¹. Calculate the fraction transmitted.
(c) Calculate the half-value thickness of the bone.
(d) Explain why PET is described as a functional rather than a structural scan.
(e) Calculate the energy in MeV of each annihilation photon, given the electron mass is 9.11 × 10⁻³¹ kg.
Mark scheme
  1. (a) Advantage: ultrasound is non-ionising, so it is safe for a fetus. Disadvantage: it gives poorer resolution and cannot image through bone or gas.One of each is required; both must be genuine comparisons.[1]
  2. (b) I/I₀ = e^(−0.60 × 4.0) = e^(−2.4)Units of µ and x are consistent in cm.[1]
  3. = 0.0907, about 9.1%Bone attenuates strongly, which is why it appears white on a radiograph.[1]
  4. (c) x½ = ln2/µ = 0.693/0.60 = 1.16 cm4.0 cm is nearly 3.5 half-value thicknesses, consistent with roughly 9% transmitted.[1]
  5. (d) The tracer accumulates where metabolic activity is highest, so the image shows how tissue is functioning rather than its shape.The contrast with structural imaging must be explicit.[1]
  6. (e) E = mc² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.20 × 10⁻¹⁴ JEach photon carries the rest energy of one electron.[1]
  7. E = 8.20 × 10⁻¹⁴ / 1.60 × 10⁻¹³ = 0.51 MeVConverting joules to MeV by dividing by 1.60 × 10⁻¹³.[1]

(b) 9.1%; (c) 1.16 cm; (e) 0.51 MeV