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A Level Physics 9702 · A Level · Topic 24 · ~5 min read

Medical physics

Ultrasound production by piezoelectric crystals, acoustic impedance and reflection at boundaries, A-scans; X-ray production, attenuation and CT scanning; and PET scanning with tracers and electron–positron annihilation.

Piezoelectric transducersAcoustic impedance Z = ρcIntensity reflection coefficientCoupling gelAttenuation I = I₀e^{−μx}X-ray productionImage sharpness & contrastCT scanning
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedDATA & FORMULAE given in the examC1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Ultrasound (above 20 kHz, typically MHz) is made and detected by a piezoelectric crystal. Echoes from boundaries between tissues build up an image.
  • Reflection depends on acoustic impedance Z = ρc. The intensity reflection coefficient is (Z₂ − Z₁)²/(Z₂ + Z₁)²: a big difference in Z means a strong echo.
  • X-rays and ultrasound are both attenuated exponentially: I = I₀e−μx.
  • PET: a β⁺ tracer emits positrons, which annihilate with electrons to produce two 511 keV γ-ray photons travelling in opposite directions.

By the end you can

  • Explain the production and detection of ultrasound with a piezoelectric transducer.
  • Use Z = ρc and the intensity reflection coefficient, and explain the need for a coupling gel.
  • Use I = I₀e−μx for ultrasound and X-rays.
  • Explain X-ray production, image sharpness and contrast, and CT scanning.
  • Explain PET scanning, including tracers and annihilation.

1Ultrasound

Production and reflection

  • Piezoelectric effect: an alternating p.d. makes the crystal vibrate (emitter); returning pressure waves make it produce a p.d. (receiver). Pulses are sent and echoes timed.
Z = ρc ; I_R/I₀ = (Z₂ − Z₁)²/(Z₂ + Z₁)² DATA & FORMULAE
c = speed of sound in the medium
Worked example

Explain, with numbers, why gel is used. Zair = 430 kg m⁻² s⁻¹, Zskin = 1.7 × 10⁶, Zgel = 1.5 × 10⁶.

  1. air–skin: (1.7 × 10⁶ − 430)²/(1.7 × 10⁶ + 430)² = 0.999, so almost all reflected C1 A1
  2. gel–skin: (0.2/3.2)² = 0.0039, so almost all transmitted A1
t / μsecho204060801000skin–fatfat–musclemuscle–boneΔt = 42 μsDepth between boundaries = v × Δt/2 (the pulse goes there and back).Smaller echoes: less reflected and more attenuated.
A-scan: each boundary gives an echo; the time delay gives the depth.
Worked example

Echoes from two boundaries are 42 μs apart. The speed of sound in tissue is 1540 m s⁻¹. Find the distance between the boundaries.

  1. distance = v × Δt/2 C1
  2. = 1540 × 42 × 10⁻⁶/2 = 3.2 cm A1
Common mistakeForgetting to halve the time (the pulse travels there and back).

2X-rays

Production and attenuation

  • Production: electrons accelerated through a large p.d. hit a metal target (tungsten); a tiny fraction of their energy becomes X-rays (braking radiation plus characteristic lines). The maximum photon energy is eV.
  • Sharpness: a small focal spot on the target. Contrast: large differences in μ between tissues (bone vs soft tissue), or a contrast medium.
I = I₀e−μx DATA & FORMULAE
μ = linear attenuation coefficient; half-value thickness x½ = ln 2/μ
thickness xintensity II₀I₀/2I₀/4x½2x½I = I₀ e^(−μx)x½ = ln 2/μX-ray attenuation: each half-value thickness halves the intensity.
Each half-value thickness halves the intensity.
Worked example

X-rays pass through 2.0 cm of bone with μ = 60 m⁻¹. Find the fraction transmitted and the half-value thickness.

  1. I/I₀ = e−60 × 0.020 = e−1.2 = 0.30 C1 A1
  2. x½ = ln 2/60 = 1.2 cm A1
  • CT scanning: an X-ray tube and detectors rotate around the patient, taking many images of a thin slice. A computer combines them into a 3D image built from voxels.

3PET scanning

Tracers and annihilation

patientγ (511 keV)γ (511 keV)ring of detectorstracer emits β⁺e⁺ + e⁻ → 2γ(annihilation)γ-rays opposite ways,detected at the same time⇒ source on that line
Two 511 keV photons leave in opposite directions; detecting both at once locates the source.
Worked example

Show that each photon from electron–positron annihilation (at rest) has energy about 0.51 MeV, and find its wavelength.

  1. 2mc² shared between two photons, so E = mc² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.2 × 10⁻¹⁴ J C1
  2. = 0.51 MeV A1
  3. λ = hc/E = 2.4 × 10⁻¹² m A1
  • Tracer: a β⁺ emitter (e.g. fluorine-18 in a glucose-like molecule) collects where cells are most active. Short half-life keeps the dose low.
  • Why two photons in opposite directions: momentum is conserved (it was about zero before annihilation).
Common mistakeSaying the γ-rays come from the tracer. The tracer emits positrons; the γ-rays come from annihilation. 2024 Paper 4 candidates often missed this step.
Linking questionExponential attenuation matches the decay equations in Nuclear physics; β⁺ decay and antiparticles are in Particle physics; and photon energies are in Quantum physics.

Test yourself — cover the answers first

Quick check

  1. What does a piezoelectric crystal do in an ultrasound probe?
  2. Define acoustic impedance.
  3. Why is coupling gel used?
  4. What particle does a PET tracer emit?
  5. Find the half-value thickness for μ = 0.35 cm⁻¹.

Exam-style

  1. Muscle has ρ = 1080 kg m⁻³ and c = 1580 m s⁻¹. Find Z.
  2. X-ray intensity falls to 20% through 3.0 cm of tissue. Find μ.
  3. An X-ray tube uses an accelerating p.d. of 80 kV. Find the minimum wavelength produced.
Show answers
  1. converts an alternating p.d. into vibrations, and echoes back into a p.d.
  2. Z = ρc (density × speed of sound)
  3. to avoid almost total reflection at the air–skin boundary
  4. a positron (β⁺)
  5. 1.98 cm
  6. 1.71 × 10⁶ kg m⁻² s⁻¹
  7. ln 5/0.030 = 54 m⁻¹
  8. E = 80 keV = 1.28 × 10⁻¹⁴ J ⇒ λ = hc/E = 1.55 × 10⁻¹¹ m
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.