Revision notes · Topic 24 of 25 · A2

Medical physics

Ultrasound, X-rays and PET: the physics behind seeing inside the body.

Syllabus 24.1 to 24.3 Tier A Level · A2 Prepared by the TheLucidSTEM team

§ 24.1 Production and use of ultrasound

Key ideas
  • A piezoelectric transducer both generates ultrasound (an alternating p.d. makes the crystal vibrate) and detects the echoes (vibration induces an e.m.f.).
  • The pulse-echo technique times reflections from tissue boundaries to find their depth.
  • The fraction reflected depends on the mismatch of acoustic impedances Z = ρc; large mismatches reflect strongly.
  • A coupling gel removes the air gap (almost total reflection at skin/air) so the pulse can enter the body.
Equations
Z = ρ cspecific acoustic impedancekg m⁻² s⁻¹
I_R/I₀ = (Z₂−Z₁)²/(Z₂+Z₁)²intensity reflection coefficientno unit
I = I₀ e^(−μx)attenuation through a thickness xno unit
transducer pulse in echo back tissue boundary
Fig. 1 · Pulse-echo: the transducer emits a pulse and then receives the reflection from a boundary; the round-trip time gives the depth as d = ½ct.
Watch out: the echo travels there and back, so the depth is d = ½ct, not ct. Forgetting the factor of a half doubles every depth.

§ 24.2 Production and use of X-rays

Key ideas
  • X-rays are produced when fast electrons are stopped by a metal target; most of their energy becomes heat, a little becomes X-ray photons.
  • Contrast comes from different tissues attenuating the beam by different amounts: dense bone absorbs far more than soft tissue.
  • CT scanning combines many X-ray images from different angles into a 3D picture.
Equations
I = I₀ e^(−μx)attenuation; μ is the linear attenuation coefficientno unit
Watch out: X-ray attenuation is exponential, not linear. Doubling the thickness does not double the absorption; it squares the transmitted fraction.

§ 24.3 PET scanning

Key ideas
  • A tracer emitting positrons (β⁺) is given to the patient and gathers where it is metabolised.
  • Each positron annihilates with an electron, producing two gamma photons that fly off in exactly opposite directions.
  • A ring of detectors records the pair in coincidence; the line joining them locates the annihilation, and many lines build the image.
Equations
e⁺ + e⁻ → 2γeach photon 0.511 MeV, emitted back-to-backMeV
γ γ detector ring annihilation
Fig. 2 · PET: a positron annihilates with an electron, sending two 0.511 MeV gamma photons in opposite directions to detectors on the ring, fixing the line they came from.
Watch out: the two gamma photons leave in opposite directions (180° apart) to conserve momentum. This back-to-back pair is what lets the scanner place the annihilation on a straight line.