24. Medical physics A2

An alternating p.d. therefore drives alternating deformation; an incoming mechanical vibration produces an alternating electrical signal.

Choose the drive frequency equal to the crystal's natural frequency in the ultrasound range (>20 kHz). Resonance gives large-amplitude vibration and a stronger ultrasound pulse.

For c=1540 m s⁻¹ and echo delay t=80 μs, d=(1540)(80×10⁻⁶)/2=6.16×10⁻² m=6.16 cm.

Echo timing locates boundaries; echo intensity gives information about impedance contrast and therefore boundary type. Repeating along many directions builds a cross-sectional image.

Coupling gel removes the air gap at skin, reducing the severe reflection that an air-tissue impedance mismatch would cause.

With ρ in kg m⁻³ and c in m s⁻¹, Z has unit kg m⁻² s⁻¹.

For tissue with ρ=1060 kg m⁻³ and c=1540 m s⁻¹, Z=(1060)(1540)=1.63×10⁶ kg m⁻² s⁻¹.

It is the contrast between Z values on the two sides of a boundary—not either density alone—that determines the reflected intensity fraction.

If Z₁=Z₂, R=0 and ideally no intensity reflects. If the impedances are very different, R approaches 1 and little intensity transmits.

For water Z₁=1.48×10⁶ and steel Z₂=40.4×10⁶ kg m⁻² s⁻¹, R=[(40.4−1.48)/(40.4+1.48)]²=0.864: about 86.4% reflects.

I₀ is intensity before a path length x, I is intensity after it, and μ is the linear attenuation coefficient. μ has reciprocal-length units matching x.

If I=0.62I₀ after x=2.1 cm, μ=−ln(0.62)/2.1=0.23 cm⁻¹.

The transmitted fraction is I/I₀; the fraction attenuated is 1−I/I₀. For μ=0.053 cm⁻¹ and x=9.3 cm, 39% is attenuated.

For an echo from depth d in one tissue, propagation covers approximately x=2d before adding boundary-reflection factors; each tissue layer contributes its own μx.

For V=58 kV, λmin=hc/eV=(6.63×10⁻³⁴)(3.00×10⁸)/[(1.60×10⁻¹⁹)(58×10³)]=2.14×10⁻¹¹ m=21.4 pm.

Most electron energy becomes thermal energy, so the target needs a high melting point and heat removal; tungsten is commonly suitable.

Bone has a much larger attenuation coefficient than many soft tissues, so fewer X-rays reach the detector behind bone. Bone and soft tissue therefore give good contrast; tissues with similar μ give poor contrast.

On traditional film, more transmitted X-rays produce greater blackening; digital displays may map detector signal to brightness differently, but contrast still comes from signal differences.

I/I₀ is the transmitted fraction. If 80% is absorbed, 20%=0.20 remains and belongs on the left of the exponential equation.

For muscle μ=0.22 cm⁻¹ and 80% absorption: 0.20=e^(−0.22x), so x=−ln(0.20)/0.22=7.3 cm.

Equal thickness x through materials with μ=3.0 and 0.22 cm⁻¹ gives I/I₀=e^(−3.22x). If I/I₀=0.13, x=0.63 cm.

A computer combines the angular attenuation data to reconstruct a two-dimensional map/image of that section.

Repeat the angular scan for successive sections along an axis, then combine/stack the reconstructed 2D section images to produce a three-dimensional representation.

Compared with one projection radiograph, CT separates overlapping structures and gives 3D localization, but multiple X-ray exposures generally increase ionising-radiation dose.

Its emitted radiation can be detected from outside the body, so the spatial distribution of tracer reveals where that labelled substance is taken up or metabolically active.

A useful tracer behaves chemically like the substance of interest, gives detectable radiation and has a half-life long enough for the scan but short enough to limit unnecessary dose.

The positron is the antiparticle needed for the later electron-positron annihilation that produces detectable gamma photons.

If the electron-positron pair has negligible total momentum, one photon alone cannot leave zero final momentum. Two equal photons travelling in opposite directions give equal and opposite momenta.

Their combined rest energy 2mec² becomes photon energy (plus any initial kinetic energy). Matter has not vanished without accounting; its mass-energy has changed form.

The electron and positron mass-energy becomes a pair of gamma-ray photons emitted in approximately opposite directions so that momentum is conserved.

Gamma photons are penetrating enough for many to leave the body and reach detectors around the patient; the scanner detects the paired event rather than the positron itself.

Momentum conservation gives two equal, opposite photons, so each photon receives half the total rest energy: Eγ=mec².

Eγ=(9.11×10⁻³¹)(3.00×10⁸)²=8.20×10⁻¹⁴ J.

Eγ=(8.20×10⁻¹⁴)/(1.60×10⁻¹⁹)=5.12×10⁵ eV=0.512 MeV (usually quoted as 0.511 MeV). Total photon energy is about 1.02 MeV.

The two hit detectors define a line of response on which the annihilation occurred. A difference in arrival times places the event closer to the detector reached first.

Processing many event lines and timing differences reconstructs event locations. Regions with more events contain a greater concentration/uptake of tracer and are displayed more strongly.

The PET image is primarily a functional map of tracer concentration or metabolic activity, not simply a direct map of tissue density.