Medical Physics

Keep the two directions separate: the piezoelectric effect is mechanical → electrical (deformation generates an e.m.f., used to DETECT); the converse piezoelectric effect is electrical → mechanical (a p.d. causes deformation, used to GENERATE) — 'explain how detected' answers that describe the converse effect (or vice versa) lose marks

It is the SAME crystal doing both jobs — one transducer alternately transmits a short pulse (converse effect) then listens for the echo (direct effect); it is not two different crystals

'Explain how ultrasound is generated' needs the FULL chain, not just 'the p.d. makes it vibrate': an alternating p.d. is applied across the crystal, this makes it change shape (vibrate) at the frequency of the applied p.d., and when that frequency equals the crystal's own natural (resonant) frequency — which lies in the ultrasound range — the crystal vibrates strongly, radiating ultrasound into the coupling medium

'Explain how ultrasound is detected' needs BOTH linked ideas for full marks: the incoming (reflected) wave makes the crystal vibrate/change shape, AND that changing shape generates an alternating e.m.f./p.d. across the crystal — stating only one half of the chain loses marks

'Define specific acoustic impedance' needs BOTH marking points: state it as the PRODUCT of density and speed, AND specify that the speed is the speed of ultrasound IN THAT MEDIUM — 'density × speed' alone, with no reference to which speed, drops the second mark

Z = ρc is a property of ONE medium on its own — it is not itself 'how much reflects'; that comparison ACROSS a boundary is §03's intensity reflection coefficient, built directly on top of Z

The units of Z are kg m⁻² s⁻¹ (density's kg m⁻³ × speed's m s⁻¹) — never quote Z in the units of density or speed alone, and never quote it with no units at all

In a time-of-flight calculation, the measured time t is for the ROUND TRIP (pulse out, echo back) — always halve c×t to get the depth of the boundary; using the full c×t as the depth is exactly double the correct answer

A LATER echo means a DEEPER boundary — different boundaries at different depths along one line of sight return their echoes at correspondingly different times, and it is exactly this ordering (not just the single-boundary formula) that a 2D B-scan image is built from

Reflection and attenuation are DIFFERENT physical causes of a weak signal — reflection needs a boundary and depends on the impedance mismatch Z₁, Z₂; attenuation happens continuously along the path (absorption + scattering) even through uniform tissue with no boundary at all; a real echo is weakened by both together

α = (Z₁−Z₂)²/(Z₁+Z₂)² always needs BOTH the top and the bottom squared — this is what makes α positive and independent of which medium is labelled 1; α = 0 when Z₁ = Z₂ (why gel works), and α → 1 for a large mismatch (why an air gap ruins a scan)

'1 − α' converts a reflection coefficient into the TRANSMITTED fraction (since I₀ = I_R + I_T) — a very common final step, easy to skip when the question only ever mentions α

For the ln-rearrangement of I = I₀e⁻ᵘˣ, always form the ratio I/I₀ FIRST, then take ln, then isolate μ or x — never take ln of I and I₀ separately or round the ln value before the final line

For multiple layers of different material, sum each layer's OWN μx product before exponentiating (μ₁x₁ + μ₂x₂ + …) — never add the bare μ values or ignore each layer's own thickness

'Percentage remaining/transmitted' (I/I₀ × 100) and 'percentage attenuated' (100 × (1 − I/I₀)) are different numbers that add to 100% — reread the question's exact wording before writing the final answer

'Explain how X-rays are produced' needs the FULL chain: electrons are accelerated through a large p.d. (gaining kinetic energy eV), they hit the metal target, they decelerate suddenly (or stop) on impact, and this loss of kinetic energy is emitted as (X-ray) photons — stopping after 'electrons hit the target' with no mention of deceleration/energy loss leaves marks unclaimed

The CONTINUOUS spectrum (a whole range of wavelengths) and the CHARACTERISTIC spikes (a few fixed wavelengths) have two DIFFERENT physical causes — a continuous-spectrum answer talks about a RANGE of decelerations/energy losses (bremsstrahlung); a characteristic-peaks answer talks about an electron KNOCKING OUT an inner-shell electron and a specific atomic energy-level transition filling the gap — never explain one using the other's mechanism

λ_min is reached in the RARE single event where ONE electron loses its ENTIRE kinetic energy eV to ONE photon — using only a fraction of eV, or treating every electron as losing all its energy, both give the wrong physical picture (only λ_min itself corresponds to the maximum-energy, single-photon case; every other wavelength in the continuous spectrum corresponds to a PARTIAL energy loss)

λ_min = hc/(eV) depends ONLY on the accelerating p.d. V — changing the target material changes the POSITIONS of the characteristic peaks (a different element, different energy levels) but does NOT change λ_min at all; conflating the two is a common wrong answer to 'explain why λ_min is unaffected by the choice of target'

Always convert an accelerating p.d. given in kV to volts (×1000) before substituting into λ_min = hc/(eV) — a missed factor of 1000 is the single most common arithmetic slip in this calculation, exactly as for any eV-based Quantum Physics calculation

Contrast is defined as a DIFFERENCE (in transmitted/detected intensity, or in degree of blackening on the image) between two adjacent regions — never quote a single intensity or a single μ value as 'the contrast' on its own

'High contrast' and 'high intensity' are NOT the same thing — a region can transmit a large intensity and still show poor contrast, if the neighbouring region transmits an almost equally large intensity; what matters is the SIZE OF THE DIFFERENCE between the two, not either value alone

A bigger μ for one tissue does not automatically mean a good image — contrast depends on how DIFFERENT that μ is from the OTHER tissue's μ; two tissues can each have a large μ and still show poor contrast if their μ values are close to each other (see the real worked example below where a factor of only 2.5 between two regions is judged POOR contrast, despite a real difference in the materials' μ)

The linear attenuation coefficient μ for X-rays depends on BOTH the material (its density and atomic/proton number — high-Z materials like bone absorb strongly) AND the energy of the X-ray photons (a more penetrating, 'harder' beam has a lower μ in the same material) — never treat μ as fixed by the material alone

Soft tissues (e.g. blood and muscle) commonly have very SIMILAR μ values to each other — this is exactly why plain X-ray images are poor at showing soft-tissue-to-soft-tissue boundaries, and exactly why a contrast agent (a substance of deliberately very different μ, introduced to create an artificial mismatch) is used when that boundary needs to be seen

For an X-ray path crossing several different materials, apply the SAME multi-layer technique built in §03 — sum each layer's own μx product into one exponent (or multiply each layer's own transmitted fraction) — never add the bare μ values or ignore any one layer's own thickness

Don't conflate CT scanning's three separate stages — (1) many angled images of the SAME slice, (2) computer-combined into ONE 2D image of that slice, (3) REPEATED for many slices and the 2D images combined into a 3D image — 'explain/outline the principles' answers are marked as this fixed checklist, and jumping straight from 'many angles' to '3D image' with no mention of the 2D-per-slice step loses marks

CT scanning does not use a different KIND of X-ray or a different physical law from plain X-ray imaging — it is still the same beam obeying the same $I=I_0e^{-\mu x}$ attenuation from §05; what's new is taking MANY images from different angles of the same slice and COMBINING them by computer, not the underlying physics of how the X-rays interact with tissue

In the exam's back-projection toy model, the 'background' reading is simply the SAME total given (or stated) for every viewing direction — it needs no separate calculation, because every direction's rays partition the same fixed grid of voxels exactly once, so their total is always the sum of all the voxels

Always subtract the background from EACH of the four given (summed/back-projected) readings BEFORE dividing by 3 — dividing first, or subtracting the background only once from a running total, both give the wrong pixel values

The 'divide by 3' step is specific to the STANDARD exam setup of a 2×2 grid viewed from 4 directions (rows, columns, two diagonals) — it comes from each voxel appearing in exactly 4 direction-readings, 3 of which also contain exactly one other voxel; a differently-shaped grid or a different number of directions needs the equivalent constant re-derived the same way, never assumed to always be 3

Read voxel/pixel LABELS carefully off the given grid before placing corrected values into it — real exam grids don't always run clockwise (e.g. A, B along the top then D, C along the bottom, so D is bottom-LEFT, not bottom-right) — a correctly-calculated number placed in the wrong labelled position still loses the mark

The standard advantage of CT over a single plain X-ray image is a genuine 3D/depth image that can distinguish tissues too similar in μ to show up on a plain X-ray (§05, poor contrast); the standard disadvantage is a MUCH higher radiation dose to the patient, since a CT scan is built from many individual X-ray exposures rather than just one — both are frequently examined together as a simple 'state one advantage and one disadvantage' question

A 'tracer' definition needs BOTH marking points: a (radioactive) substance introduced into the body, AND that it is absorbed by (or its position/emission can be detected in) the tissue being studied — 'a radioactive substance' alone, with no mention of it being taken up by the tissue of interest, is an incomplete definition

PET specifically needs a tracer that decays by β⁺ decay (positron emission) — not any radioactive tracer will do; ordinary β⁻ or α decay produces no positron, and so no annihilation event to detect at all

'Annihilation' is a precise term with two required ideas, not one: a particle interacting with its OWN antiparticle (never just 'two particles colliding'), AND mass-energy converting into (photon) energy — stating only the collision, with no mention of mass becoming energy, is an incomplete answer

Annihilation converts the FULL rest mass of BOTH the positron and the electron into photon energy — use $E=2m_ec^2$ for the TOTAL energy released, never a mass difference or the mass of only one particle

$E_\gamma=m_ec^2$ is the energy of EACH photon, not the total — the single most common slip in this whole section is writing $E_\gamma=\tfrac12m_ec^2$ (halving twice) or $E_\gamma=2m_ec^2$ (forgetting to split the total between the two photons) instead of $E_\gamma=m_ec^2$

The 'why two photons, in exactly opposite directions' explanation needs the FULL momentum-conservation chain: total momentum before annihilation is (approximately) zero (both particles approximately at rest) → a single photon can never have zero momentum ($p=E/c$, always nonzero and in one direction) → so at least two photons, travelling in EXACTLY opposite directions with EQUAL momentum, are needed so their momenta cancel — quoting only 'momentum is conserved' with no explanation of why one photon fails is an incomplete answer

Both mass-energy conservation AND momentum conservation are needed for annihilation, and they answer DIFFERENT questions — mass-energy conservation ($E=mc^2$) fixes the TOTAL energy released and (by the equal-momentum argument) how it SPLITS between the two photons; momentum conservation fixes WHY there are two photons and WHY they travel in exactly opposite directions. A common error is invoking only one of the two laws to explain a fact that actually depends on both

The annihilation event happens INSIDE the body (wherever the positron meets an electron in the tissue) — it is the resulting GAMMA PHOTONS, not the positron or the annihilation event itself, that travel outside the body and are detected by the ring of detectors

'Coincidence' detection means two detectors on the ring register a photon at (very nearly) the SAME instant — this is what tells the scanner the two photons came from the SAME annihilation event, and hence that the event lies somewhere on the straight line joining those two particular detectors (the 'line of response')

A single detected coincident pair only locates the event to somewhere ALONG a line, not to one exact point — it is combining the lines of response from MANY annihilation events (by processing their arrival times) that lets a computer build up a full image of where the tracer is concentrated

PET images FUNCTION (where in the body a tracer substance is concentrated, e.g. showing unusually high metabolic activity) — this is a fundamentally different kind of information from ultrasound (§01–§03) or X-ray/CT (§04–§06), both of which image ANATOMY (physical structure); do not describe PET as 'a clearer picture of structure' when asked what it adds over X-ray or CT imaging