Medical Physics: X-Ray, CT, Ultrasound & Tracers

Medical imaging is one of physics' quietest triumphs. A modern CT scanner rotates its X-ray tube around you in under a second and reconstructs a hundred slices; with iodine contrast it traces a blocked coronary artery without an incision. PET fused with CT overlays function on anatomy, lighting up an active tumour against the organs around it, and has reshaped how cancers are staged. Ultrasound has become the stethoscope of the twenty-first century — cheap, portable, harmless, now shrunk to a probe that plugs into a phone and images a heart at the bedside. The frontier now is computational. The physics of transmission, reflection, and emission is settled; what advances fast is the reconstruction — machine-learning algorithms that build a clean image from a lower dose, de-noise a fast ultrasound sweep, or spot a lesion a tired eye might miss. Hounsfield's revolution was to realise that an image need not be captured directly — it can be computed from indirect measurements, an idea, born in a lab funded by pop records, that still drives the field.

The body is not transparent, but it is not opaque either: different tissues let different kinds of radiation through by different amounts, and that difference is the image. A picture forms wherever one structure absorbs, reflects, or emits differently from its neighbour — contrast, not brightness, is what a radiologist reads. Imaging comes down to finding a signal that bone treats differently from muscle, or a tumour differently from healthy tissue. There are exactly four ways to get such a signal, and every scanner is one of them. Send radiation through the body and record what survives — that is transmission, what a plain X-ray film does. Send it through from many angles and let a computer reconstruct a true slice — that is computed tomography. Send a pulse in and listen for what bounces back — that is ultrasound, reflection imaging. Or put a source inside the body, as an injected tracer, and record what comes out — that is radionuclide (nuclear-medicine) imaging. Transmission, reconstruction, reflection, emission: four windows, four pieces of physics.

Two of these windows use ionising radiation — X-ray, CT, and the tracer's gamma rays carry enough energy per photon to knock electrons out of atoms, so they deliver a dose and are used with care. Two do not: ultrasound is a mechanical (sound) wave and the electrocardiogram merely listens to the heart's own voltages. This ionising-versus-not divide governs which modality a doctor reaches for — you would never CT a healthy fetus but would scan it with ultrasound, precisely because ultrasound deposits no ionising dose. It is the single most examined concept in the whole topic.

An X-ray image is a shadow. Behind the patient sits a detector (film, or a flat digital panel); in front, an X-ray tube fires a broad beam of high-energy photons — diagnostic energies of about $30$ to $120\ \text{keV}$, produced as in the previous article on X-ray production. As the beam crosses the body, every tissue removes some photons (by the photoelectric effect and Compton scattering), and dense, high–atomic-number material such as bone removes far more than soft tissue of the same thickness. So the beam reaching the detector is strong where it passed through flesh and weak where it passed through bone — and the developed image is the negative of that: bone prints white, air-filled lung dark, soft tissue a mid grey. The picture maps how much of the beam each straight-line path let through. How fast does a beam die away inside a uniform absorber? For a narrow monoenergetic beam the answer is the exponential attenuation law, which we quote — it has the same standing, and the same shape, as the radioactive-decay law $N = N_0 e^{-\lambda t}$ you already use in nuclear physics:

where: $I_0$ is the beam intensity entering the absorber, $I$ the intensity after a depth $x$, and $\mu$ the linear attenuation coefficient (units $\text{m}^{-1}$ or $\text{cm}^{-1}$), a property of both the material and the photon energy. The physical reading: each thin layer removes the same fraction of whatever beam reaches it, so the intensity decays geometrically with depth, never quite reaching zero. A large $\mu$ (bone, or a contrast agent like iodine or barium) means a beam that dies quickly and casts a strong shadow; a small $\mu$ (soft tissue, lung) leaves little contrast. This energy-dependence is deliberate clinical currency: physicians pick a beam energy that maximises the difference in $\mu$ between the structures they want to tell apart. Because equation (1) is exponential, the natural way to describe an absorber is not "how much stops the beam" but "how much halves it" — the half-value thickness (HVT, or half-value layer). Setting $I = I_0/2$ in equation (1) gives a relation identical in form to the radioactive half-life $t_{1/2} = 0.693/\lambda$:

where: $\ln 2 = 0.693$, $\mu$ is the same linear attenuation coefficient, and $n = x/\text{HVT}$ is how many half-value layers deep you have gone. The physical reading is the whole reason HVT is used: it converts an awkward exponential into simple halving. One HVT leaves $50\%$, two leaves $25\%$, three leaves $12.5\%$ — you never need a calculator, only powers of one-half. Radiographers quote shielding in HVTs for exactly this reason.

A plain X-ray has one great weakness: it collapses the whole depth of the body onto one plane, so a small tumour hidden behind a rib can be invisible. Godfrey Hounsfield's insight, built at EMI in 1971, was that if you record the transmitted beam from many directions around the patient and feed the set to a computer, the machine can solve for the attenuation coefficient at every point of a thin cross-sectional slice — a genuine map of the interior, one plane at a time. That is computed tomography, from the Greek tomos, a slice. The mathematics of reconstructing an object from all its projections had been worked out by Johann Radon in 1917; Hounsfield and, independently, Allan Cormack turned it into a machine. CT does not report raw $\mu$; it reports a calibrated number, the Hounsfield unit (HU), comparing each voxel's attenuation to water's:

where: $\mu$ is the local attenuation coefficient and $\mu_{\text{water}}$ that of water at the same beam energy. The physical reading is a clinician-friendly scale pinned to two anchors: water sits at exactly $0\ \text{HU}$ by construction, and air (with essentially no attenuation, $\mu \approx 0$) at $-1000\ \text{HU}$. Dense cortical bone, whose $\mu$ is roughly twice water's, lands near $+1000\ \text{HU}$; fat around $-100\ \text{HU}$; clotted blood around $+60\ \text{HU}$. A single CT slice measures $\mu$ everywhere; the Hounsfield unit is just how that measurement is packaged.

Switch off the radiation entirely. Ultrasound sends a pulse of sound — mechanical vibration far above hearing, typically $1$ to $15\ \text{MHz}$ — into the body from a hand-held transducer, then falls silent and listens. Wherever the pulse crosses a boundary between tissues, part reflects straight back as an echo; the transducer times each echo and, knowing the speed of sound in soft tissue ($v \approx 1540\ \text{m/s}$), converts the delay into a depth. Sweep the beam across a plane and the machine paints a real-time cross-section. Because sound is not ionising, ultrasound carries no radiation dose — the reason it, and not X-rays, is used to watch a developing baby. What decides how strong an echo a boundary gives? Not density alone and not stiffness alone, but a combination of the two called the acoustic impedance:

where: $\rho$ is the tissue's density and $c$ the speed of sound in it (units of $Z$: $\text{kg}\,\text{m}^{-2}\text{s}^{-1}$, often written as the rayl). The physical reading: acoustic impedance measures how much a medium "resists" being set into sound-vibration. Soft tissue has $Z \approx 1.63\times10^{6}$ (density about $1060\ \text{kg/m}^3$, sound speed $1540\ \text{m/s}$); air is far lower, about $430$; bone far higher, about $7.8\times10^{6}$. An echo appears only where $Z$ changes, and its strength depends on the size of the jump. The fraction of intensity reflected at a boundary between media of impedance $Z_1$ and $Z_2$ is:

where: $Z_1$ and $Z_2$ are the impedances on the two sides of the boundary, and $R$ is the reflected fraction of the incident intensity (the rest transmits onward to make deeper echoes). The physical reading holds the two facts every exam wants. First, a matched boundary ($Z_2 = Z_1$) reflects nothing — it is acoustically invisible — while boundaries between soft tissues, differing in impedance by only a percent or two, give faint echoes of about $1\%$ that are exactly what a scan is built from. Second, a mismatched boundary reflects almost everything: at a tissue–air interface the impedances differ by a factor of nearly four thousand, so $R \approx 99.9\%$ — virtually the whole pulse bounces off the first sliver of air. That is why the sonographer smears coupling gel between probe and skin: it expels the thin air layer that would otherwise mirror the beam straight back. The same near-total reflection at tissue–bone and tissue–gas boundaries is why ultrasound cannot see through lungs or behind bone.

One more number matters: resolution. Like any wave, sound cannot resolve detail finer than about one wavelength, $\lambda = v/f$. At $2\ \text{MHz}$ in tissue $\lambda = 1540/(2\times10^6) = 0.77\ \text{mm}$; at $10\ \text{MHz}$ it shrinks to $0.15\ \text{mm}$, sharpening the image. But higher frequencies are absorbed faster and penetrate less deeply, so a sonographer trades resolution against reach: a high-frequency probe for a shallow thyroid, a lower-frequency one for a deep fetus or liver.

The three modalities so far show anatomy — what is where. Nuclear medicine shows function — what is working. A radionuclide tracer is chemically tailored so the body takes it to a particular organ or process — a bone-seeking compound to the skeleton, a sugar analogue to hungry tumour cells — and, being radioactive, it broadcasts its own location. A gamma camera maps where the emitted gamma rays come from; positron-emission tomography (PET) does the same with positron emitters, detecting the two back-to-back annihilation photons. The image is not of density but of activity: a "hot spot" is a region taking up the tracer avidly. The workhorse tracer is technetium-99m, and its choice is a lesson in radiation physics. Its half-life is about $6$ hours — long enough to inject, circulate, and scan, short enough to decay away within a day and spare the dose. It emits a single $140\ \text{keV}$ gamma — energetic enough to escape the body yet low enough to capture efficiently — and, being an isomeric ("m", metastable) transition, essentially no alpha or beta, which would add dose without adding image. It is the same $\alpha$/$\beta$/$\gamma$ physics you met for radioactive decay, now chosen deliberately: a pure, medium-energy gamma emitter with a short half-life is the ideal messenger. The electrocardiogram (ECG) is the odd one out — no image, no radiation. Every heartbeat is triggered by a wave of electrical depolarisation across the heart muscle, setting up tiny voltages, of order a millivolt, on the skin. Electrodes record the familiar trace: the small P wave (atria), the sharp QRS complex (ventricles firing), the rounded T wave (ventricles resetting). From the peak spacing a cardiologist reads the heart rate — a healthy $75$ beats per minute is one beat every $0.8\ \text{s}$ — and from their shape, the health of the muscle and its conduction. It sits in medical-physics syllabuses because it is pure applied electricity: potentials, electrodes, and filtering out the $50\ \text{Hz}$ mains hum.

Board level — half-value thickness and transmitted fraction (IB-HL and A-Level pattern)

An X-ray beam passes through an absorber with $\mu = 0.693\ \text{cm}^{-1}$. (a) Find the half-value thickness. (b) What fraction is transmitted through $3.0\ \text{cm}$? (c) As a percentage?

Step 1. From equation (2): $\text{HVT} = 0.693/\mu = 0.693/0.693 = 1.0\ \text{cm}$ — one centimetre halves the beam.

Step 2. Through $3.0\ \text{cm}$ the beam crosses $n = x/\text{HVT} = 3$ half-value thicknesses, so $I/I_0 = 1/2^3 = 1/8 = 0.125$.

Step 3. That is $12.5\%$ transmitted, $87.5\%$ removed. Cross-check the raw exponential: $e^{-0.693\times3} = e^{-2.079} = 0.125$. Answer: HVT $= 1.0\ \text{cm}$; transmitted $= 0.125$, i.e. $12.5\%$.

Board level — acoustic impedance and why we use gel (A-Level pattern)

Soft tissue has density $1060\ \text{kg/m}^3$ and sound speed $1540\ \text{m/s}$. (a) Find its acoustic impedance. (b) A thin air layer ($Z_{\text{air}} \approx 430$) is trapped between probe and skin — what fraction of the pulse reflects, and what does that say about gel?

Step 1. From equation (4): $Z = \rho c = 1060 \times 1540 = 1.63\times10^{6}\ \text{kg}\,\text{m}^{-2}\text{s}^{-1}$.

Step 2. From equation (5): $R = \big((1.63\times10^{6} - 430)/(1.63\times10^{6} + 430)\big)^2 = 0.9989$.

Step 3. So about $99.9\%$ reflects off the air layer and only $0.1\%$ enters — the scan would be blank. Gel, whose impedance is close to tissue's, displaces the air and nearly matches probe to skin. Answer: $Z_{\text{tissue}} = 1.63\times10^{6}$; $R \approx 99.9\%$, so gel is essential.

Competitive level — echo depth from time of flight (A-Level and IB-HL pattern)

Speed of sound in soft tissue is $1540\ \text{m/s}$. (a) An echo returns $100\ \mu\text{s}$ after the pulse — how deep is the boundary? (b) What is the wavelength and best resolution at $2\ \text{MHz}$? (c) Why switch to $10\ \text{MHz}$, and at what cost?

Step 1. The pulse goes down and the echo comes back up, so the sound covers $2d$ in time $t$: $d = v\,t/2 = 1540 \times (100\times10^{-6})/2 = 0.077\ \text{m} = 7.7\ \text{cm}$.

Step 2. Resolution is set by wavelength: $\lambda = v/f = 1540/(2\times10^{6}) = 0.77\ \text{mm}$.

Step 3. At $10\ \text{MHz}$, $\lambda = 0.154\ \text{mm}$ — five times finer, but higher frequencies are absorbed faster and cannot reach deep, so the $10\ \text{MHz}$ probe suits shallow targets only. Answer: $d = 7.7\ \text{cm}$; $\lambda = 0.77\ \text{mm}$ at $2\ \text{MHz}$, $0.15\ \text{mm}$ at $10\ \text{MHz}$. Timing note: under a minute. Halve the time-of-flight before multiplying (the round-trip factor of two is the most-forgotten step), and $\lambda = v/f$ uses the same $1540\ \text{m/s}$ — this is sound, not light, so never reach for $3\times10^8$.

Competitive level — mixed attenuation and a tracer sanity check (A-Level pattern)

(a) An X-ray beam passes through $2.5\ \text{cm}$ of a material with $\mu = 0.693\ \text{cm}^{-1}$ — what fraction transmits? (b) A technetium-99m dose (half-life $6\ \text{h}$) is injected — what fraction remains after $24\ \text{h}$? (c) The tracer emits a $140\ \text{keV}$ gamma — estimate its wavelength.

Step 1. $n = 2.5/1.0 = 2.5$ half-value thicknesses, so $I/I_0 = 2^{-2.5} = 0.177$, about $17.7\%$ (exponential check: $e^{-0.693\times2.5} = 0.177$).

Step 2. Tracer decay is the same half-life rule: $24\ \text{h}$ is $4$ half-lives, so $(1/2)^4 = 1/16 = 0.0625$, i.e. $6.25\%$ remains — nearly all gone within a day, why such a short half-life is chosen.

Step 3. From $\lambda = hc/E$ with $hc = 1240\ \text{eV}\,\text{nm}$ and $E = 1.4\times10^{5}\ \text{eV}$: $\lambda = 1240/(1.4\times10^{5}) = 8.9\times10^{-3}\ \text{nm} = 8.9\ \text{pm}$ — a hard gamma, why it escapes the body. Answer: (a) $17.7\%$; (b) $6.25\%$; (c) $\lambda \approx 8.9\ \text{pm}$. Timing note: seconds each once you spot the structure — attenuation and radioactive decay are the SAME exponential, $2^{-x/\text{HVT}}$ and $2^{-t/t_{1/2}}$, so count half-layers or half-lives and raise one-half to that power.

Calling a plain X-ray a "slice". A radiograph is a shadow — a projection of the whole depth onto one plane. Only CT resolves depth by reconstructing from many angles. This anatomy-versus-projection point is a standard IB-HL and A-Level discrimination question.

Adding half-value thicknesses instead of multiplying. Each HVT halves what remains, so $n$ layers leave $1/2^n$, not $1 - n/2$. Three HVTs leave $12.5\%$, not "nothing" and not $25\%$. The exponential is multiplicative — this is the same trap as radioactive half-lives.

Treating $\mu$ as fixed. The linear attenuation coefficient depends on photon energy as well as material; changing the tube voltage changes the contrast. Do not quote a single $\mu$ for "bone" as if it were a constant.

Confusing ionising and non-ionising modalities. X-ray, CT, and radionuclide/PET use ionising radiation and deliver a dose; ultrasound (sound) and ECG (voltages) do not. The reason ultrasound, not CT, is used in pregnancy is precisely this. A frequent slip is to call ultrasound or MRI "a type of X-ray" — it is not.

Forgetting the factor of two in pulse-echo. The echo makes a round trip, so depth is $d = vt/2$, not $vt$. And use the speed of sound ($1540\ \text{m/s}$), never the speed of light.

Board scope — read this carefully. Medical imaging is a genuine syllabus topic only on ICSE/ISC, IGCSE 0625, IB Physics HL, and A-Level (e.g. the AQA Medical Physics option). On IGCSE it is qualitative only — you are asked about the uses of ultrasound and radiation, not to compute an HVT or an impedance. The quantitative treatment here — $I = I_0 e^{-\mu x}$, $\text{HVT} = 0.693/\mu$, $Z = \rho c$, the reflection formula — is IB-HL and A-Level material. It is not in the CBSE, JEE, NEET, or AP Physics 2 core, so if you sit those, treat this as enrichment that deepens the X-ray, wave-reflection, and radioactive-decay physics you are examined on.

Texts. IB Physics (Tsokos), the Imaging option · A-Level Physics (AQA), the Medical Physics topic (attenuation, HVT, acoustic impedance, gamma camera) · ISC Physics Class 12 for the ICSE treatment · Webb, Introduction to Biomedical Imaging, and Bushberg et al., The Essential Physics of Medical Imaging, for the full account.

Provenance. Constants from LANE_SPEC: sound in tissue $1540\ \text{m/s}$; $\ln 2 = 0.693$; $hc = 1240\ \text{eV}\,\text{nm}$. Recompute list: $\text{HVT} = 0.693/\mu = 1.0\ \text{cm}$; $1/2^n = 0.500,\,0.250,\,0.125$ for $n=1,2,3$ and $2^{-2.5} = e^{-0.693\times2.5} = 0.177$; Hounsfield units water $0$, air $-1000$, fat $-100$, bone $+1000$ from $\text{HU} = 1000(\mu-\mu_w)/\mu_w$; $Z = \rho c$ giving tissue $1.63\times10^{6}$, air $430$; $R = ((Z_2-Z_1)/(Z_2+Z_1))^2$ giving tissue–air $0.9989$ (99.9%), fat–muscle $0.011$ (1%), tissue–bone $0.43$; $\lambda = v/f = 0.77\ \text{mm}$ at $2\ \text{MHz}$, $0.15\ \text{mm}$ at $10\ \text{MHz}$; echo depth $d = vt/2 = 7.7\ \text{cm}$ for $t = 100\ \mu\text{s}$; Tc-99m half-life $6\ \text{h}$, fraction after $24\ \text{h}$ is $1/2^4 = 6.25\%$, gamma $140\ \text{keV}$ with $\lambda = hc/E = 8.9\ \text{pm}$; $75\ \text{min}^{-1}$ gives a beat every $0.8\ \text{s}$; CT/X-ray dose ratio $7/0.02 = 350$.

Prerequisites. X-Rays: Production & Spectra — where the beam this article sends through the body is made. Leads to. Ultrasound deepens the wave-reflection physics of the echo, and Radioactivity: Alpha, Beta & Gamma underpins the tracer emission of nuclear medicine. For the reconstruction mathematics behind CT (the Radon transform and its inversion, which need calculus) and the detector physics of PET, the university seam on biomedical imaging goes deeper — enrichment only, never required for the boards above.