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X-rays and CT scanning

X-rays are produced when high-energy electrons strike a metal target, and differences in attenuation through the body produce the contrast in the image. One tube, one exponential law, and a century of refinement ending in the CT scanner, which turns hundreds of shadows into slices.

Builds on The photoelectric effect and The time constant and exponential decay.

IN THIS TOPIC

  • Describe how an X-ray tube produces X-rays, and find the maximum photon energy from the tube pd.
  • Explain the continuous and the characteristic parts of the spectrum a tube emits, and what fixes the minimum wavelength.
  • For OCR, name the four attenuation mechanisms and the energies at which each matters; AQA asks for the differential absorption without the process details.
  • Use I = I₀e−μx and the half-value thickness, and explain image contrast including contrast media.
  • Say how an intensifying screen and a flat-panel detector record the beam, and why each can cut the dose.
  • Explain how a CT scanner builds cross-sectional images, what its narrow beam and detector array are for, and weigh it against a plain X-ray.

COMMON MISCONCEPTION

A CT scanner is just a sharper X-ray photograph.

Making X-rays

An X-ray tube is an electron gun aimed at a metal target. Electrons boil off a heated cathode, accelerate through a pd of tens of kilovolts, and decelerate violently in the target metal. Rapid deceleration of charge radiates, and at these energies the radiation is X-rays. Around 99 per cent of the incident electrons' kinetic energy arrives as heat in the target rather than as X-rays, and that single fact explains the design. The anode is tungsten, for its melting point, and it spins so no one spot stays under the beam.

The X-ray tube: electrons accelerate from cathode to tungsten anode, and X-rays leave through the windowheated cathodetungsten anodeelectrons, energy eVX-rays outpd V across the tube caps the photon energy at eV
FIG. 1The X-ray tube: electrons from the heated cathode accelerate across the vacuum to the angled tungsten anode, and X-rays leave through the window. Nearly all the energy arrives as heat, so the anode rotates.

The photons come out with a spread of energies, and there is a hard ceiling on that spread. No photon can carry away more than one electron brought in. An electron accelerated through pd V arrives with energy eV, so

Emax=eVE_{max} = eVNOT ON THE AQA DATA SHEET: LEARN IT

and the shortest wavelength in the beam follows from E = hc/λ.

Two spectra, one beam

Sort the photons leaving the window by energy and the beam turns out to be two spectra laid on top of each other. Both are asked for by name, and each has its own mechanism.

The broad background is the continuous spectrum, or bremsstrahlung, which is German for braking radiation. An electron flying past a tungsten nucleus is pulled off course by its charge and slowed, and decelerating charge radiates. How much energy any one electron loses in any one such encounter depends on how close it passed, and most electrons are braked in several stages on the way to rest, so the photons emerge with every energy from almost nothing upwards. That is why the background is a smooth hump and not a set of lines.

The hump ends abruptly at the short-wavelength end, and that edge is the minimum wavelength. It belongs to the rare electron that gives up the whole of its kinetic energy in a single encounter, producing one photon that carries all of it. Then hf = eV, so the edge sits at λmin=hceV\lambda_{min} = \frac{hc}{eV} with V the tube pd. Nothing about the target appears in that expression. Change the metal, the current or the exposure time and the edge does not move; change the accelerating pd and it does, which is the standard two-mark discrimination.

Sitting on the hump are a few sharp characteristic lines, and they come from the target's electrons rather than from the beam's. An arriving electron with enough energy can knock an inner-shell electron clean out of a tungsten atom. The vacancy does not last: an electron from a higher shell drops into it, and the photon emitted carries exactly the difference between the two levels. Those differences belong to tungsten alone, so the lines sit at fixed photon energies whatever the tube pd, which is where the name comes from.

The lines therefore behave quite unlike the hump. Below the pd needed to eject an inner electron they are missing altogether, and above it they only grow brighter while staying where they are. That splits the two dials on the machine cleanly. Raising the tube voltage hardens the beam, pushing the continuous ceiling up and the minimum wavelength down. Raising the tube current sends more electrons per second, so every part of the spectrum grows in proportion while the edge and the lines stay exactly where they were.

The output spectrum of an X-ray tube at two accelerating pds: each hump stops at its minimum wavelength hc/eV; the characteristic lines stand still140 kV: λ min = 8.9 pm100 kV: λ min = 12.4 pmwavelengthintensityλ min = hc/eV, so the pd sets where the hump stopsno photons at allbelow λ mincharacteristic lines ofthe tungsten target
FIG. 2The output spectrum of the tube at two accelerating pds. Each smooth bremsstrahlung hump stops dead at its own minimum wavelength with bare axis to the left of it, and raising the pd from 100 kV to 140 kV drags that edge from 12.4 pm down to 8.9 pm. The tungsten target's two characteristic lines stand at the same pair of wavelengths on both curves, taller at the higher pd but not shifted.

WORKED EXAMPLE

Where the spectrum stops

A tube runs at 100 kV. Find the shortest wavelength in its beam.

λmin = hc/(eV) = (6.63 × 10−34 × 3.00 × 108)/(1.60 × 10−19 × 1.00 × 105).

The numerator is 1.99 × 10−25 J m and the denominator is 1.60 × 10−14 J, so λmin = 1.24 × 10−11 m, about 12 pm.

Halving the tube pd would double that wavelength. The characteristic lines of the tungsten, meanwhile, would not shift by a picometre, because they are set by the target's energy levels and not by the beam.

Attenuation and contrast

Passing through matter, X-rays are absorbed and scattered, and the surviving intensity falls exponentially with thickness:

I=I0e-μxI = I_{0}e^{-\mu x}ON THE AQA DATA SHEET

where μ is the attenuation coefficient of the material, measured per metre. The same mathematics as capacitor discharge and radioactive decay, now applied to X-rays in tissue. One label belongs on the model: μ is defined for one photon energy, so the clean exponential describes a narrow monochromatic beam, which is how AQA treats it; a real tube's filtered spectrum spans many energies and clinical work quotes effective values. The booklet also gives you the mass attenuation coefficient, which divides out the density so that different materials can be compared fairly:

μm=μρ\mu_{m} = \frac{\mu}{\rho}ON THE AQA DATA SHEET

Its unit is m2 kg−1, and a question that gives you μm and a density requires you to multiply back to μ before touching the exponential.

Attenuation is exponential, and contrast is the gap: bone's larger mu drops its curve far faster than soft tissue'sthickness xIsoft tissue, small μbone, large μI = I₀ exp(−μx): the gap between the curves is the contrast
FIG. 3Attenuation through tissue and bone: both fall exponentially, but bone's larger attenuation coefficient drops its curve far faster. The gap between the curves is the contrast in the image.

Four named mechanisms do the attenuating. It is OCR that expects each one named with a rough energy against it; AQA's specification asks for the differential absorption of tissues while excluding the details of the absorption processes, so on that board the list below is background. Simple scatter matters at low photon energies. The photoelectric effect dominates the lower diagnostic range and is the source of bone contrast, because its strength climbs steeply with atomic number. Compton scattering takes over across most of the diagnostic range and above. Pair production cannot happen at all below 1.02 MeV, twice the electron rest energy, so it plays no part in diagnosis.

Image quality depends on contrast. Bone absorbs far more strongly than soft tissue, so bones throw crisp shadows. Two soft tissues with similar μ are nearly indistinguishable, and that is the problem a contrast medium solves. A patient drinks a barium compound before a gut X-ray, or is given iodine for blood vessels. Both have high atomic numbers, so both attenuate strongly and paint an outline into the image where the anatomy alone offered none.

WORKED EXAMPLE

Halving thickness for bone

The attenuation coefficient of bone for one diagnostic beam is μ = 0.60 cm−1. What thickness of bone halves the intensity?

Half means e−μx = 0.5, so x = ln 2 / μ = 0.693/0.60 = 1.2 cm.

The half-value thickness plays exactly the role half-life plays in decay. It is the exponential's own natural yardstick, and x½ = ln 2 / μ is worth memorising in that form.

Recording the shadow

The beam that survives the patient has to be turned into an image, and the receptor chosen sets how much beam the patient had to be given in the first place. Photographic film on its own is a poor X-ray detector. The emulsion is thin and made of light elements, so only a per cent or two of the arriving photons interact with it at all. The rest cross the patient, deliver their dose and blacken nothing.

An intensifying screen is the fix. It is a layer of fluorescent material, calcium tungstate in the older ones and a rare-earth phosphor in the newer, held in contact with the film and usually one on each side of it. Its atoms have high atomic numbers, so it absorbs X-rays far more readily than the emulsion does, and each X-ray it absorbs releases thousands of visible-light photons, which the film is highly sensitive to. Almost all the blackening is then done by light rather than by X-rays directly.

Follow that through to the patient and the reason for the screen appears. The same darkening now needs perhaps a fiftieth of the photons through the body, so the exposure, and with it the dose, falls by that factor. What is lost is sharpness, because the light spreads sideways from the point where the X-ray was absorbed before it reaches the emulsion, so each photon blackens a small patch instead of a point. A thicker screen absorbs more and saves more dose, and blurs more. Dose against detail is the trade-off in every receptor.

Digital flat-panel detectors have largely replaced film and screen, and they begin the same way. A scintillator layer converts each X-ray to light, and in the better panels the caesium iodide is grown as fine parallel needles so the light is channelled down its own column rather than spreading sideways, which recovers much of the sharpness lost with a screen. Under the scintillator sits a matrix of photodiodes, one per pixel, each with its own thin-film transistor. Light frees charge in the diode, the charge is stored, and the array is read out row by row as numbers.

Reading numbers instead of developing a film changes what can be done next. The response stays proportional to the intensity over a far wider range, so one exposure covers bone and soft tissue where a film would be over-exposed in one and under-exposed in the other; the contrast can then be stretched electronically to suit whatever is being looked for; and the image exists at once and can be sent anywhere. The panel also absorbs a larger fraction of the arriving photons than film, so the same image can be made at a lower dose, provided the exposure settings are actually turned down to take the benefit.

Live screening needs a different instrument, and the image intensifier supplies it. An input phosphor converts the X-rays to light, a photocathode converts that light to electrons, and those electrons are accelerated through tens of kilovolts and focused down onto a small output phosphor. Accelerating them is what supplies the gain: each arriving X-ray ends as a far brighter flash than it began, so a surgeon can watch a catheter move in real time on a beam weak enough to be run continuously.

From shadows to slices

A plain X-ray flattens the body into one shadow, stacking everything along each ray into a single darkness. A CT scanner avoids that flattening. Its tube and detectors rotate around the patient, recording attenuation along thousands of directions, and a computer then solves the inverse problem. What map of μ across this slice would produce all of these shadows at once?

Two pieces of hardware make that possible, and questions ask what each is for. The first is the shape of the beam. A collimator at the tube narrows the output to a thin fan, wide enough to cross the patient but only a few millimetres thick, and that narrowness does two jobs at once. It confines the dose to the slice being imaged, leaving the tissue above and below it alone, and it defines how thick that slice is, which sets the resolution along the length of the body.

The narrow beam also improves the reconstruction. Reconstruction assumes each reading is the attenuation along one straight ray, and a Compton-scattered photon breaks that assumption, since it arrives from the wrong direction but is counted as though it had travelled straight. Fewer photons are set scattering in the first place when only a thin slab is irradiated, and each detector carries its own collimator facing the tube, so what reaches it has come along its own ray or not at all.

The second is the detector array facing the tube across the patient, an arc of hundreds of small scintillator and photodiode elements in place of a sheet of film. Each element measures the intensity transmitted along one ray and reports it as a number, and the whole arc is read hundreds of times per rotation. That is what the computation needs, a table of transmitted intensities indexed by ray and by angle rather than a picture. The elements are small, which sets the detail across the slice, and they respond in microseconds, so a full set of angles is gathered while the patient holds one breath.

The answer is a cross-sectional image on which soft tissues are distinguishable, and stacking slices gives a three-dimensional reconstruction. Set against that, there are drawbacks. Many exposures mean a far larger radiation dose than a single plain film, and the machine is expensive, so the sharper picture has to be clinically justified. Detail against dose is the recurring theme of every ionising technique.

ASSESSMENT FOCUS

  • Tube questions score on the energy chain. Electrons gain eV crossing the tube, and the maximum photon energy equals it, so Emax = eV and then λmin = hc/Emax if asked.
  • State what most of the electron energy becomes. Heat in the target. That one word explains the rotating tungsten anode and is a routine mark.
  • Spectrum questions want both parts with a mechanism against each. The continuous background is bremsstrahlung, electrons braked by the nuclei of the target and radiating as they slow, cut off at λmin = hc/(eV) where one electron gives up everything at once. The lines are characteristic, an inner-shell electron knocked out and replaced from a higher shell, so they sit at energies fixed by the target element.
  • Asked what changes if a dial is turned, answer for both parts of the spectrum. More tube voltage lowers λmin; more tube current raises the whole curve and moves neither the edge nor the lines.
  • Receptor answers score on dose. An intensifying screen absorbs X-rays far better than the emulsion and turns each one into thousands of light photons, so far fewer photons need to cross the patient, and the drawback is a blurrier image because the light spreads sideways.
  • Attenuation working mirrors decay working. Identify μ, use I = I₀e−μx, and take logs when the thickness is wanted. If the data gives you μm instead, multiply by the density first.
  • Contrast answers must name the mechanism. Materials with different attenuation coefficients transmit different intensities, and a contrast medium adds a high atomic number absorber where the anatomy alone offers none.
  • CT against plain film needs both halves. Cross-sectional detail with no superposition of structures, but a much larger dose. If the question asks about the hardware, the narrow collimated beam fixes the slice and keeps scatter out, and the detector array reads each ray as a number the computer can reconstruct from.

CHECK YOURSELF

An X-ray tube runs at 80 kV. Find the maximum photon energy in joules, and state why photons of lower energy are also present.

Show a hint

One electron's energy after the pd caps one photon's energy.

Show the answer

Emax = eV = 1.60 × 10−19 × 80 000 = 1.3 × 10−14 J.

Most electrons lose their energy in several stages, not one. Each stage radiates a photon carrying only part of the total, so the spectrum fills in below the maximum.

The tube pd caps the photon energy, so E max equals eV, and everything else in the beam is softer.

The smooth part of the spectrum is bremsstrahlung and stops at λ min = hc/eV; the sharp lines are characteristic of the target's own energy levels.

Attenuation is exponential, and contrast is a difference in μ between neighbouring tissues.

An intensifying screen lowers the dose at the expense of a little sharpness, and a flat panel reads the beam out as numbers.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

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  • Describe how an X-ray tube produces X-rays, and find the maximum photon energy from the tube pd.
  • Explain the continuous and the characteristic parts of the spectrum a tube emits, and what fixes the minimum wavelength.
  • For OCR, name the four attenuation mechanisms and the energies at which each matters; AQA asks for the differential absorption without the process details.
  • Use I = I₀e−μx and the half-value thickness, and explain image contrast including contrast media.
  • Say how an intensifying screen and a flat-panel detector record the beam, and why each can cut the dose.
  • Explain how a CT scanner builds cross-sectional images, what its narrow beam and detector array are for, and weigh it against a plain X-ray.

Open the full revision checklist to track your progress across the whole unit.