Physics › Medical physics › The physics of diagnostic X-rays
The physics of diagnostic X-rays
Electrons accelerated through tens of kilovolts decelerate in a tungsten target, and the X-ray beam they produce carries bremsstrahlung and characteristic lines. The tube pd caps the photon energy at eV and fixes the minimum wavelength. Intensity then falls exponentially with thickness at a rate set by the attenuation coefficient μ, and the half-value thickness is the thickness that halves it.
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X-rays and CT scanning, part 1 of 2. Part 2 is X-ray detectors and CT scanning.
Builds on The photoelectric effect and the photon 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.
COMMON MISCONCEPTION
Every X-ray leaving a tube carries the energy the tube pd gave the electron that made it.
The tube pd sets the maximum photon energy, not the only one. An electron gives up its energy in many steps, most of which only heat the target; a step that radiates a photon gives it part of the total, so the beam is a continuous spectrum running from E = eV all the way down.
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 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
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 many stages on the way to rest, most of which only heat the target, so the photons that are produced 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 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.
Describe the drawing
Two X-ray spectra plotted against wavelength on shared axes. Each is a smooth hump of bremsstrahlung that falls to nothing at its own minimum wavelength and stops there, the 140 kilovolt hump taller and cut off at 8.9 picometres and the 100 kilovolt one lower and cut off at 12.4 picometres. Two narrow spikes, the characteristic lines of the tungsten target, stand on both humps at the same pair of wavelengths, taller on the higher-voltage curve 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:
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 caveat applies: 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:
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.
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, about 1–20 keV. The photoelectric effect dominates below about 100 keV, which covers most of the diagnostic range, and it is the source of bone contrast, because its strength climbs steeply with atomic number. Compton scattering dominates at higher energies, about 0.5–5 MeV, above the diagnostic range, though it still removes some photons from a diagnostic beam. 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 has moved to the next part of this lesson.
From shadows to slices has moved to the next part of this lesson.
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.
- 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.
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.
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.
Or read them with their mark schemes on the physics of diagnostic X-rays questions page.
WHERE TO GO NEXT
- Exponentials and logarithms is the maths this lesson leans on, worked through step by step.
USED LATER IN
CHECK YOUR PROGRESS
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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.
Open the full revision checklist to track your progress across the whole unit.