Questions on the physics of diagnostic X-rays

Question 1

Describe how X-rays are produced in an X-ray tube.

Accepted Answer

Electrons are released from a heated cathode by thermionic emission (1) and accelerated through a large pd towards a metal target (1). When they decelerate rapidly in the target, part of their kinetic energy is radiated as X-ray photons; most of the rest becomes heat in the target (1).

Question 2

An X-ray tube operates at 65 kV. Calculate the maximum energy of a photon in the beam.

Accepted Answer

E_max = eV = 1.60 × 10⁻¹⁹ × 65 000 (1) E_max = 1.04 × 10⁻¹⁴ J; no photon can carry more energy than one electron brought across the tube (1)

Question 3

State the equation for the attenuation of X-rays passing through matter, defining each symbol.

Accepted Answer

I = I₀e^−μx (1), where I₀ is the incident intensity, I the intensity after passing through thickness x of material, and μ the attenuation coefficient of that material (1).

Question 4

State why the inside of an X-ray tube is evacuated.

Accepted Answer

So that the electrons do not collide with gas molecules on the way to the target, and arrive carrying the full energy eV gained from the accelerating pd (1).

Question 5

The photons leaving an X-ray tube have a spread of energies below the maximum eV. Explain why photons of lower energy are present.

Accepted Answer

Most electrons are not stopped in a single event but decelerate in several stages in the target (1). Each stage radiates a photon carrying only part of that electron's energy, so the spectrum fills in below eV (1).

Question 6

Photographic film on its own absorbs only a small percentage of the X-ray photons that reach it. State what an intensifying screen does with each photon it absorbs, and state the effect of using one on the dose the patient receives.

Accepted Answer

Each X-ray photon the screen absorbs is converted into thousands of visible-light photons, and it is that light which blackens the film (1). The screen absorbs a far larger share of the beam than the emulsion does, so far fewer photons have to cross the patient for the same blackening and the dose falls (1).

Question 7

A collimator narrows the beam of a CT scanner to a thin fan before it reaches the patient. Give two reasons for making the beam this narrow.

Accepted Answer

It fixes which slice is being imaged and how thick that slice is, so the dose goes into that slab alone and the tissue above and below it is left unirradiated (1). Irradiating only a thin slab sets fewer photons scattering, and a scattered photon reaches a detector along a path the reconstruction has not allowed for (1).

Question 8

Calculate the minimum wavelength of X-rays from a tube operating at 100 kV.

Accepted Answer

E_max = eV = 1.60 × 10⁻¹⁴ J (1) λ_min = hc/E_max = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(1.60 × 10⁻¹⁴) (1) λ_min = 1.24 × 10⁻¹¹ m ≈ 1.2 × 10⁻¹¹ m (1)

Question 9

Soft tissue has an attenuation coefficient of 0.21 cm⁻¹ for one diagnostic beam. Calculate the fraction of the incident intensity transmitted through 4.0 cm of this tissue.

Accepted Answer

I/I₀ = e^−μx (1) = e^−0.21 × 4.0 (1) = 0.43, about 43% (1)

Question 10

An operator increases the tube voltage. Separately, the operator increases the tube current. State the effect of each change on the X-ray beam.

Accepted Answer

Raising the voltage raises the maximum photon energy, making the beam more penetrating, or harder (1). Raising the current increases the number of electrons per second, so more photons are produced and the beam is more intense, with the photon energies unchanged (1).

Question 11

The attenuation coefficient of bone for a certain beam is 0.60 cm⁻¹. Calculate the half-value thickness of bone, the thickness that halves the beam intensity.

Accepted Answer

e^−μx = 0.5 (1) x = ln 2/μ = 0.693/0.60 (1) x = 1.2 cm (1)

Question 12

Explain why bone shows clearly on a plain X-ray image while two neighbouring soft tissues often do not, and how a contrast medium helps.

Accepted Answer

Contrast comes from differences in attenuation coefficient. Bone's μ is much larger than soft tissue's, so bone transmits far less intensity and throws a sharp shadow (1). Neighbouring soft tissues have similar μ and transmit almost equally, so they are nearly indistinguishable (1). A contrast medium such as a barium compound is a strong absorber swallowed into the organ, painting its outline into the image (1).

Question 13

A beam of X-rays passes through 2.5 cm of muscle. The transmitted intensity is 0.35 of the incident intensity. Determine the attenuation coefficient of muscle for this beam.

Accepted Answer

e^−μx = 0.35 (1). μ = ln(1/0.35)/x = 1.05/2.5 (1) = 0.42 cm⁻¹ (1).

Question 14

Modern radiography records the transmitted beam with a digital flat-panel detector rather than photographic film. State two advantages of the digital detector.

Accepted Answer

It needs far less exposure than film, so the patient receives a smaller dose (1). The digital image can be enhanced and shared immediately (1).

Question 15

Pair production is one mechanism by which photons are absorbed in matter. State the minimum photon energy for pair production, and explain why it plays no part in imaging with a 90 kV tube.

Accepted Answer

Pair production needs a photon of at least 1.02 MeV, twice the rest energy of an electron (1). A 90 kV tube produces photons of at most 90 keV, far below this threshold, so no pair production occurs (1). (Board note: naming the attenuation mechanisms is an OCR requirement; AQA asks for differential tissue absorption while excluding the details of the absorption processes, so treat this as extension for AQA.)

Question 16

Bare photographic film absorbs 2.0% of the X-ray photons reaching it, while film held against an intensifying screen has 40% of them absorbed. The same number of absorbed photons is needed for the image in each case, and the film alone needs an entrance dose of 1.6 mGy. Calculate the factor by which the screen reduces the number of photons crossing the patient, and the entrance dose with the screen fitted. State what the screen costs the image.

Accepted Answer

The same number has to be absorbed either way, so the number arriving falls in the ratio of the absorbed fractions, 40/2.0 (1)
Factor = 20, so the dose becomes 1.6/20 = 0.080 mGy (1)
The image is less sharp: the light spreads sideways from the point at which the X-ray was absorbed, so each absorbed photon blackens a small patch of emulsion rather than a point (1)

Question 17

An image intensifier has a circular input phosphor of diameter 25 cm and a circular output phosphor of diameter 2.5 cm. Accelerating the electrons through the tube multiplies the brightness by a further factor of 60. Calculate the overall brightness gain of the instrument.

Accepted Answer

The same picture is delivered onto a face of much smaller area, so the gain from minification is the ratio of the two areas, (25/2.5)² (1) = 100 (1) Overall gain = 100 × 60 = 6.0 × 10³ (1)

Question 18

The detector array of a CT scanner is an arc of 720 elements, and the whole arc is read 1000 times during one rotation lasting 0.50 s. Calculate the number of separate measurements of transmitted intensity made in one rotation, and the time between one reading of the arc and the next. State what the scanner does with this many numbers.

Accepted Answer

Measurements = 720 × 1000 = 7.2 × 10⁵ (1)
Time between readings = 0.50/1000 = 0.50 ms (1)
Each number is the intensity transmitted along one ray at one angle, and the computer solves for the map of attenuation coefficient across the slice that fits all of them at once (1)

Question 19

Describe how a digital flat-panel detector turns the X-rays leaving the patient into a stored image.

Accepted Answer

A scintillator layer absorbs each X-ray and converts it to visible light; in the better panels the caesium iodide is grown as fine parallel needles, so the light is channelled down its own column instead of spreading sideways (1). Beneath the scintillator is a matrix of photodiodes, one to a pixel, and light reaching a diode frees charge in it (1). Each photodiode has its own thin-film transistor, and the stored charge is read out row by row as a number for every pixel (1).

Question 20

A beam passes through soft tissue with μ = 0.35 cm⁻¹. Calculate the thickness that reduces the intensity to one tenth of its incident value.

Accepted Answer

e^−μx = 0.1 (1) x = ln 10/μ = 2.303/0.35 (1) x = 6.6 cm (1)

Question 21

Explain the principle of CT scanning, and state one advantage and one disadvantage of a CT scan compared with a plain X-ray image.

Accepted Answer

The X-ray tube and detectors rotate around the patient, measuring attenuation along very many directions through one slice (1). A computer reconstructs the map of attenuation coefficient across the slice that is consistent with all the measurements, giving a cross-sectional image; stacked slices give three dimensions (1). Advantage: structures are not superimposed, and soft tissues can be distinguished (1). Disadvantage: many exposures give a much larger radiation dose than a single film (1).

Question 22

An X-ray tube runs at 120 kV with a beam current of 2.0 mA. Calculate the maximum photon energy, and the number of electrons striking the target each second.

Accepted Answer

E_max = eV = 1.60 × 10⁻¹⁹ × 120 000 = 1.92 × 10⁻¹⁴ J (1) Electrons per second = I/e = 2.0 × 10⁻³/(1.60 × 10⁻¹⁹) (1) = 1.25 × 10¹⁶ s⁻¹ (1)

Question 23

Most of the energy delivered to the target of an X-ray tube does not become X-rays. State what it becomes, and explain two design features of the anode that deal with it.

Accepted Answer

It becomes heat in the target: only around one per cent of the electron energy is radiated as X-rays (1). The anode is made of tungsten, whose very high melting point tolerates the heating (1), and it is rotated so that the electron beam spreads its heating over a ring of material rather than boring into one spot (1).

Question 24

A patient's abdomen presents a 5.0 cm path to an X-ray beam. Soft tissue along this path has attenuation coefficient 0.20 cm⁻¹. Before the gut is imaged, the patient swallows a barium meal that fills a 1.0 cm section of the path with a compound of attenuation coefficient 3.0 cm⁻¹. Deduce, with calculations, whether the barium meal gives the gut useful contrast against the surrounding tissue.

Accepted Answer

Path missing the gut: I/I₀ = e^−0.20 × 5.0 = 0.368 (1). Path through the barium: I/I₀ = e^−(0.20 × 4.0 + 3.0 × 1.0) = e^−3.8 = 0.0224 (1). The transmitted intensities differ by a factor of about 16 (1). Rays crossing the gut arrive far weaker than their neighbours, so its outline stands out clearly: the barium meal does give useful contrast (1).

Question 25

Describe the mechanisms by which photons are removed from a diagnostic X-ray beam as it passes through the body, and explain which mechanism gives bone its contrast.

Accepted Answer

Simple scatter removes photons at the lowest photon energies, about 1–20 keV (1). The photoelectric effect, in which a photon is absorbed and an electron ejected, dominates below about 100 keV, most of the diagnostic range (1). Its strength climbs steeply with atomic number, so bone absorbs far more strongly than soft tissue: this is the source of bone contrast (1). Compton scattering, in which a photon scatters off an electron and loses part of its energy, dominates at about 0.5–5 MeV, above the diagnostic range, though it still removes some photons from the beam (1). Pair production needs photons above 1.02 MeV, so it is absent from diagnostic beams (1). (Board note: OCR asks for these mechanisms by name; AQA asks for differential tissue absorption while excluding the details of the absorption processes, so treat the mechanism list as extension for AQA.)

Question 26

Some of the photons in a CT beam are Compton scattered inside the patient. Explain why these photons degrade the reconstructed slice, and describe two features of the scanner that keep most of them out of the detectors.

Accepted Answer

The reconstruction treats every reading as the attenuation suffered along one straight ray from the tube to that detector element (1). A scattered photon arrives from another direction but is counted as though it had come along that ray, so the intensity credited to the ray is too high and the attenuation coefficients the computer solves for are wrong (1). The beam is collimated at the tube into a thin fan, so only a thin slab of tissue is irradiated and fewer photons are set scattering at all (1). Each detector element carries its own collimator facing the tube, so a photon not travelling along that element's own ray is absorbed before it arrives (1).

Question 27

A surgeon watches a catheter move along an artery using an image intensifier, and a still radiograph of the same region is recorded on film held between two intensifying screens. Describe how the image intensifier produces a picture bright enough to watch continuously, and explain the effect the screens have on the sharpness of the still image.

Accepted Answer

In the intensifier an input phosphor converts the arriving X-rays to light, a photocathode converts that light to electrons, and those electrons are accelerated through tens of kilovolts (1). They are focused onto a much smaller output phosphor, so the same picture arrives on a face of a fraction of the area, and the reduction in area and the acceleration together make the output far brighter than the input, which is why a beam weak enough to be left running gives a picture that can be watched in real time (1). In the still image the screen absorbs the X-ray at one point, but the light it releases spreads sideways before it reaches the emulsion (1), so each absorbed photon blackens a small patch rather than a point and the picture is less sharp than one made by X-rays on the emulsion alone (1).