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X-rays and CT scanning questions
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.
19 original questions · 53 marks · the x-rays and ct scanning notes · Medical physics
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Describe how X-rays are produced in an X-ray tube.
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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).An X-ray tube operates at 65 kV. Calculate the maximum energy of a photon in the beam.
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Emax = eV = 1.60 × 10−19 × 65 000 (1)
Emax = 1.04 × 10−14 J; no photon can carry more energy than one electron brought across the tube (1)State the equation for the attenuation of X-rays passing through matter, defining each symbol.
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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).State why the inside of an X-ray tube is evacuated.
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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).The photons leaving an X-ray tube have a spread of energies below the maximum eV. Explain why photons of lower energy are present.
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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).Calculate the minimum wavelength of X-rays from a tube operating at 100 kV.
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Emax = eV = 1.60 × 10−14 J (1)
λmin = hc/Emax = (6.63 × 10−34 × 3.00 × 108)/(1.60 × 10−14) (1)
λmin = 1.24 × 10−11 m ≈ 1.2 × 10−11 m (1)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.
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I/I₀ = e−μx (1)
= e−0.21 × 4.0 (1)
= 0.43, about 43% (1)An operator increases the tube voltage. Separately, the operator increases the tube current. State the effect of each change on the X-ray beam.
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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).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.
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.
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.
Modern radiography records the transmitted beam with a digital flat-panel detector rather than photographic film. State two advantages of the digital detector.
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.
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.
Explain the principle of CT scanning, and state one advantage and one disadvantage of a CT scan compared with a plain X-ray image.
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.
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.
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.
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.
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