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A2 · Practice questions · Production and use of X-rays

Fast electrons in, hard light out.

Six original Cambridge-style questions on how an X-ray tube works, the minimum wavelength from hfmax = eV, exponential attenuation I = I₀e^(−μx) and half-value thickness, image contrast and CT scanning.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

eV in, hf out, then e^(−μx) through the body.

01
Analysis
[3 marks]

Describe how X-rays are produced in an X-ray tube, and explain why the metal target must be cooled.

  • Electrons emitted from a heated cathode (filament) are accelerated through a large potential difference, typically tens of kV, in an evacuated tube. ✓
  • When they strike the metal target (anode) they decelerate rapidly, and some of their kinetic energy is emitted as X-ray photons. ✓
  • Only a small fraction (about 1%) of the electrons' energy becomes X-rays; the rest becomes thermal energy in the target, so it must be cooled, for example by rotating the anode or circulating oil. ✓
02
Calculation
[3 marks]

The shortest wavelength present in the beam from an X-ray tube is 2.4 × 10⁻¹¹ m. Calculate the maximum photon energy and hence the accelerating potential difference of the tube. The tube current is 20 mA and 1.0% of the electrons' energy is converted to X-rays; calculate the rate at which thermal energy is produced in the target.

  • Emax = hc / λmin = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 2.4 × 10⁻¹¹ = 8.3 × 10⁻¹⁵ J. ✓
  • eV = Emax, so V = 8.29 × 10⁻¹⁵ / 1.60 × 10⁻¹⁹ = 5.2 × 10⁴ V (52 kV). ✓
  • Electrical power = IV = 0.020 × 5.18 × 10⁴ = 1.04 × 10³ W, of which 99% is heat: about 1.0 × 10³ W. ✓
03
Calculation
[3 marks]

An X-ray beam passes through 4.0 cm of muscle (linear attenuation coefficient 0.22 cm⁻¹) and then 2.0 cm of bone (0.55 cm⁻¹). Calculate the fraction of the incident intensity transmitted, and compare it with the fraction transmitted through 6.0 cm of muscle alone.

  • Muscle: I/I₀ = e^(−0.22 × 4.0) = e^(−0.88) = 0.415; bone: e^(−0.55 × 2.0) = e^(−1.10) = 0.333. ✓
  • The fractions multiply: 0.415 × 0.333 = 0.14 transmitted. ✓
  • Muscle alone: e^(−0.22 × 6.0) = e^(−1.32) = 0.27, nearly twice as much, so the bone casts a noticeably darker shadow on the detector and shows with good contrast. ✓
04
Analysis
[3 marks]

Describe how a computed tomography (CT) scan produces an image, and state one advantage and one disadvantage of a CT scan compared with a single conventional X-ray image.

  • An X-ray tube and detectors rotate around the patient, taking many images of a thin slice from different angles; a computer combines them into an image of the slice, and successive slices are stacked into a three-dimensional image. ✓
  • Advantage: it can distinguish small differences in attenuation, so soft tissues are seen with much better contrast, and the 3D image can be viewed from any direction without overlapping structures. ✓
  • Disadvantage: the patient receives a much larger dose of ionising radiation (and the scan takes longer and costs more). ✓
05
Calculation
[3 marks]

For a particular X-ray beam the half-value thickness of lead is 0.30 mm. Calculate the linear attenuation coefficient of lead for this beam, and the thickness of lead that reduces the beam intensity to 1.0% of its initial value.

  • μ = ln 2 / x½ = 0.693 / 0.30 = 2.3 mm⁻¹. ✓
  • I/I₀ = 0.010 = e^(−μx), so x = ln(100) / μ = 4.61 / 2.31 = 2.0 mm. ✓
  • Check: 2.0 mm is 6.6 half-value thicknesses and (½)^6.6 = 0.010, as required. ✓
06
Calculation
[4 marks]

An X-ray tube operates at 90 kV.

(a) Calculate the maximum energy of an X-ray photon, in joules. (b) Calculate the minimum wavelength of the X-rays produced. (c) The tube current is doubled while the voltage is unchanged. State and explain the effect on the minimum wavelength and on the intensity of the beam.

  • (a) Emax = eV = 1.60 × 10⁻¹⁹ × 90 × 10³ = 1.44 × 10⁻¹⁴ J. ✓
  • (b) λmin = hc / Emax = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) / 1.44 × 10⁻¹⁴. ✓
  • λmin = 1.4 × 10⁻¹¹ m. ✓
  • (c) The minimum wavelength is unchanged, since it depends only on the energy eV of each electron; the intensity doubles, because twice as many electrons per second strike the target and produce twice as many photons. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.