A Level Physics 24.2 Production and Use of X-rays MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice MCQs cover Cambridge AS & A Level Physics (9702) Section 24.2, an A Level (A2) topic in Medical Physics. They test how X-rays are produced in an X-ray tube, how the tube voltage and current control the spectrum, and how attenuation makes medical imaging possible.
Formula recap (see the A-Level Physics Formula Sheet):
- Kinetic energy of each electron reaching the target: E = eV (V = accelerating p.d.)
- Minimum wavelength of the continuous spectrum: λmin = hc / (eV)
- Attenuation: I = I0 e−μx
- Half-value thickness: x1/2 = ln 2 / μ = 0.693 / μ
- Constants: h = 6.63 × 10−34 J s, c = 3.00 × 108 m/s, e = 1.6 × 10−19 C, so 1 eV = 1.6 × 10−19 J
Producing X-rays
Q1. In an X-ray tube, the electrons that strike the target are produced at the cathode by
- (A) photoelectric emission caused by the heating current
- (B) thermionic emission from a heated filament
- (C) ionisation of the residual gas in the tube
- (D) radioactive decay of the cathode material
Show Answer
Answer: B – A current heats the cathode filament until it is hot enough for electrons to escape its surface. This is thermionic emission, the same process used in cathode-ray tubes, and the filament current controls how many electrons are available.
Q2. The target (anode) of an X-ray tube is usually made of tungsten embedded in copper mainly because tungsten
- (A) is the cheapest metal that conducts electricity well
- (B) emits electrons easily when heated
- (C) has a very high melting point and a high atomic number
- (D) is transparent to the X-rays it produces
Show Answer
Answer: C – About 99% of the electron energy becomes heat in the target, so the target material must survive very high temperatures: tungsten melts at about 3400 °C. Its high atomic number (Z = 74) also makes it efficient at slowing down electrons and producing X-ray photons.
Q3. An X-ray tube operates at a potential difference of 50 kV. The kinetic energy of each electron when it reaches the target is
- (A) 8.0 × 10−15 J
- (B) 8.0 × 10−18 J
- (C) 5.0 × 10−15 J
- (D) 3.1 × 10−12 J
Show Answer
Answer: A – E = eV = 1.6 × 10−19 × 5.0 × 104 = 8.0 × 10−15 J. This is 50 keV per electron: a useful check is that an electron accelerated through 1 volt gains exactly 1 eV.
Q4. The glass envelope of an X-ray tube is evacuated (pumped out to a vacuum) so that
- (A) the X-rays can escape from the tube without being absorbed
- (B) the high voltage cannot cause a spark between the electrodes
- (C) the filament does not burn away in the heat
- (D) the electrons travel to the target without colliding with air molecules and losing energy
Show Answer
Answer: D – The accelerating gap is several centimetres long. If air were present, the electrons would collide with gas molecules, lose energy, scatter in random directions and never arrive at the target with the full energy eV.
Q5. When the electrons strike the target, approximately 99% of their kinetic energy is converted into
- (A) X-ray photons
- (B) thermal energy (heat) in the target
- (C) light from the glowing target
- (D) kinetic energy of electrons knocked out of the target
Show Answer
Answer: B – Most collisions simply jiggle the target atoms more vigorously, heating the anode. Only about 1% of the electron energy emerges as X-radiation, which is why cooling (oil baths, rotating anodes) is a major design problem.
Q6. The X-ray photons produced by the continuous spectrum are emitted when an electron in the target
- (A) is knocked completely out of its atom
- (B) falls from an outer shell into a vacancy in an inner shell
- (C) is slowed down (deflected and decelerated) as it passes close to a target nucleus
- (D) collides with another electron and both are annihilated
Show Answer
Answer: C – A charged particle that is decelerated radiates electromagnetic energy. Each electron may lose any fraction of its energy in these encounters, giving the continuous “braking radiation” (bremsstrahlung) spectrum. Option B describes the characteristic lines instead.
Q7. Many hospital X-ray tubes have a rotating anode. The purpose of rotating the target is to
- (A) spread the heating over a larger area of the target so it does not melt
- (B) increase the kinetic energy of the electrons
- (C) make the X-ray beam sweep across the patient
- (D) reduce the attenuation of the X-rays in the tube
Show Answer
Answer: A – The electron beam hits only a small focal spot. By rotating the anode disc, the hot spot moves around a circular track, so the intense heating is shared by a much larger area of tungsten and the target survives longer exposures.
Q8. The accelerating potential of an X-ray tube is 75 kV. What is the kinetic energy of each electron at the target, expressed in joules?
- (A) 7.5 × 10−19 J
- (B) 4.7 × 10−15 J
- (C) 7.5 × 104 J
- (D) 1.2 × 10−14 J
Show Answer
Answer: D – E = eV = 1.6 × 10−19 × 7.5 × 104 = 1.2 × 10−14 J, i.e. 75 keV. Note that the numerical value in keV always matches the tube voltage in kV.
Q9. A tube runs at 40 kV with a tube current of 100 mA. The electrical power supplied to the electron beam is
- (A) 4.0 W
- (B) 4.0 × 103 W, of which about 3.96 × 103 W heats the target
- (C) 4.0 × 103 W, of which about 3.96 × 103 W becomes X-rays
- (D) 4.0 × 106 W
Show Answer
Answer: B – P = VI = 4.0 × 104 × 0.100 = 4.0 × 103 W. With only about 1% converted to X-rays, roughly 99% (3.96 × 103 W) appears as heat that must be removed from the anode.
Q10. The tube current of an X-ray tube is 200 mA. The number of electrons striking the target each second is
- (A) 3.2 × 1017
- (B) 1.25 × 1021
- (C) 1.25 × 1018
- (D) 200
Show Answer
Answer: C – Current is charge per second, so N = I / e = 0.200 / (1.6 × 10−19) = 1.25 × 1018 electrons per second. This is why raising the tube current raises the intensity of the beam: there are more electrons available to produce photons.
The X-ray Spectrum & Tube Controls
Q11. The X-ray spectrum from a tube consists of a continuous range of wavelengths with sharp peaks (lines) superimposed. The continuous part of the spectrum is produced by
- (A) electrons losing different fractions of their energy as they are decelerated in the target
- (B) electrons falling between the energy levels of the target atoms
- (C) the range of speeds of the electrons leaving the cathode
- (D) X-rays of one wavelength being scattered by different amounts in the target
Show Answer
Answer: A – In bremsstrahlung, one electron might lose 10% of its energy to a photon, another 60%, another all of it. Every energy loss up to the maximum is possible, so the photon energies (and wavelengths) form a continuous spread. The sharp lines come from option B, shell-to-shell transitions.
Q12. The sharp characteristic lines in the X-ray spectrum are produced when
- (A) an electron is decelerated to rest in a single collision
- (B) the target becomes hot enough to glow at X-ray frequencies
- (C) electrons in the beam collide with each other
- (D) an incoming electron knocks an inner-shell electron out of a target atom, and an outer electron falls into the vacancy, emitting a photon of a definite energy
Show Answer
Answer: D – The photon energy equals the difference between two atomic energy levels, so it has an exact value for that element (this is the same physics as the line spectra of Section 22.4). Because the level spacings are characteristic of tungsten, these are called characteristic lines.
Q13. An X-ray tube is operated at 40 kV. The minimum wavelength present in the continuous spectrum is
- (A) 3.1 × 10−9 m
- (B) 3.1 × 10−11 m
- (C) 5.0 × 10−11 m
- (D) 1.2 × 10−10 m
Show Answer
Answer: B – λmin = hc / (eV) = (6.63 × 10−34 × 3.00 × 108) / (1.6 × 10−19 × 4.0 × 104) = 1.99 × 10−25 / 6.4 × 10−15 = 3.1 × 10−11 m.
Q14. The shortest wavelength produced by a tube at 25 kV is approximately
- (A) 2.5 × 10−11 m
- (B) 8.0 × 10−11 m
- (C) 5.0 × 10−11 m
- (D) 1.0 × 10−9 m
Show Answer
Answer: C – λmin = hc / (eV) = 1.99 × 10−25 / (1.6 × 10−19 × 2.5 × 104) = 1.99 × 10−25 / 4.0 × 10−15 = 5.0 × 10−11 m. Lower voltage means less electron energy, so the cut-off wavelength is longer than at 40 kV.
Q15. There is a sharp cut-off at the minimum wavelength of the continuous spectrum because
- (A) a single photon cannot be given more energy than the total kinetic energy of one electron
- (B) shorter wavelengths are completely absorbed inside the target
- (C) the electrons cannot travel faster than the speed of light
- (D) the target atoms have no energy levels corresponding to shorter wavelengths
Show Answer
Answer: A – Photon energy E = hf and wavelength are linked by λ = hc / E. The most energetic possible photon is made when one electron gives up ALL of its kinetic energy eV in a single encounter, so no photon can have E > eV and no wavelength can be shorter than hc / (eV).
Q16. Increasing the potential difference across the X-ray tube (the tube voltage) while keeping the tube current the same will
- (A) increase the number of photons per second but keep their energies the same
- (B) decrease the minimum wavelength and make the beam less penetrating
- (C) increase only the intensity of the characteristic lines
- (D) increase the maximum photon energy, decrease λmin, and make the beam more penetrating (“harder”)
Show Answer
Answer: D – A higher voltage gives each electron more kinetic energy, so the most energetic photons have energy eV and λmin = hc / (eV) falls. Higher-energy photons are attenuated less, so the beam is harder and more penetrating. The tube voltage controls quality (penetration); the current controls quantity.
Q17. Increasing the tube current (mA) while keeping the tube voltage the same will
- (A) increase the maximum photon energy
- (B) increase the number of X-ray photons produced per second, so the intensity rises, but the photon energies are unchanged
- (C) decrease the minimum wavelength of the spectrum
- (D) make the beam more penetrating
Show Answer
Answer: B – A larger tube current means more electrons hit the target each second, so more photons are produced per second and the image forms faster (greater intensity/dose rate). Each electron still has energy eV, so λmin and the penetrating power do not change.
Q18. The tube current of an X-ray tube is controlled by the filament (heating) current. Increasing the filament current increases the tube current because
- (A) a hotter filament increases the accelerating voltage of the tube
- (B) a hotter filament lowers the work function of the target
- (C) a hotter filament emits more electrons per second by thermionic emission, so more charge reaches the target per second
- (D) a hotter filament reduces the resistance of the vacuum
Show Answer
Answer: C – Thermionic emission rises steeply with temperature. A larger filament current makes the cathode hotter, releasing more electrons per second, and the tube current (charge arriving at the anode per second) rises in step.
Q19. A tube is run at 30 kV. The maximum possible energy of any X-ray photon it produces is
- (A) 4.8 × 10−15 J
- (B) 3.0 × 10−15 J
- (C) 6.6 × 10−16 J
- (D) 30 J
Show Answer
Answer: A – Emax = eV = 1.6 × 10−19 × 3.0 × 104 = 4.8 × 10−15 J, which is 30 keV. No photon can exceed this, because no electron arrives with more energy than this.
Q20. An X-ray photon has a wavelength of 4.0 × 10−11 m. Its energy is approximately
- (A) 5.0 × 10−18 J (0.031 keV)
- (B) 2.5 × 10−15 J (16 keV)
- (C) 8.0 × 10−15 J (50 keV)
- (D) 5.0 × 10−15 J (31 keV)
Show Answer
Answer: D – E = hc / λ = 1.99 × 10−25 / 4.0 × 10−11 = 5.0 × 10−15 J. In electronvolts: 5.0 × 10−15 / 1.6 × 10−19 ≈ 31,000 eV = 31 keV.
Attenuation & Medical Use
Q21. A parallel X-ray beam passes through a material of attenuation coefficient μ = 0.50 cm−1. The fraction of the incident intensity transmitted by a thickness of 2.0 cm is
- (A) 0.50
- (B) 0.37
- (C) 0.25
- (D) 0.14
Show Answer
Answer: B – I / I0 = e−μx = e−(0.50 × 2.0) = e−1.0 = 0.37. Each successive centimetre absorbs the same fraction of what is left, giving the exponential fall, never a sudden cut-off.
Q22. A beam passes through a thickness of material equal to three half-value thicknesses. The fraction of the original intensity that emerges is
- (A) 1/8
- (B) 1/6
- (C) 1/3
- (D) 1/9
Show Answer
Answer: A – Each half-value thickness halves the intensity: (1/2)3 = 1/8 = 12.5% of I0 gets through.
Q23. A material has an attenuation coefficient μ = 0.35 mm−1 for a particular X-ray beam. Its half-value thickness is approximately
- (A) 0.25 mm
- (B) 0.50 mm
- (C) 2.9 mm
- (D) 2.0 mm
Show Answer
Answer: D – x1/2 = ln 2 / μ = 0.693 / 0.35 = 1.98 mm ≈ 2.0 mm. A larger μ means stronger attenuation and a smaller half-value thickness.
Q24. In a radiograph, bones appear much lighter (block far more of the beam) than the surrounding soft tissue because bone
- (A) is thicker than the tissue around it
- (B) reflects X-rays back towards the tube
- (C) has a larger attenuation coefficient, being denser and containing higher-Z elements (calcium), so it absorbs a greater fraction of the beam
- (D) emits its own X-rays when irradiated
Show Answer
Answer: C – Attenuation rises strongly with density and with atomic number Z. Bone contains calcium (Z = 20) and is much denser than soft tissue (mostly low-Z elements such as H, C, N, O), so far fewer photons pass through bone and the image shows contrast.
Q25. When imaging the blood vessels or the digestive system, the patient is given a contrast medium containing barium or iodine because these substances
- (A) have a high atomic number, so they attenuate X-rays strongly and make the structure stand out against soft tissue
- (B) are radioactive, so the structure emits radiation that can be detected
- (C) are magnetic, so they can be guided to the correct organ
- (D) have a low attenuation coefficient, so they appear darker than tissue
Show Answer
Answer: A – Soft tissues all attenuate X-rays by similar amounts, so vessels and the gut would be almost invisible. Barium (Z = 56) and iodine (Z = 53) absorb far more strongly, filling the structure with a high-attenuation material and creating clear contrast. This is the basis of the “barium meal”.
Q26. In a CT (computed tomography) scan,
- (A) a single X-ray photograph is taken and enlarged by a computer
- (B) the X-ray source and detectors rotate around the patient, taking attenuation measurements from many angles, and a computer combines them into cross-sectional (and 3D) images
- (C) the patient is placed in a strong magnetic field and radio waves are detected
- (D) gamma rays emitted by a tracer inside the patient are detected
Show Answer
Answer: B – By measuring how much the beam is attenuated along thousands of different paths through one slice of the body, software can work out the attenuation coefficient of each small volume element and build up a detailed cross-section. Repeating this for many slices gives a 3D image. Option C describes MRI and D describes PET scanning.
Q27. The beam from a medical X-ray tube is passed through an aluminium filter before it reaches the patient. This removes the low-energy (“soft”) photons in order to
- (A) increase the total intensity of the beam
- (B) make the image sharper
- (C) increase the contrast between bone and tissue
- (D) reduce the dose to the patient, because soft photons would be absorbed in the skin and surface tissue without ever reaching the detector
Show Answer
Answer: D – Attenuation is greatest for low-energy photons, so soft photons contribute almost nothing to the image but deposit their energy in the patient, increasing the dose and the risk of harm. Filtering them out “hardens” the beam and keeps the dose as low as reasonably possible.
Q28. A beam of intensity I0 enters tissue for which μ = 0.46 cm−1. The fraction of the intensity remaining after travelling 3.0 cm is
- (A) 0.75
- (B) 0.50
- (C) 0.25
- (D) 0.14
Show Answer
Answer: C – μx = 0.46 × 3.0 = 1.38, so I / I0 = e−1.38 ≈ 0.25. (Notice 1.38 ≈ ln 4, so the intensity has fallen by a factor of four, i.e. two half-value thicknesses.)
Q29. A beam has an incident intensity of 800 units. After passing through a thickness equal to two half-value thicknesses of an absorber, the transmitted intensity is
- (A) 200 units
- (B) 400 units
- (C) 100 units
- (D) 600 units
Show Answer
Answer: A – Two half-value thicknesses reduce the intensity to (1/2)2 = 1/4 of the original: 800 / 4 = 200 units.
Q30. For a beam with attenuation coefficient μ = 0.50 cm−1 in lead, the thickness of lead needed to reduce the intensity to 10% of its incident value is
- (A) 1.4 cm
- (B) 4.6 cm
- (C) 2.0 cm
- (D) 10 cm
Show Answer
Answer: B – I / I0 = e−μx = 0.10, so μx = ln 10 = 2.30 and x = 2.30 / 0.50 = 4.6 cm. This kind of calculation is used to design the lead shielding around X-ray rooms.
Keep practising: revise the whole of Medical Physics and the rest of A Level Physics on the Physics MCQs hub, and keep the key equations at hand with the A-Level Physics Formula Sheet.
These are original practice questions written for revision. They are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.