OCR A Level Physics A: X-rays, attenuation and CAT scanning (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 6.5.1 (Using X-rays), statements (a) to (g), of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Use e = 1.60 × 10⁻¹⁹ C, h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹. The rest energy of an electron is 0.511 MeV. Start with the study guide or the revision notes.

Questions

1. State the function of each of these parts of an X-ray tube: (i) the heater (cathode), (ii) the high voltage supply, (iii) the target metal, (iv) the vacuum. [4]

2. A tube operates at 70 kV.

(a) Calculate the maximum energy of an X-ray photon from the tube, in joules. [2] (b) Calculate the shortest wavelength emitted. [2]

3. The same tube has a tube current of 9.0 mA. About 1% of the electrical power supplied becomes X-rays.

(a) Calculate the number of electrons that strike the target each second. [2] (b) Calculate the thermal energy produced in the target in 2.0 s. [3] (c) Suggest one way the tube is designed to cope with this energy. [1]

4. Explain why the spectrum from an X-ray tube has (a) a smooth continuous background with a sharp minimum wavelength, and (b) narrow peaks at particular wavelengths. [5]

5. Name the attenuation mechanism described in each case.

(a) A low-energy photon changes direction without losing energy. [1] (b) A photon is completely absorbed and an inner-shell electron is ejected. [1] (c) A photon collides with a loosely bound outer electron and continues with less energy. [1] (d) A photon near a nucleus becomes an electron and a positron. [1] (e) Calculate the minimum photon energy for the process in (d), in MeV and in joules. [2]

6. A narrow X-ray beam of a single photon energy passes through 7.5 cm of a uniform material. 18% of the incident intensity is transmitted.

(a) Calculate the attenuation coefficient, in cm⁻¹. [2] (b) Calculate the half-value thickness. [2]

7. A beam passes through 3.0 cm of soft tissue (μ = 0.22 cm⁻¹) followed by 2.0 cm of bone (μ = 0.55 cm⁻¹).

(a) Calculate the fraction of the incident intensity transmitted. [3] (b) Calculate the fraction transmitted by 5.0 cm of soft tissue alone, and explain what the difference means for the image. [3]

8. A shielding material has an attenuation coefficient of 35 cm⁻¹ for a certain X-ray beam.

(a) Calculate the thickness needed to reduce the intensity to 1.0% of its incident value. [3] (b) Calculate the extra thickness needed to reduce it from 1.0% to 0.10%. [2]

9. A student measures the count rate R of a narrow beam transmitted through different thicknesses x of a metal sheet.

x / cm	0	1	2	3	4	5
R / counts per second	840	525	328	205	128	80

(a) Show that a graph of ln R against x should be a straight line, and state what its gradient represents. [2] (b) Calculate ln R for x = 0 and x = 5, and use them to find the attenuation coefficient. [3] (c) State the unit of the attenuation coefficient and give one reason why a best-fit line through all points is better than using two points. [2]

10. An artery of internal width 0.60 cm lies in soft tissue. The blood has μ = 0.21 cm⁻¹. When it contains an iodine compound, its effective attenuation coefficient is 1.9 cm⁻¹.

(a) Explain why a barium meal is used to image the stomach on a plain X-ray image. [2] (b) Explain why barium and iodine are suitable contrast media. [2] (c) Calculate the fraction of intensity transmitted across the artery without and with the iodine compound. [2]

11. Describe how a CAT scanner produces an image of a slice of the body, naming the components involved. [6]

12. A patient has had a plain X-ray image of the abdomen but the doctor suspects a small tumour behind other organs. A CAT scan is offered.

(a) Explain two advantages of a CAT scan over the plain X-ray image in this situation. [4] (b) Discuss whether a CAT scan should be used to check a minor injury that a plain X-ray image can already assess. [4]

Answers

1. (i) The heater is a hot filament that releases electrons by thermionic emission [1]. (ii) The high voltage supply accelerates the electrons from cathode to anode, giving each kinetic energy eV [1]. (iii) The target metal is where the electrons are stopped and X-ray photons are emitted [1]. (iv) The vacuum stops electrons colliding with gas molecules on the way [1]. [4] Examiner insight: Each part needs a function, not a name. “The heater heats the tube” does not say that electrons are released.

2. (a) E = eV = 1.60 × 10⁻¹⁹ × 70 000 [1] = 1.12 × 10⁻¹⁴ J [1] [2] (b) λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(1.12 × 10⁻¹⁴) [1] = 1.78 × 10⁻¹¹ m [1] [2] Examiner insight: Show the substitution in joules. Convert keV before using hc.

3. (a) n = I/e = 9.0 × 10⁻³/1.60 × 10⁻¹⁹ [1] = 5.6 × 10¹⁶ s⁻¹ [1] [2] (b) Power in = VI = 70 000 × 9.0 × 10⁻³ = 630 W [1]. Thermal power = 0.99 × 630 = 624 W [1]. Energy in 2.0 s = 624 × 2.0 = 1.2 × 10³ J [1] [3] (c) Any one of: rotating anode, copper anode to conduct heat, oil or fins for cooling [1] [1] Examiner insight: State that 99% is thermal before multiplying by 0.99.

4. (a) Electrons are decelerated by target nuclei and lose varying amounts of energy [1], so photons of all energies up to a maximum are emitted [1]. The most energetic photon takes all of an electron’s energy, eV = hc/λ_min, so there is a sharp minimum wavelength [1]. (b) An incoming electron can eject an inner-shell electron from a target atom [1]. An outer electron falls into the gap and emits a photon with energy equal to the difference between the levels, which is fixed by the target element, so a narrow peak appears [1]. [5] Examiner insight: Link each feature to its cause. A name with no mechanism rarely gains full credit.

5. (a) Simple scatter [1] (b) Photoelectric effect [1] (c) Compton effect [1] (d) Pair production [1] (e) Two particles are made, so E = 2 × 0.511 MeV = 1.02 MeV [1] = 1.022 × 10⁶ × 1.60 × 10⁻¹⁹ = 1.64 × 10⁻¹³ J [1] [6] Examiner insight: In (e) two particles are made, so use two rest energies.

6. (a) I/I₀ = e^(−μx), so ln(1/0.18) = μ × 7.5 [1]. μ = 1.715/7.5 = 0.229 cm⁻¹ [1] [2] (b) x½ = ln 2/μ = 0.693/0.2286 [1] = 3.03 cm [1] [2] Examiner insight: Use the natural logarithm, not log to base 10, and quote the unit.

7. (a) Total exponent = 0.22 × 3.0 + 0.55 × 2.0 [1] = 0.66 + 1.10 = 1.76 [1]. I/I₀ = e^(−1.76) = 0.17 [1] [3] (b) I/I₀ = e^(−0.22 × 5.0) = e^(−1.10) = 0.33 [1]. The beam through the bone path is about half as intense (0.17/0.33 = 0.52) [1], so the bone casts a shadow and gives contrast in the image [1] [3] Examiner insight: Add the exponents for layers. Adding intensities is a common wrong route.

8. (a) I/I₀ = 0.010, so μx = ln 100 = 4.605 [1]. x = 4.605/35 [1] = 0.13 cm (1.3 mm) [1] [3] (b) From 1.0% to 0.10% is a factor of 10 [1]. Extra thickness = ln 10/35 = 0.066 cm [1] [2] Examiner insight: Each factor-of-ten reduction needs the same extra thickness, so (b) is quicker than starting again.

9. (a) Taking natural logs of R = R₀e^(−μx) gives ln R = ln R₀ − μx [1], which has the form y = c + mx. The gradient is −μ (minus the attenuation coefficient) [1] [2] (b) ln 840 = 6.73 and ln 80 = 4.38 [1]. Gradient = (4.38 − 6.73)/5 = −0.47 [1]. μ = 0.47 cm⁻¹ [1] [3] (c) Unit: cm⁻¹ [1]. A best-fit line uses every reading, so random errors average out [1] [2] Examiner insight: Give the gradient its sign, then state μ as positive.

10. (a) Soft tissue in the stomach wall and its surroundings have similar attenuation coefficients, so there is little contrast [1]. Barium strongly absorbs X-rays, so the barium sulfate suspension outlines the inside of the stomach [1] [2] (b) They have high atomic numbers (56 and 53) [1], so photons are strongly absorbed, mainly by the photoelectric effect, compared with soft tissue [1] [2] (c) Without: e^(−0.21 × 0.60) = 0.88 [1]. With: e^(−1.9 × 0.60) = 0.32 [1] [2] Examiner insight: In (b) name the property (high atomic number) and the consequence (strong absorption). “They show up well” is too vague.

11. The X-ray tube rotates around the patient [1] and produces a thin fan-shaped beam that passes through one slice of the body [1]. A ring of detectors records the transmitted intensity at many positions [1]. Because the tube turns, each part of the slice is crossed by beams from many different angles [1]. Computer software combines all the readings and calculates the attenuation of each small volume in the slice [1]. The display shows this as a grey-scale cross-sectional image, with brightness set by the attenuation [1]. [6] Examiner insight: Give each component a role, in order. Names alone earn only part of the marks.

12. (a) First advantage: the CAT scan gives cross-sectional slices, so a tumour is not hidden behind overlapping organs as in a single flat image [1], and the doctor can see exactly where it lies [1]. Second advantage: small differences in attenuation between soft tissues are shown with better contrast [1], so a tumour with attenuation close to its surroundings can still be seen [1]. [4] (b) Benefit: if the plain image already answers the question, the CAT scan adds little diagnostic information [1]. Risk: a CAT scan usually gives a larger dose of ionising radiation, which can damage cells and slightly raises the chance of cancer [1]. It also costs more [1]. Conclusion: it should not be used when the plain image is sufficient, but would be justified if the plain image were unclear or the injury were serious [1] [4] Examiner insight: A “discuss” question needs both sides and a reasoned conclusion.

Where marks are usually lost

• Forgetting that about 99% of the electrical power becomes thermal energy, not X-rays.
• Describing characteristic lines or the continuous spectrum by name only, with no mechanism.
• Confusing the photoelectric effect (photon absorbed) with Compton scattering (photon survives with less energy).
• Using one electron rest energy instead of two for the pair production threshold.
• Adding intensities, or multiplying coefficients, for layers of different material.
• Using log to base 10 instead of the natural logarithm in the attenuation law.

Next steps

• Go over the revision notes and the study guide.
• Return to the OCR A Level Physics hub and tick items off the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Section 6.5.1 Using X-rays.