AQA A-Level Physics Paper 3 (3BB), June 2025: Question 4

8 marks · Medium difficulty · Short Answer

Explain focal spot size considerations, methods to limit patient X-ray exposure, and calculate the transmission percentage of X-rays through a femur cross-section.

Practise this question

Question

Figure 5 shows a schematic of a rotating-anode X-ray tube showing the cathode and the angled rotating anode target. Question 04.1 asks for an advantage of a small focal spot and why it should not be too small. Question 04.2 asks for three methods and explanations to limit exposure of a patient's healthy cells to X-rays. Figure 6 shows a scale cross-section diagram of a femur with an outer diameter of 27 mm and a hollow center, showing an X-ray beam traversing from A to B across the diameter. Question 04.3 asks to determine the percentage of X-ray photons incident at A that emerge at B, given a half-value thickness of 15 mm.
Question text

04 Figure 5 shows a rotating-anode X-ray tube used to produce an image of a bone.

Figure 5

04.1 X-rays originate from a region on the rotating anode known as the focal spot.

Explain:

• one advantage of using a small focal spot

• why the focal spot should not be too small.

[2 marks]

04.2 State and explain three methods used to limit exposure of the patient’s healthy cells

to X-rays during a diagnostic procedure.

[3 marks]

04.3 Figure 6 shows a scale diagram of a cross-section through a human bone called

a femur.

Figure 6

A dashed line has been drawn from point A to point B through the centre of the femur.

Along AB the femur has a circular cross-section with an external diameter of 27 mm.

The femur has a hollow centre.

A beam of X-ray photons is incident on the femur at A in the direction AB.

For these photons, the half-value thickness of bone is 15 mm.

Determine, using the scale diagram in Figure 6, the percentage of the X-ray photons

incident on A that emerge at B.

[3 marks]

Mark scheme

Show the mark scheme The mark scheme details marks for 04.1: sharp image/reduced penumbra (1 mark) and avoiding anode overheating (1 mark). For 04.2, up to 3 marks for matching methods to explanations across reducing target area, reducing intensity, or avoiding repeats. For 04.3, 3 marks awarded for calculating the linear attenuation coefficient or using half-value thickness, measuring bone thickness x between 17 mm and 19 mm, and calculating a final transmission percentage between 41% and 46% (e.g. 44%).

Question Answers Additional comments/Guidance Mark AO

04.1 (small size of focal spot) to increase sharpness of image 1 for 1 allow ‘to reduce (size of) penumbra’ / 2 AO1

‘reduce region of geometric unsharpness’

Condone:

• improve (spatial) resolution

• see finer detail

• reduce blurring at edges (of image)

Neutral statements include

• more precise beam / more accurate beam

• clearer image / better image

• better focused

for 2 allow:

Idea of (must not be too small otherwise) danger of

• to limit temperature increase (of anode)

overheating (the anode) 2

• to prevent it heating up too quickly

Where no other mark is scored allow Max 1 for two correct methods

04.2 3 AO1

Method Explanation Explanation for limit exposure

one correct (how it works) (why it is necessary)

method and For reducing target area

one aspect of ‘window’ defining target area / Highly attenuating material that reduces x-ray Less exposure to surrounding tissue /

a relevant tube surrounded in lead (casing) propagation in undesired direction thereby body

/ collimator / set aperture of producing a directed beam

explanation diaphragm

1 use of a lead apron on patient Highly attenuating material reduces x-rays Exposure to rest of body is reduced

(to shield area not being

imaged)

To reduce the divergence in beam Less exposure for rest of body / healthy 17

reduce distance between cells

second source and patient

correct

keep the patient still Only intended target is exposed to X-rays Less exposure (for healthy cells) in other

method and parts of body

one aspect of

relevant

For reducing X-ray intensity

explanation

Idea of (unnecessary) exposure to ‘softer’

2 use of an (aluminium) filter / % of less energetic X-rays is reduced by filter

x-rays is reduced

beam hardening

use of intensifying screen / Exposure (time) for (healthy cells) is

contrast medium / flat panel To ensure adequate image quality (from a lower minimised

detector / FTP detector intensity beam)

third correct

method and For avoiding repeats

one aspect of

relevant correct tube voltage

to obtain best / appropriate X-ray (energy) Image quality maximised without need to

explanation repeat so minimising exposure (for healthy

cells)

keep the patient still Reduces blurring in image. The x-ray doesn’t need to be retaken

therefore minimising exposure (for healthy

cells)

reduce distance between To reduce the divergence in beam Image quality maximised without need to

source and patient repeat so minimising exposure

04.3 Max Two from for (1) and (2) allow working in m or cm 3 1 × AO2

μ = 46.2 ( m-1) μ = 0.462 (cm-1) 2 × AO3

• μ = 4.6(2) × 10−2 (mm-1) (1)

ln 2

for (1) accept 𝜇𝜇 = or equivalent in m or cm

for (2)

• calculates thickness x (2) thickness x in range 17 mm to 19 mm

Expect to see a calculation based on scale

diagram

𝐼𝐼 −𝜇𝜇𝜇𝜇

• Use of = 𝑒𝑒 with their μ and their thickness x

𝐼𝐼0 In (3), their x must be a distance, condone power

OR of ten errors

1 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡ℎ𝑒𝑒𝑖𝑖𝑖𝑖𝑖𝑖𝑒𝑒𝑖𝑖𝑖𝑖 𝑥𝑥

calculates ( ) 15 (3)

2 An answer for solid bone = 29% (28.7%) is worth

2 marks.

accept an answer in range 41% to 46%

(percentage=) 44 (%) reject decimal answer

reject answer rounded to one significant figure

Total 8

How to answer it

X-Ray Production, Radiation Safety & Exponential Attenuation

📋 What this question tests

This question assesses core Medical Physics concepts: the trade-off between image sharpness and heat dissipation in rotating-anode tubes, practical clinical methods for radiation protection of healthy tissue, and quantitative exponential attenuation calculations using half-value thickness (HVT) combined with scale diagram measurement.

Part 04.1 — Anode Focal Spot Size

Explain one advantage of a small focal spot and why it should not be too small [2 marks]

✅ Model Answer

  • Advantage: Increases image sharpness / spatial resolution (reduces penumbra / geometric unsharpness). [1 mark]
  • Limitation: Risk of overheating or melting the anode due to concentrated thermal energy over a tiny area. [1 mark]

💡 Key Physics Concepts

Over 99% of kinetic energy from incident electrons is converted into heat, with less than 1% emitted as X-rays.

  • Point source vs. finite area: A smaller spot behaves closer to an ideal point source, casting narrower shadow edges (less penumbra).
  • Thermal stress: High tube current concentrated onto too small an area causes local melting, pitting, and tube failure.

🧠 Examiner Insight

Vague statements like "gives a clearer picture", "more accurate beam", or "better focused" receive no credit. Always use formal terminology: image sharpness, spatial resolution, or reduction of penumbra.

❌ Common Errors

  • Failing to mention the word heat, overheating, or temperature rise for the second mark.
  • Confusing the size of the focal spot with the overall size or width of the X-ray beam field.
Mark Allocation: 1 mark for sharpness/penumbra reduction + 1 mark for thermal limit/overheating risk.

Part 04.2 — Limiting Patient Exposure

State and explain three methods used to limit exposure of healthy cells [3 marks]

✅ Three High-Scoring Method & Explanation Pairs

Category Method (State) Explanation (How & Why it protects healthy cells)
Beam Filtration Aluminium filter (beam hardening) Absorbs low-energy ("soft") X-rays that would otherwise be absorbed by surface skin/tissue without contributing to the diagnostic image.
Beam Collimation Collimator / lead diaphragms / aperture window Restricts the beam strictly to the area of clinical interest, preventing unnecessary exposure to surrounding healthy tissues.
Detection Sensitivity Intensifying screen / flat-panel digital detector Converts X-rays to visible light / digital signal efficiently, allowing adequate image quality at a much lower exposure time and total dose.

🧠 Exam Technique: "State and Explain"

To score each mark, you must provide both the method and a scientifically valid mechanism. Simply listing three items without explaining how they protect the patient yields a maximum of 1 mark overall.

❌ What NOT to say

  • "Wear a lead apron" without specifying that it is placed over non-target areas (e.g., gonads or thyroid).
  • Stating "reduce exposure time" without explaining how (e.g., via intensifying screens or avoiding repeat exposures by immobilising the patient).
Mark Allocation: 1 mark per correct method paired with an appropriate explanation (Max 3 marks).

Part 04.3 — Scale Diagram & Bone Attenuation

Determine the percentage of incident X-ray photons that emerge at B [3 marks]

📐 Step-by-Step Calculation

Step 1: Determine bone thickness along path AB

  • External diameter = 27 mm .
  • Measure hollow core diameter from Figure 6: inner diameter ≈ 9 mm (measured accurately via scale).
  • Total bone traversed = External diameter − Inner hollow diameter:
    x = 27 mm − 9 mm = 18 mm (acceptable range: 17 mm to 19 mm ).

Step 2: Calculate the linear attenuation coefficient μ (or use half-value ratio)

Method A — Using Half-Value Thickness (HVT = 15 mm):

μ = ln(2) / HVT = 0.69315 / 15 mm = 0.0462 mm⁻¹ (or 46.2 m⁻¹)

Step 3: Calculate the transmission fraction and percentage

Using the attenuation law I / I₀ = e^(−μx) or (1/2)^(x / HVT) :

I / I₀ = e^(−0.0462 × 18) = e^(−0.8316) ≈ 0.4353

Or alternatively:

I / I₀ = (0.5)^(18 / 15) = (0.5)^1.2 ≈ 0.4353

Convert to percentage:

Percentage = 0.4353 × 100% ≈ 44%

❌ Common Traps & Marking Penalties

  • Solid bone assumption: Forgetting the hollow centre and using x = 27 mm gives (0.5)^(27/15) = 28.7% ≈ 29% . The mark scheme caps this at 2 marks maximum.
  • Rounding traps: The mark scheme strictly instructs: "reject decimal answer" (e.g. 43.5%) and "reject answer rounded to one sig fig" (e.g. 40%). You must write an integer percentage: 44% (acceptable range: 41% to 46%).

🧠 Examiner Checklist

  • Did you show the scale measurement for the inner hollow diameter?
  • Are your units consistent? If using μ in mm⁻¹ , your thickness x must be in mm .
  • Check the final format required on the answer line ( % was already printed, so give an integer).
Mark Allocation: Mark 1 for calculating μ = 0.046 mm⁻¹ OR setting up HVT ratio • Mark 2 for correct bone thickness (17–19 mm) and substitution • Mark 3 for final answer in range 41% to 46% (integer).

Topics

Optional topics · 3.10 Medical physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 3 (3BB), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.