AQA A-Level Physics Paper 3 (3BA), June 2025: Question 5

5 marks · Medium difficulty · Extended Answer

Discuss whether the conclusions of two student groups regarding Betelgeuse's revised intensity, distance, and radius could be correct, assuming constant temperature.

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Question

Question 05 shows early observational data for Betelgeuse: radius r = 700 times the radius of the Sun, distance d = 548 ly, and received intensity = I. Recent data suggest the intensity received is actually 1.15 I. Group X suggests that r is correct and d is actually 530 ly. Group Y suggests that d is correct and r is actually 750 times the radius of the Sun. Students are asked to discuss for each group whether their conclusion could be correct, assuming constant measured temperature.
Question text

05 Early observations of the star Betelgeuse indicated that:

radius r of Betelgeuse = 700 × radius of the Sun

distance d of Betelgeuse from the Earth = 548 ly

intensity received from Betelgeuse = I.

Recent data suggest that the intensity received from Betelgeuse is actually 1.15I.

Based on these recent data, two groups of students reach different conclusions.

Group X suggest that:

• r is correct

• d is actually 530 ly.

Group Y suggest that:

• d is correct

• r is actually 750 × radius of the Sun.

Discuss for each group whether their conclusion could be correct.

Assume that the measured temperature of Betelgeuse is constant.

[5 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 05 allocating up to 5 marks: 1st mark for Group X using inverse-square law (I is proportional to 1/d squared); 2nd mark for showing ratio of intensities gives I_1 = 1.07 I_2 or d = 511 ly; 3rd mark for Group Y using power P is proportional to r squared; 4th mark for calculating (750/700)^2 = 1.15 (or r = 750 solar radii); 5th mark for concluding that Group Y could be correct and Group X is not correct.

Question Answers Additional comments/Guidance Mark AO

05 Group X: Allow any subscripts, e.g. 1/2, x/y, etc. 5 AO3

Use of inverse-square law (to arrive at a value for I or d): 1

Allow conclusion to be drawn on basis of ratios

(I ∝ 2 )

d

2 r2

I1 530

(Ratio of intensities: = ) When I ∝ 2 is seen and used, award 1 and

I 5482 d

I1 = 1.07 I2 or d = 511 (ly) 2 3 . Expect k = 0.613.

Group Y:

Use of 2 (to arrive at a value of P or r) Accept I for P

P ∝ r 1 3

(A = 4πr2)

Alternative approach:

(Ratio of power:

P 7502 When 2.510.0725 = 1.07 seen, award and .

11 2

Use of = 2 or alternative)

P2 700

leading to a calculation of P1 = 1.15 P2 or r = 75(0)r• 4 Some evidence of relevant working for both X

and Y must be seen to award 5

So Group Y could be correct AND Group X not correct. 5

Total 5

How to answer it

Evaluating Stellar Brightness: Inverse-Square Law & Stefan-Boltzmann

📌 What this question tests

This 5-mark evaluation question assesses your ability to apply astronomical physics equations proportionally to test competing scientific hypotheses:

  • Inverse-Square Law for Radiation: Intensity received at Earth relates to distance by I = P / (4πd²) , meaning I ∝ 1 / d² .
  • Stefan-Boltzmann Law: Total radiant power relates to surface area and radius by P = σAT⁴ = 4πr²σT⁴ , meaning P ∝ r² at constant temperature.
  • Proportional Reasoning & Ratios: Formulating mathematical ratios to test predictions without needing absolute SI unit conversions.
  • Scientific Evaluation: Comparing calculated values against measured data (1.15I) to state a clear conclusion for each group.

Part 1: Evaluating Group X's Claim

Hypothesis: Radius r is correct; distance d is actually 530 ly (down from 548 ly).

💡 Underlying Physics

Since temperature T and stellar radius r are held constant, the star's total luminosity (power output P ) remains unchanged.

The intensity I received on Earth obeys the inverse-square law:

I ∝ 1 / d²  or  I₁ × d₁² = I₂ × d₂²

📐 Step-by-Step Calculations

Method A: Calculate resulting intensity from 530 ly
I_new / I_old = (d_old / d_new)²
I_new / I = (548 / 530)² = (1.034)² = 1.07
Predicted intensity = 1.07 I (not 1.15 I).
Method B: Calculate required distance for 1.15 I
1.15 = (548 / d_new)²
d_new = 548 / √1.15 = 548 / 1.072 = 511 ly
Required distance is 511 ly, not 530 ly.
Mark Scheme Breakdown:
[Mark 1] Use of inverse-square law ( I ∝ 1 / d² ) to set up a valid relationship.
[Mark 2] Determining that the new intensity is 1.07 I (or calculating required d = 511 ly ).

Part 2: Evaluating Group Y's Claim

Hypothesis: Distance d is correct; radius r is actually 750 r_Sun (up from 700 r_Sun).

💡 Underlying Physics

Distance d is constant. From Stefan-Boltzmann law:

P = 4πr²σT⁴

Since temperature T is stated as constant, power is directly proportional to surface area, which depends on radius squared:

P ∝ r²  and because d is constant,  I ∝ r²

📐 Step-by-Step Calculations

Method A: Calculate resulting power/intensity ratio
P_new / P_old = (r_new / r_old)²
P_new / P = (750 / 700)² = (15 / 14)² = 1.0714² = 1.148 ≈ 1.15
Predicted intensity = 1.15 I.
Method B: Calculate required radius for 1.15 I
r_new = r_old × √1.15
r_new = 700 × 1.0724 = 750.7 ≈ 750 × radius of Sun
Required radius matches Group Y's claim.
Mark Scheme Breakdown:
[Mark 3] Use of P ∝ r² (via A = 4πr² ) to set up power/intensity ratio.
[Mark 4] Determining ratio P_new = 1.15 P_old (or finding radius r ≈ 750 r_Sun ).

Part 3: Final Conclusion & Mark Strategy

✅ Final Definitive Conclusion (Mark 5)

Group Y could be correct, and Group X is not correct.

Examiner Condition: To secure this 5th mark, relevant numerical working for both Group X and Group Y must be clearly shown in your answer.

🧠 Exam Technique: Ratios Save Time

  • Do NOT convert units: Converting light years into metres or solar radii into metres wastes time and invites power-of-10 calculator errors. All conversion factors cancel out in ratio form!
  • Combined form: You can combine both laws directly as I ∝ r² / d² . When examiners see this relationship stated and used, Marks 1 and 3 are credited immediately.
  • Answer the prompt fully: The question asks to "discuss for each group whether their conclusion could be correct". Never finish your calculation without writing an explicit sentence judging each group.

❌ Common Traps & Misconceptions

  • Forgetting to square: Treating I as inversely proportional to d (linear) giving 548 / 530 = 1.03 instead of squaring to get 1.07 .
  • Inverting the distance ratio: Writing (530 / 548)² = 0.935 and incorrectly concluding intensity decreases despite being closer. Closer stars are brighter!
  • Linear radius assumption: Assuming luminosity scales directly with radius ( P ∝ r ) instead of surface area ( P ∝ r² ).

Topics

Optional topics · 3.9 Astrophysics (A-level only)

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