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

10 marks · Medium difficulty · Extended Answer

Analyze data and a light curve for a binary star system to confirm distance consistency, identify spectral absorption lines, compare temperatures, explain eclipse dips, and calculate the angular velocity.

Practise this question

Question

Question 04 contains Table 2 with data for stars Mintaka Aa1 and Aa2: Absolute magnitudes are -5.4 and -2.9, Apparent magnitudes are 2.5 and 5.0, and Spectral classes are O and B, respectively. Question 04.1 asks to show that the data are consistent with the two stars being in a binary system (3 marks). Question 04.2 asks for the most prominent absorption lines in the spectrum of Mintaka Aa2 (1 mark). Question 04.3 asks which star has the greater temperature (1 mark). Figure 2 presents a graph of apparent magnitude versus time in days from 2 to 12 days, showing periodic dips labeled P and Q corresponding to an eclipsing binary light curve. Question 04.4 asks to explain the variation in apparent magnitude shown in Figure 2 (3 marks). Question 04.5 asks to determine the angular velocity of the binary system in rad s^-1 (2 marks).
Question text

04 Mintaka Aa1 and Mintaka Aa2 are two stars in a binary system. This means that they

orbit their common centre of mass with the same period.

Table 2 gives information about the two stars.

Table 2

Star Absolute magnitude Apparent magnitude Spectral class

Mintaka Aa1 −5.4 2.5 O

Mintaka Aa2 −2.9 5.0 B

04.1 Show that the data in Table 2 are consistent with the two stars being in a binary

system.

[3 marks]

04.2 State the most prominent absorption lines in the spectrum of Mintaka Aa2.

[1 mark]

04.3 Explain which star in Table 2 has the greater temperature.

[1 mark]

*08* 10

Figure 2 shows how the apparent magnitude of the binary system varies with time.

Figure 2

04.4 Explain the variation in apparent magnitude shown in Figure 2.

In your answer you should explain:

• the changes in observed brightness at P and Q, as shown in Figure 2

• why the observed brightness at P is different from the observed brightness at Q.

[3 marks]

04.5 Determine the angular velocity of the binary system.

[2 marks]

angular velocity = rad s−1

Mark scheme

Show the mark scheme Mark scheme for Question 04: 04.1 awards 3 marks for correct substitution into m - M = 5 log(d/10) or rearrangement for d, calculating d = 380 pc for both stars (or noting m - M = 7.9 for both), and stating that their distances are equal so they are in a binary system. 04.2 awards 1 mark for 'H and He'. 04.3 awards 1 mark for Mintaka Aa1 with reference to O being hotter than B. 04.4 awards 3 marks: 1 for light intensity dipping when one star eclipses the other, 1 for brightness being highest when both are visible, and 1 for linking dip depths to temperature differences (largest dip Q occurs when the cooler star eclipses the hotter star). 04.5 awards 2 marks for determining time period T approx 5.7 days (range 5.6 to 5.8) and calculating omega = 2pi/T = 1.3 x 10^-5 rad s^-1.

Question Answers Additional comments/Guidance Mark AO

04.1 Correct substitution into 3 AO2

d

m − M = 5log

or rearrangement for d d = 380.189 pc

Calculation for Aa1 d = 380 pc AND Calculation for Aa2 d =

380 pc

OR

m − M = 7.9 stated for both stars

Statement that their two calculated distances are the same

distance from Earth / away

OR

Statement that as 𝑚𝑚 − 𝑀𝑀 is the same they must be the same

distance from Earth / away

04.2 H AND He Reject when other elements given 1 AO1

Reject when ions implied

04.3 (Mintaka Aa1) 1 AO1

Reference to O hotter than B

04.4 3 AO3

(idea that) when one star passes in front of the other the

amount of light received changes / light intensity ‘dips’.

(idea that) the brightest (lowest value of) apparent magnitude

occurs when both stars can be seen.

Max 1 from: In MP3 do not condone “size / bigger / smaller”

• The difference in the ‘dips’ suggests that the stars have etc. unless they are linked to temperature /

different temperatures / brightnesses. brightness

• The biggest change / Q is when the cooler/dimmer star

is in front of the hotter/brighter / when Aa2 is in front of Accept the converse for smallest change / P.

Aa1.

04.5 T ≈ 5.7 days. 2 AO2

Range in T: 5.6 – 5.8 days

2π

(Calculation of ω = )

T Allow MAX 1 for use of half of the time period.

2π −5 −1

14 ω = ( = ) 1.3 × 10 (rad s ) Accept (1.25 – 1.30) × 10−5 (rad s−1)

5.7×24×60×60

Total 10

How to answer it

Binary Star Systems & Light Curves (Mintaka)

📌 What This Question Tests

This question assesses your ability to apply the distance modulus equation to verify whether two stars are physically co-located, recall spectral classes and absorption features (OBAFGKM), interpret eclipsing binary light curves (linking apparent magnitude changes to eclipses of hot vs. cool stars), and extract the orbital period to compute angular velocity (ω) with accurate unit conversions.

Question 04.1

Binary System Distance Consistency [3 Marks]

📐 Step-by-Step Calculation

  1. Identify formula: m - M = 5 log(d / 10) or d = 10 × 10^((m - M) / 5)
  2. Method A (Distance Calculation):
    • For Aa1: m - M = 2.5 - (-5.4) = 7.9
       d = 10 × 10^(7.9 / 5) = 10 × 10^1.58 = 380 pc (380.189 pc)
    • For Aa2: m - M = 5.0 - (-2.9) = 7.9
       d = 10 × 10^(7.9 / 5) = 380 pc
  3. Method B (Distance Modulus comparison):
    • Show m - M = 7.9 for both stars.
  4. Conclusion: State clearly that since both calculated distances are the same (or both values of m - M are identical), both stars are at the same distance from Earth, which is consistent with them being gravitationally bound in a binary system.

✅ Mark Scheme Breakdown

  • Mark 1: Correct substitution into distance modulus equation OR correct rearrangement for d .
  • Mark 2: Calculating d = 380 pc for both stars OR explicitly stating m - M = 7.9 for both stars.
  • Mark 3: Clear concluding statement that calculated distances are the same / identical distance modulus means they are at the same distance from Earth.

🧠 Exam Technique

Always make the final link explicit! Many candidates correctly calculated 380 pc but failed to write the concluding sentence explaining why this proves they are in a binary system (i.e. they are at the same distance from Earth).

❌ Common Traps

Sign errors with negative absolute magnitudes: 2.5 - (-5.4) = +7.9 , not -2.9 . Remember that absolute magnitude M can be negative for very luminous stars.

Questions 04.2 & 04.3

Spectral Classification & Star Temperature [2 Marks]

💡 Key Knowledge: Spectral Classes

Order of stellar classes from hottest to coolest: O, B, A, F, G, K, M

  • Class O: Hottest (25,000 to 50,000 K). Prominent lines: He⁺, He, H (weak Balmer).
  • Class B: (11,000 to 25,000 K). Prominent lines: Helium (He) and Hydrogen (H).
  • Class A: (7,500 to 11,000 K). Strongest lines: Balmer lines (Hydrogen).

✅ Correct Answers & Marks

04.2 [1 Mark]:

  • H and He (Hydrogen AND Helium).

04.3 [1 Mark]:

  • Mintaka Aa1: spectral class O is hotter than B (O stars have higher surface temperatures than B stars).

❌ Examiner Watch-Outs for 04.2

  • Reject: Listing extra elements (e.g. "Hydrogen, Helium, and ionized metals") will lose the mark.
  • Reject: Stating ionized helium (He⁺) for class B—He⁺ is characteristic of Class O stars!

🧠 Exam Technique for 04.3

Don't just state "Aa1". The prompt specifically requires an explanation. Always state the rule: "O-type stars are hotter than B-type stars, therefore Mintaka Aa1 has the greater temperature."

Question 04.4

Interpreting the Light Curve [3 Marks]

💡 Physics of Eclipsing Binaries

  • Baseline brightness: Both stars are side-by-side; maximum total light reaches Earth (lowest apparent magnitude, ~2.23).
  • Primary dip (Q): Occurs when the cooler/dimmer star (Aa2) passes in front of the hotter/brighter star (Aa1), obscuring high-intensity radiation per unit area. This causes the biggest drop in brightness (highest apparent magnitude, ~2.37).
  • Secondary dip (P): Occurs when the hotter star passes in front of the cooler star. Less total light is blocked, causing a smaller dip in brightness (~2.29).

✅ Mark Scheme Breakdown (3 Marks)

  • Mark 1: When one star passes in front of the other (eclipses), the total light received decreases / light intensity dips.
  • Mark 2: Brightest state (lowest apparent magnitude value ~2.23) occurs when both stars are visible (neither is eclipsed).
  • Mark 3 (Any 1 of the following):
    • Difference in dip depth suggests stars have different temperatures / surface brightnesses.
    • The deepest dip (Q) occurs when the cooler / dimmer star (Aa2) passes in front of the hotter / brighter star (Aa1) [or converse for dip P].

❌ Crucial Examiner Guidance for Mark 3

Do NOT say merely "one star is bigger than the other". The mark scheme explicitly states: "do not condone 'size / bigger / smaller' etc. unless they are linked to temperature / brightness". Depth of minimum depends primarily on temperature difference and surface flux, not merely geometric size.

Question 04.5

Determining Angular Velocity (ω) [2 Marks]

📐 Step-by-Step Calculation

  1. Find the orbital period (T):
    A full cycle consists of two successive primary eclipses (or two successive secondary eclipses).
    • Primary minimum Q is at approx: t₁ ≈ 5.7 days
    • The system returns to this state after one full orbit:
      Peak dip P to next dip P: t ≈ 8.6 - 2.9 = 5.7 days
    • Allowed range: T = 5.6 to 5.8 days
  2. Convert Period to seconds:
    T = 5.7 × 24 × 60 × 60 = 492,480 s
  3. Calculate Angular Velocity (ω):
    ω = 2π / T
    ω = 2π / 492,480 = 1.28 × 10⁻⁵ rad s⁻¹
    • Round to 2 s.f.: 1.3 × 10⁻⁵ rad s⁻¹

✅ Mark Scheme Breakdown

  • Mark 1: Determining T ≈ 5.7 days (allow 5.6 – 5.8 days) and substituting into ω = 2π / T with time conversion to seconds.
  • Mark 2: 1.3 × 10⁻⁵ rad s⁻¹ (Accept range: 1.25 × 10⁻⁵ to 1.30 × 10⁻⁵ rad s⁻¹ ).
Partial credit: Max 1 mark if candidate used half the time period (time between dip P and dip Q, ~2.85 days).

❌ Major Pitfall: Halving the Period

The most frequent mistake on eclipsing binary questions is reading the time between P and Q (e.g. from 2.9 days to 5.7 days = 2.85 days) and calling that the period T . That is only half an orbit (T/2)! A full orbit requires star 1 to pass in front of star 2, and then star 2 to pass in front of star 1, returning to the starting configuration.

Topics

Optional topics · Physics · 3.9 Astrophysics (A-level only) · 3.6 Further mechanics and thermal physics (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.