AQA A-Level Physics: Stefan’s Law and Wien’s Displacement Law in Astrophysics — mark scheme explained

The question

A star has a surface area of 1.2 × 10 18 m 2 and a temperature of 6000 K. Calculate the power output using Stefan's law. [Paraphrased for study — not reproduced from any exam paper.]

Mark scheme, decoded

What each mark is really for — in plain English — and the wording trap that loses it.

• S1

  Identify the values: A = 1.2 × 10 18 m 2 , T = 6000 K, σ = 5.67 × 10 -8 W m -2 K -4 .

• S2

  Use the formula P = AσT 4 :

• S3

  P = (1.2 × 10 18 ) × (5.67 × 10 -8 ) × (6000) 4

• S4

  Calculate T 4 : (6000) 4 = 1.296 × 10 15

• S5

  P = (1.2 × 10 18 ) × (5.67 × 10 -8 ) × (1.296 × 10 15 )

• S6

  P ≈ 8.8 × 10 25 W

Model answer

Worked through, with each step tagged to the mark it earns.

1. S1

   Identify the values: A = 1.2 × 10 18 m 2 , T = 6000 K, σ = 5.67 × 10 -8 W m -2 K -4 .

2. S2

   Use the formula P = AσT 4 :

3. S3

   P = (1.2 × 10 18 ) × (5.67 × 10 -8 ) × (6000) 4

4. S4

   Calculate T 4 : (6000) 4 = 1.296 × 10 15

5. S5

   P = (1.2 × 10 18 ) × (5.67 × 10 -8 ) × (1.296 × 10 15 )

6. S6

   P ≈ 8.8 × 10 25 W

7. Final answer: 8.8 × 10 25 W

Common mistakes

• Forgetting to convert units when using Wien's displacement law. — Always ensure that all units are consistent. Convert wavelengths from nm to m by multiplying by 10 -9 .
• Using the wrong value for the Stefan-Boltzmann constant σ. — Memorize or look up the correct value of σ, which is 5.67 × 10 -8 W m -2 K -4 .
• Confusing the inverse square law with Stefan's law. — Understand that the inverse square law (I ∝ 1 / r 2 ) deals with how intensity decreases with distance, while Stefan's law (P = AσT 4 ) deals with the total power output based on temperature and surface area.
• Forgetting to raise the temperature to the fourth power in Stefan's law. — Always remember to raise the temperature to the fourth power when using Stefan's law: P = AσT 4 .
• Using the wrong constant in Wien's displacement law. — Memorize or look up the correct value of the constant, which is 2.9 × 10 -3 m K.
• Assuming that all stars are perfect black bodies without considering other factors. — Understand that while treating stars as black bodies simplifies calculations, it is an approximation. Real stars can have variations in temperature and other factors that affect their radiation.

Where the marks go

• Full worked solution (all marking points)5 marks

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