A star's radiation peaks at 600\ nm. Find its surface temperature (b = 2.90 × 10^-3\ m K). [2 marks]

WJEC Eduqas Eduqas 2019-style practice: The radiation from a star peaks at a wavelength of 480\ nm. Calculate the surface temperature of the star. Take Wien's constant b = 2.90 × 10^-3\ m K.

Wien's displacement law: _max T = b, so T = b_max. T = 2.90 × 10^-3480 × 10^-9 = 2.90 × 10^-34.80 × 10^-7 = 6.0 × 10^3\ K. Markers reward T = b_max, correct powers of ten, and the temperature about 6000\ K (similar to the Sun).

WJEC Eduqas Eduqas 2021-style practice: A star has a surface temperature of 8000\ K and a radius of 1.2 × 10^9\ m. Calculate its luminosity. Take the Stefan-Boltzmann constant σ = 5.67 × 10^-8\ W m^-2\ K^-4.

Stefan's law for a spherical star: L = 4π R^2 σ T^4. Surface area: 4π R^2 = 4π(1.2 × 10^9)^2 = 1.81 × 10^19\ m^2. T^4 = (8000)^4 = 4.10 × 10^15\ K^4. L = (1.81 × 10^19)(5.67 × 10^-8)(4.10 × 10^15) = 4.2 × 10^27\ W. Markers reward L = 4π R^2 σ T^4, evaluating the area and T^4, and the luminosity about 4.2 × 10^27\ W.

EnglandPhysicsSyllabus dot point

How do we work out the temperature, size and composition of a star from the radiation it emits?

Using radiation to investigate stars: black-body radiation, Wien's displacement law, Stefan's law and stellar luminosity, the inverse-square law for flux, and stellar spectra.

A focused answer to the Eduqas A-Level Physics Component 2 astrophysics content, covering black-body radiation, Wien's displacement law for stellar temperature, Stefan's law for luminosity, the inverse-square law for radiation flux, and how stellar spectra reveal composition.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

Eduqas wants you to describe a star as an approximate black body, use Wien's displacement law to find a star's surface temperature, use Stefan's law to relate luminosity to temperature and radius, use the inverse-square law for the radiation flux received at Earth, and explain how stellar spectra reveal composition.

The answer

Black-body radiation

Wien's displacement law

Stefan's law and luminosity

The inverse-square law and stellar spectra

Radius of a star from luminosity and temperature

A star has luminosity $8.0 \times 10^{26}\ \text{W}$ and surface temperature $5500\ \text{K}$ . Find its radius. Take $\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$ .

step 1: Rearrange Stefan's law for the radius

From $L = 4\pi R^2 \sigma T^4$ , $R = \sqrt{\dfrac{L}{4\pi\sigma T^4}}$ .

step 2: Evaluate the denominator

$T^4 = (5500)^4 = 9.15 \times 10^{14}$ . So $4\pi\sigma T^4 = 4\pi(5.67 \times 10^{-8})(9.15 \times 10^{14}) = 6.52 \times 10^{8}$ .

step 3: Solve

$R = \sqrt{\dfrac{8.0 \times 10^{26}}{6.52 \times 10^{8}}} = \sqrt{1.23 \times 10^{18}} = 1.1 \times 10^{9}\ \text{m}$ , a little larger than the Sun.

Examples in context

These laws let astronomers measure the temperature, size, luminosity and distance of stars they can never visit, purely from their light. Wien's law gives surface temperatures and classifies stars by colour; Stefan's law and the inverse-square law together yield distances and underpin the Hertzsprung-Russell diagram. Spectral absorption lines reveal that stars are made largely of hydrogen and helium, and the same physics measures the temperature of the cosmic microwave background.

Try this

Q1. State Wien's displacement law. [1 mark]

Cue. The peak wavelength times the absolute temperature is constant, $\lambda_\text{max} T = b$ .

Q2. A star's radiation peaks at $600\ \text{nm}$ . Find its surface temperature ( $b = 2.90 \times 10^{-3}\ \text{m K}$ ). [2 marks]

Cue. $T = \frac{b}{\lambda_\text{max}} = \frac{2.90 \times 10^{-3}}{600 \times 10^{-9}} = 4.8 \times 10^{3}\ \text{K}$ .

Q3. State how the flux received from a star depends on its distance. [1 mark]

Cue. It follows the inverse-square law, $F = \frac{L}{4\pi d^2}$ .

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20194 marksThe radiation from a star peaks at a wavelength of

480\ \text{nm}

. Calculate the surface temperature of the star. Take Wien's constant

b = 2.90 \times 10^{-3}\ \text{m K}

Show worked answer →

Wien's displacement law: $\lambda_\text{max} T = b$ , so $T = \dfrac{b}{\lambda_\text{max}}$ .

$T = \dfrac{2.90 \times 10^{-3}}{480 \times 10^{-9}} = \dfrac{2.90 \times 10^{-3}}{4.80 \times 10^{-7}} = 6.0 \times 10^{3}\ \text{K}$ .

Markers reward $T = \frac{b}{\lambda_\text{max}}$ , correct powers of ten, and the temperature about $6000\ \text{K}$ (similar to the Sun).

Eduqas 20215 marksA star has a surface temperature of

8000\ \text{K}

and a radius of

1.2 \times 10^{9}\ \text{m}

. Calculate its luminosity. Take the Stefan-Boltzmann constant

\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}

Show worked answer →

Stefan's law for a spherical star: $L = 4\pi R^2 \sigma T^4$ .

Surface area: $4\pi R^2 = 4\pi(1.2 \times 10^{9})^2 = 1.81 \times 10^{19}\ \text{m}^2$ .

$T^4 = (8000)^4 = 4.10 \times 10^{15}\ \text{K}^4$ .

$L = (1.81 \times 10^{19})(5.67 \times 10^{-8})(4.10 \times 10^{15}) = 4.2 \times 10^{27}\ \text{W}$ .

Markers reward $L = 4\pi R^2 \sigma T^4$ , evaluating the area and $T^4$ , and the luminosity about $4.2 \times 10^{27}\ \text{W}$ .

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Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)