A star's spectrum peaks at 480\,nm. Estimate its surface temperature. [2 marks]

WJEC WJEC 2020-style practice: The Sun has a surface temperature of 5800\,K and radius 7.0 × 10^8\,m. The Stefan constant is 5.67 × 10^-8\,W m^-2\,K^-4. Calculate the luminosity of the Sun and the intensity of its radiation at the Earth, which is 1.5 × 10^11\,m away.

Use Stefan's law for the total power, then the inverse-square law for the intensity at Earth. Surface area: A = 4π r^2 = 4π (7.0 × 10^8)^2 = 6.16 × 10^18\,m^2. Luminosity: P = σ A T^4 = 5.67 × 10^-8 × 6.16 × 10^18 × (5800)^4. (5800)^4 = 1.13 × 10^15, so P = 5.67 × 10^-8 × 6.16 × 10^18 × 1.13 × 10^15 = 3.9 × 10^26\,W. Intensity at Earth: I = P4π d^2 = 3.9 × 10^264π (1.5 × 10^11)^2 = 1.4 × 10^3\,W m^-2. Markers reward the surface area, Stefan's law with the fourth power, and the inverse-square law giving the solar constant near 1.4\,kW m^-2.

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How can the light from a star tell us its temperature, size and luminosity?

Black-body radiation, Wien's displacement law, Stefan's law, the inverse-square law of intensity, and stellar spectra.

A focused answer to WJEC A-Level Physics Unit 1 using radiation to investigate stars, covering black-body radiation, Wien's displacement law, Stefan's law, the inverse-square law of intensity, and how stellar spectra reveal temperature and composition.

Generated by Claude Opus 4.811 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
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What this dot point is asking

WJEC wants you to treat a star as a black body, use Wien's displacement law to find its surface temperature, use Stefan's law to link luminosity, temperature and radius, apply the inverse-square law of intensity, and interpret stellar spectra. This is a satisfying topic because three compact laws let you deduce the temperature, size and distance of an object trillions of kilometres away from nothing but its light.

The answer

Black-body radiation

A black body absorbs and emits all wavelengths perfectly. A star is a good approximation, emitting a continuous spectrum whose peak shifts with temperature. The shape of the black-body curve depends on temperature alone, which is what makes it such a powerful diagnostic.

Wien's displacement law

The wavelength of peak emission $\lambda_{\text{max}}$ is inversely proportional to the absolute temperature, so a hotter star peaks at a shorter wavelength and looks bluer; a cooler star looks redder. Measuring the peak of a star's spectrum therefore gives its surface temperature directly.

Stefan's law

Because power depends on $T^4$ , a small rise in temperature greatly increases luminosity.

Finding a star's radius

A star has a surface temperature of $10\,000\,\text{K}$ and a luminosity of $4.0 \times 10^{27}\,\text{W}$ . Take $\sigma = 5.67 \times 10^{-8}\,\text{W m}^{-2}\,\text{K}^{-4}$ . Find its radius.

step Rearrange Stefan's law

$P = 4\pi r^2 \sigma T^4$ , so $r^2 = \dfrac{P}{4\pi \sigma T^4}$ .

step Evaluate the denominator

$T^4 = (10\,000)^4 = 1.0 \times 10^{16}$ , so $4\pi \sigma T^4 = 4\pi \times 5.67 \times 10^{-8} \times 1.0 \times 10^{16} = 7.12 \times 10^{9}$ .

step Solve for the radius

$r^2 = \dfrac{4.0 \times 10^{27}}{7.12 \times 10^{9}} = 5.62 \times 10^{17}$ , so $r = 7.5 \times 10^{8}\,\text{m}$ , a little larger than the Sun.

Inverse-square law and spectra

The intensity received at distance $d$ from a star of power $P$ is $I = \dfrac{P}{4\pi d^2}$ , falling off as the inverse square of distance. The spectrum of a star shows dark absorption lines where cooler gas in its atmosphere absorbs specific wavelengths, identifying the elements present.

Examples in context

Example 1. Classifying Betelgeuse. The red supergiant Betelgeuse peaks at about $830\,\text{nm}$ , so Wien's law gives $T = 2.90 \times 10^{-3} / 830 \times 10^{-9} = 3500\,\text{K}$ , a cool surface. Despite this low temperature it is enormously luminous because its radius is hundreds of times the Sun's, and Stefan's law shows the huge surface area more than compensates for the low $T^4$ .

Example 2. The cosmic microwave background. The afterglow of the Big Bang is an almost perfect black body peaking in the microwave region. Its peak wavelength of about $1.1\,\text{mm}$ gives, via Wien's law, a temperature of $2.7\,\text{K}$ , the famous temperature of the present-day universe. The same physics that reads a star's surface also reads the temperature of all of space.

Try this

Q1. A star's spectrum peaks at $480\,\text{nm}$ . Estimate its surface temperature. [2 marks]

Cue. $T = \frac{2.90\times10^{-3}}{480\times10^{-9}} \approx 6040\,\text{K}$ .

Q2. State how the luminosity of a black-body star depends on its temperature. [1 mark]

Cue. Luminosity is proportional to the fourth power of the absolute temperature.

Exam-style practice questions

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

WJEC 20206 marksThe Sun has a surface temperature of

5800\,\text{K}

and radius

7.0 \times 10^{8}\,\text{m}

. The Stefan constant is

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

. Calculate the luminosity of the Sun and the intensity of its radiation at the Earth, which is

1.5 \times 10^{11}\,\text{m}

away.

Show worked answer →

Use Stefan's law for the total power, then the inverse-square law for the intensity at Earth.

Surface area: $A = 4\pi r^2 = 4\pi (7.0 \times 10^{8})^2 = 6.16 \times 10^{18}\,\text{m}^2$ .

Luminosity: $P = \sigma A T^4 = 5.67 \times 10^{-8} \times 6.16 \times 10^{18} \times (5800)^4$ .

$(5800)^4 = 1.13 \times 10^{15}$ , so $P = 5.67 \times 10^{-8} \times 6.16 \times 10^{18} \times 1.13 \times 10^{15} = 3.9 \times 10^{26}\,\text{W}$ .

Intensity at Earth: $I = \dfrac{P}{4\pi d^2} = \dfrac{3.9 \times 10^{26}}{4\pi (1.5 \times 10^{11})^2} = 1.4 \times 10^{3}\,\text{W m}^{-2}$ .

Markers reward the surface area, Stefan's law with the fourth power, and the inverse-square law giving the solar constant near $1.4\,\text{kW m}^{-2}$ .

WJEC 20173 marksTwo stars have the same radius, but star A appears blue and star B appears red. Deduce which star is more luminous and explain your reasoning using Wien's and Stefan's laws.

Show worked answer →

A blue star peaks at a shorter wavelength than a red star. By Wien's law, $\lambda_{\text{max}} T$ is constant, so a shorter peak wavelength means a higher surface temperature. Star A (blue) is therefore hotter than star B (red).

By Stefan's law, $P = \sigma A T^4$ . With equal radii the surface areas are equal, so luminosity depends only on $T^4$ . The hotter star A radiates much more power.

Therefore star A is more luminous, and because of the fourth power even a modest temperature difference produces a large difference in luminosity. Markers reward linking colour to temperature via Wien, then temperature to luminosity via the fourth power in Stefan's law.

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Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)