A star has a peak wavelength of 2.9 × 10^-7\ m. Find its surface temperature. Take b = 2.9 × 10^-3\ m K.

OCR OCR 2019-style practice: The Sun emits black-body radiation with a peak wavelength of 5.0 × 10^-7\ m. Calculate the surface temperature of the Sun. Take Wien's constant b = 2.9 × 10^-3\ m K.

Wien's displacement law: _ = b, so T = b_. T = 2.9 × 10^-35.0 × 10^-7 = 5.8 × 10^3\ K. Markers reward _ = b, rearranging for temperature, and the value about 5800\ K (the Sun's surface temperature).

OCR OCR 2022-style practice: A star has a radius of 7.0 × 10^8\ m and a surface temperature of 6000\ K. Calculate its luminosity. Take the Stefan-Boltzmann constant σ = 5.67 × 10^-8\ W m^-2\ K^-4.

Stefan's law: L = σ A T^4 with surface area A = 4π r^2. Area: A = 4π (7.0 × 10^8)^2 = 4π (4.9 × 10^17) = 6.16 × 10^18\ m^2. T^4 = (6000)^4 = 1.296 × 10^15\ K^4. L = (5.67 × 10^-8)(6.16 × 10^18)(1.296 × 10^15) = 4.5 × 10^26\ W. Markers reward L = σ A T^4 with A = 4π r^2, the fourth power of temperature, and the luminosity about 4.5 × 10^26\ W.

EnglandPhysicsSyllabus dot point

How do stars form, live and die, and how do we read their temperatures and luminosities?

Stars and the Hertzsprung-Russell diagram: stellar formation and evolution, Wien's displacement law and Stefan's law, luminosity and the inverse-square law, stellar spectra, and the structure of the Hertzsprung-Russell diagram.

A focused answer to the OCR H556 astrophysics content on stars, covering stellar formation and the life cycles of low-mass and high-mass stars, Wien's displacement law and Stefan's law for black-body radiation, luminosity and the inverse-square law for intensity, stellar spectra, and the regions of the Hertzsprung-Russell diagram.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

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

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Quick answer

Stars form when gravity collapses a cloud of gas and dust into a protostar hot enough to begin nuclear fusion. A low-mass star like the Sun spends most of its life on the main sequence, then becomes a red giant, sheds a planetary nebula and ends as a white dwarf. A high-mass star becomes a red supergiant, explodes as a supernova, and leaves a neutron star or black hole. A star approximates a black body: its peak wavelength gives temperature through Wien's law $\lambda_{\max}T = b$ , and its luminosity follows Stefan's law $L = \sigma A T^4$ . Observed intensity falls with the inverse square of distance, $I = \frac{L}{4\pi d^2}$ . The Hertzsprung-Russell diagram plots luminosity against temperature, with the main sequence, giants, supergiants and white dwarfs in distinct regions.

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What this dot point is asking

OCR wants you to describe stellar formation and the evolution of low-mass and high-mass stars, use Wien's displacement law and Stefan's law, define luminosity and use the inverse-square law for intensity, interpret stellar spectra, and describe the structure of the Hertzsprung-Russell diagram.

The answer

Stellar formation and evolution

Black-body radiation: Wien's law and Stefan's law

Luminosity, intensity and the inverse-square law

Stellar spectra and the Hertzsprung-Russell diagram

Comparing the radii of two stars

Two stars have the same luminosity, but star A has a surface temperature of $3000\ \text{K}$ and star B has $6000\ \text{K}$ . Find the ratio of their radii.

step 1: Equal luminosity

By Stefan's law $L = 4\pi r^2 \sigma T^4$ . Since $L_A = L_B$ : $r_A^2 T_A^4 = r_B^2 T_B^4$ .

step 2: Rearrange for the radius ratio

$\frac{r_A^2}{r_B^2} = \frac{T_B^4}{T_A^4} = \left(\frac{6000}{3000}\right)^4 = 2^4 = 16$ .

step 3: Take the square root

$\frac{r_A}{r_B} = \sqrt{16} = 4$ . The cooler star A must be four times larger to match the luminosity, which is why cool red giants are so big.

Examples in context

Wien's law is how astronomers measure the surface temperatures of stars from their colours, and it explains why the Sun peaks in visible light while cooler stars look red and hotter ones blue. Stefan's law lets the luminosity of a star be found from its temperature and size, and combined with the inverse-square law it underpins the cosmic distance ladder. The HR diagram is the central tool of stellar astrophysics, letting astronomers read a star's evolutionary stage from its position and estimate the age of star clusters.

Try this

Q1. State Wien's displacement law and define its terms. [2 marks]

Cue. $\lambda_{\max}T = b$ : the peak wavelength times the absolute temperature is a constant.

Q2. A star has a peak wavelength of $2.9 \times 10^{-7}\ \text{m}$ . Find its surface temperature. Take $b = 2.9 \times 10^{-3}\ \text{m K}$ . [2 marks]

Cue. $T = \frac{b}{\lambda_{\max}} = \frac{2.9 \times 10^{-3}}{2.9 \times 10^{-7}} = 1.0 \times 10^{4}\ \text{K}$ .

Q3. State the end point of a low-mass star such as the Sun. [1 mark]

Cue. A white dwarf (after a red giant phase and the ejection of a planetary nebula).

Exam-style practice questions

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

OCR 20193 marksThe Sun emits black-body radiation with a peak wavelength of

5.0 \times 10^{-7}\ \text{m}

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

b = 2.9 \times 10^{-3}\ \text{m K}

Show worked answer →

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

$T = \frac{2.9 \times 10^{-3}}{5.0 \times 10^{-7}} = 5.8 \times 10^{3}\ \text{K}$ .

Markers reward $\lambda_{\max}T = b$ , rearranging for temperature, and the value about $5800\ \text{K}$ (the Sun's surface temperature).

OCR 20224 marksA star has a radius of

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

and a surface temperature of

6000\ \text{K}

. 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: $L = \sigma A T^4$ with surface area $A = 4\pi r^2$ .

Area: $A = 4\pi (7.0 \times 10^{8})^2 = 4\pi (4.9 \times 10^{17}) = 6.16 \times 10^{18}\ \text{m}^2$ .

$T^4 = (6000)^4 = 1.296 \times 10^{15}\ \text{K}^4$ .

$L = (5.67 \times 10^{-8})(6.16 \times 10^{18})(1.296 \times 10^{15}) = 4.5 \times 10^{26}\ \text{W}$ .

Markers reward $L = \sigma A T^4$ with $A = 4\pi r^2$ , the fourth power of temperature, and the luminosity about $4.5 \times 10^{26}\ \text{W}$ .

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Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)