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Learn9702Paper 4 · A Level Structured QuestionsAstronomy and cosmologyStellar radii
Astronomy and cosmology · Subtopic 2 of 3

Stellar Radii

Stars radiate approximately as black bodies. Wien's displacement law: the peak wavelength is inversely proportional to the surface temperature, λmax⁡T=2.9×10−3\lambda_{\max}T=2.9\times10^{-3} m K, so hotter stars look bluer. The Stefan–Boltzmann law gives the luminosity: L=4πσr2T4L=4\pi\sigma r^2T^4, with σ=5.67×10−8\sigma=5.67\times10^{-8} W m⁻² K⁻⁴. Knowing TT (from Wien) and LL (from a standard candle method), you can find a star's radius.

Cambridge 9702 Paper 4Edexcel YPH11 WPH15AQA 7408 Paper 3B≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use Wien's law to estimate a star's surface temperature.
  • Use the Stefan–Boltzmann law L=4πσr2T4L=4\pi\sigma r^2T^4.
  • Combine both laws to estimate a star's radius.
  • Compare stars using ratios of LL, rr and TT.
Pause and think

Betelgeuse is red and Rigel is blue-white, yet both are among the most luminous stars we can see. How can a cool red star be so luminous?

Section 2 of 14

Core idea

Start hereMeasuring a star you can't resolve

Stars are points of light; no telescope shows their size directly. Yet we know some are hundreds of times wider than the Sun. A star's colour tells its temperature, and with its power that fixes its size.

The big idea

Wien: λmax⁡T=2.9×10−3\lambda_{\max}T=2.9\times10^{-3} m K. Stefan–Boltzmann: L=4πr2σT4L=4\pi r^2\sigma T^4.

1. Black-body spectra

Black-body spectra: hotter stars peak at shorter wavelengths
5001000150020000.20.40.60.810wavelength (nm)intensitytemperature Tλ max
The shaded strip is visible light. A 3000 K star peaks in infrared (looks red); a 10 000 K star peaks in UV (looks blue-white). Each curve is scaled to the same height here; really hotter stars are far brighter at every wavelength.
Worked examplePeak wavelengths

Sun 5800 K; red star 3000 K.

Goal
λmax⁡\lambda_{\max}.
Strategy
Wien.
  1. 2.9×10−35800=500 nm\frac{2.9\times10^{-3}}{5800}=500\text{ nm}
    Visible.
  2. 2.9×10−33000=970 nm\frac{2.9\times10^{-3}}{3000}=970\text{ nm}
    Infrared.
Answer500500 nm; 970970 nm.
✓ CheckCooler star peaks at longer wavelength ✓.

2. Radius from luminosity

rr from LL and TT
  1. power per m2=σT4\text{power per m}^2=\sigma T^4
    Stefan–Boltzmann for a black body.
  2. L=4πr2σT4L=4\pi r^2\sigma T^4
    × surface area.
  3. r=L4πσT4r=\sqrt{\frac L{4\pi\sigma T^4}}
    Rearrange.
Worked exampleA hot star

L=3.83×1028L=3.83\times10^{28} W (100 Suns), T=10 000T=10\,000 K.

Goal
rr.
Strategy
Formula.
  1. 4π(5.67×10−8)(104)4=7.13×1094\pi(5.67\times10^{-8})(10^4)^4=7.13\times10^9
    Denominator.
  2. r=3.83×10287.13×109=2.3×109 mr=\sqrt{\frac{3.83\times10^{28}}{7.13\times10^9}}=2.3\times10^9\text{ m}
    Square root.
Answer2.3×1092.3\times10^9 m (~3 solar radii).
✓ CheckT4T^4 dominates: hotter stars need less area ✓.
Common misconceptionT2T^2 instead of T4T^4 Power per area goes as T4T^4; so r∝LT2r\propto\frac{\sqrt L}{T^2}.

Summary

  • Wien → TT.
  • Stefan → rr.
Key words
Black body
Perfect emitter and absorber.
Black-body spectra: hotter stars peak at shorter wavelengths
0.511.520.511.522.530wavelength (μm)relative intensity6000 K4000 Kλmax = 0.48 μm
Wien’s law: λmax⁡T=2.9×10−3\lambda_{\max}T=2.9\times10^{-3} m K. The hotter star emits far more at every wavelength, since total power per unit area ∝T4\propto T^4.
HistoryPlanck's act of desperation. Classical physics predicted that a black body should emit infinite power at short wavelengths, the "ultraviolet catastrophe". In 19001900 Max Planck fixed it by assuming energy comes in quanta, E=hfE=hf, calling it "an act of desperation". It was the birth of quantum theory.
Fun factYou glow in the infrared. At 310310 K, your body is a black-body radiator with a peak around 9.49.4 μm, in the infrared. Thermal cameras pick up this glow, which is why people stand out clearly on night-vision footage.
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