A star is just a point of light in a telescope — far too distant to measure with a ruler. Yet we can work out how big it actually is. The trick is to combine two laws you already know: one that turns a star’s colour into its temperature, and one that links temperature and brightness to size. Put them together and a pinprick of light gives up its true radius.
📘 What you need to know
A star’s radius is found by combining Wien’s displacement law and the Stefan–Boltzmann law
Wien’s law:λmaxT = 2.9 × 10−3 m K — the peak wavelength gives the surface temperature
Stefan–Boltzmann law:L = 4πr2σT4 — links luminosity, radius and temperature
If you’re given the radiant flux and distance instead of luminosity, use the inverse square law: F = L ÷ (4πd2)
σ is the Stefan–Boltzmann constant, 5.67 × 10−8 W m−2 K−4
Rearrange the Stefan–Boltzmann law for r to get the star’s radius
The Two Laws You’ll Combine
Finding a stellar radius is really a two-tool job. Neither law gives you the radius on its own, but chained together they do.
Wien’s displacement law — colour to temperature
A star’s spectrum peaks at one particular wavelength, λmax. Wien’s law says this peak is inversely proportional to the surface temperature: hotter stars peak at shorter (bluer) wavelengths. Measure the peak, and you get the temperature.
Wien’s displacement lawλmaxT = 2.9 × 10−3 m K
Stefan–Boltzmann law — temperature and size to luminosity
The luminosity of a star (its total power output) depends on both how hot its surface is and how large it is. A bigger surface radiates more, and a hotter surface radiates far more — note the fourth power of temperature.
Stefan–Boltzmann lawL = 4πr2σT4
Here r is the star’s radius and σ is the Stefan–Boltzmann constant. If you already know L and can find T from Wien’s law, this equation has just one unknown left — the radius.
The fourth power on temperature is easy to overlook and it dominates the answer. Doubling a star’s temperature makes it 24 = 16 times more luminous at the same size. So when you rearrange for r, be extra careful raising T to the fourth — it’s the number-one place marks are lost.
When You’re Given Flux Instead of Luminosity
Sometimes a question doesn’t hand you the luminosity directly. Instead it gives the radiant fluxF — the power arriving per square metre at Earth — and the star’s distanced. The light has spread out over a sphere of radius d, so:
Inverse square law of fluxF = L ÷ (4πd2)
Rearrange this to get the luminosity, L = F × 4πd2, then carry on into the Stefan–Boltzmann law as before. Watch the two different distances: d is how far the star is from Earth, while r is the star’s own radius — don’t mix them up.
The full toolkit: Wien’s law fixes the temperature, the inverse square law fixes the luminosity, and the Stefan–Boltzmann law delivers the radius.
The Method Step by Step
Almost every stellar-radius question follows the same route. Learn this order and you can tackle any version.
🧭 Finding a stellar radius
Find the temperature from the peak wavelength using Wien’s law: T = 2.9 × 10−3 ÷ λmax
Find the luminosity — either it’s given, or get it from the flux and distance: L = F × 4πd2
Rearrange Stefan–Boltzmann for radius:r = √( L ÷ (4πσT4) )
Substitute and solve — take special care with T4 and the final square root
peak λ
→ Wien →
temperature T
→ with L →
Stefan–Boltzmann
→ rearrange →
radius r
WE 1
A star has a luminosity of 2.5 × 1028 W and emits radiation that peaks at a wavelength of 480 nm. Calculate the radius of the star. Take σ = 5.67 × 10−8 W m−2 K−4.
Step 1 — temperature from Wien’s law
T = (2.9 × 10−3) ÷ λmax= (2.9 × 10−3) ÷ (480 × 10−9)≈ 6040 KStep 2 — rearrange Stefan–Boltzmann for r
r = √( L ÷ (4πσT⁴) )
= √( (2.5 × 1028) ÷ (4π × 5.67×10−8 × 6040⁴) )r ≈ 5.1 × 109 mthat’s about 7 times the Sun’s radius
WE 2
A distant star lies 8.0 × 1017 m from Earth. Its radiant flux measured at Earth is 4.0 × 10−9 W m−2, and its spectrum peaks at 600 nm. Calculate the star’s radius. Take σ = 5.67 × 10−8 W m−2 K−4.
Step 1 — temperature (Wien)
T = (2.9 × 10−3) ÷ (600 × 10−9)
≈ 4830 KStep 2 — luminosity from flux (inverse square)
L = F × 4πd²
= (4.0 × 10−9) × 4π × (8.0 × 1017)²≈ 3.2 × 1028 WStep 3 — radius (Stefan–Boltzmann)
r = √( L ÷ (4πσT⁴) )
= √( (3.2 × 1028) ÷ (4π × 5.67×10−8 × 4830⁴) )r ≈ 9.1 × 109 mroughly 13 solar radii — a giant star
Quick recap: get T from Wien (λmaxT = 2.9×10−3), get L directly or from flux & distance (L = F·4πd2), then rearrange Stefan–Boltzmann (L = 4πr2σT4) for r.
💡 Top tips
Always start with Wien’s law to get the temperature — the Stefan–Boltzmann law can’t be used without T
Convert nanometres to metres before using Wien’s law: 480 nm = 480 × 10−9 m
Mind the two “r”s:d is the distance to the star, r is the star’s radius — the inverse square law uses d, Stefan–Boltzmann uses r
Handle T4 carefully and remember the final square root when solving for r
⚠ Common mistakes
Forgetting the square root at the end — the Stefan–Boltzmann law gives r2, not r
Raising T to the wrong power — it’s T4, and a small slip here throws the answer out massively
Using flux in the Stefan–Boltzmann law directly — you must convert flux to luminosity first
Leaving wavelength in nm — Wien’s law needs metres, so the temperature comes out wildly wrong
That’s the end of Fusion & Stars! You can now trace a star from a collapsing nebula, through fusion and the main sequence, to its death — and read its temperature, composition, distance and size straight from its light. This topic loves multi-step Paper 2 questions, so practise chaining these laws together under time pressure.
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