IB Physics HLTopic 5 — Fusion & StarsPaper 1 & 2two laws, one radius~16 min read
Finding Stellar Radii
Here’s something remarkable: we can work out how big a star is — a giant ball of gas we could never measure with a ruler — using just its colour and its brightness. The trick is to combine two powerful laws: Wien’s law (which turns colour into temperature) and the Stefan–Boltzmann law (which links temperature, brightness and size). Chain them together and out pops the star’s radius. This is a classic HL exam calculation, so let’s nail the method.
📚 What you need to know
A star’s radius is found by combining Wien’s law and the Stefan–Boltzmann law
Wien’s law:λmaxT = 2.9 × 10−3 m K — gives temperature from peak wavelength
Stefan–Boltzmann law:L = 4πr2σT4 — links luminosity, radius and temperature
The inverse square law of flux gives luminosity from radiant flux and distance
σ is the Stefan–Boltzmann constant = 5.67 × 10−8 W m−2 K−4
The procedure: find T (Wien) → find L (if needed) → rearrange Stefan–Boltzmann for r
The two laws you need
Finding a star’s radius rests on two equations. Let’s meet them both.
Wien’s displacement law
Wien’s law says the peak wavelength of a star’s light is inversely proportional to its temperature — hotter stars peak at shorter wavelengths. Rearranged, it gives us the surface temperature directly from the measured peak wavelength:
Wien’s displacement lawλmaxT = 2.9 × 10−3 m Kλmax = peak wavelength (m) • T = surface temperature (K)
The Stefan–Boltzmann law
The Stefan–Boltzmann law connects a star’s total power output (luminosity L) to its radius and temperature. A bigger or hotter star radiates more — and notice how strongly it depends on temperature (to the fourth power):
If you’re given the star’s radiant flux and distance instead of its luminosity, you first find L using the inverse square law of flux: F = L / (4πd2), where the light has spread out over a sphere of radius d.
The procedure
Putting it together, finding a stellar radius is a three-step chain:
Wien’s law gives the temperature, the inverse square law gives the luminosity (if needed), and the Stefan–Boltzmann law — rearranged for r — gives the radius.
The single step students forget is rearranging Stefan–Boltzmann for r. Starting from L = 4πr²σT⁴, make r² the subject, then square-root: r = √(L / 4πσT⁴). Practise that rearrangement until it’s automatic, because in the exam you’ll be under time pressure and the algebra is where marks slip away. Everything else is just substitution.
WE 1
Betelgeuse has a luminosity of 4.49 × 1031 W and emits radiation with a peak wavelength of 850 nm. Calculate the ratio of Betelgeuse’s radius to the Sun’s radius. (Sun’s radius = 6.95 × 108 m, σ = 5.67 × 10−8 W m−2 K−4)
Step 1 — temperature (Wien)T = 2.9×10⁻³ ÷ 850×10⁻⁹ = 3410 KStep 2 — rearrange Stefan–Boltzmann for rr = √( L / (4πσT⁴) )Step 3 — substituter = √( 4.49×10³¹ / (4π × 5.67×10⁻⁸ × 3410⁴) )r ≈ 6.83×10¹¹ mStep 4 — ratio to the Sunrₖ / rₓ = 6.83×10¹¹ ÷ 6.95×10⁸≈ 980 times the Sun’s radiusBetelgeuse is about 1000× wider than the Sun — a true supergiant. Follow the chain: Wien for T, then rearrange Stefan–Boltzmann for r, then take the ratio.
WE 2
A star has a surface temperature of 10 000 K and a luminosity of 6.0 × 1027 W. Calculate its radius. (σ = 5.67 × 10−8 W m−2 K−4)
Step 1 — rearrange for rr = √( L / (4πσT⁴) )Step 2 — substituter = √( 6.0×10²⁷ / (4π × 5.67×10⁻⁸ × (10000)⁴) )Step 3 — evaluate denominator4π × 5.67×10⁻⁸ × 10¹⁶ = 7.12×10⁹r = √(6.0×10²⁷ ÷ 7.12×10⁹) = √(8.43×10¹⁷)r ≈ 2.9 × 109 mWatch the T⁴ term — (10⁴)⁴ = 10¹⁶, easy to slip on. Compute the denominator fully before dividing, then square-root at the very end.
⚛ Finding a stellar radius
Temperature: Wien → T = 2.9×10−3 / λmax.
Luminosity: if given flux & distance, L = F × 4πd².
Rearrange Stefan–Boltzmann: r = √(L / 4πσT⁴).
Substitute carefully — watch the T4 term.
Ratio? Divide by the comparison star’s radius.
💡 Top tips
Chain the laws: Wien → (flux) → Stefan–Boltzmann.
Convert λmax to metres before using Wien.
Rearranging for r is the step to practise most.
Remember T is to the fourth power — a big effect.
Square-root last, after evaluating everything inside.
⚠ Common mistakes
Forgetting to convert λmax from nm to m
Errors raising T to the fourth power
Forgetting the square root when solving for r
Using flux where luminosity is needed (or vice versa)
Dropping the 4π or the constant σ
Quick recap: A star’s radius comes from combining two laws. Wien’s law (λmaxT = 2.9×10−3) gives the temperature; the inverse square law gives luminosity from flux and distance if needed; and the Stefan–Boltzmann law (L = 4πr2σT4), rearranged to r = √(L/4πσT4), gives the radius. Betelgeuse works out to about 1000× the Sun’s radius.
That completes the Fusion & Stars section — from the fusion powering a star’s core, through its birth, life and death, to measuring its distance and size from Earth. You now have the full toolkit: the reactions, the HR diagram, spectra, parallax, and the two laws that reveal a star’s radius. Revisit the worked examples until each calculation feels automatic, and you’ll be ready for anything the exam throws at you.
Stellar radius calculations catching you out?
Book a free meeting and we’ll drill Wien’s law, the Stefan–Boltzmann rearrangement, and the full radius-finding procedure step by step.
