All subjects · Physics · WPH15 (Unit 5)

Black body radiation, Stefan-Boltzmann and Wien

Stars behave approximately as black bodies. The radiation they give out tells us their temperature and, with their luminosity, their size.

Black body radiators

A black body is an idealised object that absorbs all the electromagnetic radiation that falls on it, and emits the maximum possible radiation for its temperature. Its emission spectrum depends only on its temperature.

Black body curves

A graph of intensity against wavelength is a smooth curve with a peak. As the temperature rises:

the peak moves to shorter wavelengths (hotter stars look bluer)
the curve gets higher at every wavelength, so the total power (area under the curve) increases sharply.

Wien's law

The peak wavelength is inversely proportional to the absolute temperature:

λ_max T = 2.898 x 10^-3 m K

The Sun (about 5800 K) peaks near 500 nm, in the visible region.

Stefan-Boltzmann law

The luminosity L (total power output) of a black body is

L = σAT⁴

with σ = 5.67 x 10^-8 W m^-2 K^-4. For a spherical star of radius r, A = 4πr². Doubling the temperature makes L 16 times larger.

Worked example

A star has radius 7.0 x 10⁸ m and surface temperature 5800 K. Find its luminosity and peak wavelength.

L = σ x 4πr² x T⁴ = 5.67 x 10^-8 x 4π x (7.0 x 10⁸)² x 5800⁴ = 3.95 x 10²⁶ W
λ_max = 2.898 x 10^-3 / 5800 = 5.00 x 10^-7 m

Answer: L = 4.0 x 10²⁶ W; λ_max = 500 nm

Key idea: Black body: absorbs all, emits most. λ_max T = 2.898 x 10^-3 m K. L = σAT⁴ with A = 4πr².

Check yourself

Find the peak wavelength of a star with surface temperature 5780 K, in nm, to 3 s.f.
Show answer
501 nm. λ_max = 2.898 x 10^-3 / 5780 = 5.01 x 10^-7 m = 501 nm.
A star's spectrum peaks at 350 nm. Find its surface temperature in K, to 3 s.f.
Show answer
8280 K. T = 2.898 x 10^-3 / 3.50 x 10^-7 = 8280 K.
Find the luminosity of a star of radius 7.0 x 10⁸ m and surface temperature 5800 K, in units of 10²⁶ W, to 3 s.f.
Show answer
3.95 x 10²⁶ W. L = σ4πr^2T⁴ = 5.67 x 10^-8 x 6.16 x 10¹⁸ x 1.13 x 10¹⁵ = 3.95 x 10²⁶ W.
A star has luminosity 3.8 x 10²⁸ W and surface temperature 4000 K. Find its radius, in units of 10¹⁰ m, to 3 s.f.
Show answer
1.44 x 10¹⁰ m. A = L / σT⁴ = 3.8 x 10²⁸ / (5.67 x 10^-8 x 2.56 x 10¹⁴) = 2.62 x 10²¹ m²; r = sqrt(A / 4π) = 1.44 x 10¹⁰ m.
How does the black body curve of a hotter star differ from that of a cooler star?
- Its peak is at a longer wavelength and it is lower
- Its peak is at the same wavelength but higher
- It has no peak
- Its peak is at a shorter wavelength and it is higher at all wavelengths
Show answer
Its peak is at a shorter wavelength and it is higher at all wavelengths. Wien's law moves the peak; Stefan-Boltzmann raises the total power.
A star's surface temperature doubles while its radius stays the same. What happens to its luminosity?
- It doubles
- It becomes 16 times larger
- It becomes 4 times larger
- It becomes 8 times larger
Show answer
It becomes 16 times larger. L is proportional to T⁴, and 2⁴ = 16.
What is a black body?
- An object that emits no radiation
- Any object that looks black
- An object at absolute zero
- An object that absorbs all radiation falling on it and emits the maximum possible radiation for its temperature
Show answer
An object that absorbs all radiation falling on it and emits the maximum possible radiation for its temperature. The radiation it emits depends only on its temperature.

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Last updated 7 November 2026. Spotted a mistake? Tell us.