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Black-body radiation and spectral classes

A star radiates approximately as a black body, so Wien's displacement law turns the peak wavelength of its spectrum into a surface temperature. Stefan's law P = σAT⁴ then converts that temperature and the star's surface area into a power output, and with the power known the two laws together give the radius. Absorption lines sort stars into the classes O to M.

Year 13AQA 3.9.2.3, 3.9.2.4 CIE 25.2.1, 25.2.2, 25.2.3 OCR A 5.5.2(e), 5.5.2(i), 5.5.2(j), 5.5.2(k)EDEXCEL 153, 154, 155

Builds on Energy levels and photon emission and Star brightness and magnitude.

IN THIS TOPIC

Use black-body curves, Stefan's law and Wien's law to find stellar temperatures, powers and radii.
State the assumptions made when treating a star as a black body.
Recall the spectral classes O to M with their colours, temperatures and absorption lines, and explain the Balmer condition.

COMMON MISCONCEPTION

A red-hot star is hotter than a white one.

Red stars are the coolest, at about 3000 K, not the hottest: Wien's law says a hotter star peaks at a shorter wavelength, so colour moves from red towards blue-white as temperature rises.

The black-body curve

A black body is a perfect absorber and therefore a perfect emitter. It emits a continuous spectrum whose shape depends only on temperature. Wien's law relates the peak wavelength to temperature; Stefan's law relates total radiated power to surface area and temperature. A star is, to a good approximation, a black body. Stefan's law gives the total power radiated:

P = \sigma A T^{4}

IN THE AQA DATA BOOKLET

where A is the star's surface area, T its surface temperature in kelvin, and σ the Stefan constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴. The fourth power is steep: double the temperature and the output multiplies by sixteen. Wien's displacement law locates the curve's peak:

\lambda_{max} T = 2.9 × 10^{−3}\,\text{m K}

IN THE AQA DATA BOOKLET

Three computed black-body curves on shared axes with the visible band shaded: the 6600 kelvin curve towers over the others with its peak in the blue, the Sun's 5800 kelvin curve peaks in the green, and the 5000 kelvin curve is lowest with its peak towards the red. — FIG. 1Computed black-body curves for 5000 K, 5800 K and 6600 K, drawn per square metre of emitting surface. Area for area, the hotter surface out-radiates the cooler at every wavelength, and its peak sits further into the blue; whether a hotter star outshines a cooler one overall depends on their sizes too.

so hotter stars peak at shorter wavelengths. Red stars are the cool ones and blue-white stars the hottest, which makes "red hot" the mild end of stellar temperatures. Note that the unit of Wien's constant is the metre kelvin, a product, not metres per kelvin.

WORKED EXAMPLE

Weighing the Sun's output

The Sun's surface temperature is 5800 K and its radius 6.96 × 10⁸ m. Find its peak wavelength and total power output.

Wien first. λ_max = 2.9 × 10⁻³ / 5800 = 5.0 × 10⁻⁷ m, in the green, matching the curve above.

Then the surface area, A = 4πr² = 4π × (6.96 × 10⁸)² = 6.09 × 10¹⁸ m².

Stefan finishes it. P = σAT⁴ = 5.67 × 10⁻⁸ × 6.09 × 10¹⁸ × 5800⁴ = 3.9 × 10²⁶ W.

That structure repeats in every question of this family. Wien for temperature or wavelength, geometry for area, Stefan for power, in whichever order the givens allow.

Reading a star from its light

The same three relationships also work from the measurements back to the star. The peak of its spectrum gives T through Wien's law. Its measured brightness and distance give P through the inverse square law, and Stefan's law then gives the surface area, and hence the radius. A star's size can be measured without ever resolving its disc.

Two assumptions underlie that, and a question that asks you to state them wants both. The star radiates as a black body, and the light reaches us undimmed, with nothing absorbed by dust or atmosphere on the way. Real measurements correct for both.

GUIDED PRACTICE

Two thermometers

Betelgeuse has a surface temperature of about 3500 K; Rigel runs near 12 000 K. Find each star's peak wavelength and state what colour each appears.

Show the working

Betelgeuse gives λ_max = 2.9 × 10⁻³ / 3500 = 8.3 × 10⁻⁷ m, peaking in the infrared, so the visible light that escapes is reddish.

Rigel gives λ_max = 2.9 × 10⁻³ / 12 000 = 2.4 × 10⁻⁷ m, peaking in the ultraviolet, and blue-white to the eye.

Both stars lie in Orion, and their colours indicate their surface temperatures directly.

INDEPENDENT PRACTICE

Same power, twice the temperature

Star X radiates the same total power as the Sun but has twice its surface temperature. Find the ratio of X's radius to the Sun's.

Show the working

Equal P means the σAT⁴ products match, so the areas trade against T⁴ and the area ratio is (1/2)⁴ = 1/16.

Area goes as radius squared, so the radius ratio is the square root, 1/4. A small hot star can radiate as much power as a large cool one, because the fourth power of temperature outweighs the difference in area.

The spectral classes

Starlight arrives with narrow wavelengths missing. These are the absorption lines, printed where atoms and ions in the star's cooler outer layers have soaked up their characteristic photons. Which lines appear depends almost entirely on temperature, and stars are filed by their lines into the spectral classes, hottest to coolest:

Class	Colour	Temperature / K	Prominent absorption
O	blue	25 000 to 50 000	He⁺, He, H
B	blue	11 000 to 25 000	He, H
A	blue-white	7500 to 11 000	H (strongest), ionised metals
F	white	6000 to 7500	ionised metals
G	yellow-white	5000 to 6000	ionised and neutral metals
K	orange	3500 to 5000	neutral metals
M	red	below 3500	neutral atoms, TiO

The Sun, at 5800 K, is a class G star. The classic memory aid is the sentence "Oh Be A Fine Girl, Kiss Me", and the letters' scrambled order is a fossil of an older cataloguing scheme.

A bar chart of hydrogen Balmer line strength across the spectral classes O to M: the bars rise to a clear peak at class A and fall away on both sides, because hot stars ionise their hydrogen and cool stars leave too few atoms in the second energy level. — FIG. 2Why hydrogen's visible lines peak at class A. The Balmer series absorbs from the n = 2 level: cool stars keep too few atoms excited that high, and the hottest stars ionise their hydrogen away.

One detail is examinable in depth. Hydrogen's visible absorption lines, the Balmer series, come from atoms absorbing photons while sitting in the n = 2 energy level. In cool stars almost every hydrogen atom rests in the ground state, so Balmer absorption is feeble. In the hottest stars the hydrogen has been ionised, and an atom with no bound electron gives no lines at all. The strongest Balmer lines therefore appear in the middle, at class A around 7500 to 11 000 K, hot enough to lift plenty of atoms into n = 2 and cool enough to leave the atoms intact.

ASSESSMENT FOCUS

Stefan's law needs area, not radius. Convert with A = 4πr² and keep T in kelvin. The fourth power magnifies any slip.
Wien's constant carries the unit m K, a product. Quote λ_max in metres and resist writing it "per kelvin".
Two assumptions, both wanted. The star radiates as a black body, and nothing between star and telescope absorbs the light.
Learn the class table as data. Order OBAFGKM, colours blue through red, the temperature bands, the prominent lines. This is pure recall.
The Balmer explanation must name the level. Absorption from n = 2, too few excited atoms when the star is cool, hydrogen ionised when it is hot, peak at class A. Give every clause in that chain.

CHECK YOURSELF

A white dwarf has surface temperature 25 000 K and radiates 9.5 × 10²⁴ W. Find its peak wavelength and its radius, and comment on the size.

Show a hint

Wien for the peak, then rearrange Stefan for area and get the radius from A = 4πr².

Show the answer

λ_max = 2.9 × 10⁻³ / 25 000 = 1.2 × 10⁻⁷ m, deep in the ultraviolet.

A = P/σT⁴ = 9.5 × 10²⁴ / (5.67 × 10⁻⁸ × 25 000⁴) = 4.3 × 10¹⁴ m², so r = $\sqrt{A/4\pi}$ = 5.8 × 10⁶ m.

That is about the radius of the Earth. It packs a star's worth of matter, hotter than the Sun's surface, into a planet's volume.

Wien reads the temperature off the peak of the curve.

Stefan turns temperature and area into power, and the fourth power dominates everything.

OBAFGKM files every star hot to cool.

Balmer lines peak at class A, because absorption needs atoms already sitting in n = 2.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

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WHERE TO GO NEXT

Proportional reasoning is the maths this lesson leans on, worked through step by step.

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Use black-body curves, Stefan's law and Wien's law to find stellar temperatures, powers and radii.
State the assumptions made when treating a star as a black body.
Recall the spectral classes O to M with their colours, temperatures and absorption lines, and explain the Balmer condition.

Open the full revision checklist to track your progress across the whole unit.

Black-body radiation and spectral classes

The black-body curve

Reading a star from its light

The spectral classes

Question workbook

Question workbook

Answer book