Physics › Astrophysics › Black-body radiation and spectral classes
Black-body radiation and spectral classes
A star's colour is a thermometer and its spectrum is a fingerprint. Two short laws turn the shape of starlight into a temperature and a power output, and the missing wavelengths sort every star in the sky into seven lettered classes.
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.
WHAT YOU PROBABLY THINK
A red-hot star is hotter than a white one.
The black-body curve
A black body is a perfect absorber and therefore a perfect emitter: the radiation it gives off depends only on its temperature, spread over wavelength in a lopsided hump called the black-body curve. A star is, to a good approximation, a black body, and most of stellar astronomy is built on that single assumption. Two laws summarise the curve. Stefan's law gives the total power radiated:
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 savage: double the temperature and the output multiplies by sixteen. Wien's displacement law locates the curve's peak:
so hotter stars peak at shorter wavelengths: red stars are the cool ones, blue-white stars the furnaces, and "red hot" turns out to be the mild end of stellar temperatures. Note the unit of Wien's constant, metre kelvin, a product, and never 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: λmax = 2.9 × 10−3 / 5800 = 5.0 × 10−7 m, in the green, matching the curve above.
Surface area: A = 4πr2 = 4π × (6.96 × 108)2 = 6.09 × 1018 m2.
Stefan: P = σAT4 = 5.67 × 10−8 × 6.09 × 1018 × 58004 = 3.9 × 1026 W.
The 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
Point a telescope at any star and the same three-step logic runs in reverse: the peak of its spectrum gives T through Wien, its measured brightness and distance give P through the inverse square law, and Stefan's law then hands over the surface area, hence the radius. A star's size can be measured without ever resolving its disc.
The honesty clause is the list of assumptions: 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, and questions asking you to "state the assumptions" want exactly those.
YOUR TURN
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, before opening the working.
Show the working
Betelgeuse: λmax = 2.9 × 10−3 / 3500 = 8.3 × 10−7 m, peaking in the infrared, so the visible light that escapes is reddish.
Rigel: λmax = 2.9 × 10−3 / 12 000 = 2.4 × 10−7 m, peaking in the ultraviolet: blue-white to the eye.
One constellation, Orion, carries both: a glance at their colours is a glance at their temperatures.
TRY IT UNSEEN
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: the area ratio is (1/2)4 = 1/16.
Area goes as radius squared, so the radius ratio is the square root: 1/4. Hot little stars can match cool bloated ones watt for watt: temperature does the heavy lifting through that fourth power.
The spectral classes
Starlight arrives with narrow wavelengths missing: 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.
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 is ionised: no bound electron, no lines at all. The strongest Balmer lines therefore appear in the middle, class A around 7500 to 11 000 K, hot enough to lift plenty of atoms into n = 2, cool enough to leave the atoms intact.
THE EXAM BIT
- Stefan's law needs area, not radius: convert with A = 4πr2 and keep T in kelvin. The fourth power punishes any slip twice over.
- Wien's constant carries the unit m K, a product. Quote λmax in metres and resist quoting it "per kelvin".
- The black-body assumption list is two items: the star radiates as a black body, and nothing between star and telescope absorbs the light. State both when asked.
- Learn the class table as data: order OBAFGKM, colours blue through red, the temperature bands, and the prominent lines. It is pure recall and appears often.
- The Balmer explanation must name the level: absorption from n = 2, too few excited atoms when cool, hydrogen ionised when hot, peak at class A. Four clauses, four marks.
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; rearrange Stefan for area, then radius from A = 4πr².
Show the answer
λmax = 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 = √A/4π = 5.8 × 106 m.
That is about the radius of the Earth: a star's worth of matter, hotter than the Sun's surface, packed into a planet's volume. The next lesson explains how it got that way.
Wien reads the temperature from the peak; Stefan turns temperature and area into power.
OBAFGKM files every star hot to cool, and Balmer lines peak at A because absorption needs 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.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device only.
- 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.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.