The big idea: A hot object pouring out light across all the wavelengths is behaving as a black body, a perfect absorber and emitter of radiation.
Seeing a black body
- Push a poker into a fire and watch it glow dull red, then orange, then white.
- A star, or a glowing iron bar, is a good model of one.
- Its brightness across the wavelengths makes a single humped curve called the black-body spectrum.
Animated graph
Watch the graph build step by step in study mode.
Spot it: Hotter → taller and bluer. As the temperature rises the curve gets taller (more total power) and its peak shifts to a shorter wavelength (towards blue).
That's why a heated bar glows dull red, then orange, then white as it gets hotter.
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Two given equations turn that picture into numbers. The first is the Stefan-Boltzmann law — the total power a black body radiates (its luminosity):
- luminosity — total power radiated (W)
- Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴ (given)
- surface area of the body (m²)
- absolute surface temperature (K — kelvin)
Why the T⁴ matters: Because power depends on T to the fourth power, doubling the kelvin temperature multiplies the radiated power by 2⁴ = 16.
Always put T in kelvin (K), never °C: kelvin = °C + 273.
- emissivity (no unit, 0 to 1)
- Stefan-Boltzmann constant, 5.67 × 10⁻⁸ W m⁻² K⁻⁴
- surface temperature (K)
Finding an emissivity: A satellite of area 3.0 m² is held at 300 K by a 25 W heater, and it radiates all 25 W. Then e = 25 ÷ (5.67 × 10⁻⁸ × 3.0 × 300⁴) ≈ 0.018: a shiny, poorly emitting surface.
Comparing two bodies: Power ∝ eAT⁴, and for spheres A ∝ r², so P ∝ e r² T⁴. Write the ratio factor by factor.
Twice the temperature and the same power: r² must fall by 2⁴ = 16, so the radius is one quarter.
The second is Wien's displacement law — it locates the peak of the curve (the brightest wavelength):
- wavelength of peak (brightest) emission (m)
- absolute surface temperature (K — kelvin)
Read Wien as 'they trade off': λmax and T multiply to a constant, so they move opposite ways: a hotter body (bigger T) has a smaller peak wavelength (bluer light). Rearrange to λmax = 2.9 × 10⁻³ ÷ T.
A star has a surface temperature of 5.0 × 10³ K. Find the wavelength at which it radiates most strongly.
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How this is tested — black-body radiation can appear on either paper:
Paper 1A
- How the curve changes when T rises or falls (peak moves, height changes).
- Or a one-step Stefan-Boltzmann / Wien sum.
Paper 2
- Compare two black bodies with L = σAT⁴ (e.g. find a star's radius).
- Or find the Sun's peak wavelength from its temperature.
The classic trap: Leaving T in °C instead of kelvin, or forgetting the power is T⁴ (not T).
Comparing two bodies: When two black bodies are compared, write L = σAT⁴ for each and divide one by the other — σ cancels, leaving a clean ratio of areas and temperatures. For a sphere the area is A = 4πr², so the area ratio is the radius ratio squared.
The words to know
| Word | What it means |
|---|---|
| Black body | A perfect emitter and absorber: emissivity 1 |
| Emissivity e | Power radiated per unit area ÷ σT⁴ for the same temperature (0 to 1) |
| Peak wavelength λmax | The wavelength of strongest emission; λmax T = 2.9 × 10⁻³ m K |
| Luminosity | Total power radiated, L = eσAT⁴ (e = 1 for a black body) |
Two stars behave as black bodies and radiate the same total power. Star P has surface temperature 6.0 × 10³ K and radius 7.0 × 10⁸ m. Star Q is cooler, at 3.0 × 10³ K. Determine the radius of star Q. (Each star is a sphere, area A = 4πr².)
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