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v0.1.1040
NotesPhysicsTopic 2.2Black-body radiation: Stefan-Boltzmann and Wien
Back to Physics Topics
2.2.22 min read

Black-body radiation: Stefan-Boltzmann and Wien

IB Physics • Unit 2

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Contents

  • What a black body radiates
  • The two laws
  • Exam-style question
The big idea: A black body is a perfect absorber and emitter — it soaks up every wavelength that hits it and, when hot, radiates over every wavelength too.

A hot object (a star, 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.
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.

Two given equations turn that picture into numbers. The first is the Stefan-Boltzmann law — the total power a black body radiates (its luminosity):

Stefan-Boltzmann law. Given in the data booklet. T MUST be in kelvin; the T⁴ makes power rise very fast with temperature.
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.

The second is Wien's displacement law — it locates the peak of the curve (the brightest wavelength):

Wien's displacement law. Given in the data booklet. λ_max is the peak wavelength; the product with T is a fixed constant.
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 λ_max = 2.9 × 10⁻³ ÷ T.

Worked example — peak wavelength of a star

A star has a surface temperature of 5.0 × 10³ K. Find the wavelength at which it radiates most strongly.

Solution

  1. Start with the given Wien law:
  2. Rearrange for the peak wavelength:
  3. Put in T = 5.0 × 10³ K:
  4. Work it out — keep the unit:

Final answer

λmax = 5.8 × 10⁻⁷ m (580 nm) — in the visible (yellow) range.

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How this is tested: Black-body radiation can be tested on either paper.

- Paper 1A: a quick MCQ — how the curve changes when T is raised or lowered (peak moves, height changes), or a one-step Stefan-Boltzmann / Wien sum. - Paper 2: 'determine' questions — compare two black bodies with L = σAT⁴ (e.g. find a star's radius), or find the peak wavelength of the Sun from its temperature.

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.

IB-style question — find the radius of a second star

Two stars behave as black bodies. Star P and star Q 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. (Treat each star as a sphere, area A = 4πr².)

Solution

  1. Start with the given Stefan-Boltzmann law for each star:
  2. Equal power means LP = LQ. The σ and 4π cancel:
  3. Rearrange for rQ:
  4. Put in the numbers (TP / TQ = 6.0 × 10³ ÷ 3.0 × 10³ = 2.0):
  5. Work it out — keep the unit:

Final answer

rQ = 2.8 × 10⁹ m — the cooler star must be larger to radiate the same power, because L ∝ r²T⁴ and its T is halved (T⁴ falls by 16×).

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The filament of a lamp behaves as a black body.

The lamp is dimmed, so the filament's temperature falls.

how the shape of the filament's black-body radiation curve changes as a result.
[2 marks]

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2.1.2Specific heat capacity
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2.1.4Conduction, convection and radiation
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