Blackbody Radiation: Wien Displacement Law and Stefan–Boltzmann Law
Blackbody radiation is where quantum theory began. Classical physics could not explain the shape of the light emitted by a hot object, and Planck’s idea that energy comes in packets, E = hν, solved it. Entrance exams ask short numericals on two results that follow from Planck’s law: Wien’s displacement law and the Stefan–Boltzmann law. This page gives both laws, the constants, and four computed examples.
What is a blackbody?
A blackbody is an ideal object that absorbs all radiation falling on it and, when heated, emits a spectrum that depends only on its temperature and not on its material. A small hole in a closed hot cavity is a good practical model. Hot solids such as a heated filament behave approximately like a blackbody, and so do stars to a rough approximation.
As the temperature rises, two things happen: the object emits much more total energy, and the colour of the peak shifts towards shorter wavelengths (from red towards white and blue).
The two laws
Stefan–Boltzmann law: M = σT4, σ = 5.670 × 10−8 W m−2 K−4
- λmax = the wavelength at which emission per unit wavelength is largest. Since T is in kelvin, b is in m K.
- M = total power emitted per unit surface area (all wavelengths together), in W/m2. For an object of area A, total power = σAT4.
- The energy of a photon at the peak: E = hc/λmax, with h = 6.626 × 10−34 J s and c = 2.998 × 108 m/s.
Planck’s law itself describes the full curve; Wien’s and Stefan–Boltzmann’s laws are what you get by finding the peak and by adding up the area under it.
Worked example 1: temperature from the peak wavelength
A star’s spectrum peaks at 500 nm. Assuming it is a blackbody, find its surface temperature.
T = b / λmax = 2.898 × 10−3 / (500 × 10−9) = 5796 K, about 5800 K.
Check: λmax T = 500 × 10−9 × 5796 = 2.898 × 10−3 m K. Correct.
Worked example 2: peak of a hot filament
A lamp filament is at 3000 K. Find λmax, the photon energy at the peak, and say which region it is in.
λmax = 2.898 × 10−3 / 3000 = 9.66 × 10−7 m = 966 nm. Visible light is about 400–700 nm, so this peak lies in the near-infrared. That is why filament lamps waste much of their energy as heat.
E = hc/λ = (6.626 × 10−34 × 2.998 × 108) / (9.66 × 10−7) = 2.06 × 10−19 J = 1.28 eV (dividing by 1.602 × 10−19 J/eV).
Worked example 3: total emitted power
Find the power radiated per square metre by a blackbody at 1000 K, and by one at 3000 K.
At 1000 K: M = 5.670 × 10−8 × (1000)4 = 5.670 × 10−8 × 1012 = 5.67 × 104 W/m2 = 56.7 kW/m2.
At 3000 K: temperature is 3 times larger, so power is 34 = 81 times larger: 81 × 56.7 = 4593 kW/m2 (about 4.59 MW/m2).
Worked example 4: the fourth-power rule
By what factor does the radiated power change when the absolute temperature is doubled, say from 1500 K to 3000 K? What happens to λmax?
Power ratio = (3000/1500)4 = 24 = 16.
λmax is inversely proportional to T, so it halves: 2.898 × 10−3/1500 = 1932 nm, and 2.898 × 10−3/3000 = 966 nm.
The temperature must be in kelvin. Using 1500 °C and 3000 °C directly would give a wrong ratio.
Common mistakes
- Using Celsius. Both laws need absolute temperature in kelvin.
- Forgetting the unit of b. b is in m K, so λ comes out in metres; convert to nm (× 109).
- Using T instead of T4 in the Stefan–Boltzmann law.
- Using the same b for peak frequency. The peak of the spectrum per unit frequency is at a different place; the relation is νmax = 5.879 × 1010 Hz/K × T. And c/νmax is not equal to λmax. Use the form the question asks for.
- Saying a hotter body emits only at shorter wavelengths. It emits more at all wavelengths; only the peak shifts.
- Applying the law to non-ideal surfaces without an emissivity. A real surface emits εσT4, where ε is less than 1.
Exam relevance
| Question type | Method |
|---|---|
| Find T from a peak wavelength | T = b/λmax |
| Find the colour region | Compute λmax, compare with 400–700 nm |
| Power ratio between two temperatures | (T2/T1)4 |
| Why quantum theory was needed | The classical ultraviolet catastrophe; Planck’s E = hν |
Practice values only. Constants may differ in the last digit between books, so use those provided in your paper, and check your exam’s current official syllabus.
Convert the peak wavelength to a photon energy using the Photon and de Broglie calculator.
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