B.2 · IB Physics
Blackbody radiation and Wien's law
Every hot object glows across a spectrum. Raise the temperature and the curve grows dramatically while its peak slides towards shorter wavelengths. Compare two bodies, or jump to a star preset.
Predict first
Set the Temperature T slider to 2000 K, then to 4000 K. What happens to the peak wavelength λmax?
Wien's law gives λmax × T = 2.90 × 10⁻³ m K, so λmax ∝ 1/T. Doubling T from 2000 K to 4000 K halves λmax, from 1.45 μm to 0.725 μm.
Check it in the model: Move the Temperature, T slider from 2000 K to 4000 K and compare the λmax values in the readout.
body 1 body 2 (compare) · dot = λmax from Wien's law
P/A = σT⁴, σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
I(λ) ∝ 1 ÷ [λ⁵ (e^(hc/λkT) − 1)]
Try this
- Set T = 3000 K, then 6000 K. By what factor does the peak wavelength change, and by what factor does the emitted power change?
- Press 'Sun 5800 K' and read where the peak sits relative to the shaded visible band. Then press 'Red dwarf 3000 K' and note what has happened to the visible part of the curve.
- Tick 'Compare' with T = 6000 K and T₂ = 3000 K. Check that the ratio of powers equals (T/T₂)⁴.
Exam tip: Wien's law gives the wavelength of the peak (λ ∝ 1/T) while Stefan–Boltzmann gives the total power per unit area (∝ T⁴): doubling T halves the peak wavelength and multiplies the power by 16.