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For a blackbody spectrum expressed per wavelength, the peak wavelength is inversely proportional to absolute temperature.
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Formula
λmax in m; T in K; bW ≈ 2.898×10−3 m·K
| Symbol | Meaning | Unit |
|---|---|---|
| Wavelength of the peak in spectral radiance per wavelength | m | |
| Absolute temperature of the blackbody | K | |
| Wien displacement constant | m·K |
\lambda_{\max} T = b_WThe relation λmaxT=bW applies to the maximum of Planck’s distribution per unit wavelength. A spectrum plotted per frequency has a different peak location, so the wavelength and frequency peaks cannot be interchanged by simply converting units. Real stars are only approximately blackbodies and may show absorption lines, dust effects or atmospheric features. The equation estimates a characteristic continuum temperature, not every spectral feature.
Applies to the peak of a blackbody spectrum per wavelength; real stellar spectra and frequency representations may peak elsewhere.
As temperature rises, the Planck peak shifts to shorter wavelengths.
Temperature from peak wavelength
Substitute λmax in metres.
Peak wavelength from temperature
This refers to the spectral distribution per wavelength.
A stellar continuum peaks at 500 nm. Estimate T.
T≈2.898·10−3/(500·10−9)≈5800 K, assuming a blackbody-like continuum.
How does λmax change if temperature doubles?
Since λmaxT is constant, λmax halves when temperature doubles.
Applying the law to any emission peak without a blackbody assumption.
Check the continuum shape and possible line, dust or atmospheric effects.
Treating wavelength and frequency spectra as having the same peak.
Name the spectral-density variable; dλ and dν weight the spectrum differently.
Formula cluster
The peak estimates temperature; total power also requires Stefan–Boltzmann.
If a blackbody-like stellar continuum peaks at 500 nm, T≈2.898×10−3/(500×10−9)≈5,800 K.
Estimate the characteristic temperature of stars, lamps and thermal emitters from their wavelength spectra.
Here is how to work through a typical Wien displacement law (λₘₐₓT = bW) task step by step:
Task
A stellar continuum peaks at 500 nm. Estimate T.
Solution path
T≈2.898·10−3/(500·10−9)≈5800 K, assuming a blackbody-like continuum.
Task
How does λmax change if temperature doubles?
Solution path
Since λmaxT is constant, λmax halves when temperature doubles.
It relates an ideal thermal emitter’s temperature to the wavelength where its spectral radiance per wavelength is greatest. A hotter body peaks at a shorter wavelength. It does not determine total power; emitting area and Stefan–Boltzmann are also relevant. Use temperature in kelvin and λmax in metres when using the SI constant. The law concerns a thermal blackbody continuum and spectral density per wavelength interval; real stellar spectra differ because of lines, atmospheres and dust. The peak therefore gives a characteristic temperature estimate, not a complete spectral analysis.
A spectrum can be expressed per wavelength or per frequency. These densities are related by a nonlinear variable change and have different peak locations. One cannot simply convert λmax to c/λ and claim that it is the frequency-spectrum maximum. Use temperature in kelvin and λmax in metres when using the SI constant. The law concerns a thermal blackbody continuum and spectral density per wavelength interval; real stellar spectra differ because of lines, atmospheres and dust. The peak therefore gives a characteristic temperature estimate, not a complete spectral analysis.
Use T=bW/λmax. For 500 nm, first convert to 500 nm=5×10−7 m; this gives T≈2.898×10−3/(5×10−7)≈5800 K. A common error is substituting nanometres directly with a constant given in m·K. As a check, λmax·T should be about 2.898×10−3 m·K. Use temperature in kelvin and λmax in metres when using the SI constant. The law concerns a thermal blackbody continuum and spectral density per wavelength interval; real stellar spectra differ because of lines, atmospheres and dust. The peak therefore gives a characteristic temperature estimate, not a complete spectral analysis.
No. An emission or absorption line comes from a transition between energy levels and does not generally mark the blackbody peak. The law applies to a thermal continuum. Lines can distort the measured peak and should be identified or modeled before estimating temperature. Use temperature in kelvin and λmax in metres when using the SI constant. The law concerns a thermal blackbody continuum and spectral density per wavelength interval; real stellar spectra differ because of lines, atmospheres and dust. The peak therefore gives a characteristic temperature estimate, not a complete spectral analysis.
bW is Wien’s displacement constant, approximately 2.898×10−3 m·K. Its unit makes λmaxT a length multiplied by temperature. It follows from Planck’s distribution and its maximum, not an empirical fit to one star. With consistent units, λmaxT should yield the same constant. Use temperature in kelvin and λmax in metres when using the SI constant. The law concerns a thermal blackbody continuum and spectral density per wavelength interval; real stellar spectra differ because of lines, atmospheres and dust. The peak therefore gives a characteristic temperature estimate, not a complete spectral analysis.
Quanta exam set
Curated exam set for "Wien displacement law":
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Which formula describes Wien displacement law?
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How do you rearrange Wien displacement law for Temperature from peak wavelength?
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Which common mistake happens with Wien displacement law?
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