About the Wien's Law Calculator
This Wien's law calculator uses Wien's displacement law, λmax = b / T, to find the wavelength at which a hot object — a star, a light bulb filament, a stove element or your own skin — emits most strongly. Enter a temperature in kelvin, °C or °F to get the peak wavelength in nanometres and micrometres, or enter a measured peak wavelength to work back to the temperature. It also reports the spectral band of the peak, the photon energy there, the frequency-form peak and the total power radiated per square metre.
It is aimed at astronomy and physics students, at photographers and lighting designers thinking about colour temperature, and at anyone using a thermal camera who wants to know why warm bodies glow in the infrared. The chart draws the full Planck curve around the peak so you can see how lopsided the blackbody spectrum is.
The law assumes an ideal blackbody. Real objects are approximately black over parts of the spectrum, so treat the temperature from a peak as a colour (effective) temperature. Note that the peak of the spectrum per unit wavelength and per unit frequency are at different places; both are shown.
How to use the wien's law calculator
- 1Choose whether you know the temperature or the peak wavelength.
- 2Enter the temperature (K, °C or °F) or the wavelength (nm, µm or mm).
- 3Read the peak wavelength or temperature in the answer.
- 4Check the spectral band and photon energy of the peak.
- 5Use the chart to see the shape of the blackbody spectrum.
Formula and method
Wien's displacement law follows from Planck's law for blackbody radiation: setting the derivative of spectral radiance per unit wavelength to zero gives a peak at λmax = b/T, where b is Wien's constant (exact in the 2019 SI because h, c and k are defined). Hotter objects therefore peak at shorter wavelengths — the Sun near 500 nm (green-blue visible light), a light bulb filament near 1 µm (infrared), and a human body near 9.3 µm (thermal infrared).
The spectrum plotted per unit frequency peaks at a different point, νmax = 5.879 × 10¹⁰ Hz/K × T, which does not equal c/λmax. The photon energy at the peak is E = hc/λ (1239.84 eV·nm ÷ λ), and the total power radiated by a blackbody surface is σT⁴ with σ = 5.670374419 × 10⁻⁸ W/(m²·K⁴).
- λmax
- Wavelength of peak emission
- T
- Absolute temperature (K)
- b
- Wien's displacement constant
- νmax
- Peak of the per-frequency spectrum
- σ
- Stefan–Boltzmann constant
Worked examples
The Sun (5,772 K)
λmax = 2.8978 × 10⁻³ ÷ 5772 = 5.020 × 10⁻⁷ m = 502 nm, in the blue-green part of the visible spectrum. The Sun’s surface radiates about 6.29 × 10⁷ W per square metre.
Human skin at 37 °C
37 °C is 310.15 K, so λmax = 2.8978 × 10⁻³ ÷ 310.15 = 9.34 µm. That is why thermal cameras that see people work in the 8–14 µm band.
Incandescent bulb filament at 2,700 K
λmax = 2.8978 × 10⁻³ ÷ 2700 = 1,073 nm, in the near infrared. Most of a tungsten bulb’s output is invisible heat, which is why such bulbs are so inefficient.
Temperature from a 10 µm peak
T = 2.8978 × 10⁻³ ÷ (10 × 10⁻⁶) = 289.8 K, about 16.6 °C — typical of Earth’s surface, which radiates mainly near 10 µm.
Frequently asked questions
What is Wien's displacement law?+
Wien's law states that a blackbody's spectrum peaks at a wavelength inversely proportional to its temperature: λmax = b/T, with b = 2.898 × 10⁻³ m·K. Double the temperature and the peak wavelength halves.
What is the value of Wien's constant?+
The Wien wavelength displacement constant is b = 2.897771955 × 10⁻³ m·K (about 2,898 µm·K). Since the 2019 SI redefinition it is an exact value derived from the Planck, Boltzmann and light-speed constants.
Why does the Sun look white if it peaks in green?+
The solar spectrum is broad and covers the whole visible range with fairly even intensity, so our eyes perceive the mixture as white. The peak only marks the single strongest wavelength, not the perceived colour.
How do astronomers use Wien’s law?+
By measuring where a star’s spectrum peaks, astronomers estimate its surface temperature. Red stars peak in the infrared and are cooler (about 3,000 K), while blue stars peak in the ultraviolet and can exceed 20,000 K.
Why is the frequency peak different from the wavelength peak?+
Spectral radiance per unit wavelength and per unit frequency are different distributions, so their maxima fall at different places. For the Sun the frequency peak is near 339 THz, which corresponds to about 884 nm, not 502 nm.