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Astronomy · Stars & Light

Wien’s Law Calculator

Convert between a black body’s temperature and its peak wavelength with λ_peak = b ÷ T.

K
Absolute temperature of the emitter, in kelvin.
Real emitters — tap to load
Peak wavelength (λ_peak)
501.5nmVisible — green

That is 0.5015 µm — Visible — green. A black body this hot appears yellow-white. b = 2.897771955×10⁻³ m·K.

Wien’s law gives a black body’s peak wavelength as λ_peak = b ÷ T, with b = 2.898×10⁻³ m·K. The Sun (T ≈ 5778 K) peaks near 500 nm — green light — yet looks white because it radiates across the whole visible range. Hotter bodies peak at shorter, bluer wavelengths; cooler ones peak in the red and infrared.

Blackbody radiation and Wien’s law

Any object above absolute zero glows with thermal (blackbody) radiation, emitting across a broad range of wavelengths. The spectrum has a single hump, and the wavelength of that hump — the peak wavelength — depends only on temperature. Wien’s displacement law pins it down: λ_peak = b ÷ T, where b = 2.897771955×10⁻³ m·K is Wien’s displacement constant, T is the absolute temperature in kelvin, and λ_peak is the peak wavelength in metres.

Hotter means bluer. Because temperature sits in the denominator, raising T pushes the peak to shorter wavelengths — toward blue, then ultraviolet and beyond. A cool 3000 K ember peaks in the near-infrared and glows dull red, while a 25,000 K star peaks in the ultraviolet and looks blue-white. This is why astronomers can read a star’s temperature straight from its colour.

λ_peak = b ÷ T  ·  T = b ÷ λ_peak

b = 2.898×10⁻³ m·K (Wien’s displacement constant), T = temperature in kelvin, λ_peak = peak wavelength in metres

Worked example

At what wavelength does the Sun’s surface (T ≈ 5778 K) radiate most strongly?

  1. 1
    Make sure the temperature is in kelvin. Wien’s law needs absolute temperature: here T = 5778 K.
  2. 2
    Write the law. λ_peak = b ÷ T with b = 2.897771955×10⁻³ m·K.
  3. 3
    Divide the constant by the temperature. λ_peak = 2.898×10⁻³ ÷ 5778 ≈ 5.01×10⁻⁷ m.
  4. 4
    Convert to nanometres (× 1×10⁹). 5.01×10⁻⁷ m × 10⁹ ≈ 501 nm — green light, in the middle of the visible band.

Peak wavelength of some objects

Peak wavelengths from λ_peak = b ÷ T. A star’s perceived colour follows its temperature, not the single peak wavelength.

ObjectTemperaturePeak wavelengthRegion / colour
The CMB2.725 K1.06 mmMicrowave
Cool red star3500 K828 nmNear-infrared (looks red)
The Sun5778 K502 nmGreen peak (looks white)
A-type star (Sirius)9940 K292 nmUltraviolet (blue-white)
Hot blue star25,000 K116 nmUltraviolet (looks blue)

Reading the result

The peak is not the whole story. Wien’s law finds where the spectrum peaks, but a black body emits at every wavelength. The Sun peaks in green yet appears white because our eyes blend its strong red, green, and blue output together — no single colour dominates.

A peak outside the visible band is normal. Many familiar emitters peak in the infrared or ultraviolet: a 3500 K star peaks at about 828 nm (near-infrared) but still looks red, because plenty of its light spills into the visible red. Only the visible tail of the spectrum reaches your eye.

Use kelvin. Temperature must be absolute. Feeding Celsius or Fahrenheit into λ_peak = b ÷ T gives a meaningless answer — convert first with K = °C + 273.15.

Why does the Sun look yellow-white if it peaks in green?
Wien’s law puts the Sun’s peak near 500 nm — green — but a black body radiates strongly across the whole visible range. Our eyes combine its red, green, and blue light into white; from space the Sun looks white, not green.
What is Wien’s displacement constant?
It is b = 2.897771955×10⁻³ m·K, the fixed proportionality in λ_peak = b ÷ T. Multiply it by 1 ÷ T (with T in kelvin) to get the peak wavelength in metres, then × 10⁹ for nanometres.
How is Wien’s law different from the Stefan-Boltzmann law?
Wien’s law tells you where a black body’s spectrum peaks (λ_peak = b ÷ T). The Stefan-Boltzmann law tells you how much total power it radiates (P = εσAT⁴). One gives colour, the other gives brightness — both from the same temperature.
Why do hotter objects glow bluer?
Temperature is in the denominator of λ_peak = b ÷ T, so a higher T gives a shorter peak wavelength. Short wavelengths are toward the blue and ultraviolet end, so hotter bodies shift from red to white to blue.
Do I have to use kelvin?
Yes. Wien’s law needs absolute temperature, because the ratio b ÷ T only makes sense from absolute zero. Convert Celsius with K = °C + 273.15 (and Fahrenheit to Celsius first) before entering T.
Why does the calculator sometimes report an infrared or ultraviolet peak?
Because the peak wavelength often lies outside visible light. A 3500 K star peaks near 828 nm in the near-infrared, and a 25,000 K star peaks around 116 nm in the ultraviolet — yet each still shows a visible colour from the part of its spectrum that reaches your eye.
What temperature peaks in the middle of visible light?
Setting λ_peak ≈ 550 nm and solving T = b ÷ λ_peak gives about 5270 K. Stars near this temperature, like our Sun at 5778 K, peak within the visible band — which is roughly why our eyes evolved to be most sensitive there.