Peak wavelength λmax = b/T, inverse temperature solver, peak-frequency form, and wavelength converter — for blackbody thermal radiation.
Tap an object to see where its thermal glow peaks.
| Quantity | Value |
|---|---|
| Wien's displacement constant b (CODATA 2018) | 2.897771955×10⁻³ m·K |
| Peak-frequency constant νmax/T | 5.87892576×10¹⁰ Hz/K |
| Wavelength root xλ of x = 5(1−e−x) | 4.9651142317 |
| Frequency root xν of x = 3(1−e−x) | 2.8214393721 |
| Fixed ratio νmax / (c/λmax) = xν/xλ | 0.5683 |
| Object | Temperature | Peak wavelength | Band |
|---|---|---|---|
| Sun (photosphere) | 5778 K | 501.52 nm | visible (green) |
| Tungsten bulb | 3000 K | 965.92 nm | near-infrared |
| Molten lava | 1500 K | 1932 nm (1.93 µm) | infrared |
| Human body | 310 K | 9.35 µm | far infrared |
| Cosmic microwave background | 2.725 K | 1.063 mm | microwave |
All values computed from λmax = b/T with b = 2.897771955×10⁻³ m·K.
Wien's displacement law states that the wavelength where a blackbody radiates most intensely is inversely proportional to its absolute temperature: λmax = b / T, where b = 2.897771955×10⁻³ m·K (CODATA 2018). Hotter objects peak at shorter wavelengths — that is why the Sun (5778 K) peaks in visible light at 501.52 nm while the human body (310 K) peaks deep in the infrared at 9.35 µm.
Wien's displacement constant is b = 2.897771955×10⁻³ m·K (CODATA 2018). It equals hc/(k·x) where x ≈ 4.9651142317 is the root of x = 5(1−e⁻ˣ), from differentiating Planck's law. Use meters and kelvin with this value; if you work in nm·K the constant is 2,897,771.955 nm·K.
The Sun's photosphere is about 5778 K, so λmax = b/T = 2.897771955×10⁻³ / 5778 ≈ 501.52 nm — green visible light (0.502 µm). Sunlight looks white rather than green because the Sun emits a broad spectrum around that peak, not a single wavelength.
At a body temperature of 310 K (37 °C), λmax = b/310 ≈ 9.35 µm (9348 nm) — far infrared, invisible to the eye but exactly what thermal cameras detect. This is why night-vision and fever-screening cameras work in the 8–14 µm band.
Invert the law: T = b / λmax. For example, a spectrum peaking at 500 nm gives T = 2.897771955×10⁻³ / (500×10⁻⁹) ≈ 5795.54 K. Use the “Temperature from peak” tab above — enter the wavelength in nm, µm, mm or m and get the temperature in K, °C and °F.
Per unit frequency, the Planck spectrum peaks at νmax/T = 5.87892576×10¹⁰ Hz/K, where the constant is x·k/h with x ≈ 2.8214393721 (root of x = 3(1−e⁻ˣ)). For the Sun that is 339.68 THz. Crucially, this is NOT the same as c/λmax.
Because λmax and νmax are peaks of two different distributions (per-unit-wavelength vs per-unit-frequency); a peak does not survive a nonlinear change of variable. For the Sun, c/λmax ≈ 597.77 THz while the true per-frequency peak is νmax ≈ 339.68 THz — a fixed ratio of 0.5683 (xν/xλ) at every temperature. Calculators that just divide c by λmax report the wrong peak frequency.
Peak wavelength sets the dominant hue only loosely — real objects emit a broad spectrum. A tungsten bulb at 3000 K peaks at 965.92 nm (0.966 µm, near-infrared), so most of its energy is invisible heat and the visible tail looks orange. Molten lava near 1500 K peaks at 1932 nm (1.93 µm) and glows deep red-orange.
Yes. The CMB is the most perfect blackbody known, at 2.725 K, so λmax = b/2.725 ≈ 1.063 mm — in the microwave band, which is why it is called the microwave background. Penzias and Wilson's 1965 discovery measured exactly this glow.
With b = 2.897771955×10⁻³ m·K, temperature must be in kelvin (absolute — never °C or °F directly) and the result comes out in meters. The converter tab above handles nm, µm, mm and m; the temperature inputs accept K, °C and °F and convert to kelvin first.
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