Wien's Law Peak Wavelength Tool

Enter a temperature in kelvin to find the peak emission wavelength via Wien's displacement law.

Worked example — Wien's Law Peak Wavelength Tool

Peak wavelength (nanometers)	502.039 nm
Frequency at peak	5.971 × 10¹⁴ Hz
Photon energy at peak	2.47 eV
Closest reference	The Sun
The Sun	5772 K, 0 K away
Input blackbody temperature	5772 K
same peak in common astronomy units	502.039 nm · 0.502 µm · 5020.395 Å

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

The constant hides a number nobody can write down 🖖

λ_max = b/T looks like it needs a measured constant, but b is not measured — it is derived, and it contains a transcendental root. Maximising the Planck curve leads to x = 5(1 − e^−x), whose non-zero solution is x = 4.965114232…, a number with no closed form. Then b = hc/(x·k_B), which lands on 2.897772 × 10⁻³ m·K — the value this tool uses. So the awkward 2.8978 is not an accident of units: it is Planck's constant times the speed of light, divided by Boltzmann's constant and by a root nobody can write out exactly.

Why a star's color reveals its heat 🖖

This tool turns the everyday link between color and temperature into a number. Everyday intuition runs backwards: on a tap, red means hot and blue means cold, but for glowing objects it is the reverse. A cool body glows dull red; heat it and the peak slides through orange, yellow, and finally blue-white. That is why the red star Betelgeuse (~3500 K) is far cooler than blue-white Rigel (~12000 K).

The peak depends on how you plot it 🖖

Wien's law has a hidden trap: the wavelength where a blackbody peaks and the frequency where it peaks do not describe the same photon. The Sun's spectrum peaks near 500 nm in wavelength (green), yet its frequency spectrum peaks around 880 nm (infrared). They differ because squeezing wavelength into frequency stretches the axis unevenly. So the 'peak frequency' this tool reports, ν = c/λ_max, is not where the frequency curve actually maxes out.

Problem solved in full

The colour the Sun peaks at and why a greenhouse works 5 steps
Find the colour the Sun peaks at, and then use the same one-line law to explain why a greenhouse works — and why night-vision cameras see warm bodies at all. Sun: T = 5772 K.
1. Everything with a temperature radiates, and the wavelength it radiates most strongly at is inversely proportional to temperature. Hotter means bluer — that is the whole content of Wien’s law, and the constant is fixed by nature.
  
  $λ_{m a x} = \frac{b}{T}, b = 2.898 \times 1 0^{- 3} m K$
2. Substitute the Sun’s effective temperature. The calculator above prints this figure, and it lands in green — near the middle of the visible band, and very close to where the human eye is most sensitive.
  
  $λ_{m a x} = \frac{2.898 \times 1 0 ^{- 3}}{5772} = 5.020 \times 1 0^{- 7} m = 502.039 nm$
3. The frequency follows immediately from the wave relation, and the tool prints this too.
  
  $f = \frac{c}{λ _{m a x}} = \frac{2.998 \times 1 0 ^{8}}{5.020 \times 1 0 ^{- 7}} = 5.971 \times 1 0^{14} Hz$
4. Now apply the identical law to something that is not a star: Earth’s surface at 288 K.
  
  $T = 288 K : λ_{m a x} = \frac{2.898 \times 1 0 ^{- 3}}{288} = 1.006 \times 1 0^{- 5} m = 10.06 μ m$
5. Compare the two peaks. Same law, same constant, temperatures differing by a factor of 20 — so wavelengths differ by a factor of 20 in the other direction.
  
  $\frac{λ _{Earth}}{λ _{Sun}} = \frac{10.06 μ m}{0.502 μ m} = 20$
Answer

502 nm, in the green. Earth peaks at 10.06 μm, deep in the infrared, and that factor of twenty is the entire greenhouse mechanism: glass and CO₂ are transparent at half a micron and opaque at ten, so sunlight arrives freely and the re-radiated heat cannot leave by the same route. It is also why thermal cameras work in the dark — a 310 K body peaks near 9.4 μm and glows constantly at a wavelength your eye cannot see. One inverse proportionality, three unrelated-looking phenomena.

$λ_{m a x} T = b 502 nm (Sun) v s 10.06 μ m (Earth)$

References (2)

The distribution whose maximum gives the displacement law: M. Planck, "Ueber das Gesetz der Energieverteilung im Normalspectrum." Annalen der Physik 309(3), 553–563, 1901.
The value of the displacement constant b, and of h, c and k_B behind it: E. Tiesinga, P. J. Mohr, D. B. Newell and B. N. Taylor, "CODATA recommended values of the fundamental physical constants: 2018." Reviews of Modern Physics 93(2), 025010, 2021.

By the 42Math team · Updated 2026-07-03

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Astronomy & Space lessons →

Example problems

Room temp - Room temperature objects peak deep in infrared, not visible light.
Sun-like - The Sun peaks near visible wavelengths, around green-yellow.
Hot blue star - Hot stars peak in UV, which is why they look blue-white in visible light.
X-ray plasma - Extremely hot plasmas can peak in X-ray wavelengths.