Wien's Displacement Law

Equivalent forms

b \approx 2.8977719 \times 10^{-3}\,\text{m}\cdot\text{K}

\nu_{\max} = \alpha k_B T / h,\ \alpha \approx 2.821

One constant tells you the color of any glowing object — from a stove to a quasar.

Symbol	Name	SI unit	Dimension	Range
$\lambda_{\max}$	Peak wavelengthoutput Wavelength of maximum spectral radiance	m	L	1e-9 – 0.001
$T$	Temperature Absolute temperature	K	Θ	3 – 10000000
$b$	Wien displacement constant Universal blackbody peak constant	m·K	L·Θ	0.0028977719 – 0.0028977719

Unit systems

Symbol	SI	CGS	Imperial
$\lambda_{\max}$	m	cm	in
$T$	K	K	°R
$b$	m·K	cm·K	in·°R

Where it holds

Applies to thermal blackbody (or graybody) spectra at equilibrium. Different definitions of 'peak'

(B(\lambda ) vs B(\nu ) vs

energy-per-log-frequency) give different numerical constants — be consistent.

Dimensional analysis

$[\lambda \cdot T] = L\cdot \Theta = [b] \checkmark$

Discovery

Wilhelm Wien · 1893

Wien empirically observed that the peak of the blackbody spectrum shifts inversely with T. He derived the displacement law from thermodynamics before quantum theory existed. Awarded the 1911 Nobel Prize for his work on heat radiation.

Try this

Why does a 5778 K Sun look yellow but a 3000 K bulb look orange?

Hot objects glow with a characteristic color. As temperature rises, the peak shifts to shorter wavelengths. What law governs this — and why is the product λ_max × T a universal constant?

Research status: stable

Real-world applications

Pyrometry — measure T of furnaces, foundries, lava from $\lambda _\max$ .
Astronomy — convert star color → temperature → mass/age.
Thermal imaging — design IR cameras around 8– $14 \mu m$ (body temperature peak).
Greenhouse science — Earth radiates IR (peak $\sim 10 \mu m)$ absorbed by $CO_{2}/H_{2}O$ .

Common misconceptions

Wien's law gives the 'color' of a hot object — only loosely; human color perception integrates across the spectrum.
$\lambda _\max \cdot T = b$ is the same constant whether you $use B(\lambda )$ or $B(\nu )$ — false; $\nu _\max /T$ uses a different constant (2.821 k_B/h).
Wien's law is exact at all temperatures — it's exact for ideal blackbodies; real surfaces with wavelength-dependent emissivity shift the apparent peak.

Experimental verification

Stellar spectra: Sun peaks

at \sim 500\,\mathrm{nm}

matching

T=5778\,\mathrm{K}

. Vega

(T\approx 9600\,\mathrm{K})

peaks

at \sim 300\,\mathrm{nm}

. Tungsten lamps

(T\sim 2800\,\mathrm{K})

peak

at \sim 1 \mu m

(most output is IR — bulbs are inefficient). CMB peak at 1.06 mm confirms

T = 2.725\,\mathrm{K}

to 5 digits.

Derivation

Set dB(\lambda

,

T)/d\lambda = 0

in Planck's law.

Substituting

x = hc/(\lambda kT)

yields transcendental equation

(5 - x) = 5e^{-}

ˣ, with solution

x \approx 4.9651

.

Then

\lambda _\max \cdot T = hc/(x\cdot k_B) = b \approx 2.898e-3 m\cdot K

.

Limiting cases

T \to \infty

⟶

\lambda _\max \to 0

Extremely hot bodies peak in X-rays or gamma rays (accretion disks, black-hole coronae).

T \to 0

⟶

\lambda _\max \to \infty

Cold objects glow at long radio wavelengths (effectively dark, since power scales as

T^{4})

.

T = 2.725\,\mathrm{K} (CMB)

⟶

\lambda _\max \approx 1.06\,\mathrm{mm}

Cosmic microwave background peaks in mm-wave — directly measured by COBE/Planck satellites.

What if…

What if Earth were at 1000 K instead of 288 K?

$\lambda _\max$ would drop from $\sim 10 \mu m$ $to \sim 2.9 \mu m$ — atmospheric IR windows would shift; current greenhouse gases couldn't trap heat the same way.

What if a star

had T = 30

,000 K (O-type)?

$\lambda _\max \approx 97\,\mathrm{nm}$ — peak in EUV. Stars ionize hydrogen in surrounding gas, lighting up nebulae and ending stellar nurseries.

1

Body temperature thermal peak

Given ·

T:: 310
b:: 0.0028977719

Find · \lambda_{\max}

Steps

$\lambda _\max = b/T = 2.898e-3 / 310 \approx 9.35e-6\,\mathrm{m}$
This is why thermal cameras operate at 8– $14 \mu m$ ; matches human body emission.

Result ·

\lambda _\max \approx 9.35 \mu m

— mid-IR

2

Color of a 3000 K incandescent bulb

Given ·

T:: 3000
b:: 0.0028977719

Find · \lambda_{\max}

Steps

$\lambda _\max = 2.898e-3 / 3000 \approx 9.66e-7\,\mathrm{m} = 966\,\mathrm{nm}$
Peak is in IR! Visible 'warm' light is the small high-wavelength tail — bulbs waste $\sim 95$ % of energy as heat.

Result ·

\lambda _\max \approx 966\,\mathrm{nm}

— near-infrared

Planck Radiation Law→Stefan-Boltzmann Law→Rydberg Formula→