What is Planck's law?

The formula giving how much energy a blackbody — an ideal absorber of all radiation — emits at each wavelength as a function of temperature. Integrated over all wavelengths it gives the Stefan–Boltzmann law σT⁴; numerical integration matched at 300 K, 1,000 K and 5,800 K. Hotter bodies emit more and shift to shorter wavelengths.

How is the peak wavelength found?

From Wien's displacement law, λ_max·T = 2,898 µm·K: 9.7 µm infrared at room temperature (300 K), 2.9 µm at 1,000 K, 1.0 µm for an incandescent filament (2,800 K) and 0.50 µm — green light — for the sun (5,800 K).

What are band fractions used for?

For light-source efficiency and window or sensor design. A 2,800 K incandescent lamp puts only 9.7% of its emission in the visible (0.38–0.78 µm); the rest is infrared heat. 38% of room-temperature emission lies in 8–14 µm, which thermal cameras use. Glass transmits only up to about 2.7 µm, so sunlight (97% below 2.7 µm) enters while indoor thermal radiation barely escapes.

Planck's Law

Category: Thermal – Radiation | Revised 2026-10-01

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Planck's Law

Theory: the blackbody spectrum

Overview

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Radiation calculations often just use σT⁴. When do we need Planck's law itself?

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When material properties vary with wavelength. Glass passes visible light but blocks long infrared; paint and metal emissivities vary with wavelength; sensors and cameras see only certain bands. In these cases you need the energy split by wavelength. Let's integrate Planck's law numerically, confirm the total is σT⁴, and see where emission concentrates at familiar temperatures.

Planck's formula

$$ E_{b\lambda}(\lambda,T) = \frac{C_1}{\lambda^5\left[\exp\!\left(\dfrac{C_2}{\lambda T}\right) - 1\right]},\quad C_1 = 3.742\times10^8\ \mathrm{W\,\mu m^4/m^2},\ C_2 = 14{,}388\ \mathrm{\mu m\,K} $$

Total, peak and band fraction

$$ \int_0^\infty E_{b\lambda}\,d\lambda = \sigma T^4,\qquad \lambda_{max}T = 2{,}898\ \mathrm{\mu m\,K},\qquad F_{\lambda_1\to\lambda_2} = \frac{1}{\sigma T^4}\int_{\lambda_1}^{\lambda_2}E_{b\lambda}\,d\lambda $$

Integrated over all wavelengths it gives σT⁴; peak wavelength is inversely proportional to temperature. The band fraction F gives the share of emission within a band.

Coffee Break Trivia

The birth of the quantum

In 1900 Germany's Max Planck found a formula matching measured blackbody spectra. Deriving it required assuming energy came only in discrete multiples of hν. Planck regarded this as a mathematical device, but it became the starting point of quantum theory. Radiation thermometers that measure hot objects without contact are also based on Planck's formula.

Worked examples

Example 1: numerical integral and σT⁴

Temperature	Numerical integral	σT⁴
300 K	459.3 W/m²	459.3 W/m²
1,000 K	56.7 kW/m²	56.7 kW/m²
5,800 K	64.2 MW/m²	64.2 MW/m²

Example 2: peak wavelength

Temperature	Example	Peak wavelength
300 K	Room-temperature object	9.66 µm
1,000 K	Red-hot iron	2.90 µm
1,500 K	Near molten steel	1.93 µm
2,800 K	Incandescent filament	1.03 µm
5,800 K	Sun's surface	0.50 µm

Example 3: band fractions

Band	Temperature	Fraction
Visible 0.38–0.78 µm	2,800 K	9.7%
Visible 0.38–0.78 µm	5,800 K	46.6%
Thermal camera 8–14 µm	300 K	37.6%
Glass-transmitted, below 2.7 µm	5,800 K	97.2%
Glass-transmitted, below 2.7 µm	300 K	0.002%
Above 4 µm	1,000 K	51.9%

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An incandescent lamp turns less than a tenth of its emission into light. And glass passing sunlight but trapping indoor heat is also about wavelength.

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At a 2,800 K filament the emission peaks at 1 µm in the near infrared, and visible wavelengths are just the tail. The sun at 5,800 K peaks mid-visible, so nearly half becomes light. Glass is the same story: 97% of sunlight lies below 2.7 µm and passes through, but room-temperature emission is almost all longer infrared, absorbed by glass and hardly escaping. That trapped heat is how a greenhouse works. Low-e window coatings exploit this by reflecting long infrared to cut heat transfer.

Use in analysis

If emissivity or transmissivity varies with wavelength, compute by bands (band model).
Get each band's blackbody fraction from integrals of Planck's formula (F functions).
When temperatures differ greatly, as with sunlight and room-temperature emission, use at least short- and long-wave bands.
For radiation thermometers and thermal cameras, use emissivity in the measured band.
Gas radiation (CO₂, water vapour) occurs only at specific wavelengths, requiring spectral treatment.

Coffee Break Trivia

“A thermal camera read the metal too cold”

At a plant, a thermal camera read heated metal parts over 100 K below thermocouple values. The metal's emissivity in the long-wave 8–14 µm band was only about 0.1, while the camera remained set to 0.95. Painting a reference spot with black high-temperature paint and matching emissivity there gave correct temperatures. Without attention to spectral emissivity, radiation thermometry can be badly off.

Common mistakes

Mistakes and fixes

Mistake	Effect	Fix
Treating glass as gray	Solar gain and loss wrong	Two or more bands
Ignoring spectral emissivity	Large thermometry errors	Use in-band values
Unit mix-up (µm vs m)	Orders-of-magnitude errors	Check constant units
Using °C in T⁴	Large radiation error	Use kelvin
Gray gas assumption	Furnace errors	Spectral models

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Planck's Law

Theory: the blackbody spectrum

Overview

Planck's formula

Total, peak and band fraction

The birth of the quantum

Worked examples

Example 1: numerical integral and σT⁴

Example 2: peak wavelength

Example 3: band fractions

Use in analysis

“A thermal camera read the metal too cold”

Common mistakes

Mistakes and fixes

Related Topics

Related fields