Planck's Law Calculator
Blackbody spectral radiance and exitance at a chosen wavelength and temperature, plus the Wien peak.
What it does
A blackbody is an ideal emitter that absorbs everything falling on it and radiates according to its temperature alone. Planck's law is the formula that describes that radiation, and it is one of the foundations of modern physics. It says how much energy leaves a surface at each wavelength, with no adjustable parameters beyond the temperature.
The law came from a problem. Classical physics predicted that a hot body should radiate ever more strongly at shorter wavelengths, an absurd result known as the ultraviolet catastrophe. Planck fixed it in 1900 by assuming that energy is exchanged in discrete packets, and the constant that carries his name came out of the fit. That constant is now fixed exactly in the SI, along with the speed of light and Boltzmann's constant.
Two quantities come out of the formula and they are easy to confuse. Spectral radiance is the power per unit area, per unit solid angle, per unit wavelength, measured in watts per square metre per steradian per metre. Spectral exitance drops the solid angle and integrates over the hemisphere, which multiplies the radiance by pi for a flat Lambertian surface. The tool shows both.
The peak of the curve follows a simple rule of its own. Wien's displacement law says the peak wavelength times the temperature is a constant, 2.897771955e-3 metre kelvin, so a hotter body peaks at a shorter wavelength. The Sun at 5778 K peaks near 501 nm, which is why daylight looks white, while a 300 K wall peaks in the thermal infrared near 9.7 um, which is why thermal cameras see it.
The numbers here are the exact SI constants, so the output matches a physics reference to the digits shown. Use the tool for a quick spectrum check, a thermal imaging estimate or a teaching example, and remember that real surfaces are rarely perfect blackbodies. Emissivity scales the whole curve down, and it varies with wavelength for most materials.
How to use
- Enter the absolute temperature in kelvin. Add 273.15 to a Celsius figure to get there.
- Enter the wavelength and choose its unit. Nanometres suit visible light, micrometres suit thermal infrared.
- Read the spectral radiance first, since it is the quantity most tables quote.
- Compare the spectral exitance, which is the same curve scaled by pi for a flat surface.
- Check the peak wavelength to see where the body radiates most strongly, and the band line to place that peak in the ultraviolet, visible or infrared.
- Multiply by the emissivity of your surface if it is not close to a blackbody, since the formula assumes an ideal emitter.
Formula
- B
- Spectral radiance, in watts per square metre per steradian per metre
- M
- Spectral exitance, in watts per square metre per metre of wavelength
- h
- Planck constant, 6.62607015e-34 joule seconds, exact
- c
- Speed of light, 299792458 metres per second, exact
- k
- Boltzmann constant, 1.380649e-23 joules per kelvin, exact
- lambda
- Wavelength in metres
- T
- Absolute temperature in kelvin
The exponent h c / (lambda k T) decides everything: when it is small the wavelength is in the classical limit and the curve follows the Rayleigh-Jeans approximation, and when it is large the exponential dominates and the curve falls away. The pi factor between radiance and exitance holds for a flat surface radiating into a hemisphere with the same radiance in every direction, which is the usual Lambertian assumption.
Worked example
Example: the Sun at 5778 K and 500 nm
Take a blackbody at the Sun's effective temperature of 5778 K and look at 500 nm, which is blue green light.
- The exponent is h c / (lambda k T) = 4.9802, so the exponential term is about 145.5 and the bracket is 144.5.
- Spectral radiance comes out at 2.638e13 W/(m2 sr m).
- Multiply by pi for a flat surface and the spectral exitance is 8.286e13 W/m2 per m of wavelength.
- Wien's law puts the peak at 2.897771955e-3 / 5778 = 501.5 nm, squarely in the visible.
- The frequency at 500 nm is c / lambda = 5.996e14 Hz.
- At 300 K the peak moves to about 9.66 um in the thermal infrared, and at 5800 K and 1000 nm the radiance is 1.088e13 W/(m2 sr m).
Reference table
| Object | Temperature (K) | Peak wavelength |
|---|---|---|
| Sun's photosphere | 5778 | 501.5 nm, visible |
| Incandescent filament | 2800 | 1.035 um, near infrared |
| Human skin | 307 | 9.44 um, thermal infrared |
| Room temperature wall | 300 | 9.66 um, thermal infrared |
| Liquid nitrogen | 77 | 37.6 um, far infrared |
Limits
- The formula assumes a perfect blackbody, and real surfaces emit less, with an emissivity that changes with wavelength.
- It gives the spectrum of a body in thermal equilibrium, so it does not describe fluorescent, LED or laser light.
- The tool works in wavelength, not frequency, so comparisons with radio or X-ray data in frequency units need a conversion.
- Values become extremely small or large at the edges of the range, and the reported three significant figures are all that is reliable there.
Common mistakes
- Using Celsius or Fahrenheit in the formula, which requires kelvin.
- Confusing spectral radiance with the total emitted power, since radiance is per unit wavelength and must be integrated to give a total.
- Forgetting the pi factor when comparing a radiance value with an exitance table.
- Entering the wavelength in nanometres but treating it as metres in a hand calculation, which shifts the answer by many orders of magnitude.
- Expecting the peak of the curve to be where most of the energy in a band lies, when a wide band away from the peak can still carry a large share.
Try it yourself
Quick checks. Answers sit tucked under each question.
Where does a blackbody at 300 K peak?
About 9.66 um. Wien's law gives 2.897771955e-3 / 300 = 9.659e-6 m, which is in the thermal infrared.
What is the spectral radiance at 1000 nm for a 5800 K blackbody?
1.088e13 W/(m2 sr m). The same constants give that figure, and the peak for 5800 K is 499.6 nm, so 1000 nm sits well down the long wavelength side.
How much more radiance does the Sun give at 500 nm than a 300 K surface?
Far more than the ratio of temperatures suggests, because the exponential in Planck's law grows quickly as the wavelength moves away from the peak. That is why sunlight and room temperature surfaces look so different in the visible.
Questions
What is a blackbody?
An ideal object that absorbs all radiation falling on it and emits a spectrum that depends only on its temperature. Stars, cavity openings and, roughly, human skin are close approximations.
What is the difference between radiance and exitance?
Radiance is measured per unit solid angle, so it describes brightness in a direction. Exitance is the total leaving a flat surface into the hemisphere above it, which is pi times the radiance for a Lambertian surface.
Why does the peak move to shorter wavelengths as temperature rises?
Because Wien's law ties peak wavelength inversely to temperature. Hotter bodies have more energy per photon, and shorter wavelengths carry more energy.
Can I use this for a thermal camera calculation?
Yes, as a first estimate. Real surfaces have emissivity below 1, often around 0.9 to 0.98 for skin and most non-metals, so scale the result by that figure.
Why are the constants exact?
The 2019 revision of the SI fixed the numerical values of h, c and k by definition, so they carry no measurement uncertainty. The formula's output is limited only by the arithmetic.
Does the law apply to the whole spectrum?
It applies at every wavelength for an ideal blackbody, but real bodies deviate, and at very short wavelengths the output is so small that other emission processes can dominate.
Sources
- OpenStax University Physics Volume 3, section 6.1: blackbody radiation and Planck's law
- NIST CODATA values of the fundamental physical constants