Planck's Law

In one sentence: Planck’s law says how brightly a hot object glows at each color, and the mix of colors it gives off depends on how hot the object is.

$B_\lambda(\lambda, T) = \frac{2 h c^2}{\lambda^5} \cdot \frac{1}{e^{h c / (\lambda k_B T)} - 1}$

Read it out loud: “B sub lambda of lambda and T equals two h c squared, over lambda to the fifth power, times one over the quantity e to the power of h c over lambda k sub B T, minus one. B sub lambda is the brightness of the light at wavelength lambda. Lambda, the Greek letter said LAM-duh, is the wavelength. The letter T is the absolute temperature in kelvin. H is Planck's constant, c is the speed of light, k sub B is the Boltzmann constant, and e is the number about 2.718, the base of natural growth and decay.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, like $10^{-7}$ and $10^{-23}$: Very large and very small numbers are written as a short number times a power of ten. The small raised number, the exponent, counts how many times to multiply by 10, and a minus sign means divide instead. So $10^{-23}$ is a decimal point, then 22 zeros, then a 1. When you raise a power of ten to a power, multiply the exponents, so $(10^{-7})^{5} = 10^{-35}$ . This way of writing is called scientific notation.
Blackbody: A blackbody is an idealized object that absorbs all the light that falls on it. Its glow comes only from its temperature. No real object is perfect, but the inside of a hot furnace and the surface of a star come close. Every hot object gives off light across many wavelengths, and a blackbody’s glow depends only on its temperature, not on what it is made of.
Absolute temperature and the kelvin (K): The kelvin, written K, is the temperature scale used in physics formulas. Zero kelvin, called absolute zero, is the lowest temperature possible, and it equals $-273.15$ degrees Celsius. One kelvin is the same size as one degree Celsius, so room temperature of about 20 degrees Celsius is about 293 K. Formulas like this one need kelvin, because the scale starts at true zero.
The number $e$ and exponentials, $e^{x}$: The number $e$ is about 2.71828, a fixed number that appears whenever something grows or shrinks at a rate proportional to its own size. $e^{x}$ is called an exponential, and it is also written $\exp(x)$ . For a large positive $x$ , $e^{x}$ grows very fast, and for a large negative $x$ it shrinks toward zero. In Planck’s law the exponential sits in the denominator, so a large exponent makes the brightness small.
Spectral radiance and the units W sr $^{-1}$ m $^{-3}$: Spectral radiance measures how much power a surface sends out, per unit area, per unit solid angle, and per unit of wavelength. Its unit, W sr $^{-1}$ m $^{-3}$ , reads watts per steradian per cubic meter. A watt is one joule per second. A steradian, written sr, is the three-dimensional version of an angle, measuring how wide a set of directions is as seen from the source.
The Boltzmann constant $k_B$: The Boltzmann constant, $k_B = 1.380649 \times 10^{-23}$ J/K, links temperature to energy. Multiplying it by a temperature in kelvin gives an energy scale for heat. At room temperature, $k_B T$ is about $4.1 \times 10^{-21}$ J. Its unit, joules per kelvin, means energy per degree of temperature.

What each symbol means

Symbol	Meaning
$B_\lambda$	The spectral radiance: the power emitted per unit area, per steradian, per meter of wavelength, in watts per steradian per cubic meter (W sr $^{-1}$ m $^{-3}$ ). In the example, a 5772 K surface gives about $2.624 \times 10^{13}$ W sr $^{-1}$ m $^{-3}$ at 500 nm.
$\lambda$	The wavelength of the light, in meters (m). Visible light runs from about $4.0 \times 10^{-7}$ m to $7.0 \times 10^{-7}$ m. The formula needs meters, so 500 nm must become $5.00 \times 10^{-7}$ m before any calculation.
$T$	The absolute temperature of the object, in kelvin (K), counted from absolute zero. The Sun’s surface is about 5772 K, a hot oven is about 500 K, and a comfortable room is about 293 K. Celsius cannot be used directly.
$h$	Planck’s constant, $6.62607015 \times 10^{-34}$ J s, with the unit joule seconds, which is energy multiplied by time. It is a fixed number of nature. In this law it sets the size of the energy steps in the light.
$c$	The speed of light in empty space, $2.99792458 \times 10^{8}$ m/s, exact by definition. In the formula, $hc$ combines with the wavelength to give the energy of one photon at that wavelength.
$k_B$	The Boltzmann constant, $1.380649 \times 10^{-23}$ J/K. It turns a temperature in kelvin into an energy. Comparing that heat energy with a photon’s energy is what sets the shape of the glow.
$e$	Euler’s number, about 2.71828, the base of natural logarithms. It is a fixed number. In this formula it is raised to the power $\frac{hc}{\lambda k_B T}$ , and the result is subtracted by 1.

The formula piece by piece

$\lambda$: Start with the wavelength, measured in meters. The law gives a brightness for each wavelength, so the full glow is a curve drawn over many values of $\lambda$ .
$T$: Next is the temperature in kelvin. For a fixed wavelength, a hotter object is brighter, and the temperature also decides where the peak of the curve sits.
$\frac{h c}{\lambda k_B T}$: This is the exponent that controls the shape. It compares the energy of one photon at this wavelength, $hc/\lambda$ , with the heat energy $k_B T$ . When it is large, the brightness is very small. When it is small, the brightness follows a simpler pattern.
$e^{h c / (\lambda k_B T)} - 1$: Raise $e$ to that exponent and subtract 1. For short wavelengths or cool objects this number is huge, and dividing by it makes the brightness drop sharply, which is what gives the curve its falling tail.
$\frac{2 h c^2}{\lambda^5}$: The prefactor sets the overall scale. The 2 comes from the two polarization directions of light, $c^2$ and $h$ are constants, and dividing by the fifth power of the wavelength makes the brightness fall fast at short wavelengths.
$B_\lambda(\lambda, T) = \frac{2 h c^2}{\lambda^5} \cdot \frac{1}{e^{h c / (\lambda k_B T)} - 1}$: The whole formula in one sentence: the brightness of a blackbody at a wavelength equals a scale factor times one over $e$ raised to the photon-to-heat energy ratio, minus one. Give it a wavelength and a temperature, and it returns the brightness.

What the formula is doing, and why

What goes in. A wavelength $\lambda$ and a temperature $T$ in kelvin. What happens. The formula compares the energy of one photon at that wavelength, $hc/\lambda$ , with the heat energy $k_B T$ . Their ratio sets the exponent. What comes out. The brightness at that wavelength, the spectral radiance $B_\lambda$ . Repeating the calculation for many wavelengths draws the whole glow.

Where it came from. Max Planck derived this law in 1900 by assuming that light is exchanged in fixed steps of size $hf$ , where $f$ is the frequency. That assumption introduced Planck’s constant and began quantum theory. Older formulas fit only part of the measured glow of hot objects. Planck’s formula fits all of it, which is why it was accepted.

Why the exponent controls the shape. When $\frac{hc}{\lambda k_B T}$ is large, which happens at short wavelengths or low temperatures, the denominator is huge, so the brightness falls sharply. When that quantity is small, which happens at long wavelengths or high temperatures, the formula approaches the older Rayleigh-Jeans law. That simpler law works well only at long wavelengths.

Why the glow changes with temperature. Raising the temperature increases the brightness at every wavelength, and the peak of the curve moves toward shorter wavelengths. The peak position is given by Wien’s displacement law. Adding up the brightness over all wavelengths gives the total power, which is the Stefan-Boltzmann law.

When it applies. The law describes the ideal blackbody exactly. Real surfaces emit less than this ideal curve, and their emission can vary with wavelength, so for them the formula is only an approximation. Most solids that glow with heat are close enough to a blackbody that the law is a useful guide.

An everyday picture

Think of an iron poker left in a fire. At first it glows a dull red, then orange, then white as it heats up. Each temperature has its own mix of colors and brightness at each color. Planck’s law is the recipe that gives the brightness of every color for a given temperature, in the same way a recipe gives the amount of each ingredient for a given number of servings.

Worked example: Brightness at 500 nm for a 5772 K surface

The Sun’s effective surface temperature is about 5772 K, the temperature of a blackbody that gives off the same total power per area as the Sun. We find the spectral radiance at $\lambda = 500$ nm, which is $5.00 \times 10^{-7}$ m. Use $h = 6.62607015 \times 10^{-34}$ J s, $c = 2.99792458 \times 10^{8}$ m/s and $k_B = 1.380649 \times 10^{-23}$ J/K. At the end we compare with a 3000 K lamp filament at the same wavelength.

Find the exponent. The exponent is $x = \frac{hc}{\lambda k_B T}$ . The top is $hc = 6.62607015 \times 10^{-34} \times 2.99792458 \times 10^{8} \approx 1.98645 \times 10^{-25}$ J m. The bottom is $\lambda k_B T = (5.00 \times 10^{-7})(1.380649 \times 10^{-23})(5772) \approx 3.9846 \times 10^{-26}$ J m. Dividing the front numbers gives $1.98645 / 3.9846 \approx 0.49851$ , and subtracting the exponents, $-25 - (-26) = 1$ , gives $x \approx 4.985$ , a plain number with no units.
Find the denominator. We need $e^{x} - 1 = e^{4.985} - 1$ . The number $e$ raised to the power 4.985 is about 146.26, and subtracting 1 gives $e^{x} - 1 \approx 145.26$ . This is the number the brightness is divided by.
Find the prefactor. The prefactor is $\frac{2 h c^2}{\lambda^5}$ . Square the speed of light, $c^2 \approx 8.98755 \times 10^{16}$ m $^2$ /s $^2$ , then multiply by $2h$ to get $2hc^2 \approx 1.19104 \times 10^{-16}$ . For the wavelength, $(5.00 \times 10^{-7})^5$ means $5^5 = 3125$ and exponent $-7 \times 5 = -35$ , so $\lambda^5 = 3.125 \times 10^{-32}$ m $^5$ . Dividing gives $\frac{1.19104 \times 10^{-16}}{3.125 \times 10^{-32}} \approx 3.8113 \times 10^{15}$ W sr $^{-1}$ m $^{-3}$ .
Divide the prefactor by the denominator. $B_\lambda = \frac{3.8113 \times 10^{15}}{145.26}$ . The 145.26 equals $1.4526 \times 10^{2}$ , so the exponents subtract, $15 - 2 = 13$ , and the front numbers divide to $3.8113 / 1.4526 \approx 2.624$ . The result is $B_\lambda \approx 2.624 \times 10^{13}$ W sr $^{-1}$ m $^{-3}$ .
Compare with a 3000 K lamp filament at the same wavelength. For 3000 K the exponent is $x = \frac{1.98645 \times 10^{-25}}{5.00 \times 10^{-7} \times 1.380649 \times 10^{-23} \times 3000} \approx 9.592$ . Then $e^{x} - 1$ is about 14,644, and the brightness is $\frac{3.8113 \times 10^{15}}{\text{14,644}} \approx 2.603 \times 10^{11}$ W sr $^{-1}$ m $^{-3}$ . The hotter surface is about 100 times brighter at this wavelength.
Read the comparison. A hotter object is brighter at 500 nm, and this is because the exponent falls quickly when the temperature rises, so the denominator shrinks. The 3000 K filament is not dim overall. Its peak lies in the infrared, far beyond 500 nm, and most of its power comes out at longer wavelengths.

Answer: At 500 nm a 5772 K surface emits a spectral radiance of about $2.624 \times 10^{13}$ W sr $^{-1}$ m $^{-3}$ . A 3000 K filament emits about 100 times less at this same wavelength. This is close to the peak region of the Sun’s spectrum, which is why sunlight contains a lot of green and blue light.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Using degrees Celsius for $T$ . Convert to kelvin first. Adding 273.15 to a Celsius value gives kelvin, so 25 degrees Celsius is 298.15 K.
Writing $\lambda$ in nanometers in the formula. Convert to meters first, so 500 nm becomes $5.00 \times 10^{-7}$ m. Otherwise the fifth power of the wavelength is wrong by an enormous factor.
Expecting the peak to stay in the same place when the object heats up. The peak moves to shorter wavelengths as $T$ rises.
Treating every glowing object as a perfect blackbody. Real surfaces emit less than the ideal curve, and their emission can vary with wavelength, so the result is an estimate for them.

Where you will meet it

Estimating the surface temperature of stars from the shape of their spectra, since starlight is close to a blackbody curve.
Designing furnaces, infrared heaters and incandescent lamps, where the emitted spectrum depends on the temperature of the hot part.
Measuring high temperatures with pyrometers, which compare the brightness of a hot body at chosen wavelengths.
Cosmology, where the cosmic microwave background follows a nearly perfect blackbody curve at about 2.7 K.

Questions people ask

Why is it called a blackbody?

An ideal blackbody absorbs all the light that falls on it, so it looks black when it is cold. When it is hot, it emits light with the spectrum that Planck’s law describes, and its color shifts as its temperature rises.

Why is the Sun's spectrum not a perfect blackbody?

The Sun’s outer layers absorb light at specific wavelengths, which carves dark lines into the spectrum. Overall the shape is close to a blackbody at about 5772 K.

Does this law describe ordinary light bulbs?

Roughly. A glowing filament is close to a blackbody, but it is not perfect, so its brightness differs somewhat from the ideal curve.

Wien's Displacement Law: Wien’s law says a hotter object glows brightest at a shorter wavelength, and its brightest wavelength equals a fixed number divided by its temperature.
Stefan-Boltzmann Law: A hot object gives off energy as heat and light, and doubling its temperature on the kelvin scale makes it give off 16 times as much.
Planck-Einstein Relation (E = hf): E = hf says each packet of light carries energy in proportion to its frequency: higher frequency means more energy per packet, and brightness does not change it.