# 6.2: Blackbody Radiation

- Page ID
- 4520

LEARNING OBJECTIVES

By the end of this section you will be able to:

- Apply Wien’s and Stefan’s laws to analyze radiation emitted by a blackbody
- Explain Planck’s hypothesis of energy quanta

All bodies emit electromagnetic radiation over a range of
wavelengths. In an earlier chapter, we learned that a cooler body
radiates less energy than a warmer body. We also know by
observation that when a body is heated and its temperature rises,
the perceived wavelength of its emitted radiation changes from
infrared to red, and then from red to orange, and so forth. As its
temperature rises, the body glows with the colors corresponding to
ever-smaller wavelengths of the electromagnetic spectrum. This is
the underlying principle of the incandescent light bulb: A hot
metal filament glows red, and when heating continues, its glow
eventually covers the entire visible portion of the electromagnetic
spectrum. The temperature (*T*) of the object that emits
radiation, or the **emitter**, determines the
wavelength at which the radiated energy is at its maximum. For
example, the Sun, whose surface temperature is in the range between
5000 K and 6000 K, radiates most strongly in a range of wavelengths
about 560 nm in the visible part of the electromagnetic spectrum.
Your body, when at its normal temperature of about 300 K, radiates
most strongly in the infrared part of the spectrum.

Radiation that is incident on an object is partially absorbed
and partially reflected. At thermodynamic equilibrium, the rate at
which an object absorbs radiation is the same as the rate at which
it emits it. Therefore, a good absorber of radiation (any object
that absorbs radiation) is also a good emitter. A perfect
**absorber** absorbs all electromagnetic radiation
incident on it; such an object is called a
**blackbody**.

Although the blackbody is an idealization, because no physical
object absorbs 100% of incident radiation, we can construct a close
realization of a blackbody in the form of a small hole in the wall
of a sealed enclosure known as a cavity radiator, as shown in
Figure \(\PageIndex{1}\). The inside walls of a cavity radiator are
rough and blackened so that any radiation that enters through a
tiny hole in the cavity wall becomes trapped inside the cavity. At
thermodynamic equilibrium (at temperature *T*), the cavity
walls absorb exactly as much radiation as they emit. Furthermore,
inside the cavity, the radiation entering the hole is balanced by
the radiation leaving it. The emission spectrum of a blackbody can
be obtained by analyzing the light radiating from the hole.
Electromagnetic waves emitted by a blackbody are called
**blackbody radiation**.

The intensity \(I(\lambda, T)\) of blackbody radiation depends
on the wavelength \(\lambda\) of the emitted radiation and on the
temperature *T* of the blackbody (Figure \(\PageIndex{2}\)).
The function \(I(\lambda, T)\) is the **power**
intensity that is radiated per unit wavelength; in other words, it
is the power radiated per unit area of the hole in a cavity
radiator per unit wavelength. According to this definition,
\(I(\lambda, T)d\lambda\) is the power per unit area that is
emitted in the wavelength interval from \(\lambda\) to \(\lambda +
d\lambda\). The intensity distribution among wavelengths of
radiation emitted by cavities was studied experimentally at the end
of the nineteenth century. Generally, radiation emitted by
materials only approximately follows the blackbody radiation curve
(Figure \(\PageIndex{3}\)); however, spectra of common stars do
follow the blackbody radiation curve very closely.

Two important laws summarize the experimental findings of
blackbody radiation: ** Wien’s displacement
law** and

*Stefan’s law*. Wien’s displacement law is illustrated in Figure \(\PageIndex{2}\) by the curve connecting the maxima on the intensity curves. In these curves, we see that the hotter the body, the shorter the wavelength corresponding to the emission peak in the radiation curve. Quantitatively, Wien’s law reads

\[\lambda_{max}T = 2.898 \times 10^{-3} m \cdot K \label{Wien}\]

where \(\lambda_{max}\) is the position of the maximum in the
radiation curve. In other words, \(\lambda_{max}\) is the
wavelength at which a blackbody radiates most strongly at a given
temperature *T*. Note that in Equation \ref{Wien}, the
temperature is in kelvins. Wien’s displacement law allows us to
estimate the temperatures of distant stars by measuring the
wavelength of radiation they emit.

Example \(\PageIndex{1}\): Temperatures of Distant Stars

On a clear evening during the winter months, if you happen to be in the Northern Hemisphere and look up at the sky, you can see the constellation Orion (The Hunter). One star in this constellation, Rigel, flickers in a blue color and another star, Betelgeuse, has a reddish color, as shown in Figure \(\PageIndex{4}\). Which of these two stars is cooler, Betelgeuse or Rigel?

**Strategy**

We treat each star as a blackbody. Then according to Wien’s law, its temperature is inversely proportional to the wavelength of its peak intensity. The wavelength \(\lambda_{max}^{(blue)}\) of blue light is shorter than the wavelength \(\lambda_{max}^{(red)}\) of red light. Even if we do not know the precise wavelengths, we can still set up a proportion.

**Solution**

Writing Wien’s law for the blue star and for the red star, we have

\[\begin {align*} \lambda_{max}^{(red)}T_{(red)} &= 2.898 \times 10^{-3} m \cdot K \\[5pt] &= \lambda_{max}^{(blue)} T_{(blue)} \end{align*}\]

When simplified, this gives

\[T_{(red)} = \dfrac{\lambda_{max}^{(blue)}}{\lambda_{max}^{(red)}}T_{(blue)} < T_{(blue)} \nonumber \]

Therefore, Betelgeuse is cooler than Rigel.

**Significance**

Note that Wien’s displacement law tells us that the higher the temperature of an emitting body, the shorter the wavelength of the radiation it emits. The qualitative analysis presented in this example is generally valid for any emitting body, whether it is a big object such as a star or a small object such as the glowing filament in an incandescent lightbulb.

Exercise \(\PageIndex{1}\)

The flame of a peach-scented candle has a yellowish color and the flame of a Bunsen’s burner in a chemistry lab has a bluish color. Which flame has a higher temperature?

**Answer**

The second experimental relation is **Stefan’s
law,** which concerns the total power of blackbody radiation
emitted across the entire spectrum of wavelengths at a given
temperature. In \(\PageIndex{2}\) , this total power is represented
by the area under the blackbody radiation curve for a given
*T*. As the temperature of a blackbody increases, the total
emitted power also increases. Quantitatively, Stefan’s law
expresses this relation as

\[P(T) = \sigma AT^4\]

where \(A\) is the surface area of a blackbody, \(T\) is its
temperature (in kelvins), and \(σ\) is the **Stefan–Boltzmann
constant**, \(\sigma = 5.670 \times 10^{-8} W/(m^2 \cdot
K^4)\). Stefan’s law enables us to estimate how much energy a star
is radiating by remotely measuring its temperature.

Example \(\PageIndex{2}\): Power Radiated by Stars

A star such as our Sun will eventually evolve to a “red giant” star and then to a “white dwarf” star. A typical white dwarf is approximately the size of Earth, and its surface temperature is about \(2.5 \times 10^4 K\). A typical red giant has a surface temperature of \(3.0 \times 10^3 K\) and a radius ~100,000 times larger than that of a white dwarf. What is the average radiated power per unit area and the total power radiated by each of these types of stars? How do they compare?

**Strategy**

If we treat the star as a blackbody, then according to Stefan’s
law, the total power that the star radiates is proportional to the
fourth power of its temperature. To find the power radiated per
unit area of the surface, we do not need to make any assumptions
about the shape of the star because *P*/*A* depends
only on temperature. However, to compute the total power, we need
to make an assumption that the energy radiates through a spherical
surface enclosing the star, so that the surface area is \(A = 4\pi
R^2\), where *R* is its radius.

**Solution**

A simple proportion based on Stefan’s law gives

\[\dfrac{P_{dwarf}/A_{dwarf}}{P_{giant}/A_{giant}} = \dfrac{\sigma T_{dwarf}^4}{\sigma T_{giant}^4} = \left(\dfrac{T_{dwarf}}{T_{giant}}\right)^4 = 4820 \label{6.5}\]

The power emitted per unit area by a white dwarf is about 5000 times that the power emitted by a red giant. Denoting this ratio by \(a=4.8×10^3\), Equation \ref{6.5} gives

\[\dfrac{P_{dwarf}}{P_{giant}} = \alpha \dfrac{A_{dwarf}}{A_{giant}} = \alpha \dfrac{4\pi R_{dwarf}^2}{4\pi R_{giant}^2} = \alpha \left( \dfrac{R_{dwarf}}{R_{giant}} \right)^2 = 4.8 \times 10^{−7}\]

We see that the total power emitted by a white dwarf is a tiny fraction of the total power emitted by a red giant. Despite its relatively lower temperature, the overall power radiated by a red giant far exceeds that of the white dwarf because the red giant has a much larger surface area. To estimate the absolute value of the emitted power per unit area, we again use Stefan’s law. For the white dwarf, we obtain

\[\dfrac{P_{dwarf}}{A_{dwarf}} = \sigma T_{dwarf}^4 = 5.670 \times 10^{-8} \dfrac{W}{m^2 \cdot K^4} (2.5 \times 10^4 \, K)^4 = 2.2 \times 10^{10} \dfrac{W}{m^2} \label{6.7}\]

The analogous result for the red giant is obtained by scaling the result for a white dwarf:

\[\dfrac{P_{giant}}{A_{giant}} = \dfrac{2.2 \times 10^{10}}{4.82 \times 10^3} \dfrac{W}{m^2} = 4.56 \times 10^6 \dfrac{W}{m^2} \cong 4.6 \times 10^{-6} \dfrac{W}{m^2}\]

**Significance**

To estimate the total power emitted by a white dwarf, in principle, we could use Equation \ref{6.7}. However, to find its surface area, we need to know the average radius, which is not given in this example. Therefore, the solution stops here. The same is also true for the red giant star.

Exercise \(\PageIndex{2A}\)

An iron poker is being heated. As its temperature rises, the poker begins to glow—first dull red, then bright red, then orange, and then yellow. Use either the blackbody radiation curve or Wien’s law to explain these changes in the color of the glow.

**Answer**-
The wavelength of the radiation maximum decreases with increasing temperature.

Exercise \(\PageIndex{2B}\)

Suppose that two stars, \(α\) and \(β\), radiate exactly the same total power. If the radius of star \(α\) is three times that of star \(β\), what is the ratio of the surface temperatures of these stars? Which one is hotter?

**Answer**-
\(T_{\alpha}/T_{\beta} = 1/\sqrt{3} \cong 0.58\), so the star \(\beta\) is hotter.

The term “blackbody” was coined by Gustav R. Kirchhoff in 1862.
The blackbody radiation curve was known experimentally, but its
shape eluded physical explanation until the year 1900. The physical
model of a blackbody at temperature *T* is that of the
electromagnetic waves enclosed in a cavity (Figure
\(\PageIndex{1}\)) and at thermodynamic equilibrium with the cavity
walls. The waves can exchange energy with the walls. The objective
here is to find the energy density distribution among various modes
of vibration at various wavelengths (or frequencies). In other
words, we want to know how much energy is carried by a single
wavelength or a band of wavelengths. Once we know the energy
distribution, we can use standard statistical methods (similar to
those studied in a previous chapter) to obtain the blackbody
radiation curve, Stefan’s law, and Wien’s displacement law. When
the physical model is correct, the theoretical predictions should
be the same as the experimental curves.

In a classical approach to the blackbody radiation problem, in
which radiation is treated as waves (as you have studied in
previous chapters), the modes of electromagnetic waves trapped in
the cavity are in equilibrium and continually exchange their
energies with the cavity walls. There is no physical reason why a
wave should do otherwise: Any amount of energy can be exchanged,
either by being transferred from the wave to the material in the
wall or by being received by the wave from the material in the
wall. This classical picture is the basis of the model developed by
Lord Rayleigh and, independently, by Sir James Jeans. The result of
this classical model for blackbody radiation curves is known as the
*Rayleigh–Jeans law*. However, as shown in Figure
\(\PageIndex{5}\), the **Rayleigh–Jeans law** fails to
correctly reproduce experimental results. In the limit of short
wavelengths, the Rayleigh–Jeans law predicts infinite radiation
intensity, which is inconsistent with the experimental results in
which radiation intensity has finite values in the ultraviolet
region of the spectrum. This divergence between the results of
classical theory and experiments, which came to be called the
** ultraviolet catastrophe**, shows how
classical physics fails to explain the mechanism of blackbody
radiation.

The blackbody radiation problem was solved in 1900 by Max
**Planck**. Planck used the same idea as the
Rayleigh–Jeans model in the sense that he treated the
electromagnetic waves between the walls inside the cavity
classically, and assumed that the radiation is in equilibrium with
the cavity walls. The innovative idea that Planck introduced in his
model is the assumption that the cavity radiation originates from
atomic oscillations inside the cavity walls, and that these
oscillations can have only *discrete* values of energy.
Therefore, the radiation trapped inside the cavity walls can
exchange energy with the walls only in discrete amounts. Planck’s
hypothesis of discrete energy values, which he called
*quanta*, assumes that the oscillators inside the cavity
walls have **quantized energies**. This was a brand
new idea that went beyond the classical physics of the nineteenth
century because, as you learned in a previous chapter, in the
classical picture, the energy of an oscillator can take on any
continuous value. Planck assumed that the energy of an oscillator
(\(E_n\)) can have only discrete, or quantized, values:

\[E_n = nhf, \, where \, n = 1,2,3, \ldots \label{eq30}\]

In Equation \ref{eq30}, \(f\) is the frequency of Planck’s
oscillator. The natural number \(n\) that enumerates these discrete
energies is called a quantum number. The physical constant \(h\) is
called ** Planck’s constant**:

\[h = 6.626 \times 10^{-34} J \cdot s = 4.136 \times 10^{-15} eV \cdot s \label{6.10}\]

Each discrete energy value corresponds to a **quantum
state of a Planck oscillator**. Quantum states are
enumerated by quantum numbers. For example, when Planck’s
oscillator is in its first \(n 1\) quantum state, its energy is
\(E_1 = hf\); when it is in the \(n = 2\) quantum state, its energy
is \(E_2 = 2hf\); when it is in the \(n = 3\) quantum state, \(E_3
= 3hf\); and so on.

Note that Equation \ref{eq30} shows that there are infinitely
many quantum states, which can be represented as a sequence
{*hf*, 2*hf*, 3*hf*,…, (*n* –
1)*hf*, *nhf*, (*n* + 1)*hf*,…}. Each
two consecutive quantum states in this sequence are separated by an
energy jump, \(\delta E = hf\). An oscillator in the wall can
receive energy from the radiation in the cavity (absorption), or it
can give away energy to the radiation in the cavity (emission). The
absorption process sends the oscillator to a higher quantum state,
and the emission process sends the oscillator to a lower quantum
state. Whichever way this exchange of energy goes, the smallest
amount of energy that can be exchanged is *hf*. There is no
upper limit to how much energy can be exchanged, but whatever is
exchanged must be an integer multiple of *hf*. If the energy
packet does not have this exact amount, it is neither absorbed nor
emitted at the wall of the blackbody.

PLANCK’S QUANTUM HYPOTHESIS

*Planck’s hypothesis of energy quanta* states that the
amount of energy emitted by the oscillator is carried by the
quantum of radiation, \(\Delta E\):

\[\Delta E = hf \label{planck}\]

Recall that the frequency of electromagnetic radiation is related to its wavelength and to the speed of light by the fundamental relation \(f\lambda = c\). This means that we can express Equation \ref{6.10} equivalently in terms of wavelength \(\lambda\). When included in the computation of the energy density of a blackbody, Planck’s hypothesis gives the following theoretical expression for the power intensity of emitted radiation per unit wavelength:

\[I(\lambda, T) = \dfrac{2\pi hc^2}{\lambda^5} \dfrac{1}{e^{hc/\lambda k_BT} - 1} \label{6.11}\]

where *c* is the speed of light in vacuum and kBkB is
Boltzmann’s constant, \(k_B = 1.380 \times 10^{-23} J/K\). The
theoretical formula expressed in Equation \ref{6.11} is called
*Planck’s blackbody radiation law*. This law is in agreement
with the experimental blackbody radiation curve (Figure
\(\PageIndex{2}\)). In addition, Wien’s displacement law and
Stefan’s law can both be derived from Equation \ref{6.11}. To
derive Wien’s displacement law, we use differential calculus to
find the maximum of the radiation intensity curve \(I(\lambda,
T)\). To derive Stefan’s law and find the value of the
Stefan–Boltzmann constant, we use integral calculus and integrate
\(I(\lambda, T)\) to find the total power radiated by a blackbody
at one temperature in the entire spectrum of wavelengths from
\(\lambda = 0\) to \(\lambda = \infty\). This derivation is left as
an exercise later in this chapter.

Example \(\PageIndex{3}\): Planck’s Quantum Oscillator

A quantum oscillator in the cavity wall in Figure Figure \(\PageIndex{1}\) is vibrating at a frequency of \(5.0 \times 10^{14} Hz\). Calculate the spacing between its energy levels.

**Strategy**

Energy states of a quantum oscillator are given by Equation
\ref{eq30}. The energy spacing \(\Delta E\) is obtained by finding
the energy difference between two adjacent quantum states for
quantum numbers *n* + 1 and *n*.

**Solution**

We can substitute the given frequency and Planck’s constant directly into the equation:

\[\begin {align*} \Delta E &= E_{n + 1} − En = (n + 1)hf − nhf \\[5pt] &= hf \\[5pt] &= (6.626 \times 10^{−34} \, J \cdot s)(5.0 \times 10^{14} \, Hz) \\[5pt] &= 3.3 \times 10^{− 19} \, J \end{align*}\]

**Significance**

Note that we do not specify what kind of material was used to build the cavity. Here, a quantum oscillator is a theoretical model of an atom or molecule of material in the wall.

Exercise \(\PageIndex{3}\)

A molecule is vibrating at a frequency of \(5.0 \times 10^{14}\, Hz\). What is the smallest spacing between its vibrational energy levels?

**Answer**-
\(3.3 \times 10^{-19} J\)

Example \(\PageIndex{4}\): Quantum Theory Applied to a Classical Oscillator

A 1.0-kg mass oscillates at the end of a spring with a spring constant of 1000 N/m. The amplitude of these oscillations is 0.10 m. Use the concept of quantization to find the energy spacing for this classical oscillator. Is the energy quantization significant for macroscopic systems, such as this oscillator?

**Strategy**

We use Equation \ref{planck} as though the system were a quantum
oscillator, but with the frequency *f* of the mass vibrating
on a spring. To evaluate whether or not quantization has a
significant effect, we compare the quantum energy spacing with the
macroscopic total energy of this classical oscillator.

**Solution**

For the spring constant, \(k = 1.0 \times 10^3 N/m\), the
frequency *f* of the mass, \(m = 1.0 \, kg\), is

\[f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}} = \dfrac{1}{2\pi} \sqrt{\dfrac{1.0 \times 10^3 N/m}{1.0 \, kg}} \simeq 5.0 \, Hz \nonumber\]

The energy quantum that corresponds to this frequency is

\[\Delta E = hf = (6.626 \times 10^{-34} J \cdot s)(5.0 \, Hz) = 3.3 \times 10^{-33} J \nonumber\]

When vibrations have amplitude \(A = 0.10 \, m\), the energy of oscillations is

\[E = \dfrac{1}{2} kA^2 = \dfrac{1}{2}(1000 \, N/m)(0.1 \, m)^2 = 5.0 \, J \nonumber\]

**Significance**

Thus, for a classical oscillator, we have \(\Delta E/E \approx 10^{-34}\). We see that the separation of the energy levels is immeasurably small. Therefore, for all practical purposes, the energy of a classical oscillator takes on continuous values. This is why classical principles may be applied to macroscopic systems encountered in everyday life without loss of accuracy.

Exercise \(\PageIndex{4}\)

Would the result in Example \(\PageIndex{4}\) be different if
the mass were not 1.0 kg but a tiny mass of 1.0 *µ*g, and
the amplitude of vibrations were 0.10 *µ*m?

**Answer**-
No, because then \(\Delta E /E \approx 10^{-21}\)

When Planck first published his result, the hypothesis of energy quanta was not taken seriously by the physics community because it did not follow from any established physics theory at that time. It was perceived, even by Planck himself, as a useful mathematical trick that led to a good theoretical “fit” to the experimental curve. This perception was changed in 1905 when Einstein published his explanation of the photoelectric effect, in which he gave Planck’s energy quantum a new meaning: that of a particle of light.

## Contributors and Attributions

Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).