In this explainer, we will learn how to describe the change with temperature of an ideal blackbody’s color and explain how this shows the quantization of light.

A blackbody is an idealized physical object. Even though no perfect blackbody exists, the notion of a blackbody is still useful as an approximation of many observed phenomena.

### Definition: Blackbody

A blackbody is an object that absorbs all electromagnetic radiation (light) incident on it. An object that absorbs all visible light appears black, which is how blackbodies get their name.

Blackbodies also emit radiation, depending on their temperature.

We know that any object that absorbs energy will heat up; it will increase in temperature. The object then has an opportunity to radiate some of that absorbed energy away. Blackbodies not only absorb energy, but they also radiate it, and they do this in a very specific way.

Consider the electric heating element shown below.

This heating coil is emitting radiation. We see that part of the coil is literally “red-hot.” Other parts of the coil are not quite so hot and appear dull red in color. As the coil heats, the wavelength (and therefore the energy) of the light emitted changes.

This heating coil is a good approximation of a blackbody. Blackbodies also emit light depending on their temperature.

Keeping in mind that many wavelengths of light are invisible to our eyes, consider a graph showing how light is emitted by blackbodies at different temperatures.

This graph shows light intensity on the vertical axis and wavelength on the horizontal axis. The four curves plotted are for blackbodies heated to temperatures of 3 000, 4 000, 5 000, and 6 000 kelvins respectively. 6 000 K is the approximate temperature of the surface of the Sun.

Regarding how these blackbodies would appear to us, note that up to a temperature of roughly 5 000 K, the peak wavelength of visible light emitted by a blackbody looks red. At higher temperatures, the perceived color of the blackbody will change.

As the graph indicates, the way a blackbody radiates energy depends on its temperature. We can better understand why these radiation curves have the specific shapes they do by thinking of a blackbody as a cavity.

Light that enters a cavity will bounce around inside, losing energy at each reflection through absorption. We can say that the cavity effectively absorbs all the light incident on it, meaning none is reflected back out.

This energy absorption heats the cavity, which then begins emitting energy as radiation.

Recalling that light is a wave, we note that the dimensions of the cavity enforce certain boundary conditions on the waves the cavity produces. Some wavelengths—those that fit within the cavity boundary conditions—are “allowed,” while waves that do not fit these conditions do not form.

Specifically, waves that can exist in the cavity are those whose displacement equals zero at the cavity walls.

Waves of this type—and there are many more possible beyond the ones shown—are those the cavity can emit as blackbody radiation.

The important point here is not that a blackbody is a cavity whose physical dimensions determine the light it can emit. Rather, it is that a blackbody, as a system, constrains the levels of energy of the system so that they come in discrete amounts.

By way of example, consider the energy of an object located on a staircase.

Regardless of the step on which the object rests, its energy will be some integer multiple of , the object’s gravitational potential energy when it is on the first step. The object cannot have, for example, an energy of , , or any other noninteger multiple of this value.

When a system’s energy can only take on certain discrete amounts, we say that it is “quantized.”

A physicist named Max Planck studied blackbody systems. He compared how blackbodies would radiate if their energies were not quantized to how they would radiate if they were. A graph of these two models is shown below.

Planck noted that the emission curve based on the assumption that blackbody energies are quantized agreed with experimental measurements.

He proposed a mathematical relationship between the energy levels of a blackbody and the wavelengths (or equivalently, frequencies) of its radiation.

### Formula: Energy, Frequency, and Number of Quanta Emitted by a Blackbody

where is the number of light quanta in the system possessing a frequency and is a constant called Planck’s constant with an approximate value of m^{2}⋅kg/s.

Note that for light waves, wave speed is the speed of light in vacuum . Since wave speed in general equals the frequency of the wave times its wavelength we can write

This equation helps us understand why the assumption of energy quantization creates a blackbody radiation curve that goes to zero as wavelength gets smaller and smaller.

In a cavity blackbody system, even though the number of resonant modes increases as wavelength decreases, the amount of energy needed to actually produce a quantum of light at such short wavelengths is very large—so large that it is unlikely for the system to do so.

At the other end of the spectrum, when wavelengths become large, there are fewer and fewer modes that meet the boundary conditions of a given cavity. Therefore, fewer light quanta are emitted, explaining why the emission curve tails off as it does.

### Example 1: Identifying Allowed Resonant Modes in a Cavity

Which of the electromagnetic waves shown in the diagram of a cavity corresponds to a possible resonant mode of the cavity?

### Answer

The resonant modes of a cavity are those waves that can fit between the endpoints of the wave such that the wave’s initial and final displacement is zero. We can think of these as boundary conditions imposed on all standing waves in a cavity.

Studying answer choice (A), we see the wave’s displacement (and therefore its electric field) is zero at the left end but not at the right end. Such a wave therefore cannot exist between those two points on the cavity walls.

Option (B), on the other hand, shows a wave with zero displacement at both endpoints. This fits the cavity boundary conditions and is therefore a possible resonant mode.

### Example 2: Identifying Boundary Conditions for Resonant Modes in a Cavity

Which word would correctly complete the following statement about the resonant modes of electromagnetic waves in a cavity?

An electromagnetic wave that is a resonant mode for a cavity must have displacement at the cavity boundaries.

- maximum
- zero
- nonzero

### Answer

To fill in this sentence blank correctly, it would be helpful to recall that a resonant mode in a cavity satisfies the same boundary conditions as a standing wave on a string fixed at both ends.

Standing waves with both ends fixed in place must have a displacement of zero at each end. For electromagnetic waves in cavities, the cavity walls serve as fixed endpoints. Therefore, any resonant modes must have a displacement of zero at these boundaries. We choose for our answer option (B).

### Example 3: Determining How Wavelength Affects Number of Cavity Modes

The diagrams (a), (b), and (c) show the possible resonant cavity modes for electromagnetic waves that are emitted from a point in a cavity. Considering the wavelength of a wave that can form a resonant cavity mode and the number of modes with that wavelength that can exist in a particular cavity, would increasing the wavelength increase, decrease, or not affect the number of modes?

### Answer

We see in these diagrams three snapshots of a given cavity. In all three, radiation of a specific wavelength is emitted from a fixed point on the cavity’s left side. In each case, the wavelengths differ.

Diagram (c) shows the longest wavelength. Given its start point and the dimensions of the cavity, there is only one location on the cavity walls where this wavelength can have a displacement of zero at both ends. Therefore, this wave has one resonant mode.

In diagram (b), a wave with slightly shorter wavelength is emitted from the same point. Again, given the cavity dimensions, there are now two places where this wave can meet a wall having a displacement of zero. Therefore, this wave has two resonant modes.

Diagram (a) shows the shortest wavelength of all, which “fits into” the cavity in three different ways. Each of these represents a separate resonant mode.

The trend we observe is that the shorter the wavelength of the waves in the cavity, the more resonant modes those waves possess. Shorter-wavelength waves have more ways to fit into a given cavity by satisfying the boundary conditions compared to longer-wavelength waves.

In answer to our question, then, we can say that increasing the wavelength of a wave in a cavity decreases the number of resonant modes for that wave.

### Example 4: Identifying the Curve Corresponding to Quantized Blackbody Energy Levels

The diagram shows two curves, each of which represents the predicted electromagnetic radiation emission spectrum of a blackbody according to a different model of blackbody radiation.

- Which of the curves better corresponds to a model of blackbody radiation in which the number of electromagnetic waves emitted by a blackbody increases as the wavelength of the waves decreases and the number of such waves is
**not**affected by any other factors? - For wavelengths greater than the peak wavelength of the spectrum shown by the blue curve, how does the difference between the intensities at given wavelengths predicted by the two models change as the wavelength increases?
- The difference in intensity decreases.
- The difference in intensity increases.

### Answer

Considering the given diagram, we see the blue and purple curves, each representing a different model of blackbody radiation. The main difference between the curves is that as wavelength gets close to zero, the blue curve approaches zero while the purple one appears to increase without limit.

**Part 1**

Part one of the question asks which curve better corresponds to a radiation model in which the number of waves emitted increases as the wavelength decreases, with no other factor affecting that number.

We have seen that with decreasing wavelength, the blue curve goes to zero. Therefore, the number of waves emitted at progressively shorter wavelengths must decrease as well.

On the other hand, since the intensity (and therefore energy) of the purple curve increases with decreasing wavelength, that curve better corresponds to a model in which the number of waves emitted goes up as the wavelength of the waves goes down. We choose the purple curve as our answer for part one.

**Part 2**

In part two, we want to compare the difference between the blue and purple curves as wavelength increases. Revisiting the diagram, we see that these curves get closer and closer together as wavelength increases. Therefore, the difference in intensity between the two decreases, and we choose option (A) as our answer.

### Example 5: Comparing Blackbody Emission to Blackbody Absorption

Pick the word that would correctly complete the following statement about the emission of electromagnetic radiation by an ideal blackbody:

The electromagnetic-radiation-emission spectrum of an ideal blackbody is the spectrum of the radiation that the blackbody absorbs.

- independent of
- the inverse of
- the same as

### Answer

In this question, we are comparing the way blackbodies emit radiation to the way they absorb it.

We recall from the definition of a blackbody that blackbodies are idealized objects that absorb absolutely all the electromagnetic radiation incident on them. In this way, they have no preference or limit in the type of radiation absorbed.

When they emit radiation, blackbodies radiate neither exactly the wavelengths of light they have absorbed nor equally at all wavelengths.

Instead, they emit radiation according to a functional form that varies only with the blackbody’s temperature.

Since the temperature of a given blackbody does not indicate what sorts of radiation the blackbody has absorbed, we can say the radiation-emission spectrum of an ideal blackbody is independent of the spectrum of the radiation that the blackbody absorbs. This is indicated by answer choice (A).

### Key Points

- A blackbody is an idealized object that absorbs all incident radiation and emits radiation based on its temperature.
- Blackbodies can be modeled as cavities whose boundary conditions allow certain emitted waves and disallow others.
- Understanding how blackbodies radiate energy requires the assumption that wave energy is discrete or “quantized.”
- The energy of light quanta emitted by a blackbody is described by Planck’s equation , where is the number of quanta possessing a frequency and is Planck’s constant with an approximate value of m
^{2}⋅kg/s.