37.2.1 Continuous Spectra and Blackbody Radiation

Cool lava is black, but lava heated to a high temperature glows red and, if hot enough, yellow. A tungsten wire, dark gray at room temperature, emits bright white light when heated by a current—thus becoming the bright filament in an incandescent lightbulb. Hot, self-luminous objects, such as the lava or the lightbulb, form a rainbow-like continuous spectrum in which light is emitted at every possible wavelength. FIGURE 37.2 is a continuous spectrum.

The figure illustrates the continuous spectrum of an incandescent light bulb.

FIGURE 37.2 Full Alternative Text

This temperature-dependent emission of electromagnetic waves was called thermal radiation when we studied it as the mechanism of heat transfer in 《 SECTION 19.8. Recall that an object with surface area A and absolute temperature T radiates heat energy at the rate

\frac{d Q}{d t} = e σ A T^{4}

(37.1)

where $σ = 5.67 \times 10^{- 8} {W/m}^{2} K^{4}$ is the Stefan-Boltzmann constant. Notice the very strong fourth-power dependence on temperature.

A photograph of hot glowing lava oozing out from a rock. — Black lava glows brightly when hot.

The parameter e in Equation 37.1 is the emissivity of the surface, a measure of how effectively it radiates. A perfectly absorbing—and thus perfectly emitting—object with $e = 1$ is called a blackbody, and the thermal radiation emitted by a blackbody is called blackbody radiation. Charcoal is an excellent approximation of a blackbody.

Our interest in Chapter 19 was the amount of energy radiated. Now we want to examine the spectrum of that radiation. If we measure the spectrum of a blackbody at three temperatures, 3500 K, 4500 K, and 5500 K, the data appear as in FIGURE 37.3. These continuous curves are called blackbody spectra. There are four important features of the spectra:

All blackbodies at the same temperature emit exactly the same spectrum. The spectrum depends on only an object’s temperature, not the material of which it is made.
Increasing the temperature increases the radiated intensity at all wavelengths. Making the object hotter causes it to emit more radiation across the entire spectrum.
Increasing the temperature causes the peak intensity to shift toward shorter wavelengths. The higher the temperature, the shorter the wavelength of the peak of the spectrum.
The visible light that we see is only a small portion of the continuous blackbody spectrum. Much of the emission is infrared. Extremely hot objects, such as stars, emit a significant fraction of their radiation at ultraviolet wavelengths.

The figure illustrates an intensity versus wavelength graph of blackbody radiation spectra.

FIGURE 37.3 Full Alternative Text

The wavelength corresponding to the peak of the intensity graph is given by

λ_{peak} (in nm) = \frac{2.90 \times 10^{6} nm K}{T}

(37.2)

where T must be in kelvin. Equation 37.2 is known as Wien’s law.

EXAMPLE 37.1 Finding peak wavelengths

What are the peak wavelengths and the corresponding spectral regions for thermal radiation from the sun, a glowing ball of gas with a surface temperature of 5800 K, and from the earth, whose average surface temperature is $15 ° C ?$

MODEL The sun and the earth are well approximated as blackbodies.

SOLVE The sun’s wavelength of peak intensity is given by Wien’s law:

λ_{peak} = \frac{2.90 \times 10^{6} nm K}{5800 K} = 500 nm

This is right in the middle of the visible spectrum. The earth’s wavelength of peak intensity is

λ_{peak} = \frac{2.90 \times 10^{6} nm K}{288 K} = 10, 000 nm

where we converted the surface temperature to kelvin before computing. This is rather far into the infrared portion of the spectrum, which is not surprising because we don’t “see” the earth glowing.

REVIEW The difference between these two wavelengths is quite important for understanding the earth’s greenhouse effect. Most of the energy from the sun—its spectrum is much like the highest curve in Figure 37.3—arrives as visible light. The earth’s atmosphere is transparent to visible wavelengths, so this energy reaches the ground and is absorbed. The earth must radiate an equal amount of energy back to space, but it does so with long-wavelength infrared radiation. These wavelengths are strongly absorbed by some gases in the atmosphere, so the atmosphere acts as a blanket to keep the earth’s surface warmer than it would be otherwise.

That all blackbodies at the same temperature emit the same spectrum was an unexpected discovery. Why should this be? It seemed that a combination of thermodynamics and Maxwell’s new theory of electromagnetic waves ought to provide a convincing explanation, but scientists of the late 19th century failed to come up with a theoretical justification for the curves seen in Figure 37.3.

37.2.1 Continuous Spectra and Blackbody Radiation

FIGURE 37.2 The continuous spectrum of an incandescent lightbulb.

FIGURE 37.3 Blackbody radiation spectra.

EXAMPLE 37.1 Finding peak wavelengths