2.4: Characterizing Emission

Where does this electromagnetic energy come from? It surrounds us every moment of every day in many forms. In fact, any object warmer than absolute zero (0 K) emits radiant energy. In order to estimate the amount of radiant energy an object emits, a common simplification is needed: we assume an object behaves as a blackbody. A blackbody is an object that emits and absorbs the maximum amount of radiation possible given its temperature.

Planck’s Curves

Planck’s curves are used to show the amount of emitted radiation and primary wavelengths of electromagnetic energy that a black body emits given its temperature. The diagram below shows multiple Planck function curves for various temperature black bodies. The red line denotes an object that has a temperature of 3000 K, the green line is 4000 K, and the blue line is 5000 K. As the black body becomes hotter, it emits at shorter wavelengths and greater intensities.

The spectral radiance (y-axis, emitted radiation) vs. the wavelength (x-axis) of radiation emitted from black bodies of various temperatures. As the temperature of the black body increases, the amount of emitted radiation increases, and the radiation is emitted at shorter and shorter wavelengths (Public Domain).

Wien’s Law

Planck’s function tells us the amount of emitted radiation and wavelengths over which it is emitted given the temperature of the black body. We can use another law to determine the maximum wavelength emitted by a black body. Wien’s Law states that the shorter the wavelength emitted, the hotter (more kinetic energy) the object is. In Wien’s equation, sometimes the numerator is given as “a”, a constant equal to 2897.

\[\text { Maximum Wavelength }=\frac{a}{\text { Temperature }}=\frac{2897}{T} \nonumber \]

\[\lambda_{\max }=\frac{2897}{T} \nonumber \]

Using Wein’s equation to find wavelength gives an answer in microns, µm. One micron is equal to 10^-6 meters.

Example \(\PageIndex{1}\): Solar Radiation Wavelength

Our sun is a yellow dwarf star with a surface temperature near 5800 K. What is the maximum wavelength of the radiation it emits? Where does that radiation fall on the electromagnetic spectrum?

Solution

Wein’s Law gives us an equation that relates temperature to the maximum wavelength of radiation emitted by an object:

\[\begin{gathered}
\lambda_{\max }=\frac{2897}{T} \\
\lambda_{\max }=\frac{2897}{5800 K}=0.499
\end{gathered} \nonumber \]

This radiation, 0.499 µm, is in the range of visible light, given as 0.5 × 10^-6 m by the chart earlier this chapter.

Stefan-Boltzmann Law

One last very important law for radiation will be discussed here. The Stefan-Boltzmann Law relates the total radiation emitted (total emitted power per area) to the area under Planck’s curve. This can be used to show that the hotter the object, the more energy it radiates per unit area.

\[\text { Emission }=\frac{\text { Power }}{\text { Area }}=\text { Stefan }- \text { Boltzmann Constant } * \text { Temperature }{ }^4 \nonumber \]

\[E=\sigma * T^4 \nonumber \]

In the above equation the amount of radiance emitted per area is equal to the temperature of the black body raised to the 4th power. This relationship is extremely important. It shows that the amount of radiation emitted depends heavily on temperature such that small temperature fluctuations result in large changes in emittance.

The Stefan-Boltzmann constant is 5.67×10^–8 W·m^–2·K^–4 and the symbol sigma, \(\sigma\), is used.

Pro Tip: Whenever you see the Stefan-Boltzmann constant being used (σ), you know that the assumption of a black body has been made.