# 6.8: Kirchhoff’s Law explains why nobody is perfect

Remember that when radiation encounters matter it may be absorbed or transmitted or scattered (including reflected). For an object acting as a perfect Planck distribution function, it must absorb all radiation completely with no scattering and no transmission. Some objects absorb very well at some wavelengths but not at others. For instance, water vapor absorbs little visible radiation but absorbs infrared radiation at some wavelengths very well.

At the same time, the sun, like other objects, does not radiate perfectly according to the Planck distribution function spectral irradiance, but instead radiates at a fraction of it at some wavelengths. This fraction, which goes from 0 to 1, is called the emissivity and is denoted by ε. How is an object’s emissivity related to its absorptivity?

Kirchhoff’s Law states that at any given wavelength, an object’s emissivity ε is equal to its absorptivity, that is:

$\varepsilon(\lambda)=\alpha(\lambda)$

Thus, if an object has some wavelengths at which radiation is scattered or reflected, then the object will have an emissivity less than 1 at the wavelength, and the fraction that is absorbed will be equal to the emissivity at each wavelength.

Thus, when we integrate the Planck distribution function spectral irradiance over wavelength to obtain the irradiance emitted by the object, it first has to be multiplied by the wavelength-dependent emissivity, thus leading to the modified form of the Stefan-Boltzmann law:

$F=\varepsilon \sigma T^{4}$

where we understand that ε is some form of averaged emissivity.

Watch the following video (1:07), where the Stefan–Boltzmann Law is described in greater detail:

Stefan-Boltzmann Law

Click here for transcript of the Stefan-Boltzmann law.

This formula, Stefan-Boltzmann law, is the one that we will use the most. Note that for a perfect emitter, epsilon equals to one, the total irradiance submitted into a hemisphere equals the product of the Stefan-Boltzmann constant, sigma, and the temperature to the fourth power. However, the irradiance is modified by the emissivity, which equals the [INAUDIBLE]. Note that this emissivity here is some sort of average over all emissivities for different wavelengths, and we've seen that emissivity can vary a lot with wavelengths. Water vapor, for example, has a very low emissivity invisible, but a very strong one in the infrared. The [INAUDIBLE] depends on the composition of matter, but it also depends on the number concentration of gaseous matter. And the pass length through that matter. Go back and look at Beer's Law of Absorption to see this dependence. With this form of the stuff on Boltzmann law, we can compare the irradiances of two different bodies of matter at different temperatures or different emissivities.

#### Note

Some typical average emissivities are listed in the table below. These are emissivities averaged over all wavelengths. At any particular wavelength, the emissivity may be greater or less than the average.

Wavelength-Averaged Emissivity of Some Common Materials

Material Emissivity, ε
ice 0.97
pure water 0.96
snow snow
trees (oak, beech, maple, pine) 0.97-0.98
grass 0.98
soil 0.93
aluminum foil 0.03
asphalt 0.88-0.94

What about gases? Gases absorb and thus emit like all other matter. To know more about the emissivity of all objects, we need to know more about the absorption of objects.

#### Quiz 6-2: Thank You, Planck

1. Find Practice Quiz 6-2 in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz.
2. When you feel you are ready, take Quiz 6-2. You will be allowed to take this quiz only once. Good luck!