Blackbody radiation and greenhouse effect
Much of the light we see comes from atoms and molecules, which are so hot they vibrate and collide with high-energy motion. Every time an atom changes the direction it is moving, the acceleration of electrons at the exterior of the atom causes light to be emitted. Objects at temperatures like the surface of the sun emit light in the visible part of the spectrum.
For a better understanding of what blackbody radiation is, read section 8.6 of the textbook (pp. 310–313), paying particular attention to the formula for the wavelength that is given the maximum power (p. 311) and example 8.4 which uses it.
To see the relationship between temperature and frequency of sunlight and light emitted by the earth itself, open the simulation Blackbody Spectrum:
The graph shows the distribution of light in terms of wavelength. In this chart, the farther to the left, the higher the frequency. The vertical direction represents intensity of light. (See Figure 8.38 on p. 312.)
Check the boxes labeled “Graph Values” and “Labels” to see the specific wavelength of the peak intensity and in what part of the electromagnetic spectrum the wavelength falls.
Set the temperature slider so the temperature is 5800 K, about the temperature of the surface of the sun. This should be the default temperature when the simulation opens. If you’ve set the “Graph Values” check box, the simulation should identify the wavelength of the peak intensity. But we can also calculate this temperature (see Example 8.3).
Record the value here: Sun: λ = _______________ µm
Convert this to meters: Sun: λ = _______________m. (Use scientific notation).
Use this to calculate the frequency of the light: ; If wavelength is in meters, frequency should come out in Hertz (Hz). See Example 8.2 on p. 303 for an example of how to do this.
Sun: f = _________________
Adjust the temperature to 300 K, ground temperature on a warm day.
At first glance, it appears nothing is graphed for this low temperature. But if you adjust the scale of the graph, you can see the curve. At the bottom right of the graph are two magnifying glass icons, one + and one -. Click on the “minus magnifier” three times. Then click on the “plus magnifier” at the upper left of the graph until you can see the curve clearly (9 times).
Record the wavelength and calculate the frequency as before.
Earth: λ = ___________________ µm = ________________________ m
Earth: f = ____________________
Record the results into the following table.
Review the infographic below. The entire spectrum is depicted, with illustrations as well, indicating their wavelengths when compared to physical objects. Also note the illustration indicating the regions of spectrum to which the atmosphere is mostly transparent but also to which its opaque (at the bottom). [See next page.]
Electromagnetic Spectrum: The entire electromagnetic spectrum, running from long-wavelength, low-frequency radio waves to short-wavelength, high-frequency gamma rays. (Chaisson, 20130909, p. 66)
A word about units: 1 micron = 1 micrometer = 1 μm = 1 millionth of a meter = 1 x 10–6 m.
Graphic copyright © 2019 Pickcomfort.com All Rights Reserved.
Visible light wavelengths are usually represented in nanometers, which are about the size of an individual atom. 1 nm = 1 x 10–9 m. 1 μm = 1000 nm and so 0.5 μm = 500 nm. Visible light ranges from 400 nm (violet) to 700 nm (red).
Opacity is the opposite of transparency. The effect of atmospheric opacity is that there are only a few windows at certain points in the EM spectrum where Earth’s atmosphere is transparent. Component gases that make up parts of the Earth’s atmosphere absorb radiation very efficiently at some wavelengths. Water vapor (H2O) and carbon dioxide (CO2) are strong absorbers of infrared radiation. (Chaisson, 20130909, p. 68)
For a simulation of the interaction of different kinds of light and important atmospheric gases, see this link:
In what spectral region does the major wavelength emitted from the Sun appear? [Choices: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma ray.]
In what spectral region does the major wavelength emitted from the earth appear?
Does it seem reasonable that the earth’s atmosphere will allow the solar radiation to pass?
Does it seem reasonable that the earth’s atmosphere will allow the blackbody radiation due earth’s temperature to pass?
What atmospheric gases will absorb infrared light? (These are sometimes called “Greenhouse Gases”.)
Sunlight carries energy which gets absorbed by the earth. Infrared light also carries energy back out into space from the earth. If gases in the atmosphere absorb more infrared light before the light can leave the atmosphere, what does conservation of energy say should change about the gases that absorb the energy?
What does this indicate is a likely outcome if the amount of greenhouse gases in the atmosphere increases? [See Section 8.7b, pp. 317–319 for data.]
Reference: Chaisson, E., McMillan, S. (20130909). Astronomy Today, 8th Edition