When ultraviolet and visible wavelength photons get sent from the sun they seem to not interact with CO₂ and other greenhouse gases. When that electromagnetic energy gets to the earth, the earth radiates energy back out in the form of infrared radiation which does interact with greenhouse gases causing those gas molecules to vibrate which causes some of this energy to be sent back to earth.

My question is why does smaller wavelength radiation not interact with CO₂ but larger wavelengths do?

I know that the smaller wavelengths would likely cause the electrons in the gas molecules to go to a more excited state and eventually release a photon instead of just vibrating the molecule but it seems like this doesn't even occur and the smaller wavelength radiation just passes through the greenhouse gases like they aren't even there.

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    $\begingroup$ I think the answer may be that greenhouse gases and other gases do indeed interact with UV light! However there's not a lot of heating power in the Sun's UV and there's almost no UV light in the Earth's thermal radiation, which has most of its power at wavelengths longer than 5 microns. UV is very important to the chemistry of the Earth's atmosphere, including ozone's role in protecting us from harmful UV, but that's a different question. You'll find some discussion of Earth's atmosphere and photochemistry here, but also in Earth Science SE. $\endgroup$
    – uhoh
    Commented Dec 21, 2019 at 6:45

2 Answers 2


CO$_2$, CH$_4$ and H$_{2\!}$O absorb at infrared frequencies because the electrical field acts on the polar chemical bonds. These will excite vibrational and rotational movements of the molecule, which then also radiate at these infrared frequencies (different charges moving).

Infrared does not act on O$_2$ and N$_2$ molecules because these bonds are not polar. IR cannot do anything with the argon atoms in the atmosphere. So that is why the small amounts of CO$_2$ govern the Earth's climate.

Visible light is not absorbed because it does not match any electronic excitation energies in these molecules.

UV light is absorbed when the energies are high enough to excite the electrons - UV C and vacuum-UV.


Every molecule has a unique band of frequencies that will affect its electrons courtesy of chemical bonds. To elaborate on why this is, the following is a direct quote fro "Organic Structures from Spectra: Fourth Edition by L.D. Field, et al, 2008, Wiley and Sons, p. 2-3":

"An absorption band can be characterised [sic] primarily by two parameters:

(a) the wavelength at which maximum absorption occurs

(b) the intensity of absorption at this wavelength compared to the base-line (or background) absorption. A spectroscopic transition takes a molecule from oen state to a state of a higher energy. For any spectroscopic transition between energy states...the change in energy...is given by E=hv where h is planks constant and v is the frequency of the electromagnetic energy absorbed... A spectrum consists of distinct bands or transitions because the absorption (or emission of energy is quantised [sic]. The energy gap of a transition is a molecular property and is characteristic of molecular structure."

So for organic molecules, and greenhouse gases, the infrared radiation is extremely useful, both for probing the chemicals, and to provide them energy, which they can store/absorb and release later. This is one of the reasons infrared spectroscopy is so useful for probing organic compounds.

You can find out more information on infrared absorption of greenhouse gases here:


and more about infrared spectroscopy here:


  • $\begingroup$ Could you elaborate on "Every molecule has a unique band of frequencies that will affect its electrons courtesy of chemical bonds". How do the chemical bonds effect the band of frequencies that effect it? $\endgroup$
    – Corey
    Commented Dec 21, 2019 at 7:45
  • $\begingroup$ It's after midnight here, so please give me until tomorrow to dig up my old chemistry textbook for the section on spectroscopy in general, and then I can answer you in a bit more detail. $\endgroup$ Commented Dec 21, 2019 at 8:32
  • $\begingroup$ ok Corey. I've elucidated as best as I can. I hope the quote helps. $\endgroup$ Commented Dec 21, 2019 at 21:10

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