Maybe I am not understanding something simple, but longer wavelengths go through things easier correct? So why is it that infrared heat is REFLECTED by the clouds and contributes to global warming? It seems to me it would just go straight through clouds compared to being absorbed. After all, isn't this why we have red sunsets is because that's the only light that hasn't be reflected away?
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1$\begingroup$ "longer wavelengths go through things easier correct? " No, this is not correct. x-rays goes through your body (including bones) and visible light doesn't. So starting from a false premise, the question is not well posed. $\endgroup$– nasuCommented May 1, 2017 at 16:00
1 Answer
I think you're confusing scattering with absorption. The atmosphere's Rayleigh scattering has a straightforward wavelength dependence, but absorption is a more complicated issue that varies by molecule, each having its own "absorption spectrum".
Molecules such as $\text{CO}_2,\,\text{CH}_4,\,\text{H}_2\text{O}$ absorb infrared strongly, whereas molecules such as $\text{N}_2,\,\text{O}_2$ do not. The rule of thumb is that molecules with a central atom attached to multiple identical covalent bonds have their strongest absorption in the infrared spectrum, whereas single-bond molecules do not.
This is because absorption depends on the energy spectrum of the molecule depending on its electrons' quantum numbers. In single-bond molecules, we expect the photon wavelengths most strongly be absorbed to be on the order of $1/R_\infty$, $R_\infty$ being the Rydberg constant. (This estimate is about 100 nm, about one fifth of the correct peak wavelength for oxygen's absorption spectrum, but this is a back-of-the-envelope explanation.) By contrast, comparably weak interactions between covalent bonds in molecules such as $\text{CO}_2$ create a spectrum containing smaller typical energies for absorption, so that infrared is most affected.
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$\begingroup$ Can you expand on that? Does c02 do something special with infrared unlike other light? $\endgroup$ Commented May 1, 2017 at 14:49
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$\begingroup$ @user3466986 I've tried to make it clearer in an edit. $\endgroup$– J.G.Commented May 1, 2017 at 15:47