Does CO2 emit infrared photons at the same frequency in which it absorbs them?

I started answering your previous question. I noticed it because it had a couple of stars. But you chose best answer before I finished. So I will combine that answer in to this one.

But before I get going, my hat is off to ∂/∂x and his superlative job in explaining how quantum mechanic and vibration/rotation are linked to allow infrared absorption and emission in the previous question.

Now some atmospheric physics:

1. Take a view from Space. Earth is in equilibrium (or at least very close). The same amount of energy coming in from the sun = energy going out. The black body temperature of the earth is 255K. The surface temperature 285K. Moreover, roughly 30% of the surface IR goes straight up and out and this swath is at the center of the 285K black body spectrum. How can this possibly work?

What happens is that the atmosphere gets colder as you go up to the boundary between the stratosphere and the troposphere. This boundary is roughly -55C or 218K. Part of this is due to convection in the lower atmosphere, but a significant reason for this drop is the absorption and emissions of IR by CO2 and H2O. The effective temperature of the CO2 emission into space is well below earth's 255K black body temperature. End result: There is far less IR emissions into space in the CO2 spectrum than one would expect based on either earth's surface temperature or black body temperature.

2. There will be a slight tilt towards lower frequencies as you go higher and get colder -- but not much. The tilt is based on Wein's law, which says the frequency of the median blackbody emission varies directly with temperature. Drop the temperature and you have a small increase in the lower frequency photons.

But the big drop comes from less emissions as you go up and get colder. This drops off with T^4. A rule of thumb is that in the temperature range we are dealing with a 1° drop in temperature results in a 1.2% to 1.8% drop in emissions.

3. There is a big transmission of energy to other gas molecules. Actually, only a very small number of CO2 molecules have the necessary energy to emit a photon. Mostly, they just bang around against other gas molecules and if they do capture a photon, that quickly turns into general kinetic energy and temperature rise in the surrounding gas.

[The surrounding gas acts as a big heat reservoir. It mediates and dampens the CO2 emissions]

But every once in a while, a CO2 molecule will get a hard knock, and before it gets knocked again, it fires off a photon cooling off itself and the surrounding gas.

You can work this out using P = e^(-E/kt) and E = c h/λ

Just for your information, the relaxation time for a CO2 molecule after absorbing a 15μm photon is 10μs. But the time between collisions at atmospheric pressure is 0.27 ns. Accordingly a CO2 molecule will get knocked on average 370,000 times before its mean emission time.

4. Clouds slow everything down (they also reflect a large amount of light back into space). If you have clouds, you have a lot of water vapor in the air. This decreases the mean free path for a photon considerably.

Moreover, the water drops are great at reflecting back IR in the bandwidth that normally escapes the earth unabsorbed. Ergo, you stay warmer at night with clouds.

In the desert their is very little water vapor above you. And there is no reflection of IR back from the clouds. CO2 covers a much smaller percentage of the IR spectrum than water. (let's assume 15%). So you are looking up at a very cold sky. Microwave background ~3K. Let's do a thermal radiative balance:

Assume desert temperature = 300 K

Assume CO2 radiates back to desert at 300K

Assume no H2O absorption. (this, of course, is not true. It is just that little water vapor means that the mean free path for photons that are absorbed by H20 is quite long.

Radiation from the surface to the the cup = σ 300K^4 * 1/2

Infrared radiation from CO2 back to cup = σ 300K^4 * 1/2 * .15

Radiation from CMB = σ 3k^4 * 1/2

T = (σ 300K^4 * 1/2 + σ 300K^4 * 1/2 * .15 + σ 3k^4 * 1/2)^1/4 / σ

T = 261 K ... effective temperature water "sees"

yep, it will freeze.

5. Instead of rms, I'd get the mean free path for the size of the shell which would be 1 m/ln2 =1.44m

Given atm pressure is 100 kPa, g= 9.8 m/s and density is 1.2 kg/m^3, this is the equivalent of 8500 m (This sounds right since atm at the top of everest is 1/3 atm sealevel). That means it is 5905 mean free path lengths.

*****First time ever. I ran out of room on an answer *******

Here is

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Co2 Emission Spectrum

Hello,

You can get Photon for free here http://j.mp/1k3PJ3c

Good free solution