Will CO2 absorb photon in all directions?

[Billy T]

... That You Tube on Ozone production if watched will tell you the answer we were looking for. There was momentarily a high energy ozone that was unstable. It either broke up or collided with another atom and transferred its excess energy. So the energy package was transferred by contact with another atom.

Yes that animation is very good aid for one who does not know all that already. My only fauting of it is it did not explain the * was used to indicate the atom or molecule it was attached to was in and excited state. If it is the nucleus that is in an excited state, typically that is indicated by an m following the atomic symbol. For example the meta-stable isotope of T99, which is used to make whole body scan for cancer spread into the bones is written T99m.

... Had you ever heard of this {non-radiative de-excitation of an excited state} before?

Of course. I mentioned as qualifying notes, that the energy radiation could carry away might be removed instead by a collision. In fact that is the ONLY way radiation can heat a gas. I.e. the excitation energy is first shaired by the two colliding atoms and they with many more collision, with others, so that the average RANDOM kinetic energy is very slghtly increased - i.e. the gas is warmed to a higher temperature.

If every gas atom (or molecule) that is excited by a photon is only in the excited state (typically for less than a ms) before it radiates that energy again, etc. then there is no heating of the gas.

... Reactions of this type could be unbalanced on either side of the planet Venus so on one side the energy remains in the form of broken bond energy (it takes energy to break bonds) and on the other kinetic energy (when the ozone remains stable after transferring excess kinetic energy). The energy side driving the wind (for the momentum of the photon is not lost).

No in the top of the atmosphere were O2* or O3 or O3* is formed by UV it typically will return to the ground state (O2, or O3) in tiny fracion of a second by collison or radiation. Note the energy transfered by collsion has a random velocity (Is thermal) and the re-radiated photon has equal probability to be traveling in any direction - no memory of what was the direction of travel of the photon that was absorbed to make the excited state. The only imbalance in absorption torques is the one I mentioned earlier. There are very slightly more UV photons that can ionize on the limb moving towards the sun as some are "blue shifted" to have just barely the ionization potential energy.

 

[Robittybob1]

.... Note the energy transfered by collsion has a random velocity (Is thermal) and the re-radiated photon has equal probability to be traveling in any direction - no memory of what was the direction of travel of the photon that was absorbed to make the excited state.

Isn't that a serious error? The incoming UV photon has momentum of a certain direction. That momentum is transferred to the O3 with a direction so the initial impact has a direction of motion still influenced by the photon.
Energy is transferred and so the random nature of further collisions make the situation chaotic but since the initial collision was directed that ever so slight momentum is never lost (conservation of momentum).
We are talking of winds that can take billions of years to develop the whole intensified circulation, so the slightest imbalance might never disappear especially since Venus doesn't have a moon to moderate the situation.

 

[Emil]

NOTE: Before I comment on the above quote(s), let me admit that in a recent reply in another thread, I stated that as a light sphere expands from its point of origin, it does appear to do so essentially as a 3D wave front and there does not appear to be any widening distance between individual photons, as the total area of the light sphere increases. (I really thought someone would jump on that one and even considered several times countering my own post.) While this may be as things appear, it is not entirely consistent with our experience of light and photons.

To this I add my comment from Post #44 of this thread,


This last part does raise some issue with how we interpret the results we obtain in a number of experiments involving the wave vs particle nature of EM radiation and even electrons. We are, as I said to some extent fish in a fish bowl. Our understanding of what is happening is unavoidably influenced by the quantifiable way in which light, or photons, interact with atoms, or matter.

There is a question that continually comes up for me where experiments as described above are involved. They are almost always conducted using a light source composed of many photons and yet the conclusions are often, as above projected as descriptions of the nature and character of individual photons.

When the same or similar experiments are conducted with single photons or single electrons, the result invariably appears as a particle event, which if repeated using again single photons or electrons, over time seems to result in a pattern consistent with the interference pattern associated with the classical experiment and a concentrated "beam" of light.

How are we to know with any certainty that even when we are using a "beam" of light the interference pattern is not just a statistical representation of particle behavior, as opposed to a true wave interference?... And if our understanding of EM radiations as photons is an artifact of the quantifiable manner in which an EM "wave" interacts with atoms or matter, how are we with certainty, to know that the photon actually exists? You see on one hand what we "see" are photons and on the other we cannot with certainty know that the photons we see, are not just artifacts of the process of observation and measurement.

I know this is sliding a long way into the philosophical, but it is just such seemingly paradoxical situations that, suck me down the rabbit hole.

I have the same opinion.

Also I find interesting the following.

Speed of electricity :

Electromagnetic waves
The velocity and the electrical resistance outside of conduction is often assumed by the propagation speed of an electromagnetic wave. In simplified systems, the speed of electricity is given as the electromagnetic wave which conveys information (data), not the movement of electrons. Electromagnetic wave propagation is fast and depends on the dielectric constant of the material. In a vacuum the wave travels at the speed of light and almost that fast in air. Propagation speed is affected by insulation, such that in an unshielded copper conductor range 95 to 97% that of the speed of light, while in a typical coaxial cable it is about 66% of the speed of light.

Electric drift
The drift velocity deals with the average velocity that a particle, such as an electron, attains due to an electric field. In general, an electron will 'rattle around' in a conductor at the Fermi velocity randomly. Free electrons in a conductor vibrate randomly, but without the presence of an electric field there is no net velocity. When a DC voltage is applied the electrons will increase in speed proportional to the strength of the electric field. These speeds are on the order of millimeters per hour. AC voltages cause no net movement; the electrons oscillate back and forth in response to the alternating electric field.

Defined return circuit
For an infinite distance of the return conductor or a conductor without return conductor, a finite length of an infinitely long conductor without return conductor would have an infinite inductance and inversely, zero capacity. The magnetic field is assumed as instantaneous, that is, the velocity of propagation of the magnetic field is neglected.
Considering, however, the finite velocity of the magnetic field, the magnetic field at a distance from the conductor and at a time which corresponds to the current in the conductor at the time, the time required for the electric field to travel the distance; or, the magnetic field at distance and time corresponds to the current in the conductor at the time.

So, what is the nature of the magnetic field?

 

[Billy T]

... I stated that as a light sphere expands from its point of origin, it does appear to do so essentially as a 3D wave front and there does not appear to be any widening distance between individual photons, as the total area of the light sphere increases. ...

And you are fully correct as speaking of zillions of randomly directed photons traveling away from a brief "point" source. (Say a flash bulb in practice.)

I.e. there is spherical shell of light moving away from that source point of thickness ct where c is the speed of light and t is the duration of the flash. Only thing one can question is your statement: "does not appear to be any widening distance between individual photons" which is true ONLY because no one is looking at photons or measuring their average separation, but you know it is increaces with time.

For example consider a tiny cone with apex at the point source that at one meter from the source point has 1 cm square cross section with a 100 million photons passing thru that 1cm^2. Those same photons will pass thru a 4 cm^2 when 2 meters from the source point. There, after 2 meters of travel the density will be only 25 million / cm^2 - I.e. on average the photons are four times farther apart at 2m than they were at 1m.
------
Einstiein said "physic is the same in all reference frames." I.e there is no special, unique reference frame. One part of "being the same" is that speed of light, c, is the same for all frames. Another part is that the spherical shell of light expanding from a flash bulb is spherical in all frames.

This is very hard for most people to accept if they think about it, but true. For example imagine on the side of a fast moving train the are many concentric circles of photo detectors connected to time of light flash arrival recorder. Exactly the same set up of light detectors with recorders is mounted on large vertical pannel which the train almost hits (just 1 cm miss as it speeds by)

Also the flash bulb is ground mounted at the center of the concentric detector circles of the large pannel, but has a wire sticking 0.6cm over towards the train. The train too has an 0.6cm wire, contected to ground via the steel wheels extending over towards the large pannel. At some common, t=0 for both train and the wire from the pannel these two 0.6cm wires touch, completing the flash bulb´s circuit and it flashes.

Later the physicist on the train and the one with the ground mounted pannel get together and compare their records. Both agree the bulb flased at t = 0 and that at time 2/c the expanding shell of light reached all the detectors on the radius 2m detection circle at the same time by their recording clocks, (like wise the shell arrived at the 4 meter detector radius at exactly 4/c on their clocks. etc. I.e. physics was the same for both -an expanding spherical shell of light traveling at same c m/s.

The physicist on the ground also had a high speed movie camera taking pictures of the circular ring of light on the side of the train, The frame of this film that corresponded to the instant of time when his 2 meter radius ring of detectors detected the spherical shell of light cleary showed the circular ring of light on the side of the train had already passed (was closser to the rear of the train) the train´s two meter radius ring of detectors. Obviously due to fact that after t=0 those nearer to rear of train detectors had moved ahead to met the on coming light. Likewise that frame of the film showed the detectors nearer to the front of the train on the 2 m ring were not yet illuminated by the circular ring of light as they were running away from the expanding circle of light, but not fast enough to keep it from eventually getting to them.

The ground physicist, who was not well versed in the theory of special relativity, belived he had hard photographic evidence that the train born physicist was lying when he said the all the detectors on the 2 meter ring had detected the light at the same time - i.e. it was an expanding circle of light that arrived at them all at exactly 2/c time on his recording clocks.

The resolution of this paradox is the although t=0 is simultaneous for both, one´s time 2/c is not simultaneous with the others 2/c. -I.e Although each of them measures time with idential atomic clocks, but by the clock of one frame, the other´s clock is running too slow.

Most people never get fully happy with the truth of special relativity.

 

[Robittybob1]

Can we go back to the OP please? In particular any effect that might cause an imbalance in the forces energy and momentum on either side of a planet to cause a wind.

 

[Billy T]

... The incoming UV photon has momentum of a certain direction. That momentum is transferred to the O3 with a direction so the initial impact has a direction of motion still influenced by the photon.
Energy is transferred and so the random nature of further collisions make the situation chaotic but since the initial collision was directed that ever so slight momentum is never lost (conservation of momentum).
We are talking of winds that can take billions of years to develop the whole intensified circulation, so the slightest imbalance might never disappear especially since Venus doesn't have a moon to moderate the situation.

That is true, but very small and the wind does not steadily gain an accumulating momentum as these randomly moving (with a tiny net bias in the velocity) are also colliding with the ground, trees and even tall buildings.

Thus as the limb traveling towards the sun is perhaps getting slightly more momentum to the west added than the other limb is getting to the east, there is in the end a slight slowing of the earth´s spin - days growing longer and to keep "high noon" sun time same as 12 o’clock noon, we do a every decades or so add a "leap second" to our clocks.

This differential torque by absorbed UV in the two limbs is tiny compared to the tidal slowing of the earth´s spin rates and does not accumulate as a wind even it there were no moon. It accumulates as a slowing of the earth´s (or Venus´s) spin, not as ever increasingly stronger winds. Note also that the solar tides would still be there for all planets and are much larger than the differential limb torques.

SUMMARY: Yes, in principle, but undetectable in practice, part of the reason why the Earth´s rate of rotation is slowing is the differential limb torques caused by high altitude absorption of UV in the two limbs, but as this net torque is soon applied to the solid earth it is the spin rate of the Earth that is slowed (undetectably more than for example by the torque of just the solar tides) not an ever increasingly stronger wind that accumulates for millions of years.

 

[Robittybob1]

That is true, but very small and the wind does not steadily gain an accumulating momentum as these randomly moving (with a tiny net bias in the velocity) are also colliding with the ground.....

Thus as the limb traveling towards the sun is perhaps getting slightly more momentum to the west added than the other limb is getting to the east, there is in the end a slight slowing of the earth´s spin - days growing longer and to keep "high noon" sun time same as 12 o’clock noon, we do a every decades or so add a "leap second" to our clocks.

I'll reply to your post in two sections. Momentum added to the ground is as good as momentum added the wind. For in the end they rotate in lock-step with each other. A fast spinning planet has the same effect on the incoming UV as would a super rotating atmosphere moving in the same direction.

I did calculate that even 1 microsecond per year increment from any of these effects would have major change to day length over the life of the planet. Your off the cuff quote of "add a leap second every decade or so", if that was possible you would end up with a planet spinning that fast it would fly apart.
I'll calculate that effect add edit the post soon.

If there would be more than 1 sec change in 62500 years the day length would be less than 4 hours and that gets into the region where the centrifugal forces overcome the gravitational forces. In other words the planets are sensitive to any increase in rotational velocity.

 

[Billy T]

I'll reply to your post in two sections. Momentum added to the ground is as good as momentum added the wind. For in the end they rotate in lock-step with each other.

Are you forgetting that "wind" is the DIFFERENCE between ground speed and air speed? If both are in "lock step" then the wind speed is zero. At the earth´s equator the ground speed is quite close to 1000 mph (I forget the exact value) towards the east. There the trade winds blow towards the west. If we assume a 20mph average WIND speed for the trade winds, then the average air speed is to the east at 980mph.

The rest of your post I could not follow. If you have some point, please make it again.

 

[Robittybob1]

....
Thus as the limb traveling towards the sun is perhaps getting slightly more momentum to the west added than the other limb is getting to the east, there is in the end a slight slowing of the earth´s spin - days growing longer and to keep "high noon" sun time same as 12 o’clock noon, we do a every decades or so add a "leap second" to our clocks.

This differential torque by absorbed UV in the two limbs is tiny compared to the tidal slowing of the earth´s spin rates and does not accumulate as a wind even it there were no moon. ....

Sorry about truncating your post but the bit I left is the bit I want you to double check. I worked out the momentum would be increasing the rotation rather than slowing it.
Remember the post a couple of days ago I showed how the absorption of the light by hydrogen atoms can make a slowing effect of the atmosphere, but in other situations where there is no release of a photon the effect is other way around.
So could you take me through this bit slower just to make sure we have the momentum and energy levels correct please.

 

[Billy T]

... So could you take me through this bit slower just to make sure we have the momentum and energy levels correct please.

I HAVE DONE THAT several times already, so briefly because of the blue shift, slightly more photons have the energy to ionize or disassociate 02 -> 0 + 0 etc. I.e. the limb moving towards the sun absorbes slightly more momentum AWAY from the sun than the other limb. This imblance reduces the earth´s rate of rotation. (But by such a tiny amount compared to just the sun´s tidal (gravitational) torque it would not be detectable.)

 

[Robittybob1]

Are you forgetting that "wind" is the DIFFERENCE between ground speed and air speed? If both are in "lock step" then the wind speed is zero. At the earth´s equator the ground speed is quite close to 1000 mph (I forget the exact value) towards the east. There the trade winds blow towards the west. If we assume a 20mph average WIND speed for the trade winds, then the average air speed is to the east at 980mph.

The rest of your post I could not follow. If you have some point, please make it again.

This first part is agreed - the way the incoming photon interacts with the atmosphere is related to the net air speed. So a -20 mph wind on the planet spinning at 1000 mph is the same as a 980 mph wind on a stationary planet.

The next bit worked out like this. (Now I am looking at an effect which speeds up rotation)
Life of planet = 4.5 billion years
Present day length = 24 hour
max. rotation speed of Earth is 1 REVOLUTION/4 hours
so how many years per second change in day length will it take
20 hour X 60 X 60 = number of seconds in the 20 hour period = 72000 sec
divide 4.5 billion years by 72,000 = 6.25E+04 years per second change.

 

[RealityCheck]

I HAVE DONE THAT several times already, so briefly because of the blue shift, slightly more photons have the energy to ionize or disassociate 02 -> 0 + 0 etc. I.e. the limb moving towards the sun absorbes slightly more momentum AWAY from the sun than the other limb. This imblance reduces the earth´s rate of rotation. (But by such a tiny amount compared to just the sun´s tidal (gravitational) torque it would not be detectable.)

Hi Billy T.

I would like to check something with you about the dissociation and momentum transfer aspects?

If, as you say, the relative blueshifting makes the incoming radiation more 'ionising', then the energy is transferred in a sort of 'inelastic' interaction where the energy is absorbed in breaking the O-O and creating maybe O-O-O bonds. Would the light momentum be transferred to the molecules 'thermodynamically' as a whole (as kinetic movement of the overall 'products' in the other direction), or is the momentum energy absorbed 'chemically' via the bond creation/breaking route?

Is there a net net kinetic thrust to the extent you say? Or is there a random thermal motion when bonds are created/destroyed and then equally random re-radiation of the excess energy received? I hope you get what I'm meaning/asking here?

Just want to clarify these aspects with you just so I am not misunderstanding the ongoing conversation; which is very interesting in many ways and byways!

Back tomorrow, mate. Cheers!

.

PS: Billy T, Robbitybob1, eeveryone: I just remembered something else I wanted to add. If ionisation occurs from the input photon energy, the charged particle will immediately interact with the Earth's magnetic field lines? If so, then the energy/motion of such ionised particles will be coupling to the Earth via that magnetic field, so some of the molecular motion will be determined by the field as much as the ionisation photon-molecule 'collision' itself? Your thoughts, guys? Back tomorrow.

.

 

[Robittybob1]

....
PS: Billy T, Robbitybob1, eeveryone: I just remembered something else I wanted to add. If ionisation occurs from the input photon energy, the charged particle will immediately interact with the Earth's magnetic field lines? If so, then the energy/motion of such ionised particles will be coupling to the Earth via that magnetic field, so some of the molecular motion will be determined by the field as much as the ionisation photon-molecule 'collision' itself? Your thoughts, guys? Back tomorrow.

When a molecule is ionised there are two parts at least with opposite charges.
So there will be equal amounts of positive and negative charges on the ions, and even if they are of different masses the ideal gas law (PV = NRT) the pressure from the lighter molecules matches that of the heavier particles.
So from my limited knowledge of magnetism there should not be any torque from the ionised gas molecules.

 

[Robittybob1]

I HAVE DONE THAT several times already, so briefly because of the blue shift, slightly more photons have the energy to ionize or disassociate 02 -> 0 + 0 etc. I.e. the limb moving towards the sun absorbes slightly more momentum AWAY from the sun than the other limb. This imblance reduces the earth´s rate of rotation. (But by such a tiny amount compared to just the sun´s tidal (gravitational) torque it would not be detectable.)

Well this is the interesting bit, in that when I first thought up the concept of the radiation being absorbed in a slightly different ways on each side of the planet, there is at least someone else agreeing that a similar event can occur.

Personally I still think it still has some involvement of the greenhouse gases for the super rotating winds are occurring on planets with reasonable high levels of greenhouse gases.
But if we can show any absorption of radiation and it subsequent release could result in a torque this is a start.

You have not commented at all on the photoelectric absorption of the Hydrogen atoms. The interesting thing here is the momentum of the photon is lost (maybe) because the release of the photon is in all directions.

Now to look at the possible reasons we both see the same event causing opposite effects. To repeat your view:

because of the blue shift, slightly more photons have the energy to ionize or disassociate 02 -> 0 + 0 etc. I.e. the limb moving towards the sun absorbes slightly more momentum AWAY from the sun than the other limb.

Firstly, you have made assumptions as to what was the limiting factor.
So if we eliminate this and say equal number of ionizations occur on both limbs, the main difference I see is that on one side a photon of lower energy and momentum does the same job a higher energy photon does on the other side. Higher energy implies higher momentum, so more momentum is added to the planet whereas on the blue shift side less momentum is added for the ionization is using the gases own energy and momentum to do some of the ionization work.
So I still think at this stage, all but tentatively, that the higher energy photon doing the task on the red shift limb will be adding greater momentum and energy to the planet's rotational momentum (pro-grade wind).

 

[Robittybob1]

Co2 will flood the Earth in a half billion years if im not mistaken? The suns increasing brightness.

We'll just have to wait and see!

 

[Billy T]

... You have not commented at all on the photoelectric absorption of the Hydrogen atoms.

That is because absorption of photon is not normally called "photoelectric absorption" so not sure what you were speaking of. I will tell you about hydrogen’s interaction with photons, assuming the atomic hydrogen is initially in the ground state (lowest energy state).

If the photon has 13.6eV of energy (or more) only thing it can do ionize (strip off the bound electron) the H. If that photon had energy E, the electron will leave with very slightly less than (E-13.6)eV of energy. (The slightly less is due to the tiny KE in the recoil of the proton.) I think less (probably much less) than 1% of the solar photon have 13.6 eV of energy - that is pretty harsh UV.

Most photons with less than 13.6eV can not be absorbed by H. Here are the energies that H can absorb:

Photon energy E must be: E = 13.6{1/(n^2) - 1/(m^2)}eV, where n & m are integers.

n=1 is the ground state and m = 2 or higher is an excited state. If the H has been excited to m=2 and radiatively decays, the energy of the photon emitted is thus 13.6{1 -0.25} That same energy is precisely the energy a photon MUST have to excite normal ground state H to the first excited level. If the transition were between n=1 & m=3, the energy of those photons is greater and must be 13.6(1- 1/9} There is a whole set of these excited to ground state /excited state transition and the set of spectral lines is called the Lyman series. The least energy Lyman line is the one with 13.6(0.75)eV and it is relatively harsh UV - I.e. humans can not see any of the Lyman lines. (But I think honey bees may be able see this weakest one only.)

The visible radiation from H that human can see is called the Balmer series. For it n = 2 always. The transition between m=3 and n=2 is with energy 13.6(1/4 - 1/9)eV. This is a deep red line with wavelength 6563Angstroms. The transition m = 7 to n = 2 has wavelength 3970 Angstroms which bees can certainly see, but humans can not as it (and from all larger m values) is in the UV.

SUMMARY: Only very specific energy photons from the sun can be absorbed by atomic hydrogen. This is true of all gases found in the atmosphere - only specific photon energys can lift their ground state electrons to excited states.

I don´t think ANY visible light solar photons, can be absorbed by the dominate atmospheric gasses (O2 & N2) as these symmetric molecules (and also symmetric H2,) have lowest excited state needing UV to excite. Also they will not even absorb solar IR (or "earthshine") as being symmetric no matter how they change vibration or rotation levels (as they can in collisions) they do not have any change in electric dipoles - that is required for radiation or absorption of radiation.

If the photon has more than the ionization (or disassociation of molecule) energy then it can be absorbed, even by symmetric molecules. When one item is split off (for example an electron throw out of atom or molecule) the radiation / absorption allowed selection rules are different from those that apply to excitation only.

... Firstly, you have made assumptions as to what was the limiting factor. ...

No, that is not an "assumption." I just stated the facts about what nature allows wrt photon absorptions. Some of these facts I gave above for specific case of atomic hydrogen. It would be impossible to give others to you as you don´t know how electron orbitals are described in terms of the quantum specifications. To you I can only tell facts that follow from energy conservation.

 

[Robittybob1]

Is the whole process of taking in a photon and moving through the different energy level just called "absorption"? I was thinking what is the process called and it was blank so I thought "photoelectric effect" but that is more when an electron is knocked right out of the energy levels. So going up the levels heading toward the free state is that not part of the photoelectric effect too?

 

[Robittybob1]

.... I just stated the facts about what nature allows wrt photon absorptions. Some of these facts I gave above for specific case of atomic hydrogen. It would be impossible to give others to you as you don´t know how electron orbitals are described in terms of the quantum specifications. To you I can only tell facts that follow from energy conservation.

Say Billy - Give me a little taste of this quantum specifications you are talking about?

 

[Robittybob1]

That is because absorption of photon is not normally called "photoelectric absorption" so not sure what you were speaking of. I will tell you about hydrogen’s interaction with photons, assuming the atomic hydrogen is initially in the ground state (lowest energy state).

If the photon has 13.6eV of energy (or more) only thing it can do ionize (strip off the bound electron) the H. If that photon had energy E, the electron will leave with very slightly less than (E-13.6)eV of energy. (The slightly less is due to the tiny KE in the recoil of the proton.) I think less (probably much less) than 1% of the solar photon have 13.6 eV of energy - that is pretty harsh UV.

Most photons with less than 13.6eV can not be absorbed by H. Here are the energies that H can absorb:

Photon energy E must be: E = 13.6{1/(n^2) - 1/(m^2)}eV, where n & m are integers.

n=1 is the ground state and m = 2 or higher is an excited state. If the H has been excited to m=2 and radiatively decays, the energy of the photon emitted is thus 13.6{1 -0.25} That same energy is precisely the energy a photon MUST have to excite normal ground state H to the first excited level. If the transition were between n=1 & m=3, the energy of those photons is greater and must be 13.6(1- 1/9} There is a whole set of these excited to ground state /excited state transition and the set of spectral lines is called the Lyman series. The least energy Lyman line is the one with 13.6(0.75)eV and it is relatively harsh UV - I.e. humans can not see any of the Lyman lines. (But I think honey bees may be able see this weakest one only.)

The visible radiation from H that human can see is called the Balmer series. For it n = 2 always. The transition between m=3 and n=2 is with energy 13.6(1/4 - 1/9)eV. This is a deep red line with wavelength 6563Angstroms. The transition m = 7 to n = 2 has wavelength 3970 Angstroms which bees can certainly see, but humans can not as it (and from all larger m values) is in the UV.

SUMMARY: Only very specific energy photons from the sun can be absorbed by atomic hydrogen. This is true of all gases found in the atmosphere - only specific photon energys can lift their ground state electrons to excited states.

I don´t think ANY visible light solar photons, can be absorbed by the dominate atmospheric gasses (O2 & N2) as these symmetric molecules (and also symmetric H2,) have lowest excited state needing UV to excite. Also they will not even absorb solar IR (or "earthshine") as being symmetric no matter how they change vibration or rotation levels (as they can in collisions) they do not have any change in electric dipoles - that is required for radiation or absorption of radiation.

If the photon has more than the ionization (or disassociation of molecule) energy then it can be absorbed, even by symmetric molecules. When one item is split off (for example an electron throw out of atom or molecule) the radiation / absorption allowed selection rules are different from those that apply to excitation only.No, that is not an "assumption." I just stated the facts about what nature allows wrt photon absorptions. Some of these facts I gave above for specific case of atomic hydrogen. It would be impossible to give others to you as you don´t know how electron orbitals are described in terms of the quantum specifications. To you I can only tell facts that follow from energy conservation.

You missed the point completely. I was looking to see if you could think a way of the Sun powering up winds through the absorptions and release of the photons at random angles.
I'll have to go back a few days and find my post on that.

 

[Billy T]

Say Billy - Give me a little taste of this quantum specifications you are talking about?

No, I´m not going to try to teach you any quantum physics, but here is link to discussion of quantum numbers of bound electrons:
http://en.wikipedia.org/wiki/Quantum_number
That is place you can start and them seach on some of the terms found there.