Emission of photons by vibrational relaxation of molecules

[ns_phonon]

I have read that relaxation of a rotational or vibrational mode in gases occurs through collisions of molecules , transferring energy of an excited vibrational mode into heat.

But isn't it possible that an excited vibrational mode in molecules relaxes directly by emission of radiation...?

I think that is what happens in thermal radiation heat transfer.

According to Wikipedia, "Thermal radiation is the emission of electromagnetic waves from all matter that has a temperature greater than absolute zero.[3] It represents a conversion of thermal energy into electromagnetic energy. Thermal energy results in kinetic energy in the random movements of atoms and molecules in matter. All matter with a temperature by definition is composed of particles which have kinetic energy, and which interact with each other. These atoms and molecules are composed of charged particles, i.e., protons and electrons, and kinetic interactions among matter particles result in charge-acceleration and dipole-oscillation. This results in the electrodynamic generation of coupled electric and magnetic fields, resulting in the emission of photons, radiating energy away from the body through its surface boundary.".

So I believe that in case of thermal radiation, some vibrational states directly convert their vibrational energy into electromagnetic energy, without lowering vibrational energy through molecular collisions, and generating heat.

Am I correct..?

 

[DrClaude]

I have read that relaxation of a rotational or vibrational mode in gases occurs through collisions of molecules , transferring energy of an excited vibrational mode into heat.

But isn't it possible that an excited vibrational mode in molecules relaxes directly by emission of radiation...?

Yes, but there are many things to consider. As you are mentioning gases, I imagine that you are concerned about small molecules, like di- and triatomic.

The relaxation of rotation by emission of radiation, which corresponds to microwave radiation, can be relatively slow. In many cases, collisional relaxation will happen before emission of radiation.

Relaxation of vibrations of results in the emission of infrared radiation, which is the principle on which IR cameras operate. But in homonuclear diatomic molecules, like N₂ and O₂, vibrational excitation cannot decay by direct emission of photons. The same is true of some vibrational modes in molecules like CO₂ and CH₄. Notice that you don't have to worry about air when using an IR camera: it is mostly transparent to IR radiation.

 

[ns_phonon]

But in homonuclear diatomic molecules, like N₂ and O₂, vibrational excitation cannot decay by direct emission of photons. The same is true of some vibrational modes in molecules like CO₂ and CH₄.

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First of all thanks for replying DrClaude. But can u please specify the reason as to why some vibrational modes cannot decay by direct emission of photons.

I believe that in molecules like CO₂ , whenever vibration takes place, it leads to acceleration/oscillation of molecular dipole moment. So a vibrating molecule(electric dipole), "no matter in what energy state it is vibrating" must radiate energy in the form of radiation, as we are familiar with the principle of electrodynamics that a vibrating/accelerating charge is an emitter of radiation.

But as you say here, ''The same is true of some vibrational modes in molecules like CO₂ and CH₄. ''. I don't get it completely.

It would be a great help to me of you if u explain, "why some vibrational modes cannot decay by direct emission ?" .

"THANKS FOR READING"​

 

[DrClaude]

But can u please specify the reason as to why some vibrational modes cannot decay by direct emission of photons.

Using quantum mechanics, you can calculate the transition rates for different processes in a molecules. Some of these rates turn out to be exactly zero, and this is related to something called selection rules. To put it simply, in the case at hand, for a vibrational transition to be possible ("allowed"), the dipole moment of the molecule must be different in the initial vibrational state and the final vibrational state. Take for instance CO, which has a permanent dipole moment due to the fact that oxygen is more electronegative than carbon. The magnitude of this dipole moment depends on the distance between the two atoms. If CO is in the first excited vibrational state, it is slightly more elongated, and has a higher dipole moment, than in the ground state. Th decay of vibrational excitation can thus happen by simple emission of a photon.

In the case of a homonuclear diatomic molecule (in the ground electronic state), the dipole moment is strictly zero, due to the symmetry, whatever the vibrational state. Therefore, the dipole cannot change, and the molecule cannot emit a photon.

I believe that in molecules like CO₂ , whenever vibration takes place, it leads to acceleration/oscillation of molecular dipole moment. So a vibrating molecule(electric dipole), "no matter in what energy state it is vibrating" must radiate energy in the form of radiation, as we are familiar with the principle of electrodynamics that a vibrating/accelerating charge is an emitter of radiation.

That is a classical picture, which doesn't hold in the quantum world of atoms and molecules. When it is in a defined vibrational state, the atoms in the molecule do not oscillate in the classical sense. As such, the molecule as a fixed dipole moment when it is in a given state. But your analogy is not completely bad, as I explained above. Emission of light is only possible when there is a change in the dipole moment.

But as you say here, ''The same is true of some vibrational modes in molecules like CO₂ and CH₄. ''. I don't get it completely.

It is simpler to explain for CO₂. In its ground state, the molecule is linear and symmetric, and therefore there is no dipole moment. When it is excited, it can vibrate in three different ways: (1) the two CO bonds can stretch and contract in phase; (2) one bond can stretch while the other contracts, and vice versa; (3) the angle between the bonds can change.

In case (1), because the symmetry is preserve at all times, the dipole moment is always zero, and this mode can't decay (or be excited) by radiation. For the other two cases, the dipole moment will change, because the symmetry is broken and the distribution of charges changes such that a dipole is created. Therefore, vibrational transitions in these modes lead to changes in the dipole moment and radiation can be emitted.

For CH₄, the thinking is similar: only asymmetric vibrational modes will be infrared active.

 

[ns_phonon]

"Great explanation DrClaude. Now I understand completely what u meant to tell. Thanks...! I don't normally appreciate but this time I have to. U took time to read this thread and explained so well...!"
Fantastic knowledge u have...

THANKS A LOT...!​

:thumbs::thumbs:

 

[Bill_K]

Emission of light is only possible when there is a change in the dipole moment.

Dipole radiation is not the only kind. Light can also be emitted as quadrupole or higher.

 

[DrClaude]

Dipole radiation is not the only kind. Light can also be emitted as quadrupole or higher.

Indeed. I didn't want to go into too much details of other processes, as the original question focussed on collisional vs radiative relaxation, and higher-order emission processes have very long relaxation times.

 

[ns_phonon]

I was thinking about what you said. I have one question that do symmetric molecules like N₂ AND O₂ also absorb electromagnetic radiation,as they are symmetric molecules and not dipoles. So, how can they also absorb radiation and convert absorbed energy into vibrations(heat) in the same way like asymmetric molecules like CO₂?

 

[DrClaude]

I was thinking about what you said. I have one question that do symmetric molecules like N₂ AND O₂ also absorb electromagnetic radiation,as they are symmetric molecules and not dipoles. So, how can they also absorb radiation and convert absorbed energy into vibrations(heat) in the same way like asymmetric molecules like CO₂?

They can't. The calculations are the same for emission and absorption, so if one is zero, so is the other. (Again, with the caveat that higher-order processes are possible, but rare.) That is the reason why N₂ and O₂ are not greenhouse gases: they can't capture the IR emitted by the ground, while H₂O, CO₂ and CH₄ all are.

 

[ns_phonon]

But don't N₂ and O₂ absorb u.v radiation by going to higher electronic level and convert energy of electronic excitation that into heat(vibration of molecules) instead of simply re-emitting the absorbed photon by falling back again to ground electronic level

 

[DrClaude]

But don't N₂ and O₂ absorb u.v radiation by going to higher electronic level and convert energy of electronic excitation that into heat(vibration of molecules) instead of simply re-emitting the absorbed photon by falling back again to ground electronic level

I didn't understand that you were considering also electronic transitions now, not just pure vibrations.

A molecule can indeed absorb UV light, and end up in excited electronic states, with some vibrational excitation. The molecule will then relax, by emitting a photon, to the ground electronic state, where it will most probably not be in the ground vibrational state, but some excited state.

By the way, I din't point this out before, but you seem to equate heat with vibration of molecules. It is more complicated than that, with heat corresponding to all possible motion: translation, rotation, and vibration. At room temperature, most small molecules don't have enough energy to be vibrationally excited.

 

[ns_phonon]

Oh I understand now. I asked because I had read that most of the molecules have a tendency to convert energy of electronic excited state into vibrations of molecules. In that article it was also written that unlike some isolated atoms find in rarefied gases, most molecules convert energy of electronic excitation in vibrations.

Thanks anyway.I have now understood almost everything. I will try to send a link of that material if I can find that material as I have misplaced that material. It is a very simple and easy material which though not completely but explains almost all the basics of "How light Interacts With Matter"

 

[ns_phonon]

Yay DrClaude. I found that file.It was hidden in my computer. It is a nice article, if you wish to read.Google has now removed this article.