Why molecules absorb light?

  • #1
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Main Question or Discussion Point

We know that molecules absorb light based on their electronical, rotational, and vibrational transitions and the governed transition rules (so that each molecule has its own "fingerprints"). But I do not know why it happens?

Lets simplify it a little:

A methane molecule, for example, is charge-less (the net summation of p+e=0) while we know CH4 absorbs the photons with 3.3um wavelength. Now the question is why/how this happens? Why electromagnetic wave (photons) interacts with a charge-less "thing"?

Any help and/or advice is appreciated,
S
 

Answers and Replies

  • #2
DrClaude
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Molecules are not charge-less even though they are globally electrically neutral. They are still composed of positive nuclei and negative electrons. Changing the vibrational or electronic state will change the charge distribution, and therefore these can couple with the electromagnetic field.

The example of CH4 is a good one to explore some subtleties of this. Generally speaking, a vibrational transition will occur when it leads to a change in the dipole moment. If a molecule is polar (e.g., CO), then it can absorb IR light and be excited vibrationally, as this will necessarily change the dipole. However, a non-polar molecule (e.g., N2, O2) will not absorb in the infrared, as the charge symmetry is preserved even with vibrational excitation (the dipole moment is always 0). In the case of CH4, which is non-polar, certain modes will be active in the IR, such as an symmetric stretch, where some bonds contract at the same time others lengthen. But the purely symmetric stretch (breathing mode), where all bonds vibrate in phase, is not active (there will be no absorption of photons that can excite this mode).

Note that this is all true to first order only. There can be additional transition mechanisms (such as electric quadrupole, magnetic dipole, ...) or other processes (e.g., Raman) that can excite such vibrations. But dipole transitions are by far the strongest transitions in molecules, and the main reason why a molecule will absorb (or not) photons at a certain wavelength.

Rotations are a bit different, as they depend on the relative orientation of the molecule and the field. But it is again due to the coupling of the dipole (or induced dipole) of the molecule with the EM field.
 
  • #3
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Molecules are not charge-less even though they are globally electrically neutral. They are still composed of positive nuclei and negative electrons. Changing the vibrational or electronic state will change the charge distribution, and therefore these can couple with the electromagnetic field.

The example of CH4 is a good one to explore some subtleties of this. Generally speaking, a vibrational transition will occur when it leads to a change in the dipole moment. If a molecule is polar (e.g., CO), then it can absorb IR light and be excited vibrationally, as this will necessarily change the dipole. However, a non-polar molecule (e.g., N2, O2) will not absorb in the infrared, as the charge symmetry is preserved even with vibrational excitation (the dipole moment is always 0). In the case of CH4, which is non-polar, certain modes will be active in the IR, such as an symmetric stretch, where some bonds contract at the same time others lengthen. But the purely symmetric stretch (breathing mode), where all bonds vibrate in phase, is not active (there will be no absorption of photons that can excite this mode).

Note that this is all true to first order only. There can be additional transition mechanisms (such as electric quadrupole, magnetic dipole, ...) or other processes (e.g., Raman) that can excite such vibrations. But dipole transitions are by far the strongest transitions in molecules, and the main reason why a molecule will absorb (or not) photons at a certain wavelength.

Rotations are a bit different, as they depend on the relative orientation of the molecule and the field. But it is again due to the coupling of the dipole (or induced dipole) of the molecule with the EM field.

Thank you for all this helpful information.

Just one more question: How can we relate absorption phenomenon to the "resonance" which happens when the frequency of applied external source is very close to the natural frequency of the vibrating objects. In the case of CH4 , for example, can we say the CH4 molecule is vibrating with the frequency of c/3.3um (c is the speed of light) and when the light with this frequency hits the molecule, resonance occurs and so the energy of the molecule increases?
 
  • #4
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Can anyone please give me an advice on the above/below question:

Can we relate the molecular vibration to resonance effect? i.e. Does it mean that if CH4 absorbs 3.3 um light, it is really vibrating with the frequency of c/3.3 um (c is the speed of light) and that is why higher or lower wavelengths are not absorbed?
 
  • #5
hilbert2
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A substance that is in the gas phase absorbs EM radiation of frequencies that are equal to the vibration frequencies of its normal modes. In liquid or solid phase there aren't similar sharp spikes in the absorption spectrum, because the interaction between the molecules in condensed matter affects the absorption process. Sometimes there is also rovibrational coupling which means that molecules that are in different rotation states vibrate at slightly different frequencies, this can happen even in the gas phase.

A photon of a certain wavelength can also cause stimulated emission of a photon of same frequency from the molecule.
 
  • #6
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Thank you for your explanation. With all due respect, I think this is not the answer of my question though.

Maybe I should rephrase it:

All I am asking is if CH4 molecule (for example) is vibrating with the frequency of light (in this case, c/3.3 um=9.1*10^13 Hz) that is absorbed or light frequency and vibration frequency are two different aspects of absorption process?
 
  • #7
hilbert2
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All I am asking is if CH4 molecule (for example) is vibrating with the frequency of light (in this case, c/3.3 um=9.1*10^13 Hz) that is absorbed or light frequency and vibration frequency are two different aspects of absorption process?
Yes it is. That's what I said.
 
  • #8
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