Talk:Absorption spectroscopy

Not all absorption techniques excite electrons; IR, for instance, only changes a molecule's vibrational energy level. It sounds like the author is also trying to include X-ray crystallography within absorption spectroscopy as well; that is also incorrect. Only IR spectra contain the "fingerprint" region mentioned in this article; UV spectra really don't provide such unique information. The analogy between calibration curves and Hooke's law is tenuous at best, and would be very confusing to someone not familiar with both topics. This page is listed under "pages needing expert attention," so my little list of inaccuracies might be a good place to start. Mihovil 19:03, 30 August 2006 (UTC)[reply]

I think that all absorption techniques affect the electrons. Unless one is trying to involve the nucleus, which would require quite high energy radiation, most of the techniques of general interest affect the electrons. The energy levels between which the electron switches are determined by the various energy modes available to the molecule (these include the rotational, vibrational and electronic energy level). The way the EM radiation interacts with the molecule is by affecting the electron cloud distribution around the nucleus (which according to Born-Oppenheimer approximation are fixed).

As far as "fingerprint" is concerned. Well the "author" hasn't made it specific. He is refering to the spectrum on the whole of which IR spectrum is a part. So I guess there is no problem in that. Also UV spectra is usually determined by the electronic configuration of the molecule under investigation. Since the configuration can change, how can you say that the UV spectra does not provide any unique information.

I agree with the fact that the analogy is bit confusing and can be put it in simple terms. Thank you for your suggestions. 69.180.9.141 00:37, 31 August 2006 (UTC)[reply]

Of course all spectroscopies "affect" the electron cloud; I never disputed that. But IR radiation absolutely does not promote electrons to higher energy levels, UV does. When a photon of infrared radiation is absorbed by the absorbing species, said species is promoted from its (usually) ground vibrational state to a higher vibrational state. Yes, the electron cloud will be polarized, but that is a matter for the Raman spectroscopist, and Raman in not an absorption technique. UV spectra of most organic molecules are basically broad Gaussian blobs, and it is very common for different molecules to have UV absorption spectra which are centered around the same wavelength. This means that it will be difficult or impossible to distinguish them on the basis of their UV spectra alone. For this reason, UV is not considered to give any "fingerprint" information. Mihovil 02:05, 31 August 2006 (UTC)[reply]

Well even in case of IR, it's only the electron cloud that is affected. Only difference is that in this case we are not concerned with the atomic orbital, but the molecular orbitals. Also IR does not have enough energy to affect the nucleus of most of the atoms, so one cannot say that the nucleus is absorbing the energy from the radiation. One can also think of it as absorption of radiation in IR range affects the vibration mode of the molecule. When the molecule absorbs the IR radiation, it starts vibrating at higher amplitudes. This is because the electrons are displayed to higher energy orbitals, so there is more repulsion between the nucleus and hence the amplitude is higher.

When we talk of UV exciting the electrons to higher energy levels, it is usually the transfer between the atomic orbitals, where the energy difference between the levels is more compared to molecular orbitals.

Also when we talk of broad absortion septrum, it is because we are not resolving it enough. Given ehough resolution power, we can distinguish between the different lines, only limitation will be due to the line broadining mechanisms. So even rotational spectrum can distinguish between species, but the problem with that is multiple species can have overlapping rotational spectrum or some of them have no rotational spectrum at all, so one needs the vibrational spectrum or the vibrotational spectrum to make sure. I agree that just using one kind of spectra may not be sufficient to given enough information about the molecule. 69.180.9.141 03:56, 2 September 2006 (UTC)[reply]

At this point, I'm really not sure if you are actually reponding to any of my comments, or just trying to demonstrate your knowledge of spectroscopy. Mihovil 00:24, 4 September 2006 (UTC)[reply]