|WikiProject Physics||(Rated Start-class, Mid-importance)|
|WikiProject Spectroscopy||(Rated Stub-class, Mid-importance)|
How about Lorentian?
Why the profile of the spectral line of absorption is so asymmetric? Shouldn't it be Lorentian, or Gaussian or Foight's profile? dima 01:11, 24 July 2007 (UTC)
I think the picture is of a generic molecular absorption and its fluorescence. The reason they aren't both a more recognizable lineshape is that we are not seeing all of the fine structure. We see a broad envelope of an electronic transition from one electronic state to another. In reality the system is changing electronic, vibrational, and rotational states at the same time. So if the spectra were taken with a higher resolution spectrometer we would see hundreds of lines, each of which had a recognizable line shape. Man It's So Loud In Here 21:16, 16 August 2007 (UTC)
The electrons remain in the excited state for about 10−8 seconds. This number varies over several orders of magnitude. In view of the second sentence, what does the first mean? Is that number at the bottom, middle, or top of that huge range? I would fix it myself if I knew the answer. --Vaughan Pratt (talk) 04:26, 3 December 2010 (UTC)
- 10−8 seconds is standard for fluorescence. Seeing as phosphorescence can last seconds, and internal conversion is normally treated as many times faster than fluorescence (Kasha's rule), I'd say we're talking mid-range here. Tomásdearg92 (talk) 18:55, 8 February 2014 (UTC)
WikiProject class rating
This article was automatically assessed because at least one WikiProject had rated the article as stub, and the rating on other projects was brought up to Stub class. BetacommandBot 10:04, 10 November 2007 (UTC)
only the anti-stokes portion even mentions phonon coupling as a source/sink of that gain/loss of energy in an (anti)stokes shift. There should at least be a mention of the usual topics, vibronic excitation, vibrational relaxation to electronically excited state's ground vibronic state (the ground vibrational state of the excited electronic manifold). radiative deactivation from this ground vibrational state down to (usually) some vibrationally excited state of the GROUND electronic state. the overall loss of energy is associated with those internal conversions between the excited vibrational states within each electronic manifold. basically the system transmits a small percentage of the overall energy into the surrounding medium by coupling vibrational modes (phonons) and exciting the surrounding medium. photon of energy E excites the system, which transfers a small percentage in the excited state, and decays to the ground electronic state (but not the ground vibrational state, assuming usual born-oppenheimer/frank condon principles), which also then transmits some of its energy to the surrounding medium. the overall effect is that the emitted photon is of a tiny bit less energy (usually the difference is very small, and is comparable to a far infrared or even radio photon). even further clarification can be given to engineers, more knowledgeable about the terminology of semiconductors. stokes shifts are extremely similar to Indirect-Bandgap transitions, in which the lattice must donate (usually) a quantum of lattice momentum to ensure a transition at a given temperature. in the same fashion, a stokes shift in dense media occurs when the system DONATES lattice momentum, and an anti-stokes shift is literally almost an identical process (absorption of lattice momentum), but studied from a different perspective.220.127.116.11 (talk) 21:48, 5 April 2011 (UTC)
Self excitation and anti-shift
Looking at this page and at Kasha's rule, it seems that in general emission spectrum shouldn't depend on excitation frequency. Does it follow that the emission resulting from self-excitation (excitation due to emission in the overlap of the two curves in the graph) will generally have the same spectrum as emission from a higher excitation frequency? —Ben FrantzDale (talk) 20:44, 9 August 2011 (UTC)
- Kasha's rule states that the emission will occur in appreciable yield only from the lowest energy excited electronic state. What it doesn't say is what vibrational or rotational state the system will end up in immediately after emission. Under the Born-Oppenheimer approximation the various types of energies - electronic, vibrational, rotational, etc. - of molecules are treated separately. Because the patterns of vibrational energy levels of different electronic states are similar, the patterns of energy levels would also be similar. The absorption and fluorescence curves should, therefore, appear as mirror images. Tomásdearg92 (talk)
Relaxing by emitting a photon vs. losing heat?
The 2nd paragraph says: "One way for the system to relax is to emit a photon, thus losing its energy (another method would be the loss of heat energy)"