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I am currently studying computational chemistry and I am trying to understand the UV-VIS absorption of molecules. I know that UV-VIS absorption is electronic excitation and requires a specific energy.

What I don't understand is why a molecule can not absorb a shorter wavelength as it has more energy? A slightly higher energy causes vibrational excitations which is then released through vibrational relaxation (releasing heat). Isn't it the case when a very shorter wavelength is used so that the electrons in the molecule are excited from more lower MOs (lower energy) to the higher ones? Isn't it still absorption? It is supposed to be by definition?


absorption

Molecule can absorb a photon that has "red" amount of energy, it can also absorb "green" and "blue" amounts. But why can't it absorb anywhere between red and green or more basically isn't it supposed to absorb anything more than "red"? Just to be clear, by red I mean the energy difference between $n = 1$ and $n = 2$.

Also, I am already aware of the fact that, for example, any energy between $n = 2$ and $n = 3$ can not be emitted fully as some of them will be lost for rotational and vibrational transitions until the energy hits the $n = 2$ state but I am not expecting this to stop a molecule from absorbing that light.

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  • $\begingroup$ You should quantify what you mean for shorter wavelength... $\endgroup$ – G M Mar 25 '14 at 9:25
  • $\begingroup$ Lets say the molecule has maximum absorptivity at 400nm and I am sending a light beam having a 300nm wavelength. $\endgroup$ – theGD Mar 25 '14 at 19:19
  • $\begingroup$ You can have (depending on the symmetry) Raman spectroscopy, where light of a much shorter wavelength excites a vibrational change and gets scattered (and shifted in wavelength) as a result. $\endgroup$ – Nick Mar 26 '14 at 16:46
  • $\begingroup$ If I understand this correctly, you're wondering why the system can't absorb a photon that's slightly larger than the n1->n2 gap, then vibrationally relax to n2? I think the answer is that it can happen. I overheard something about electronic Raman scattering the other day, which would be something like what you described, but trying to read the material on it just hurts my head too much for me to delve into it right now. $\endgroup$ – chipbuster Mar 27 '14 at 2:56
  • $\begingroup$ @chipbuster Yes I am wondering that. $\endgroup$ – theGD Mar 27 '14 at 23:27
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Classical approach

Absorption occurs when there are "oscillators" that can resonate with the frequency of the electromagnetic wave. In classical weave physics, these frequencies are called natural frequencies.

These oscillators can involve electrons (UV, VIS, XR), bonds (MIR, FIR), nuclei (gamma rays). They are able to absorb electromagnetic waves changing their spins(NMR, ESR) orientation (microwave), configuration (infrared) nuclear configuration ($\gamma$ rays) and finally electronic distribution (VIS, UV, X-Ray).

Of course, not all the materials have oscillators that absorb in all the spectrum. For wavelength between 10 nm and 100 pm the oscillators, that mainly involve electrons in these regions, have to absorb energy between $10^{7}$ and $10^{9}$ J/mol without being ionized, in this case only the inner electrons of some atoms (almost always metals) absorbs these wavelengths see XAS, so in our organic world seems that most compounds are X-Ray transparent but in fact there are a lot of inorganic compounds that absorb X-Ray.

When radiation is absorbed there are many ways to release his energy(see figure 1). We can divide them into two great categories radiative relaxation and non-radiative relaxation. In fact radiative relaxation is most common when we are dealing with shorter wavelength because with more energy is possible to excite an electron to a higher level and this can sometimes re-emit radiation when it goes from a higher level to the ground level, this as you know, is called fluorescence and is very common when you deal with UV, and X-Ray (see X-Ray Fluorescence spectroscopy). Eventually, there is also non-radiative relaxation process that are the main processes that occur from VIS (note that however is possible a VIS-NIR luminescence for some compounds) to the other spectral region with lower wavelength. enter image description here

Quantum approach

Compound energy levels are quantized, so compounds can absorb only some discrete energy values. These values are determinate by transitions that respect strictly selection rules. If you know the energy of the transition, from Plank relation:

$$E=\frac{ch}{\lambda}$$

you can see that for every energy value there is a specific wavelength, so this means that compounds can absorb only at specific discrete wavelength. However if you see a UV-VIS spectrum you will see a continuous spectrum because roto-vibrational changes affect the energy levels of the compounds involved.

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  • $\begingroup$ I have passed all the courses including Quantum Chemistry, Instrumental Analysis, Introduction to Computational Chemistry etc... I am quite familiar with those concepts but I have never been thought about resonance part of the story. So still I don't understand why a resonance is necessary since diffrent waves can destructively and constructively interfere. $\endgroup$ – theGD Mar 25 '14 at 19:32
  • $\begingroup$ Am I missing something very fundamental? $\endgroup$ – theGD Mar 25 '14 at 19:33
  • $\begingroup$ @GunDeniz I don't know, I'm not a physicist and I never done such courses, but how can you explain absorption without resonance? If scattering is due to the constructive and destructive interference how can you distinguish it from absorption? If you notice Hydrogen emission spectrum is the "opposite" of Hydrogen absorption spectrum, this mean that absorption occur at some frequencies, we could say "natural frequencies", that are involved in the process of emission. So normally are allowed only oscillation with these freq. for H2 atoms, can you think that these oscillator can act backward? $\endgroup$ – G M Mar 25 '14 at 21:35
  • $\begingroup$ I can distunguish because while absorption is the part where the molecule receives energy, emission is another story because after the energy is absorbed the molecule undergoes a lot of things like singlet to singlet or singlet to triplet transitions. $\endgroup$ – theGD Mar 26 '14 at 9:09
  • $\begingroup$ Emission makes perfect sense about a spesific wavelength but absorption doesn't. $\endgroup$ – theGD Mar 26 '14 at 9:11
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A cryogenic atom or molecule in vacuum is a narrow bandwidth oscillator (particle in a box). If you do not hit the right note it will not resonantly electronically ring for you.

If it is polyatomic, warming it will introduce rotational (microwave) then vibrational (infrared) degrees of freedom, broadening lines and introducing new ones. Put it in a fluid medium and the random jingle jangle of collisions broadens things further. Put it in a solid and modes may be physically eliminated, e.g., whole molecule rotation. Water's pale blue color is summed infrared vibronic modes, the only visible color caused by vibration absorption.

The higher the frequency, the more localized the photon. At very high energies photon collision is literally physical. Things get collisionally knocked out rather than resonantly pumped up.

A heavy atomic nucleus has density ~$2.3{×10^{14} \ce{g/cm^3}}$. Add a muon 206.768 times the mass of an electron into U-238. It decays to orbit at a fair fraction of lightspeed just inside the nuclear surface. Selection rules for hadrons and leptons are different. They may not see each other. Triprismane, $\ce{C6H6}$, sits 90+ kcal/mole above benzene at room temp. It won't fragment until increased temperature adds another 33 kcal/mole - and it doesn't crack to benzene. Reaction path is defined by conservation of orbital symmetry. Tetra-tert-butyl tetrahedrane does not thermally fragment. The insanely strained molecule is so symmetric that its lowest energy path is synchronous transformation into tetra-tert-butyl cyclobutadiene' at 130°C!

enter image description here

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