Heisenberg Principle in Context of Absorption/Emission Spectroscopy

[DDTea]

Why do we get line spectra from absorption/emission of atoms, but band spectra from absorption/emission of molecules?

As I understand the Heisenberg Uncertainty Principle mathematically, some operators do not commute and as such, the order of measurements determines the results. In regard to spectroscopy, then, the energy of electronic transitions cannot be measured with arbitrary precision over a finite time period. That makes sense with say, UV/Vis spectroscopy of organic molecules like benzene: we see a big band around ~250 nm corresponding to the pi-->pi* shift. We can determine the energy of that shift more precisely by taking multiple measurements and averaging them.

What about atoms, though? Why do we see discrete line spectra? Wouldn't that violate the Heisenberg uncertainty principle to know *exactly* the energy of their electronic transitions over a finite period? Or rather, how exactly *do* we know the energy of their transitions? Are they the results of theoretical calculations or actual instrumental measurements?

 

[Ygggdrasil]

Molecules show much broader absorption and emission spectra because, in addition to electronic transitions, molecules have vibrational and rotational transitions as well. See the image here. The diagram shows two electronic states (S₀ and S₁). Within those states are shown different rotational/vibrational states. Now, at normal temperatures, nearly all of the molecules will be in the ground electronic and vibrational states. Upon absorption of a photon, the molecule will be excited to one of the vibrational states in S₁. As you can see, the energy of the transition depends on which vibrational state the molecule occupies after excitation. If the molecule goes into the v=1 state, the transition will have used a higher energy photon than a transition to the v = 0 state. The vibrational state that the molecule ends up in after absorption of a photon is a random process with probabilities dictated by the Franck-Condon principle. Therefore, because the molecule can be excited into any number of vibrational states with different energies, the molecule can absorb photons with a fairly wide range of energies.

After excitation, the molecule relaxes quickly to the ground vibrational state of S₁. This generally occurs through a non-radiative relaxation process called internal conversion. When the molecule emits a photon to return to the ground electronic state, the molecule can fall into any number of vibrational states in S₀. So, if the molecule ends up in the v=0 state after emission it will have emitted a photon of higher energy than a molecule that ends up in the v=1 state. Therefore, because the molecule can end up in a number of different vibrational states after emission of a photon and these vibrational states have different energies, molecules can emit photons with a relatively broad range of energies.

These broad spectra are not observed with atoms because atoms are symmetric and have no vibrational or rotational degrees of freedom.

Although the emission and absorption spectra of atoms are described as line spectra, they do still have finite width (dependent on the time over which emission/absorption are observed) and obey the Heisenberg uncertainty principle. The term "line spectra" merely reflects that the absorption/emission spectra of atoms are much narrower than the broad absorption/emission spectra of molecules.

 

[tkjtkj]

Molecules show much broader absorption and emission spectra because, in addition to electronic transitions, molecules have vibrational and rotational transitions as well.

Yours is the most clear, efficient, convincingly knowledgeable explanation i have ever encountered here on physorg.
We are grateful for your interest and skills in our world of science.
Thank you.