De-excitation of a molecule

Question

In our chemistry lecture today on UV/vis-spectroscopy, we discussed the origin of colour in conjugated molecules due to electronic transitions from lower-energy molecular orbitals to higher-energy molecular orbitals. One of the examples we talked about was the origin of the orange colour of carrot. The lecturer mentioned that carrots appear orange as it contains conjugated molecules that absorbs blue light, resulting in the complementary orange colour being emitted. This blue light is used to excite electrons from π MOs to π^* MOs in the molecule. One of my classmates asked what happens to the excited electron when it falls back to the ground state, wouldn't it cause blue-light photons to be emitted?

To this, the lecturer replied that perhaps, there is de-excitation of the electrons through other pathways, such as through molecular vibrations (i.e. kinetic energy of the molecules). However, he also was not so sure about the matter. Thus, I would like to seek clarification on how these excited electrons de-excite after initial excitation.

Answer 1

After radiation is absorbed and the electron is at an excited state in the molecule there are several pathways for de-excitation to occur (see fig.). The pathway of choice depends on its rate, ie how fast it can happen. It turns out that the fastest de-excitation pathways are radiationless (wavy arrows in fig) such as internal conversion (IC) that happens through vibrational relaxation of the molecule and external conversion (EC) where the molecule returns to its ground state through collision with solvent molecules. This is why matter usually doenst emit light after absorbing it. There are instances however where this is actually possible because it happens that the radiationless pathways are not the most efficient way for the molecule to return to its ground state and that is when fluorescence (FL) and phosphorescence (Ph) occur.

Answer 2

Simply put, the reason the emitted color is different from the absorbed one is because there is a loss of vibrational energy that takes place in addition to the photon emitted by "falling back to ground state". This site has a good detailed explanation: https://micro.magnet.fsu.edu/primer/lightandcolor/fluoroexcitation.html

Essentially, the "π MOs to π∗ MOs" description is a simplification that omits the existence of vibrational states. For another visual, here is a diagram taken from Wikipedia. The loss of vibrational energy is shown as the red arrows.

Answer 3

What your lecturer (as many lecturers, unfortunately) did not tell you is that the de-excitation pathways depend on the substance and the wavelengths you consider.

You can't know beforehand if a substance will be fluorescent, phosphorescent, or de-excitates with or without radiations. That is why only some substances can be used to make lasers (they need to support stimulated emission of radiations)