Once the system absorbs a photon of light and successfully excites an electron to an excited state, what happens to this excited electron? As with any system, molecules will slowly change from a high energy state to a lower stable state. The extra energy in this excited state can be lost in several ways, which include: a chemical reaction, non-radiative relaxation, and radiative relaxation.
In the case of a chemical reaction, the molecule is prone to undergoing a chemical reaction since it is already in an excited state. For non-radiative relaxation, the molecule collides with other molecules in its surroundings and energy is slowly transferred away. In the previous module we saw how the molecule converts from an excited state to a ground state in a process called internal conversion.
In cases of radiative relaxation, the energy between an excited state and the ground state is lost as a photon. This photon is always of lower energy than the photon that was absorbed. One example of radiative relaxation is the process of fluorescence.
Fluorescence begins with absorption of a photon resulting in an excited state. The electron will then fall to the lowest vibrational energy within that excited state. This happens relatively quickly (10-11 s). The transition from the excited state to the ground state through the loss of a photon happens on the order of nanoseconds (10-9 s), which is much slower than other processes. An example of an absorption spectrum from a fluorescent event is given below.
Photon
When moving from the excited state to the ground state, the electron follows the Franck-Condon principle to determine which vibrational state it enters. Essentially, the emission of the photon is nearly instantaneous compared to molecular vibration.
Click the system to the left to start, you can choose when to promote the electron and relax the electron. How does the photon change with respect to the vibrational state it falls too? How does this compare to its excitation photon?
