Chemistry Reference
In-Depth Information
electrons. This will alter the bonding forces discussed above; in particular, the optimum
geometry will be different in the electronic excited state to that in the ground state. The
excited state is higher in energy than the ground state, and so the excited-state Morse
potential is drawn above the ground state in Figure 6.9a.
The difference in energy levels for the electronic states is usually much greater than
that for the molecular vibrational states; hence the need for higher energy photons for
the excitation! This also means that the two Morse curves are separated further than the
spacing between the vibrational energy levels they contain. We think of each electronic
state as having a subset of vibrational states. To define the state of the molecule fully we
would need to specify the electronic and the vibrational states.
The effect of excitation on the geometry of the molecule is also represented in
Figure 6.9a. In the electronic excited state the nuclei will be less strongly bound than
the ground state, so that the optimum structure will tend to be expanded by the excitation.
This is shown in the figure by placing the optimum of the potential in the excited state to
the right of that in the ground state.
The Absorption Process
Since electrons have much lower mass than the nuclei of the molecule, the photon absorp-
tion and electron transition can occur much faster than the molecular geometry can
respond. Hence, the system will 'arrive' in the excited state at the ground-state molec-
ular geometry, as shown by the vertical arrow in Figure 6.9a; indeed, such a process is
often referred to as a vertical transition. The idea that electronic transitions occur without
an initial change in the molecular geometry is called the
Franck-Condon principle
.
A treatment of the excitation process based on a quantum model is given in Appendix 7.
Here, we use the result that, because the excited state Morse curve is shifted to the right
in Figure 6.9, the vertical transition implies that it is relatively easy for the molecule to
have a vibrational state with a quantum number greater than zero in the electronic excited
state. After absorption, the electronic structure will relax back to the ground state and
must release the excess energy. The most probable process is to emit a single photon that
allows relaxation directly back to the ground state. It is easy to see that this photon will
have the same energy, and so the same frequency, as the absorbed photon (Figure 6.9a).
This process is referred to as Rayleigh scattering, and the direction in which the photon is
emitted is random.
Raman spectroscopy depends on alternative relaxation processes to Rayleigh scattering,
for which the energy released as a photon by a molecule during relaxation back to the
ground state is different to the energy of the exciting photon. The difference occurs because
emission is accompanied by a change in the vibrational state of the molecule.
In Figure 6.9b we see that the molecule starts in an
n
= 0 vibrational state but is returned
to the
n
1 state. The difference between the exciting photon energy and the photon emit-
ted in the relaxation process is simply the difference in energy between the two vibrational
levels;
h
=
ν
v
is used to emphasize that this is the frequency of
the molecular vibrational mode. In the illustration of Figure 6.9b the emitted photon is
lower in energy than the exciting photon. This is observed as a band shifted downward in
frequency relative to the Rayleigh scattered light; such processes are referred to as Stokes
absorptions.
ν
v
. Here the subscript 'v' in
