Thursday, August 20, 2009

Trying to shed light on organic LED's?

The system below is just one example of an organometallic phosphorescent system which is the basis of some LED's.

Some of the questions that interest me are:

Suppose we ignore spin-orbit coupling and the environment (either a solvent or thin film)
what are the quantum numbers and energies of the different excited states?

If the complex has C3v symmetry excited states can be labelled by A1, A2, and E (the irreps of the group) . Spin rotational invariance means the states will be either singlets or triplets.

What is their oscillator strength? Which states can be identified with features in the optical absorption spectrum? and with the emission spectrum?

How much mixing between the ligand centred (LC) and metal-to-ligand-charge transfer (MLCT) states occurs due to hybridization of metal orbitals and ligand pi and pi* orbitals?

How does the large spin-0rbit coupling on the metal core change things?
This will mix singlet and triplet states, particularly states with significant metal character.
It will also allow intersystem crossing (i.e., transitions between singlet and triplet states and visa versa) and will give "triplet" states oscillator strength.

How does the environment change things?
It breaks the C3v symmetry and tends to localise excitations on the ligands.

What determines the quantum efficiency of phosphorescence? What processes compete with phosphorescence? What is the dynamics following photoexcitation?

It must be something like:

A1 (ground state) + photon ->
A1 or E (depending on polarisation of light) "singlet" ->
"MLCT" "singlet" via internal conversion (and a conical intersection), symmetry? ->
"MLCT" "triplet" via spin-orbit coupling (i.e., intersystem crossing), symmetry? ->
"LC" "triplet" with some "singlet" character from spin-orbit coupling-> A1 (ground state) + photon

There are well-defined selection rules from group theory for spin-orbit coupling and internal conversion. This may help identify the symmetry of the different states.

Can it be made any simpler?

1 comment:

  1. Just a small point, the complex shown has C_3 symmetry, not C_3v. That means there are only two irreps, A and E.