Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equilibrium bond lengths for excited molecules

  1. Mar 31, 2013 #1
    When making the Morse Potential graphs of diatomic molecules going through a transition, the excited state sometimes has a different equilibrium bond length. A visual example is here: http://en.wikipedia.org/wiki/File:Franck-Condon-diagram.png

    I was wondering how the average bond length of the excited state can be calculated ab-initio, or how it can be determined from spectroscopic data.
     
  2. jcsd
  3. Apr 1, 2013 #2

    cgk

    User Avatar
    Science Advisor

    In the ab initio world, bond lengths of excited states are basically determined exactly as bond lengths of ground states: You take some initial geometry, calculate a wave function, and based on that the gradient of the excited state total energy with respect to the nuclear positions. You then update the positions to reduce this gradient. This process is iterated until the gradient becomes sufficiently close to zero; you then have your equilibirum geometry from which the bond lengths can be directly read off. The only difference to a ground state geometry optimization is that the energy and gradient of an excited state need to be taken, not of the ground state. This often requires using some different electronic structure methods, because not all methods can describe excited states (e.g., TD-DFT instead of Kohn-Sham DFT, CC2 instead of MP2, MCSCF instead of HF, etc). But conceptually there is no difference.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook