Download Notes on Class QM Foerster SpectraII - Optical Spectroscopy | PHYS 552 and more Study notes Optics in PDF only on Docsity! QMFoersterII 1 ClassQMFoersterSpectraII How about the intensities of the transition moments, 0v , 1P v′′ ′ (refer to the lecture: “ClassQMFoersterSpectra”)? Which terms contribute to the amplitude of the transition? There are two main points to take into account. Frank Condon 1) Only a small number of the 0v , 1P v′′ ′ terms contribute because of what is called the “Frank Condon” effect. The electron transition is much faster than the possible motion of the nuclei. So the “end state” of the transition has the nuclei in the wrong position. There are only a few “end states” with the nuclei in the same approximate position as in the “beginning state”. There is a narrow distribution of end states that are possible due to the uncertainty principle of quantum theory. Boltzmann Occupation in Beginning State 2) The occupation probabilities, (refer to ClassQMFoersterSpectraI) v and vg g′′ ′ , are Boltzmann distributions. The vibration energies do not exceed kT by very much, so the higher vibrational energy levels are very sparsely populated. In absorption, only the 0v′ ≠ values appear. In emission, only the 0v′′ ≠ values appear. After reaching 0v′′ > and 0v′ > states, they very rapidly lose vibrational energy and decay in 10-12 sec to 0v′′ = or 0v′ = states. Through higher temperature, you set overlapping of the absorption and emission spectrum. **Homework**: Draw the diagram representing the absorption and emission spectra for higher temperatures (i.e., when kT is larger than the difference between the vibrational levels.). QMFoersterII 2 Also, the more the electronic configurations of the two states differ (excited and ground), the more the higher vibrational states of the excited electron state will be produced in the electronic transition. This follows from the Frank-Condon principle. The thermal equilibrium to the lowest vibrational state will always take place unless the sample is a very, very dilute gas, with no collisions (vacuum). Essentially, this is always true for the ground state with absorption. In gases one can get emission from higher vibrational levels of the excited state. This means the emission depends in the excitation energies. Then it can lead to emission of the same wavelength as the absorbed light (Resonance Fluorescence Spectrum). Rotational levels also then start to split the vibrational levels. [Calculate the Absorption Coefficient] Derivation from the equations for ( )an v of the absorption coefficient. ( ) ( ) ( )-M, v,x 10 v dxv x dx εσ σ+ = position from the front of the cuvettex = ( )vσ = the spectral light energy density mol literM ≡ ( ) ( ) ( ) , ln -ln10 M ln v,x v x dx dv dx σ σ σε σ σ ⎡ ⎤+ +⎡ ⎤= =⎡ ⎤⎢ ⎥ ⎣ ⎦ ⎢ ⎥⎣ ⎦⎣ ⎦ ( )ln ln ln lnd dd σ σσ σ σ σ σ σ σ = + − = + − = So, [ ]ln10 Md dxσ ε σ= Note we have dropped writing the frequency dependence of ( )σ ν and ( )ε ν .
