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Electronic Transitions

In the Born-Oppenheimer approximation the energy of a molecule can be presented as sum of electronic energy $E_{el}$, vibrational energy $E_{vib}$ and rotational energy $E_{rot}$ energy.
\begin{displaymath}
E = E_{el} + E_{vib} + E_{rot}
\end{displaymath} (65)

As a zero-th approximation, we describe the energy $E=\hbar\omega$ of a diatomic molecule as a sum of electronic, vibrational and rotational energies

\begin{displaymath}
\omega = T_{e} + \omega_e\left(v+\frac{1}{2}\right) + B_eJ(J+1),
\end{displaymath} (66)

where $T_e$ is the minimum of the potential curve. Each of the three terms in eq. (66) may be different for the lower and higher energy states and may be changes during an optical transition. The total transition frequency can be written as
$\displaystyle \Delta\omega$ $\textstyle =$ $\displaystyle T_e'-T_{e}'' + \omega_e'\left(v'+\frac{1}{2}\right)-
\omega_e''\left(v''+\frac{1}{2}\right) + B_e'J'(J'+1)-B_e''J''(J''+1)$  
  $\textstyle =$ $\displaystyle \omega_{el}+\omega_{vib}+\omega_{rot},$ (67)

where the lower energy state is denoted by a double prime (''), while the higher energy state is denoted by prime (').

Note, that the energy difference corresponding to the excitation of electrons $\Delta E_{el} =
\hbar \omega_{el}$ in this approximation is much larger that the energy difference corresponding to the molecular vibration $\Delta E_{vib} = \hbar \omega_{vib}$ which is large that the energy difference corresponding to the molecular rotations $\Delta E_{rot} = \hbar \omega_{rot}$:

\begin{displaymath}
\omega_{el} \gg \omega_{vib}\gg \omega_{rot}
\end{displaymath} (68)



Subsections
next up previous contents
Next: Selection Rules for Electronic Up: Molecular Spectroscopy Previous: Vibrational and Vibrational-Rotational Spectra   Contents
Markus Hiereth 2005-01-20

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