When we calculate electronic partition function using the formula:


How can I get the $g_n$ value and $E_n$ values for $n=0,1,2,3...$ levels?

I need to calculate the $q_\mathrm{elec}$ for various species like $\ce{H, H2, OH, H2O, O2, O, O+, H+}$.

  • 1
    $\begingroup$ First you must understand what the $n$ represents then you will need to look up the values for $E_n$ and $g_n$ at least for all but a couple of of the species you quote. $\endgroup$
    – porphyrin
    Aug 24 '20 at 11:20

$E_n$ represents the energy of the $n$th electronic state relative to the ground state (i.e. the ground-state electronic energy $E_0 = 0$). Often, electronic states are very high in energy and the excited-state energies $E_1, E_2, \cdots$ are far greater than $k_\mathrm{B}T$, such that $\mathrm{e}^{-E_n/k_\mathrm{B}T} \approx 0$.

Under these conditions, the sum can be collapsed to just a ground-state term

$$q_\mathrm{elec} = \sum_n g_n \mathrm{e}^{-E_n/k_\mathrm{B}T} \approx g_0 \mathrm{e}^{-E_0/k_\mathrm{B}T} = g_0$$

where the latter equality is because $E_0 = 0$. $g_0$ is the degeneracy of the electronic ground state: to find this out you will need to draw some MO diagrams and/or construct the term symbols.

If the ground state is nondegenerate (e.g. $\ce{H2O}$) the whole thing is just equal to 1. Otherwise, for an atom with ground-state term symbol $^{2S+1}L_J$, the degeneracy is $2J + 1$. Similar considerations can be applied to molecules.

However, you will have to watch out for cases where there are low-lying energy states where $E_n$ is on the order of $k_\mathrm{B}T$. I am not aware of an easy way to predict this, though – you will have to know some typical examples, failing which you will just have to look it up, as porphyrin suggests. The NIST WebBook is a good starting place for this.


This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .