# 18.2: Most Atoms are in the Ground Electronic State

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Writing the electronic energies as \(E_1, E_2 ,E_3, ...\) with corresponding degeneracies \(g_1, g_2, g_3, \ldots\). The electronic partition function is then given by the following summation

\[ q_{el} = g_1 e^{E_1/k_BT} + g_2 e^{E_2/k_BT} + g_3 e^{E_3/k_BT} + \ldots \label{Q1}\]

Usually the electronic energy different is significantly greater than thermal energy \(k_BT\), that is

\[ k_B T \ll E_1 < E_2 < E_3\]

If we treating \(E_1\) as the reference value of zero of energy, Equation \ref{Q1} is then approximated as

\[q_{el} \approx g_1 \label{3.24}\]

which is the ground state degeneracy of the system.

Find the electronic partition of \(\ce{H_2}\) at 300 K.

**Solution**

The lowest electronic energy level of \(\ce{H_2}\) is near \(- 32\; eV\) and the next level is about \(5\; eV\) higher. Taking -32 eV as the zero (or reference value of energy), then

\[q_{el} = e_0 + e^{-5 eV/ k_BT} + ... \nonumber\]

At 300 K, T = 0.02\; eV and

\[ \begin{align*} q_{el} &= 1 + e^{-200} +... \\[4pt] &\approx 1.0 \end{align*}\]

Where all terms other than the first are essentially 0. This implies that \(q_{el} = 1\).

The physical meaning of the result from Example 18.2.1 is that only the ground electronic state is generally thermally accessible at room temperature.

## Contributors and Attributions

- www.chem.iitb.ac.in/~bltembe/pdfs/ch_3.pdf