# 18.2: Most Atoms are in the Ground Electronic State

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.

Example $$\PageIndex{1}$$

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 $$\PageIndex{1}$$ is that only the ground electronic state is generally thermally accessible at room temperature.