Question 22.1: In analogy with the characteristic vibrational and rotationa...
In analogy with the characteristic vibrational and rotational temperatures, \theta_{v i b} and \theta_{\text {rot }}, respectively, one can define a characteristic electronic temperature by:
\theta_{e l e c, j}=\varepsilon_{e j} / k_{B}
where \varepsilon_{e j} is the energy of the j th excited electronic state relative to the ground state.
(a) If one defines the ground electronic state to be the zero of energy, derive an expression for the electronic partition function, q_{\text {elec }}, expressed in terms of \theta_{\text {elec, } j}.
(b) The first ( j = 1) and the second ( j = 2) excited electronic states of O(g) lie 158.2 cm ^{-1} \text { and } 226.5 cm ^{-1} above the ground electronic state ( j = 0). Given the degeneracies, g_{e 0}=5, g_{e 1}=3, \text { and } g_{e 2}=1, calculate the values of \theta_{e l e c, 1}, \theta_{e l e c, 2}, and q_{\text {elec }} (ignoring any higher excited electronic states) for O(g) at 5000 K.
(c) Calculate the fraction of O(g) atoms in the ground, first, and second electronic states at 5000 K. What is the fraction of O(g) atoms in all the remaining excited electronic states?
Note: When energies are given in cm ^{-1} (which is typical in the case of electronic states, it is convenient to use \left.k_{B}=0.69509 cm ^{-1} K ^{-1}\right).
Part (a)
As discussed in Part III, if we define the ground electronic state ( j = 0) to be the zero of energy (such that \left.\epsilon_{e 0}=0\right), the electronic partition function can be written as follows:
q_{e l e c}=\sum_{j=0}^{\infty} g_{e j} \exp \left(\frac{-\epsilon_{e j}}{k_{B} T}\right)=\sum_{j=0}^{\infty} g_{e j} \exp \left(\frac{-\theta_{\text {elec }, j}}{T}\right) (1)
Note that j in Eq. (1) represents energy levels. Applying Eq. (1) to the given case:
q_{e l e c}=g_{e 0} \exp (0)+\sum_{j=1}^{\infty} g_{e j} \exp \left(\frac{-\Theta_{e l e c, j}}{T}\right) (2)
q_{\text {elec }}=g_{e 0}+\sum_{j=1}^{\infty} g_{e j} \exp \left(\frac{-\theta_{\text {elec }, j}}{T}\right) (3)
Part (b)
Using the definition of \Theta_{e l e c, j}, we can compute \Theta_{e l e c, 1} \text { and } \Theta_{e l e c, 2} as follows:
\begin{aligned}&\Theta_{\text {elec }, 1}=\frac{\epsilon_{e 1}}{k_{B}}=\frac{158.2 cm ^{-1}}{0.69509 cm ^{-1} K ^{-1}}=227.6 K \\&\Theta_{e l e c, 2}=\frac{\epsilon_{e 2}}{k_{B}}=\frac{226.5 cm ^{-1}}{0.69509 cm ^{-1} K ^{-1}}=325.8 K\end{aligned}
Ignoring excited electronic states with energy levels j > 2, we can calculate q_{\text {elec }} using Eq. (3) and the data given in the Problem Statement, as follows:
q_{\text {elec }}=g_{e 0}+g_{e 1} \exp \left(\frac{-\Theta_{e l e c, 1}}{T}\right)+g_{e 2} \exp \left(\frac{-\Theta_{\text {elec }, 2}}{T}\right)
q_{e l e c}=5+3 \exp \left(\frac{-227.6}{5000}\right)+1 \exp \left(\frac{-325.8}{5000}\right) (4)
q_{e l e c}=8.8034 (5)
Part (c)
The various fractions can be calculated using the following expression:
f_{j}=\frac{g_{e j} \exp \left(-\frac{\epsilon_{j}}{k_{B} T}\right)}{q_{e l e c}} (6)
Recall that j in Eq. (6) denotes energy levels. Because each energy level can be g_{e j} degenerate, the g_{e j} factor appears in the expression for f_{j}. Using Eq. (6) for j = 0, 1, and 2, we obtain:
f_{0}=\frac{g_{e 0}}{q_{\text {elec }}}=\frac{5}{8.8034}=0.5679 (7)
f_{1}=\frac{g_{e 1} \exp \left(-\frac{\epsilon_{1}}{k_{B} T}\right)}{q_{\text {elec }}}=\frac{3 \exp \left(-\frac{227.6}{5000}\right)}{8.8034}=0.3256 (8)
f_{2}=\frac{g_{e 2} \exp \left(-\frac{\epsilon_{2}}{k_{B} T}\right)}{q_{\text {elec }}}=\frac{1 \exp \left(-\frac{325.8}{5000}\right)}{8.8034}=0.1064 (9)
Because higher electronic states were ignored in the calculation of q_{\text {elec }}, the fraction of O(g) in the higher states will be 0. However, if the contributions of the higher-energy states are included in the calculation of q_{\text {elec }}, the fraction will turn out to be a very small positive quantity.