Question 3.6: Atomic oxygen: from Table 2.3:

Atomic oxygen : from Table 2.3:

Table 2.3 Lowest Lying Electronic States of Air Atoms

Atom Configuration Term Multiplicity Energy (eV) Degeneracy
N 2s² 2p³ ^4S_{3/2} 1 0 4
N 2s² 2p³ ^2D_{5/2,3/2} 2 2.39 10
N 2s² 2p³ ^2P_{3/2,1/2} 2 3.58 6
N 2s² 2p³ 3s ^4P_{5/2,3/2,1/2} 3 10.33 12
N 2s² 2p³ 3s ^2P_{3/2,1/2} 2 10.68 6
N 2s 2p^{4} ^4P_{5/2,3/2,1/2} 3 10.93 12
N 2s² 2p² 3p ^2S_{1/2} 1 11.60 2
N 2s² 2p² 3p ^4D_{7/2,5/2,3/2,1/2} 4 11.75 20
N 2s² 2p² 3p ^4P_{5/2,3/2,1/2} 3 11.84 12
N 2s² 2p² 3p ^4S_{3/2} 1 12.00 4
N 2s² 2p² 3p ^2D_{5/2,3/2} 2 12.00 10
N 2s² 2p² 3p ^2P_{3/2,1/2} 2 12.12 6
N 2s² 2p² 3s ^2D_{5/2,3/2} 2 12.36 10
0 2s² 2p^{4} ^3P2 1 0 5
0 2s² 2p^{4} ^3P1 1 0.0196 3
0 2s² 2p^{4} ^3P0 1 0.0281 1
0 2s² 2p^{4} ^1D2 1 1.97 5
0 2s² 2p^{4} ^1S0 1 4.20 1
0 2s² 2 3s ^5S2 1 9.16 5
0 2s² 2p³ 3s ^3S1 1 9.54 3
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Q^{O}_{el}  =  \underset{\underset{^{3}P_{2}}{\uparrow}}{5}  + 3  exp(\underset{\underset{^{3}P_{1}}{\uparrow}}{-}\frac{228}{T}) +   exp(\underset{\underset{^{3}P_{0}}{\uparrow} }{-}\frac{326}{T} )O \left[exp(- \frac{23,000}{T} )\right]

The first three states are often combined in different ways, depending on the temperature range of interest:

Q^{O}_{el}\approx 5+ 4  exp\left(- \frac{270}{T} \right)\approx 9           (3.119)

\therefore e^{O}_{el}=0=(C_{v} )^{O}_{el}

Atomic nitrogen: again, from Table 2.3:

Q^{N}_{el}=\underset{\underset{^{4}S}{\uparrow}}{4} + O\left[exp\left(- \frac{28,000}{T} \right) \right]

 

\therefore e^{N}_{el}=0 =(C_{v} )^{N}_{el}

Thus, for temperatures typical of aerospace applications, of a few thousand
degrees at the most, for both oxygen and nitrogen atoms, the partition function is simply a number, and there are no contributions to the internal energy and specific heats from the electronic mode. We will find later on, however, that the numerical values of the atomic electronic partition functions are important in determining chemical composition.

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