Nitrogen Ground State: 2s^2 2p^3 ^4 S_{3/2}

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SUMMARY

The ground state electronic configuration of the nitrogen atom is 2s2 2p3, with the ground state term symbol being ^4S3/2. The first two excited states are represented by the term symbols ^2D5/2, 3/2 and ^2P3/2, 1/2. The confusion arises from the misconception that excited states are solely determined by electron configuration; however, angular momentum terms also play a crucial role. The lowest energy absorption transition occurs to the 2p23s configuration, but the excited states can still be observed in the emission spectrum despite selection rules.

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I am looking over my notes for quantum mechanics, particularly for electronic state classification for nitrogen.
At ground state, nitrogen atom's 1s and 2s shells are fully filled and the 2p shell is half filled, thus the configuration is:
##2s^2 2p^3##
There are 3 different angular momentum terms:
##^4 S_{3/2}##, ##^2 D_{5/2, 3/2}##, ##^2 P_{3/2, 1/2}##

In my notes it says that the ground state is at ##^4 S_{3/2}##
and the first 2 excited states are at ##^2 D_{5/2, 3/2}##, ##^2 P_{3/2, 1/2}##.

I am confused. I thought excited states don't have anything to do with the angular momentum terms, but instead, it only has to do with the configuration. My understanding for excited state is that one of the electrons from the 2p subshell occupies a higher state. So wouldn't that make ##2s^2 2p^2 3s## the first excited electronic state?
 
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The ground electronic configuration can have different energy levels. In a regular absorption spectrum, however, you will not see lines corresponding to those levels, as the selection rule ##\Delta l = \pm1## is not satisfied. The lowest energy absorption transition will be to the 2p23s, as you correctly noted.

Nevertheless, the ##^2 D_{5/2, 3/2}##, and ##^2 P_{3/2, 1/2}## levels are there, and may be observed in the emission spectrum.
 

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