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Since there are 2$p$ electrons $\ell_1 = 1, \ell_2 = 1 \rightarrow L=0,1,2$ and $s_1=1/2, s_2=1/2 \rightarrow S=0,1$. As you said, ignoring the Pauli principle results in $^3\text{D}, ^1\text{D},^3\text{P},^1\text{P},^3\text{S}$, and $^1\text{S}$ terms. The easiest way to impose the Pauli principle is to make a table as below where the first row and column ...

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I worked on a program two decades ago where we were to determine if nations "XYZ" were building nuclear weapons based on intelligence "ABC." Before I did this I took a DOE course that amounted to intermediate level (210) nuclear weapons. A lot of data on this is classified, and one needs CWDI Q-clearance to know this stuff. A lot of effort has gone into ...

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I actually think I have figured it out: if we consider the situation when the principle quantum number for, let's say, excited helium is different, then we can have spin-up electron in $1S_1$ state and spin-up electron in $2S_1$ state. This is going to be a triplet. Please, correct me if I am wrong.

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This is a very good book which covers atomic spectra and Laser Physics, with good Physical intuition and fairly rigorous Mathematical proofs. https://www.amazon.com/Atomic-Physics-Oxford-Master-Optical/dp/0198506961

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The figure in Sparkler's answer described the energy levels but it needs some further explanation. First this picture only applies in the condensed phase or if in the gas phase only at extremely high pressure. When a photon is absorbed from the ground state, provided the electronic transition is allowed, transitions from v=0 in the ground state to v'=0,1,2 ...

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In addition, glass prism will experience its resonance frequency absorption at certain wavelength (eg: some glasses absorb UV). That why it is not a brilliant choice to disperse the light spectrum in most of the professional spectroscopy.

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