Transition from ground state to excited state

Question

I don't understand how to correctly find the term symbol of an excited state. I know that the first two permitted transitions for the Ne atom are 16.8eV and 19.8eV and I want to find the electronic configuration and the term symbol.

I assume that from the $2p$ orbital I'll have a transition to the $3s$ orbital (first excited state) and to the $4s$ orbital (second excited state). The selection rule $\Delta l=\pm1$ is obeyed. But what about the total angular momentum $j$? To find $j$ I have to know something about the spin $s$. Is it something about singlet and triplet states?

Answer 1

First you need to have the ground state electron configuration, for Neon this is 1s$^2$2s$^2$2p$^6$ or what ever it is for your atom. This will give you the term symbol for the ground state. The excited state has one electron promoted, say p to a d orbital, then 1s$^2$2s$^2$2p$^5$3d$^1$.

A term symbol has the form $^{2S+1}L_J$, where S is the total spin ($2S+1$ is the multiplicity) , L the orbital angular momentum and J the total angular momentum.

Because angular momentum is added vectorially it is necessary to follow some rules to get all the values. Each type of angular momentum forms a series of values (the Clebsch-Gordon series). If total spin angular momentum is S and for each electron the spin quantum number is s1 and s2 then the series is $S= |s1+s2|..|s1-s2|$ with steps of 1 between values. Note that all values are positive and there may only be one value in the series. The same type of series applies to L and J total spin and orbital angular momentum.

A p$^5$ can be considered as the same as p$1$ as only 1 electron is unpaired. Thus using the series if there are two electrons each with spin quantum numbers s1, s2 = 1/2 each the the total spin for two electrons is S = 0 or 1 and the spin multiplicity is 1 (a singlet) or 3 a triplet. The series is |s1+s2| = 1, and |s1-s2| =0.

The L values is calculated with l1=0 (s orbital) and l2=1 (p orbital) thus L only has 1 value = 1. Thus far the possible term symbols are $^1P$ and $^3P$.

The total angular momentum J is found in the same way as S and L values were using the Clebsch-Gordon series, $$J= |L+S|...|L-S| $$ Again this means find the maximum value, then the minimum make it positive, and separate any other values by 1. Sometimes there is only 1 value so be careful. There are $2J+1$ degenerate J levels, these are split in an external magnetic field into separate states.

This for the triplet state $S=1$ and so $J=|1+1|..|1-1|$, or 2, 1 and 0. (note three values from the series). Thus the term symbols are $^3P_2$ , $^3P_1$, $^3P_0$. But there is also a singlet term so the J values for this are $J=|1+0|..|1-0| = 1$ (one term only). Thus the last term symbol is $^1P_1$. Although this looks rather complicated it becomes quite quick with a bit of practice.

Once you have all the term symbols you can then apply selection rules to see which transitions are allowed.

Finally if the configuration contains, say, p$^2$ then one has also to consider the Pauli principle and it is more tricky. The way to do this is to list all s1, s2, l1, l2 values and remove any with the same set of quantum numbers. Finally when you have the term symbol Hund's rules are used to determine which levels and states are lowest in energy. (Many text books work through the p$^2$ case, See for example Steinfeld, 'Molecules & radiation')

Answer 2

Yes, it does have something to do with that: doublet states. Both spin orientations are possible. However, the energy difference (through spin-orbit interaction) between $j = 1/2$ and $j = 3/2$ is way smaller (around 0.002 eV) than the excitation energies you are looking at here.

One way to indicate that j does play a role in the configuration but not on the order of the energies you are dealing with, is to add a "J" as the total angular momentum quantum number to the letter of the orbital angular momentum quantum number, e.g., $2^2P_J$.