Question

consider the neon atom in all possible n+1, l-1 excited states that relax down to the ground state and emit light in the process. (a) Write the atomic term symbol for each of these states including the ground state (hint, use the Clebsch-Gordon series, there should be 8 total term symbols). Show your work in how you arrive at these term symbols. (b) Draw an energy level diagram showing decay of these states to the ground state by listing them in the proper order of there energies and indicate which states are allowed transitions by using an arrow to show the decay of energy from one state to the next. This is a qualitative exercise, no need to look up the values or draw the energy levels accurately - just the proper order of energy will suffice.

Answer 1

The term symbol is defined as (2S+1) L (J), where the first parenthesis is a superscript and represents the multiplicity and the second parenthesis is a subscript and represents the total angular momentum of the system.
S = sum of all the spins
L = sum of magnetic moments
J = L + S, L +S -1,...l L-S l
Now let's begin.
The electronic configuration of scandium should be [Ar]4s2 3d1.
Only unpaired electrons contribute to the spin; therefore, S = 1/2
L = 2 because the lowest energy state is that which maximizes L. Remember that there are 5 d orbitals with magnetic moment values of 2, 1, 0, -1, -2. The first electron occupies the highest one, the next one the following and so on.
So, J = 5/2 and 3/2 (5/2 = 2 +1/2; 3/2 = (5/2 -1) = (2 - 1/2))
The terms symbols therefore would be:
(2) D (5/2) and (2) D (3/2)
Because this system is less than half-filled, the state with lowest energy corresponds to the one with lowest J value.

For He*
S = 1 or 0
L = 1 (since ml = 0 for an s-orbital)
J = 2, 1, 0 for S =1
J = 1 for S =0
Therefore, all terms would be
(3) P (2), (3) P (1), (3) P (0) for the large J.
(1) P (1) for small J

If you are only considering the lowest excited state then S will probably be zero because the electron on the p orbital would have come from the s orbital. Since the Pauli-exclusion principle dictates that 2 electrons must have different spin, the lowest transition would be the one that does not change the spin of the electron.