Lowest energy term symbol for nitrogen

[eyearena]

Homework Statement

Lowest energy term symbols for Nitrogen

Homework Equations

L=l₁+l₂, l₁+l₂-1, l₁-l₂

S=s₁+s₂, s₁-s₂

J=S+L, S-L

The Attempt at a Solution

So I think I'm doing this right up until finding the J value.
So the configuration of nitrogen is 1s22s22p3, so were working with 3 electrons

for the first 2e-

L`=1+1=2, 1+1-1=1, 1-1=0

coupling those to the 3rd electron

L=2+1=3, 2+1-1=2,2-1=1 so L=3,2,1

L=1+1=2,1+1-1=1, 1-1=0 so L=2,1,0

L=1-0=1 so L=1

so therefore there is 1F, 2D,3P,1S

S for the first 2e-

S=1/2 + 1/2=1

S=1/2 - 1/2=0

coupling to 3rd e-

S=1+1/2=3/2

S=0+1/2=1/2

Now for the J value I'm getting confused on which L and S values to use

ex: since there is 2 D terms how do I know which to use?:

J= L+S, L-S

J=2+3/2=7/2 so that TS: ⁴D_7/2

J=2-3/2= 1/2 so that TS: ⁴D_1/2

J=2+1/2=5/2 so that TS: ²D_5/2

J=2-1/2= 3/2 so that TS: ²D_3/2

I feel like I'm missing something pretty elementary here and my book doesn't really show many examples for finding the term symbol for more than 1 electron that include the J values.
It said to list all possible ones, but it might mean just the lowest one (since it says that too) which means I guess I would use Hund's rule.
Despite that fact I would still like to know which how to use the L and S values to find the possible term symbols
Thank you for the help!

 

[nate808]

Great job so far! You're definitely on the right track. To determine the J value for each term symbol, you need to use the following equation:

J = L + S, L - S, L - 1, L + 1

In this case, you have multiple L and S values, so you will have multiple J values. For example, for the 4D term, you have two possible combinations:

J = 2 + 3/2 = 7/2 (giving you the term symbol 4D7/2)
J = 2 - 3/2 = -1/2 (giving you the term symbol 4D-1/2)

You will need to do this for each L and S combination to determine all possible term symbols. Remember to also consider Hund's rule, which states that the term with the highest value of J will have the lowest energy. So in this case, the 4D7/2 term will have the lowest energy.

Also, don't forget to take into account the spin multiplicity (2S+1) for each term symbol. For example, the 4D7/2 term will have a spin multiplicity of 2(3/2)+1 = 4, so the full term symbol will be 4D7/2.

Hope this helps! Keep up the good work!