Jj coupling scheme and electronic configuration

[vst98]

Homework Statement

Write all possible terms (n1 j1, n2 j2, …) J
for these electronic configurations (considering jj coupling scheme):
1s²
1s2s
1s²2s²2p
1s²2s²2p⁵

The Attempt at a Solution

1s²
I have two electrons in the s sub-shell,
since |l-s|<=j<=l+s then j1=1/2 and j2=1/2.
This gives possible J values of J=0,1 and the terms are

(1/2,1/2)₀ (1/2,1/2)₁


1s2s
one electron in 1s and one in 2s.
Again using a rule |l-s|<=j<=l+s gives j1=j2=1/2 for both electrons.
j1 and j2 can couple together go give total J=0,1 so the terms are again

(1/2,1/2)₀ (1/2,1/2)₁


So far I'm not sure that this is right, do I have to do this calculations with all electrons
or can I ignore completely filled shells ? But what about 1s2s then ? Also is my designation
of these terms like (j1,j2)_J correct ?

 

[member 553083]

1s22s22pOne electron in 1s, two in 2s and two in 2p.Again using a rule |l-s|<=j<=l+s gives j1=1/2, j2,j3,j4=1 for all electrons in 2s and j5,j6=3/2 for electrons in 2p.These six can couple together to give total J=0,1,2,3 so the terms are(1/2,1,1,1,3/2,3/2)0 (1/2,1,1,1,3/2,3/2)1 (1/2,1,1,1,3/2,3/2)2 (1/2,1,1,1,3/2,3/2)31s22s22p5One electron in 1s, two in 2s, two in 2p and one in 5s.Again using a rule |l-s|<=j<=l+s gives j1=1/2, j2,j3,j4=1 for all electrons in 2s and j5,j6=3/2 for electrons in 2p and j7=1/2 for electron in 5s.These seven can couple together to give total J=0,1,2,3,4,5 so the terms are(1/2,1,1,1,3/2,3/2,1/2)0 (1/2,1,1,1,3/2,3/2,1/2)1 (1/2,1,1,1,3/2,3/2,1/2)2 (1/2,1,1,1,3/2,3/2,1/2)3 (1/2,1,1,1,3/2,3/2,1/2)4 (1/2,1,1,1,3/2,3/2,1/2)5