1. Will the orbital energies for multielectron atoms depend on their angular momentum quantum number ℓ?
(A) In the H atom, the orbital energies depend on the principal quantum
number n only, if there are no external fields present. This will be the
same in multielectron atoms.  E depends only on n.
(B) The repulsion between different electrons is different for
different ℓ (e.g. s or p orbitals), because their spatial distributions
are different. Therefore, E depends on n and ℓ.

2. Will the orbital energies for multielectron atoms depend on their angular momentum quantum number ℓ?
(A) In the H atom, the orbital energies depend on the principal quantum
number n only, if there are no external fields present. This will be the
same in multielectron atoms.  E depends only on n.
(B) The repulsion between different electrons is different for
different ℓ (e.g. s or p orbitals), because their spatial distributions
are different. Therefore, E depends on n and ℓ.

3. For constant n, how will the orbital energies for multielectron
atoms depend on ℓ?
(A) Higher ℓ → higher angular momentum → more “circular” behavior → more screening of nuclear charge → for the same n, E will increase with increasing ℓ
(B) Higher ℓ → higher angular momentum → more “circular” behavior → less overlap with orbitals of “core” electrons → less Coulomb repulsion → for the same n, E will decrease with increasing ℓ

4. For constant n, how will the orbital energies for multielectron
atoms depend on ℓ?
(A) Higher ℓ → higher angular momentum → more “circular” behavior → more screening of nuclear charge → for the same n, E will increase with increasing ℓ
(B) Higher ℓ → higher angular momentum → more “circular” behavior → less overlap with orbitals of “core” electrons → less Coulomb repulsion
→ for the same n, E will decrease with increasing ℓ

5. Consider the H atom in its ground state: H(1s). What is the term
symbol?
(A) 1S0
(B) 2S1/2
(C) 3P2
(D) 3S0
(E) 1S1/2
2S+1LJ

6. Consider the H atom in its ground state: H(1s). What is the term
symbol?
(A) 1S0
(B) 2S1/2 correct: S=1/2; L=0; J can only be 1/2
(C) 3P2
(D) 3S0 not possible, not with any electron configuration!
(E) 1S1/2 not possible, not with any electron configuration!
2S+1LJ