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Will the orbital energies for multielectron atoms depend on their angular momentum quantum number ℓ?

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

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Will the orbital energies for multielectron atoms depend on their angular momentum quantum number ℓ?

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  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) 1S1/2not possible, not with any electron configuration! 2S+1LJ

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