Give the electronic configuration for the Boron atom (Z =
5). Then enumerate the allowed term symbols ^{2S+1}L_{J} for
the ground state from the point of view of angular momenta alone.

Solution:

- Concepts:

The energy levels of multi-electron atoms - Reasoning:

We use the Pauli exclusion principle and the rules for adding angular momentum to find the allowed terms^{2S+1}L. - Details of the calculation:

The electronic configuration is (1s)^{2}, (2s)^{2}, 2p.

The 1s and 2s shells are full and have L = 0, S = 0, J = 0.

The one 2p electron determines S, L, and J.

s = ½, l = 1, j = ½, 3/2.

The allowed terms are^{2}P_{½ }and^{ 2}P_{3/2}.

Hund's rule: The component with the smallest value of J (J = ½) has the lowest energy.

Aluminum (Al) is element No. 13 in the periodic table

(a) Write down the electronic shell configuration of Al.

(b) Find the total L (orbital angular momentum), total S (spin
angular momentum), and J (total angular momentum) for the ground state of
the aluminum atom.

(c) The Al atom is singly ionized by photoionization of the 2p level.

Find the “term symbol”
^{2S+1}L_{J} for the open 2p shell,
assuming that the ion is in its lowest energy state.

(d)
Couple the orbital angular momentum you found in (b) with the total
orbital angular momentum you found in (c). What possible values of L do
you get?

(e) Couple the total spin angular momentum you found in (b) with the total
spin angular momentum you found in (c). What possible values of S do you
get?

(f) Find all possible term symbols for the Al ion.

Solution:

- Concepts:

The energy levels of multi-electron atoms, angular momentum coupling, Hund's rules - Reasoning:

We use the Pauli exclusion principle and the rules for adding angular momentum to find the allowed terms^{2S+1}L. -
Details of the calculation:

(a) Electronic configuration, 1s^{2}2s^{2}2p^{6}3s^{2}3p^{1}= [Ne] 3s^{2}3p^{1 }(b) We have to only consider the open shell. L = 1, S = ½.

Possible values for J: J = ½, J = 3/2.

Ground State: J = ½, (Hund’s 3^{rd}rule), Term symbol:^{2}P_{½ }(c) For the open 2p shell (2p^{5}) we have L = 1, S = ½.

Possible values for J: J = ½, J = 3/2.

Lowest energy term: J = 3/2, (Hund’s 3^{rd}rule), Term symbol:^{2}P_{3/2 }(d) Electronic configuration, 1s^{2}2s^{2}2p^{5}3s^{2}3p^{1 }Possible values for L: L = 0, 1, 2

(e) Possible values for S: S = 0, 1

(f)^{1}S_{0}^{3}S_{1}^{1}P_{1}^{3}P_{2,0,1}^{1}D_{2}^{3}D_{3,2,1< }

(a) A configuration is an enumeration of the values of n and l
for all electrons of an atom. Give the ground and the lowest two excited
configurations of atomic Helium.

(b)
Give the spectroscopic terms in Russell-Saunders (LS) notation.

[The term specifies the total spin S and the total orbital angular
momentum L.]

(c) Which of the excited states has the shortest radiative lifetime?
Why?

Solution:

- Concepts:

An atom with two electrons, coupling of angular momentum, dipole selection rules - Reasoning:

We use the rules for adding angular momentum to find the allowed terms^{2S+1}L. -
Details of the calculation:

(a) H_{0 }= p_{1}^{2}/(2m) + p_{2}^{2}/(2m) - 2e^{2}/r_{1 }- 2e^{2}/r_{2 }+ U_{c}(r_{1}) + U_{c}(r_{2})

is the Hamiltonian in the central field approximation.

W = e^{2}/r_{12 }- U_{c}(r_{1}) - U_{c}(r_{2})

is the dominant correction term. It is due to the non-central part of the electron-electron interaction.

The eigenvalues of H_{0}depend on n_{1}, l_{1}and n_{2}, l_{2}.

Configuration of the ground state: 1s^{2 }Configuration of the first excited state: 1s2s

Configuration of the second excited state: 1s2p

(b) 1s^{2}:^{2S+1}L_{J}=^{1}S_{0 }1s2s:^{2S+1}L_{J}=^{1}S_{0},^{3}S_{1}.

1s2p:^{2S+1}L_{J}=^{1}P_{1},^{3}P_{0,1,2}.

(c) Selection rules for allowed transitions: ∆j = 0, ±1, (except j_{i }= j_{f }= 0), ∆l = ±1, ∆m_{j }= 0, ±1.

∆l = ±1 is only satisfied for the 1s2p state.

The transition^{1}P_{1}-->^{1}S_{0 }is allowed.

Among the values of L and
S obtained from the general rules for
addition of angular momenta are those which correspond to states forbidden by
the Pauli principle. Consider the configuration np^{2}, i. e. two
electrons with the same principal quantum number n and the same angular momentum
quantum number l = 1.

Find the allowed terms ^{2S+1}L_{J}.

Solution:

- Concepts:

The Pauli exclusion principle, energy levels of multi-electron atoms - Reasoning:

The Pauli exclusion principle states that no two identical fermions can have exactly the same set of quantum numbers. In the LS coupling scheme, the energy level of a multi-electron atom in the absence of external fields is characterized by the quantum numbers n, L, S, and J. In the LS coupling scheme the notation for the angular momentum is the term^{2S+1}L_{J}. Because of the Pauli exclusion principle not all terms one obtains by simply adding the angular momentum and spin of all the electrons are allowed. -
Details of the calculation:

For each electron the following values are possible:

m_{l }= 1, 0, -1, m_{s }=½,- ½.

Combining the different values of m_{l}and m_{s}, we obtain the following possible states:m _{l }= 1m _{s }= ½(1 ^{+})m _{l }= 0m _{s }= ½(0 ^{+})m _{l }= -1m _{s }= ½(-1 ^{+})m _{l }= 1m _{s }= -½(1 ^{-})m _{l }= 0m _{s }= -½(0 ^{-})m _{l }= -1m _{s }= -½(-1 ^{-})In each of these states there cannot be more than one electron.

The following states with non-negative values of M_{L}and M_{s}are possible if the two electrons are not allowed to have exactly the same quantum numbers.State M _{L}M _{s}(1 ^{+})(0^{+})1 1 (1 ^{+})(-1^{+})0 1 (1 ^{+})(1^{-})2 0 (1 ^{+})(0^{-})1 0 (1 ^{+})(-1^{-})0 0 (0 ^{+})(1^{-})1 0 (0 ^{+})(0^{-})0 0 (-1 ^{+})(1^{-})0 0 The presence of the M

_{L }= 2, M_{s }= 0 term shows that a^{1}D term is among the possible terms. To this term we must further assign states with M_{L }= 1, M_{s }= 0 and M_{L }= 0, M_{s }= 0.

Among the states left is a state with M_{L }= 1, M_{S }= 1. This and states with M_{L }= 1, M_{S }= 0, and M_{L }= 0, M_{S }= 1, and M_{L}= 0, M_{S }= 0 yield the^{3}P term. Four M_{L}, M_{S}terms are associated with the^{3}P term.

The only remaining state has M_{L }= 0, M_{S }= 0. It corresponds to the^{1}S term.

For the^{1}D and^{1}S term we have S = 0, and therefore J = L.

For the^{3}P term we have S = 1, L = 1.

L = 1, S = 1 is associated with the 9 possible terms (1,1), (1,0), (1,-1), (0,1), (0,0), (0,-1), (-1,1), (-1,0), (-1,-1). The state with M_{L }= 1, M_{S }= 1 requires M_{J}= M_{L}+ M_{S}= 2 and therefore the term^{3}P_{2}is allowed. We must have 5 terms associated with J = 2 for the 5 possible values of M_{J}. This leaves 4 terms associated with J = 1 and J = 0.

The allowed terms^{2S+1}L_{J}are^{1}D_{2},^{3}P_{2,1,0}, and^{1}S_{0}.

For the Nitrogen atom (Z = 7) find the three terms in the LS coupling scheme which lie lowest in energy.

Solution:

- Concepts:

The energy levels of multi-electron atoms, angular momentum coupling, Hund's rules - Reasoning:

We use the Pauli exclusion principle and the rules for adding angular momentum to find the allowed terms^{2S+1}L. - Details of the calculation:

The Nitrogen atom has 7 electrons.

Electronic configuration, 1s^{2}2s^{2}2p^{3}= [He] 2s^{2}2p^{3}

The 1s and 2s shells are filled and have L = 0, S = 0.

The 2p shell is half filled.

The possible terms^{2S+1}L for 3 equivalent p electrons (np)^{3}are^{4}S,^{2}P, and^{2}D.

The p-electron can have:

m _{l }=1 0 -1 1 0 -1 m _{s}=½ ½ ½ -½ -½ -½ label state as 1 2 3 4 5 6 To satisfy the Pauli exclusion principle a configuration must not contain identical electrons.

Let us write down all possible configurations with non-negative M_{L}= ∑m_{l}and M_{S}= ∑m_{s}.

State M _{L}M _{S}123 0 3/2 124 2 ½ 125 1 ½ 126 0 ½ 134 1 ½ 135 0 ½ 234 0 ½ Pick the states starting with the largest M

_{L}and the largest M_{S}.

M_{L}= 2, M_{s}= ½. The presence of the M_{L }= 2, M_{s }= ½ term shows that a^{2}D term is among the possible terms. To this term we must further assign states with M_{L }= 1,0 , M_{S }= ½.

What is left?

M_{L}= 1, M_{s}= ½. The presence of the M_{L }= 1, M_{s }= ½ term shows that a^{2}P term is among the possible terms. To this term we must further assign states with M_{L }= 0, M_{S }= ½.

What is left?

M_{L}= 0, M_{s}= 3/2. The presence of the M_{L }= 0, M_{s }= 3/2 term shows that a^{4}S term is among the possible terms. To this term we must further assign states with M_{L }= 0, M_{S }= ½.

Hund's rule:

The level with the largest multiplicity has the lowest energy.

For a given multiplicity, the level with the largest value of L has the lowest energy.

When a shell is half filled, there is no multiplet splitting.

E(^{4}S) < E(^{2}D) < E(^{2}P).

For an atom with 3 valence electrons, find the spectroscopic terms ^{2S+1}L that
characterize the system when the electron distribution is:

(a) (3p) (4p) (5p)

(b) (3p)^{2} (4p)

The electrons are specified by (nl).

Solution:

- Concepts:

The Pauli exclusion principle, energy levels of multi-electron atoms, addition of angular momentum - Reasoning:

The Pauli exclusion principle states that no two identical fermions can have exactly the same set of quantum numbers. We use it to find the allowed terms^{2S+1}L. -
Details of the calculation:

(a) No electrons have identical quantum numbers. We do not have to worry about the Pauli exclusion principle.

We add angular momentum. First add the angular momentum of the 3p and 4p electrons.(3p)(4p) Possible terms L _{12}S _{12}0 0, 1 ^{1}S,^{3}S1 0, 1 ^{1}P,^{3}P2 0, 1 ^{1}D,^{3}DNow add to this the angular momentum of the 5p electron.

(3p)(4p)(5p) Possible terms L _{12}S _{12}l s L, ( **L**=**L**_{12}+**l**)S, ( **S**=**S**_{12}+**s**)0 0 1 ½ 1 ½ ^{2}P1 0 1 ½ 0, 1, 2 ½ ^{2}S,^{2}P,^{2}D2 0 1 ½ 1, 2, 3 ½ ^{2}P,^{2}D,^{2}F0 1 1 ½ 1 ½, 3/2 ^{2}P,^{4}P1 1 1 ½ 0, 1, 2 ½, 3/2 ^{2}S,^{2}P,^{2}D,^{4}S,^{4}P,^{4}D2 1 1 ½ 1, 2, 3 ½, 3/2 ^{2}P,^{2}D,^{2}F,^{4}P,^{4}D,^{4}FSpectroscopic terms:

^{2S+1}L^{2}S^{4}S^{2}P^{4}P^{2}D^{4}D^{2}F^{4}F# of terms 2 1 6 3 4 2 2 1 (b) For identical fermions the total wave function must be antisymmetric under exchange of any two particles.

Let (o) denote antisymmetric or odd, (e) denote symmetric or even.

First add the angular momentum of the two 3p electrons.(3p) ^{2}Possible terms L _{12}S _{12}0 (e) 0 (o) ^{1}S1 (o) 1 (e) ^{3}P2 (e) 0 (o) ^{1}DNow add to this the angular momentum of the 4p electron.

(3p) ^{2}(4p)Possible terms L _{12}S _{12}l s L, ( **L**=**L**_{12}+**l**)S, ( **S**=**S**_{12}+**s**)0 0 1 ½ 1 ½ ^{2}P1 1 1 ½ 0, 1, 2 ½, 3/2 ^{2}S,^{2}P,^{2}D,^{4}S,^{4}P,^{4}D2 0 1 ½ 1, 2, 3 ½ ^{2}P,^{2}D,^{2}FSpectroscopic terms:

^{2S+1}L^{2}S^{4}S^{2}P^{4}P^{2}D^{4}D^{2}F# of terms 1 1 3 1 2 1 1

If the spin of the electron were (3/2)ħ,what spectroscopic terms ^{2S+1}L_{J} would exist for
the electron configuration (2p)^{2}?

Solution:

- Concepts:

The Pauli exclusion principle, addition of angular momenta - Reasoning:

The Pauli exclusion principle states that no two identical fermions can have exactly the same set of quantum numbers. We use it to find the allowed terms^{2S+1}L. -
Details of the calculation:

For each electron the following values are possible:

m_{l }= 1, 0, -1, m_{s }= 3/2, ½, -½, -3/2

Combining the different values of m_{l}and m_{s}, we obtain the following possible states:m _{l }= 1m _{s }= ½(1 ^{+})m _{l }= 0m _{s }= ½(0 ^{+})m _{l }= -1m _{s }= ½(-1 ^{+})m _{l }= 1m _{s }= -½(1 ^{-})m _{l }= 0m _{s}=-½(0 ^{-})m _{l }= -1m _{s }= -½(-1 ^{-})m _{l }= 1m _{s }= 3/2(1 ^{++})m _{l }= 0m _{s }= 3/2(0 ^{++})m _{l }= -1m _{s }= 3/2(-1 ^{++})m _{l }= 1m _{s }= -3/2(1 ^{--})m _{l }= 0m _{s}=-3/2(0 ^{--})m _{l }= -1m _{s }= -3/2(-1 ^{--})In each of these states there cannot be more than one electron.

The following states with non-negative values of M_{L}and M_{s}are possible if the two electrons are not allowed to have exactly the same quantum numbers.State M _{L}M _{S}State M _{L}M _{S}State M _{L}M _{S}State M _{L}M _{S}(1 ^{+})(0 ^{+})1 1 (0 ^{+})(1 ^{-})1 0 (1 ^{-})(-1 ^{++})0 1 (1 ^{++})(-1 ^{--})0 0 (1 ^{+})(-1 ^{+})0 1 (0 ^{+})(0 ^{-})0 0 (0 ^{-})(1 ^{++})1 1 (0 ^{++})(1 ^{--})1 0 (1 ^{+})(1 ^{-})2 0 (0 ^{+})(1 ^{++})1 2 (0 ^{-})(0 ^{++})0 1 (0 ^{++})(0 ^{--})0 0 (1 ^{+})(0 ^{-})1 0 (0 ^{+})(0 ^{++})0 2 (-1 ^{-})(1 ^{++})0 1 (-1 ^{++})(1 ^{--})0 0 (1 ^{+})(-1 ^{-})0 0 (-1 ^{+})(1 ^{-})0 0 (1 ^{++})(0 ^{++})1 3 (1 ^{+})(1 ^{++})2 2 (-1 ^{+})(1 ^{++})0 2 (1 ^{++})(-1 ^{++})0 3 (1 ^{+})(0 ^{++})1 2 (1 ^{-})(1 ^{++})2 1 (1 ^{++})(1 ^{--})2 0 (1 ^{+})(-1 ^{++})0 2 (1 ^{-})(0 ^{++})1 1 (1 ^{++})(0 ^{--})1 0 The presence of the M

_{L }= 1, M_{s }= 3 term shows that a^{7}P term is among the possible terms. To this term we must further assign states with M_{L }= 1, M_{S }= 2, 1, 0 and M_{L }= 0, M_{s }= 3, 2, 1, 0. There are 8 states with nonzero M_{J}= M_{L}+ M_{S}associated with the^{7}P term.

M_{L }= 1, M_{S }= 3 requires M_{J}= 4, and therefore theis present. It accounts for 9 of the possible 21 combinations of the^{7}P_{4}^{7}P term. Theaccounts for 7 terms and the^{7}P_{3}^{7}P_{2}accounts for 5 terms and theAmong the states left is a state with M

_{L }= 2, M_{S }= 2. This and states with M_{L }= 2, M_{S }= 1, 0, and M_{L }= 1, M_{S }= 2, 1, 0, and M_{L}= 0, M_{S }= 2, 1, 0 yield the^{5}D term. There are 9 states with nonzero M_{J}= M_{L}+ M_{S}associated with the^{5}D term.

M_{L }= 2, M_{S }= 2 requires M_{J}= 4, and therefore theis present. It accounts for 9 of the possible 25 combinations of the^{5}D_{4}^{5}D term. Theaccounts for 7 terms, the^{5}D_{3}^{5}D_{2}accounts for 5 terms, theaccounts for 3 terms, and the^{5}D_{1}accounts for 1 term.^{5}D_{0}Among the states left is a state with M

_{L }= 2, M_{S }= 0. This and and states with M_{L }= 1, M_{S }= 0, and M_{L }= 0, M_{S }= 0, yield the^{1}D term. There are 3 states with nonzero M_{J}= M_{L}+ M_{S}associated with the^{1}D term. Therefore theterm is is present.^{1}D_{2}Among the states left is a state with M

_{L }= 0, M_{S }= 2. This and and states with M_{L }= 0, M_{S }= 1, 0 a yield the^{5}S term. Three M_{L}, M_{S}terms are associated with the^{5}S_{2}Among the states left is a state with M

_{L }= 1, M_{S }= 1. This and and states with M_{L }= 1, M_{S }= 0, and M_{L }= 0, M_{S }= 1, 0 yield the^{3}P term. There are 4 states with nonzero M_{J}= M_{L}+ M_{S}associated with the^{3}P term.

M_{L }= 1, M_{S }= 1 requires M_{J}= 2, and therefore the^{3}**P**_{2}is present. It accounts for 5 of the possible 9 combinations of the^{3}P term. The^{3}P_{1}accounts for 3 terms, and the^{3}**P**accounts for 1 term._{0}The only remaining state has M

_{L }= 0, M_{S }= 0. It corresponds to the^{1}S_{0}

The allowed terms^{2S+1}L_{J}are^{5}D_{4,3,2,1,0},^{1}D_{2},^{7}P_{4,3,2},^{3}P_{2,1,0}, and^{1}S_{0}.