### Multi-electron atoms

#### Problem:

Give the electronic configuration for the Boron atom (Z = 5).  Then enumerate the allowed term symbols 2S+1LJ 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+1L.
• 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 2P½ and 2P3/2.
Hund's rule:  The component with the smallest value of J (J = ½) has the lowest energy.

#### Problem:

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+1LJ  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+1L.
• Details of the calculation:
(a)  Electronic configuration, 1s2 2s2 2p6 3s2 3p1 = [Ne] 3s2 3p1
(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 3rd  rule),  Term symbol: 2P½
(c)  For the open 2p shell (2p5) we have L = 1, S = ½.
Possible values for J:  J = ½, J = 3/2.
Lowest energy term:  J = 3/2, (Hund's 3rd  rule),  Term symbol: 2P3/2
(d)  Electronic configuration, 1s2 2s2 2p5 3s2 3p1
Possible values for L:  L = 0, 1, 2
(e)  Possible values for S:  S = 0, 1
(f)  1S0 3S1 1P1 3P2,0,1 1D2 3D3,2,1<

#### Problem:

(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+1L.
• Details of the calculation:
(a)  H0 = p12/(2m) + p22/(2m) - 2e2/r1 - 2e2/r2 + Uc(r1) + Uc(r2)
is the Hamiltonian in the central field approximation.
W = e2/r12 - Uc(r1) - Uc(r2)
is the dominant correction term.  It is due to the non-central part of the electron-electron interaction.
The eigenvalues of H0 depend on n1, l1 and n2, l2
Configuration of the ground state: 1s2
Configuration of the first excited state: 1s2s
Configuration of the second excited state: 1s2p
(b)  1s22S+1LJ = 1S0
1s2s:  2S+1LJ = 1S0, 3S1.
1s2p:  2S+1LJ = 1P1, 3P0,1,2.
(c)  Selection rules for allowed transitions:  ∆j = 0, ±1, (except ji = jf = 0), ∆l = ±1, ∆mj = 0, ±1.
∆l = ±1 is only satisfied for the 1s2p state.
The transition 1P1 --> 1S0 is allowed.

#### Problem:

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 np2, 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+1LJ.

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+1LJ.  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:
ml = 1, 0, -1,  ms = ½,- ½.
Combining the different values of ml and ms, we obtain the following possible states:

 ml = 1 ms = ½ (1+) ml = 0 ms = ½ (0+) ml = -1 ms = ½ (-1+) ml = 1 ms = -½ (1-) ml = 0 ms = -½ (0-) ml = -1 ms = -½ (-1-)

In each of these states there cannot be more than one electron.
The following states with non-negative values of ML and Ms are possible if the two electrons are not allowed to have exactly the same quantum numbers.

 State ML Ms (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 ML = 2, Ms = 0 term shows that a 1D term is among the possible terms.  To this term we must further assign states with ML = 1, Ms = 0 and ML = 0, Ms = 0.
Among the states left is a state with ML = 1, MS = 1.  This and states with  ML = 1, MS = 0, and ML = 0, MS = 1, and ML= 0, MS = 0 yield the 3P term.  Four ML, MS terms are associated with the 3P term.
The only remaining state has ML = 0, MS = 0.  It corresponds to the 1S term.
For the 1D and 1S term we have S = 0, and therefore J = L.
For the 3P 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 ML = 1, MS = 1 requires MJ = ML + MS = 2 and therefore the term 3P2 is allowed.  We must have 5 terms associated with J = 2 for the 5 possible values of MJ.  This leaves 4 terms associated with J = 1 and J = 0.
The allowed terms 2S+1LJ are 1D2, 3P2,1,0, and 1S0.

#### Problem:

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+1L.
• Details of the calculation:
The Nitrogen atom has 7 electrons.
Electronic configuration, 1s2 2s2 2p3 = [He] 2s2 2p3
The 1s and 2s shells are filled and have L = 0, S = 0.
The 2p shell is half filled.
The possible terms 2S+1L for 3 equivalent p electrons (np)3 are 4S, 2P, and 2D.
The p-electron can have:
 ml = 1 0 -1 1 0 -1 ms = 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 ML = ∑ml and MS = ∑ms.

 State ML MS 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 ML and the largest MS.
ML = 2, Ms = ½.  The presence of the ML = 2, Ms = ½ term shows that a 2D term is among the possible terms.  To this term we must further assign states with ML = 1,0 , MS = ½.
What is left?
ML = 1, Ms = ½.  The presence of the ML = 1, Ms = ½ term shows that a 2P term is among the possible terms.  To this term we must further assign states with ML = 0, MS = ½.
What is left?
ML = 0, Ms = 3/2.  The presence of the ML = 0, Ms = 3/2 term shows that a 4S term is among the possible terms.  To this term we must further assign states with ML = 0, MS = 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(4S) < E(2D) < E(2P).

#### Problem:

For an atom with 3 valence electrons, find the spectroscopic terms 2S+1L 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+1L.
• 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 L12 S12 0 0, 1 1S, 3S 1 0, 1 1P, 3P 2 0, 1 1D, 3D

Now add to this the angular momentum of the 5p electron.

 (3p)(4p)(5p) Possible terms L12 S12 l s L, (L = L12 + l) S, (S = S12 + s) 0 0 1 ½ 1 ½ 2P 1 0 1 ½ 0, 1, 2 ½ 2S, 2P, 2D 2 0 1 ½ 1, 2, 3 ½ 2P, 2D, 2F 0 1 1 ½ 1 ½, 3/2 2P, 4P 1 1 1 ½ 0, 1, 2 ½, 3/2 2S, 2P, 2D, 4S, 4P, 4D 2 1 1 ½ 1, 2, 3 ½, 3/2 2P, 2D, 2F, 4P, 4D, 4F

Spectroscopic terms:

 2S+1L 2S 4S 2P 4P 2D 4D 2F 4F # 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 L12 S12 0 (e) 0 (o) 1S 1 (o) 1 (e) 3P 2 (e) 0 (o) 1D

Now add to this the angular momentum of the 4p electron.

 (3p)2(4p) Possible terms L12 S12 l s L, (L = L12 + l) S, (S = S12 + s) 0 0 1 ½ 1 ½ 2P 1 1 1 ½ 0, 1, 2 ½, 3/2 2S, 2P, 2D, 4S, 4P, 4D 2 0 1 ½ 1, 2, 3 ½ 2P, 2D, 2F

Spectroscopic terms:

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

#### Problem:

If the spin of the electron were (3/2)ħ,what spectroscopic terms 2S+1LJ 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+1L.
• Details of the calculation:
For each electron the following values are possible:
ml = 1, 0, -1,  ms = 3/2, ½, -½, -3/2
Combining the different values of ml and ms, we obtain the following possible states:

 ml = 1 ms = ½ (1+) ml = 0 ms = ½ (0+) ml = -1 ms = ½ (-1+) ml = 1 ms = -½ (1-) ml = 0 ms=-½ (0-) ml = -1 ms = -½ (-1-)
 ml = 1 ms = 3/2 (1++) ml = 0 ms = 3/2 (0++) ml = -1 ms = 3/2 (-1++) ml = 1 ms = -3/2 (1--) ml = 0 ms=-3/2 (0--) ml = -1 ms = -3/2 (-1--)

In each of these states there cannot be more than one electron.
The following states with non-negative values of ML and Ms are possible if the two electrons are not allowed to have exactly the same quantum numbers.

 State ML MS State ML MS State ML MS State ML MS (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 ML = 1, Ms = 3 term shows that a 7P term is among the possible terms.  To this term we must further assign states with ML = 1, MS = 2, 1, 0 and ML = 0, Ms = 3, 2, 1, 0.  There are 8 states with nonzero MJ = ML + MS associated with the 7P term.
ML = 1, MS = 3 requires MJ = 4, and therefore the 7P4 is present.  It accounts for 9 of the possible 21 combinations of the 7P term.  The 7P3 accounts for 7 terms and the 7P2 accounts for 5 terms and the

Among the states left is a state with ML = 2, MS = 2.  This and states with  ML = 2, MS = 1, 0, and ML = 1, MS = 2, 1, 0, and ML= 0, MS = 2, 1, 0 yield the 5D term.  There are 9 states with nonzero MJ = ML + MS associated with the 5D term.
ML = 2, MS = 2 requires MJ = 4, and therefore the 5D4  is present.  It accounts for 9 of the possible 25 combinations of the 5D term.  The 5D3 accounts for 7 terms, the 5D2 accounts for 5 terms, the 5D1 accounts for 3 terms, and the 5D0 accounts for 1 term.

Among the states left is a state with ML = 2, MS = 0.  This and and states with  ML = 1, MS = 0, and ML = 0, MS = 0, yield the 1D term.  There are 3 states with nonzero MJ = ML + MS associated with the 1D term.  Therefore the 1D2 term is is present.

Among the states left is a state with ML = 0, MS = 2.  This and and states with  ML = 0, MS = 1, 0 a yield the 5S term.  Three ML, MS terms are associated with the 5S2 term.

Among the states left is a state with ML = 1, MS = 1.  This and and states with  ML = 1, MS =  0, and ML = 0, MS = 1, 0 yield the 3P term.  There are 4 states with nonzero MJ = ML + MS associated with the 3P term.
ML = 1, MS = 1 requires MJ = 2, and therefore the 3P2  is present.  It accounts for 5 of the possible 9 combinations of the 3P term.  The 3P1 accounts for 3 terms, and the 3P0 accounts for 1 term.

The only remaining state has ML = 0, MS = 0.  It corresponds to the 1S0 term.
The allowed terms 2S+1LJ are 5D4,3,2,1,01D2, 7P4,3,2, 3P2,1,0, and 1S0.