Multi-electron atoms

Problem:

Give the electronic configuration for the Boron atom (Z = 5).  Then enumerate the allowed term symbols ^2S+1L_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+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 ²P_½ and ²P_3/2.
  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+1L_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+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 3^rd  rule),  Term symbol: ²P_½
  (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 3^rd  rule),  Term symbol: ²P_3/2
  (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 ³P_2,0,1 ¹D₂ ³D_3,2,1<

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 + U_c(r₁) + U_c(r₂)
  is the Hamiltonian in the central field approximation.
  W = e²/r₁₂ - U_c(r₁) - U_c(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)  1s²:  ^2S+1L_J = ¹S₀
  1s2s:  ^2S+1L_J = ¹S₀, ³S₁.
  1s2p:  ^2S+1L_J = ¹P₁, ³P_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 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+1L_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+1L_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:

  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 M_L and M_s 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 M_L = 2, M_s = 0 term shows that a ¹D 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 ³P term.  Four M_L, M_S terms are associated with the ³P term. 
  The only remaining state has M_L = 0, M_S = 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 M_L = 1, M_S = 1 requires M_J = M_L + M_S = 2 and therefore the term ³P₂ 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+1L_J are ¹D₂, ³P_2,1,0, 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 M_L = ∑m_l and M_S = ∑m_s.
   

  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 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 ²D 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 ²P 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 ⁴S 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(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+1L_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+1L.

• 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:

  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 M_L and M_s 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 M_L = 1, M_s = 3 term shows that a ⁷P 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 ⁷P term.
  M_L = 1, M_S = 3 requires M_J = 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 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 ⁵D term.  There are 9 states with nonzero M_J = M_L + M_S associated with the ⁵D term.
  M_L = 2, M_S = 2 requires M_J = 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 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 ¹D term.  There are 3 states with nonzero M_J = M_L + M_S associated with the ¹D term.  Therefore the ¹D₂ term is is present.

  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 ⁵S term.  Three M_L, M_S terms are associated with the ⁵S₂ term.

  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 ³P term.  There are 4 states with nonzero M_J = M_L + M_S associated with the ³P term.
  M_L = 1, M_S = 1 requires M_J = 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 M_L = 0, M_S = 0.  It corresponds to the ¹S₀ term.
  The allowed terms ^2S+1L_J are ⁵D_4,3,2,1,0,  ¹D₂, ⁷P_4,3,2, ³P_2,1,0, ⁵S₂ and ¹S₀.