Chapter 10 - MULTIELECTRON ATOMS
Problems with solutions
10.1 Find the LS terms that arise from the following configurations:
a) nsnp b) npnd c) (np)2ns
Solution:
a) 3P2,1,0 & 1P1
b) 1F3 ; 1F4,3,2,1 ; 1D2 ; 3D3,2,1 ; 1P1 ; 3P2,1,0
c) For two non-equivalent electrons we may have singlets and triplets D, P and S, six terms. We
may use a table of equivalent electrons to eliminate possible states, or we may use the following
rule: For two equivalent electrons the sum L + S must be even.
We are left with (np)2  3P, 1S and 1D.
Then
3P + s  2P
4
1/2,3/2 and P1/2.3/2.5/2
1D + s  2P
3/2,5/2
1S + s  2S
1/2
10.2 Write the complete ground state term in LS notation, i.e. 2 s 1 LJ , for the elements in the first
row of the periodic table, that is Li through Ne.
Solution:
Li: electron configuration 1s 2 2 s . L  0; S  1/ 2; J  1/ 2  2 S1/ 2
Be: electron configuration 1s 2 2s 2 . L  0; S  0; J  0  1 S 0
B: electron configuration 1s 2 2s 2 2 p . L  1; S  1/ 2; J  3 / 2,1/ 2  2 P1/ 2
C: electron configuration 1s 2 2s 2 2 p 2 . From the table of equivalent electrons we find the possible
terms are 3 P , 1 D , 1 S By Hund's rule, the lowest-lying must be the triplet. Since the p-shell is
less than half full the lowest J lies lowest.  3 P0
N: electron configuration 1s 2 2s 2 2 p3 . From the table of equivalent electrons we find the possible
terms are 4 S , 2 D , 2 P . The quartet will lie lowest. Since L  0 there is only one value of J for
this state.  4 S 3/ 2
O: electron configuration 1s 2 2s 2 2 p 4 . From the table of equivalent electrons we find the possible
terms are 3 P , 1 D , 1 S (same as carbon). By Hund's rule, the lowest-lying must be the triplet.
Since the p-shell is more than half full the highest J lies lowest.  3 P2
F: electron configuration 1s 2 2s 2 2 p5 . Possible states are the same as boron, except the highest J
will lie lowest.  2 P3/ 2
Ne: electron configuration 1s 2 2s 2 2 p 6 . Closed shell configuration so the ground state is the same
as helium, 1 S 0 .
10.3 The total number of states for given values of L and S, including M J states, is the sum of all
M J states for each possible value of J. Show that this number is  2S  1 2L  1 and verify that
it is true for 4 D states and 4 P .
Solution:
The total number of states, call it N is given by
Solution to Chapter 10 problems
page 1
Chapter 10 - MULTIELECTRON ATOMS
Problems with solutions
N
L S
  2 J  1
J LS
This assumes that L  S for which we will work it out. If L  S we merely interchange L and S
and we will get the same result. J is all possible combinations of L and S so, since it is assumed
that L  S we may write
N
LS
S
S
S
S
J LS
i  S
i  S
i  S
i  S
  2 J  1   2  L  i   1  2L  1  2  i   1
The first and last sums are  2S  1 [even if S is half-integral] and the middle one vanishes
because of symmetry so
N  2L  2S  1  0   2S  1   2S  1 2L  1
For 4 D states we have S  3 / 2 ; L  2 so the possible states are 4 D7/ 2,5/ 2,3/ 2/1/ 2 . The total
number of M J states is N  8  6  4  2  20 . Also
N   2S  1 2L  1  4 5  20
For 4 P states we have S  3 / 2 ; L  1 so the possible states are 4 P5/ 2,3/ 2/1/ 2 . The total number
of M J states is N  6  4  2  12 . Also
N   2S  1 2L  1  4 3  12
10.4 a) Write all terms for the electron configuration npn ' p in both the LS- and jj-coupling
notation.
b) Make a diagram similar to Figure 3 for the jj-coupling states showing the effects of spin-orbit
interaction and exchange and electrostatic repulsion. Put all terms in proper order.
Solution:
a) LS: S  0,1; L  0,1, 2
Possible states: 3 D3,2,1 ; 3 P2,1,0 ; 3S1 ; 1D2 ; 1P1 ; 1S0
3 1
3 1
jj: j1  , ; j2  , ; J  3, 2,1, 0
2 2
2 2
Possible states:
 1 1  1 1  1 3  1 3  3 1  3 1  3 3  3 3  3 3  3 3

 ;
 ;
 ;
 ;
 ;
 ;
 ;
 ;
 ;

 2 2 0  2 2 1  2 2 2  2 2 1  2 2 2  2 2 1  2 2 3  2 2  2  2 2 1  2 2 0
b)
Solution to Chapter 10 problems
page 2
Chapter 10 - MULTIELECTRON ATOMS
Problems with solutions
10.5 a) Write all terms for the electron configuration np 2 in both LS- and jj-coupling notation.
b) Make a diagram similar to Figure 4 showing the transition from LS- to jj-coupling. Put all
terms in proper order.
Solution:
a) LS: From the table of equivalent electrons, Table 2, we know that for LS-coupling the states
are 3 P2,1,0 ; 1D2 ; 1S0 .
jj: From the previous problem, the states for non-equivalent electrons are:
 1 1  1 1  1 3  1 3  3 1  3 1  3 3  3 3  3 3  3 3

 ;
 ;
 ;
 ;
 ;
 ;
 ;
 ;
 ;

 2 2 0  2 2 1  2 2 2  2 2 1  2 2 2  2 2 1  2 2 3  2 2  2  2 2 1  2 2 0
First, we eliminate the "duplicate" states for which j1  j2 . Arbitrarily, we keep only the ones
with lowest j2 . We thus have left:
 1 1  1 1  3 1  3 3  3 3  3 3  3 3

 ;
 ;
 ;
 ;
 ;
 ;

 2 2 0  2 2 1  2 2 1  2 2 3  2 2 2  2 2 1  2 2 0
Now we use the rule for the possible J's:
2.
If
j1  j2 then the allowed values of J are
J   2 j 1 ,  2 j  3 ,  2 j  5 ... until J becomes negative.
For the
For the
 3 3


 2 2 2
Values of J for which J   2 j  ,  2 j  2 ,  2 j  4 ... are
forbidden.
1 1
1
1 1

 states we have only J  2    1  0 so this eliminates 
.
 2 2
 2
 2 2 1
 3
 3
 3 3

 states we have J  2    1  2 or J  2    3  0 so we retain only
 2
 2
 2 2
 3 3
1 1  3 1  3 1  3 3 3 3
and 
 . We are left with 
 ;
 ;
 ;
 ;
 .
 2 2 0
 2 2 0  2 2 2  2 2 1  2 2 2  2 2 0
Solution to Chapter 10 problems
page 3
Chapter 10 - MULTIELECTRON ATOMS
Problems with solutions
10.6 An excited configuration of the Ca atom is:  Ar   3d 
a) What are the allowed LS terms?
b) A particular multiplet of a Ca atom having the above electron configuration is observed to
have the energy spacing between adjacent J levels as follows:
4
EJ  EJ 1  E0 and EJ 1  EJ 2  E0 where E0 is a constant. What is J? What is the LS term
3
designation 2 S 1 L of the multiplet for which the energy levels are as shown?
Solution:
J
2
4E
3 0
J–1
E0
J–2
a) The possible values of L are 2+2, 2+2–1,...,2–2.
So L = 4, 3, 2, 1, 0. We may therefore have G, F, D, P & S states.
The total spin can be S = 0 or 1.
If these were not equivalent electrons we would thus have the following terms:
1S
1P
1D
1F
1G and
3S
3P
3D
3F
3G
But, for two equivalent electrons the sum L + S must be even.
Therefore, we may strike out those shown.
1S
1P
1D
1F
1G and
3S
3P
3D
3F
3G
b) Since there is more than one J-state the singlets are immediately eliminated. The multiplet
must be either the 3P (for which J = 0, 1, or 2) or 3F (for which J = 1, 2, or 3). Moreover, the
levels shown are all of the levels since the highest multiplicity is 3.
Solution to Chapter 10 problems
page 4
Chapter 10 - MULTIELECTRON ATOMS
Problems with solutions
Using the Landé interval rule we may write the following equations
4
E J  E J 1  KJ  E0
3
E J 1  E J 2  K J  1  E0
where J is the total angular momentum of the highest J -state.
Dividing,
4
E0
KJ
J
4
3

 3J  4 J  4  J  4
so that
J 1 3
K  J  1 E 0
Now, the highest J-state for the 3P multiplet is J = 2. This eliminates the 3P multiplet. On the
other hand, the highest J-state for the 3F multiplet is J = 4 so the multiplet in question is the 3F.
Solution to Chapter 10 problems
page 5