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) 3P2,1,01P1
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 sumL + S must be even.
We are left with (np)2 3P,1S and1D.
Then
3P + s2P1/2,3/2and4P1/2.3/2.5/2
1D + s2P3/2,5/2
1S + s2S1/2
10.2 Write the complete ground state term in LS notation, i.e. , for the elements in the first row of the periodic table, that is Li through Ne.
Solution:
Li: electron configuration .
Be: electron configuration .
B: electron configuration .
C: electron configuration . From the table of equivalent electrons we find the possible terms are , , 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.
N: electron configuration . From the table of equivalent electrons we find the possible terms are , , . The quartet will lie lowest. Since there is only one value of J for this state.
O: electron configuration . From the table of equivalent electrons we find the possible terms are , , (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.
F: electron configuration . Possible states are the same as boron, except the highest J will lie lowest.
Ne: electron configuration . Closed shell configuration so the ground state is the same as helium, .
10.3 The total number of states for given values of L and S, including states, is the sum of all states for each possible value of J. Show that this number is and verify that it is true for states and .
Solution:
The total number of states, call it N is given by
This assumes that for which we will work it out. If 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 we may write
The first and last sums are [even if S is half-integral] and the middle one vanishes because of symmetry so
For states we have so the possible states are. The total number of states is . Also
For states we have so the possible states are . The total number of states is . Also
10.4a) Write all terms for the electron configuration 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:
Possible states:
jj:
Possible states:
b)
10.5a) Write all terms for the electron configuration 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 .
jj: From the previous problem, the states for non-equivalent electrons are:
First, we eliminate the "duplicate" states for which . Arbitrarily, we keep only the ones with lowest . We thus have left:
Now we use the rule for the possible J's:
2. If then the allowed values of J are until J becomes negative. Values of J for which are forbidden.
For the states we have only so this eliminates .
For the states we have so we retain only
and . We are left with .
10.6An excited configuration of the Ca atom is:
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:
whereE0 is a constant. What is J? What is the LS term designationof the multiplet for which the energy levels are as shown?
Solution:
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.
Using the Landé interval rule we may write the following equations
where J is the total angular momentum of the highest J -state.
Dividing,
so that
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 1