Download Answers/solutions

Assignment 3—Answers/solutions
11.1 How many electrons can be put in each of the following：(a) a shell with principal quantum number
n ; (b) a subshell with quantum number n and l ; (c) an orbital; (d) a spin-orbital?
Answer:
(a) The total number of states in the shell with principal quantum number n can be calculated with the following diagram:
l=
0,
1,
2, …
m=
0
-1, 0, 1,
-2, -1, 0, 1, 2….
–n+1, -n+2,
ms= -1,1
-1,1 -1,1 -1,1 -1,1 -1,1, -1,1, -1,1 -1,1 -1,1
-1,1
Number of states 2+6+10+…+2(2n-1)=2(1+3+5+…+2n-1)=2(2n-1+1)n/2=2n2.
n-1
n-2, n-1
-1,1 –1,1
-1,1
(b) When l is also fixed, the number of the states in the sub-shell is 2(2l+1)=4l+2.
(c) When n,l and m are fixed (orbital), the number of state is 2
(d) When all quantum numbers n, l, m and ms are all fixed (spin orbital), the number of state is 1.
11.2 If R (r1) is the radial factor in function t1 in the Hartree differential equation (11.9), write the
differential equation satisfied by R.
Answer:
From equation (6.17), we have

2
2me
[
 2 R ( r1)
 2r1

2
r1
R(r1 )]  V (r1 )R(r1 )   1 R(r1 )
11.3 Which STOs have the same form as hydrogenlike AOs?
Answer:
The angular part for both are the same because they both use spherical harmonics.
The only difference is in the radial part. Compare the hydrogen-like AOs [equation (6.100), also see Table
6.1, Table 6.2] and STOs [equation (11.14)], we readily find that the STOs with n≧1 have the same form
as hydrogen-like AOs.
11.4 Estimate the nonrelativistic 1s orbital energy in Ar; check with Fig.11.2.
Answer:
The nonrelativistic energy of the 1s orbital of Ar is given by
E1s(Ar)=182EH [E1s(Ar)/EH ]1/2=18.
Comparing with Fig. 11.2 which gives [E1s(Ar)/EH ]1/2～ 17.4, we find above estimate was lower. That is
expected because we didn’t include inter-electronic repulsions which contribute a positive value to
the energy.
11.5 At what atomic number does the second crossing of the 3d and 4s orbital energies occur in Fig. 11.2?
Take account of the logarithmic scale. (Atomic spectral data show that the crossing actually occurs
between 20 and21.)
Answer:
Measure the 3d/4s crossing and we find it occurs at atomic number=22. This means the HF
calculations made by Latter either overestimated the energy of 3d or underestimated the energy of 4s,
or both.. this means inter-electronic repulsions in 3d calculation were overestimated or that in 4s
calculation underestimated or both. Considering the limitation of HF method which normally does
not include configuration interactions, this type of error is possible.
11.6 Verify the angular-momentum commutation relations (11.32).
Answer:
[Mx,M12]=[ M1x+M2x, M12]
=[M1x,M12] + [M2x,M12]
=0+0 (using equation 11.29)
=0.
The other equations can be proved analogously.
11.7 Prove that m1+m2=MJ【Eq.(11.34)】as follows: Apply Mz=M1z+M2z to (11.33), substitute (11.33) in
the resulting equation, combine term, and use the linear independence of the function involved to
deduce (11.34).
Answer:
Applying Mz to equation (11.34), we have
the right hand side = Mz|j1,j2,J,MJ>=MJ|j1,j2,J,MJ>=MJΣC(j1,j2,J,MJ;m1,m2)|j1m1>|j2m2> (1)
and
the left hand side = MzΣC(j1,j2,J,MJ;m1,m2)|j1m1>|j2m2>=ΣC(j1,j2,J,MJ;m1,m2)(m1+m2)|j1m1>|j2m2>
=(m1+m2) ΣC(j1,j2,J,MJ;m1,m2)|j1m1>|j2m2>
(2)
Comparing equations 1and 2, since {|j1m1>|j2m2>} are independent functions, the equality of (1) and (2)
necessarily leads to
MJ=(m1+m2)
11.8 Show that [M 2, M1z]=2iħ (M1x M2y - M1y M2x), where M=M1+M2
Answer:
[M 2, M1z]= [(M1 +M2) 2, M1z]= [M12+M2 2+2M1M2 , M1z]
=2[M1M2 , M1z]
=2{[M1xM2x , M1z]+[M1yM2y , M1z]+[ M1zM2z , M1z]}
=2(-iħ M1yM2x+iħ M1xM2y)
=2iħ (M1x M2y - M1y M2x)
11.9 Give the possible values of the total-angular-momentum quantum number J that result from the
addition of angular momenta with quantum numbers (a) 3/2 and 4 ;(b) 2, 3,and 1/2.
Answer:
(a) 4+3/2, 4+3/2-1, 4+3/2-2, 4+3/2-3 (=4-3/2)11/2, 9/2, 7/2, 5/2.
(b) Addition of the first two momenta: 5, 4, 3, 2, 1.
Added to by the third one, we have the following possible total angular momenta:
11/2, 9/2;
9/2, 7/2;
7/2, 5/2;
5/2, 3/2;
3/2, 1/2.
11.11 Find the terms that arise from each of the following electron configuration:(a) 1s22s22p63s23p5g;(b)
1s22s22p3p3d;(c) 1s22s22p44d.You may use Table 11.2a for part(c).
Answer:
(a) L1=1, S1=1/2, L2=4, S2=1/2.
L=5,4,3; S=1,0.Terms: 3H,3G,3F and 1H,1G,1F.
(b) L1=1, S1=1/2, L2=1, S2=1/2, L3=2, S3=1/2.
L1+L2: 2, 1, 0;
S1+S2: 1, 0;
L1+L2+L3: 4,3,2,1,0; 3,2,1; 2; S1+S2+S3: 3/2, 1/2; 1/2.
4S, 4P, 4D, 4D, 4D, 4F, 4F,4G; 2S, 2P, 2D, 2D, 2D, 2F, 2F,2G; 2S, 2P, 2D, 2D, 2D, 2F, 2F,2G.
(c) Table 11.2a tells us the terms arising from p4 configuration are
3 1
P, D 1S
3
P: L1=1, S1=1; L2=2, S2=1/2.
Possible L: 3, 2, 1;
Possible S: 3/2, 1/2.
The terms are:
4 4
P, D, 4F; 2P, 2D, 2F.
(1)
1
D: L1=2, S1=0; L2=2, S2=1/2.
Possible L: 4,3, 2, 1,0;
Possible S: 1/2.
The terms are:
2
S, 2P, 2D, 2F, 2G.
(2)
1
S: L1=0, S1=0; L2=2, S2=1/2.
Possible L: 2;
Possible S: 1/2.
The terms are:
2
D
(3)
Summarizing (1-3), the possible terms are: 2D, 2S, 2P, 2D, 2F, 2G, 4P, 4D, 4F; 2P, 2D, 2F.
11.12 Which of the following electron configurations will contribute configuration functions to a CI
calculation of the ground state of He? (a) 1s2s; (b) 1s2p; (c) 2s2;
(d) 2s2p;(e) 2p2;(f) 3d2.
Answer:
The configuration functions that would contribute to a CI should have the same angular momentum
as that of the state to be corrected. The ground state of He has zero angular momentum. Thus 1s2s,
2s2, 2p2 and 3d2 have contributions (to that particular state).
11.13 Verify the spin-eigenfunction equations (11.52) to (11.54), (11.56), and the three equations
following (11.56).
Answer:
(Take ħ as the unit of angular momentum).
Szβ(1)β(2)= (S1z + S2z)β(1)β(2)=(-1/2β(1)β(2)-1/2β(1)β(2)=- β(1)β(2).
S2[α(1)β(2)+β(1)α(2)]=[2α(1)β(2)+2β(1)α(2)]=2[α(1)β(2)+β(1)α(2)].
The reminder can be proved exactly the same way.
11.14 (a) Calculate the angle in Fig. 11.3 between the z axis and S for the spin function α(1) α(2).(b)
Calculate the angle between S1 and S2 for each of the functions (11.57) to (11.60). 〔Hint :One
approach is to use the law of cosines. A second approach is to use S·S=(S1+S2)·(S1+S2). 〕(c)If a
vector A has components (Ax, Ay , Az), what are the components of the projection of A in the xy
plane ? Use the answer to this question to find the angle between the projection S1 and S2 in the xy
plane for the function α(1) α(2).
Answer:
(a) Sz=1, |S2|1/2=21/2angle=cos-1(2)1/2=450.
(b) Using law of cosines: S1·S2= |S1||S2|cos(θ12)
cos(θ12)= S1·S2/|S1||S2|=-[S2-S12-S22]/2[|S12||S22|]1/2
For (11.57): S2=2, S12=3/4, S12=3/4 cos(θ12)=1/3 θ12=70.30
For (11.58): S2=2, S12=3/4, S12=3/4 cos(θ12)=1/3 θ12=70.30
The same for (11.59).
For (11.60), S2=0, S12=3/4, S12=3/4 cos(θ12)=-1 θ12=1800
(d) The xy-plane component can be calculated from Sxy2=S2-Sz2=3/4-1/4=1/2
|S1xy|=1/√2, |S2xy|=1/√2.  possible values:
(S1x, S1y)=(±1/2,±1/2), (S2x, S2y)=(±1/2,±1/2),
Possible angles between S1xy and S2xy are 00, 900, 1800,2700.
Because they are not co-incidental, the angle between the two components is 900(or 2700).
11.15 Find the Ŝ2 and Ŝz eigenvalues for the spin function
3-1/2〔α(1) α(2) β(3)+ α(1)β(2) α(3)+ β(1) α(2) α(3)〕
Answer:
Ŝ23-1/2〔α(1) α(2) β(3)+ α(1)β(2) α(3)+ β(1) α(2) α(3)〕=3-1/2〔Ŝ2α(1) α(2) β(3)+ Ŝ2α(1)β(2) α(3)+ Ŝ2 β(1)
α(2) α(3)〕
Using 2Ŝ1Ŝ2 =Ŝ1+Ŝ2- +Ŝ1-Ŝ2++2Ŝ1zŜ2z
2Ŝ1Ŝ3 =Ŝ1+Ŝ3- +Ŝ1-Ŝ3++2Ŝ1zŜ3z
2Ŝ2Ŝ3 =Ŝ2+Ŝ3- +Ŝ2-Ŝ3++2Ŝ2zŜ3z
we have:
Ŝ2=(Ŝ21+Ŝ22 +Ŝ23+2Ŝ1Ŝ2 +2Ŝ1Ŝ3+2Ŝ2Ŝ3)
=(Ŝ21+Ŝ22 +Ŝ23+ Ŝ1+Ŝ2- +Ŝ1-Ŝ2++2Ŝ1zŜ2z+ Ŝ1+Ŝ3- +Ŝ1-Ŝ3++2Ŝ1zŜ3z +Ŝ2+Ŝ3- +Ŝ3+Ŝ2-+2Ŝ2zŜ3z)]
Ŝ2α(1) α(2) β(3)= (Ŝ21+Ŝ22 +Ŝ23+ Ŝ1+Ŝ2- +Ŝ1-Ŝ2++2Ŝ1zŜ2z+ Ŝ1+Ŝ3- +Ŝ1-Ŝ3++2Ŝ1zŜ3z +Ŝ2+Ŝ3- +Ŝ2-Ŝ3++2Ŝ2zŜ3z)
α(1) α(2) β(3)
= ħ2 [3/4α(1) α(2) β(3)+3/4α(1) α(2) β(3)+3/4α(1) α(2) β(3)+0α(1) α(2) β(3)+0α(1) α(2) β(3)+1/2α(1)
α(2) β(3)+0α(1) α(2) β(3)+ α(3) α(2) β(1)-1/2α(1) α(2) β(3)+0α(1) α(2) β(3)
+α(1) α(3) β(2)-1/2α(1) α(2) β(3)]
= ħ2[7/4α(1) α(2) β(3)+ α(3) α(2) β(1)+ α(1) α(3) β(2)]
(1)
Similarly, we can find that
Ŝ2α(1)β(2) α(3)= (Ŝ21+Ŝ22 +Ŝ23+ Ŝ1+Ŝ2- +Ŝ1-Ŝ2++2Ŝ1zŜ2z+ Ŝ1+Ŝ3- +Ŝ1-Ŝ3++2Ŝ1zŜ3z +Ŝ2+Ŝ3+Ŝ2-Ŝ3++2Ŝ2zŜ3z)α(1)β(2) α(3)
= ħ2[7/4α(1)β(2) α(3)+ α(3) α(2) β(1)+ α(1) α(2) β(3)]
(2)
2
2
2
2
Ŝ β(1) α(2) α(3)= (Ŝ 1+Ŝ 2 +Ŝ 3+ Ŝ1+Ŝ2- +Ŝ1-Ŝ2++2Ŝ1zŜ2z+ Ŝ1+Ŝ3- +Ŝ1-Ŝ3++2Ŝ1zŜ3z +Ŝ2+Ŝ3- +Ŝ2-Ŝ3++2Ŝ2zŜ3z)
β(1) α(2) α(3)
= ħ2[7/4β(1) α(2) α(3)+ α(3) α(1) β(2)+ α(1) α(2) β(3)]
(3)
Adding up (1,2,3), we get
Ŝ23-1/2〔α(1) α(2) β(3)+ α(1)β(2) α(3)+ β(1) α(2) α(3)〕=3-1/2〔Ŝ2α(1) α(2) β(3)+ Ŝ2α(1)β(2) α(3)+ Ŝ2 β(1)
α(2) α(3)〕= 15ħ2/4 3-1/2〔α(1) α(2) β(3)+ α(1)β(2) α(3)+ β(1) α(2) α(3)〕=3-1/2〔Ŝ2α(1) α(2) β(3)+
Ŝ2α(1)β(2) α(3)+ Ŝ2 β(1) α(2) α(3)〕The eigenvalue for Ŝ2 in this state is 15ħ2/4.
Ŝz3-1/2〔α(1) α(2) β(3)+ α(1)β(2) α(3)+ β(1) α(2) α(3)〕
=3-1/2 [Ŝzα(1) α(2) β(3)+ Ŝzα(1)β(2) α(3)+ Ŝzβ(1) α(2) α(3)]
[using Ŝzα(1) α(2) β(3)= ħ(1/2+1/2-1/2) α(1) α(2) β(3)= ħ/2 α(1) α(2) β(3);
Ŝzα(1)β(2) α(3)= ħ/2α(1)β(2) α(3); Ŝzβ(1) α(2) α(3) = ħ/2β(1) α(2) α(3)]
= ħ/2 3-1/2 [α(1) α(2) β(3)+ α(1)β(2) α(3)+ β(1) α(2) α(3)]
The eigenvalue for Ŝz is ħ/2.
11.16 (a) If Ŝ2=( Ŝ1+ Ŝ2+…)· ( Ŝ1+ Ŝ2+…),show that〔Ŝ2,Pik〕=0.(b) Show that〔L2,Pik〕=0,where L is the
total electronic orbital angular momentum.
Answer:
Take an arbitrary wavefunction for the n-spin system |a1a2..ai…ak…an> which can be expanded in
terms of the eigenfunctions of Ŝ2:
Ŝ2|a1a2..ai…ak…an>=ΣC1,2,…i,,,,k, n Ŝ2|b1b2..bi…bk…bn >=ΣC1,2,…i,,,,k, n S2(b1b2..bi…bk…bn ) |b2..bi…bk…bn
>
[Pik,Ŝ2]|a1a2..ai…ak…an>= Pik ( Ŝ1+ Ŝ2+…Ŝi+…+ Ŝk+…+Ŝn)·( Ŝ1+ Ŝ2+…Ŝi+…+ Ŝk+…+Ŝn)
|a1a2..ai…ak…an >-( Ŝ1+ Ŝ2+…Ŝi+…+ Ŝk+…+Ŝn)·( Ŝ1+ Ŝ2+…Ŝi+…+ Ŝk+…+Ŝn) Pik|a1a2..ai…ak…an>
=Pik ΣC1,2,…i,,,,k, n S2(b1b2..bi…bk…bn)|b1b2..bi…bk…bn >-ΣC1,2….k, .i…n S2(b1b2..bk…bi, bn)
|b1b2..bk…bi…bn >
=ΣC1,2….k, .i…n S2(b1b2..bk…bi, bn) |b1b2..bk…bi…bn >-ΣC1,2….k, .i…n S2(b1b2..bk…bi, bn) b1b2..bk…bi…bn
>
=0〔Ŝ2,Pik〕=0.
[An alternative, less exact method:
PikŜ2Pik-1 = Pik (Ŝ1+ Ŝ 2+…Ŝi+…+ Ŝk+…+Ŝn)·( Ŝ1+ Ŝ2+…Ŝi+…+ Ŝk+…+Ŝn)Pik-1
=(Ŝ1+ Ŝ 2+…Ŝi+…+ Ŝk+…+Ŝn)·( Ŝ1+ Ŝ2+…Ŝk+…+ Ŝi+…+Ŝn)
= Ŝ2  〔Ŝ2,Pik〕=0.]
(b). Follow the same procedure as (a).
11.17 Of the atoms with Z≦10,which have ground states of odd parity?
Answer:
Parity=the sum of the orbital angular momentum quantum numbers of the electrons
Z=1-4 all have l=0, Z=5, l=1odd parity (B), Z=7 (N) and Z=9 (F).
11.18 Give the number of states that belong to each the following terms:
(a) 4F;(b) 1S;(c) 3P;(d) 2D.
Answer:
(a) 4F 4×7=28.
(b) 1S1×1=1.
(c) 3P3×3=9.
(d) 2D2×5=10.
11.19 How many states belong to each of the following carbon configurations?
(a) 1s22s22p2;(b) 1s22s22p3p.
Answer:
(e) Terms: 3P ,1D,1S9+5+1=15.
(f) L=2,1,0,S=1,0,Terms: 3D ,3P, 3S, 1D ,1P, 1S 15+9+3+5+3+1=36.
11.20 Give the possible spin multiplicities of the terms that arise from each of the following electron
configuration: (a) f;(b) f2;(c) f3;(d) f7;(e) f12;(f) f13.
Answer:
Multiplicity=2S+1.
(a) S=1/2, multiplicity =2.
(b) S=1,0, multiplicity =3,1.
(c) S can be found by adding (a) and (b)S=3/2,1/2, multiplicity =4,2.
(d) From (b), two two-electron systems (four electrons in total) add up to S=2,1,0; which is then
added by a three-electron system S=7/2,5/2,3/2,1/2multiplicity=8,6,4,2.
For half-filled subshell, the ground state is S.term.
(e) The same as (b).
(f) The same as (a).
11.21 Give the levels that arise from each of he following term, and give the degeneracy of each level: (a)
1
S; (b) 2S; (c) 3F; (d) 4D.
Answer:
J=L+S,L+S-1, … |L-S|.
Degeneracy =(2L+1)×(2S+1) or ∑J=L+S J=|L-S| (2J+1):
(a) L=0, S=0, (J=0)  level: 1S0 degeneracy =1.
(b) L=0, S=1/2, (J=1/2)  level: 1S1/2 degeneracy =2.
(c) L=3, S=1, (J=4,3,2,1)  levels: 3F4,3,2,1 degeneracies =9,7,5,3.
(d) L=2, S=3/2, (J=7/2,5/2,3/2,1/2)  levels: 3D7/2,5/2,3/2,1/2 degeneracies =8,6,4,2.
11.22 For a state belonging to a 3D3 level, give the magnitude of (a) the total electronic orbital angular
momentum; (b) the total electronic spin angular momentum; (c) the total electronic angular
momentum.
Answer:
L=2,S=1,J=3 (a) |L|=[2(2+1)] 1/2 ħ
(b) |S|=[1(1+1)] 1/2 ħ
(c) |J|=[3(3+1)]1/2 ħ
11.23 Give the symbol for the ground level of each of the atoms with Z≦10.
Answer:
Z=1, ground configuration: 1s1
L=0, S=1/2 J=1/2 the only term
2
S1/2.
Z=2, ground configuration: 1s2
The equivalent electrons s2 are found from Table 11.2a: 1S, there is only one possible value for J (=0).
Therefore the ground state symbol is 1S0.
Z=3, ground configuration: 1s22s
L=0, S=1/2 J=1/2 the only term 2S1/2,.
Z=4, ground configuration: 1s22s2
Same as Z=2.
Z=5 ground configuration: 1s22s22p
2
P term (L=1,S=1/2J=3/2,1/2). The subshell is less than half-filled, ground state symbol: 2P1/2.
Z=6 ground configuration: 1s22s22p2
Table 11.2a + Hund’s rule3P. +less than half-filled  ground state symbol: 3P0.
Z=7 ground configuration: 1s22s22p3
Table 11.2a + Hund’s rule4S (only one J value: 3/2). ground state symbol: 4S3/2.
Z=8 ground configuration: 1s22s22p4
Table 11.2a + Hund’s rule3P. +more than half-filled  ground state symbol: 3P2.
Z=9 ground configuration: 1s22s22p5
Table 11.2a 2P. +more than half-filled  ground state symbol: 2P3/2.
Z=10 ground configuration: 1s22s22p6
Table 11.2a 1S. ground state symbol: 1S0.
11.24 Give the symbol for the ground level of each of the atoms with 21≦Z≦30.Which one of these
atoms has the most degenerate ground level?
Answer:
Z=21, ground configuration: 1s22s22p63s23p63d3
The equivalent electrons 3d3 are found from Table 11.2a: 4F, 4P, 2H, 2G, 2F, 2D(2), 2P,
Hund’s rule gives the lowest term 4F which has states 4F9/2 , 4F7/2 , 4F5/2 , 4F3/2 .
Because it is less than half-filled (inverted), the lowest state therefore is 4F3/2.
Z=22, ground configuration: 1s22s22p63s23p63d4
The equivalent electrons 3d4 are found from Table 11.2a. Hund’s rule gives the lowest term 5D which has
states 5D4 , 5D3 , 5D2 , 5D1. 5D0. Because it is less than half-filled (inverted), the lowest state therefore is
5
D0.
Z=23, ground configuration: 1s22s22p63s23p63d5
Table 11.2a + Hund’s rule  6S which has state 6S5/2 . (Half-filled).
Z=24, ground configuration: 1s22s22p63s23p63d6
Table 11.2a + Hund’s rule5D. +more than half-filled  ground state symbol: 5D4
Z=25, ground configuration: 1s22s22p63s23p63d7
Table 11.2a + Hund’s rule4F. +more than half-filled  ground state symbol: 4P9/2
Z=26, ground configuration: 1s22s22p63s23p63d8
Table 11.2a + Hund’s rule3F. +more than half-filled  ground state symbol: 3F4
Z=27, ground configuration: 1s22s22p63s23p63d9
Table 11.2a + Hund’s rule2D. +more than half-filled  ground state symbol: 2D5/2
Z=28, ground configuration: 1s22s22p63s23p63d10
Table 11.2a 1S. ground state symbol: 1S0
Z=29, ground configuration: 1s22s22p63s23p63d104s1
L=0, S=1/2 J=1/2 the only term 2S1/2.
Z=24, ground configuration: 1s22s22p63s23p63d64s2
Table 11.2a 1S. ground state symbol: 1S0
Obviously Z=25 element (Mn) has largest degeneracy in its ground state ( 4P9/2): 10.
11.25 (a) Use a diagram like (11.48) to show that the lowest term of a single half-filled-subshell
configuration has L=0.(b) How many levels arise from a term with L=0? Explain why no rule is
needed to find the lowest level of the lowest term of a half-filled-subshell configuration.
Answer:
According to Hund’s rule, half-filled subshells necessarily have the following spin are arranged
as follows
s
p
d
f
|↑
|↑ |↑
|↑ |↑ |↑
|↑ |↑ |↑ |↑
|
|↑ |
|↑ |↑ |
|↑ |↑ |↑ |
Consult Table 11.2 and we find half-filled subshells satisfying above spin arrangement are always S terms.
For s1, p3, d5, they are 2S, 4S and 6S, respectively. i.e, they all have L=0, J=S meaning spin-orbital
interaction has no effect on these terms. Thus no rule is needed to order the terms for half-filled
subshells when spin-orbital interaction is considered.
11.26 Verify that if E=1eV then E/hc=8065.5 cm-1.
Answer:
11.27 Consult Moore´s table of atomic energy levels (Section 11.5) to find at least three electron
configurations of the neutral carbon atom for which Hund´s rule does not correctly predict the
lowest term.
Answer:
Hund’s rule might be violated for excited states. One example is 1s22s2p3 where the terms are
observed to be arranged in the following way
5
S<3D<3P<1D<3S<1P.
[Moore’s book Atomic Energy Levels contains various examples.]
11.28 Selection rules for spectral transitions of atoms where Russell-Saunders coupling is valid are (Bethe
and Jackiw, Chapter 11) ∆L=0, ±1; ∆S=0; ∆J=0; ±1(but J=0 to J=0 is forbidden); ∆ (∑ili)=
±1,meaning that the change in configuration must change the sum of the l values of the electrons
by ±1.For most atomic spectral lines, only one electron changes its subshell; here the ∆ (∑ili) = ±1
rule becomes ∆l= ±1 for the electron making the transition.
For the carbon atom, the levels that arise from the 1s22s22p2 configuration are
Level
(E/hc)/cm-1
3
P0
0
3
P1
16.4
3
P2
43.4
1
D2
10192.6
1
S0
21648.0
and the energy levels of the 1s22s2p3 configuration are
5
3
3
3
3
3
3
1
1
S2
D3
D1
D2
P1
P2
P0
D2 3S1
P1
33735.2 64086.9 64089.8 64090.9 75254.0 75255.3 75256.1 97878 105798.7 119878
Use the above selection rules to find the wavenumbers of all the transitions that are allowed
between pairs of these 15 levels.
Answer:
The selection rule ∆S=0 means only the green transitions are allowed:
3
P ↔ 5S; 3P ↔ 1D; 3P ↔ 1P; 3P ↔ 3D; 3P ↔ 3P; 3P ↔ 3S;
1
D ↔ 5S; 1D ↔ 3D; 1D ↔ 3P; 1D ↔ 3S; 1D ↔ 1D; 1D ↔ 1P;
1
S ↔ 5S; 1S ↔ 3D; 1S ↔ 3P; 1S ↔ 3S; 1S ↔ 1D; 1S ↔ 1P;
The selection rule ∆L=0, ±1 rules out the transition 1S ↔ 1D.
The selection rule ∆J=0,±1 rules out the transition 3P0 ↔ 3P2; 3P0 ↔ 3D2; 3P0 ↔ 3D3; 3P1 ↔ 3D3;
The selection rule ∆J=0 with J≠0 rules out the transition 3P0 ↔ 3P0;
The allowed transitions then are as follows
3
P0,1,2 ↔ 3D1,2,3; six transitions
3
P0,1,2 ↔ 3P0,1,2 six transitions
3
P0,1,2 ↔ 3D1 three transitions
1
D2 ↔ 1D2; one transition
1
D1 ↔ 1P1 one transition
1
S1 ↔ 1P1 one transition.
The calculation of the transition frequencies is straightforward.
11.29 Does Fig. 11.6 contain a violation of the rule give in Section 11.6 for determining whether a
multiplet is regular or inverted?
Answer:
No.
11.30 Draw a diagram similar to Fig. 11.6 for the carbon 1s22s22p2 configuration. (The 1S term is the
highest.)
Answer:
Two p electrons: L=2,1,0, S=1,
0
J=3,2,1;2,1,0;1;
2, 1, 0
3
3
3
1
D, P, S,
D, 1P, 1S,
The red symbols are for 2p2 configuration (they are equivalent electrons, c.f. Table 11.2a). Without
considering spin-orbital couplings, the energy arrangement of the three terms can simply be
determined by Hund’s rule: 3P is the lowest and 1S is the highest. 3P<1D<1S.
When relativistic effects are considered, the degeracy of the states in a term with different J is
removed Thus we have three sets of states:
3
P2, 3P1, 3P0 ;
1
D1;
1
S0
Only the degeracy in 3P term is (partially) removed. Because this configuration is less than half-filled
(inverted), the energy levels for these states are arranged as
3
P0< 3P1<3P2 .
If there is an external (magnetic) field applied to the system, all the degeneracies are removed and we get
the following energy levels:
3
3
3
P2,(MJ=2,1,0,-1,-2)
P1(MJ=1,0,-1)
P0 ;
1
D1(MJ=2,1,0,-1,-2)
1
S0
The state with highest MJ has highest energy.
In conclusion, we get the following diagram of energy levels:
Configurations
Terms
Levels
States
__1S0
__1S0__
__1S__
__1D2____
__1D2____MJ=2
__1D2____MJ=1_
__1D2____MJ=0
__1D2___MJ=-1
__1D2___MJ=-2_
__1D__
__3P2__
__3P2__MJ=2
__3P2__MJ=1
__3P2__MJ=0
__3P2__MJ=-1
__3P2__MJ=-2
__3P1__
__3P1__MJ=1
__3P1__MJ=0
__3P1__MJ=-1
__3P__
__3P0__ ---------------- __3P0__MJ=0
1s22s22p2
H0
H0+Hrep
H0+Hrep+HS.O.
H0+Hrep+HS.O.+HB
(The perturbations from electron spin-spin dipolar couplings, finite size of nucleus, the motion of nucleus,
the speed-dependent mass of (inner shell, heavy atoms) electrons and the dipolar coupling between
nuclear and electronic spins have been ignored.)
11.31 Use Eq. (11.66) to calculate the separation between the 2P3/2 and 2P1/2 levels of the hydrogen-atom
2p configuration. (Because of other relativistic effects, the result will not agree accurately with
experiment.)
Answer:
For hydrogen atom, we have
V(r)=-e2/rThe spin-orbital coupling constant is
ξ(r)=e2/2mec2r3
A=<2P3/2|ξ(r)|2P3/2>=<2P1/2|ξ(r)|2P1/2>
=<2p|ξ(r)|2p>
=∫r2dr[(1/2)6-1/2(1/a)5/2re-r/2a(e2/2mec2r3)(1/2)6-1/2(1/a)5/2re-r/2a]
= (e2/48mec2a5)∫re-r/a dr
= e2/48mec2a3
[using ∫rne-r/a dr=an+1n!]
=2.56×10-38/(48×9.1×10-31×9×1016(5.3×10-11)3)
=2.38×1044 s-2J3
The energy difference between the two states is
3A ħ =7.21×10
2
-24
J=4.5×10-5 eV=1.09×10 10 Hz=10.9 GHz (microwave)
11.32 Use Eq. (11.74) to calculate the energy separation between the MJ=1/2 and MJ=-1/2 state of the2p
2
P1/2 hydrogen-atom level, if a magnetic field of 0.200T is applied.
Answer:
J=1/2, L=1, S=1/2
g1/2=2/3,
ΔE=Bgβe(M1/2-M-1/2)= Bg(eħ/2me)
=0.2 T ×(2/3) ×(1.665×10-19×1.05×10-34/2/9.1×10-31)
= 1.232×10-24 J
=7.7×10-6 eV
=1.86×109 Hz=1.86 GHz (microwave)
11.33 Derive Eqs. (11.78) and (11.79).
Answer:
Because the wavefunctions D and D’ differ by no spin orbitals in equations (11.78) and (11.79),
we should use the first row of Table 11.3 to calculate those matrix elements. The ‘single-particle’ energy
n
n
i 1
i 1
 D |  fi | D    ui (1) | f1 | ui (1) 
n

1
   i (1) ms (1) | f1 |  i (1) ms (1) 
i 1 m s1  1
n
*
1
    i (1) | f1 |  i (1) 
i 1
1
For the inter-electronic correlation terms we have
n 1
n 1
 D |   g i | D    [ u i (1)u j (2) | g12 | u i (1)u j (2) 
j
i  1i  j
i  1i  j
  u i (1)u j (2) | g12 | u j (1)u i (2) ]
n 1
1/ 2
   [
  i (1) *msi (1) j (2) msj * (2) | g12 |  i (1) msi (1) j (2) msj (2) 

i  1 i  j msi , msj  1 / 2
   i (1) *msi (1) j (2) msj * (2) | g12 |  j (1) msj (1) i (2) msi (2) ]
n 1
   [  i (1) j (2) | g12 |  i (1) j (2) 
i  1i  j
   i (1) j (2) | g12 |  j (1) i (2) ]
11.34 For a close-subshell configuration, (a) show that the double sum in (11.80) equals
n/2
n/2
∑ ∑ (4Jij-2Kij)+∑ Jii
j＞i
i=1
i=1
where the Coulomb and exchange integrals are defined in term of the n/2 different spatial orbitals
φi ;(b)use (11.84) and the result of part (a) to derive (11.83).
Answer:
For a closed subshell, we have an even number of electrons (2,6,10, etc.), half of which are in
the spin ‘up’ state and half ‘down’ state.
Useful diagram:
1,
2,
3,
…..
n/2
spin ‘up’
n/2+1
n/2+2 n/2+3 …...
n
spin ‘down’
r1
r2
r3
rn/2
…….
[Be aware that above numbering is arbitrary].
Possible Coulomb pairs Jij from ri≠rj (cross terms):
J12=J1 n/2+2=Jn/2+1 2=Jn/2+1 n/2+2.
J13=J1 n/2+3=Jn/2+1 3=Jn/2+1 n/2+3.
etc.(4×[1+2+…+n/2-1] terms in total)
Possible Coulomb pairs Jij from ri=rj,, (self-terms):
J1 n/2+1=J11
J2 n/2+2 =J22
etc. (4×n/2 terms in total).
Possible exchange terms can only happen in the spin groups with the same spin numbers
because of the δmsimsj term in equation (11.80). So for ‘cross terms’, we have
K12=Kn/2+1 n/2+2≠K2 n/2+1=K1,n/2+2.=0.
K13=Kn/2+1 n/2+3≠K3 n/2+1=K1 n/2+3.=0
etc. (2×[1+2+…n/2-1] terms in total).
The ‘self-terms’ Kii:are
K1 n/2+1 =K11
Etc, but they don’t contribute to equation (11.80) because spin 1 and spin n/2+1 have opposite spin
directions.
Therefore equation (11.80) can be rewritten in terms of the spatial coordinate indices 1, 2,
3…, n/2 as follows:
n/2
n/2
∑ ∑ (4Jij-2Kij)+∑ Jii
j＞i
i=1
i=1
[You should check the number of each term].
11.35 Use the Condon-Slater rules to prove the orthonormality of two n-electron Slater determinants of
orthonormal spin-orbitals.
Answer:
The problem we want to solve is to prove the relationship
<ψi(1,2,3…n)|ψj(1,2,3,…n)>=δij
Using the Condon-Slater rules listed in Table 11.3 (here we choose f1=1), we have
<ψi(1,2,3…n)|ψj(1,2,3,…n)>=0 if ψi and ψj differ by more than one spin orbitals
<ψi(1,2,3…n)|ψj(1,2,3,…n)>=<ui’|ui> if ψi and ψj differ by one spin orbital. However,
<ui’|ui>=0 because of the orthogonality of the spin orbitals.
Only when all spin orbitals are the same for ψi and ψj, can the inner product
<ψi(1,2,3…n)|ψj(1,2,3,…n)> be non-zero. When all spin orbitals are equal, it is obviously equal to
1 because of the normalization of each spin orbital.
11.36 Explain why it would be incorrect to calculate the experimental ground-state energy of lithium by
taking E2s+2E1s, where E2s is the experimental energy needed to remove the 2s electron from
lithium and E1s is the experimental energy needed to remove the 1s electron from lithium.
Answer:
This type of reasoning does not consider the correlation among electrons. The energy to free the
first 1s electron is not equal to the energy needed to remove the second 1s electron from an atom..
Due to the inter-electronic repulsion, it is not hard to conclude that for a given atom, the second
ionization energy is larger than the first one and the third one is larger than the second one and so
on.
11.37 The total magnetic moment m of an atom contains contributions from the magnetic moments
mL,mS,and mI associated with the total electronic orbital angular momentum L, the total electronic
spin angular momentum S, and the nuclear spin angular momentum I.(a) For an atom with L,S,and
I all nonzero, which one of these three contributions to m is by far the smallest?(b)For the 87Rb
ground electronic state (used in the Bose-Einstein condensation experiment of Section 10.5), which
one of these three contributions is zero?
Answer:
(a) The nuclear spin angular momentum contributes the smallest contribution to the magnetic
moment because that contribution is inversely proportional to the mass of the spin [equation
(10.57)].
(b) The 87Rb has 87 nucleons and 37 electrons. The electron configuration is
1s22s22p63s23p63d104s24p64d1. The ground state atomic term is 2D (2D5/2, 2D3/2). The nuclear
spin quantum number for 87Rb is 3/2 [37 protons and 50 neutrons create such a nuclear
ground state or just look for NMR books]. Therefore the related quantum numbers are
S=1/2, L=2 and I=3/2, meaning each of these angular momenta has contribution to the total
magnetic moment.
11.38 True or false? (a) The spin multiplicity of every term of an atom with an odd number of electrons
must be an even number. (b) The spin multiplicity of every term of an atom with an even number
of electrons must be an odd number. (c) The spin multiplicity of a term is always equal to the
number of levels of that term. (d) In the Hartree SCF method, the energy of an atom equals the sum
of the orbital energies of the electron. (e) The Hartree-Fock method is capable of giving the exact
nonrelativisitic energy of a many-electron atom.
Answer:
(a) True. [Odd number of electrons means there are odd-number of ‘unpaired’ electrons. Therefore,
the possible total spin angular momentum quantum numbers are 1/2, 3/2, 5/2, …. multiplicity
2S+1=2, 4, 6,…all even numbers.]
(b) True. [The unpaired electrons in this case must be an even number, meaning the possible total
spin angular momentum quantum numbers are 0, 1, 2, …. multiplicity 2S+1=1, 3, 5,…all
even numbers.]
(c) False. [This statement is true only for S terms with J=0.]
(d) False. [The inter-electronic interaction energies are calculated twice in Hatree SCF method.
Those energies should be discounted from the total of the orbital energies, as shown in equation
(11.10).]
(e) False. [HF is by nature a mean-field theory and cannot take into count all instantaneous
electron correlations.]
Back to assignment index
Back to lecture note index
Back to quantum chemistry II home
Questions?
Back to Ding’s home