Physical Chemistry Third Edition

18.6 The Lithium Atom 783

Exercise 18.7

Use the rule of Eq. (B-99) of Appendix B for expanding a three-by-three determinant to show

that the function of Eq. (18.6-6) is the same as that of Eq. (18.6-5):

Two properties of determinants presented in Appendix B relate to the properties of

antisymmetrized orbital wave functions:

1. If one exchanges two columns or two rows of a determinant, the resulting determinant
   is the negative of the original determinant. Exchanging the locations of two particles
   is equivalent to exchanging two columns, so that the Slater determinant exhibits the
   necessary antisymmetry.


2. If two rows or two columns of a determinant are identical, the determinant van-
   ishes. If two electrons occupy identical spin orbitals, two rows of the determinant
   in Eq. (18.6-6) are identical, and the determinant vanishes, in agreement with the
   Pauli exclusion principle.

For an antisymmetrized orbital wave function for the Li atom we must choose three

different spin orbitals. For the ground state, we choose orbitals with the minimum

possible sum of orbital energies. This practice is called theAufbau principle, from the

German word for “building up.” We choose the two 1sspin orbitals and one spin orbital

from the second shell. We anticipate the fact that higher-order calculations will give a

lower energy for the 2sspin orbitals than for the 2pspin orbitals, and choose one of

the 2sspin orbitals. It does not matter whether we choose the spin-up or the spin-down

spin orbital. From Eq. (18.6-4) the zero-order energy of the ground state is

E(0)gsE(0) 1 s 1 s 2 s 2 E 1 (HL)+E 2 (HL)

(− 13 .60 eV)

(

2

32

12

+

32

22

)

− 275 .4 eV (18.6-7)

This value is seriously in error, as was the zero-order value for helium. It differs from

the experimental value of− 203 .5 eV by 35%.

Using space orbitals and spin functions a zero-order Slater determinant for the

ground state is

Ψ(0)

1

√

6

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

ψ 1 s(1)α(1) ψ 1 s(2)α(2) ψ 1 s(3)α(3)

ψ 1 s(1)β(1) ψ 1 s(2)β(2) ψ 1 s(3)β(3)

ψ 2 s(1)α(1) ψ 2 s(2)α(2) ψ 2 s(3)α(3)

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

(18.6-8)

The spin orbitalψ 2 sβcould have been chosen instead ofψ 2 sα. We therefore have two

states of equal energy. This doubly degenerate ground level corresponds toS 1 /2(a

doublet term), since the possible values ofMSare+ 1 /2 and− 1 /2. SinceML0, the

value ofLis 0, the only value ofJis 1/2, and the ground term symbol of lithium is

(^2) S 1 / 2.
Excited states of the lithium atom can correspond to various choices of orbitals. The
values ofMLandMSfor these excited states can be calculated by algebraic addition.
Using the rules that MLranges from+Lto−Land thatMSranges from+Sto−S, one
can deduce the values ofLandSthat occur and can assign term symbols. Higher-order
calculations must be used to determine the order of the energies of the excited states.