Physical Chemistry Third Edition

772 18 The Electronic States of Atoms. II. The Zero-Order Approximation for Multielectron Atoms

to show that

̂S^2 [α(1)β(2)+β(1)α(2)] 2 h ̄^2 [α(1)β(2)+β(1)α(2)]

b. Show that for a two-electron system

̂S^2 [α(1)β(2)−β(1)α(2)] 0

18.4 Excited States of the Helium Atom

States of higher energy than the ground state are calledexcited states. For orbital wave

functions there are two cases: (1) both electrons occupy the same space orbital with

different spin functions, as is the case with the ground state; (2) the two electrons occupy

different space orbitals, either with the same or different spin functions. A statement

of which space orbitals are occupied is called theelectron configuration. Thedetailed

configurationis specified by writing the designation of each occupied space orbital

with a right superscript giving the number of electrons occupying that space orbital

(either 1 or 2). Thesubshell configurationis specified by writing the designation of each

occupied subshell with a right superscript giving the number of electrons occupying

orbitals of that subshell. Unoccupied space orbitals or subshells are omitted from the

configuration. The maximum value of this superscript is 2 for anssubshell, 6 for ap

subshell, 10 for adsubshell, and 14 for anfsubshell. A superscript equal to unity is

usually omitted, so that (1s)(2s) is the same as (1s)^1 (2s)^1.

If both electrons occupy the same space orbital, the wave function for an excited

state is similar to that of the ground state. For the configuration (2s)^2 :

Ψ 2 s 2 s

1

√

2

ψ 2 s(1)ψ 2 s(2)[α(1)β(2)−β(1)α(2)] (18.4-1)

The probability density for two electrons and the electric charge density are analogous

to those of the ground state.

If the electrons occupy different space orbitals there are four possible states for each

pair of space orbitals. For the (1s)(2s) configuration we can write four antisymmetric

wave functions that are products of a space factor and a spin factor:

Ψ 1 

1

√

2

[ψ 1 s(1)ψ 2 s(2)−ψ 2 s(1)ψ 1 s(2)]α(1)α(2) (18.4-2a)

Ψ 2 

1

√

2

[ψ 1 s(1)ψ 2 s(2)−ψ 2 s(1)ψ 1 s(2)]β(1)β(2) (18.4-2b)

Ψ 3 

1

√

2

[ψ 1 s(1)ψ 2 s(2)−ψ 2 s(1)ψ 1 s(2)]

1

√

2

[α(1)β(2)+β(1)α(2)] (18.4-2c)

Ψ 4 

1

√

2

[ψ 1 s(1)ψ 2 s(2)+ψ 2 s(1)ψ 1 s(2)]

1

√

2

[α(1)β(2)−β(1)α(2)] (18.4-2d)

The number of states can be determined by counting the number of ways of arranging

two spins: up–up, down–down, up–down, and down–up. As shown in Eq. (18.4-2), the

up–down and down–up arrangements combine in two ways, corresponding to symmet-

ric and antisymmetric spin factors.

Exercise 18.2

a.Show thatΨ 3 andΨ 4 satisfy the zero-order Schrödinger equation and find the energy

eigenvalue.

b.Show that these functions are normalized if the orbitals are normalized.