Concise Physical Chemistry

c15 JWBS043-Rogers September 13, 2010 11:28 Printer Name: Yet to Come

PROBLEMS AND EXAMPLES 243

brilliant step in quantum theory. It permits one to calculate a better approximation to

V 1 andV 2 , which leads to improved values ofZeff, which leads to improved values

ofψ 1 andψ 2 , which leads to betterV 1 andV 2 , and so on. This iterative process is

continued until the energies do not change from one iteration to the next. When that

condition has been met, the energies calculated fromV 1 andV 2 areself-consistent.

The Hartree equations areeigenvalueequations, hence they lead to discrete (scalar)

valuesε 1 andε 2 for the twoone-electronSchrodinger equations. The energies are ̈

not exact, but they are the best we can get from assumed hydrogen-like orbitalsψ 1

andψ 2 by the Hartree procedure.

Conveniently, the applied mathematics ofself-consistent field(SCF) iterations had

been worked out by astronomers calculating the orbits of planets, prior to the advent

of quantum mechanics. They called it thevariational method.

15.5 A DIGRESSION ON ATOMIC UNITS

We can measure length in any unit we want: m, mm, furlongs, and so on. It is perfectly

legitimate to use the radius of the first Bohr orbit as our unit of distance, replacing

the meter but still related to it:

1 a 0 ≡ 5. 292 × 10 −^11 m

Likewise, we can define units of mass, charge, and angular momentum:

1 me≡ 9. 109 × 10 −^31 kg

1 e≡ 1. 602 × 10 −^19 C

1 
≡ 1. 055 × 10 −^34 Js

These definitions lead to considerable simplification in the equations of quantum

chemistry (see Problems and Examples).

PROBLEMS AND EXAMPLES

Example 15.1 A Hartree Fock Solution for He

The limiting solution for a Hartree–Fock procedure to the problem of the helium

atom is−2.862 hartrees, where the unit hartree (Eh) is defined as exactly half the

energy of the electron in the ground state of hydrogen,Eh≡ 627 .51 kcal mol−^1 =

2625 kJ mol−^1. The crude unshielded solution for the helium ionization energy is

Z^2 = 22 =4 hartrees. By our trial-and-error method, we gotZeff= 1 .6 which leads

to about( 1. 6 )^2 = 2. 56 Eh. The experimental value is 2.903Eh.