Concise Physical Chemistry

c17 JWBS043-Rogers September 13, 2010 11:28 Printer Name: Yet to Come

HELIUM 275

These energies can be measured experimentally with considerable accuracy. Since

they are the energy necessary to pull an electron away from the helium atom, they

are equal and opposite in sign to the binding energy of the ionized electron. In this

way we have a measure of both (a) the first or “outside” electron attracted to the He+

system and (b) the inside electron attracted to the He^2 +nucleus. This second energy

IP 2 can be calculated exactly because He+is a one-electron (two-particle) system.

The calculated second ionization potential IP 2 is exactly 2.0Eh.

If we compare the calculated total ionization potential, IP=4.00Eh, with the

experimental value, IP=2.904 hartrees, the result is very bad. The magnitude of the

disaster is even more obvious if we subtract the known second ionization potential,

IP 2 = 2 .0, from the total IP to find thefirstionization potential, IP 1 :

IP 1 =IPtotal−IP 2 = 2. 904 − 2. 000 = 0. 904 Eh

Under the approximations we have made, IP 1 (calculated)= 2. 000 Ehis about 110%

in error. Clearly, we cannot ignore interelectronic repulsion.

17.4.1 An SCF Variational Ionization Potential for Helium

One approach to the problem of ther 12 term is a variationalself-consistent field

approximation. We shall start from Hartree’sorbital approximation, assuming that

the orbital of helium is separable into two one-electron orbitals (1,2)=φ 1 (1)φ 2 (2).

It is reasonable to use the H atom as our starting point with the same kinetic energy

operator for each starting orbitalφ 1 (1) andφ 2 (2)and− 2 /ras the potential energy

part for attraction of each electron to a nuclear charge of 2:

Hˆ =−^1

2 r^2

d

dr

r^2

d

dr

−

2

r

Although we are solving for one-electron orbitals,φ 1 andφ 2 , we do not want to fall

into the trap of the last calculation. This time we shall include a potential energy term

V 1 to account for the repulsion between the negative charge on the electron arbitrarily

designated electron 1, exerted by the other electron which we shall call electron 2.

We don’t know where electron 2 is, so we must integrate over all possible locations

dτ:

V 1 =

∫∞

0

φ 2

1

r 12

φ 2 dτ

The entire Hamiltonian for electron 1 is

Hˆ 1 =−^1

2 r 12

d

dr 1

r 12

d

dr 1

−

2

r 1

+

∫∞

0

φ 2

1

r 12

φ 2 dτ