Biomedical Engineering Reference
In-Depth Information
1. Nothing from any shell outside the one considered.
2. An amount 0.35 from each other electron in the group considered (except the 1s group,
where 0.30 is used instead).
3. If the shell considered is an s, p shell, an amount 0.85 from each electron with total
quantum number less ny one, and an amount 1.00 from every electron still further in.
But if the shell isadorf,anamount 1.00 from every electron inside it.
Slater gives three examples, reproduced in Table 14.8, which are worth quoting; C, Fe and
Fe cation lacking a K-electron so that there is only 1 @1s electron.
Slater's rules are still widely quoted in atomic and molecular structure theory.
Table 14.8
Slater
Z
−
s
values
C
Fe
Fe
+
(1s
1
)
1s
5.70
=
6
−
0.30
25.70
=
26
−
0.30
26.00
2s, 2p
3.25
=
6
−
3
(
0.35
)
−
2
(
0.85
)
21.85
=
26
−
7
(
0.35
)
−
2
(
0.85
)
22.70
3s, 3p
14.75
=
26
−
7
(
0.35
)
−
8
(
0.85
)
15.75
−
2
(
1.00
)
3d
6.25
=
26
−
5
(
0.35
)
−
18
(
1.00
)
7.25
4s
3.75
=
26
−
1
(
0.35
)
−
14
(
0.85
)
4.75
−
18
(
0.85
)
−
10
(
1.00
)
14.12 Koopmans' Theorem
Consider the ionization process
Ne(1s
2
2s
2
2p
6
)
Ne
+
(1s
2
2s
1
2p
6
)
→
where I have ionized a 2s electron from neon (Figure 14.5). Suppose that the energy level
diagram represents a HF calculation on neon before ionization, and that the orbitals do not
relax in any way after ionization. That is, the neutral atom and the ionized cation have the
same HF orbitals. According to Koopmans' theorem, the ionization energy for the process
is the negative of the HF orbital energy. The theorem holds for all HF orbitals.
Koopmans' Theorem
Ionization from HF orbital
ψ
i
Ionization energy
=−
orbital energy
Ionization energies can be measured accurately by modern-day versions of the photo-
electric effect. For valence shell ionization we use ultraviolet photoelectron spectroscopy
(UVPES) and, as the name suggests, ultraviolet photons are the ionizing source. Inner shell
electrons need X-rays for ionization.
