Chemistry Reference
In-Depth Information
namely:
c
0
c
0
5
8
c
0
¼
c
0
2 Z
5
16
«
w
¼
2
2
Zc
0
þ
ð
4
:
27
Þ
In (4.26), the first term on the right is the one-electron matrix element
already found for H, while the second is the two-electron repulsion
integral between the two spherical 1s charge distributions written in
charge density notation:
ðð
d
r
1
d
r
2
½
1s
ðr
2
Þ
1s
ðr
2
Þ
r
12
h
1s
1
1s
2
j
r
1
1s
ðr
1
Þ
1
s
ðr
1
Þ
¼ ð
1s
2
j
1s
2
12
j
1s
1
1s
2
i¼
½
Þ
ð
4
:
28
Þ
This new two-electron integral is evaluated in terms of the purely radial
electrostatic potential J
1s
(r
1
):
ð
d
r
2
½
1s
ðr
2
Þ
1s
ðr
2
Þ
r
12
exp
ð
2c
0
r
1
Þ
r
1
1
r
1
J
1s
ð
r
1
Þ¼
¼
ð
1
þ
c
0
r
1
Þ
4
:
29
Þ
when use is made of the one-centre Neumann expansion for 1
=
r
12
(Magnasco, 2007). Then, integral (4.28) is readily obtained by integration
in spherical coordinates:
ð
d
r
1
J
1s
ð
r
1
Þ½
1s
ðr
1
Þ
1s
ðr
1
Þ ¼
5
8
c
0
ð
1s
2
j
1s
2
Þ¼
ð
4
:
30
Þ
Optimization of (4.27) with respect to c
0
gives
5
16
c
0
¼
Z
ð
4
:
31
Þ
as best value for the orbital exponent and
2
5
16
«
w
ð
best
Þ¼
Z
:
ð
4
:
32
Þ
as best energy for the S(1s
2
) configuration of the He-like atom.