Chemistry Reference
In-Depth Information
, the average radial position for an electron in an H-like atom 1s
orbital, can be found using the simple operator
r
and the appropriate atomic orbital
functions. To calculate the expectation value we would write
For example,
r
100
π
π
∞
π
π
∞
χ
100
r
χ
100
r
2
sin(
A
2
r
3
exp(
r
100
=
θ
)d
r
d
θ
d
φ
=
d
φ
sin(
θ
)d
θ
−
ρ
r
)d
r
−
π
0
0
−
π
0
0
(A9.61)
As we saw earlier, the limits for the angular integrals give the surface of a sphere; the
additional radial integral then ensures that we visit every point in space. Note that; although
this integral is out to infinity; the decaying exponential in the integrand ensures a finite
result is obtained.
The angular integrals, again, give a factor of 4
:
π
A
2
∞
r
100
=
4
r
3
exp(
−
ρ
r
)d
r
(A9.62)
π
0
The radial integral is less straightforward, but does crop up regularly in these sorts of
problem and can be solved by repeated application of the integration by parts method. The
general result (for arbitrary power of
r
) is given in Table A9.4, which also lists the first few
cases. Taking the value of the integral from the table:
24
Z
eff
a
0
3
A
2
ρ
a
0
3
2
a
0
Z
eff
r
100
=
=
16
Z
eff
=
4
π
(A9.63)
4
Table A9.4
The first few integrals of the type required for the integration of
the radial functions. The n
0 case is straightforward and the other solutions
are obtained by repeated application of the method of integration by parts.
Note with the limits shown the result is I
n
=
=
n
/ρ
n
+
1
.
0
∞
Solutions for integrals of the type
I
n
=
r
n
e
−
ρr
d
r
n
Analytical integral
Application of limits
e
−
ρr
ρ
1
ρ
0
−
r
)
e
−
ρr
ρ
1
ρ
1
−
(1
+
ρ
2
2
2
r
2
)
e
−
ρr
ρ
2
ρ
2
−
(2
+
2
ρ
r
+
ρ
3
3
3
r
3
)
e
−
ρr
ρ
6
ρ
3
−
(6
+
6
ρ
r
+
3
ρ
2
r
2
+
ρ
4
4
4
r
4
)
e
−
ρr
ρ
24
ρ
4
−
(24
+
24
ρ
r
+
12
ρ
2
r
2
+
4
ρ
3
r
3
+
ρ
5
5
