17.5 Expectation Values in the Hydrogen Atom 751
Solution
Since the wave function is normalized we can omit the denominator in the formula for the
expectation value shown in Eq. (16.4-1). We can factor the integral:
〈 1
r
〉
∫∞
0
R∗ 10
1
r
R 10 r^2 dr
∫π
0
Θ∗ 00 Θ 00 sin(θ)dθ
∫ 2 π
0
Φ∗ 0 Φ 0 dφ (17.5-7)
By our separate normalizations, the second and third integrals both equal unity, so that
〈 1
r
〉
∫∞
0
R∗ 10
1
r
R 10 r^2 dr 4
(
Z
a
) 3 ∫∞
0
e−^2 Zr/ardr 4
(
Z
a
) 3 (
a
2 Z
) 2 ∫∞
0
e−uudu
4
(
Z
a
) 3 (
a
2 Z
) 2
(1)
Z
a
(17.5-8)
where we looked up the integral in Appendix C. Since theθand theφfunctions are both
separately normalized, we could have omitted them from the outset and used only the
radial factorRin calculating〈 1 /r〉. The same thing is true for the expectation value of any
function ofr.
We can use the result for〈 1 /r〉to calculate〈V〉:
〈V〉−
Ze^2
4 πε 0
〈
1
r
〉
−
Z^2 e^2
4 πε 0 a
2 E 1 s (17.5-9)
Note that〈 1 /r〉is proportional toZand〈V〉is proportional toZ^2.
Exercise 17.13
Obtain numerical values for〈 1 /r〉and〈V〉for a hydrogen atom (Z1) in the 1sstate.
The expectation value of the potential energy of a hydrogen-like atom equals twice
the total energy. Therefore, the kinetic energy is half as large as the magnitude of the
potential energy, and is equal in magnitude to the total energy (the kinetic energy must
be positive while the total energy and the potential energy are negative). This behavior
occurs in all systems of particles interacting only with the Coulomb potential energy,
and is a consequence of thevirial theoremof mechanics.^4
The Radial Distribution Function
The radial distribution function,fr(r), is defined as the probability per unit value ofr
for finding the electron at a given distance from the nucleus. Ifr′is a particular value
ofr, then
⎛
⎝
probability that the distance
from the nucleus to the electron
lies betweenr′andr′+dr
⎞
⎠fr(r′)dr (17.5-10)
(^4) I. N. Levine,op. cit., p. 459ff (note 2).