Page 107 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch08 Elementary Physical Chemistry

92 Elementary Physical Chemistry

8.6. The Schr¨odinger Equation

Schr¨odinger’s treatment is mathematically the simplest, and is most often
used in calculations. Basically, it is a differential equation which has to be
solved to produce the properties of the system under consideration. For
o
example, the Schr¨dinger equation of a particle moving in one dimension,
having a mass m,is
2
2
2
2
−(h /8π m)d ψ/dx + Vψ = Eψ (8.4)
where V is the potential energy, E is the energy of the particle and ψ is the
wave-function. Solving this equation produces two types of information:
(1) A set of energy levels: E 1 , E 2,etc.
(2) A set of wave-functions, ψ n (x), associated with each energy state.
Originally, Schr¨dinger believed that the wave function described the
o
position of the particle. However, Bohr, who invited him to Copenhagen,
quickly convinced him that that was not the case. The wave-function ψ
has no physical significance. Today’s accepted interpretation (due to Born)
is that ψ ψ represents a probability density,and that ψ ψδv denotes the
∗
∗
probability of finding the particle within the volume element δv.[Thewave-
function is in general complex and the ψ denotes the complex conjugate.
∗
The product of the wave-function and its complex conjugate is always real.]
+
Example 8.1. The ground state of He is
3
3 1/2
ψ =(8/πa ) exp (−2r/a ) (8.5)
0
0
where r is the distance of the electron from the nucleus, and
a 0 =52.9 × 10 −12 m=(52.9pm) (8.6)
(a) What is the probability of finding the electron within a volume of ∆v =
3
1pm around the nucleus?
(b) What is the probability of finding the electron within a volume of 1 pm 3
at a distance a 0 from the nucleus?

Solution
The probability expression is (ψ is real)
3 3 2
Prob = ψψdv = (8/πa ) {exp(−2r/a )} dv
0 0
3
In this problem, the r’s are constant and dv =1 pm .