Page 72 - Electrical Properties of Materials

The hydrogen atom 55

We now have to work out the partial derivatives of r. Since
2
2
2 1/2
r =(x + y + z ) , (4.8)
we get
∂r x
= (4.9)
2 1/2
2
2
∂x (x + y + z )
and
2
∂ r 1 x 2
= – , (4.10)
2
2 1/2
2
2
2
2 3/2
∂x 2 (x + y + z ) (x + y + z )
and similar results for the derivatives by y and z. Substituting all of them in
eqn (4.7), we get
2
∂ ψ x 2 y 2 z 2
2
∇ ψ = + +
2
2
2
2
2
2
∂r 2 x + y + z 2 x + y + z 2 x + y + z 2
2
∂ψ 3 x
+ – (4.11)
2
2
2 1/2
2 3/2
2
2
∂r (x + y + z ) (x + y + z )
2
y 2 z 2
∂ ψ 2 ∂ψ
– – = + .
2
2
2
2
2 3/2
2 3/2
(x + y + z ) (x + y + z ) ∂r 2 r ∂r
Thus, for the spherically symmetrical case of the hydrogen atom the
Schrödinger equation takes the form,
2
2 ∂ ψ 2 ∂ψ e 2
+ + E + ψ = 0. (4.12)
2m ∂r 2 r ∂r 4π 0 r
It may be seen by inspection that a solution of this differential equation is
ψ =e –c 0 r . (4.13)
The constant, c 0 , can be determined by substituting eqn (4.13) in eqn (4.12)
2 2 –c 0 r 2 –c 0 r
e 2 –c 0 r
c e + (–c 0 e ) + E + e = 0. (4.14)
0
2m r 4π 0 r
The above equation must be valid for every value of r, that is the coefficient of
exp(–c 0 r) and that of (1/r)exp(–c 0 r) must vanish. This condition is satisfied if
2 2
c
E =– 0 (4.15)
2m
and
2 2
c 0 e
= . (4.16)
m 4π 0
From eqn (4.16),
2
e m
c 0 = 2 , (4.17)
4π 0