Concise Physical Chemistry

c15 JWBS043-Rogers September 13, 2010 11:28 Printer Name: Yet to Come

PROBLEMS AND EXAMPLES 245

or

d^2 e−αr

dr^2

+

2

r

de−αr

dr

+

2 me


^2

(

E+

e^2

4 π^2 ε 0 r

)

e−αr= 0

We can carry out the differentiations, divide through bye−αr, and segregate the

terms into those that depend onr(terms 2 and 4) and those that do not (terms

1 and 3):

α^2 −

2

r

α+

2 me


^2

(

E+

e^2

4 π^2 ε 0 r

)

= 0

α^2 −

2

r

α+

2 meE


^2

+

2 me


^2

e^2

4 π^2 ε 0 r

= 0

so

α^2 +

2 meE


^2

=

2

r

α−

2 me


^2

e^2

4 π^2 ε 0 r

The basis! functionφ=e−αris a negative exponential, so there must be some value

ofrfor which the right-hand side of the equation becomes zero. But the left-hand

side of the equation is a group of constants, which add up to a single constant. If this

constant is zero for one value ofr, it must be zero for all. Thus,

α^2 +

2 meE


^2

= 0

where
= 2 hπThe consequences of this simple equation are very important:

α^2 =−

2 meE


^2

E(r)=−

^2

2 me

α^2

and

2

r

α−

2 me


^2

e^2

4 π^2 ε 0 r

= 0

2

r

α=

2 me


^2

(

e^2

4 π^2 ε 0

) 2

1

r