Physical Chemistry , 1st ed.

Example 12.11 shows how variation theory works with a variable that al-

lows one to minimize the energy of a system. It uses the idea that in a multi-

electron atom, the electrons may not experience the full nuclear charge because

of the presence of other electrons. This concept is known as shielding.It also

illustrates a necessity in approaching higher-level quantum mechanics: the

need to be able to evaluate many different types of integrals. A good handbook

of integrals and the ability to determine integrals by numerical methods be-

come very helpful when mathematically evaluating even simple problems in

quantum mechanics.

Example 12.11

In the helium atom, assume that each electron does not experience the full

2 nuclear charge but instead, due to shielding from the other electron, ex-

periences an effective nuclear charge ofZ. Use Zas a variable in the form

of the hydrogen-like 1sorbital to obtain the lowest average energy of helium.

Solution

Using normalized hydrogen-like 1sorbitals as trial wavefunctions, the two-

electron is



Z

a

^3

 3 e

Zr 1 /aeZr 2 /a

where ahas been defined in Chapter 11 as 4

 0 ^2 /e^2. Assuming that the nu-

clear charge experienced by each electron is Z, the initial Hamiltonian for

the helium atom is

 2





2

^21 

4

Z





e

0

2

r 1



2





2

^22 

4

Z





e

0

2

r 2



4

e



2

0 r 12



However, this Hamiltonian is not complete. If the first electron is experienc-

ing an attractive potential ofZe^2 /4

 0 r 1 , then the other electron must be ex-

periencing a potential of [(2 Z)e^2 ]/4

 0 r 2. A term like this should appear

for both electrons. The complete Hamiltonian is therefore

 2





2

^21 

4

Z





e

0

2

r 1



2





2

^22 

4

Z





e

0

2

r 2





(2

4





Z

0 r



1

)e^2



(2

4





Z

0 r



2

)e^2



4

e



2

0 r 12



The terms can be rearranged, and the average energy can be calculated. Since

we are presuming a one-electron system in the first two parenthetical terms,

we can use the energy eigenvalues from the hydrogen atom. We get

Etrial 2 



8

Z





(^20)
2
h
e
2
(^4) 


Z

2



a

6

 6 e

Zr 1 /aeZr 2 /a

(2 Z)

4

e



2

0 r 1

(2 Z)

4

e



2

0 r 2



4

e



2

0 r 12

 eZr^1 /aeZr^2 /ad

where dis with respect to electron 1 and electron 2; that is,ddr 1 dr 2.

Using the proper substitutions and manipulations (which will be omitted

here), the integral can be evaluated analytically. Its value is

^8 Z

1

8

0

Z



8 

e

(^20)
4
h



 2

396 CHAPTER 12 Atoms and Molecules