Physical Chemistry , 1st ed.

same level that one can know such observables for the hydrogen atom—if one

has the tools.

Example 12.3

Assume that the helium wavefunction is a product of two hydrogen-like

wavefunctions (that is, neglect the term for the repulsion between the elec-

trons) in the n1 principal quantum shell. Determine the electronic energy

of the helium atom and compare it to the experimentally determined energy

of1.265    10 ^17 J. (Total energies are determined experimentally by mea-

suring how much energy it takes to remove all of the electrons from an atom.)

Solution

Using equation 12.6 and neglecting the electron-repulsion term by assuming

that the wavefunction is the product of two hydrogen-like wavefunctions:

HeH,1    H,2

the Schrödinger equation for the helium atom can be approximated as

 2





2

^21 

4

2



e

0

2

r 1



2





2

^22 

4

2



e

0

2

r 2


H,1H,2 EHeH,1H,2

where EHeis the energy of the helium atom. Because the first term in brack-

ets is a function of only electron 1 and the second term in the brackets is a

function of only electron 2, this Schrödinger equation can be separated just

like a two-dimensional particle-in-a-box can be separated. Understanding

this, we can separate the Schrödinger equation above into two parts:

 2





2

^21 

4

2



e

0

2

r 1

H,1E 1 H,1

 2





2

^22 

4

2



e

0

2

r 2

H,2E 2 H,2

where EHe E 1 E 2. These expressions are simply the one-electron

Schrödinger equations for a hydrogen-like atom where the nuclear charge

equals 2. An expression for the energy eigenvalue for such a system is known.

From the previous chapter, it is

E

8

Z

^20

2

h

e^4

2



n^2



for each hydrogen-like energy. For this approximation, we are assuming that

helium is the sum of two hydrogen-like energies. Therefore,

EHeEH,1EH,2



8

2



2

(^20)
e
h
4
2



n^2



8

2



2

(^20)
e
h
4
2



n^2



^20

e

h

4

2



n^2



where we get the final term by combining the two terms to the left. Keep in

mind that is the reduced mass for an electron about a helium nucleus, and

that the principal quantum number is 1 for both terms. Substituting the val-

ues of the various constants, along with the value for the reduced mass of the

electron-helium nucleus system (9.108   10 ^31 kg), we get

EHe1.743  10 ^17 J

376 CHAPTER 12 Atoms and Molecules