Physical Chemistry , 1st ed.

Example 11.9

The hydrogen molecule vibrates at a frequency of about 1.32 

1014 Hz.

Calculate the following:

a.The force constant of the H–H bond

b.The change in energy that accompanies a transition from the n1 to

n2 vibrational level, assuming that the hydrogen molecule is acting as an

ideal harmonic oscillator

Solution

a.The mass of a single hydrogen atom, in kilograms, is 1.674 

10 ^27 kg.

Therefore, the reduced mass of a hydrogen molecule is

 8.370 

10 ^28 kg

Using the rearranged equation 11.5 in terms ofkand remembering to use the

reduced mass in place of the mass, we find

k 4 ^2 (1.32 

1014 s^1 )^2 (8.370 

10 ^28 kg)

k575 kg/s^2

which, as explained earlier, is equal to 575 N/m or 5.75 mdyn/Å.

b.According to equation 11.16, the energy of a harmonic oscillator is

E(n^12 )h

For n1 and 2, the energies are

E(n1) (1 ^12 )(6.626 

10 ^34 J s)(1.32 

1014 s^1 ) 1.31 

10 ^19 J

E(n2) (2 ^12 )(6.626 

10 ^34 J s)(1.32 

1014 s^1 ) 2.19 

10 ^19 J

The difference in energy is 2.19 

10 ^19 J minus 1.31^10 ^19 J, or

8.8 

10 ^20 J.

Example 11.10

The HF molecule has a harmonic vibrational frequency of 1.241 

1014 Hz.

a.Determine its force constant using the reduced mass of HF.

b.Assume that the F atom doesn’t move and that the vibration is due solely

to the motion of the H atom. Using the mass of the H atom and the force

constant just calculated, what is the expected frequency of the atom?

Comment on the difference.

Solution

a.Using the masses of H and F as 1.674 

10 ^27 kg and 3.154 

10 ^26 kg

respectively, the reduced mass can be calculated as

 1.590 

10 ^27 kg

Substituting into the same expression as in the previous example, we find

for k:

k 4 ^2 (1.241 

1014 s^1 )^2 (1.590 

10 ^27 kg) 966.7 kg/s^2

(1.674 

10 ^27 kg)(3.154 

10 ^26 kg)



1.674 

10 ^27 kg 3.154 

10 ^26 kg

(1.674 

10 ^27 kg)(1.674 

10 ^27 kg)



1.674 

10 ^27 kg 1.674 

10 ^27 kg

332 CHAPTER 11 Quantum Mechanics: Model Systems and the Hydrogen Atom