To show
(A12)
Again we solve the wave equation for two molecules undergoing oscillation about an equilibrium position x = 0. The potential energy is shown below as a function of the displacement from the equilibrium position x = 0 (A5p402)
The uncertainty principle says that we cannot know exactly where the particle is located. Therefore zero frequency of vibration in the ground state, i.e. u = 0 is not an option (A5p402 and pA22). When vo is the frequency of vibration, the ground state energy is
(V1)
Harmonic oscillator (A5p402)
Spring Force potential energy from equilibrium position x = 0
the solution is of the form for t=0 then x=0
where
The potential energy is
(V2)
We now want to show
(V3)
We now solve the wave equation:
(V4)
to find the allowable energies, .
Let , , where , , and
With these changes of variables Eqn. (A15) becomes
(V5)
The solutions to this equation (A5 pA22, i.e., Appendix 8)will go to infinity unless
= 2+1
= 0, 1, 2, 3 . . .
[c = speed of light]
= wave length
(V6)
Measuring energy relative to the zero point vibration frequency, i.e., = 0
Substituting for in the partition function summation
|
(V7)
For , we can make the approximation
(V8)
For m multiple frequencies of vibration
Order of Magnitude and Representative Values
For H2O we have three vibrational frequencies with corresponding wave numbers, .
and
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