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 = 017
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. ν = 0 is not an option18. When ν0 is the frequency of vibration, the ground state energy is
(V1) | |||||
Harmonic oscillator 19 | |||||
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, ε. | |||||
Let , , where , , and | |||||
With these changes of variables Eqn. (A15) becomes | |||||
(V5) | |||||
The solutions to this equation20 will go to infinity unless | |||||
[c = speed of light]21 | |||||
(V6) | |||||
Measuring energy relative to the zero point vibration frequency, (i.e., ν = 0) gives | |||||
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 | |||||
17P. W. Atkins, Physical Chemistry, 5th ed. (New York: Freeman, 1994), p. 402.
18 P. W. Atkins, Physical Chemistry, 5th ed. (New York: Freeman, 1994), pp. 22, 402.
19P. W. Atkins, Physical Chemistry, 5th ed. (New York: Freeman, 1994), p. 402.
20P. W. Atkins, Physical Chemistry, 5th ed. (New York: Freeman, 1994), p. 22, Appendix 8.
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