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Topic: Finding the vibrational wavenumber for a harmonic oscillator  (Read 2769 times)

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Offline bmu123

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Finding the vibrational wavenumber for a harmonic oscillator
« on: April 29, 2015, 09:57:02 AM »
If I'm given a Deslandres table like this:

                 v''
           0           1 
   0      -         23361.9

v' 1  27988.7   25781.7

   2  30408.6   28201.6

and asked to work out the vibrational wavenumber assuming the molecule behaves as a harmonic oscillator, how do I go about this?
With I've got the equation E = (v +1/2)ω so if the 1 :larrow: 0 band is 27988.7cm-1 this equals 1/2ω and the 1  :larrow: 1 band is 25781.7cm-1 that would be equal to 3/2ω, so would the difference between those, 2207cm-1 = ω?

Offline mjc123

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Re: Finding the vibrational wavenumber for a harmonic oscillator
« Reply #1 on: April 29, 2015, 05:11:53 PM »
Right answer, bad reasoning. It is obvious that this
Quote
if the 1 :larrow: 0 band is 27988.7cm-1 this equals 1/2ω and the 1  :larrow: 1 band is 25781.7cm-1 that would be equal to 3/2ω, so would the difference between those, 2207cm-1 = ω?
is nonsense arithmetically. The full expression for the energy of a state (not a transition) is E = Eel + (v+1/2)ω. Write an expression for the energies of the two states involved in a transition, and thus for the transition energy. What is the difference between the two transition energies you have chosen? It should be ω.

Offline bmu123

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Re: Finding the vibrational wavenumber for a harmonic oscillator
« Reply #2 on: May 04, 2015, 04:28:56 AM »
Right OK so it's correct to assume that in the table I'm given in the question, the difference in rows = ω' and difference in columns = ω'' but my reasoning isn't right.
 
Bit confused about the reasoning
So for v'' = 0 E = Eel + (1/2ω) and for v' = 1 E = Eel + (3/2ω)
So for the transition energy between the 2 it would be ΔE = 2Eel + 2ω

Then for another transition v'' = 1 E = Eel + (3/2ω) and for v' = 1 E = Eel + (3/2ω)
ΔE = 2Eel + (3ω)

And so the difference between the two transitions = ω?

Offline mjc123

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Re: Finding the vibrational wavenumber for a harmonic oscillator
« Reply #3 on: May 05, 2015, 05:20:31 AM »
Quote
So for v'' = 0 E = Eel + (1/2ω) and for v' = 1 E = Eel + (3/2ω)
So for the transition energy between the 2 it would be ΔE = 2Eel + 2ω
The transition energy is the difference, not the sum!
And the electronic energy is different in the two states - call them E(0) and E(1)
So for v'' = 0 E = E(0) + 1/2 ω'' and for v' = 1 E = E(1) + 3/2 ω'
The difference is E(1) - E(0) + 3/2 ω' - 1/2 ω''. This is the 1 :larrow:0 transition energy.
Now do the same for 1 :larrow:1. What is the difference between the two transition energies?
[Note: you say "in the table I'm given in the question, the difference in rows = ω' and difference in columns = ω'' ", but in your calculations you just use ω. Why?]

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