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Topic: Rotational Spectroscopy  (Read 8996 times)

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Offline Adamcp898

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Rotational Spectroscopy
« on: January 15, 2009, 08:38:10 AM »
I have an exam coming up in spectroscopy and well I just can't seem to get it fully or even partially straight in my head for that matter so I'm doing some problems to see if they can help me understand it a bit more and I was wondering if I could get a bit of help and advice on them,

The fundamental vibrational band of 14N16O exhibits rotational fine structure. The fine structure consists of equally spaced peaks separated by 3.41018 cm-1. Using this datum, calculate the equilibrium internuclear distance in 14N16O.
Assuming that the internuclear distance is unaffected by isotopic enrichment, calculate the rotational constant B for 14N18O
.

To calculate the internuclear distance I would us the formula 

Only I don't have B, so to get it I would use

And to that I would have to calculate the moment of inertia, I, using I = M1R12 + M2R22

But I don't know what to use as the R values, the distance each atom is from the centre of mass.

Should I be doing something different or using the value 3.41018cm-1 which separate the peaks as outlined in the questions.

Apologies for the length, any help is greatly appreciated.

Adam

P.S. I don't know if I should have posted this here or in the Undergraduate Sub-Forum, apologies if it's in the wrong one.

Offline Adamcp898

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Re: Rotational Spectroscopy
« Reply #1 on: January 15, 2009, 09:13:47 AM »
Upon further thinking...

Could I take it that the 3.41018 cm-1 figure is the wavenumber for the measurement of the energy difference between transitions states, say the J=0 and J=1 states and so equate that

VJ :larrow: J+1 = 2B(J+1)

(This is taken from the selection rules for rotational transitions given by the Schrodinger Wave Equation)

and if J=0 then 2B would equal 3.41018 cm-1 and B would equal 1.70509 cm-1

I could then use this figure for B to find the equilibrium internuclear distance in 14N160

Does any of this make any sense or am I being totally erratic now?

Offline Yggdrasil

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Re: Rotational Spectroscopy
« Reply #2 on: January 15, 2009, 09:46:47 AM »
Yes, you are on the right track with your second post.

Offline Adamcp898

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Re: Rotational Spectroscopy
« Reply #3 on: January 15, 2009, 10:10:19 AM »
Yes, you are on the right track with your second post.

Thank you very much!

More than likely more questions on spectroscopy to follow from me.....

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