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Topic: Infrared spectrum of CO  (Read 1218 times)

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Offline bmu123

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Infrared spectrum of CO
« on: February 17, 2015, 12:18:06 PM »
Lines near the band origin in the high-resolution infrared spectrum of 12C16O are observed at 2131.58, 2135.50, 2139.38, 2147.04, 2150.81 and
2154.55 cm-1. Assign Jinitial and Jfinal quantum numbers to each transition
and, hence, determine the rotational constant of this molecule in both the v = 1 and the v = 0 vibrational levels. Determine the band origin.

OK so, in the P branch:
2131.58  J=2  :larrow: J=3
2135.50  J=1  :larrow: J=2
2139.38  J=0  :larrow: J=1


R branch:
2147.04  J=1  :larrow: J=0
2150.81  J=2  :larrow: J=1
2154.55  J=3  :larrow: J=2

R branch: ΔE=hν0+hB[J(J+1)−J′(J′+1)]
ΔE =hν0 +hB [(J+1)(J+2)-J(J+1)]
ΔE =hν0+2hB(J+1)

These are the equations I've found but I'm guessing that they're not right as I don't have v0, but I can't do the rotational constant from the line spacing because they're all different? Just need a point in the right direction, not the answer, thanks.

Offline Corribus

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Re: Infrared spectrum of CO
« Reply #1 on: February 17, 2015, 12:46:56 PM »
Those equations assume the rotational constant for the two vibrational states are equivalent. In some cases this assumption is ok, but by the wording of the question clearly it isn't here.

Have you seen this website?


(There is a procedure there, near the bottom under "real spectra", that can be used to get the correct answers using your input values.)
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline bmu123

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Re: Infrared spectrum of CO
« Reply #2 on: February 17, 2015, 01:25:26 PM »
Ah yes I remember doing this before, didn't think about that! Thanks so much :)

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