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    Calculate the frequency of J=4 < --- 3 transition in the pure rotational spectrum of 14N16O. the eqm bond length is 115.

    For this do you calculate the energy difference then just convert to frequency? im not sure as I did this the first time and got it wrong
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    I have no idea how to do it so maybe these people can help. These are just people I've seen help out before.

    @JoinedUp
    @mik1a
    @pangol





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    How did you work out the energy difference?
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    (Original post by alow)
    How did you work out the energy difference?
    I = ur^2 then Ej = (h/(2pi*2I)) * J(J+1) worked it out for 4 and 3 then got the difference
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    (Original post by jacksonmeg)
    I = ur^2 then Ej = (h/(2pi*2I)) * J(J+1) worked it out for 4 and 3 then got the difference
    IMO using wavenumbers and doing it in steps is less messy:

    \widetilde{\nu}_{J=3\rightarrow 4} = \widetilde{E}_4-\widetilde{E}_3 = \widetilde{B}\cdot 4 (4+1) - \widetilde{B}\cdot 3 (3+1) = 8\widetilde{B}

    Therefore, all you need to work out is:

    8\widetilde{B}=8\cdot \dfrac{h}{8\pi^2\widetilde{c}I}
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    (Original post by alow)
    IMO using wavenumbers and doing it in steps is less messy:

    \widetilde{\nu}_{J=3\rightarrow 4} = \widetilde{E}_4-\widetilde{E}_3 = \widetilde{B}\cdot 4 (4+1) - \widetilde{B}\cdot 3 (3+1) = 8\widetilde{B}

    Therefore, all you need to work out is:

    8\widetilde{B}=8\cdot \dfrac{h}{8\pi^2\widetilde{c}I}
    how do i get speed of light in wavenumbers?
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    (Original post by jacksonmeg)
    how do i get speed of light in wavenumbers?
    \widetilde{c} just means the speed of light in cms-1.
 
 
 
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