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Spacing/population of rotational states

  1. Jan 7, 2013 #1
    1. The problem statement, all variables and given/known data

    This is a two part question I can do about half of each but get a little lost when trying to finish.

    I have written all the values below but just in case the full question is here (sorry about clarity) - http://screencast.com/t/jHQTMFnYOhp [Broken]

    λ = 308nm
    T = 2000K

    population of ground state (j=0) is 3.6*1010 cm

    (i) So we are trying to calculate the spacing (in cm-1) between the ground rotational state and the j =6.

    (ii) The population of the rotational state j =6


    2. Relevant equations

    (i)
    εj = BJ(J+1)
    gj = 2J+1

    (ii)
    Nj/Nd = gj exp( -εj / KT)


    3. The attempt at a solution

    (i)
    gj = (2*6) + 1 = 13 but this is only degeneracy

    can also try:

    wave number = 1 /λ = 1/308*10-9 but this is only the wave number of j=0 and not he distance between j = 0 --> 6?


    (ii)
    use:
    Nj/Nd = gj exp( -εj / KT)

    We know gj, K, T Just need εj but how as we don't have a value for B?

    Sorry if this is confusing and please let me know if I can clear anything up!
    Thanks
    David
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 7, 2013 #2

    TSny

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    Hello, dscot. If you know the bond length of OH then you can calculate B. Then you could get the energy spacing between J = 0 and J = 6.
     
  4. Jan 7, 2013 #3
    Hi TSny =)

    I think I understand what you are saying but it doesn't look like the bond length was provided in the question? I'm just guessing here but maybe a clue is that its an OH radical? I could try google but this was a past exam question so I think there must be a way to figure it out?

    Thanks!
     
  5. Jan 7, 2013 #4

    TSny

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    I can't see a way to get B without knowing the bond length.

    I also don't see the relevance of the 308 nm wavelength. That seems to me to be the wavelength corresponding to some electronic excitation of OH. I don't see how to use it to help answer the question.

    Maybe someone else can help.
     
  6. Jan 7, 2013 #5
    Hi TSny,

    This does seem to be a tricky one, I'll keep trying and hopefully someone else might know :)

    Thanks for trying!
    David
     
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