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[PoM] Lennard-Jones potential parameters

  1. Nov 24, 2016 #1

    BRN

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    Hi guys! I need your help!

    1. The problem statement, all variables and given/known data


    Use the function of Lennard-Jones V (R) = ε [(σ / R)12 - (σ / R)6] as model for the adiabatic potential energy in function of the separation between the 11B boron and nitrogen nuclei 14N. You determine the parameters ε and σ to reproduce the spectroscopic values of the vibrational quantum, ν = 1514.6 cm-1, and the separation of 1,666 cm-1 between the lines of rotational BN molecule.

    3. The attempt at a solution

    I have

    ΔErot2/I=1.666 cm-1=3.3091*10-23 J I=ħ2/ΔErot=3.3603*10-46 Kgm2

    and

    R0=√(I/μ)=1.8125*10-10 m

    At this point, I calculating the minimum oh the potential:

    V(0)=∂V(R)/∂R =0 ⇒ R06=2σ6 ⇒ σ=6√(R06/2)=1.6147*10-10 m


    For ε, I calculating k by:

    k=∂2V(R)/∂R2=6σ6εR-14(26σ6-7R6)

    but here I'm lost...

    Who can help me? Please!
     
  2. jcsd
  3. Nov 24, 2016 #2

    DrClaude

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    Staff: Mentor

    You need an equation to relate k to ##\nu##, then the only unknown in there will be ε.
     
  4. Nov 24, 2016 #3

    BRN

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    Tanks for your answer,
    the only equation i know to relate k to ν is:

    ν=1/(2πc)√(k/μ)

    so, i have (using R=R0):

    ε=[μ(2πcν)2σ7(6√2)14]/(72σ6)

    but is dimensional incorrect...
     
  5. Nov 24, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    When you let R = Ro, you should then be able to simplify to get a nice expression relating ##k## to ##\varepsilon## and ##\sigma##. This might make it a little easier to get the correct expression for ##\varepsilon## in terms of ##\nu##.
     
    Last edited: Nov 24, 2016
  6. Nov 24, 2016 #5

    TSny

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    Homework Helper
    Gold Member

    I think your expression for ##\varepsilon## is correct except that you appear to be off by a factor of ##\sigma##. Try to simplify the expression as much as possible.
     
  7. Nov 24, 2016 #6

    BRN

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    Oh Damn!
    yes, the equation is correct (apart the simplifications), but υ must be convert to m-1 and not to J!

    So:

    ε=[μ(2πcν)2σ2(6√2)14]/72=1.5192*10-18 J

    Now it's ok!

    Tanks at all!
     
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