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Lennard-Jones Work

  1. Feb 27, 2017 #1
    1. The problem statement, all variables and given/known data
    I need to calculate the work donde by the Lennard-Jones Law, considering:

    F(r)=F0 [2(σ/r)13-(σ/r)7]

    when approximating two atoms from infinity to the equilibrium position between both atoms

    2. Relevant equations

    First thing I don't know how to calculate is the equilibrium position (x0) between two arbitrary atoms


    3. The attempt at a solution

    I just integrated this force, considering

    WAB=∫F⋅dr ; r=A to r=B

    As I don't know the x0, I just integrate from A to B, to later analyze the result. Assuming A=-∞ and B=x0

    W= ∫ F0 [2(σ/r)13-(σ/r)7] dr ; r=A to r=B
    = integrating...
    = 2F0σ13 ( r-12 /-12) - F0σ7 (r-6/-6) ; still need to evaluate

    Considering that A=-∞, I finally obtain:

    W= F0σ7/6 (1/x06)-F0σ13/6(1/x012)

    Is it right until here?
     
  2. jcsd
  3. Feb 28, 2017 #2

    kuruman

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    What is the force at the equilibrium position. What does "equilibrium" mean?
    What are your limits of integration? Consider particle A already at the origin. Then calculate the work done by the force on particle B as it is brought from infinity to x0.
     
  4. Feb 28, 2017 #3

    The statement of this problems is:

    In diatomic molecules, the constituent atoms exert attractive forces between themselves at great distances and repulsive forces at short distances. For many molecules the Lennard-Jones law is a good approximation for the modulus of these forces.

    $$ F(r)=F_0 \cdot [2(\frac{\sigma}{r})^{13}-(\frac{\sigma}{r})^7] $$

    where r is the distance between the center of the nucleus of the atoms, ## \sigma ## a length parameter and ## F_0 ## a constant.

    Determine the work to be performed by this force or by an external agent to approximate two atoms from infinity to the position of balance between the two.



    - So, what I mean to equilibrium position is the position of balance between the two atoms, but there is no more information about that.

    - The limits of integration, as I don't know the position of balance, are ## A=-\infty ## to ## B=x_0 ## (the so called position of balance).

    - Didn't catch up the last advice
     
    Last edited: Feb 28, 2017
  5. Feb 28, 2017 #4

    kuruman

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    How big is the force on molecule A when it is at equilibrium or, as you say, in balance with molecule B?
     
  6. Feb 28, 2017 #5
    I guess it is zero right? I that way it is in equilibrium.
     
  7. Feb 28, 2017 #6

    kuruman

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    Exactly. So at what inter-molecular distance r0 is the force zero?
     
  8. Mar 2, 2017 #7
    I don't know :C I took a while to answer cause of that. I think it should be at a huge distance, right?
     
  9. Mar 2, 2017 #8

    kuruman

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    It is simpler than you think. The force in general is
    $$F(r)=F_0 \cdot [2(\frac{\sigma}{r})^{13}-(\frac{\sigma}{r})^7]$$
    At r = r0 you have
    $$0=F(r_0)=F_0 \cdot [2(\frac{\sigma}{r_0})^{13}-(\frac{\sigma}{r_0})^7]$$
    Do you see what you have to do?
     
  10. Mar 2, 2017 #9
    yes! with $$F(r_0)=0$$ I can actually find $$r_0$$ and later evaluate that value in the integral.
     
  11. Mar 2, 2017 #10

    kuruman

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    Go for it! :smile:
     
  12. Mar 2, 2017 #11
    thank you very much!:biggrin:
     
  13. Mar 27, 2017 #12
     
  14. Mar 27, 2017 #13
     
  15. Mar 27, 2017 #14

    kuruman

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    Is there a question you wish to ask?
     
  16. Mar 27, 2017 #15
    No thanks! I just made a mistake editing this.
     
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