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Abhisar

  • one year ago

If the potential energy of two molecules is given by \(\sf U=\frac{A}{r^{12}} - \frac{B}{r^6}\)then at equilibrium position, its potential energy is equal to

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  1. Frostbite
    • one year ago
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    We define the equilibrium position as the position in which the attractive force and the repulsion force is equal to 0. that is: \[\Large F_{total}=F_{repulsion}+F_{attractive}=0\] The expression you want to evaluate is therefore: \[\large F=\frac{ dV }{ dr }=0\] I have just noted \(V\) as the potential energy. For your function we then get: \[\large \frac{ d }{ dr }\left( \frac{ A }{ r^n }-\frac{ B }{ r^m } \right)=Bmr ^{-m-1}-Anr ^{-n-1}=0\] Solve for the the distance \(r\): \[\large r=\exp \left( \frac{ -\log(A)+\log(B)+\log(m)-\log(n) }{ m-n } \right)\] Insert the values for \(n\) and \(m\) we can evaluate the distance to: \[\Large r=\left( \frac{ 2a }{ b } \right)^{\frac{ 1 }{ 6 }}\] I have discarded the negative solution as it does not make since to talk about negative distances. Or it kinda does if you think about it like this: |dw:1440933102863:dw| Insert the result into your function and evaluate the potential energy: \[\Large U=\frac{ A }{ \left( \sqrt[6]{\frac{ 2A }{ B }} \right)^{12} }-\frac{ B }{ \left( \sqrt[6]{\frac{ 2A }{ B }} \right)^{6} }\] I have tried to evaluate the expression to get: \[\Large U=\frac{ -B^2 }{ 4A }\] Unsure if I've got the last evaluation right, but I am kinda confident. Hope this is what you were looking for..

  2. Frostbite
    • one year ago
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    When I wrote \[\Large r=\left( \frac{ 2a }{ b } \right)^{\frac{ 1 }{ 6 }}\] I meant: \[\Large r=\left( \frac{ 2A }{ B } \right)^{\frac{ 1 }{ 6 }}\] Sorry.

  3. Abhisar
    • one year ago
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    Wow! Thanks c:

  4. Frostbite
    • one year ago
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    No problem. the values of A and B I can obviously not help with. :)

  5. Abhisar
    • one year ago
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    No, actually I wanted in terms of A and B only...

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