Lennard-Jones Potential: Derive an expression for equilibrium constant

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SUMMARY

The discussion focuses on deriving the equilibrium separation \( r_0 \) of atoms using the Lennard-Jones potential. The cohesive energy equation is expressed as \( E = 6.065Br^{-12}_{0} - 7.225Ar^{-6}_{0} \). By differentiating this equation and setting the derivative equal to zero, the relationship \( r^{6}_{0} = \frac{72.78}{43.35}\frac{B}{A} \) is established. The final expression for \( r_0 \) is given as \( r_0 = \left(\frac{72.78}{43.35}\frac{B}{A}\right)^{\frac{1}{6}} \).

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Flucky
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Edit: well this is frustrating, not sure why the itex things aren't working..

Edit2: I've attached a picture of the question at the bottom as well.

Homework Statement


The cohesive energy of a solid noble gas may be written as

(See picture below)

Derive an expression for the equilibrium separation r0 of the atoms.


The Attempt at a Solution



At Ecohesive, r=r0 and dE/dr0 = 0

So first I rearranged the main equation into something nicer to differentiate:

E = 6.065Br^{-12}_{0} - 7.225Ar^{-6}_{0}

Then differentiated:

dE/dr0 = -72.78\frac{B}{r^{13}_{0}} + 43.35\frac{A}{r^{7}_{0}} = 0

43.35\frac{A}{r^{7}_{0}} = 72.78\frac{B}{r^{13}_{0}}

r^{6}_{0} = \frac{72.78}{43.35}\frac{B}{A}

So r0 = (\frac{72.78}{43.35}\frac{B}{A})\frac{1}{6}

But this looks pretty horrible, is it any good?

Cheers.

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Last edited:
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Finally found a similar question in the depths of the internet and looks like it is right, so all is good here.
 

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