Lennard-Jones Potential: Derive an expression for equilibrium constant

  • Thread starter Flucky
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  • #1
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[itex]^{-12}_{0}[/itex] - 7.225Ar[itex]^{-6}_{0}[/itex]

Then differentiated:

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

43.35[itex]\frac{A}{r^{7}_{0}}[/itex] = 72.78[itex]\frac{B}{r^{13}_{0}}[/itex]

r[itex]^{6}_{0}[/itex] = [itex]\frac{72.78}{43.35}[/itex][itex]\frac{B}{A}[/itex]

So r0 = ([itex]\frac{72.78}{43.35}[/itex][itex]\frac{B}{A}[/itex])[itex]\frac{1}{6}[/itex]

But this looks pretty horrible, is it any good?

Cheers.

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vfe5wm.png
 
Last edited:

Answers and Replies

  • #2
Flucky
95
1
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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