- #1

Flucky

- 95

- 1

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.

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

(See picture below)

Derive an expression for the equilibrium separation r

At E

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/dr

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 r

But this looks pretty horrible, is it any good?

Cheers.

-------------------------------------------------------------------------------------------

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 r

_{0}of the atoms.## The Attempt at a Solution

At E

_{cohesive}, r=r_{0}and dE/dr_{0}= 0So 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/dr

_{0}= -72.78[itex]\frac{B}{r^{13}_{0}}[/itex] + 43.35[itex]\frac{A}{r^{7}_{0}}[/itex] = 043.35[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 r

_{0}= ([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.

-------------------------------------------------------------------------------------------

Last edited: