Relationship between binding energy of atoms in a solid and compressibility defined by bulk modulus

Consider an atom which has η nearest neighbours. The equilibrium binding energy $$L_0$$ of $$N$$ atoms is: $$L_0 = \frac{Nηϵ}{2}$$ The bulk modulus $$K$$ of a solid of volume $$V$$ compressed under pressure $$P$$ is defined as : $$-V\frac {dP}{dV} = K$$ If we consider the work done on the solid that is compressed we determine that: $$K=V\frac {d^2E}{dV^2}$$

As atoms are forced away from equilibrium positions the interatomic spacing $$r$$ decreases and since we know that $$V = Nr^3$$ we can deduce that near equilibrium: $$\frac {d^2E}{dV^2}=\frac {1}{9N^2a^4} \frac {d^2E}{dr^2}$$ where $$a$$ is the equilibrium separation.

I am trying to show that the bulk modulus is related to the binding energy whereby $$K = 8\frac{L_0}{V_0}$$ and $$V_0$$ is the equilibrium volume.

Any help would be appreciated.

• I doubt that is possible. One can try to model an ionic salt or a VanderWaals-solid. – Pieter Oct 30 '18 at 19:41
• I forgot to mention that we are considering a solid whose atoms interact via the Lennard Jones potential – David Abraham Oct 30 '18 at 20:09
• You might try incorporating a pair potential. Right now you’re stuck trying to relate an energy to a second derivative of energy. – Chemomechanics Oct 30 '18 at 20:15

You set $$V=Nr^3$$, but this is only true if the atoms lie on a simple cubic lattice. This is actually incorrect for the minimum-energy Lennard-Jones solid, in which the atoms lie on a face centred cubic (fcc) lattice. Fortunately it is not necessary to know the lattice, if you have made the approximation above. All you need is $$V\propto r^3$$, which implies $$\frac{dV}{V} = 3\frac{dr}{r} \qquad\text{or}\qquad V\frac{d}{dV} = \frac{1}{3}\, r \frac{d}{dr}$$ Then, provided $$dE/dV=0$$ at $$V=V_0$$ (which also means $$dE/dr=0$$ at $$r=a$$) we can write $$V_0 K = \left . V^2 \frac{d^2E}{dV^2} \right|_{V=V_0} = \frac{1}{9}\left . r^2 \frac{d^2E}{dr^2} \right|_{r=a} = \frac{a^2}{9}\left . \frac{d^2E}{dr^2} \right|_{r=a}$$ which is basically your equation, but without assuming $$V=Nr^3$$. Note that there would be extra terms here if we had not been able to set the first derivative of $$E$$ equal to zero, so this equation holds only at equilibrium.
Finally, it is most convenient to write the Lennard-Jones pair potential in the form parametrized by the well depth, $$\epsilon$$, and the position of the minimum, $$a$$, thus: $$v(r) = \epsilon \left[ \left(\frac{a}{r}\right)^{12}- 2\left(\frac{a}{r}\right)^{6} \right]$$ and the total energy as a function of nearest-neighbour distance $$r$$ is $$E(r) = \frac{1}{2}N\eta \, v(r) = L_0 \left[ \left(\frac{a}{r}\right)^{12}- 2\left(\frac{a}{r}\right)^{6}\right]$$ When $$r=a$$, $$E(a)=-L_0$$ as you would expect. You can verify yourself that $$dE/dr=0$$ at $$r=a$$.
I'll leave you to finish the calculation. Notice that you can derive the result $$V_0 K=8L_0$$ without knowing the lattice, or even $$\eta$$.
If you don't make the approximation of neglecting nearest neighbour interactions, then you need to know what lattice the atoms lie on. The long range attractions in the Lennard-Jones potential mean that the crystal energy will be a minimum when the nearest neighbour distance is somewhat less than the minimum in the pair potential, $$a$$. However, all the distances scale with volume in a known way, and hence the dependence of the two terms in the LJ potential on $$r$$ is known, from numerically calculated lattice sums.