# 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

Looking at your question, it looks like some hints would be more appropriate than a full answer.

Your comment clarifies that you are assuming the Lennard-Jones potential, but it is also clear from your question that you are making the approximation of neglecting all interactions beyond nearest neighbours. This simplifies the calculation quite a bit: it means that the minimum energy of the solid occurs when all of the nearest neighbour distances are at the minimum of the LJ potential. I'll come back to this approximation at the end. Note that we are also implicitly assuming that the equilibrium state corresponds to zero pressure, and of course thermal effects are neglected, so we are doing the calculation for a low temperature solid.

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.