Exploring the LJ Cohesive Energy of a 4 Atom Square

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Homework Statement

i) Write down an expression for the LJ cohesive energy of a molecule consisting of four atoms lying on the corners of a square of side a.
ii) Deduce the equilibrium value of the nearest-neighbour separation, a, assuming the molecule retains its square shape
iii) What other configuration could the molecule adopt that would lower its LJ energy further?

Relevant equation:

U$_{tot}$=2N$\epsilon$[A$_{12}$($\frac{σ}{a}$)$^{12}$-A$_{6}$($\frac{σ}{a}$$)^{6}$]

The Attempt at a Solution

I have done parts i) and ii). It is part iii) that I am stuck on. Because bcc, hcp and fcc all have higher N, number of atoms per unit cell, and higher values of A$_{12}$, making the cohesive energy higher - this is the case, isn't it? So what other configuration could the molecule adopt?

 

[ardie]

yes and no. you have to realize that U is a potential energy and therefore negative in nature. you have to do work to remove these molecules from their square setup into something else. in a simple cubic structure ( 2 sets of these squares make a simple cubic of 8 atoms), the cohesive energy between the molecules is stronger, and so the the potential U is more negative at the equilibrium position, making these bonds more stable.so yes N increases, and maybe even A12, A6 increase, but U is inherently negative making the cohesive energies stronger.