Crystals of neon have a ccp structure. The potential energy of the crystal has been measured as \(-1.88 \mathrm{~kJ} \mathrm{~mol}^{-1}\), and the lattice parameter \(a\) is \(4.466 \AA\). Assuming that we only need to consider nearest- neighbour interactions, and that the bonding can be described by the Lennard- Jones potential (eqn 5.23), use these data to obtain numeric values for the parameters \(\epsilon\) and \(\sigma\).

The Lennard-Jones potential is given by the equation: \[ V(r) = 4\epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6} \right] \] Here, \(V(r)\) is the potential energy, \(\epsilon\) is the depth of the potential well, \(\sigma\) is the finite distance at which the inter-particle potential is zero, and \(r\) is the distance between particles.

In a face-centered cubic (ccp) structure, the nearest-neighbor distance \(r\) is given by: \[ r = \frac{a}{\sqrt{2}} \] Given that the lattice parameter \(a\) is \(4.466 \text{\AA}\), convert this to meters: \[ a = 4.466 \times 10^{-10} \text{ m} \] Then, calculate \(r\): \[ r = \frac{4.466 \times 10^{-10}}{\sqrt{2}} \approx 3.158 \times 10^{-10} \text{ m} \]

The potential energy for one mole of neon atoms in the crystal is measured as \(-1.88 \text{kJ/mol}\). For simplicity, convert this to joules: \[ -1.88 \text{kJ/mol} = -1.88 \times 10^3 \text{ J/mol} \] The number of nearest neighbors in a ccp structure is 12. Thus, the net potential energy per bond is: \[ \text{Net potential energy per bond} = \frac{-1.88 \times 10^3 \text{ J/mol}}{12} \approx -1.567 \times 10^2 \text{ J/mol} \] Note that this is the total energy divided by Avogadro's number \(N_A\) to find per bond: \[ \frac{-1.567 \times 10^2}{6.022 \times 10^{23}} \approx -2.6 \times 10^{-21} \text{ J} \] Set this equal to the Lennard-Jones potential evaluated at \(r\): \[ -2.6 \times 10^{-21} = 4\epsilon \left[ \left( \frac{\sigma}{3.158 \times 10^{-10}} \right)^{12} - \left( \frac{\sigma}{3.158 \times 10^{-10}} \right)^{6} \right] \]

Assuming the minimum energy corresponds to \(r = 2^{1/6}\sigma\), replace \(r\) with \(3.158 \times 10^{-10}\) and solve: \[ 3.158 \times 10^{-10} = 2^{1/6} \sigma \implies \sigma \approx \frac{3.158 \times 10^{-10}}{1.122} \approx 2.816 \times 10^{-10} \text{ m} \] Knowing \(\sigma\), substitute \(r\) and \(\sigma\) back into the potential formula and solve for \(\epsilon\): \[ -2.6 \times 10^{-21} = 4\epsilon \left[ \left( \frac{2.816 \times 10^{-10}}{3.158 \times 10^{-10}} \right)^{12} - \left( \frac{2.816 \times 10^{-10}}{3.158 \times 10^{-10}} \right)^{6} \right] \] Simplifying the ratio: \( \approx 0.89\) \[ -2.6 \times 10^{-21} = 4\epsilon \left( 0.2528 - 0.5645 \right) \implies \epsilon \approx \frac{-2.6 \times 10^{-21}}{4(-0.3117)} \approx \frac{2.6 \times 10^{-21}}{1.2468} \approx 2.08 \times 10^{-21}\text{ J} \] Convert \(\epsilon\) back to kJ/mol: \[ \epsilon \approx 2.08 \times 10^{-21} \text{ J} \times 6.022 \times 10^{23} = 1.25 \times 10^{3} \text{ J/mol} \approx 1.25 \text{kJ/mol} \]