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Year 4 of the natural sciences course of the University of Cambridge

Chemistry course of the year 4 of the natural sciences course of the University of Cambridge

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge

2022

Paper 2

Section E

A triangular lattice has area per particle $\sqrt3\ell_{\rm hex}^2/2$, so $\ell_{\rm hex}=\sqrt{2A/(\sqrt3N)}$. Inserting this into the supplied accessible area yields
$$
\frac{a_c^{\rm hex}}{\Lambda^2}=\left[\frac2\Lambda\frac{\sqrt A-\sigma\sqrt{\sqrt3N/2}}{\sqrt N}\right]^2.
$$
Therefore
$$
\boxed{F_{\rm hex}=-2Nk_BT\ln\left[\frac2\Lambda\frac{\sqrt A-\sqrt{B_{\rm hex}}}{\sqrt N}\right],\qquad B_{\rm hex}=\frac{\sqrt3}{2}N\sigma^2.}
$$
The prefactor $C$ remains $2/\Lambda$. The lattice is triangular in terms of particle centres; “hexagonal” here describes its sixfold coordination and cell symmetry.


Solution

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section E / 26 / d / i / Solution

Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section E / 26 / d / i / Solution

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