Let $\Lambda$ be a Bravais lattice with basis vectors $\mathbf{a}_{1}, \mathbf{a}_{2}, \mathbf{a}_{3}$. Define the reciprocal lattice $\Lambda^{*}$ and write down basis vectors $\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}$ for $\Lambda^{*}$ in terms of the basis for $\Lambda$.

A finite crystal consists of identical atoms at sites of $\Lambda$ given by

$$\ell=n_{1} \mathbf{a}_{1}+n_{2} \mathbf{a}_{2}+n_{3} \mathbf{a}_{3} \quad \text { with } \quad 0 \leqslant n_{i}<N_{i}$$

A particle of mass $m$ scatters off the crystal; its wavevector is $\mathbf{k}$ before scattering and $\mathbf{k}^{\prime}$ after scattering, with $|\mathbf{k}|=\left|\mathbf{k}^{\prime}\right|$. Show that the scattering amplitude in the Born approximation has the form

$$-\frac{m}{2 \pi \hbar^{2}} \Delta(\mathbf{q}) \tilde{U}(\mathbf{q}), \quad \mathbf{q}=\mathbf{k}^{\prime}-\mathbf{k}$$

where $U(\mathbf{x})$ is the potential due to a single atom at the origin and $\Delta(\mathbf{q})$ depends on the crystal structure. [You may assume that in the Born approximation the amplitude for scattering off a potential $V(\mathbf{x})$ is $-\left(m / 2 \pi \hbar^{2}\right) \tilde{V}(\mathbf{q})$ where tilde denotes the Fourier transform.]

Derive an expression for $|\Delta(\mathbf{q})|$ that is valid when $e^{-i \mathbf{q} \cdot \mathbf{a}_{i}} \neq 1$. Show also that when $\mathbf{q}$ is a reciprocal lattice vector $|\Delta(\mathbf{q})|$ is equal to the total number of atoms in the crystal. Comment briefly on the significance of these results.

Now suppose that $\Lambda$ is a face-centred-cubic lattice:

$$\mathbf{a}_{1}=\frac{a}{2}(\hat{\mathbf{y}}+\hat{\mathbf{z}}), \quad \mathbf{a}_{2}=\frac{a}{2}(\hat{\mathbf{z}}+\hat{\mathbf{x}}), \quad \mathbf{a}_{3}=\frac{a}{2}(\hat{\mathbf{x}}+\hat{\mathbf{y}})$$

where $a$ is a constant. Show that for a particle incident with $|\mathbf{k}|>2 \pi / a$, enhanced scattering is possible for at least two values of the scattering angle, $\theta_{1}$ and $\theta_{2}$, related by

$$\frac{\sin \left(\theta_{1} / 2\right)}{\sin \left(\theta_{2} / 2\right)}=\frac{\sqrt{3}}{2}$$