Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 33B / Solution

A Bravais lattice is the set of all integer combinations of linearly independent primitive translation vectors. Its reciprocal lattice consists of $g$ satisfying $g\cdot l\in 2\pi \mathbb{Z}$ for every translation $l$; equivalently $e^{ig\cdot l}=1$.

The original PDF's third family has a bare $j$, not $n_{2}j$. Taken literally, the listed set is not the claimed Bravais lattice: the claimed primitive vector $a(i+k)/2$ does not occur. Its half-integer first and third coordinates force it into that third family, whose second coordinate is always $a$, not zero. The intended repair is $j\mapsto n_{2}j$ in the third family. The following conclusions apply to that corrected set.

The four corrected families are exactlyThe allowed parity patterns are all even, or exactly two odd coordinates. An integer combination of the proposed primitive vectors has coordinates $(a/2)(m_2+m_3,m_1+m_3,m_1+m_2)$, so lies in this set. Conversely, solveAll three are integers precisely when $h+k+l$ is even. This proves that the corrected set is the face-centered cubic lattice with the proposed primitive basis.

Direct dot products give $a_i\cdot b_j=2\pi {\delta }_{ij}$ forThese vectors generate the entire reciprocal lattice, since the coefficients of an arbitrary reciprocal vector in this basis are its integer pairings with the $a_i$. Its vectors are $(2\pi /a)(h,k,l)$ with all three coordinates of the same parity, the body-centered cubic lattice.

In Bragg scattering, translating a scattering atom by $l$ changes the amplitude by $e^{i(k-k^{\prime })\cdot l}$. Constructive interference from every lattice translate requires this factor to be one, so $k^{\prime }-k=g$ is a reciprocal lattice vector. For elastic scattering, $|k^{\prime }|=|k|$; the chord between these two vectors has length $2|k|\sin (\theta /2)$. ThereforeThe shortest nonzero reciprocal vectors are the $(111)$ family of length $(2\pi /a)\sqrt{3}$, followed by the $(200)$ family of length $(2\pi /a)2$. Identical atoms on this Bravais lattice introduce no additional basis extinction. If both families are accessible at the incident wavelength, their scattering angles obey $\boxed{\sin ({\theta }_1/2)/\sin ({\theta }_2/2)=\sqrt{3}/2}$.