 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/34a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice is

Λ^{*} = {G : G \cdot R \in 2 π Z for every R \in Λ},

and the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume is

Ω = a_{1} \cdot (a_{2} \times a_{3}) = \frac{a ^{3}}{4} .

b_{1} = \frac{2 π}{a} (- e_{1} + e_{2} + e_{3}), b_{2} = \frac{2 π}{a} (e_{1} - e_{2} + e_{3}), b_{3} = \frac{2 π}{a} (e_{1} + e_{2} - e_{3}) .

A reciprocal primitive cell, and hence the Brillouin zone, has volume

\frac{( 2 π ) ^{3}}{Ω} = \frac{32 π ^{3}}{a ^{3}} .

q = k - k^{'}

f_{V} (k; k^{'}) = - \frac{m}{2 π ℏ ^{2}} \int_{R^{3}} e^{i q \cdot r} V (r) d^{3} r .

For the spherical shell, align the polar axis with

q

and use its radial Dirac delta function:

\int e^{i q \cdot r} V_{0} δ (r - d) d^{3} r = 4 π V_{0} d^{2} \frac{s i n ( q d )}{q d} .

Therefore

f_{V} (k; k^{'}) = - \frac{2 m V _{0} d ^{2}}{ℏ ^{2}} \frac{sin ( q d )}{q d} .

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor gives

f_{V_{Λ}} (k; k^{'}) = f_{V} (k; k^{'}) \sum_{R \in Λ} e^{i q \cdot R} .

\sum_{R \in Λ} e^{i q \cdot R} = \frac{( 2 π ) ^{3}}{Ω} \sum_{G \in Λ^{*}} δ^{(3)} (q - G) .

∣ k ∣ = ∣ k^{'} ∣

and hence

∣ q ∣ = 2 k sin \frac{θ}{2} .

The two shortest relevant reciprocal-vector lengths of this FCC lattice are

∣ G_{2} ∣ = \frac{2 π 3}{a}, ∣ G_{1} ∣ = \frac{4 π}{a} .

Both are kinematically available when

k > 2 π / a

. Assigning their scattering angles

θ_{2}

and

θ_{1}

, respectively, gives

\frac{sin ( θ _{1} /2 )}{sin ( θ _{2} /2 )} = \frac{∣ G _{1} ∣}{∣ G _{2} ∣} = \frac{2}{3} .

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