Past exam of the materials science course of the University of Cambridge / 2015 / Paper 1 / 4

For $N$ distinguishable lattice sites occupied by $N_i$ mutually indistinguishable atoms of each species, the number of configurations isUsing Stirling's approximation,For one mole, $N=N_A$ and $k_B{N}_A=R$, soEquiatomic configurational entropy follows when the composition has $X_i=1/n$, givingThe reference is the unmixed configurational arrangement, and the model assumes random site occupation without short-range order. It counts configurational entropy, not all vibrational, electronic or magnetic entropy. A large number of elements does not by itself guarantee a stable single solid solution: competing thermodynamic phases and the enthalpy term also enter the Gibbs free energy.

A Shockley partial dislocation in an FCC slip plane has Burgers vector of the family $a⟨112⟩/6$. Its vector must lie in the specified plane, and the two vectors must sum to the perfect dislocation vector.

For the first plane normal ${\mathbf{n}}_1=[\bar{1}11]$, one allowed dissociation isThe vectors add to $a[303]/6$, and ${\mathbf{n}}_1\cdot [211]=-2+1+1=0$, ${\mathbf{n}}_1\cdot [1\bar{1}2]=-1-1+2=0$.

For ${\mathbf{n}}_2=[111]$,Their sum is $a[\bar{3}30]/6$, and both have zero dot product with $[111]$. Interchanging leading and trailing labels does not change these vector identities.

Each perfect vector has squared magnitude $a^2/2$, whereas the sum of the two partial squared magnitudes is $a^2/3$. The line-energy estimate proportional to $G{b}^2$ therefore favors separation, opposed by the stacking-fault energy of the ribbon between the partials.

Use the two partials on each plane as obtained above. The four pairwise sums, in units of $a/6$, areThe shortest resultant gives the largest line-energy reduction in the $G{b}^2$ approximation:The initial summed squared magnitude is $a^2/3$ and the final value is $a^2/18$, a sixfold reduction in that part of the energy estimate.

The line of intersection of the two original planes is parallel toIts dot product with the resultant Burgers vector is zero, so the junction is edge in character. A plane containing both line and Burgers vector has normal $[100]$, rather than an FCC close-packed $\{111\}$ glide normal. The resulting stair-rod dislocation is therefore sessile for ordinary glide and can form a Lomer-Cottrell lock. It obstructs further dislocation motion and promotes work hardening.

This energetic comparison assumes comparable line lengths and elastic prefactors. Geometry, line character and stacking-fault geometry affect a complete junction-energy calculation, but the small resultant identifies the favorable candidate requested by the vector model.