Past exam of the condensed matter physics major option course of the University of Oxford / 2015 / 1 / d / Solution

Choose orthorhombic basal axes along the tetragonal diagonals, $\mathbf{A}=\mathbf{a}+\mathbf{b}$ and $\mathbf{B}=\mathbf{a}-\mathbf{b}$, and allow their lengths to differ slightly. Two orthorhombic twins interchange $A$ and $B$. This corresponds to a small shear of the original tetragonal basal cell, or to unequal nearest-neighbor Fe–Fe distances along its two diagonal directions; it is not simply stretching the original tetragonal $x$ and $y$ axes unequally.

The tetragonal indices map to $H=h+k$, $K=h-k$. Thus tetragonal $110$ becomes orthorhombic $200$ in one twin and $020$ in the rotated twin. Their spacings are $A/2$ and $B/2$, producing the two peaks. Equal populations of the symmetry-related twins explain their equal intensities.

In contrast, tetragonal $200$ becomes orthorhombic $220$. Its inverse spacing squared is $4/A^2+4/B^2$, invariant under exchanging $A$ and $B$. It therefore does not split between those twins, exactly as observed. The diffraction facts determine the distortion's orientation and twin equivalence, but not its numerical strain amplitude without the measured peak separation.