X-rays of wavelength \(\lambda\), when incident on the \((1\,0\,1)\) plane of a cubic lattice with lattice constant \(a\), produce a first-order Bragg reflection at \(\theta = 30^{\circ}\). The unit cell is then compressed along the z-axis so that its x and y edges keep length \(a\) while its z edge shortens to \(a/\sqrt{3}\), as depicted in the figure below.
\(\theta = 45^{\circ}\)
Step 1 - Original (cubic) case, extract \(\lambda\). For a cubic lattice \(d_{hkl}=a/\sqrt{h^{2}+k^{2}+l^{2}}\), so
\[d_{101}^{\text{cubic}}=\frac{a}{\sqrt{1^{2}+0^{2}+1^{2}}}=\frac{a}{\sqrt{2}}.\]First-order Bragg (\(n=1\)) with \(\theta=30^{\circ}\) gives
\[\lambda=2\,d_{101}^{\text{cubic}}\sin 30^{\circ}=2\cdot\frac{a}{\sqrt{2}}\cdot\frac{1}{2}=\frac{a}{\sqrt{2}}.\]Step 2 - Compressed (tetragonal) spacing for \((1\,0\,1)\). With \((a,a,c)\) and \(c=a/\sqrt{3}\),
\[\frac{1}{d_{hkl}^{2}}=\frac{h^{2}+k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}}\ \Longrightarrow\ \frac{1}{d_{101}^{2}}=\frac{1}{a^{2}}+\frac{1}{(a/\sqrt{3})^{2}}=\frac{1}{a^{2}}+\frac{3}{a^{2}}=\frac{4}{a^{2}},\]so \(d_{101}^{\text{new}}=a/2\).
Step 3 - New Bragg angle. Same wavelength \(\lambda=a/\sqrt{2}\):
\[\sin\theta_{\text{new}}=\frac{\lambda}{2\,d_{101}^{\text{new}}}=\frac{a/\sqrt{2}}{2\cdot a/2}=\frac{1}{\sqrt{2}}\ \Longrightarrow\ \theta_{\text{new}}=45^{\circ}.\]Answer: (A) \(\theta=45^{\circ}\).
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