Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC chromium (Cr) when monochromatic radiation of wavelength \(0.0711 \mathrm{~nm}\) is used.

The lattice constant for BCC Chromium (Cr) is known to be \(a = 2.910 \mathrm{~nm}\).

In a BCC structure, the equation for interplanar spacing \(d_{hkl}\) is given as: $$ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} $$ For the (310) planes, we have \(h = 3\), \(k = 1\), and \(l = 0\). By plugging these values and the lattice constant (from step 1) into the formula, we can calculate the interplanar spacing. $$ d_{310} = \frac{2.910 \mathrm{~nm}}{\sqrt{3^2 + 1^2}} = \frac{2.910 \mathrm{~nm}}{\sqrt{10}} \approx 0.919 \mathrm{~nm} $$

Bragg's law relates the diffraction angle \(\theta\), the interplanar spacing \(d_{hkl}\), and the wavelength \(\lambda\) of the incident radiation: $$ n\lambda = 2d_{hkl} \sin{\theta} $$ We are looking for the first-order reflection, so \(n=1\), and we are given the wavelength \(\lambda = 0.0711 \mathrm{~nm}\). Plugging the values of \(d_{310}\) and \(\lambda\) into Bragg's law: $$ 1 \cdot 0.0711 \mathrm{~nm} = 2 \cdot 0.919 \mathrm{~nm} \cdot \sin{\theta} $$ Now, we can solve for the diffraction angle \(\theta\): $$ \sin{\theta} = \frac{1 \cdot 0.0711 \mathrm{~nm}}{2 \cdot 0.919 \mathrm{~nm}} \approx 0.03873 $$ $$ \theta = \arcsin{(0.03873)} \approx 2.22^\circ $$ Thus, the expected diffraction angle for the first-order reflection from the (310) set of planes for BCC Chromium with a monochromatic radiation wavelength of \(0.0711 \mathrm{~nm}\) is approximately \(2.22^\circ\).