Discussion
A lattice is characterized by the following primitive vectors \(\vec{a} = 2(\hat{i} + \hat{j})\), \(\vec{b} = 2(\hat{j} + \hat{k})\), \(\vec{c} = 2(\hat{k} + \hat{i})\). The reciprocal lattice corresponding to the above is :
Answer: B
The reciprocal lattice is bcc with cube edge π.
The vectors \(2(\hat{i} + \hat{j})\), \(2(\hat{j} + \hat{k})\), \(2(\hat{k} + \hat{i})\) are the primitive vectors of an fcc lattice of cube edge a = 4, since fcc primitive vectors are \(\frac{a}{2}(\hat{i} + \hat{j})\) and so on. The reciprocal of an fcc lattice of edge a is a bcc lattice of cube edge \(4\pi/a\):
\[ \frac{4\pi}{a} = \frac{4\pi}{4} = \pi \]
Direct calculation agrees: \(\vec{b}_{1} = 2\pi(\vec{b}\times\vec{c})/V\) with V = 16 gives \(\frac{\pi}{2}(1, 1, -1)\), a bcc primitive vector of cube edge \(\pi\). The fcc options mistake the reciprocal for the direct lattice type, and an edge of \(2\pi\) comes from taking a = 2.
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