Discussion
The primitive translational vectors of the body centered lattice are given by
\[ \vec{a} = \frac{a}{2}(\hat{i} + \hat{j} - \hat{k}), \]
\[ \vec{b} = \frac{a}{2}(-\hat{i} + \hat{j} + \hat{k}), \]
\[ \vec{c} = \frac{a}{2}(\hat{i} - \hat{j} + \hat{k}), \]
where a is the side of the conventional unit cube and \(\hat{i}, \hat{j}, \hat{k}\) are orthogonal unit vectors parallel to the cube edges. The volume of the reciprocal lattice of bcc cell is :
Answer: B
The volume is \(2(2\pi/a)^{3}\).
The volume of the primitive cell is the scalar triple product of the given vectors. For the bcc lattice it is half the conventional cube:
\[ \begin{aligned} V_{c} &= \vec{a}\cdot(\vec{b} \times \vec{c}) \\ &= \frac{a^{3}}{2} \end{aligned} \]
The reciprocal lattice cell has volume \((2\pi)^{3}/V_{c}\):
\[ \begin{aligned} V^{*} &= \frac{(2\pi)^{3}}{a^{3}/2} \\ &= 2\left(\frac{2\pi}{a}\right)^{3} \end{aligned} \]
The option \(\frac{1}{2}a^{3}\) is the direct-lattice primitive cell, not its reciprocal. The reciprocal of bcc is an fcc lattice, and the other options do not give its cell volume.
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