Discussion
In a BCC lattice with lattice constant 'a', the body-centered position from the origin is at a distance of :
Answer: C
The distance is \(\frac{\sqrt{3}}{2}a\).
The body centre of a cubic cell of side \(a\) is at \((a/2, a/2, a/2)\) from the corner taken as origin:
\[ \begin{aligned} d &= \sqrt{\left(\tfrac{a}{2}\right)^{2} \times 3} \\ &= \frac{\sqrt{3}}{2}a \end{aligned} \]
This is half the body diagonal \(\sqrt{3}a\), and it is the nearest-neighbour distance in BCC. \(\frac{\sqrt{3}}{4}a\) is the atomic radius (half that distance). \(\sqrt{2a}\) and \(\sqrt{3/2}\,a\) do not correspond to this point.
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