Discussion
Number of atoms per unit area of the (110) plane of a body centered cubic crystal, with lattice parameter ‘\(a\)’, is
Answer: B
The planar density is \(\frac{\sqrt{2}}{a^{2}}\).
Within one unit cell the (110) plane is a rectangle a wide and \(a\sqrt{2}\) long, the face diagonal. It contains the four corner atoms, each shared by four such rectangles, and the body-centre atom, which lies in the plane:
\[ \begin{aligned} n &= 4 \times \frac{1}{4} + 1 = 2 \\ \frac{n}{\text{area}} &= \frac{2}{\sqrt{2}a^{2}} = \frac{\sqrt{2}}{a^{2}} \end{aligned} \]
\(1/a^{2}\) counts the atoms of a (100) face. \(1/(\sqrt{2}a^{2})\) forgets the body-centre atom, and \(1/(\sqrt{3}a^{2})\) is the density of the BCC (111) plane.
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