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For plane (1 1 1) of SC having a lattice parameter ‘a’, planar atomic density is given by?

The triangle is made up of six hexagonal carbon molecules joined along their edges to form a triangle - an unusual arrangement that leaves two unpaired electrons unable to form a stable bond.

The atomic arrangement on the triangle (1 1 1) in a simple cubic (SC) unit cell indicates

3 corners × 1/6 per corner atom = 0.5 atom

Area of Plane

\(\begin{array}{l} \dfrac{1}{2} \times BC \times AD\\ \\ = \dfrac{1}{2} \times \sqrt 2 a \times \dfrac{{\sqrt 3 }}{2} \times \sqrt 2 a\\ \\ = \dfrac{{\sqrt 3 }}{2}{a^2} = 0.866{a^2} \end{array}\)

Therefore, the number of atoms per square inch which is nothing but its planar density = \(\dfrac{{0.5}}{{0.866{a^2}}} = \dfrac{{0.577}}{{{a^2}}}\)