Atoms of electrons B form hcp lattice and those of the element A occu…

Given that the atoms of \(B\) forms hcp lattice. So, the atoms at the corners are shared by \(6\) unit cells. So, its contribution is \(\frac{1}{6}\)th, fcc contributes \(1\) atom each.

So effective number of atoms in the unit cell

\(= \frac{1}{6} × 12 + \frac{1}{2} × 2 + 1 × 3\)

\(= 2 + 1 + 3 = 6\)

Number of tetrahedral voids

\(= 2 × \text{Number of atoms forming the lattice}\)

\(= 2 × 6 = 12\)

So, the number of atoms of A

\(\frac{2}{3} × 12 = 8\)

Number of atoms of B  \(= 6\)

So, \(A : B ⇒ 8 : 6\)

or, \(A : B = 4 : 3\)

So, the formula of the compound is \(A_4B_3\)