Question

The hexagonal close-packed lattice can be represented by figure, if $C = a \sqrt{\frac{3}{3}}$. There is an atom at each corner of the unit cell and another atom which can be located by moving one third the distance along the diagonal of the rhombus base, and moving perpendicularly upward by $C / 2$. A metal crystallizes in this lattice and has a density of $2.10 \,g / cc$. Find out the distance between the nearest atoms in $\, \mathring{A}. (\sqrt{2}=1.4$ and $\left.N _{ A }=6 \times 10^{23}\right)$

Answer 1

Volume of the cell $=\frac{\text { mass of cell }}{\text { density }}$
$=\frac{24 \times 2}{6 \times 10^{28} \times 2.10}$
$=38.12 \,\, \mathring{A}^{3}$
$3 \times 38.10=6 \times \frac{\sqrt{3}}{4} a ^{2} \times a \sqrt{\frac{8}{3}}$
$a =3 \, \mathring{A}$