1. Hexagonal close packed lattice.
2. a = 288 pm = 288 × 10-10 cm
Volume of the unit cell = (288 × 10-10cm)3
= 23.9 × 10-24 cm3
Volume of 208 g of the element,
\(\frac{208g}{7.2gcm^{-3}}\) = 28.88 cm3
∴ Number of units cells in 208 g = \(\frac{28.88cm^3}{23.9\times10^{-24}cm^3}\)
= 1.208 × 1024
For bcc structure,
Number of atoms in one unit cell = 2
∴ Number of atoms in 208 g = 2 × 1.208 × 1024
= 2.416 × 1024