Question

An element occurs in bcc structure with a cell edge of $288pm$. The density of metal is $$7.2\ g\ cm^{-3}$.Howmanyatomsdoes$208\ g$ of the element contain?

Answer 1

Volume of the unit cell $=(288\times {10}^{-10}{)}^3$$ $=23.9\times {10}^{-24}c{m}^3$
Volume of $208g$ of the element $=\frac{208}{7.2}=28.88c{m}^3$
Number of unit cells in $28.88c{m}^3=\frac{28.88}{23.9\times {10}^{-24}}$
$=12.08\times {10}^{23}$ unit cells
Each bcc structure contains $2$ atoms.
So, total atoms in $12.08\times {10}^{23}$ unit cells
$=2\times 12.08\times {10}^{23}$
$=24.16\times {10}^{23}$