Question

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2g/c$m^3$. How many atoms present in 208g of the element.

Answer 1

Volume of unit cell $=(288pm{)}^3=(288\times {10}^{-10}cm{)}^3=2.389\times {10}^{-23}c{m}^3$

Volume of 208 g of the element $=\frac{Mass}{Density}=\frac{208}{7.2}=28.89c{m}^3$

Number of unit cells $=\frac{TotalVolume}{Volumeofaunitcell}=\frac{28.89}{2.389\times {10}^{-23}}=12.09\times {10}^{23}$

For a BCC structure, number of atoms per unit cell $=2$

$\therefore$ Number of atoms present in 208 g $=$ No. of atoms per unit cell $\times$ No. of unit cells

$=2\times 12.09\times {10}^{23}$

$=24.18\times {10}^{23}$

$=2.418\times {10}^{24}$