Question

The density of iron crystal is $8.54$ gram $c{m}^{-3}$. If the edge length of unit cell is $2.8$ $A^o$ and atomic mass is $56$ gram $mo{l}^{-1}$, find the number of atoms in the unit cell. (Given: Avogadro's number $=6.022\times {10}^{23}$, $|A^o=1\times {10}^{-8}$cm).

Answer 1

The edge length $a=2.8A^o\times (1\times {10}^{-8}cm/A^o)=2.8\times {10}^{-8}cm$

Density $\rho =\frac{ZM}{N_0{a}^3}$

$8.54g/c{m}^3=\frac{Z\times 56g/mol}{6.022\times {10}^{23}/mol\times (2.8\times {10}^{-8}cm{)}^3}$

$Z=2$
The number of atoms in the unit cell is $2$.