Lithium has a  b c c structure. Its density is $530 \mathrm{kg} {\mathrm{m}}^{-3}$  and its atomic mass is $6.94 \mathrm{g} {\mathrm{mol}}^{-1}$ . Calculate the edge length of a unit cell of lithium metal. $\left({\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}{ \mathrm{mol}}^{-1}\right)$

$\mathrm{For} \mathrm{bcc},\mathrm{Z}=2,\mathrm{\rho }=530 \mathrm{kg}{ \mathrm{m}}^{-3},\mathrm{at}.\mathrm{mass} \mathrm{of} \mathrm{Li}=6.94 \mathrm{g}{ \mathrm{mol}}^{-1},$

${\mathrm{N}}_{\mathrm{A}}=6.02\times {10}^{23}{ \mathrm{mol}}^{-1}$

$\rho =530 \mathrm{kg}{ \mathrm{m}}^{-3}=\frac{530\times 1000 \mathrm{g}}{1\times (100{)}^3{ \mathrm{cm}}^3}=0.53 \mathrm{g}{ \mathrm{cm}}^{-3}$

$\mathrm{\rho }=\frac{\mathrm{Z}\times \text{ At. mass }}{{\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3}$

$\begin{array}{l}{\mathrm{a}}^3=\frac{\mathrm{Z}\times \text{ At. mass }}{{\mathrm{N}}_{\mathrm{A}}\times \mathrm{\rho }}=\frac{2\times 6.94}{6.02\times {10}^{23}\times 0.53}=43.5\times {10}^{-24}{ \mathrm{cm}}^3\\ \mathrm{a}=352\times {10}^{-10} \mathrm{cm}=352\mathrm{pm}\end{array}$