Question

Density of Li atom is $0.53$ $g/c{m}^3$. The edge length of unit cell of Li crystal is $3.5$A. Find out the number of Li atoms in a unit cell. $(N_A=6.0\times {10}^{23}mo{l}^{-1},M=6.94gmo{l}^{-1})$.

Answer 1

The density of the crystal is given by

$\rho =\frac{nM}{N_A{a}^3}g/c{m}^3$
where symbols have usual meaning

Given, $\rho =0.53g/c{m}^3$

$a=3.5\times {10}^{-8}cm$

$M=6.94g/mol$

$N_A=6.022\times {10}^{23}$

$\Rightarrow n=\frac{0.53\times 6.022\times {10}^{2}3\times (3.5{)}^3\times ({10}^{-8}{)}^3}{6.94}$

$n=2$ (Body Centered Cubic (BCC))