Question

Density of Li atom is $0.53$ g cm${}^3$. The edge length of Li is $3.5\mathring{A}$. Find out the number of Li atoms in a unit cell.

[Given: $N_A=6.02\times {10}^{23},M=6.94$ g mol${}^{-1}$]

Answer 1

The expression for density is as follows:

$d=\frac{n\times m}{a^3\times N_A}$
$d=0.53$ g cm${}^{-3}$
$a=3.5\mathring{A}=3.5\times {10}^{-8}$ cm
$M=6.94$ g mol${}^{-1}$
$n=\frac{d\times a^3\times N_A}{M}=\frac{0.53\times (3.5\times {10}^{-8}{)}^3\times 6.023\times {10}^{23}}{6.94}$ $⟹n=2$