Question

Copper crystallizes with face centred cubic unit cell. If the radius of copper atom is $127.8$ pm, calculate the density of copper metal?
(Atomic mass of Cu = $63.55$u and Avogadro's number $N_A=6.02\times {10}^{23}mo{l}^{-1}$).

Answer 1

$r=127.8pm$
$r=127.8\times {10}^{-10}cm$
$r=1.278\times {10}^{-8}cm$
$N_A=6.02\times {10}^{23}mo{l}^{-1}$
$M=63.55u$
From the formula, $d=\frac{Z\times M}{a^3\times N_A}$
In fcc lattice, Z = 4 atoms
For fcc lattic for copper, $a=2\sqrt{2r}$
$a^3=(2\sqrt{2r}{)}^3\Rightarrow a^3=8\times 2\sqrt{2}(1.278\times {10}^{-8}cm{)}^3$
$\Rightarrow a^3=4.723\times {10}^{-23}c{m}^3$

$d=\frac{4\times 63.55gmo{l}^{-1}}{4.723\times {10}^{-23}c{m}^3\times 6.02\times {10}^{23}mo{l}^{-1}}$

$\Rightarrow d=\frac{254.2gmo{l}^{-1}}{28.43}=8.94g{m}^{-3}$