Question

An element (relative atomic mass $=60)$ having face centred cubic crystal has a density of $6.25{\text{g cm}}^{-3}$. What is the edge length of the unit cell?
(Take Avogadro's constant, $N_A=6\times {10}^{23}{\text{mol}}^{-1}$)

Answer 1

Since element has fcc structure, hence there are $94$ atoms in a unit cell (or $Z=4$).
Given, reative atomic mass is $60$ (or $M=60{\text{g mol}}^{-1}$)
$N_A=6.0\times {10}^{23}\text{and density,}d=6.25{\text{g cm}}^{-3}$

$\text{We know That}d=\frac{Z\times M}{a^3\times N_A}\Rightarrow a^3=\frac{Z\times M}{d\times N_A}$
$=\frac{\left(4\right)\left(60\text{g mol}{}^{-1}\right)}{\left(6.25{\text{g cm}}^{-3}\right)(6.0\times {10}^{23}\text{mol}{}^{-1})}$
$=6.4\times {10}^{-23}{\text{cm}}^3$
$\therefore a^3=64\times {10}^{-24}{\text{cm}}^3$
$\Rightarrow a=4\times {10}^{-8}\text{cm}$