Question

A metal crystal of atomic mass $40{\text{g mol}}^{-1}$ has a density of $\frac{100}{24}{\text{g/cm}}^3$. If the metal crystal exists as face centered cubic then calculate edge length of the unit cell (in $\stackrel{o}{A}$). (Given, Avogadro's number $N_A=6\times {10}^{23}$).

Answer 1

We know,
$d=\frac{Z\times M}{a^3\times N_A}$
For FCC, $Z=4$
$\Rightarrow d=\frac{4\times \frac{40}{N_A}}{a^3}=\frac{100}{24}$
$\Rightarrow a^3=\frac{4\times 40}{6\times {10}^{23}}\times \frac{24}{100}$
$\therefore a=4\times {10}^{-8}\text{cm}=4A^{\circ }$