Question
A metal crystallises into two cubic faces namely face centered (FCC) and body centered (BCC), whose unit cell edge lengths are 3.5 Å and 3.0 Å respectively. Find the ratio of the densities of FCC and BCC.

Answer

Given : Edge length of unit cell of fcc metal $=3.5 
 Å$
$=3.5 \times 10^{-8} cm$

Edge length of unit cell of bcc metal $=3 Å=3 \times 10^{-8} cm$
Density $d=\frac{n \times M }{a^3 \times N _{ A }}$
where, $n =$ Number of $Fe$ atoms in the unit cell
$M=$ Atomic mass of metal
$a=$ Edge length of unit cell
$N _{ A }=$ Avogadro number
$\therefore$ For fcc unit cell $= n =4$
For bcc unit cell $= n =2$
$\begin{aligned}
& \therefore \frac{\text { Density of fcc unit cell }}{\text { Density of bcc unit cell }}=\frac{d_{\text {fcc }}}{d_{ bcc }} \\
& =\frac{\frac{n_{\text {fcc }} \times M }{a_{ fcc }^3 \times N _{ A }}}{\frac{n_{ bcc } \times M }{a_{\text {bcc }}^3 \times N _{ A }}} \\
& =\frac{n_{ fcc }}{n_{ bcc }}\left(\frac{a_{ bcc }}{a_{ fcc }}\right)^3 \\
& =\frac{4}{2}\left(\frac{3 \times 10^{-8}}{3.5 \times 10^{-8}}\right)^3 \\
\end{aligned}$
$=1.26$
Ans. Ratio of densities, $\frac{d_{ fcc }}{d_{\text {bcc }}}= 1 . 2 6$.

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