Question

A metal crystallises into two cubic phases, face centred cubic $(fcc)$ and body centred cubic $(bcc)$, whose unit cell lengths are $3.5$ $?$ and $3.0$ $?$, respectively. The ratio of densities of $fcc$ and $bcc$ is

• A. $1.259 : 1$
• B. $1 : 1.259$
• C. $3 : 2$
• D. $1.142 : 1$

Answer 1

The Correct Option is A

Unit cell length for $fcc=3.5\,?$ Unit cell length for $bcc = 3.0 \,?$ $\therefore$ Density in $fcc=\frac{n_{1}\times at.wt.}{V_{1}\times N_{A}}$ Density in $bcc=\frac{n_{2}\times at.wt.}{V_{2}\times N_{A}}$ or $\frac{\text{Density} \left(fcc\right)}{\text{Density} \left(bcc\right)}=\frac{n_{1}}{n_{2}}\times\frac{V_{2}}{V_{1}}$ $=\frac{4}{2}\times\frac{V_{2}}{V_{1}}$$\begin{bmatrix}\therefore for \,fcc,&n_{1}=4\\ \therefore for \,bcc,&n_{2}=2\end{bmatrix}$ Volume for $fcc=V_{1}=a^{3}=\left(3.5\times10^{-8}\right)^{3}\,cm^{3}$ and Volume for $bcc=V_{2}=a^{3}=\left(3.0\times10^{-8}\right)^{3}\,cm^{3}$ $\therefore \frac{\text{Density} \left(fcc\right)}{\text{Density} \left(bcc\right)}$ $=\frac{4\times\left(3.0\times10^{-8}\right)^{3}}{2\times\left(3.5\times10^{-8}\right)^{3}}$ $=1.259$ or $1.259 : 1$