Question

A CCP of $n$ spheres of radius $r$, $n$ spheres of radius $0.414r$ and $2n$ spheres of radius $0.225r$ are introduced in the octahedral and tetrahedral void respectively. Therefore packing of the lattice is:

• A. $0.74$
• B. $0.81$
• C. $0.86$
• D. $0.79$

Answer 1

Answer: (B)

$0.81$

$\begin{array}{c}\text{Given: - CCP or FCC packing radius=r}\\ \text{Zeffective}=\frac{1}{8}\times 8+\frac{1}{2}\times 6=4\end{array}$

$\begin{array}{c}\text{Let radius of sphere of atom of Fcc be r}\\ \text{and side of}Fcc\text{be a}\end{array}$

$\begin{array}{cc}\text{Then}& :a\sqrt{2}=4r\\ & \Rightarrow a=2\sqrt{2}r\end{array}$

$\begin{array}{c}\because \text{There are n spheres,}80\text{total number of}\\ \text{unit cells =}\frac{n}{4}\\ \text{volume of Lattice}=\frac{n}{4}\times (a{)}^3=\frac{n}{4}(2\sqrt{2}r{)}^3\\ =4\sqrt{2}n{r}^3\end{array}$

$\begin{array}{cc}\text{volume of atoms}=& n\times \frac{4}{3}\times \pi r^3+n\times \frac{4}{3}\times \pi \times (0.414r{)}^9\\ +2n\times \frac{4}{3}\times \pi (0.225{)}^3& \\ =& \frac{n\times 4}{3}\times r^3\times \pi [1+(0.414{)}^3+\left(0.225{)}^9\right]\\ =& 4.534{nr}^3\end{array}$

$\begin{array}{c}\therefore \text{packing of lattice}=\frac{4.534{nr}^9}{4\sqrt{2}{nr}^3}=0.81\\ \text{So, option (B) is correct.}\end{array}$