In a FCC unit cell of A atom (At wt.= 150) having side length $10\stackrel{°}{\mathrm{A}}$. Number of atom per unit cell is Z, number. of next nearest neighbour is X and packing efficiency is y % find value of $\frac{\mathrm{yz}}{\left({\mathrm{x}}^2+1\right)}$.

 FCC lattice $\to$8 corners $\to$ Atom 

                           6 face $\to$ Atom 

Corners contribution $=\frac{1}{8}\times 8=1$

 Face contribution $=\frac{1}{2}\times 6=3$

$\frac{\mathrm{No}. \mathrm{of} \mathrm{atoms}}{\mathrm{unit} \mathrm{cell}}=4$ 

Next nearest neighour $\to$ 6

Total area occupied by sphere 

Total area $\mathrm{FCC}\to \sqrt{2}\mathrm{a}=4\mathrm{r}$

$=\mathrm{a}=\frac{4\mathrm{r}}{\sqrt{2\mathrm{a}}}$

$=\frac{4\times \frac{4}{3}\pi r^3}{{{\displaystyle \mathrm{a}}}^3}=\frac{\frac{16}{3}\times \pi r^3}{2\sqrt{2}{r}^3}$

$=74\%$

Packing efficiency $=74\to \mathrm{y}$

$\text{ Value of }\frac{yz}{\left(x^2+1\right)}$

$=\frac{74\times 4}{37}=8$