What will be approximate packing efficiency of the crystal which forms FCC with half the tetrahedral and octahedral voids occupied without disturbing the lattice ?

$$\begin{gathered} 4\text{ tetrahedral voids occupied, }{r}_t=0.225\mathrm{R} \\ 2\text{ octahedral voids occupied, }{r}_0=0.414R \end{gathered}$$

The required percentage,

$=\frac{4\times \frac{4}{3}\pi R^3+4\times \frac{4}{3}\pi (0.225R{)}^3+2\times \frac{4}{3}\pi (0.414R{)}^3}{(2\sqrt{2}R{)}^3}$

$=\frac{\frac{8\pi }{3}{R}^2\left(2+2\times (0.225{)}^3+(0.414{)}^3\right)}{16\sqrt{2}{R}^x}$

$=\frac{2.092\times 3.14}{6\times 1.414}\times 100$

$=0.773\times 100=77.3\mathrm{\%}$