Question

In CCP of n spheres of radius r, 2n spheres of radius 0.225 r is introduced in the tetrahedral void. Therefore, packing fraction of the lattice is
$({0.225}^3=0.0114)$

• A. 0.9
• B. 0.81
• C. 0.68
• D. 0.76

Answer 1

The correct option is D 0.76

The cubic closed packing, the number of atoms per unit cell is 4
$\therefore n=4$
To this packing arrangement, 8 spheres of radius 0.225r is introduced in voids.

We know for cubic close packing,
$4r=\sqrt{2}a$
$a=2\sqrt{2}r$
$\text{Volune of unit cell is}=(2\sqrt{2}r{)}^3$

$\text{Packing fraction}=\frac{\text{Total volume occupied by spheres}}{\text{Volume of unit cell}}$
$\therefore$
$P.F.=\frac{4\times \frac{4}{3}\pi r^3+8\times \frac{4}{3}\pi \times (0.225r{)}^3}{(2\sqrt{2}r{)}^3}$
$P.F.=\frac{4\times \frac{4}{3}\pi r^3[1+2\times (0.225{)}^3]}{16\sqrt{2}{r}^3}$
$P.F.=\frac{\pi }{3\sqrt{2}}[1+2\times (0.225{)}^3$
$P.F.=\frac{\pi }{3\sqrt{2}}(1+0.0228)P.F.=0.757\simeq 0.76$

Thus, option (d) is correct.