Question

Volume occupied for a cubic close packed lattice of sphere is

• A. 60%
• B. 52.4%
• C. 68%
• D. 74%

Answer 1

Answer: (D)

In $c.c.p.$ or $f.c.c.$ arrangement, number of sphere (atoms) per unit cell is $4$.
Let $a$ be the edge length of the cube.
$\therefore$ Volume of the cube, $V=a^3..(1)$
Let $r$ be the radius of the sphere
For $fcc$ arrangement $r=\frac{a\sqrt{2}}{4}$
Volume occupied by $4$ sphere, $V^{\prime }$
$=4\times \frac{4}{3}\pi r^3$
$=\frac{4\times 4\pi }{3}\times {\left(a\frac{\sqrt{2}}{4}\right)}^3$
$=\frac{4\times 4\pi \times a^3\times 2\sqrt{2}}{3\times 4\times 4\times 4}$
$=\frac{\frac{22}{7}\times a^3\times \sqrt{2}}{3\times 2}$
$\therefore$ Ratio of volume occupied by the spheres to the volume of the cube
$\frac{V^{\prime }}{V}=\frac{22}{7}\times \frac{a^3\times \sqrt{2}}{3\times 2\times a^3}$
$=0.74$ or $74\%$