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Q.
Volume occupied for a cubic close packed lattice of sphere is
The Solid State
Solution:
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, $V'$
$= 4 \times \frac{4}{3}\pi r^3$
$= \frac{4\times 4 \pi}{3}\times \left(a \frac{\sqrt{2}}{4}\right)^{3} $
$ = \frac{4\times 4\pi \times a^{3} \times2\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'}{V} = \frac{22}{7} \times \frac{a^{3}\times \sqrt{2}}{3\times 2\times a^{3}} $
$= 0.74$ or $74\%$