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The fraction of total volume occupied by the atoms present in a simple cube is
A
$\dfrac{\pi}{3\sqrt{2}}$
B
$\dfrac{\pi}{4\sqrt{2}}$
C
$\dfrac{\pi}{4}$
D
$\dfrac{\pi}{6}$
Detailed Solution
In a simple cubic unit cell the atoms are present only at the eight corners.
Number of atoms per unit cell $= 8 \times \dfrac{1}{8} = 1$
The atoms touch each other along the edges, hence $a = 2r$, i.e. $r = \dfrac{a}{2}$ ($r$ = radius of the atom and $a$ = edge length).
Volume occupied by the atom $= \dfrac{4}{3}\pi r^3$
Volume of the unit cell $= a^3 = (2r)^3 = 8r^3$
Therefore the packing fraction $= \dfrac{\frac{4}{3}\pi r^3}{(2r)^3} = \dfrac{\pi}{6}$
$= 0.52$, i.e. 52 percent of the volume is occupied.
Number of atoms per unit cell $= 8 \times \dfrac{1}{8} = 1$
The atoms touch each other along the edges, hence $a = 2r$, i.e. $r = \dfrac{a}{2}$ ($r$ = radius of the atom and $a$ = edge length).
Volume occupied by the atom $= \dfrac{4}{3}\pi r^3$
Volume of the unit cell $= a^3 = (2r)^3 = 8r^3$
Therefore the packing fraction $= \dfrac{\frac{4}{3}\pi r^3}{(2r)^3} = \dfrac{\pi}{6}$
$= 0.52$, i.e. 52 percent of the volume is occupied.