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# Calculate the percentage of void volume in the primitive

$\begin{array}{1 1} 47.6\% \\ 1.143\% \\ 8.4\% \\ 9.5\% \end{array}$

Answer :$47.6\%$
In the primitive units cell, atoms are present only at the cube .
Each of the eight atoms is shared among eight cells.
Hence , the number of atoms present in one unit cell is one.
Moreover , atoms will touch each other along the edges of the cube.
Volume of the unit cell $=a^3=8r^3$
Volume occupied by the atoms $=1 \bigg( \large\frac{4}{3} $$\pi r^3 \bigg) Fraction of volume occupied by the atoms = \large\frac{(4/3)\pi r^3}{8r^3} \qquad= \large\frac{\pi}{6} \qquad= \large\frac{22}{42} Fraction of volume unoccupied by atoms = \large\frac{20}{42} Percentage of void volume =\large\frac{20}{42}$$\times 100 = 47.6 \%$