$\begin{array}{1 1} 32 \% \\ 1.143\% \\ 8.4\% \\ 9.5\% \end{array} $

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Answer :$32 \%$

In the body - centred cubic unit cell, atoms are present at the corners and the body centre of the cube.

There will be two atoms per unit cells and also atoms touch each other along diagonal of the cube.

Hence $4r= \sqrt 3 a$

Now, Volume of the cell $=a^3 =\bigg( \large\frac{4}{\sqrt 3} $$r \bigg)^3$

Volume occupied by atoms $=2 \bigg( \large\frac{4}{3} $$ \pi r^3\bigg)$

fraction of volume occupied $=\large\frac{2(4/3) \pi r^3}{(4/\sqrt 3 )^3 r^3}= \large\frac{\sqrt 3 \pi}{8}$$=0.68 $

fraction of volume unoccupied by atoms $=0.32$

Percentage void volume $=32 \%$

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