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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

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 \%$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...