Chemistry - Solid State Question with Solution | TestHub

In a simple cubic unit cell, the number of particles per unit cell is $1$.

The edge length $a$ and the radius of the particle $r$ are related by $a=2r$.

Volume of the unit cell, $V=a^3=(2r{)}^3=8r^3$.

Volume occupied by one particle, $v=\frac{4}{3}\pi r^3$.

The ratio of the volume of the unit cell to the volume of the particle is:

$\frac{V}{v}=\frac{8r^3}{\frac{4}{3}\pi r^3}=\frac{6}{\pi }$

Given that $v=2.1\times 10^{-23}{\text{ cm}}^3$, we have:

$V=\frac{6}{\pi }\times 2.1\times 10^{-23}{\text{ cm}}^3$

$V=\frac{6}{3.14}\times 2.1\times 10^{-23}{\text{ cm}}^3\approx 4.01\times 10^{-23}{\text{ cm}}^3$

Comparing this with $x\times 10^{-23}{\text{ cm}}^3$, we get $x\approx 4.0$.

Answer: $4.0$