Chemistry - Solid State Question with Solution | TestHub

For a face-centered cubic (fcc) unit cell, the particles touch each other along the face diagonal.

The relationship between the radius of the particle $r$ and the edge length $a$ is given by:

$4r=\sqrt{2}a$

$r=\frac{\sqrt{2}a}{4}=\frac{a}{2\sqrt{2}}$

Given $a=200\text{ pm}=200\times 10^{-10}\text{ cm}=2\times 10^{-8}\text{ cm}$.

Substituting the value of $a$:

$r=\frac{2\times 10^{-8}}{2\sqrt{2}}\text{ cm}$

$r=\frac{10^{-8}}{\sqrt{2}}\text{ cm}$

$r=\frac{\sqrt{2}}{2}\times 10^{-8}\text{ cm}$

$r=0.707\times 10^{-8}\text{ cm}=7.07\times 10^{-9}\text{ cm}$

Answer: $7.07\times 10^{-9}\text{ cm}$