Chemistry - Solid State Question with Solution | TestHub

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

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

$\sqrt{2}a=4r$

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

Given $a=360$ pm:

$r=\frac{360}{2\sqrt{2}}=90\sqrt{2}$

Using $\sqrt{2}=1.414$:

$r=90\times 1.414=127.26$ pm

Answer: $127.26$ pm