Chemistry - Solid State Question with Solution | TestHub

For a face-centered cubic (fcc) unit cell, the relationship between the edge length $a$ and the radius of the atom $r$ is given by:

$4r=\sqrt{2}a$

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

Given $a=408$ pm:

$r=\frac{408}{2\sqrt{2}}=\frac{204}{\sqrt{2}}=102\sqrt{2}$ pm

Substituting $\sqrt{2}=1.414$:

$r=102\times 1.414=144.228$ pm $\approx 144.2$ pm

Answer: $144.2$ pm