Chemistry - Solid State Question with Solution | TestHub

Given edge length of the unit cell, $a=4.3\times 10^{-8}\mathrm{cm}$

Converting the edge length to pm:
$a=4.3\times 10^{-8}\times 10^{10}\mathrm{pm}=430\mathrm{pm}$

For a body-centered cubic (bcc) unit cell, the relationship between the atomic radius $r$ and the edge length $a$ is:
$r=\frac{\sqrt{3}}{4}a$

Substituting the value of $a$:
$r=\frac{1.732}{4}\times 430$

$r=0.433\times 430=186.19\mathrm{pm}$

Rounding off to one decimal place, we get $186.2\mathrm{pm}$.

Answer: 186.2