Chemistry - Solid State Question with Solution | TestHub

For a body-centered cubic (bcc) unit cell, the atoms touch each other along the body diagonal. The relationship between the edge length $a$ and the atomic radius $r$ is given by:

$4r=\sqrt{3}a$

$r=\frac{\sqrt{3}a}{4}$

Given the edge length $a=522$ pm.

Substituting the value of $a$:

$r=\frac{\sqrt{3}\times 522}{4}$

$r=\frac{1.732\times 522}{4}$

$r=1.732\times 130.5$

$r=226.026$ pm

Rounding to the nearest integer, the radius of the potassium atom is $226$ pm.

Answer: 226 pm