Potassium forms bcc unit cell structure with edge length of 522 pm....

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:

$4r = \sqrt{3}a$

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

Given the edge length $a = 522$ pm.

Substituting the value of $a$:

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

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

$r = 1.732 \times 130.5$

$r = 226.026$ pm

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

Answer: 226 pm