Calculate the radius of atom of metal forming fcc unit cell having edge length $360$ pm

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:

$\sqrt{2}a = 4r$

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

Given $a = 360$ pm:

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

Using $\sqrt{2} = 1.414$:

$r = 90 \times 1.414 = 127.26$ pm

Answer: $127.26$ pm