Question

Xenon crystallize in the face centred cubic lattice and the edge of the unit cell is 620 pm . what is the nearest neighbour distance and what is the radius of xenon atom

Answer 1

1) For fcc, edge = √8r
Therefore, 620 = 2.82*r
i.e., r = 620/2.82 = 219.86 pm
2) For fcc, packing efficiency = 74%

Answer 2

Hey !!

Here,  

        a = 620 pm , d = ? , r = ?

For the face-centered cubic d = a / √2 = 620 / 1.414

             = 438.5 pm

  r = d / 2  

       = 438.5 / 2

       = 219.25 pm

Hope it helps you !!