Question

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

Answer 1

We are given that Xenon crystallises in FCC lattice. So, in FCC lattice, we have the formula for radius as:

4r = √2a

where a = edge length and r = radius

so, r = √2a / 4

Here value of a = 620 pm

So, putting this in our formula, we get

$r = \frac{\sqrt{2(620)}}{4}$

r = 219.13 pm

In FCC lattice, the nearest neighbour atom is calculated by the formula:

d = a / √2

where d is distance

d = 620 / √2

d = 438.40 pm

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 !!