Question

Xenon crystallizes in fcc lattice and the edge of the unit cell is 620 pm the radius of cenon atom is

Answer 1

Answer:

a=620pm

r=?

For fcc lattice,

r=a/√2

=620/√2

r=438.47pm

Answer 2

Hey Dear,

◆ Answer -

r = 219 pm

● Explaination -

We know that - in FCC lattice, edge length of unit cell is related to radius of atom as -

a = 2√2 r

Rearranging this we get,

r = a / 2√2

r = 620 / 2√2

r = 219 pm

Therefore, radius of Xenon atom is 219 pm.

Hope this is useful...