Xenon crystallises in the fcc lattice and the edge of the unit cell is 620 pm. The nearest neighbour distance of Xe is

Question

Xenon crystallises in the fcc lattice and the edge of the unit cell is 620 pm. The nearest neighbour distance of Xe is (A) 438.5 pm (B) 219.25 pm (C) 420 pm (D) 261.5 pm

Tardigrade Admin · Accepted Answer

As xenon (Xe) crystallises in fcc lattice, nearest neighbouring distance (d)=(a/√2)=(620/√2) =(620/1.414)=438.5 pm

Q. Xenon crystallises in the $f cc$ lattice and the edge of the unit cell is $620 p m$ . The nearest neighbour distance of $X e$ is

$438.5 p m$

$219.25 p m$

$420 p m$

$261.5 p m$

As xenon $(X e)$ crystallises in $f cc$ lattice, nearest neighbouring distance
$(d) = \frac{a}{2} = \frac{620}{2}$
$= \frac{620}{1.414} = 438.5 p m$

Q. Xenon crystallises in the fcc lattice and the edge of the unit cell is 620pm. The nearest neighbour distance of Xe is

438.5pm

219.25pm

420pm

261.5pm