question_answer

Atomic radius of fee is :

A) \[\frac{a}{2}\]                                   

B)  \[\frac{a}{2\sqrt{2}}\]

C) \[\frac{\sqrt{3}}{4}a\]                                   

D) \[\frac{\sqrt{3}}{2}a\]

Correct Answer: B

Solution :

                In a fee lattice, the lattice points are at corners plus one at the centre of each face of the cube. In fee lattice the spheres touch each other along a face diagonal. Since, atomic radius is the distance from the atomic nucleus to the outermost stable electron orbital in an atom that is at equilibrium. Hence,    \[ur={{(2)}^{1/2}}a\]                 \[r=\frac{\sqrt{2}a}{4}=\frac{a}{2\sqrt{2}}\]