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, \[ur={{(2)}^{1/2}}a\] \[r=\frac{\sqrt{2}a}{4}=\frac{a}{2\sqrt{2}}\]You need to login to perform this action.
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