[J & K CET 2001]
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
For the fcc structure $ 4r={{(a^{2}+a^{2})}^{1/2}} $
$ =a\sqrt{2} $
Therefore $ r=\frac{a\sqrt{2}}{4}=\frac{a}{2\sqrt{2}} $
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