question_answer

Atomic radius of fcc is                                            [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

Solution :

                   For the fcc structure \[4r={{({{a}^{2}}+{{a}^{2}})}^{1/2}}\]\[=a\sqrt{2}\] Þ \[r=\frac{a\sqrt{2}}{4}=\frac{a}{2\sqrt{2}}\]