A) 288 pm
B) 408 pm
C) 144 pm
D) 204 pm
Correct Answer: A
Solution :
[a] For fcc lattice, |
\[4r=\sqrt{2}a\] |
\[r=\frac{\sqrt{2}}{4}a=\frac{a}{2\sqrt{2}}\] |
\[=\frac{408}{2\sqrt{2}}\] |
\[=144\,pm\] |
diameter d = 2r = 2 x 144 pm |
= 288pm |
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