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A metal crystallizes in $fcc$ arrangement with edge length equal to 400 nm. Calculate maximum possible radius of the atom which can fit in the voids without disturbing the crystal structure?

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165.6 nm

71.71 nm

39 nm

58.55 nm

answer is D.

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Detailed Solution

For tetrahedral voids, $\frac{r}{R} = 0.225$

For octahedral voids, $\frac{r}{R} = 0.414$

For FCC,

$\begin{array}{l} R = \frac{a}{2 \sqrt{2}} = \frac{400}{2 \sqrt{2}} \\ \frac{r}{R} = 0.414 \\ \Rightarrow r = 0.414 \times \frac{400}{2 \sqrt{2}} = 58.55 pm \end{array}$

Diameter is 2r=2×58.55=117.1 pm

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