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A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is $408 \mathrm{pm}$. The diameter of the metal atom is
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Verified Answer
The correct answer is:
$288 \mathrm{pm}$
For fcc lattice,
$\begin{aligned}
& 4 r=\sqrt{2} a \\
& r=\frac{\sqrt{2}}{4} a=\frac{a}{2 \sqrt{2}} \\
& =\frac{408}{2 \sqrt{2}} \\
& =144 \mathrm{pm} \\
& \text {diameter } d=2 r=2 \times 144 \mathrm{pm} \\
& =288 \mathrm{pm}
\end{aligned}$
$\begin{aligned}
& 4 r=\sqrt{2} a \\
& r=\frac{\sqrt{2}}{4} a=\frac{a}{2 \sqrt{2}} \\
& =\frac{408}{2 \sqrt{2}} \\
& =144 \mathrm{pm} \\
& \text {diameter } d=2 r=2 \times 144 \mathrm{pm} \\
& =288 \mathrm{pm}
\end{aligned}$
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