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If $a$ be the edge length of the unit cell and $r$ be the radius of an atom then for fcc arrangement, the correct relation is
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Verified Answer
The correct answer is:
$4 r=\sqrt{2} a$
Given, the edge length of the unit cell $=a$ and, radius of the sphere $=r$
Consider $\triangle A B C$,
$\begin{aligned}
A C^2 & =A B^2+A C^2 \\
& =a^2+a^2 \\
& =2 a^2 \\
\therefore \quad A C & =\sqrt{2} \cdot a
\end{aligned}$
But, from the figure, $A C=4 r$
$\therefore \quad 4 r=\sqrt{2} \cdot a$
Consider $\triangle A B C$,
$\begin{aligned}
A C^2 & =A B^2+A C^2 \\
& =a^2+a^2 \\
& =2 a^2 \\
\therefore \quad A C & =\sqrt{2} \cdot a
\end{aligned}$
But, from the figure, $A C=4 r$
$\therefore \quad 4 r=\sqrt{2} \cdot a$
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