The atomic radius of an fcc unit cell is: | vedclass

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(B) In a face-centered cubic $(fcc)$ unit cell,the atoms touch each other along the face diagonal.
Let $a$ be the edge length of the unit cell and $r

Vedclass · Accepted Answer

$\frac{a}{2\sqrt{2}}$

Vedclass · Answer

(B) In a face-centered cubic $(fcc)$ unit cell,the atoms touch each other along the face diagonal.
Let $a$ be the edge length of the unit cell and $r$ be the atomic radius.
The length of the face diagonal is given by $\sqrt{a^2 + a^2} = a\sqrt{2}$.
Along this diagonal,there are four atomic radii (one radius from each corner atom and the full diameter of the face-centered atom),so $4r = a\sqrt{2}$.
Therefore,$r = \frac{a\sqrt{2}}{4} = \frac{a}{2\sqrt{2}}$.

The atomic radius of an $fcc$ unit cell is:

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