In a face-centered cubic (FCC) unit cell (or | vedclass

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

Vedclass · Accepted Answer

$a = 2\sqrt{2}r$

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(B) In a face-centered cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.Let the edge length of the unit cell be $a$ and the radius of the atom be $r$.The length of the face diagonal is given by $\sqrt{2}a$.Since the atoms touch along the face diagonal,the length of the face diagonal is also equal to $4r$ (one radius from each corner atom and two radii from the face-centered atom).Therefore,$\sqrt{2}a = 4r$.Solving for $a$,we get $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$.

In a face-centered cubic $(FCC)$ unit cell (or cubic close packing $(CCP)$ unit cell),the edge length $a$ is equal to:

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