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(B) In a face-centred cubic $(FCC)$ unit cell,the atoms touch along the face diagonal.The length of the face diagonal is $\sqrt{2} a$,where $a$ is the

Vedclass · Accepted Answer

$\frac{4}{\sqrt{2}} r$

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(B) In a face-centred cubic $(FCC)$ unit cell,the atoms touch along the face diagonal.The length of the face diagonal is $\sqrt{2} a$,where $a$ is the edge length.Since the face diagonal consists of four atomic radii $(r + 2r + r = 4r)$,we have the relation: $4r = \sqrt{2} a$.Rearranging for the edge length $a$,we get: $a = \frac{4r}{\sqrt{2}}$.

In a face-centred cubic $(FCC)$ unit cell,the edge length $(a)$ in terms of atomic radius $(r)$ is:

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