A metal crystallises in a face-centred cubic | vedclass

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(C) In a face-centred cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.Let the radius of each atom be '$r$'.The length of th

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

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(C) In a face-centred cubic $(FCC)$ unit cell,the atoms touch each other along the face diagonal.Let the radius of each atom be '$r$'.The length of the face diagonal is given by $\sqrt{2}a$.Since the atoms along the face diagonal touch each other,the length of the face diagonal is also equal to $4r$.Therefore,$4r = \sqrt{2}a$.The closest approach between two atoms is the distance between their centers,which is $2r$.From $4r = \sqrt{2}a$,we get $2r = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.

$A$ metal crystallises in a face-centred cubic $(FCC)$ structure. If the edge length of its unit cell is '$a$',the closest approach between two atoms in the metallic crystal will be:

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