A metal crystallises in a face-centred cubic | vedclass

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(D) In a face-centred cubic $(FCC)$ unit cell,the atoms touch along the face diagonal.Let the radius of the atoms be $r$ and the edge length be $a$.Th

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) In a face-centred cubic $(FCC)$ unit cell,the atoms touch along the face diagonal.Let the radius of the atoms be $r$ and the edge length be $a$.The face diagonal $d$ is given by $d = \sqrt{a^2 + a^2} = \sqrt{2} a$.Since the atoms touch along the face diagonal,$d = 4r$,where $r$ is the radius of the atom.Therefore,$4r = \sqrt{2} a$,which implies $r = \frac{\sqrt{2} a}{4} = \frac{a}{2\sqrt{2}}$.The closest approach between two atoms is the distance between their centers,which is $2r$.Thus,$2r = 2 	imes \left( \frac{a}{2\sqrt{2}} ight) = \frac{a}{\sqrt{2}}$.

$A$ 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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