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 each other along the face diagonal.Let the radius of the atoms be $r$ and the edge lengt

Vedclass · Accepted Answer

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

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(D) In a face-centred cubic $(fcc)$ unit cell,the atoms touch each other along the face diagonal.Let the radius of the atoms be $r$ and the edge length of the unit cell be $a$.The face diagonal of the cube is given by $\sqrt{a^2 + a^2} = \sqrt{2} a$.Since the atoms touch along the face diagonal,the length of the face diagonal is equal to $4r$.Therefore,$4r = \sqrt{2} a$,which gives $r = \frac{\sqrt{2} a}{4} = \frac{a}{2\sqrt{2}}$.The closest approach between two atoms is the distance between their centers,which is $2r$.Thus,$2r = 2 	imes \frac{a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}$.

$A$ 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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