A metal crystallises in an FCC lattice with A | vedclass

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(C) In an $FCC$ lattice,the distance between nearest neighbours is $d = \frac{a_{edge}}{\sqrt{2}} = a$. Thus,the edge length of the unit cell is $a_{e

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

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(C) In an $FCC$ lattice,the distance between nearest neighbours is $d = \frac{a_{edge}}{\sqrt{2}} = a$. Thus,the edge length of the unit cell is $a_{edge} = a\sqrt{2}$.The triangular voids in $ABCABC...$ packing are the tetrahedral voids $(THV)$.The distance between the centres of the nearest triangular voids of two consecutive layers corresponds to the distance between two adjacent tetrahedral voids.In an $FCC$ unit cell,the distance between two adjacent tetrahedral voids is $\frac{a_{edge}}{2}$.Substituting $a_{edge} = a\sqrt{2}$,we get: $	ext{Distance} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.

$A$ metal crystallises in an $FCC$ lattice with $ABCABC...$ packing. If the distance between the nearest neighbours is '$a$',then the distance between the centres of the nearest triangular voids of two consecutive layers is:

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