An element has a face-centred cubic (fcc) str | vedclass

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Question

(A) In an $fcc$ lattice,there are $8$ tetrahedral voids,each located on the body diagonals at a distance of $\frac{1}{4}$ of the body diagonal length

Vedclass · Accepted Answer

$\frac{a}{2}$

Vedclass · Answer

(A) In an $fcc$ lattice,there are $8$ tetrahedral voids,each located on the body diagonals at a distance of $\frac{1}{4}$ of the body diagonal length from each corner.Each body diagonal has a length of $\sqrt{3}a$.The distance of each tetrahedral void from the nearest corner is $\frac{\sqrt{3}a}{4}$.There are two tetrahedral voids on each body diagonal.The distance between these two nearest tetrahedral voids is $\frac{\sqrt{3}a}{4} + \frac{\sqrt{3}a}{4} = \frac{\sqrt{3}a}{2}$ is incorrect; let us re-evaluate.The distance between two nearest tetrahedral voids is the distance between the two voids located on the same body diagonal,which is $\frac{\sqrt{3}a}{2}$ is not correct. Actually,the distance between two nearest tetrahedral voids is $\frac{a}{2}$.

An element has a face-centred cubic $(fcc)$ structure with a cell edge of $a$. The distance between the centres of two nearest tetrahedral voids in the lattice is

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