In the face-centred cubic (FCC) lattice struc | vedclass

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(B) In a face-centred cubic $(FCC)$ lattice, the atoms touch each other along the face diagonal. The length of the face diagonal is $a \sqrt{2}$. Sinc

Vedclass · Accepted Answer

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

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(B) In a face-centred cubic $(FCC)$ lattice, the atoms touch each other along the face diagonal. The length of the face diagonal is $a \sqrt{2}$. Since the face diagonal consists of two radii of the corner atoms and one full diameter of the face-centred atom, the distance between the centres of two nearest atoms (the closest distance) is half of the face diagonal. Therefore, the closest distance $d = \frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.

In the face-centred cubic $(FCC)$ lattice structure of gold, the closest distance between gold atoms is ('a' being the edge length of the cubic unit cell).

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