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In the face-centred cubic 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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The correct answer is:
$\frac{a}{\sqrt{2}}$
Hint:
Closest distance between two gold atoms in fcc lattice of gold is $\frac{a}{\sqrt{2}} .$ This is the distance between the corner atom and the closest face centre atom.
Closest distance between two gold atoms in fcc lattice of gold is $\frac{a}{\sqrt{2}} .$ This is the distance between the corner atom and the closest face centre atom.
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