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)

a \sqrt{2}

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

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

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2 \sqrt{2}

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Solution

The correct option is B $\frac{a}{\sqrt{2}}$
For, FCC, the closet distance between two atoms is 2R.
Taking 'a' is the edge length
$(4 R)^{2} = a^{2} + a^{2}$
$\Rightarrow 16 R^{2} = 2 a^{2}$
$\Rightarrow R = \frac{a}{2 \sqrt{2}}$
$\Rightarrow 2 R = \frac{a}{\sqrt{2}}$

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