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Q.
Copper has face-centered cubic $(fcc)$ lattice with interatomic spacing equal to $2.54 \,\mathring{A}$. The value of lattice constant for this lattice is
Interatomic spacing for a fcc lattice
$r=\left[\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{2}\right)^{2}+(0)^{2}\right]^{1 / 2}=\frac{a}{\sqrt{2}}$
a being lattice constant.
$a=\sqrt{2} r=\sqrt{2} \times 254=3.59 \,\mathring{A}$ NOTE : Interatomic spacing is just the nearest neighbours distance.