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(C) For an $fcc$ unit cell,the relationship between the edge length $a$ and the atomic radius $r$ is given by $4r = \sqrt{2}a$.The distance between tw

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$2.87$

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(C) For an $fcc$ unit cell,the relationship between the edge length $a$ and the atomic radius $r$ is given by $4r = \sqrt{2}a$.The distance between two nearest atoms in an $fcc$ lattice is equal to $2r$.From the relation $4r = \sqrt{2}a$,we get $2r = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}}$.Given $a = 4.07 \ \mathring{A}$,the distance $d = \frac{4.07}{1.414} \approx 2.878 \ \mathring{A}$.Thus,the distance is approximately $2.87 \ \mathring{A}$.

$Au$ crystallizes in an $fcc$ lattice. If the edge length of the unit cell is $4.07 \ \mathring{A}$,then the distance between two nearest $Au$ atoms is ............... $\mathring{A}$.

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