A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be:-

Options

A. `(a)/(sqrt(2))`
B. 2a
C. `2sqrt(2)`a
D. `sqrt(2)`a

Explanation

In FCC distance of closest approach between two atoms = 2r
In FCC, atoms are in close contact along face diagonal of FCC unit cell.
`therefore " " 4r = sqrt(2) a " " rArr 2r = (sqrt(2)a)/(2) = ((a)/(sqrt(2)))`