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A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is:
A
204 pm
B
288 pm
C
408 pm
D
144 pm
Detailed Solution
In an fcc lattice the atoms touch along the face diagonal: $\sqrt2a = 4r$
Diameter $= 2r = \frac{\sqrt2a}{2} = \frac{a}{\sqrt2}$
$2r = \frac{408}{1.414} = 288$ pm
Diameter $= 2r = \frac{\sqrt2a}{2} = \frac{a}{\sqrt2}$
$2r = \frac{408}{1.414} = 288$ pm