Question

The edge length of an fcc lattice is x times the atomic radius. Value of x is

Answer 1

Solution :

: In FCC unit cell, the atoms at face diagonal touch each other; thus diagonal is four times the atomic radius(r).

Diagonal of the square faces is also equal to × times of the edge length(a). Hence, 2–√a=4r.

Answer 2

The relation between edge length (a) and radius of atom (r) for FCC lattice is √(2a) = 4r .the value of x is $4r$

Explanation:

• A face-centered unit cell or fcc structure is the arrangement where an atom lies in the middle of each face of the cube.
• Here  one face of a face-centered cube is represented by the square
• When applying Pythagoras theorem we get
• $a^{2} +a^{2}$ =$(r+2r+r)^{2}$
• ⇒$2$$a^{2}$= $4r$
• √2a=4r
• The relation between edge length (a) and radius of atom (r) for FCC lattice is √(2a) = 4r .

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