Question

Copper crystallizes in a face centered cubic lattice. if the edge of the unit cell is 351 pm what is the radius of the copper atom

Answer 1

Density of unit cell is mathematically expressed as
D = $\frac{\text{Z X Atomic Weight}}{\text{Avagadro's Number X a^3}}$

where, Z = number of atoms/unit cell = 4 (For FCC structure)
Atomic weight of Cu = 63.5 g
a = edge length = 351 pm = 351 X 10^-10 cm
Avagadro's number = 6.023 X 10^23

∴ Density of unit cell = $\frac{4X63.5}{6.023X10^2^3X(351X10^-^1^0)^3}$
                                 = 9.752 g/cm3

Now, for FCC structure a = √8 r
where r = radius of Cu
∴ r = a/√8 = (351 X 10^-10)/√8 = 1.24 X 10^-8 cm = 124 pm