Question

Copper which crystallizes as a face – centred cubic lattice has a density of 8.93g/cm3 at 20o C. calculate the length of the unit cell.

Answer 1

Answer : The length of the unit cell is, 361.47 pm

Solution : Given,

Number of atom in unit cell of FCC (Z) = 4

Density = $8.93g/cm^3$

Atomic mass of copper(M) = 63.5 g/mole

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get

$8.93g/cm^3=\frac{4\times (63.5g/mol)}{(6.022\times 10^{23}mol^{-1}) \times a^3}$

$a=3.6147\times 10^{-8}cm=361.47pm$

$(1pm=10^{-10}cm)$

Therefore, the length of the unit cell is, 361.47 pm