Question

Calculate the theoretical density of iron, and then determine the number of vacancies per cm3 needed for a BCC iron crystal to have a density of 7.874 g/cm3 . The lattice parameter of iron is 2.866 * 10-8 cm.

Answer 1

Answer:

Therefore the theoretical density of iron is 7.877 g/cm³ .

Therefore the number of vacancy per cm³ is $3.27 \times 10^{20}$

Explanation:

BCC structure contains 2 atoms per cell.

BCC : Body centered cubic.

Atomic mass of iron = 55.845 gram/mole.

Atomic mass of a chemical element is the mass of 1 mole of the chemical element.

Density: Density of a matter is the ratio of mass of the matter and volume of the matter.

Avogrdro Number is number of atoms per 1 mole.

Avogrdro Number= 6.023×10²³

Lattice parameter of iron = 2.866×10⁻⁸ cm

The theoretical density:

$\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}$

 $=\frac{2\times 55.845 }{(6.023\times 10^{23})\times (2.866\times 10^{-8})^3}$   g/cm³      [x= 2,Since it is BCC structure]

=7.877 g/cm³

Therefore the theoretical density of iron is 7.877 g/cm³ .

Now we have to find out the number of unit cell of iron crystal having density  7.874 g/cm³.

$\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}$

$\Rightarrow 7.874 =\frac{x\times 55.845}{6.023\times 10^{23}\times (2.866\times 10^{-8})^3}$

$\Rightarrow x= \frac{7.874\times 6.023\times 10^{23}\times (2.866\times 10^{-8})^3}{55.845}$

⇒ x =1.9923 atom/cell

Therefore the vacancy of atom per cell = (2- 1.9923)=0.0077

$Vacancy\ per\ cm^3=\frac{\textrm{Vacancy per cm} ^3 }{\textrm{( lattice parameter)} ^3}$

                          $=\frac{0.0077}{(2.866\times 10^-8)^3}$

                           $=3.27 \times 10^{20}$

Therefore the number of vacancy per cm³ is $3.27 \times 10^{20}$