Question

FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate (a) the density; and (b) the number of vacancies per gram of Pb.

Answer 1

Explanation:

(a)   Number of atoms present in FCC are 4 atoms. Hence, number of atoms present in the given cell are as follows.

       No. of atoms/cell = $4 \times \frac{499}{500}$

                                     = 4

Density will be calculated as follows.

          Density = $\frac{ZM}{Na^{3}}$

                = $\frac{4 \times 207}{6.023 \times 10^{23}} \times (0.4949 \times 10^{-7})^{3}$

                        = 11.34 $g/cm^{3}$

Therefore, density of the given substance is 11.34 $g/cm^{3}$.

(b)   Now, there is 1 vacancy present for every 500 lead (Pb) atoms. Therefore, 500 Pb atoms occupied by four lattice points is as follows.

           $\frac{500}{4}$

            = 125 unit cells

 $\frac{\frac{1}{125}}{(0.4949 \times 10^{-7})^{3}} \times \frac{1}{11.34 g/cm^{3}}$

             = $5.82 \times 10^{18}$

Thus, we can conclude that number of vacancies present per gram are $5.82 \times 10^{18}$.