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slega
2 months ago
10

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.
Chemistry
1 answer:
VMariaS [2.9K]2 months ago
5 0

Explanation:

(a)   In a Face-Centered Cubic (FCC) structure, there are 4 atoms present. Hence, the number of atoms in the specified cell is given by:

       Number of atoms/cell = 4 \times \frac{499}{500}

                                     = 4

The density can be computed 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}

Consequently, the density of the specified substance amounts to 11.34 g/cm^{3}.

(b)   There exists 1 vacancy for every 500 lead (Pb) atoms. As such, the number of Pb atoms occupying four lattice points can be calculated 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 arrive at the conclusion that the number of vacancies per gram is 5.82 \times 10^{18}.

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