Question

Calculate the energy (in eV/atom) for vacancy formation in some metal, M, given that the equilibrium number of vacancies at 296oC is 9.19 × 1023 m-3. The density and atomic weight (at 296°C) for this metal are 8.85 g/cm3 and 51.40 g/mol, respectively.

Answer 1

Explanation:

The given data is as follows.

       Temperature of metal = $296^{o}C$ = (296 + 273) K

                                            = 569 K

     Density of the metal = 8.85 $g/cm^{3}$ = $8.85 \times 10^{-6} g/m^{3}$      (as $1 cm^{3} = 10^{-6} m^{3}$)

     Atomic mass = 51.40 g/mol

    Vacancies = $9.19 \times 10^{23} m^{-3}$

Formula to calculate the number of atomic sites is as follows.

           n = $\frac{\rho \times N_{A}}{\text{atomic weight}}$

              = $\frac{8.85 \times 10^{-6} \times 6.022 \times 10^{23}}{51.40 g/mol}$

              = $1.036 \times 10^{17} atom/m^{3}$

Now, we will calculate the energy as follows.

                E = $-KT \times ln (\frac{\text{no. of vacancies}}{\text{no. of atomic sites}})$

where,    K = $8.62 \times 10^{-5}$

         E = $-8.62 \times 10^{-5} \times 569 K \times ln (\frac{9.19 \times 10^{23}}{1.036 \times 10^{17} atom/m^{3}})$

               = $78.46 eV/atom$

Therefore, we can conclude that energy (in eV/atom) for vacancy formation in given metal, M, is $78.46 eV/atom$.