Question

The activation energy (Q_v) vacancy in pure Ag is 1.762 x 10^-19 J/atom. The atomic weight and density for Ag are 107.870 g/mol and 10.5 g/cm³ respectively. Also given the Avogadro’s number is 6.022 x 10²³ atom/mol and Boltzmann’s constant is 1.38 x 10^-23 J/atom.K.

1.Calculate the value of N, the total number of atomic sites per cubic meter in Ag.

2.Calculate the equilibrium concentration of vacancies (N_v) per cubic meter in pure Ag at 750^oC

Answer 1

1) The equilibrium concentration of vacancies per cubic meter is given as:

$N_v = Ne^{\frac{-Q_v}{kT}}$

where $N= {\frac{N_A\rho_{(Ag)}}{A_{Ag}}}$

Given $\rho_{Ag} = 10.5 g/cm^3$

$N_A = 6.022\times 10^{23} atoms/mol$ , and K = Boltzmann’s constant = 1.38 x 10^-23J/atom.K.

Hence $N= \frac{6.022\times 10^{23}\times (10.5)}{107.87} = 5.86 \times 10^{22 } cm^{-3}$

$N= \frac{6.022\times 10^{23}\times (10.5)}{107.87} = 5.86 \times 10^{28 } m^{-3}$

2) Equilibrium concentration of vacancies (Nv):

$N_v = Ne^{\frac{-Q_v}{kT}}$

$N_v = (5.86\times 10^{28})e^{\frac{1.762\times 10^{-19}}{(1.38\times 10^{-23})(750+273)}}$

$N_v = 2.23\times 10^{23}m^{-3}$