Question

(a) Calculate the activation energy, QV, for vacancy formation in silver, given that the equilibrium number of vacancies at 1073°K is 3.6 x 10²³ m^-3. The atomic weight and density (at 1073°K) of silver are, respectively, 107.9 g/mol and 9.5 g/cm³ .

(b) Suppose that CaO is added as an impurity to Li₂O. If the Ca2+ substitutes for Li+ , what kind of vacancies would you expect to form? How many of these vacancies are created for every Ca2+ added?

Answer 1

activation energy, QV

can be obtained from the equation

Qv = -kT x ln(Nv/N)

Where Nv is the number of vacancies and N = Total number of atomic sites

Total number of atomic sites/m3 can be calculated as

N = (N_A x )/A

where N_A = Avagadro number, = density and A = atomic weight

N = (6.022 x 10²³ atoms/mol)x(9.5 g / cm3) /107.9 g/mol = 5.30 × 10²² atoms/cm³ = 5.30 × 10²⁸ atoms/m³

Qv = -kT x ln(Nv/N)

Qv = - (8.62 x10^-5 eV/atom- K)(1073 K) ln (3.6 x 10²³ atoms m³/5.30 × 10²⁸ atoms/m³)

Qv = 1.10 eV/atom

b)  For Ca²⁺ substituting for Li⁺ in Li₂O, lithium vacancies would be created. For each Ca²⁺substituting for Li⁺ one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca²⁺ ion added, one lithium vacancy is formed.