Question

Calculate the energy (in eV/atom) for vacancy formation in some metal, M, given that the equilibrium number of vacancies at 476oC is 3.61E+23 m-3. The density and atomic weight for this metal are 20.1 g/cm3 and 92.40 g/mol, respectively.

Answer 1

N, the total number of atomic sites. However, N is related to the density, (ρM), Avogadro's number (NA), and the atomic weight (A_m)

N = N_aρM / A_m = (6.022 * 10^23 atoms/mol)(20.1 g / cm3) / 92.40 g /mol = 1.30 X 10^23 atom/cm³ = 1.30* 10^29

ln N = ln N - Qv/ kT

3.61E+23 m-3 = K

T = 476+273 = 749k

Q = - kT ln(N_v/N) = ((8.62*10-5 eV/atom- K)* 749k ln( 3.61X 10^23/6.022 X10^29) = ln(5.99X 10^-7 )*8.62*10^-5*749 =

            energy for vacancy formation = 14.3 *8.62*10^-5*749 = 0.92eV