Answer
$Q_{v} = 1.40 eV/atom $
Work Step by Step
Given:
Nickel (Ni) where the equilibrium number of vacancies at 1123 K is $4.7 \times 10^{22} m^{-3}$
Atomic weight - 58.69 g/mol
density - 8.80 g/$cm^{3}$
Required:
energy for vacancy formation
Solution:
Using Equation 4.2:
$N = \frac{N_{A}p_{Ni}}{A_{Ni}} = {\frac{(6.022 \times 10^{23} atoms/mol)(8.80 g/cm^{3})}{58.69 g/mol}} = 9.03 \times 10^{22} atoms/cm^{3} = 9.03 \times 10^{28} atoms/m^{3} $
Using Equation 4.1 and taking its natural logarithm:
$ ln N_{v} = ln N (-\frac{Q_{v}}{kT}) $
$Q_{v} = -kT ln(\frac{N_{v}}{N}) = -(8.62 \times 10^{-5} eV/atom-K)(1123 K)ln [\frac{4.7 \times 10^{22} m^{-3}}{9.03 \times 10^{28}}] = 1.40 eV/atom $
