Answer:
vacancy formation energy of Ni is 1.400 eV
Explanation:
given data
number of vacancies in Ni = 4.7 x [tex]10^{22}[/tex] [tex]m^{-3}[/tex]
atomic weight = 58.69 g/mol
density = 8.8 g/cm³
solution
we get here N that is
N = [tex]\frac{N_A \times \rho}{A}[/tex] ...........1
N = [tex]\frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}[/tex]
N = [tex]9.030 \times 10^{28}[/tex]
and here no of vacancy will be
Nv = [tex]N \times e^{\frac{-Qv}{kT}}[/tex] .................2
put here value
[tex]4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}[/tex]
[tex]10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}[/tex]
take ln both side
[tex]ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})[/tex]
[tex]-14.468 = \frac{-Qv}{0.0968}[/tex]
Qv = 1.400 eV
so vacancy formation energy of Ni is 1.400 eV
