Question

1. The equilibrium number of vacancies in Ni at 1123 K is 4.7x1022m-3. The atomic weight and density of Ni at 1123 K are 58.69 g/mol and 8.8 g/cm3. What is the vacancy formation energy of Ni?

Answer 1

Answer:

vacancy formation energy of Ni is 1.400 eV

Explanation:

given data

number of vacancies in Ni = 4.7 x $10^{22}$  $m^{-3}$

atomic weight = 58.69 g/mol

density = 8.8 g/cm³  

solution

we get here N that is

N  = $\frac{N_A \times \rho}{A}$   ...........1

N = $\frac{6.023\times 10^{23} \times 8.8 \times 10^6}{58.69}$

N = $9.030 \times 10^{28}$  

and here no of vacancy will be

Nv = $N \times e^{\frac{-Qv}{kT}}$  .................2

put here value

$4.7 \times 10^{22} = 9.030 \times 10^28 \times e^{\frac{-Qv}{8.62\times 10^{-5}\times 1123}}$  

$10^{-7} \times 5.20487 = e^{\frac{-Qv}{0.0968}}$

take ln both side

$ln (10^{-7} \times 5.20487 ) = ln (e^{\frac{-Qv}{0.0968}})$

$-14.468 = \frac{-Qv}{0.0968}$  

Qv = 1.400 eV

so vacancy formation energy of Ni is 1.400 eV