Question

Calculate the number of vacancies per cubic meter in iron at 850 °C. The energy for vacancy formation is 1.08 eV/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm3 and 55.85 g/mol, respectively.

Answer 1

Answer:

The number of vacancies per cubic meter is 1.18 X 10²⁴ m⁻³

Explanation:

$N_v = N*e[^{-\frac{Q_v}{KT}}] = \frac{N_A*\rho _F_e}{A_F_e}e[^-\frac{Q_v}{KT}}]$

where;

N$_A$ is the number of atoms in iron = 6.022 X 10²³ atoms/mol

ρFe is the density of iron = 7.65 g/cm3

AFe is the atomic weight of iron = 55.85 g/mol

Qv is the energy vacancy formation = 1.08 eV/atom

K is Boltzmann constant = 8.62 X 10⁻⁶ k⁻¹

T is the temperature = 850 °C = 1123 k

Substituting these values in the above equation, gives

$N_v = \frac{6.022 X 10^{23}*7.65}{55.85}e[^-\frac{1.08}{8.62 X10^{-5}*1123}}]\\\\N_v = 8.2486X10^{22}*e^{(-11.1567)}\\\\N_v = 8.2486X10^{22}*1.4279 X 10^{-5}\\\\N_v = 1.18 X 10^{18}cm^{-3} = 1.18 X 10^{24}m^{-3}$

Therefore, the number of vacancies per cubic meter is 1.18 X 10²⁴ m⁻³

Answer 2

The number of vacancies will be "1.18 × 10²⁴ m⁻³".

Vacancy formation

According to the question,

Number of atoms, $N_A$ = 6.022 × 10²³ atoms/mol

Iron's density, ρFe = 7.65 g/cm³

Iron's atomic weight, AFe = 55.85 g/mol

Energy vacancy formation, Qv = 1.08 eV/atom

Boltzmann constant, K = 8.62 × 10⁻⁶ k⁻¹

Temperature, T = 850°C or, 1123 K

We know the formula,

→ $N_v$ = N × e [$-\frac{Qv}{KT}$]

        = $\frac{N_A\times \rho Fe}{AFe}$ e [$-\frac{Qv}{KT}$]

By substituting the above values, we get

        = $\frac{6.022\times 10^{23}\times 7.65}{55.85}$ e [$- \frac{1.08}{8.62\times 10^{-5}\times 1123}$]

        = 8.2486 × 10²² × $e^{(-11.1567)}$

        = 8.2486 × 10²² × 1.4279 × 10⁻⁵

        = 1.18 × 10¹⁸ cm⁻³ or,

        = 1.18 × 10²⁴ m⁻³

Thus the answer above is correct.

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