Question

Calculate the equilibrium number of vacancies per cubic meter for copper at 1000K. The energy for vacancy formation is 0.9eV/atom; the atomic weight and density at 1000K for copper are 63.5g/mol and 8.45g/cm3, respectively.

Answer 1

Answer:

Therefore the equilibrium number of vacancies per unit cubic meter =2.34×10²⁴ vacancies/ mole

Explanation:

The equilibrium number of of vacancies is denoted by $N_v$.

It is depends on

• total no. of atomic number(N)
• energy required for vacancy
• Boltzmann's constant (k)= 8.62×10⁻⁵ev K⁻¹
• temperature (T).

$N_v=Ne^{-\frac{Q_v}{kT} }$

To find  equilibrium number of of vacancies we have find N.

$N=\frac{N_A\ \rho}{A_{cu}}$

Here ρ= 8.45 g/cm³  =8.45 ×10⁶m³

$N_A$= Avogadro Number = 6.023×10²³

$A_{Cu}$= 63.5 g/mole

$N=\frac{6.023\times 10^{23}\times 8.45\times 10^{6}}{63.5}$

   $=8.01\times 10^{28$ g/mole

Here $Q_v$=0.9 ev/atom , T= 1000k

Therefore the equilibrium number of vacancies per unit cubic meter,

$N_v=( 8.01\times 10^{28}) e^{-(\frac{0.9}{8.62\times10^{-5}\times 1000})$

   =2.34×10²⁴ vacancies/ mole

Answer 2

The equilibrium number of vacancies per cubic meter for copper at 1000K is equal to $2.34 \times 10^{24}\;vacancies/m^3$

Given the following data:

• Temperature = 1000 K.

• Energy = 0.9 eV/atom.

• Atomic weight of copper at 1000K = 63.5 g/mol.

• Density of copper at 1000K = 8.45 $g/cm^3$.

Scientific data:

• Avogadro's number = $6.02 \times 10^{23}$

• Boltzmann's constant, k = $8.62 \times 10^{-5}\;eV/K$

How to calculate the equilibrium number of vacancies.

First of all, we would determine the number of atomic sites per cubic meter (N) for copper by using this formula:

$N=\frac{N_A\rho}{A_{cu}} \\\\N=\frac{6.02 \times 10^{23} \times 8.45 \times 10^6}{63.5}$

N = $8.0 \times 10^{28}\;atoms/m^3$

Now, we can calculate the number of vacancies as follows:

$N_v = Nexp({-\frac{E}{kT} })\\\\N_v = 8.0 \times 10^{28}\times e(-\frac{0.9}{8.62 \times 10^{-5} \times 1000}) \\\\N_v = 8.0 \times 10^{28}\times e(-10.441)\\\\N_v = 8.0 \times 10^{28}\times 2.92 \times 10^{-5}\\\\N_v =2.34 \times 10^{24}\;vacancies/m^3$

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