Calculate the number of Frenkel defects per cubic meter in silver chloride at \(350^{\circ} \mathrm{C}\). The energy for defect formation is \(1.1 \mathrm{eV}\), while the density for AgCl is \(5.50 \mathrm{g} / \mathrm{cm}^{3}\) at \(\left(350^{\circ} \mathrm{C}\right)\).

Short Answer

Expert verified
Based on the given information and calculations, there are approximately \(1.70 \times 10^{18}\) Frenkel defects per cubic meter in silver chloride (AgCl) at \(350^{\circ} \mathrm{C}\).

Step by step solution

01

Convert the temperature to Kelvin

To convert the temperature from Celsius to Kelvin, we simply add 273.15. Therefore, the temperature in Kelvin can be calculated as: $$ T = 350 + 273.15 = 623.15 \mathrm{K} $$
02

Calculate the molar mass of AgCl

To find the molar mass of AgCl, we need to add the atomic mass of silver (Ag) and chlorine (Cl). The atomic masses are: - Ag: \(107.87 \mathrm{g/mol}\) - Cl: \(35.45 \mathrm{g/mol}\) So, the molar mass of AgCl is: $$ M_{AgCl} = 107.87 + 35.45 = 143.32 \mathrm{g/mol} $$
03

Calculate the total number of atomic sites in the crystal lattice (\(n_0\))

The density of AgCl (\(5.50 \mathrm{g} / \mathrm{cm}^{3}\)) can be used to find the total number of atomic sites in the lattice. We can convert the density to the number of moles per cubic centimeter: $$ \frac{5.50 \mathrm{g}}{\mathrm{cm}^{3}} \times \frac{1 \mathrm{mol}}{143.32 \mathrm{g}} = \frac{0.0383 \mathrm{mol}}{\mathrm{cm}^{3}} $$ Since there are two atoms (Ag and Cl) per unit cell, the total number of atomic sites (\(n_0\)) will be double the number of moles per unit volume: $$ n_0 = 2 \times \frac{0.0383 \mathrm{mol}}{\mathrm{cm}^{3}} \times \frac{6.022 \times 10^{23} \mathrm{atoms}}{\mathrm{mol}} = 4.61 \times 10^{22} \mathrm{atoms / cm^3} $$
04

Calculate the defect concentration (\(n\))

Now we have all the necessary information to calculate the defect concentration using the formula: $$ n = n_0 \exp{\left(-\frac{E}{kT}\right)} $$ Plugging in the values, we get: $$ n = 4.61 \times 10^{22} \mathrm{atoms / cm^3} \times \exp{\left(-\frac{1.1 \mathrm{eV}}{8.617 \times 10^{-5} \mathrm{eV/K} \times 623.15 \mathrm{K}}\right)} $$ $$ n = 4.61 \times 10^{22} \mathrm{atoms / cm^3} \times \exp{(-20.92)} $$ $$ n = 1.70 \times 10^{6} \mathrm{atoms / cm^3} $$
05

Convert the defect concentration to defects per cubic meter

Finally, we need to convert the defect concentration from atoms per cubic centimeter to atoms per cubic meter by using the conversion factor: $$ 1 \mathrm{cm^3} = 10^{-6} \mathrm{m^3} $$ So, the defect concentration in defects per cubic meter is: $$ n = 1.70 \times 10^{6} \mathrm{atoms / cm^3} \times \frac{1 \mathrm{m^3}}{10^{-6} \mathrm{cm^3}} = 1.70 \times 10^{18} \mathrm{defects / m^3} $$ Thus, there are approximately \(1.70 \times 10^{18}\) Frenkel defects per cubic meter in silver chloride at \(350^{\circ} \mathrm{C}\).

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