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Calculate the number of Frenkel defects per cubic meter in zinc oxide at 904˚C. The energy for defect formation is 2.51 eV, while the density for ZnO is 5.55 g/cm3 at this temperature. The atomic weights of zinc and oxygen are 65.41 g/mol and 16.00 g/mol, respectively.

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Answer:

[tex]1.02858429*10^{24}[/tex] defects/m³

Explanation:

Let's first determine the number of lattice sites per cubic meter; and which can be denoted as:

[tex]N = \frac{N_{A} \rho}{A_{zn}+A_o}[/tex]

[tex]N= \frac{6.022*10^{23}*5.55(10^6cm^3/m^3)}{(65.41/mol+16.00g/mol)}[/tex]

[tex]N = 4.11*10^{28}[/tex] lattice sites/m³

Now, to Calculate the number of Frenkel defects per cubic meter in zinc oxide at 904˚C, we have:

[tex]N_{fr}=Nexp(\frac{-Q_{fr}}{2KT})[/tex]

[tex]N_{fr}=4.11*10^{28}exp(\frac{-2.15eV}{2*8.62*10^{-5}eV*(904+273)K} )[/tex]

[tex]N_{fr}=4.11*10^{28}exp(-10.59558002)[/tex]

[tex]N_{fr}=4.11*10^{28}*2.50263818*10^{-5}[/tex]

[tex]N_{fr}=1.02858429*10^{24}[/tex] defects/m³

Hence, the  number of Frenkel defects per cubic meter in zinc oxide at 904˚C  = [tex]1.02858429*10^{24}[/tex] defects/m³

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