Question

Calculate the number of Frenkel defects per cubic meter in zinc oxide at 935˚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.

Answer 1

Answer : The number of Frenkel defects per cubic meter in zinc oxide is, $1.78\times 10^{23}$

Explanation :

First we have to calculate the number of lattice sites per cubic meter.

Formula used :

$N=\frac{N_A\rho}{A_{Zn}+A_O}$

where,

$N_A$ = Avogadro's constant = $6.022\times 10^{23}$

$A_{Zn}$ = atomic mass of zinc = 65.41 g/mol

$A_{O}$ = atomic mass of oxygen = 16 g/mol

$\rho$ = density of ZnO = $5.55g/cm^3$

Now put all the given values in the above formula, we get:

$N=\frac{(6.022\times 10^{23})\times (5.55)}{(65.41)+(16)}$

$N=4.105\times 10^{28}\text{Lattice site}/m^3$

Now we have to calculate the Frenkel defects per cubic meter.

Formula used :

$N_f=N\times \exp \left(\frac{-Q_f}{2kT} \right)$

where,

$Q_f$ = energy for defect formation = 2.51 eV

k = Boltzmann constant = $8.62\times 10^{-5}$

T = temperature = $935^oC=273+935=1208K$

Now put all the given values in the above formula, we get:

$N_f=(4.105\times 10^{28})\times \exp \left(\frac{-2.51}{2\times (8.62\times 10^{-5})\times (1208)} \right)$

$N_f=1.78\times 10^{23}$

Thus, the number of Frenkel defects per cubic meter in zinc oxide is, $1.78\times 10^{23}$