The equilibrium concentration of Schottky defects in metal o

The equilibrium concentration of Schottky defects in metal oxide having a formula MO is 5.7 times 10^9/m^3 at 750 degree C and 5.8 times 10^17/m^3 at 1500 degree C. The density of die oxide at these temperatures is found to be 3.5 g/cm^3 at 750 degree C and 3.4 g/cm^3 at 1500 degree C. Determine the activation energy for defect formation. What would the equilibrium concentration of Shottky defects at 1000 degree C when the oxide has a density of 3.45 g/cm^3?

Solution

T(C) ...................... density(p) (g/cm3) ..................................Ns (m-3)

750 ....................... 3.5 ....................5.7 x 10⁹

1000 ..................... 3.45 ................................................. ?

1500 ....................... 3.4 ................................................5.8 x 10¹⁷

This problem provides for some oxide ceramic, at temperatures of 750°C and 1500°C, values for density and the number of Schottky defects per cubic meter.

(a)

first combine a modified form of Equation 1 and Equation 2 as

Ns = N exp (-Qs / 2KT) ...................1

= (N_A p / A_M + A_O) exp(-Qs / 2KT).......2

In as much as this is a hypothetical oxide material, we don\'t know the atomic weight of metal M, nor the value of Qs in the above equation. Therefore, let us write equations of the above form for two temperatures, T1 and T2 . These are as follows:

Ns1 =   (N_A p1 / A_M + A_O) exp(-Qs / 2KT1)...................3

Ns2 = (N_A p2 / A_M + A_O) exp(-Qs / 2KT2).........4

Dividing the first of these equations by the second leads to

Ns1 / Ns2 =   (N_A p1 / A_M + A_O) exp(-Qs / 2KT1) / (N_A p2 / A_M + A_O) exp(-Qs / 2KT2) ..................5

which, after some algebraic manipulation, reduces to the form

Ns1 / Ns2 = p1 / p2 exp [ - Qs / 2K (1/T1 - 1/T2) ] ...........6

Now, taking natural logarithms of both sides of this equation gives

ln ( Ns1 / Ns2) = ln (p1/p2) - Qs/2K (1/T1 - 1/T2) ............7

and solving for Qs leads to the expression

-Qs = -2K [ ln(Ns1/Ns2) - ln(p1/p2) ] / (1/T1 -1/T2) ............8

Let us take T1 = 750°C and T2 = 1500°C, and we may compute the value of Qs as

Qs = [ -2(8.62x10^-5 eV/K) [ ln (5.7 x10⁹ / 5.8 x10¹⁷) - ln(3.5 / 3.4 ) ] / [1/ (750 + 273) - 1/(1500 +273) ]

Qs = 7.7 ev

(b) It is now possible to solve for Ns at 1000°C using Equation 6 above. This time let\'s take T1 = 1000°C and T2 = 750°C. Thus, solving for Ns1, substituting values provided in the problem statement and Qs determined above yields

Ns1 = (Ns2 x p1 / p2) exp [ -Qs / 2K (1/T1 - 1/T2) ].............9

= (5.7 x 10⁹ (3.45) / 3.5) exp [ - 7.7 / (2x 8.62 x10^-5 ) ) ( 1/1000+273 - 1/750 +273) ]

= 2.97 x 10¹³ /m³