Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
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Chapter 4, Problem 4.20P
To determine

The activation enthalpy and pre-exponential constant for the diffusion of oxygen (O18) into ZnO.

Expert Solution & Answer
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Answer to Problem 4.20P

The activation enthalpy is 3.83×105J/moleofOatoms and the pre exponential constant is 0.40×103m2/s.

Explanation of Solution

Formula used:

The diffusion coefficient of point 1 is given by,

D0ZnO1=D0ZnO0exp(ΔH D 0ZnO kT1)   ....... (I)

Here, D0Zn01 is the diffusion coefficient of point 1, D0ZnO is the pre-exponential constant for oxygen diffusion into ZnO, k is the Boltzmann’s constant, T1 is the temperature at point 1 and ΔHD0ZnO is the activation enthalpy for oxygen into ZnO.

The diffusion coefficient of point 2 is given by,

D0ZnO2=D0ZnO0exp(ΔH D 0ZnO kT2)   ....... (II)

Here, D0Zn02 is the diffusion coefficient of point 1 and T2 is the temperature at point 2.

The ratio of diffusion coefficient of points 1 and 2 is given by,

D 0 ZnO 1 D 0 ZnO 2 =D 0 ZnO 0 exp( Δ H D 0ZnO k T 1 )D 0 ZnO 0 exp( Δ H D 0ZnO k T 2 )=exp( Δ H D 0ZnO k T 1 )exp( Δ H D 0ZnO k T 2 )=exp[( Δ H D 0ZnO k)( 1 T 1 1 T 2 )]   ....... (III)

Calculation:

The activation enthalpy is calculated as,

Substitute 1×1012cm2/s for D0ZnO1, 1×1018cm2/s for D0ZnO2, 8.62×105eV/atomK for k, 1587K for T1 and 1075K for T2 in equation (III).

1× 10 12 cm 2/s1× 10 18 cm 2/s=exp[( Δ H D 0ZnO 8.62× 10 5 eV/ atom K)( 1 1587K 1 1075K)]106=exp[ΔH D 0ZnO(3.48 atom/ eV)]ln( 106)=ΔHD 0ZnO(3.48atom/eV)ΔHD 0ZnO=ln( 10 6 )3.48atom/eV

Solve further,

ΔHD 0ZnO=13.823.48atom/eV=(3.97eV/atom× 96.521× 10 3 J/ moleofOatoms 1 eV/ atom )=3.83×105J/moleofOatoms

The diffusion coefficient is calculated as,

Substitute 1×1012cm2/s for D0ZnO1, 8.62×105eV/atomK for k, 1587K for T1 and 3.97eV/atom for ΔHD0ZnO in equation (I).

1×1012cm2/s=D0 ZnO0exp( 3.97 eV/ atom ( 8.62× 10 5 eV/ atom K )( 1587K ))D0 ZnO0=( 1× 10 12 cm 2 /s × 10 4 m 2 1 cm 2 )exp( 29.02)=1× 10 16 m 2/s2.32× 10 13=0.40×103m2/s

Conclusion:

Therefore, the activation enthalpy is 3.83×105J/moleofOatoms and the pre exponential constant is 0.40×103m2/s.

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