Steady-state diffusion is diffusion in which the concentration of the diffusing atoms at

Chapter 7 What is steady state diffusion? Steady-state diffusion is diffusion in which the concentration of the diffusing atoms at any point, x, and hence the concentration gradient at x, in the solid, remains constant. Steady-state diffusion occurs when a gas at a constant pressure permeates through a metal foil or thin-walled tube. Hydrogen gas, for example, can be purified by allowing it to diffuse through a palladium thimble. The same steady state conditions can arise when oxygen diffuses through a plastic wrapping film covering food. Under steady-state conditions, the diffusion coefficient is obtained using Fick s first law. This is written: J i = -D i (dc i / dx) Where D i is the diffusion coefficient of atom type i, c i is the concentration of these atoms and x is the position in the solid. J i is called the flux of atoms of type i, that is, the net flow of these atoms through the solid. When the steady state has been reached, the diffusion coefficient across the foil, D, will be given by: D = J l / (c 1 c 2 ) where the concentrations on each side of the foil are c 1 and c 2 and the foil thickness is l. What are diffusion coefficients? Movement through the body of a solid is called volume, lattice or bulk diffusion. Atoms can also diffuse along surfaces and grain boundaries or along dislocations. Diffusion by way of these pathways is collectively referred to as short-circuit diffusion. Diffusion is studied by measuring the concentration of the atoms at different distances from the release point after a given time has elapsed. Raw experimental data thus consists of concentration and distance values. The speed at which atoms or ions

move is expressed in terms of a diffusion coefficient, D, which is obtained from the experimental data by use of two diffusion equations. In general, it has been found that D depends on position and concentration, although in many experiments to measure the diffusion coefficient it is assumed that the diffusion coefficient does not depend upon concentration or position and is constant. What is a correlation factor? When an atom is diffusing a given jump direction may be correlated with the direction of the previous jump. For example when a vacancy diffuses it can follow a truly random path. However, the diffusion of a tracer atom by the mechanism of vacancy diffusion is different because the tracer atom can only move when it is next to a vacancy, and the most probable next jump is back to the same place that it started from, as this is now occupied by a vacancy. When these processes are considered over many jumps, the mean square displacement of the tracer will be less than that of the vacancy, even though both have taken the same number of jumps. Therefore the observed diffusion coefficient of the tracer will be less than that of the vacancy. The same is true of some other diffusion mechanisms. The correlation factor, for any mechanism, is given by the ratio of the values of the mean square displacement of the atom (often the tracer) moving in a correlated motion to that of the atom (or vacancy) moving by a random walk process. If the number of jumps considered is large, the correlation factor, f, can be written as: f = <x²> c / <x²> r = D* / D r that is: D* = f D r where <x²> c represents the mean square displacement of a correlated walk by the diffusing atom and <x²> r the mean square displacement for a truly random diffusion process with the

same number of jumps, and D r and D* are the random walk and tracer diffusion coefficients. The correlation factor will then be a function of the diffusion mechanism and the crystal structure matrix in which the diffusion occurs. Quick quiz 1. b; 2. c; 3. a; 4. b; 5. c; 6. b; 7. b; 8. c; 9. c; 10. a; 11. b; 12. a; 13. c; 14. c; 15. c; 16. a; 17. a. Calculations and questions 3. 1.56 x 10-14 m 2 s -1. 4. 1.3 x 10-13 m 2 s -1. 5. ~2.0 x 10-10 m 2 s -1. 6. 0.29 wt %. 7. 3.77 hr. 8. 0.134 mm. 9. 4.19 x 10-6 kg. 10. 0.53 kg per hour. 11. 460 kj mol -1. 12. 1350 C. 13. 423 kj mol -1. 14. 1230 C. 15. 162 kj mol -1. 16. 289 kj mol -1. 17. 158 kj mol -1.

18. 3.29 x 10-14 m 2 s -1. 19. 2.10 x 10-9 m 2 s -1. 20. 4.64 x 10-16 m 2 s -1. 21. 1.95 x 10-18 m 2 s -1. 22. 1.12 x 10-12 m 2 s -1. 23. 1.9 m. 24. 250 s. 25. 2.63 x 10-9, (500 C); 6.15 x 10-6 (1000 C). 26. 1.86 x 10 4 C 2 J -1 m -3. 27. 1.39 x 10-14 m 2 s -1. 28. 1.09 x 10-14 m 2 s -1. Solutions 1 Show that the units of the diffusion coefficient are m 2 s -1. Fick s first law is: J = -D (dc / dx) Inserting the units: atoms m -2 s -1 = -D (atoms m -3 / m) Hence the units of D are m 2 s -1 2 Show that the two equations are equivalent a c c x s c c 0 0 1 erf 2 x Dt b cs cx c c s 0 erf 2 x Dt From (a) erf (z) = 1 [(c x c 0 ) / (c s c 0 )]

= [(c s c 0 ) - c x c 0 )] / (c s c 0 ) = (c s c x ) / (c s c 0 ) 3 Radioactive nickel-63 was coated onto a crystal of CoO and made into a diffusion couple. The sample was heated for 30 min at 953 C. The radioactivity perpendicular to the surface is given in the Table. Calculate the impurity tracer diffusion coefficient of nickel-63 in CoO. Activity / counts s -1 Distance / m 80 6 50 10 20 14 6 18 5 20 ln c x = constant - x 2 / 4 D* t Plot ln c x versus x 2 / 4 D* t; slope = -1 / 4 D* t Note that as activity is proportional to concentration, a plot of ln (activity) will not alter the slope. Activity / counts s -1 ln activity Distance / 10-6 m x 2 / m 2 80 4.38 6 3.6 x 10-11 50 3.91 10 1.0 x 10-10 20 3.00 14 1.96 x 10-10 6 1.79 18 3.24 x 10-10

5 1.61 20 4.0 x 10-10 From the graph: slope -4.8 / 5.4 x 10-10 = -1 / 4 D* t t = 30 min = 1800 s D* = 5.4 x 10-10 / 4 x 1800 x 4.8 = 1.56 x 10-14 m 2 s -1 4 Radioactive cobalt-60 was coated onto a crystal of CoO and made into a diffusion couple. The sample was heated for 30 min at 953 C. The radioactivity perpendicular to the surface is given in the Table. Calculate the tracer diffusion coefficient of cobalt-60 in CoO. Activity / counts s -1 Distance / m 110 10 70 20 39 30 23 40 9 50 ln c x = constant - x 2 / 4 D* t

Plot ln c x versus x 2 / 4 D* t; slope = -1 / 4 D* t Note that as activity is proportional to concentration, a plot of ln (activity) will not alter the slope. Activity / counts s -1 ln activity Distance / 10-6 m x 2 / m 2 110 4.70 10 1.0 x 10-10 70 4.25 20 4.0 x 10-10 39 3.66 30 9.0 x 10-10 23 3.14 40 1.6 x 10-9 9 2.20 50 2.5 x 10-9 From the graph: slope -2.5 / 24 x 10-10 = -1 / 4 D* t t = 30 min = 1800 s D* = 24 x 10-10 / 4 x 1800 x 2.5 = 1.33 x 10-13 m 2 s -1 5 Radioactive iron-59 was coated onto the (001) face of a single crystal of TiO 2 (rutile) (tetragonal) and made into a diffusion couple. The sample was heated for 300 s at 800 C. The radioactivity perpendicular to the surface is given in the Table. Calculate the impurity tracer diffusion coefficient of iron-59 parallel to the c-axis in rutile.

Activity / counts s -1 Distance / m x 10-4 520 3 400 4 270 5 185 6 130 6.5 90 7 53 8 ln c x = constant - x 2 / 4 D* t Plot ln c x versus x 2 / 4 D* t; slope = -1 / 4 D* t Note that as activity is proportional to concentration, a plot of ln (activity) will not alter the slope. Activity / counts s -1 ln activity Distance / m x 2 / m 2 520 6.25 300 9.0 x 10-8 400 5.99 400 1.6 x 10-7 270 5.60 500 2.5 x 10-7 185 5.22 600 3.6 x 10-7 130 4.87 650 4.23 x 10-7 90 4.50 700 4.9 x 10-7 53 3.97 800 6.4 x 10-7

From the graph: slope 1-3.6 / 8.8 x 10-7 = -1 / 4 D* t t = 300 s D* = 8.8 x 10-7 / 4 x 300 x 3.6 = 2.04 x 10-10 m 2 s -1 From the graph: slope 2-3.7 / 8.4 x 10-7 = -1 / 4 D* t t = 300 s D* = 8.4 x 10-7 / 4 x 300 x 3.7 = 1.89 x 10-10 m 2 s -1 The two values give an indication of the range of answer to be anticipated. A computer best fit would give a superior estimate. 6. Carbon-14 is diffused into pure -iron from a gas atmosphere of CO + CO 2. The gas pressures are arranged to give a constant surface concentration of 0.75 wt % C. The diffusivity of 14 C into -Fe is 9.5 x 10-11 m 2 s -1 at 827 C. Calculate the concentration of 14 C 1 mm below the surface after a heating time of 2 hours. Assuming that the C in the Fe can be ignored, so that c 0 = 0: c x = c s [1 - erf (x /2(Dt) ½ ] c x = 0.75 [1 erf (0.001 / (2 x {9.5 x 10-11 x 2 x 60 x 60} ½ )] = 0.75 [1 erf (0.0005 / 0.000827)] = 0.75 [1 erf (0.60456)]

The value of the error function must be obtained from a table or a computer routine. In this case: z = 0.60456; erf (z) = 0.607436 1 erf (0.60456) = 0.39256 c x = 0.75 x 0.39256 = 0.29 wt % C 7 Using the data in Question 6, how long would it take to make the carbon content 0.40 wt % at 1 mm below the surface? c x = c s [1 - erf (x /2(Dt) ½ ] 0.4 = 0.75 [1 erf (0.001 / (2 x {9.5 x 10-11 x t} ½ )] 0.4 / 0.75 = 1 erf (0.0005 / (9.5 x 10-11 x t} ½ ] 0.46667 = erf (0.0005 / (9.5 x 10-11 x t} ½ ] = erf (51.2989 / t ½ ) The value of the error function must be obtained from a table or a computer routine. In this case: erf (51.2989 / t ½ ) = erf (z) = 0.46667 z = (51.2989 / t ½ ) = 0.44048 t ½ = 51.2989 / 0.44048 = 116.46 s ½ t = 3,563 s = 226 min 8 An ingot of pure titanium metal is heated at 1000 C in an atmosphere of ammonia, so that nitrogen atoms diffuse into the bulk. The diffusivity of nitrogen in -Ti, the stable structure at 1000 C, is 5.51 x10-12 m 2 s -1 at this temperature. What is the thickness of the surface layer of titanium that contains a concentration of nitrogen atoms greater than 0.25 at % after heating for 1 hour?

c x = c s [1 - erf (x /2(Dt) ½ ] Suppose c s = 0.5 at % and c x is given as 0.25 at % 0.25 = 0.5 [1 erf (x / (2 x {5.51 x 10-12 x 1 x 60 x 60} ½ )] 0.5 = 1 erf (x / 0.000282) erf (x / 0.000282) = 0.5 The value of the error function must be obtained from a table or a computer routine. In this case: erf (x / 0.000282) = erf (z) = 0.5 z = x / 0.000282 = 0.4769 x = 0.134 mm Note: it is interesting to compare this with the penetration depth: x P = (Dt) ½ = {5.51 x 10-12 x 1 x 60 x 60} ½ = 0.141 mm 9 Zircalloy is a zirconium alloy used to clad nuclear fuel. How much oxygen will diffuse through each square metre of casing in a day under steady state diffusion conditions, at 1000 C, if the following apply: diffusivity of oxygen in zircalloy at 1000 C, 9.89 x 10-13 m 2 s - 1, concentration of oxygen on the inside of the container, 0.5 kg m -3, oxygen concentration on the outside of the container, 0.01 kg m -3, container thickness 1 cm. Using Fick s first law: J = -D (dc / dx) = -D (c 1 c 2 ) / (x 1 x 2 ) J = 9.89 x 10-13 x (0.5 0.01) / (1 x 10-2 ) = 4.85 x 10-11 kg m 2 s -1

Hence, in 1 day through each square metre: J = 4.85 x 10-11 x 24 x 60 x60 = 4.19 x 10-6 kg 10 Pure hydrogen is made by diffusion through a Pd-20%Ag alloy thimble. What is the mass of hydrogen prepared per hour if the total area of the thimbles used is 10 m 2, the thickness of each is 5 mm, and the operating temperature is 500 C? The equilibrium alloy in the surface of the thimble on the hydrogen rich side has a composition PdH 0.05, and on the further side, the hydrogen is swept away rapidly so that the surface is essentially pure metal. The diffusion coefficient of hydrogen in the alloy at 500 C is 1.3 x 10-8 m 2 s -1. Pd has the A1 (face-centred cubic) structure, with lattice parameter a = 0.389 nm. Each unit cell in the hydrogen-rich surface has an overall composition of 4 x (PhH0.05) because a unit cell of the A1 structure contains 4 atoms of Pd. Hence the surface concentration of H is: c = (4 x 0.05 x molar mass H) / a 3 x N A = (4 x 0.05 x 1.008 x 10-3 kg) / [(0.389 x 10-9) 3 x 6.02214 x 10 23 ] kg m -3 = 5.687 kg m -3 Using Fick s first law: J = -D (dc / dx) = -D (c 1 c 2 ) / (x 1 x 2 ) J = 1.3 x 10-8 x (5.687 0) / (5 x 10-3 ) = 1.48 x 10-5 kg m 2 s -1 Hence in 1 hr across 10 m 2 : J = 1.48 x 10-5 x 60 x 60 x 10 kg hr -1 = 0.53 kg hr -1

11 The radioactive tracer diffusion coefficient of silicon atoms in silicon single crystals is given in the Table. Estimate the activation energy for diffusion. T / C D* / m 2 s -1 1150 8.82 x 10-19 1200 3.40 x 10-18 1250 1.20 x 10-17 1300 3.90 x 10-17 1350 1.18 x 10-16 1400 3.35 x 10-16 Use the Arrhenius equation: D = D 0 exp [-E / RT] ln D - ln D 0 - E / RT A plot of ln D versus 1 / T has a slope E / R T / C T / K 1/T / K -1 D* / m 2 s -1 ln D* 1150 1423 7.03 x 10-4 8.82 x 10-19 -41.57 1200 1473 6.79 x 10-4 3.40 x 10-18 -40.22 1250 1523 6.57 x 10-4 1.20 x 10-17 -38.96 1300 1573 6.36 x 10-4 3.90 x 10-17 -37.78 1350 1623 6.16 x 10-4 1.18 x 10-16 -36.68 1400 1673 5.98 x 10-4 3.35 x 10-16 -35.63

from the graph, slope -5.5 / 0.0001 = -E / 8.31451 E 460 kj mol -1 A computer best fit would give a superior estimate. 12 Using the data in question 11, at what temperature, ( C) will the penetration depth be 2 m after 10 hours of heating? x P = (D*t) ½ (2 x 10-6 ) 2 = D* x 10 x 60 x 60 D* = (2 x 10-6 ) 2 / 10 x 60 x 60 = 1.11 x 10-16 m 2 s -1 Using the graph in the previous question: ln D* = -36.74, hence 1 / T 6.16 x 10-4 T 1623 K = 1350 C 13 The diffusion coefficient of carbon impurities in a silicon single crystal is given in the Table. Estimate the activation energy for diffusion. T / C D* / m 2 s -1 900 1.0 x 10-17 1000 3.0 x 10-16

1100 4.0 x 10-15 1200 5.0 x 10-14 1300 9.0 x 10-13 1400 5.0 x 10-12 Use the Arrhenius equation: D = D 0 exp [-E / RT] ln D - ln D 0 - E / RT A plot of ln D versus 1 / T has a slope E / R T / C T / K 1/T / K -1 D* / m 2 s -1 ln D* 900 1173 8.53 x 10-4 1 x 10-17 -39.14 1000 1273 7.86 x 10-4 3 x 10-16 -35.74 1100 1373 7.28 x 10-4 4 x 10-15 -33.15 1200 1473 6.79 x 10-4 5 x 10-14 -30.63 1300 1573 6.36 x 10-4 9 x 10-13 -27.74 1400 1673 5.98 x 10-4 5 x 10-12 -26.02

from the graph, slope -13.0 / 2.55 x 10-4 = -E / 8.31451 E 423 kj mol -1 A computer best fit would give a superior estimate. 14 Using the data in question 13, at what temperature, ( C) will the penetration depth be 10-4 m after 20 hours of heating? x P = (D*t) ½ (1 x 10-4 ) 2 = D* x 20 x 60 x 60 D* = (1 x 10-4 ) 2 / 20 x 60 x 60 = 1.39 x 10-13 m 2 s -1 Using the graph in the previous question: ln D* = -29.6, hence 1 / T 6.65 x 10-4 T 1524 K = 1251 C 15 The diffusion coefficient of radioactive Co 2+ tracers in a single crystal of cobalt oxide, CoO, is given in the Table. Estimate the activation energy for diffusion. T / C D* / m 2 s -1 1000 1.0 x 10-13 1100 3.5 x 10-13 1200 9.0 x 10-13 1300 2.0 x 10-12 1400 4.0 x 10-12 1500 8.0 x 10-12 1600 1.5 x 10-11

Use the Arrhenius equation: D = D 0 exp [-E / RT] ln D* - ln D 0 - E / RT A plot of ln D versus 1 / T has a slope E / R T / C T / K 1/T / K -1 D* / m 2 s -1 ln D* 1000 1273 7.86 x 10-4 1 x 10-13 -29.93 1100 1373 7.28 x 10-4 3.5 x 10-13 -28.68 1200 1473 6.79 x 10-4 9 x 10-13 -27.74 1300 1573 6.36 x 10-4 2 x 10-12 -26.94 1400 1673 5.98 x 10-4 4 x 10-12 -26.24 1500 1773 5.64 x 10-4 8 x 10-12 -25.55 1600 1873 5.34 x 10-4 1.5 x 10-11 -24.92 from the graph, slope -5.0 / 2.57 x 10-4 = -E / 8.31451 E 162 kj mol -1 A computer best fit would give a superior estimate.

16 The diffusion coefficient of radioactive Cr 3+ tracers in Cr 2 O 3 is given in the Table. Estimate the activation energy for diffusion. T / C D* / m 2 s -1 1050 1.0 x 10-15 1100 4.6 x 10-15 1200 1.05 x 10-14 1300 6.2 x 10-14 1400 2.7 x 10-13 1500 6.5 x 10-13 Use the Arrhenius equation: D = D 0 exp [-E / RT] ln D - ln D 0 - E / RT A plot of ln D versus 1 / T has a slope E / R T / C T / K 1/T / K -1 D* / m 2 s -1 ln D* 1050 1323 7.56 x 10-4 1 x 10-15 -34.54 1100 1373 7.28 x 10-4 4.6 x 10-15 -33.01 1200 1473 6.79 x 10-4 1.05 x 10-14 -32.19 1300 1573 6.36 x 10-4 6.2 x 10-14 -30.41 1400 1673 5.98 x 10-4 2.7 x 10-13 -28.94 1500 1773 5.64 x 10-4 6.5 x 10-13 -28.06 from the graph, slope -8.0 / 2.3 x 10-4 = -E / 8.31451 E 289 kj mol -1 A computer best fit would give a superior estimate.

17 The impurity diffusion coefficient of Fe 2+ impurities in a magnesium oxide single crystal is given in the Table. Estimate the activation energy for diffusion. T / C D* / m 2 s -1 1150 2.0 x 10-14 1200 3.2 x 10-14 1250 5.0 x 10-14 1300 7.5 x 10-14 1350 1.0 x 10-13 Use the Arrhenius equation: D = D 0 exp [-E / RT] ln D - ln D 0 - E / RT A plot of ln D versus 1 / T has a slope E / R T / C T / K 1/T / K -1 D* / m 2 s -1 ln D* 1150 1423 7.03 x 10-4 2 x 10-14 -31.54 1200 1473 6.79 x 10-4 3.2 x 10-14 -31.07 1250 1523 6.57 x 10-4 5 x 10-14 -30.63

1300 1573 6.36 x 10-4 7.5 x 10-14 -30.22 1350 1623 6.16 x 10-4 1 x 10-13 -29.93 from the graph, slope -1.9 / 1 x 10-4 = -E / 8.31451 E 158 kj mol -1 A computer best fit would give a superior estimate. 18 Calculate the diffusivity of 51 Cr in titanium metal at 1000 C. D 0 = 1 x 10-7 m 2 s -1, E = 158 kj mol -1. Use the Arrhenius equation: D = D 0 exp [-E / RT] = 1 x 10-7 exp [-158000 / (8.31451 x 1273) = 3.29 x 10-14 m 2 s -1 19 Calculate the diffusivity of 51 Cr in a titanium 18 wt % Cr alloy at 1000 C. D 0 = 9 x 10-2 m 2 s -1, E = 186 kj mol -1. Use the Arrhenius equation:

D = D 0 exp [-E / RT] = 9 x 10-2 exp [-186000 / (8.31451 x 1273) = 2.10 x 10-9 m 2 s -1 20 Calculate the diffusivity of 55 Fe in forsterite, Mg 2 SiO 4, at 1150 C. D 0 = 4.17 x 10-10 m 2 s -1, E = 162.2 kj mol -1. Use the Arrhenius equation: D = D 0 exp [-E / RT] = 4.17 x 10-10 exp [-162200 / (8.31451 x 1473) = 4.64 x 10-16 m 2 s -1 21 Calculate the diffusivity of 18 O in Co 2 SiO 4 at 1250 C. D 0 = 8.5 x 10-3 m 2 s -1, E = 456 kj mol -1. Use the Arrhenius equation: D = D 0 exp [-E / RT] = 8.5 x 10-3 exp [-456000 / (8.31451 x 1523) = 1.95 x 10-18 m 2 s -1 22 Calculate the diffusivity of Li in quartz, SiO 2, parallel to the c-axis, at 500 C. D 0 = 6.9 x 10-7 m 2 s -1, E = 85.7 kj mol -1. Use the Arrhenius equation: D = D 0 exp [-E / RT] = 6.9 x 10-7 exp [-85700 / (8.31451 x 773) = 1.12 x 10-12 m 2 s -1

23 The diffusion coefficient of Ni 2+ tracers in NiO is 1 x 10-15 m 2 s -1 at 1100 C. Estimate the penetration depth of the radioactive Ni 2+ ions into a crystal of NiO after heating for 1 hour at 1100 C. x P = (Dt) ½ = {1 x 10-15 x 1 x 60 x 60} ½ = 1.90 x 10-6 m = 1.90 m 24 Ge is diffused into silica glass for fibre optic light guides. How long should a fibre of 0.1 mm diameter be annealed at 1000 C to be sure that Ge has diffused into the centre of the fibre? The diffusion coefficient of Ge in SiO 2 glass is 1 x 10-11 m 2 s -1. Use the penetration depth to gain an idea of the time required. x P = (Dt) ½ (0.05 x 10-3 ) 2 = {1 x 10-11 x t} t = 250 s 25 What is the probability of a diffusing atom jumping from one site to another at 500 and 1000 C if the activation energy for diffusion is 127 kj mol -1. p = exp [-E / RT] At 500 C: p = exp [-127000 / (8.31451 x 773) = 2.63 x 10-9 At 500 C: p = exp [-127000 / (8.31451 x 1273) = 6.15 x 10-6 26 Estimate the ratio of the ionic conductivity to diffusion coefficient for a monovalent ion diffusing in an ionic solid. Take a typical value for the number of mobile diffusing ions as the number of vacancies present, approximately 10 22 defects m -3, and T as 1000K.

The ratio required, / D, is given by: / D = (n Z 2 e 2 ) / k B T For a monovalent ion with Z = 1, n = 1 x 10 22, T = 1000 K / D = [1 x 10 22 x (1.602 x 10-19 ) 2 ] / (1.38066 x 10-23 x 1000) = 1.86 x 10 4 C 2 J -1 m -3 27 The ionic conductivity of F - ions in the fast ionic conductor Pb 0.9 In 0.1 F 2.1 at 423 K is 1 x 10-4 -1 m -1. The cubic unit cell (fluorite type) has a cell parameter 0.625 nm, and there are on average, 0.4 mobile F- anions per unit cell. Estimate the diffusion coefficient of F - at 423 K. Use / D = (n Z 2 e 2 ) / k B T D = k B T / n Z 2 e 2 For F -, Z = -1 To calculate n, a unit cell has volume a 3 and contains 0.4 mobile F- ions, so: n = 0.4 / (0.625 x 10-9 ) 3 m -3 = 1.64 x 10 27 m -3 D = (1 x 10-4 x 1.38066 x 10-23 x 423) / [1.64 x 10 27 x (1.602 x 10-19 ) 2 ] = 1.39 x 10-14 m 2 s -1 28 The conductivity of SrO, (halite (B1) structure, a = 0.5160 nm), depends upon oxygen partial pressure. The value of the ionic conductivity is 2 x 10-3 -1 m -1 at 900 C under 0.1 atm O 2. Assuming the ionic conductivity is due to vacancy diffusion of Sr 2+ ions estimate the diffusion coefficient of Sr 2+ at 900 C.

D = k B T / n Z 2 e 2 For Sr 2+, Z = 2 To calculate n, a unit cell has volume a 3 and contains 4 mobile Sr 2+ ions, so: n = 4 / (0.5160 x 10-9 ) 3 m -3 = 2.91 x 10 28 m -3 D = (2 x 10-3 x 1.38066 x 10-23 x 1173) / [2.91 x 10 28 x 2 2 x (1.602 x 10-19 ) 2 ] = 1.09 x 10-14 m 2 s -1