3.022 Microstructural Evolution in Materials

Transcription

1 3.022 Microstructural Evolution in Materials D EPARTMENT OF M ATERIALS S CIENCE AND E NGINEERING MASSACHUSETTS INSTITUTE OF TECHNOLOGY SPRING 2011, M.J. CIMA Exam 1 March 15 th, :30 to 9:00 PM Please read the exam through before starting. Clearly show all work in the space provided below each question. Feel free to use the additional pages at the end of this packet if necessary. Calculators and one double-sided 8 ½ x 11 page of notes are permitted. You must provide your solution on the pages of the exam and turn in your notes to get credit for the exam. It is suggested that you outline the solution to your problem wherever possible. Partial credit will be given for answers which follow a logical thought process. Good luck! Name Question 1 (out of 15) Question 2 (out of 25) Question 3 (out of 30) Question 4 (out of 30) Total (out of 100) 1

2 Problem 1 Vaksman et al doped ZnSe single crystals with Ni by diffusion. Metallic nickel was deposited on the surface of the crystal and then heat treated for five hours. The nickel dopant is colored, which means concentration can be determined by light absorption methods. The nickel impurity diffusion profile was determined by measuring the relative optical density of crystals in the visible spectral region. The data is shown below: Vaksman et al, Semiconductors, Vol. 44[2] (2010) Estimate the diffusivity of Ni in ZnSe at 1173 K. Also, estimate the activation energy for diffusion. Estimating diffusivity at 1173 K: x = 200 µm at T = 1173 K, t = 5 hrs. Use engineering approximation: x 2 /4Dt = 1 to find D: D = (200 µm) 2 /(4*5 hrs) = 2000 µm 2 /hr = 5.56 x 10-9 cm 2 /s. Estimating activation energy for diffusion: x = 130 µm at T = 1123 K, t = 5 hrs. Use engineering approximation: x 2 /4Dt = 1 to find D: D = (130 µm) 2 /(4*5 hrs) = 845 µm 2 /hr = 2.35 x 10-9 cm 2 /s. 2

3 Take the ratio of the two diffusivities D 1123 and D 1173, where D = Do exp (-Ea/RT), and R is the gas constant ( J/mol K). Do in both expressions cancel out, and Ea may be calculated: ln (D 1123 /D 1173 ) = ln (2.35/5.56) = Ea (1/1173R 1/1123R) = -4.6x10-6 *Ea Ea = activation energy = kj/mol 3

4 Problem 1 cont. 4

5 Problem 2 Titanium dioxide is an oxide semiconductor used in applications including photocatalysis and solar cells. Reducing conditions for titania (TiO 2 ) produces doubly ionized oxygen vacancies. Titania can be doped by chromia (Cr 2 O 3 ). Chromium incorporation into titanium also produces doubly charged vacancies in oxygen sites. The log conductivity dependence on log P(O 2 ) for low P(O 2 ) conditions is shown for undoped titania (TiO 2 ) and Cr 2 O 3 -doped TiO 2. Bak et al, J. Phys. Chem C, Vol (2008) a) Write the electroneutrality equation for undoped TiO 2 O O x = ½O 2 (g) + V O + 2eʹ b) At low P(O 2 ), what is the Brouwer approximation for undoped TiO 2? 2[V O ] = n c) How does the log conductivity vary with log P(O 2 ) for undoped TiO 2? K R = n 2 [V O ]P(O 2 ) 1/2 K R = ½*n 3 P(O 2 ) 1/2 n 3 is proportional to P(O 2 ) -1/2 n is proportional to P(O 2 ) -1/6 5

6 log conductivity is proportional to log n, which in turn is proportional to -1/6. This is slope a. d) Write the electroneutrality equation for Cr 2 O 3 -doped TiO 2 Cr 2 O 3 into TiO 2 = 2Cr Tiʹ + 3O O x + V O e) At low P(O 2 ), what is the Brouwer approximation for Cr 2 O 3 -doped TiO 2? Assume that Cr dopant concentration is much greater than the concentration of conduction electrons. [Cr Tiʹ] >> n So the Brouwer approximation is [Cr Tiʹ] = 2[V O ] f) How does the log conductivity vary with log P(O 2 ) for doped TiO 2? K R = n 2 [V O ]P(O 2 ) 1/2 K R = n 2* ½[Cr Tiʹ]P(O 2 ) 1/2 [Cr Tiʹ] only depends on the amount of dopant, not on P(O 2 ) n 2 is proportional to P(O 2 ) -1/2 n is proportional to P(O 2 ) -1/4 log conductivity is proportional to log n, which in turn is proportional to -1/4. This is slope b. 6

7 Problem 2 cont. 7

8 Problem 2 cont. 8

9 Problem 3 Consider the interdiffusion between Ni and Mo (V. D. Divya et al. / ScriptaMaterialia 62 (2010) ). Investigators prepared diffusion couples composed of sheets of pure Ni and Mo with a small amount of titanium dioxide particles at the interface between the two sheets. The particles served as markers for the position of the initial interface. Shown below are the Ni-Mo phase diagram and microstructural observations of the interface region after heat treatment. a) Why is there no composition gradient in the Mo phase? Limited solubility of Ni in Mo at 1150ºC b) What do you think the relative magnitude of the Ni and Mo diffusivities are in the Ni and σ phases? Same order of magnitude no Kirkendall effect is observed; the markers and interface still coincide after 9 hours of annealing. 9

10 c) How do you think the position of the σ-mo interface will change with time? With respect to K, will move away (to the right) as t 1/2. 10

11 Problem 3 cont 11

12 Problem 4 Au-Pt diffusion couples were studied by Bolk (ActaMetallurgica Vol. 9 July 1961 p632). The phase diagram is shown below as well as a micrograph of a diffusion couple cross section after heat treatment at 1000ºC for 100hr. The initial interface between Au and Pt sheets was decorated with silica fiber, shown as small circles in the cross section. a) Sketch the composition versus position graph corresponding to the situation in figure 7 b. Label point c in your graph. The composition graph will have a discontinuity (vertical line) with two different compositions as its end points, at point/interface c. b) What is the composition at point c? The moving interface, c, has a discontinuity in composition: one composition is 12%Au/88%Pt and the other composition is 68%Au/32%Pt. c) The distance between points c and d is observed to vary with time according to the equation, where t is the time. What is the value of n? n = 1/2 d) Which species has the higher diffusivity? Au 12

13 Shown below is a lower magnification of the entire diffusion couple. Porosity is clearly noted. Also shown is a graph of the pore density as a function of position from point c in one particular couple. Why are these pores present? Kirkendall effect; Au diffusivity is higher than Pt diffusivity. 13

14 Problem 4 cont. 14

15 Cont. 15