PROPERTIES OF MATERIALS IN ELECTRICAL ENGINEERING MIME262 MID-TERM EXAM

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1 MID-TERM EXAM, Oct 16 th, 008 DURATION: 50 mins Version CLOSED BOOK McGill ID Name All students must have one of the following types of calculators: CASIO fx-115, CASIO fx-991, CASIO fx-570ms SHARP EL-50, or the SHARP EL-546. NON-REGULATION CALCULATORS WILL BE REMOVED AND NO REPLACEMENT CALCULATOR WILL BE PROVIDED. Leave your calculator cover in your book-bag. You may take only your calculator, a ruler, a pen and your student ID to your desk. NO equation sheets. Rulers are subject to inspection. McGill University Dept. of Mining and Materials Engineering PROPERTIES OF MATERIALS IN ELECTRICAL ENGINEERING MIME6 MID-TERM EXAM Examiner: Prof. R.R. Chromik Date: Friday, October 16 th, 009 Time: 50 mins Instructions 1. Read the instructions carefully.. Put your name and ID# on this page. Put your NAME on pages -7 and ii. (Worth 1 pt) 3. Read each question completely before working on each part, (the parts may be interrelated). 4. Write your final solutions, showing your work, in the space provided below the relevant question. 5. Any work you wish to be ignored should be clearly indicated as such. 6. Draw any diagrams as large as possible in the space provided. 7. Label all diagrams, axes, curves etc. 8. For concept type questions, the space provided roughly indicates the level of detail expected. 9. Phase diagram(s), a periodic table, and other useful information are attached at the end of this examination book (pages i through iii). Official USE ONLY P1 P P3 P4 1

2 1. Short Answer Section (4 pts each) Follow the instructions below carefully. Overall, you will answer 7 of the 10 questions in this section. Answer any two of questions i iii. i. What is the lowest energy state of an atom? The lowest energy state of an atom (the ground state) occurs when we fill the electron orbitals starting from the lowest energy up until all of the electrons are used up. ii. Two atoms are brought together, what is the lowest energy state? The lowest energy state will be achieved through bonding of some type, where the atoms try to donate or share electrons to achieve a stable electron configuration (usually one similar to that of the noble gases). iii. Why do metals exhibit close-packed structures? This is primarily due to the fact that metallic bonding is omnidirectional. Answer any two of questions iv vi. iv. Sketch a (00) plane in the monoclinic unit cell below. Indicate your origin and label axes. O v. Sketch the [111] direction in the monoclinic unit cell below. Indicate your origin and label axes. O

3 vi. On the axes below, sketch and label bonding curves for material A and material B, where material B has a higher melting point and lower thermal expansion. Energy A Bonding distance B Answer any two of questions vii ix. vii. Hume-Rothery rule #1 states: Compare crystal structure If different, complete solubility is not possible. Why? If the crystal structures are different, at some point as we add element A to B, the crystal structure will have to change. For this to happen, we need to change from one phase to another. Thus, there needs to be a region of insolubility (remember 1 phase phase 1 phase, etc. as we go across a phase diagram). viii. Hume-Rothery rule # states: Calculate radii difference If less than 15%, some solubility is likely If greater than 15%, solubility will be limited. Why? If the atomic radii are too different, there will be too much strain induced by the substitutional atoms. Solubility will not be possible beyond some point. Also, if the difference is even larger, interstitial solutions would be preferred. ix. Hume-Rothery rule #3 states: Compare electronegativity If less than about 0.4, solubility is favored If greater than about 0.4, formation of new compounds is favored. Why? When electronegativity difference becomes great, ionic bonds are formed. This would favor new compounds. In terms of solubility, this maintains metallic bonds. Thus, if the electronegativity difference is too high, solubility will be reduced in favor compounds. 3

4 For the following question, please refer to the Pd-Ti phase diagram, included in the end-pages of the exam booklet. x. A lamellar microstructure is shown below, where the two phases are both solids. Provide a composition and temperature at which one might observe this microstructure. Also indicate what two phases would exist at the conditions you have chosen. You needed to find a eutectic or eutectoid point. Determine the eutect-ic or oid composition and provide a temperature anywhere below the eutect-ic or oid temperature and still in the -phase region. DO NOT WRITE BELOW THIS LINE ON THIS PAGE 4

5 . Pd-Ti Phase Diagram (4 pts) Magneto and Professor Xavier are preparing alloys of Pd-Ti for the device Cerebro. They consult you as a professional materials engineer to help them with the following. a) Identify any three of the five invariant points labeled on the diagram. Point Type of invariant point Shorthand Notation (if applicable) i Eutectic Pd 3 Ti Liquid βpdti ii Congruent melting of Pd 3 Ti N/A iii Peritectic Pd Ti Liquid iv Peritectoid PdTi Pd 5 Ti 3 αti v Eutectoid PdTi PdTi 3 βti αti b) Magneto and Prof. Xavier process an alloy with 80at% Ti at 100 K to equilibrium and then cool it very slowly. Complete the table below regarding the composition of this alloy at various temperatures. Show your work either on the diagram, or in the space below the table. Temperature (K) Name of at% Ti in mol fraction of alloy that Name of at% Ti in Phase 1 Phase 1 is Phase 1 Phase Phase 100 βti (i.e. 100%) n/a n/a n/a mol fraction of alloy that is Phase 1000 PdTi βti = PdTi 3 75 αti = 0.0 5

6 3. Radiotracer Diffusion ( pts) A film of Technetium isotope is applied to one side of a Technetium disc. The sample is raised to 900 C and the isotope begins to diffuse into the disc. After 1 hr the sample is quenched to room temperature. The concentration of the isotope at a depth of 10 microns into the disc is found to be 5.5x10-5 atom fraction. At an 80 micron depth, the isotope concentration is 1.3x10-6 atom fraction. What is the diffusion coefficient? T = 900 C = 1173 K, t = 1 hr = 3600 s. x 1 = 10x10-6 m, C 1 = 5.5x10-5 atom fraction, x = 80x10-6 m, C = 1.3x10-6 atom fraction C ϕ ( x, t) = exp D t 4D t x ln C = ln ϕ 1 D t x1 4D t ln C = ln ϕ D t x 4D t Combining leads to: = 1.x10-13 m /s Given an activation energy of diffusion of 1.14 ev, what is the pre-exponential factor, D o? D = D o exp (-Q / RT) D o = D exp (Q / RT) = 9.x10-9 m /s 6

7 4. Silicon (5 pts) Si has an atomic radius of nm and a diamond cubic structure (shown below) with lattice parameter, a = nm. a) In the structure shown above label each atom as a corner atom (C), face-centered atom (F) or entirely in the unit cell (I). b) Based on what you have determined in part (a), how many atoms are in the unit cell of Si? c) Calculate the APF for Si. Is this close packed? APF = # atoms * volume of atoms / volume of cell APF = 8 atoms/cell * (4/3 π (0.118x10-9 m) 3 ) / (0.543x10-9 m) 3 = 0.34 No this is not close packed. Close packed structures had APF = 0.74 d) Calculate the density of Si. ρ = = (8 atoms/cell * g/mol) / ((0.543x10-9 m) 3 * 6.0x10 3 atoms/mol) ρ =.33x10 6 g / m 3 =.33 g/cm 3 7

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9 Please hand in his page with your exam. Name: For any questions which require this phase diagram, assume that the dashed lines are true, experimentally confirmed phase boundaries. ii iii i v iv ii

10 Other possibly useful constants, equations and relations: The gas constant, R = J/mol/K. Boltzmann s constant, k b = 8.6x10-5 ev/k = 1.38x10-3 J/K. Avogadro s number is 6.0x10 3 e =.7188 and π = D = D o exp (-Q / RT) D = D o exp (-Q / k b T) ln (a*b) = ln (a) +ln (b) log b a = 1 / (log a b) N v = N exp (-Q v / RT) N v = N exp (-Q v / k b T) ln (a / b) = ln(a) ln(b) E A = A/r and E R = B/r n ϕ x ( ) C x, t = exp D t 4D t %ionic character = {1-exp[-(0.5)(X A X B ) ]} x 100 iii