TRINITY COLLEGE ENGINEERING AND COMPUTER SCIENCE SPRING 1996 232L ENGINEERING MATERIALS INSTRUCTOR: E. GUTIERREZ-MIRAVETE ernesto@hgc.edu (860) 548-2464 MWF 12:00-12:50 P.M. COURSE OUTLINE REV. 1/16/95 -------------- WEEK DATES TOPIC READING ---- ----- ----- ------- 1. 1/17,19 INTRODUCTION CH. 1 2. 1/22,24,26 ATOMIC STRUCTURE OF MATERIALS CH. 2 3. 1/29,31, 2/2 CRYSTAL STRUCTURE OF MATERIALS CH. 3 4. 2/5,7,9 CRYSTAL IMPERFECTIONS AND ATOMIC MOVEMENTS IN MATERIALS CH. 4 5. 2/12,14,16 PHASE DIAGRAMS CH. 8 (EXAM 1) 6. 2/19,21,23 READING WEEK CH. 9 7. 2/26,28, 3/1 ELECTRICAL PROPERTIES CH. 5 8. 3/4,6,8 MECHANICAL PROPERTIES CH. 6 9. 3/11,13,15 POLYMERIC MATERIALS CH. 7 10. 3/18,20,22 CERAMIC MATERIALS CH. 10 11. 3/25,27,29 SPRING VACATION 12. 4/1,3,5 MAGNETIC PROPERTIES CH. 11 (EXAM 2) 13. 4/8,10,12 OPTICAL PROPERTIES CH. 14 14. 4/15,17,19 CORROSION CH. 12 15. 4/22,24,26 COMPOSITE MATERIALS CH. 13 (EXAM 3) 16. 5/4-11 FINAL EXAM TRINITY COLLEGE ENGINEERING AND COMPUTER SCIENCE SPRING 1996 232L ENGINEERING MATERIALS INSTRUCTOR: E. GUTIERREZ-MIRAVETE ernesto@hgc.edu (860) 548-2464 MWF 12:00-12:50 P.M. COURSE POLICY ------------- OBJECTIVE OF THE COURSE: ------------------------ TO DEVELOP FAMILIARITY WITH THE NATURE AND CHARACTERISTICS OF ENGINEERING MATERIALS. COURSE MATERIAL: ---------------- A BROAD COVERAGE OF THE SCIENCE AND ENGINEERING OF MATERIALS. EMPHASIS ON THE STRUCTURE-PROPERTIES PARADIGM. THE COURSE COVERS ATOMIC, CRYSTAL AND MICROSTRUCTURE OF MATERIALS; ELECTRICAL, MAGNETIC, OPTICAL, MECHANICAL AND CHEMICAL PROPERTIES AND THEIR ORIGINS. READING ASSIGNMENTS: -------------------- MAINLY FROM THE TEXT AS INDICATED IN THE COURSE OUTLINE. READINGS MUST BE PERFORMED PRIOR TO THE DAY IN WHICH THE TOPIC WILL BE DISCUSSED IN CLASS. WRITTEN ASSIGNMENTS: -------------------- HOMEWORK WILL BE ASSIGNED REGULARLY. ANSWERS TO HOMEWORK QUESTIONS WILL BE SUBMITTED BY STUDENTS FOR GRADING VIA E-MAIL. EXAMS: ----- EXAM 1: ATOMIC STRUCTURE, CRYSTAL STRUCTURE, MICROSTRUCTURE, ATOM MOVEMENTS, AND PHASE DIAGRAMS EXAM 2: ELECTRICAL AND MECHANICAL PROPERTIES, METALS, POLYMERS, CERAMICS EXAM 3: MAGNETIC AND OPTICAL PROPERTIES, CORROSION, COMPOSITES FINAL EXAM : ALL TOPICS LABORATORY: ---------- THIS CLASS INCLUDES LABORATORY SESSIONS. SEE ATTACHED SCHEDULE. GRADING: -------- EXAM 1: 20% EXAM 2: 20% EXAM 3: 20% FINAL EXAM: 20% HOMEWORK: 20% THE GRADE OBTAINED IN THE LABORATORY WILL ALSO CONTRIBUTE TO THE FINAL GRADE. DETAILS WILL BE ANNOUNCED SOON. OFFICE HOURS: ------------- OPEN BUT BY APPOINTMENT ONLY. CALL ME OR EVEN BETTER, USE E-MAIL TO ARRANGE MEETINGS. TRINITY COLLEGE ENGINEERING AND COMPUTER SCIENCE SPRING 1996 232L ENGINEERING MATERIALS INSTRUCTOR: E. GUTIERREZ-MIRAVETE ernesto@hgc.edu TEXT: WILLIAM F. SMITH PRINCIPLES OF MATERIALS SCIENCE AND ENGINEERING, 3RD EDITION MCGRAW-HILL PUBLISHING CO., NEW YORK, 1995, ISBN 059 241-1 TRINITY COLLEGE ENGINEERING AND COMPUTER SCIENCE SPRING 1996 232L ENGINEERING MATERIALS HOMEWORK ASSIGNMENTS -------------------- 1.- SELECT 20 OBJECTS OR THINGS AROUND YOU. IDENTIFY THE MATERIAL(S) WHICH MAKE UP EACH OBJECT AS METAL, PLASTIC, CERAMIC OR COMPOSITE. BRIEFLY EXPLAIN THE REASON(S) WHY EACH OBJECT IS MADE UP OF THE MATERIAL IT IS MADE UP. Out: 1/17/96 ; Due: 1/22/96 2.- CONSIDER BONDING IN THE NaCl MOLECULE. a) WRITE DOWN AN EXPRESSION FOR THE TOTAL BONDING FORCE F IN THE FORM F = Fa + Fr = C1/x^a + C2/x^b (Note: x^a MEANS "x TO THE POWER a"; C1 AND C2 ARE CONSTANTS) FEEL FREE TO USE MAPLE OR SIMILAR TO PLOT AND TABULATE YOUR RESULTS. b) COMPUTE THE EQUILIBRIUM INTERATOMIC SPACING. c) SUBMIT YOUR RESULTS IN THE FORM OF A TABLE OF INTERATOMIC FORCES F, Fa AND Fr AGAINST INTERATOMIC SEPARATION. Out: 1/26/96 : Due: 1/29/96 3.- CONSIDER THE ASSEMBLY OF COPPER ATOMS, ONE AT A TIME, TO FORM SOLID COPPER. A FIRST CLOSE-PACKED PLANE OF ATOMS (A) IS FORMED BY ARRANGING INDIVIDUAL ATOMS IN AN HEXAGONAL PATTERN. A SECOND PLANE (B), IDENTICAL TO (A) IS THEN PLACED WITH THE ATOMS OF (B) LYING IN THE VALLEYS LEFT BY THE ATOMS IN (A). A THIRD PLANE (C) IS THEN PLACED ON TOP OF PLANE (B). ATOMS IN PLANE (C) ARE PLACED ON THOSE VALLEYS LEFT BY PLANE (B) ATOMS WHICH ARE NOT DIRECTLY ABOVE ATOM SITES IN PLANE (A). THE THREE LAYER PACKING IN THIS CASE CAN BE DESCRIBED AS ABCABCABC ... a) FIND THE FACE CENTERED UNIT CELL OF COPPER BY EXAMINING THE THREE LAYER PACKING OF ATOMIC PLANES (A), (B), (C). HINT: SIX OF THE 14 ATOMS OF THE FCC UNIT CELL ARE IN LAYER (B) b) COMPUTE THE VOLUME DENSITY OF ATOMS IN COPPER. c) HOW MANY UNIT CELLS OF COPPER ARE THERE IN A CUBE 1 CM IN SIDE? d) COMPUTE THE AREAL DENSITY OF ATOMS ON THE FOLLOWING ATOMIC PLANES: (111) (110) (100) e) COMPUTE THE LINEAL DENSITY OF ATOMS ON THE FOLLOWING LATTICE DIRECTIONS: [111] [110] [100] f) REPEAT THE COMPUTATIONS IN (b)-(e) ABOVE BUT THIS TIME FOR VANADIUM WHICH HAS A BODY CENTERED CUBIC (BCC) UNIT CELL STRUCTURE. g) OF THE THREE PLANES AND DIRECTIONS IN (d) AND (e) WHICH ONES HAVE THE HIGHEST ATOMIC DENSITY IN THE CASE OF COPPER (FCC)? h) OF THE THREE PLANES AND DIRECTIONS IN (d) AND (e) WHICH ONES HAVE THE HIGHEST ATOMIC DENSITY IN THE CASE OF VANADIUM (BCC)? i) TITANIUM HAS AN HEXAGONAL CLOSE PACKED (HCP) UNIT CELL. CONSIDER CLOSE PACKING OF Ti ATOMS TO FORM A CLOSED PACKED PLANE OF ATOMS (A). PLACE A SECOND CLOSE PACKED PLANE (B) ON THE VALLEYS LEFT BY ATOMS IN (A). FINALLY, PLACE A THIRD CLOSE PACKED PLANE OF ATOMS (C) ON TOP OF PLANE (B) BUT IN SUCH A WAY THAT ATOMS IN (C) ARE DIRECTLY ABOVE ATOMS IN THE FIRST PLANE (A). THE THREE-LAYER PACKING CAN THEN BE DESCRIBED AS ABABABAB ... FIND THE HEXAGONAL UNIT CELL IN THIS ARRANGEMENT. Out: 2/1/96 ; Due: 2/7/96 4.- ANSWER THE FOLLOWING QUESTIONS CONCERNING CRYSTAL DEFECTS. a) AT WHAT TEMPERATURE WILL THERE BE 1 EMPTY LATTICE SITE FOR EVERY 100,000 SITES IN A SAMPLE OF COPPER? b) WHAT IS THE FRACTION OF VACANT SITES IN A SAMPLE OF SOLID COPPER ONE DEGREE BELOW THE MELTING POINT ( Tm = 1336 K) c) WHILE COPPER ALLOWS ZINC ATOMS TO EASILY BECOME SUBSTITUTIONAL IMPURITIES IN THE LATTICE, IT DOES NOT ALLOW LEAD ATOMS DO THE SAME. EXPLAIN WHY. d) SKETCH THE ARRANGEMENT OF ATOMS ON A (100) PLANE OF A SIMPLE CUBIC LATTICE CONTAINING AN EDGE DISLOCATION WITH BURGERS VECTOR [010] e) AN ELECTRON MICROSCOPE STUDY OF A PIECE OF GOLD SHOWS DISLOCATION LINES RUNNING ALL PARALLEL TO EACH OTHER AND UNIFORMLY SPACED BY A SPACING OF 10^(-6) cm. WHAT IS THE DENSITY OF DISLOCATIONS (i.e. NUMBER OF DISLOCATION LINES CROSSING AN AREA OF 1 cm^2)? f) CONSIDER AN ATOM OF COPPER ON A (111) PLANE. HOW MANY NEAREST NEIGHBORS DOES THE ATOM HAVE? NOW, IF THE (111) PLANE IS THE OUTSIDE SURFACE OF THE SAMPLE, HOW MANY NEAREST NEIGHBORS DOES THE ATOM HAVE? g) CALCULATE THE NUMBER OF GRAINS IN THE TOP SURFACE OF THE HEAD OF A PIN PINS ARE MADE OF STEEL WITH ASTM GRAIN SIZE OF 6. Out: 2/8/96 ; Due: 2/14/96 5.- ATOM MOVEMENTS a) USING THE "DIFFUSION DISTANCE" x = 2 (D t)^0.5 , CALCULATE THE DISTANCE TRAVELLED BY IMPURITY ATOMS AFTER 1000 SEC AT 1000 K, FOR THE FOLLOWING SYSTEMS: IMPURITY MATRIX CARBON IRON (FCC) MANGANESE IRON (FCC) ALUMINUM SILICON ARSENIC SILICON EXPLAIN THE DIFFERENCES ENCOUNTERED IN YOUR RESULTS. 6.- PHASE DIAGRAMS WHAT IS THE MAIN DIFFERENCE IN THE MELTING/FREEZING BEHAVIOR BETWEEN PURE COPPER AND A 50%COPPER-50%NICKEL ALLOY? 7.- ELECTRICAL PROPERTIES CONSIDER THIN WIRES (RADIUS = 0.0005 m ; LENGTH = 1 m ), MADE OF THE FOLLOWING MATERIALS AND AT ROOM TEMPERATURE: PURE COPPER PURE SILICON SILICON DOPED WITH 10^21 ATOMS OF PHOSPHOROUS/m^3 a) CALCULATE THE RESISTIVITY OF EACH WIRE b) IF A 1 V BATTERY IS CONNECTED TO THE ENDS OF EACH WIRE, CALCULATE THE CURRENT DENSITY IN EACH CASE c) HOW MANY CHARGED PARTICLES CROSS THE CROSS SECTION OF WIRE PER SECOND IN EACH CASE? 8.- MECHANICAL PROPERTIES CONSIDER THE FOLLOWING MECHANICAL PROPERTY DATA 304 STAINLESS STEEL MODULUS OF ELASTICITY E = 200 GPa = 200*10^9 (Pa) TRUE STRESS-TRUE STRAIN CURVE sigma = 1,450*10^6*(epsilon)^0.6 (Pa) PURE COPPER MODULUS OF ELASTICITY E = 130 GPa = 130*10^9 (Pa) TRUE STRESS-TRUE STRAIN CURVE sigma = 450*10^6*(epsilon)^0.33 (Pa) a) TABULATE AND DRAW THE COMPLETE TRUE STRESS-TRUE STRAIN CURVES FOR BOTH MATERIALS FOR VALUES OF epsilon BETWEEN 0 AND 1 b) ESTIMATE THE FORCE REQUIRED TO STRETCH THIN WIRES OF THE TWO MATERIALS IN THE FOLLOWING MANNER INITIAL WIRE RADIUS = 0.0005 m INITIAL WIRE LENGTH = 0.1 m FINAL WIRE LENGTH = 0.2 m c) HOW LARGE ARE THE COMPUTED FORCES COMPARED, SAY, WITH YOUR OWN WEIGHT? d) CONSIDERING THE VALUES OF ULTIMATE TENSILE STRENGTH GIVEN BELOW, WHAT WOULD BE THE MAXIMUM LENGTH OF THE STRETCHED WIRES JUST BEFORE NECKING AND THE TRUE STRESS AT THAT POINT (TRUE STRESS AT MAXIMUM LOAD)? e) CONSIDERING THE VALUES PERCENT ELONGATION (TO FRACTURE) GIVEN BELOW, WHAT WOULD BE THE MAXIMUM POSSSIBLE STRETCHED LENGTH OF EACH WIRE AND THE ENGINEERING STRESS AT THAT POINT (ENG FRACTURE STRESS, s_f)? f) FROM YOUR RESULTS IN (e), WHAT WOULD BE THE TRUE STRAIN TO FRACTURE (epsilon_f) AND THE TRUE FRACTURE STRESS (sigma_f) FOR EACH OF THE TWO WIRES? (Hint: THIS IS A TRICKY QUESTION SINCE IT CAN'T BE ANSWERED WITH NUMBERS USING THE DATA GIVEN HERE; WHY?) ADDITIONAL MECHANICAL PROPERTY DATA: ULTIMATE TENSILE STRENGTH = MAXIMUM LOAD SUSTAINED BY MATERIAL/INITIAL CROSS SECTIONAL AREA = S_u S_u(Copper) = 216*10^6 Pa S_u(304SS) = 927*10^6 Pa PERCENT ELONGATION (TO FRACTURE) = e_f = [(LENGTH AT FRACTURE - INITIAL LENGTH)/INITIAL LENGTH]*100 e_f(Copper) = 48% e_f(304SS) = 13% 9.- CONSTRUCT COLUMNS RANKING THE FOLLOWING POLYMERS ON THE BASIS OF THEIR A) MELTING POINT (INCLUDE VALUE) B) MECHANICAL STRENGTH IN EACH CASE, WRITE THE CHEMICAL FORMULA OF THE POLYMER USING THE STICK NOTATION DISCUSSED IN CLASS AND ALSO ADD A SHORT NOTE EXPLAINING YOUR REASONS FOR SELECTING A SPECIFIC PLACE IN THE RANKING. POLYETHYLENE POLYVINYLCHLORIDE POLYPROPYLENE POLYSTYRENE POLYACRYLONITRILE POLYVINYL ACETATE POLIVINYLIDENE CHLORIDE POLYMETHYL METHACRYLATE POLYBUTADIENE ABS POLYTETRAFLUOROETHYLENE NYLON 6 POLYOXYMETHYLENE POLYBUTYLENE TEREPHTHALATE CROSS LINKED POLYESTHER 10.- ANSWER THE FOLLOWING QUESTIONS ABOUT CERAMIC MATERIALS A) COMPUTE THE PLANAR ATOMIC DENSITIES OF Mg AND O ATOMS IN THE (100) PLANES OF MgO . B) COMPUTE THE VOLUMETRIC ATOMIC DENSITY OF Ca and F ATOMS IN A CaF2 CRYSTAL C) COMPUTE THE VOLUMETRIC ATOMIC DENSITIES OF Ca, Ti AND O ATOMS IN A CaTiO3 CRYSTAL. D) COMPUTE THE NUMBER OF Si AND O ATOMS IN A UNIT CELL OF CRISTOBALITE. CALCULATE ALSO THE VOLUMETRIC DENSITY OF Si AND O ATOMS. E) IF MOLTEN SiO2 IS COOLED VERY, VERY SLOWLY, IT EXHIBITS A CLEAR FREEZING POINT AND THE SOLID PRODUCT IS CRISTOBALITE. ON THE OTHER HAND, IF THE COOLING IS RAPID, THE SOLID PRODUCT IS ORDINARY WINDOW GLASS. EXPLAIN THE THE REASON FOR THIS BEHAVIOR. 11.- MAGNETIC PROPERTIES A) USING THE B-H CURVE FOR ALNICO V, FIND THE MAXIMUM ENERGY PRODUCT (BH)max. B) THE MAGNETIC DOMAIN STRUCTURE OF A FERROMAGNET IN THE UNMAGNETIZED STATE (B = 0) CONSIST OF MANY DOMAINS. WHY ISN'T A FERROMAGNET IN SUCH CONDITION MAGNETIZED? C) WHAT HAPPENS TO THE DOMAIN STRUCTURE OF THE MATERIAL IN (B) ABOVE WHEN IT IS PLACED IN THE GAP OF A COIL THROUGH WHICH INCREASING CURRENT IS PASSED? 12.- OPTICAL PROPERTIES A) DESCRIBE THE BASIC ELEMENTS OF A FIBER-OPTICS COMMUNICATIONS SYSTEM. B) EXPLAIN THE FUNCTIONING OF THE DEVICES ENCOUNTERED IN THE TRANSMITTING AND RECEIVING ENDS OF THE SYSTEM. C) DESCRIBE THE KEY CHARACTERISTICS REQUIRED OF AN OPTICAL FIBER. 13.- CHEMICAL PROPERTIES 14.- COMPOSITES TRINITY COLLEGE ENGINEERING AND COMPUTER SCIENCE SPRING 1996 232L ENGINEERING MATERIALS KEY TO SOLUTIONS TO HOMEWORK ASSIGNMENTS ---------------------------------------- 1.- SELECT 20 OBJECTS OR THINGS AROUND YOU. IDENTIFY THE MATERIAL(S) WHICH MAKE UP EACH OBJECT AS METAL, PLASTIC, CERAMIC OR COMPOSITE. BRIEFLY EXPLAIN THE REASON(S) WHY EACH OBJECT IS MADE UP OF THE MATERIAL IT IS MADE UP. SODA CAN METAL (ALUMINUM) -> LEAKPROOF, EASE OF MANUFACTURE, CHEAP WRITING PAD PAPER (WOOD) -> ABSORBING OF INK OR CHARCOAL PC CASE FIBER REINFORCED PLASTIC -> STRONG, DURABLE, STAIN RESISTANT, CHEAP DESK METAL + PRESSED WOOD CHIPS -> STRENGTH, DURABILITY, CHEAP 2.- CONSIDER BONDING IN THE NaCl MOLECULE. a) WRITE DOWN AN EXPRESSION FOR THE TOTAL BONDING FORCE F IN THE FORM F = Fa + Fr = C1/x^a + C2/x^b ATTRACTION FORCE (EQ. 2.5, p. 40 TEXT) Fa = - Z1*Z2*e^2/(4*3.1416*(8.85*10^-12)*x^2) = 2.3*10^-28 /x^2 REPULSION FORCE (EQ 2.6, P. 41, TEXT, ALSO, P. 44) Fr = - n*b/x^(n+1) = - 7.73*10^-105 /x^10 SO F = 2.3*10^-28 /x^2 - 7.73*10^-105 /x^10 b) COMPUTE THE EQUILIBRIUM INTERATOMIC SPACING. EQUILIBRIUM SPACING (x = x_eq) WHEN F = 0 , I.E. 2.3*10^-28 /x_eq^2 = 7.73*10^-105 /x_eq^10 SOLVING FOR x_eq x_eq = 2.76*10^-10 m (MATCHES WITH RESULT ON P. 41, TEXT) c) SUBMIT YOUR RESULTS IN THE FORM OF A TABLE OF INTERATOMIC FORCES F, Fa AND Fr AGAINST INTERATOMIC SEPARATION. x Fa Fr F 2.5 x 10^-10 3.68 x 10^-9 8.105 x 10^-9 -4.425 x 10^-9 2.76 x 10^-10 3.019 x 10^-9 3.014 x 10^-9 ~0 3.0 x 10^-10 2.556 x 10^-9 1.309 x 10^-9 1.247 x 10^-9 3.5 x 10^-10 1.878 x 10^-9 2.802 x 10^-10 1.5979 x 10^-9 4.0 x 10^-10 1.438 x 10^-9 7.37 x 10^-11 1.431 x 10^-9 4.5 x 10^-10 1.136 x 10^-9 2.27 x 10^-11 1.134 x 10^-9 5.0 x 10^-10 0.92 x 10^-9 7.916 x 10^-12 0.92 x 10^-9 3.- CONSIDER THE ASSEMBLY OF COPPER ATOMS, ONE AT A TIME, TO FORM SOLID COPPER. A FIRST CLOSE-PACKED PLANE OF ATOMS (A) IS FORMED BY ARRANGING INDIVIDUAL ATOMS IN AN HEXAGONAL PATTERN. A SECOND PLANE (B), IDENTICAL TO (A) IS THEN PLACED WITH THE ATOMS OF (B) LYING IN THE VALLEYS LEFT BY THE ATOMS IN (A). A THIRD PLANE (C) IS THEN PLACED ON TOP OF PLANE (B). ATOMS IN PLANE (C) ARE PLACED ON THOSE VALLEYS LEFT BY PLANE (B) ATOMS WHICH ARE NOT DIRECTLY ABOVE ATOM SITES IN PLANE (A). THE THREE LAYER PACKING IN THIS CASE CAN BE DESCRIBED AS ABCABCABC ... a) FIND THE FACE CENTERED UNIT CELL OF COPPER BY EXAMINING THE THREE LAYER PACKING OF ATOMIC PLANES (A), (B), (C). HINT: SIX OF THE 14 ATOMS OF THE FCC UNIT CELL ARE IN LAYER (B) THE FCC UNIT CELL HAS ITS MAIN CUBE DIAGONAL NORMAL TO PLANES ABC. PLANE B CONTAINS SIX ATOMS OF THE UNIT CELL IN A TRIANGULAR PATERN PLANES A AND C CONTAIN THREE ATOMS EACH THE TWO REMAINING ATOMS LIE RESPECTIVELY, BELOW AND ABOVE PLANES A AND C b) COMPUTE THE VOLUME DENSITY OF ATOMS IN COPPER. THERE ARE 4 WHOLE Cu ATOMS PER UNIT CELL. THE MASS OF EACH ATOM IS 63.64/6.02*10^23 = 1.05*10^-22 g THE ATOMIC RADIUS IS R = 0.128 nm (TABLE 3.3, P. 78, TEXT) THE UNIT CELL EDGE IS a = 4*R/SQRT(2) = 0.3615*10^-7 cm (EQN 3.3, P. 77-78, TEXT) THE UNIT CELL VOLUME IS a^3 = 4.72*10^-23 cm^3 DENSITY = 4*1.05*10^-22 g / 4.72*10^-23 cm^3 = 8.89 g/cm^3 c) HOW MANY UNIT CELLS OF COPPER ARE THERE IN A CUBE 1 CM IN SIDE? # CELLS = 1 cm^3/4.72*10^-23 cm^3 = 2.11*10^22 d) COMPUTE THE AREAL DENSITY OF ATOMS ON THE FOLLOWING ATOMIC PLANES: (111) AREA = sqrt(2)*a*(sqrt(2)*a*cos(30))/2 = a^2*cos(30) = 0.1131 nm^2 # ATOMS ON PLANE = 2 PLANAR DENSITY = 2/0.1131 = 17.6 ATOMS/nm^2 (110) AREA = sqrt(2)*a*a = sqrt(2)*a^2 = 0.1848 nm^2 # ATOMS ON PLANE = 2 PLANAR DENSITY = 2/0.1848 = 10.8 ATOMS/nm^2 (100) AREA = a*a = a^2 = 0.1306 nm^2 # ATOMS ON PLANE = 2 PLANAR DENSITY = 2/0.1306 = 15.3 ATOMS/nm^2 e) COMPUTE THE LINEAL DENSITY OF ATOMS ON THE FOLLOWING LATTICE DIRECTIONS: [111] LENGTH = sqrt(24)*R = 0.627 nm # ATOMS ALONG LINE = 2 LINEAR DENSITY = 2/0.627 = 3.18 ATOMS/ nm [110] LENGTH = 4*R = 0.512 nm # ATOMS ALONG LINE = 2 LINEAR DENSITY = 2/0.512 = 3.90 ATOMS/ nm [100] LENGTH = a = 0.3615 nm # ATOMS ALONG LINE = 1 LINEAR DENSITY = 1/0.3615 = 2.76 ATOMS/ nm f) REPEAT THE COMPUTATIONS IN (b)-(e) ABOVE BUT THIS TIME FOR VANADIUM WHICH HAS A BODY CENTERED CUBIC (BCC) UNIT CELL STRUCTURE. THERE ARE 2 WHOLE V ATOMS PER UNIT CELL. THE MASS OF EACH ATOM IS 50.94/6.02*10^23 = 8.46*10-23 g THE ATOMIC RADIUS IS R = 0.132 nm (TABLE 3.2, P. 75, TEXT) THE UNIT CELL EDGE IS a = 4*R/SQRT(3) = 0.304*10^-7 cm (TABLE 3.2, P. 75, TEXT) THE UNIT CELL VOLUME IS a^3 = 2.80*10^-23 cm^3 DENSITY = 2*8.46*10-23 g / 2.80*10^-23 cm^3 = 6.02 g/cm^3 HOW MANY UNIT CELLS OF COPPER ARE THERE IN A CUBE 1 CM IN SIDE? # CELLS = 1 cm^3/2.8*10^-23 cm^3 = 3.57*10^22 d) COMPUTE THE AREAL DENSITY OF ATOMS ON THE FOLLOWING ATOMIC PLANES: (111) AREA = sqrt(2)*a*(sqrt(2)*a*cos(30))/2 = a^2*cos(30) = 0.08 nm^2 # ATOMS ON PLANE = 0.5 PLANAR DENSITY = 0.5/0.08 = 6.25 ATOMS/nm^2 (110) AREA = sqrt(2)*a*a = sqrt(2)*a^2 = 0.1306 nm^2 # ATOMS ON PLANE = 2 PLANAR DENSITY = 2/0.1306 = 15.3 ATOMS/nm^2 (100) AREA = a*a = a^2 = 0.092 nm^2 # ATOMS ON PLANE = 1 PLANAR DENSITY = 1/0.092 = 10.8 ATOMS/nm^2 e) COMPUTE THE LINEAL DENSITY OF ATOMS ON THE FOLLOWING LATTICE DIRECTIONS: [111] LENGTH = 4*R = 0.528 nm (FIG. 3.5, P. 74, TEXT) # ATOMS ALONG LINE = 2 LINEAR DENSITY = 2/0.528 = 3.78 ATOMS/ nm [110] LENGTH = sqrt(2)*a = 0.43 nm (FIG. 3.5, P. 74, TEXT) # ATOMS ALONG LINE = 1 LINEAR DENSITY = 1/0.43 = 2.32 ATOMS/ nm [100] LENGTH = a = 0.304 nm # ATOMS ALONG LINE = 1 LINEAR DENSITY = 1/0.304 = 3.3 ATOMS/ nm g) OF THE THREE PLANES AND DIRECTIONS IN (d) AND (e) WHICH ONES HAVE THE HIGHEST ATOMIC DENSITY IN THE CASE OF COPPER (FCC)? (111) IS THE CLOSEST-PACKED PLANE [110] IS THE CLOSEST-PACKED DIRECTION h) OF THE THREE PLANES AND DIRECTIONS IN (d) AND (e) WHICH ONES HAVE THE HIGHEST ATOMIC DENSITY IN THE CASE OF VANADIUM (BCC)? (110) IS THE CLOSEST-PACKED PLANE [111] IS THE CLOSEST-PACKED DIRECTION i) TITANIUM HAS AN HEXAGONAL CLOSE PACKED (HCP) UNIT CELL. CONSIDER CLOSE PACKING OF Ti ATOMS TO FORM A CLOSED PACKED PLANE OF ATOMS (A). PLACE A SECOND CLOSE PACKED PLANE (B) ON THE VALLEYS LEFT BY ATOMS IN (A). FINALLY, PLACE A THIRD CLOSE PACKED PLANE OF ATOMS (C) ON TOP OF PLANE (B) BUT IN SUCH A WAY THAT ATOMS IN (C) ARE DIRECTLY ABOVE ATOMS IN THE FIRST PLANE (A). THE THREE-LAYER PACKING CAN THEN BE DESCRIBED AS ABABABAB ... FIND THE HEXAGONAL UNIT CELL IN THIS ARRANGEMENT. THE UNIT CELL IS A PRISM WITH HEXAGONAL BASE. THE HEXAGONAL BASE CONTAINS 7 ATOMS AND LIES IN PLANE A THE HEXAGONAL TOP CONTAINS 7 ATOMS AND LIES IN PLANE A ALSO INSIDE THE PRISM THERE ARE 3 MORE ATOMS Out: 2/1/96 ; Due: 2/7/96 4.- ANSWER THE FOLLOWING QUESTIONS CONCERNING CRYSTAL DEFECTS. a) AT WHAT TEMPERATURE WILL THERE BE 1 EMPTY LATTICE SITE FOR EVERY 100,000 SITES IN A SAMPLE OF COPPER? n_v/N = 1/100,000 = 10^-5 = exp( - E/kT ) (EQN 4.10a, P. 155 TEXT) BUT E = 0.9 eV AND k = 8.62*10^-5 eV/K SO - (k/E)*ln(10^-5) = 1/T --> T = 906 K b) WHAT IS THE FRACTION OF VACANT SITES IN A SAMPLE OF SOLID COPPER ONE DEGREE BELOW THE MELTING POINT ( Tm = 1336 K) x = n_v/N = exp(- 0.9/((8.62*10^-5)*(1336))) = 4*10^-4 c) WHILE COPPER ALLOWS ZINC ATOMS TO EASILY BECOME SUBSTITUTIONAL IMPURITIES IN THE LATTICE, IT DOES NOT ALLOW LEAD ATOMS DO THE SAME. EXPLAIN WHY. FROM ATOMIC RADIUS DATA (FIG. 2.6, P.32 TEXT) R(Cu) = 0.128 nm R(Zn) = 0.137 nm R(Pb) = 0.175 nm THE RADIUS DIFFERENCE BETWEEN Cu AND Pb IS TOO LARGE d) SKETCH THE ARRANGEMENT OF ATOMS ON A (100) PLANE OF A SIMPLE CUBIC LATTICE CONTAINING AN EDGE DISLOCATION WITH BURGERS VECTOR [010] o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o e) AN ELECTRON MICROSCOPE STUDY OF A PIECE OF GOLD SHOWS DISLOCATION LINES RUNNING ALL PARALLEL TO EACH OTHER AND UNIFORMLY SPACED BY A SPACING OF 10^(-6) cm. WHAT IS THE DENSITY OF DISLOCATIONS (i.e. NUMBER OF DISLOCATION LINES CROSSING AN AREA OF 1 cm^2)? ASSUME THE DISLOCATIONS ARE LIKE STRAIGTH THREADS ALL ALIGNED IN THE SAME DIRECTION. IMAGINE A 1 cm LINE DRAWN IN THE MATERIAL PERPENDICULAR TO THE DISLOCATIONS, THERE WILL BE 10^6 DISLOCATIONS ON THAT LINE IMAGINE NOW A SQUARE 1 cm IN SIDE ALSO PERPENDICUALR TO THE DISLOCATIONS, THERE WILL BE 10^12 DISLOCATIONS CUTTING THE PLANE. DISLOCATION DENSITY = 10^12/cm^2 f) CONSIDER AN ATOM OF COPPER ON A (111) PLANE. HOW MANY NEAREST NEIGHBORS DOES THE ATOM HAVE? NOW, IF THE (111) PLANE IS THE OUTSIDE SURFACE OF THE SAMPLE, HOW MANY NEAREST NEIGHBORS DOES THE ATOM HAVE? g) CALCULATE THE NUMBER OF GRAINS IN THE TOP SURFACE OF THE HEAD OF A PIN PINS ARE MADE OF STEEL WITH ASTM GRAIN SIZE OF 6. Out: 2/8/96 ; Due: 2/14/96 5.- ATOM MOVEMENTS a) USING THE "DIFFUSION DISTANCE" x = 2 (D t)^0.5 , CALCULATE THE DISTANCE TRAVELLED BY IMPURITY ATOMS AFTER 1000 SEC AT 1000 K, FOR THE FOLLOWING SYSTEMS: IMPURITY MATRIX CARBON IRON (FCC) MANGANESE IRON (FCC) ALUMINUM SILICON ARSENIC SILICON EXPLAIN THE DIFFERENCES ENCOUNTERED IN YOUR RESULTS. DIFFUSIVITY DATA (FIGS. 4.46 AND 4.47, PP. 174-175, TEXT) D(C IN FCC Fe) ~ 10^-12 m^2/s D(Mn in FCC Fe) ~ 10^-19 m^2/s D(Al IN Si) = Do*exp(- Q/kT) = 0.0008*exp( -3.47/(8.62*10^-5*1000)) = 2.6*10^-21 m^2/s D(As IN Si) = Do*exp(- Q/kT) = 0.00003*exp( -3.56/(8.62*10^-5*1000)) = 3.5*10^-23 m^2/s DISTANCE TRAVELLED C IN Fe x = 6.3*10^-5 m = 63 micrometer Mn IN Fe x = 2*10^-8 m = 20 nm Al IN Si x = 3.2*10^-9 m = 3 nm As IN Si x = 3.7*10^-10 m = 0.3 nm NOTE: CARBON MOVES RELATIVELY EASILY THROUGH THE FCC LATTICE OF Fe ARSENIC, BECAUSE OF ITS LARGER SIZE COMPARED TO Si MOVES THE SLOWEST 6.- PHASE DIAGRAMS WHAT IS THE MAIN DIFFERENCE IN THE MELTING/FREEZING BEHAVIOR BETWEEN PURE COPPER AND A 50%COPPER-50%NICKEL ALLOY? PURE COPPER MELTS/FREEZES AT A CONSTANT TEMPERATURE THE COPPER-NICKEL ALLOY MELTS/FREEZES OVER A TEMPERATURE RANGE 7.- ELECTRICAL PROPERTIES CONSIDER THIN WIRES (RADIUS = 0.0005 m ; LENGTH = 1 m ), MADE OF THE FOLLOWING MATERIALS AND AT ROOM TEMPERATURE: PURE COPPER PURE SILICON SILICON DOPED WITH 10^21 ATOMS OF PHOSPHOROUS/m^3 a) CALCULATE THE RESISTIVITY OF EACH WIRE * PURE COPPER (TABLE 5.1, P. 189 TEXT) RESISTIVITY = 1/CONDUCTIVITY = 1/5.8*10^7 = 1.7*10^-8 Ohm.m * PURE SILICON (EQUATION 5.13, TABLE 5.4, P. 203 TEXT) CONDUCTIVITY = n_i*q*(mu_n + mu_p) = 1.5*10^16*(1.6*10^-19)*(0.135 + 0.048) = 4.39*10^-4 RESISTIVITY = 1/CONDUCTIVITY = 1/4.39*10^-4 = 2276 Ohm.m * SILICON DOPED WITH 10^21 ATOMS OF PHOSPHOROUS/m^3 MATERIAL BECOMES n-TYPE SEMICONDUCTOR (P. 214, TEXT) CONDUCTIVITY = n_P*q*mu_n = 1*10^21*(1.6*10^-19)*0.135 = 21.6 RESISTIVITY = 1/CONDUCTIVITY = 1/21.6 = 4.6*10^-2 Ohm.m b) IF A 1 V BATTERY IS CONNECTED TO THE ENDS OF EACH WIRE, CALCULATE THE CURRENT DENSITY IN EACH CASE * PURE COPPER (EQ. 5.10, P. 203) J = E/RESISTIVITY = 1/1.7*10^-8 = 5.8*10^7 A/m^2 * PURE SILICON J = E/RESISTIVITY = 1/2276 = 4.39*10^-4 A/m^2 * SILICON DOPED WITH 10^21 ATOMS OF PHOSPHOROUS/m^3 J = E/RESISTIVITY = 14.6*10^-2/ = 21.6 A/m^2 c) HOW MANY CHARGED PARTICLES CROSS THE CROSS SECTION OF WIRE PER SECOND IN EACH CASE? * PURE COPPER (EQ. 5.11, P. 203) CHARGED PARTICLE FLUX = J/q = 5.8*10^7/1.6*10^-19 = 3.62*10^26 * PURE SILICON CHARGED PARTICLE FLUX = J/q = 4.39*10^-4/1.6*10^-19 = 2.74*10^15 * SILICON DOPED WITH 10^21 ATOMS OF PHOSPHOROUS/m^3 CHARGED PARTICLE FLUX = J/q = 21.6/1.6*10^-19 = 1.35*10^20 8.- MECHANICAL PROPERTIES CONSIDER THE FOLLOWING MECHANICAL PROPERTY DATA 304 STAINLESS STEEL MODULUS OF ELASTICITY E = 200 GPa = 200*10^9 (Pa) TRUE STRESS-TRUE STRAIN CURVE sigma = 1,450*10^6*(epsilon)^0.6 (Pa) PURE COPPER MODULUS OF ELASTICITY E = 130 GPa = 130*10^9 (Pa) TRUE STRESS-TRUE STRAIN CURVE sigma = 450*10^6*(epsilon)^0.33 (Pa) a) TABULATE AND DRAW THE COMPLETE TRUE STRESS-TRUE STRAIN CURVES FOR BOTH MATERIALS FOR VALUES OF epsilon BETWEEN 0 AND 1 epsilon sigma Steel sigma Copper 0.0 0.0 0.0 0.1 3.64 E8 2.10 E8 0.2 5.52 E8 2.64 E8 0.3 7.04 E8 3.02 E8 0.4 8.37 E8 3.32 E8 0.5 9.57 E8 3.58 E8 0.6 1.07 E9 3.80 E8 0.7 1.17 E9 4.00 E8 0.8 1.27 E9 4.18 E8 0.9 1.36 E9 4.35 E8 1.0 1.45 E9 4.50 E8 b) ESTIMATE THE FORCE REQUIRED TO STRETCH THIN WIRES OF THE TWO MATERIALS IN THE FOLLOWING MANNER INITIAL WIRE RADIUS = 0.0005 m INITIAL WIRE LENGTH = 0.1 m FINAL WIRE LENGTH = 0.2 m CALCULATED FINAL WIRE RADIUS (VOLUME CONSERVATION) ~ 0.00035 m d(epsilon) = dL/L --> epsilon = ln(A) + const epsilon_f = ln(A_f/A_i) ~ 0.69 sigma(Steel) ~ 1170 MPa sigma(Cu) ~ 400 MPa ASSUME AVERAGE RADIUS DURING DEFORMATION = 0.000426 m --> AVERAGE AREA DURING DEFORMATION = A_ave = 5.7*10^-7 m^2 FORCE (STEEL) = sigma(Steel)*A_ave = 669 N FORCE (CU) = sigma(Cu)*A_ave = 228 N c) HOW LARGE ARE THE COMPUTED FORCES COMPARED, SAY, WITH YOUR OWN WEIGHT? MY WEIGHT IS 800 N, SO FORCES ARE SMALLER THAN MY OWN WEIGHT d) CONSIDERING THE VALUES OF ULTIMATE TENSILE STRENGTH GIVEN BELOW, WHAT WOULD BE THE MAXIMUM LENGTH OF THE STRETCHED WIRES JUST BEFORE NECKING AND THE TRUE STRESS AT THAT POINT (TRUE STRESS AT MAXIMUM LOAD)? AT NECKING, THE LOAD IS MAXIMAL, I.E. dP/d epsilon = 0 = d(sigma*A)/d epsilon = sigma*dA/depsilon + A*d sigma/depsilon = 0 OR (1/A)*sigma*dA/d epsilon + d sigma/d epsilon = 0 BUT dA/A = - d epsilon SO sigma = d sigma/d epsilon BUT sigma = K*(epsilon)^n --> d sigma/d epsilon = K*n*(epsilon)^(n-1) FINALLY sigma = d sigma/d epsilon => epsilon = n (AT MAXIMUM LOAD; NECKING) THUS epsilon(NECKING, STEEL) = 0.6 sigma(NECKING, STEEL) = 1070 MPa epsilon(NECKING, Cu) = 0.33 sigma(NECKING, Cu) ~ 310 MPa THE STRETCHING MENTIONED IN PART (b) ABOVE IS IMPOSSIBLE WITHOUT NECKING! L_f(MAX,WITHOUT NECKING) = L_i*exp(epsilon(NECKING)) L_f(MAX, STEEL) = 0.18 m L_f(MAX, Cu) = 0.14 m e) CONSIDERING THE VALUES PERCENT ELONGATION (TO FRACTURE) GIVEN BELOW, WHAT WOULD BE THE MAXIMUM POSSSIBLE STRETCHED LENGTH OF EACH WIRE AND THE ENGINEERING STRESS AT THAT POINT (ENG FRACTURE STRESS, s_f)? THE GIVEN VALUES ARE MINIMUM ACCEPTABLE VALUES. WITH CARE, LARGER STRETCHINGS ARE POSSIBLE f) FROM YOUR RESULTS IN (e), WHAT WOULD BE THE TRUE STRAIN TO FRACTURE (epsilon_f) AND THE TRUE FRACTURE STRESS (sigma_f) FOR EACH OF THE TWO WIRES? (Hint: THIS IS A TRICKY QUESTION SINCE IT CAN'T BE ANSWERED WITH NUMBERS USING THE DATA GIVEN HERE; WHY?) AS THE MATERIAL NECKS, THE STRAIN STATE AT THE NECK BECOMES MORE COMPLEX THATN THE SIMPLE UNIDIRECTIONAL STRAINING PRIOR TO NECKING ADDITIONAL MECHANICAL PROPERTY DATA (REVISED): ULTIMATE TENSILE STRENGTH = MAXIMUM LOAD SUSTAINED BY MATERIAL/INITIAL CROSS SECTIONAL AREA = S_u S_u(Copper) = 216 MPa S_u(304SS) = 585 MPa NOTE THAT SINCE THESE VALUES CORRESPOND TO MAXIMU LOAD (NECKING) S_u = P_max/A_o BUT sigma(NECKING) = P_max/A_n ELIMINATING P_max sigma(NECKING) = S_u*exp( epsilon(NECKING) CHECK!!!! PERCENT ELONGATION (TO FRACTURE) = e_f = [(LENGTH AT FRACTURE - INITIAL LENGTH)/INITIAL LENGTH]*100 e_f(Copper) = 48% e_f(304SS) = 40% 9.- CONSTRUCT COLUMNS RANKING THE FOLLOWING POLYMERS ON THE BASIS OF THEIR A) MELTING POINT (INCLUDE VALUE DEGREES C) SEE TEXTBOOK AND ALSO "HANDBOOK OF PLASTIC MATERIALS AND TECHNOLOGY" BY RUBIN POLYETHYLENE 110-137 POLYSTYRENE 150-243 POLYMETHYL METHACRYLATE 160 POLYPROPYLENE 165-177 POLYOXYMETHYLENE 175 POLYVINYLCHLORIDE 175 POLIVINYLIDENE CHLORIDE 177 NYLON 6 216-225 POLYBUTYLENE TEREPHTHALATE 225 POLYTETRAFLUOROETHYLENE 327 POLYVINYL ACETATE AMORPHOUS, Tg ~ 30 ABS AMORPHOUS, Tg ~ 100-110 POLYBUTADIENE CROSS LINKED RUBBER, Tg ~ -90 POLYACRYLONITRILE DECOMPOSES BEFORE MELTING @ ~ 300 CROSS LINKED POLYESTHER THERMOSET B) MECHANICAL STRENGTH (TENSILE STRENGTH, MPa) POLYETHYLENE 7-40 POLYTETRAFLUOROETHYLENE 20-40 POLYPROPYLENE 27 POLYSTYRENE 32-56 ABS 33-43 POLIVINYLIDENE CHLORIDE 34 POLYVINYLCHLORIDE 41-52 POLYBUTYLENE TEREPHTHALATE 55 POLYMETHYL METHACRYLATE 55-76 POLYOXYMETHYLENE 69 NYLON 6 80 POLYVINYL ACETATE NA POLYACRYLONITRILE NA POLYBUTADIENE NA CROSS LINKED POLYESTHER NA IN EACH CASE, WRITE THE CHEMICAL FORMULA OF THE POLYMER USING THE STICK NOTATION DISCUSSED IN CLASS AND ALSO ADD A SHORT NOTE EXPLAINING YOUR REASONS FOR SELECTING A SPECIFIC PLACE IN THE RANKING. STICK NOTATION CHEMICAL FORMULAE POLYETHYLENE (P. 361, TEXT) POLYVINYLCHLORIDE (P. 363, TEXT) POLYPROPYLENE (P. 365, TEXT) POLYSTYRENE (P. 366, TEXT) POLYACRYLONITRILE (P. 367, TEXT) POLYVINYL ACETATE (P. 339, TEXT) POLIVINYLIDENE CHLORIDE (P. 340, TEXT) POLYMETHYL METHACRYLATE (P. 370, TEXT) POLYBUTADIENE (P. 367, TEXT) ABS (P. 368, TEXT) POLYTETRAFLUOROETHYLENE (P. 371, TEXT) NYLON 6 (P. 375, TEXT) POLYOXYMETHYLENE (P. 380, TEXT) POLYBUTYLENE TEREPHTHALATE CROSS LINKED POLYESTHER (P. 392, TEXT) 10.- ANSWER THE FOLLOWING QUESTIONS ABOUT CERAMIC MATERIALS A) COMPUTE THE PLANAR ATOMIC DENSITIES OF Mg AND O ATOMS IN THE (100) PLANES OF MgO . FROM PP. 590-591 AND 32 , TEXT UNIT CELL EDGE = 2*(R_Mg+2 + R_O-2) = 2*( 0.065 + 0.140) = 0.41 nm AREA OF (100) PLANE = A = a^2 = 0.168 nm^2 NUMBER OF Mg+2 IONS = 2 (FIG. 10.8, P. 590, TEXT) NUMBER OF O-2 IONS = 2 PLANAR DENSITY OF Mg+2 IONS = 2/0.168 = 11.9 IONS/nm^2 PLANAR DENSITY OF O-2 IONS = 2/0.168 = 11.9 IONS/nm^2 B) COMPUTE THE VOLUMETRIC ATOMIC DENSITY OF Ca and F ATOMS IN A CaF2 CRYSTAL UNIT CELL IN FIG. 10.14, P. 596 TEXT NUMBER OF Ca+2 IONS = 4 NUMBER OF F- IONS = 8 UNIT CELL EDGE = a = 4*R_Ca+2 / sqrt(2) (FIG. 3.7, P. 78, TEXT) BUT R_Ca+2 = 0.099 nm --> a = 0.28 nm UNIT CELL VOLUME = a^3 = 0.022 nm^3 VOLUME DENSITY OF Ca+2 = 4/0.022 = 182 IONS/nm^3 VOLUME DENSITY OF F- = 8/0.022 = 364 IONS/nm^3 C) COMPUTE THE VOLUMETRIC ATOMIC DENSITIES OF Ca, Ti AND O ATOMS IN A CaTiO3 CRYSTAL. UNIT CELL IN FIG. 10.16, P. 599, TEXT NUMBER OF Ca+2 IONS = 1 NUMBER OF O-2 IONS = 3 NUMBER OF Ti+4 IONS = 1 UNIT CELL EDGE = a = 2*(R_Ca+2 + R_O-2) / sqrt(2) (FIG. 3.7, P. 78, TEXT) BUT R_Ca+2 = 0.099 nm R_O-2 = 0.140 nm --> a = 0.34 nm UNIT CELL VOLUME = a^3 = 0.038 nm^3 VOLUME DENSITY OF Ca+2 = 1/0.038 = 25.9 IONS/nm^3 VOLUME DENSITY OF O-2 = 3/0.038 = 77.7 IONS/nm^3 VOLUME DENSITY OF Ti4+ = 1/0.038 = 25.9 IONS/nm^3 D) COMPUTE THE NUMBER OF Si AND O ATOMS IN A UNIT CELL OF CRISTOBALITE. CALCULATE ALSO THE VOLUMETRIC DENSITY OF Si AND O ATOMS. UNIT CELL IN FIG. 10.22, P. 603, TEXT NUMBER OF Si ATOMS = 8 NUMBER OF O ATOMS = 16 UNIT CELL EDGE = a = 4*R_Si/ sqrt(2) (FIG. 3.7, P. 78, TEXT) = 4*(0.117)/sqrt(2) = 0.33 nm UNIT CELL VOLUME = 0.036 nm^2 VOLUME DENSITY OF Si = 8/ 0.036 = 220 ATOMS/nm^2 VOLUME DENSITY OF O = 16/ 0.036 = 440 ATOMS/nm^2 E) IF MOLTEN SiO2 IS COOLED VERY, VERY SLOWLY, IT EXHIBITS A CLEAR FREEZING POINT AND THE SOLID PRODUCT IS CRISTOBALITE. ON THE OTHER HAND, IF THE COOLING IS RAPID, THE SOLID PRODUCT IS ORDINARY WINDOW GLASS. EXPLAIN THE THE REASON FOR THIS BEHAVIOR. THE FASTER THE COOLING THE HARDER IT IS FOR THE SILICATE CHAINS TO ARRANGE INTO A CRYSTALLINE STRUCTURE 11.- MAGNETIC PROPERTIES A) USING THE B-H CURVE FOR ALNICO V, FIND THE MAXIMUM ENERGY PRODUCT (BH)max. FROM FIG. 11.24, P. 686 AND P. 697, TEXT TRY 1: B = 0.5 , H = 85.7 --> BH = 42.86 kJ/m^3 TRY 2: B = 0.3, H = 120 --> BH = 36 " TRY 3: B = 0.4, H = 113.3 --> BH = 45.3 " SO (BH)_max ~ 45.3 kJ/m^3 B) THE MAGNETIC DOMAIN STRUCTURE OF A FERROMAGNET IN THE UNMAGNETIZED STATE (B = 0) CONSIST OF MANY DOMAINS. WHY ISN'T A FERROMAGNET IN SUCH CONDITION MAGNETIZED? THE MAGNETIC MOMENTS OF THE DIFFERENT DOMAINS CANCEL EACH OTHER OUT (SEE P. 671, TEXT) C) WHAT HAPPENS TO THE DOMAIN STRUCTURE OF THE MATERIAL IN (B) ABOVE WHEN IT IS PLACED IN THE GAP OF A COIL THROUGH WHICH INCREASING CURRENT IS PASSED? MAGNETIC DOMAINS ORIENTED IN THE DIRECTION OF THE APPLIED FIELD WILL GROW, WHILE UNFAVORABLY ORIENTED DOMAINS WILL SHRINK AND DISSAPEAR. (SEE PP. 671-672, TEXT) 12.- OPTICAL PROPERTIES A) DESCRIBE THE BASIC ELEMENTS OF A FIBER-OPTICS COMMUNICATIONS SYSTEM. SEE PP. 850-855, AND FIG. 14.17, TEXT * InGaAsP TRANSMITTER * OPTICAL FIBER * PIN DETECTOR B) EXPLAIN THE FUNCTIONING OF THE DEVICES ENCOUNTERED IN THE TRANSMITTING AND RECEIVING ENDS OF THE SYSTEM. DEVICES ARE SEMICONDUCTOR LASERS . ELECTRONS ARE EXCITED BY BIASING VOLTAGE AND COHERENT PHOTONS ARE EMITTED. (SEE P. 849, TEXT) C) DESCRIBE THE KEY CHARACTERISTICS REQUIRED OF AN OPTICAL FIBER. LOW LOSS (LIGHT SIGNAL SHOULD TRAVEL LONG DISTANCES) EASY TO FABRICATE EASY TO JOIN