enmaexam1 - Quantum Numbers n – Principal Quantum...

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Unformatted text preview: Quantum Numbers n – Principal Quantum number  ­ size l – n  ­1 (angular)  ­ shape mp  ­  ­l to l (magnetic)  ­ orientation ms  ­ +1/2 or  ­1/2 (spin) Pauli Exclusion principle: For an individual atom, at most two electrons, which necessarily have opposite spins, can occupy the same state. X’s are the electronegativies. 2 % ionic character= (1 ­exp( ­(0.25)(XA ­XB) ))x100 Where XA> XB Covalently bonded substances have low MP due to weak intermolecular bonds (Van der waals) FCC Coordination number 12 (n=4atoms/cell) ! = 2! 2, apf =.74 BCC coordination number 8 (n=2 atoms/cell) ! = (4!)/ 3, apf =.68 Simple cubic coordination number 6 ! = 2! !"#$%&!"#$% !"# = !"#$%&!"#$!"## APF = Vs/Vc 3 Vs = n*4/3*pi*R 3 Vc = a HCP (n =6 atoms per cell) !3 3 ! ! =  ! !"#$#! = 2! 2 Density !! 3 ! = ! (g/cm ) !! !! !′( !! + !! ) ! !! ! Sum of Cations + Sum of Anions Atomic weights 23 Na=6.022*10 atoms/mol Aw=atomic weight #!"!"#$%!"#$"%"& !"!ℎ!!"#$%& !"#$%&!"#$%&' = !"#$%ℎ!"!"#$%&"'(!"#$%& Linear Density = # atoms / length (g/cm) BCCld110 = 2 atoms BCCld111 = 2 atoms distance 4R FCC110 = 2 atoms FCC100 = 1 atom FCC110 = 3 atoms FCC111 = 2 atoms 2 Planar Density = # atoms/ area (g/cm ) BCC100 = 1 atom BCC110 = 2 atoms 2 BCC111= 1 atom; formula = 3*1R /16sqrt(3) 2 FCC111 = 2 atom; formula = 1/2R sqrt(3) The larger the ion, the larger the coordination number Coordination number Cation anion radius ratio 2 <.155 3 .155 ­.225 4 .225 ­.414 6 .414 ­.732 8 .732 ­1.0 Structur Structur Anion Catio Anion Exampl e Name e type packin n # # es g Rock AX FCC 6 6 NaCl, salt MgO, FeO CsCl AX Simpl 8 8 CsCl e cublic Zinc AX FCC 4 4 ZnS, SiC Blende CaF2 (Ca Fluorite AX2 Simpl 8 4 is in e every cubic other != cube Perovski te Spinel ABX3 FCC AB2X4 FCC 12(A) 6(B) 4(A) 6(B) 6 BaTiO3 4 MgAl2O Rock salt ! = 2!!"# + 2!!" Seven crystal systems <111> family of directions {111} family of planes Chapter 4 (Polymers) Avg Molecular Weight  ­ Mn = !! ∗ !"  ­Mw = !! ∗ !" Mi = Mean mol weight xi fraction of the total number chains wi weight avg ! !" =  ! where m is average mol weight of repeat unit ! 4 i.e CH4 m = (1*12g/mol + 4*1g/mol) ! (! ! ! ) %!!"# = ! ! ! C is crystalline, a is totally amorphous, and !! (!! !!! ) s is density tbd • Crosslinking occurs at double bonds in the main chain • Crosslinking increases E, σuts, and oxidation resistance • Useful rubbers are 1 ­5 w/o sulfur, higher crosslinked rubbers are too stiff, hard, and brittle. Linear Polymers – Mer units are joined together end ­to ­end in single chains. – The long chains are flexible and may be thought of as spaghetti. – Extensive vanderWaals and H ­bonding – Examples: HDPE, PVC, PTFE, PMMA, polystyrene • Branched Polymers – Long side branch chains are connected to the main ones – Reduced chain packing efficiency (lowers strength and density) – Examples: LDPE • Crosslinked Polymers – Adjacent linear chains are joined to one another at various positions along their length by covalent bonds. – Crosslinking is accomplished by additive atoms or molecules – Examples: Vulcanized Rubbers, Elastomers • Network Polymers – Trifunctional mer units having three active covalent bonds form threedimensional networks. – Three dimensional networks also possible in highly crosslinked materials – Examples: Epoxies and phenol ­formaldehyde (bakelite). Linear Polymers – Mer units are joined together end ­to ­end in single chains. – The long chains are flexible and may be thought of as spaghetti. – Extensive vanderWaals and H ­bonding – Examples: HDPE, PVC, PTFE, PMMA, polystyrene • Branched Polymers – Long side branch chains are connected to the main ones – Reduced chain packing efficiency (lowers strength and density) – Examples: LDPE • Crosslinked Polymers – Adjacent linear chains are joined to one another at various positions along their length by covalent bonds. – Crosslinking is accomplished by additive atoms or molecules – Examples: Vulcanized Rubbers, Elastomers • Network Polymers – Trifunctional mer units having three active covalent bonds form threedimensional networks. – Three dimensional networks also possible in highly crosslinked materials – Examples: Epoxies and phenol ­f ormaldehyde (bakelite). Chapter 5 !  ­23  ­5 !! = !"#$ − ! k=  ­1.38x10 J/atom*k = 8.62x10 !" ev/atom*k !! ! != !! Schottky/Frenkel !! 100 100 !! = !"#$ − ; !!"# =  !! !!  ; !!"# =  !! !! !" + + !! !! !! !! C1 is wt% C1 = m1/(m1+m2) 100 ! !! = 2!!! ( ) ! !" Shear stress to move dislocation ! = ! ! ! ! τ = shear stress required to move a dislocation C and K = material constants. d = interplanar spacing b = burgers vector Impurities can be interstitial or substitutional (i.e., filling normally unoccupied spaces in the lattice or substituting for lattice, the host (matrix) atoms on lattice sites). Frenkel defect: cation vacancy/cation interstitial pair Schottky defect: cation vacancy/anion vacancy pair Hume ­Rothery Rules for Substitutional Solid Solution • The solute should have an atomic radius within 15% of that of the solvent. If the radii are very different, the solute may dissolve interstitially or separate into a distinct phase to avoid stressing the lattice. • The crystal structures of the solute and the solvent MUST be the same, or separate phases will be formed. Ie. BCC = BCC • The electronegativity of the solute should be close to that of the solvent. A large ΔEN will lead to the formation of an intermetallic compound. • The solute should have the same or lower valence than the solvent. Otherwise it’s a Intersitital Solid Solution – where the impurity atom fills the voids or the interstices In ceramics impurity ions can make subsititutational or interstitial solid solutions and must maintain electroneutraility Edge dislocation(tilt): displacement is perpendicular to axis  ­| ­ Screw dislocation(twist): displacement is parallel to axis (loop symbol) FCC{111} BCC{110} HCP{0001} As dislocations move, two adjacent planes slide, or slip past one another. Slip: The process by which plastic deformation is produced in a crystal by dislocation motion. Slip direction: the direction of dislocation movement, usually the most closely packed direction in the crystal. Chapter 6 ! ! !" 2 2 ! =  =  units of are J kg/m *s or atoms/m *s !" ! !" M is number of atoms !" !! − !! ! =  −! =  −! !" !! − !! !! − !! ! = 1 − erf( ) !! − !! 2 !" !! !! ! 2 ! = !! !"#(− ! ) units of D are m /s !" !" Determining Indices: Head (x1,y1,z1)  ­ Tail (x2,y2,z2)  ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ i.e (0,1/2  ­1) * 2  ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ ­ [0,1,2bar] Sketching Indices Given [0,1,2bar] Divide by the biggest # = 0 ½  ­1, trace and draw Polymer Crystallinity after adding MFM Crystallinity will decrease. The multifunctional monomer (MFM) creates a network that limits the relative motion of the linear chain segments. The MFM will prevent the efficient packing of the chains that is necessary to generate crystalline domains. The MFM restricts the large amplitude motion of the polymer chains making the material more rigid or glass ­like. As a result, the glass transition temperature will increase. Higher temperatures will be needed in order to soften the polymer. The tensile strength is related to the bonding between the polymer chains. In the linear polymer, chains slide past each other easily. The tensile strength is related to the weak secondary bonds. The MFM creates covalent bonds between the chains, making it more difficult to deform and break the polymer. Thus the tensile strength will increase. Two types of Point Defects Vacancy Self ­Interstitial – atom crowded into interstitial site Frenkel Defect – cation leaves normal position moves to interstitial site Schottky Defect ­ !" = ...
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