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1 Midterm Examination - Thursday, February 5, 8:00 9:5 AM Place all answers in a 8.5" x " Bluebook Allowed:, double-sided 8.5" x " page of notes must return with exam. Required: Picture ID when returning the completed exam. ote: 5 questions (front and back of this page). Miller Indices (20 points) (a) Draw the ( 2) plane in a cubic unit cell. Clearly label the axes. (b) Draw two directions in the ( 2) plane, which are separated by 90. Properly label the directions. a) plane (-2) is outlined b) Any two directions which satisfy: a. The scalar product of the two directions is 0 b. The scalar product of each direction with the normal to the plane ( - 2) is 0 Z [- ] ( - 2) Y X [ 0]

2 2. Atomic Bonding (5 points) (a) Calculate the atomic packing factor for SC (simple cubic) (b) Calculate the metallic radius of a polonium (Po) atom given that Po has a SC structure, a unit cell dimension of.52 Å, and an atomic weight of 209 g/mol. Calculate the density of Po. a.) Atomic Packing Factor for Simple Cubic Atoms contact along unit cell edges. 2R atoms along edge length a. Volume of Atoms 4/ π R Atom per unit cell Unit Cell volume a =8R Packing Factor = (4/ π R )/(8R ) = π/6 = 0.52% b.) Metallic Radius of Polonium: Unit Cell.52 Å` radius =.676 Å Density: 209 g/mol 209 (g/mol) / V (mol/atom) = g/atom atom per unit cell g/unit cell Unit Cell Volume =.52 = Å ( ) / ( x 0-24 ) = 9.26 g/

3 . Point Defects (25 points) a. Calculate the equilibrium number of vacancies per cubic centimeter ( ) in aluminum (FCC packing) at 600 C and at 298 C. The vacancy formation energy in Al is ev/atom. b. Make a simple estimate of the average spacing between vacancies at 298 C c. Calculate the number of interstitials per cubic centimeter ( ) in aluminum (FCC packing) at 600 C and at 298 C. The interstitial formation energy in Al is.58 ev/atom. d. Make a simple estimate of the average spacing between interstitials at 298 C The density and AW for aluminum are 2.70 g/ and g/mol, respectively. ote: Boltzmann s constant = 8.674x0-5 ev K - ote: Avogrado s # = x 0 2 /mol a.) = ρ M A w = g atoms e2 mole g mole atoms = e22 Q = k BT e 5* exp = 6.026e22exp ( ) vacancies (298 =.84e = 6.026e22exp 8.674e 5* ( )

4 vacancies (600 =.49e9 b.).84 e7 / / D =.76e 6 D = 76nm

5 c.) Q = exp k BT.58 ( ) = 6.026e22exp 8.674e 5* (298 = 6.90e8.58 = 6.026e22exp 8.674e 5* (600 = 4.59e ( ) d.) / 6.90 e8 / D =.e D =.µ m

6 4. Polymers (20 points) The weight average molecular weight of polystyrene (PS) is 600,000 g/mol. (a) Compute the weight average degree of polymerization. See the mer unit for PS below: ote AWs: H: g/mol; C: 2.0 g/mol; U: g/mol (b) Do you expect PS to be a thermoplastic polymer or a thermosetting polymer? (a) For each polystyrene unit there are 2 carbon atoms, hydrogen atoms and benzene Ring (or equivalently, 5 more hydrogen atoms and 6 more carbon atoms). Therefore the mer molecular weight is given as +5 m= 8( A + 8( AH) 2 = g/mol g/mol ( ) ( ) = g/mol ow the degree of polymerization (DP) can be computed from Equation 4.6, M w , 000 g/mol DP = = = m g/mol Where the weight average molecular weight (b) M w was specified in the question. Based on the weight average molecular weight we would expect polystyrene to be a thermoplastic this is because most often such materials are high molecular weight polymers. Moreover, polystyrene is also a relatively soft material and forms relatively linear chains with minimal branching, which make it quite flexible. +5

7 5. Mechanical Properties (20 points) A cylindrical specimen of a hypothetical metal alloy is stressed in compression (what is the sign of the stress?). This material, Goletum, has a Young s modulus of 00 GPa and a shear modulus of 40 GPa. If the original and final diameters are 00.0 mm and 00.0 mm, respectively, and its final length is 25.5 mm, compute the following: Compressive stress Poisson s ratio Bulk modulus Original length (assuming purely elastic deformation) Solution

N = N A Pb A Pb. = ln N Q v kt. = kt ln v N

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