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Problem # 1 Select the correct answer by putting [9] in the appropriate box. (a) Materials' properties are strongly related to: [ (b) ] [x] atomic structure [ ] atomic weight [ ] none of the above Which primary bonding is directional: [ ] (d) (e) ionic [x] covalent [ metallic ] [] (c) atomic number all of the above Which material has the highest range of electrical conductivities: [x ] metals [ ] ceramics [ ] polymers [ ] all of the above Which one of the following statements in not correct for HCP crystals: [ ] coordination number is 12 [ ] {0001} is a close packed plane [ ] closed-packed planes are arranged as ABABABAB … [x ] there are 4 atoms per unit cell (e) The APF for simple cubic structure is: ] π/8 [ ] π/4 [ x] π/6 [ ] π/2 [ (20 Points) 2 (f) The relationship between the unit cell size and the atomic radius R in FCC is: [ (g) (h) ] a = 2R/√2 [x] a = 4R/√2 [ ] a = R√2 [ ] a = 4R√2 The linear density along [110] in BCC is: [ ] √3/(√2R), where R is the atomic radius [ ] √2/(4√3R) [x] √3/(4√2R) [ √2/(√2R) ] The planar density in (100) plane in simple cubic is: [ ] 1/(2R2) [ ] 2/(R2) [ ] 1/(R2) [x] (i) Vacancies are created when: [ ] electrons are missing from lattice sites [x] atoms are missing from lattice sites [ atoms are missing from interstitial sites ] [ ] (j) 1/(4R2) none of the above The number of vacancies in metals: [ ] decreases with increasing temperature [ ] do not depend on the total number of atomic sites [ ] is independent of temperature [x] none of the above 3 (k) (l) (m) (n) (o) The energy required to form a vacancy: [ ] decreases with decreasing temperature [ ] increases with increasing temperature [x] is independent of temperature [ ] none of the above An atom occupying a position that is not a normal lattice site is: [ ] a vacancy [ ] a substitution atom [x] an interstitial atom [ a dislocation ] The crystallographic direction shown below is: [ ] [101] [ ] [111] [x] [111] [ [111] ] −− −− The crystallographic plane shown below is: [ ] ⎡ −⎤ ⎢⎣3 21⎥⎦ [ ] (2 31) [ ] < 3 21 > [x] none of the above − − A grain boundary: [ ] is a point defect [ ] is a linear defect [x] separates two grains or crystals having deferent crystallographic orientation [ ] none of the above 4 (p) (q) (r) (s) (t) The dislocation line in screw dislocation is: [ ] not related to Burger’s vector [ ] normal (perpendicular) to Burger’s vector [x] parallel to Burger’s vector [ ] none of the above In a substitutional solid solution: [ ] impurity atoms fill interstitial positions [ ] a new phase with different crystal structure will form [ ] the two elements are not soluble in each other [x] none of the above Two metallic elements will form a solid solution with substantial solubility if they have: [ ] different crystal structures [ ] large difference in electronegativities [ ] a difference in atomic radii much more than 15 % [x] none of the above Impurities often segregate along grain boundaries because: [ ] grain boundaries are less chemically reactive compared to grains [x] grain boundaries have higher energy state compared to grains [ ] grain boundaries have lower energy state compared to grains [ ] none of the above Single crystal materials have: [ ] small number of grain boundaries [ ] large number of grain boundaries [ ] random distribution of grains [x] none of the above 5 Problem # 2 (10 Points) Calcium has a cubic crystal structure, the atomic radius is 0.197 nm, the density is 1.55 g/cm3, and the atomic weight is 40.08 g/mol. Determine whether it has simple, FCC, or BCC structure. (Navg = 6.023 x 1023 atoms/mol) Solution R = 0.197 nm, ρ = 1.55 g/cm3, A = 40.08 g/mol ρ= nA nA = 3 Vc N a N If SC structure: n = 1 atom and a=2R ρ= 1atom × 40.08 g / mol ( 2 × 0.197 ×10−7 cm ) 3 6.023 × 10−7 atom / mol 40.08 g g = 1.09 cm 3 36.84cm 3 = If BCC structure: n = 2 atoms and a=4R/√3 ρ= 2atoms × 40.08 g / mol 3 ⎛ 4 ⎞ −7 −7 × 0.197 × 10 cm ⎜ ⎟ 6.023 × 10 atom / mol ⎝ 3 ⎠ = 80.16 g g = 1.41 3 3 56.72cm cm If FCC structure: n = 4 atoms and a=2√2xR ρ= ( 4atoms × 40.08 g / mol 2 2 × 0.197 × 10 −7 cm ) 3 6.023 × 10 −7 atom / mol Therefore, FCC is the appropriate structure of Calcium. = 160.32 g g = 1.54 104.20cm 3 cm 3 6 Problem # 3 (10 Points) a) (8 points): Calculate the relative volume change of a piece of iron (Fe) when its crystal structure changes from BCC to FCC upon heating through 912°C. The BCC unit cell is 0.293 nm and the FCC unit cell is 0.363 nm at the transformation temperature. V −V Initial ) (The relative volume change = Final V Initial Solution There are different ways for soving this problem. One of them is to assume a number of atoms in the material. For example 100 atoms. 1) Number of unit cells of BCC= 100/2 = 50 The volume of the sample with BCC structure = 50*(0.293nm)^3 = 1.258nm3 2) Number of unit cells of FCC= 100/4 = 25 The volume of the sample with FCC structure = 25*(0.363nm)^3 = 1.196nm3 V Final −V Initial 1.196 − 1.258 = = −0.0493 or 1.258 V Initial - 4.93% An alternative way is shown below. The following equations will be used for solving this problem. V Final −V Initial M nA (1), ρ= (2), ρ= V V Initial V c N avg (3) At the transformation temperature the mass of the material is not changing; therefore, M = ρ BCCV Initial = ρ FCCV Final Using equation (3): Therefore, ρ BCC ρ FCC ⇒ V Final = ρ BCC V ρ FCC Initial n BCC A V BCC N avg n BCCV FCC = = n FCC A n FCCV BCC V FCC N avg ρ BCC V −V Initial ρ FCC Initial V Final −V Initial n BCCV FCC 2 × 0.3633 = = −1 = − 1 = −0.0492 or 4 × 0.2933 V Initial V Initial n FCCV BCC The advantage here is that only one caculation which means that less error. b) (2 points): Is the change in volume considered as expansion or contraction? The negative sign indicates that the volume is contracting. − 4.92% 7 Problem # 4 (10 Points) The numbers of vacancies present in silver at 500°C and 900°C are 3.59 x1021 m−3 and 1024 m−3, respectively. Calculate the energy for vacancy formation in silver. Assume that the density at both temperatures is the same. (k = 1.38 x 10-23 J/atom-K or 8.62 x 10-5 eV/atom-K). Solution At T1 = 500 + 273 = 773 K NV 1 = N exp( − QV ) …………1 KT 1 At T2 = 900 + 273 = 1173 K NV 2 = N exp( − QV ) …………2 KT 2 Taking the ratio and ln of both sides and rearranging gives: N v1 Nv2 ⎛ Q ⎞ N s exp ⎜ − v ⎟ ⎝ kT 1 ⎠ = exp ⎛ − Q v = ⎜⎜ ⎛ Qv ⎞ ⎝ k1 N s exp ⎜ − ⎟ ⎝ kT 2 ⎠ ⎛ 1 1 ⎞⎞ ⎜ − ⎟ ⎟⎟ ⎝ T1 T 2 ⎠ ⎠ ⎛ T ×T ⎞ ⎛ N ⎞ Qv = − k ⎜ 1 2 ⎟ ln ⎜ v 1 ⎟ = − 8.62 × 10 −5 eV / atom .K ⎝ T 2 −T 1 ⎠ ⎝ N v 2 ⎠ QV = 1.10 eV/atom. 21 ⎛ 773 × 1173 ⎞ ⎛ 3.59 × 10 ⎞ ⎟ ⎜ ⎟ ln ⎜ 24 ⎝ 1173 − 773 ⎠ ⎝ 10 ⎠ 8 Problem # 5 (10 Points) Two metallic elements A and B having densities of 0.534 g/cm3 and 2.71 g/cm3 respectively are available. Compute the concentration of element B (in wt.%) that is required to develop an alloy with a density of 2.47 g/cm3. Solution To compute the concentration of elemnt A (in wt%) that, when added to element B, will yield a density of 2.47 g/cm3 we use the average density: V T otal = V A +V B ρ ave = mass M M = = total volume V Total V A +V B ⇒ Assume 100g of material. MA + MB = 100g ρ AV E = MB ρB + 100 g 100 g − M B ρA 100 ρ B ( ρ A − ρ AV E ) ρ AV E ( ρ A − ρ B ) 100 g × 2.71g / cm 3 × (0.534 g / cm 3 − 2.47 g / cm 3 ) MB = = 97.62 g 3 3 3 2.47 g / cm (0.534 g / cm − 2.71g / cm ) MB = CB = 97.62 g = 0.9762 or 100 g 97.62wt %