Engr Materials Science and Engineering TEST 4  Sample Solution


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1 Engr Materials Science and Engineering TEST 4  Sample Solution Part : Solve each of the following problems completely. 1. In this problem we are asked to show that the minimum cationtoanion radius ratio for a coordination number of four is 0.5. If lines are drawn from the centers of the anions, then a tetrahedron is formed. The tetrahedron may be inscribed within a cube as shown below. The spheres at the apexes of the tetrahedron are drawn at the corners of the cube, and designated as positions A, B, C, and D. (These are reduced in size for the sake of clarity.) The cation resides at the center of the cube, which is designated as point E. Let us now express the cation and anion radii in terms of the cube edge length, designated as a. The spheres located at positions A and B touch each other along the bottom face diagonal. Thus, AB r A But (AB) a + a a or AB a r A And a r A There will also be an anion located at the corner, point F (not drawn), and the cube diagonal AEF will be related to the ionic radii as AEF r A + r C ( ) (The line AEF has not been drawn to avoid confusion.) From the triangle ABF (AB) + (FB) (AEF) But,
2 FB a r A and from above. Thus, ( ) + r A r A AB r A [ ( r A + r C )] Solving for the r C /r A ratio leads to r C r A A hypothetical AX type or ceramic material is known to have a density of 3.55 g/cm 3, and a unit cell of cubic symmetry with a cell edge length of nm. The atomic weights of the A and X elements are 60. and 16.0 g/mol, respectively. On the basis of this information, which of the following crystal structures would you expect this ceramic compound to have: sodium chloride, cesium chloride, zinc blende, diamond cubic, fluorite, and perovskite? Justify your choice(s). (See table 1.4) Answer: None of the suggested crystal structure (n 3.0 formula units/units cell) Using Equation 1.1 and solving for n ρvc N A n' A + A C (3.55)(4.75 n ' A 8 ( ) n 3.0 formula units/unit cell. 3 ) (6.03 For the AX type ceramics, sodium chloride and zinc blende have n 4, while cesium chloride has n 1. Therefore, none of the suggested crystal structures would work. 3 ) 3. Consider two hypothetical ions A + and X  which have radii of 0.15 and nm, respectively. (See Table 1.) a. What is the coordination number for each A + in the compound A X? Answer: 8 b. Assuming that the crystal structure for this compound belongs to the cubic crystal system, describe or draw a unit cell for A X. Answer: A variant of fluorite crystal structure. A unit cell would have eight interconnected cubes with X  ions positioned at all cube corners; half of the cube centers would be occupied by A + ions. r + A 0.15 a) Compute the cationanion radius ratio: r X The coordination number is eight (Table 1.).
3 b) The crystal structure would be a variation of the fluorite crystal structure (Figure 1.5). A unit cell would have eight interconnected cubes with X  ions positioned at all cube corners; half of the cube centers would be occupied by A + ions. 4. The energy gap of Germanium is 0.66 ev. The electron mobility if m /Vs. The hole mobility is m /Vs. The intrinsic hole concentration at room temperature (300 K) is.3 19 holes/m 3. a) Calculate the electrical conductivity of germanium (intrinsic) at room temperature. b) Calculate the electrical conductivity of germanium (intrinsic) at 0 C. ( e C, and the k ev/k) c) Germanium is to be doped with Antimony to produce an extrinsic ntype semiconductor. What atomic density (in atoms/m 3 ) is required to yield a conductivity of 4 (Ωm) 1. (Neglect the conductivity of intrinsic germanium.) Answers: a).04 (Ωm) 1 ; b) 4.9 (Ωm) 1 a) σ n e ( μ + μ ) e h b) σ ( ( Ωm) 1 )(1.6 E g 1 19 σ 1 exp σ 1 k T T1 ln σ 0.66 σ ( ) σ 4.9 (Ωm) 1 )( ) Calculate the resistance of a 150m long, 3.1mm diameter copper wire at an ambient temperature of 0 C. The electrical conductivity of the wire is (Ωm) 1. How much current will result if a voltage of V is applied at the ends of the wire? Ans: 0.33 Ω. L 1 L R ρ A σ A R : 0.33 Ω. I V/R / A 6. Is it possible to have a random poly(ethylenepropylene) copolymer that has number and weightaverage degrees of polymerization of 500 and 000, respectively, and number and weightaverage molecular weights of 8,500 and 67,00 g/mol? Why or why not? Answer: no. Mer molecular weights computed from weightaverage and numberaverage data are different 7. Molecular weight data from some polymer are tabulated below. Compute the (a) numberaverage molecular weight, and (b) the weightaverage molecular weight. (c) If it is known
4 that the material s numberaverage degree of polymerization is 477, which one of the polymers listed in Table 14.3 is this polymer? Why? Molecular wt.range x i 8,0000, ,0003, ,00044, ,00056, ,00068, ,00080, ,0009, (a) From the tabulated data, we are asked to compute M n, the numberaverage molecular weight. This is carried out below. Molecular wt. Range Mean M i x i xm i i 8,0000,000 14, ,0003,000 6, ,00044,000 38, ,00056,000 50, ,000 56,00068,000 6, ,160 68,00080,000 74, ,0009,000 86, M n x i M i 47,70 g/mol (b) From the tabulated data, we are asked to compute M w, the weight average molecular weight. This determination is performed as follows: Molecular wt. Range Mean M i w i wm i i 8,0000,000 14, ,0003,000 6, ,00044,000 38, ,00056,000 50, ,500 56,00068,000 6, ,60
5 68,00080,000 74, ,840 80,0009,000 86, M w w i M i 53,70 g/mol (c) We are now asked if the numberaverage degree of polymerization is 477, which of the polymers in Table 14.3 is this material? It is necessary to compute m in Equation (14.4a) as m M n 47,70 g/mol n 477 n 0.04 g/mol The mer molecular weights of the polymers listed in Table 14.3 are as follows: Polyethylene g/mol Polyvinyl chloride g/mol Polytetrafluoroethylene0.0 g/mol Polypropylene g/mol Polystyrene g/mol Polymethyl methacrylate g/mol Phenolformaldehyde g/mol Nylon 6, g/mol PET g/mol Polycarbonate g/mol Therefore, polytetrafluoroethylene is the material since its mer molecular weight is closest to that calculated above.
For the M state, n = 3, and 18 states are possible. Possible l values are 0, 1, and 2; possible m l. values are 0, ±1, and ±2; and possible m s
.6 Allowed values for the quantum numbers of electrons are as follows: n 1,,,... l 0, 1,,,..., n 1 m l 0, ±1, ±, ±,..., ±l m s ± 1 The relationships between n and the shell designations are noted in Table.1.
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