FINAL_EXAM_SOLUTIONS_Fall2007

FINAL_EXAM_SOLUTIONS_Fall2007 - MECT 3345 FINAL EXAMINATION...

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Unformatted text preview: MECT 3345 FINAL EXAMINATION 12/10/07 PRINT NAME: ______________SOLUTIONS__________________________________ Instructions: Only pen, calculator and these sheets allowed on desk. Time limit 3 hours. Attempt to solve all problems. 1 CONSTANTS: Avogadro’s Number: N A = 6.022 x 10 23 Boltzmann Constant: k = 1.38 x 10-23 J/K = 8.617 x 10-5 eV/K Universal Gas Constant: R = 8.314 J/mol K Temperature Conversion: T [K] ≈ T[ o C] + 273 Problem 1 (10 points) The total potential energy between two particular adjacent ions turns out to be of the form: ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + − = 3 2 1 C r exp C r C E TOTAL Where C 1 , C 2 and C 3 are constants and r is the interionic separation. a.) At equilibrium separation, find an expression for C 1 in terms of C 2 and C 3 and equilibrium interionic separation r o . b.) Derive an expression for total potential energy as a function of interionic separation and C 2 , C 3 , r o . a.) The derivative of total energy with respect to interionic separation is zero at equilibrium separation r o . ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = = 3 3 2 2 1 3 3 2 2 1 exp exp C r C r C C C r C C r C dr dE o o o o ro r TOTAL b.) Use the expression derived in a.) to eliminate C 1 from the expression for total energy. That will give you an expression for total energy in terms of C 2 , C 3 and r o . ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − + − = 3 2 3 3 2 2 3 2 1 exp exp exp C r C C r r C r C C r C r C E o o TOTAL 2 Problem 2 (20 points) The density of potassium at room temperature is 0.862g cm-3 , and the atomic weight is 39.1g/mol. a.) How many atoms are there in one cubic centimeter of solid potassium at room temperature? b.) Potassium has body-centered cubic (BCC) crystal structures. Use this fact, together with your answer from part a) to determine the lattice parameter of solid potassium at room temperature. If you did not get the answer to part a), use an atomic density of 2 × 10 22 atoms/cm 3 c.) Use your answer from part (b) to calculate the radius of potassium atom. If you did not get an answer to part (b), use a lattice parameter of 0.4nm. d.) Calculate the atomic packing factor (APF) for Potassium. a.) Density V m = ρ . Mass in one cubic centimeter m= ρ V=(0.862 g/cm 3 )x(1cm 3 )=0.862 g The corresponding number of moles contained in 0.862g is mol mol g g w m N 022 . / 1 . 39 862 . = = = Number of atoms in one mole N A = 6.022 x 10 23 . Therefore, number of atoms in 0.022 mol is atoms x mol x mol atoms x N N n A 22 23 10 325 . 1 022 . 10 022 . 6 = = = b.) BCC – 2 atoms per unit cell. Volume of unit cell V=a 3 . Density of unit cell: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × × × ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = = = unitcell g itcell volumeofun a mol atoms N mol g a unitcell atoms V m A 23 3 3 10 022 . 6 1 . 39 2 1 1 1 . 39 / 2 ρ where 39.1g/mol / N A atoms/mol gives you the weight per potassium atom....
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This test prep was uploaded on 04/15/2008 for the course MECE 3345 taught by Professor Goran during the Spring '08 term at University of Houston.

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FINAL_EXAM_SOLUTIONS_Fall2007 - MECT 3345 FINAL EXAMINATION...

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