FinalExam2009-sol

FinalExam2009-sol - EMSE 312 Final Exam – Solutions to...

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Unformatted text preview: EMSE 312 Final Exam – Solutions to selected problems FALL 2009 KPDL: 12-15-09 1 1. Crystallography: Explain briefly each of the following. Use a sketch if convenient. (a) (2 p) A 6 3 screw axis (b) (2 p) The difference between a d glide plane and a n glide plane (c) (2 p) The seven (7) different crystal systems (d) (2 p) The m3 point group (e) (2 p) The relationship between a “Point Group” and a “Bravais Lattice” (10 p) (a) Six-fold rotation followed by a translation along the rotation axis. The total translation after 6 rotation operations corresponds to three (3) unit cell translations along the rotation axis. 3/6 1/ 2 = (b) A d glide plane is a “diamond” glide plane and has a partial translation of , or / 4 / 4 a b + / 4 / 4 a c + , or in the symmetry plane whereas a n glide plane is a “diagonal” glide plane and has a partial translation of , or , or / 4 / 4 b c + / 2 / 2 a b + / 2 / 2 a c + / 2 / 2 b c + in the symmetry plane. In each case, the translation vector [ ] , , u v w and the symmetry plane are orthogonal, i.e., ( , , h k l ) h u k v l w ⋅ + ⋅ + ⋅ = (c) The seven crystal system represent the seven different combinations of lattice vectors needed to have unit cells compatible with each of the 32 Point Groups; i.e. 1) Triclinic- no symmetry: ; a b c α β γ ≠ ≠ ≠ ≠ 2) Monoclinic (1 st setting) – one 2-fold axis: ; 90 a b c α β γ ≠ ≠ = = ≠ 3) Orthorhombic – three 2-fold axes: ; 90 a b c α β γ ≠ ≠ = = = 4) Tetragonal – one 4-fold axis: ; 90 a b c α β γ = ≠ = = = 5) Cubic – four 3-fold axes: ; 90 a b c α β γ = = = = = 6) Trigonal – one 3-fold axis: ; 90 a b c α β γ = = = = ≠ 7) Hexagonal – one 6-fold axis: ; 90 ; 120 a b c α β γ = ≠ = = = (d) The m3 point group is a combination of a mirror plane ( ) m and a three fold rotation axis . When the 3- fold rotation axis lies in the mirror plane, the point group is usually called . The point group is in fact a cubic point group in which the 3-fold rotation axis makes an angle to the mirror plane. The locations of the symmetry elements for the two cases are shown in the figures below. ( ) 3 3 m 3 m 54.7 ∼ ( Please note: There is no point deduction if you have described the point group 3 rather than ). m 3 m (e) The Point Group, being a specific combination of Point Symmetries, defines the crystal system. The Bravais Lattice represents the possible combinations of a crystal system. The primitive Bravais Lattice has one lattice point per cell whereas a non-primitive Bravais Lattice has more than one lattice point per cell but is still compatible with the Point Symmetry of the Point Group. 2. (20 p) Optical Diffraction EMSE 312 Final Exam – Solutions to selected problems FALL 2009 KPDL: 12-15-09 2 (a) (5 p) Draw the most intense features of an optical diffractogram that would be obtained from one of the square apertures shown below....
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This note was uploaded on 12/01/2011 for the course EMSE 312 taught by Professor Lagerlof,p during the Spring '08 term at Case Western.

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FinalExam2009-sol - EMSE 312 Final Exam – Solutions to...

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