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HW2Solutions

# HW2Solutions - α =(2 π/6 and thus α/2 = π/6 As seen in...

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MSE 102 Fall 2011 Problem Set 2 Solutions James Mastandrea Problem Set 2 Due Date: 9/8/2011 Note: To distinguish between lines and the mirror planes, mirror planes are highlighted with a blue line. 1

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Question 1 Associated lattice points, a set of lattice vectors, and a Wigner-Seitz primitive unit cell. 2
All rotations and mirrors that bring the structure into itself. For ease of viewing the solutions, the rotations and mirrors were drawn on a primitive unit cell. 3

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Question 2 We are told that the given lattice has perpendicular 6-fold axis symmetry and translational symmetry. Thus we can create a periodic array of lattice points, where each lattice point has a 6-fold symmetry axis normal to the lattice. Therefore we can create the following 2D lattice that has 6-fold symmetry on every lattice point. We can also prove to ourselves that this is indeed the case by using Buerger’s construction to find other locations of 6-fold symmetry. For 6-fold symmetry,
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Unformatted text preview: α = (2 π )/6 and thus α /2 = π /6. As seen in the ﬁgure above, the location that was found from Buerger’s construction, location 4, is the location of a 6-fold symmetry axis, and this location is already a lattice point. A 6-fold axis of rotation inherently has 3 and 2-fold symmetry. Using Buerger’s construction we can ﬁnd the location of the 3-fold symmetry axis. This is displayed below. For 3-fold symmetry α = (2 π )/3 and thus α /2 = π /3. 4 As mentioned above, 6-fold symmetry has 2-fold symmetry as well. The location of the 2-fold axis was found using Buerger’s construction; see the ﬁgure below. For 2-fold symmetry α = (2 π )/2 and thus α /2 = π /2. Using the information obtained throughout this process, one can display all the symmetry operations on a primitive unit cell for a lattice that has perpendicular 6-fold axis symmetry and translational symmetry. 5...
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HW2Solutions - α =(2 π/6 and thus α/2 = π/6 As seen in...

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