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Unformatted text preview: Molecular Dynamics Simulations 2011 Exercise 2: Suggested solutions Eero Holmstr¨ om September 21, 2011 1. Let’s first construct the orthorhombic unit cell of the HCP structure with side vectors a ′ 1 , a ′ 2 , and a ′ 3 . From Fig. 1 we get the unit vectors in the xyplane. Let a = a 1 = a 2 . We see that a ′ 1 = a tan π/ 3 = √ 3 a (1) and trivially that a ′ 2 = a . Also, a ′ 3 = a 3 = radicalbig 8 / 3 a . Now, to make the (111) planes of the HCP lattice and the FCC lattice match, we set a to the nearestneighbor distance of FCC, a = a / √ 2, where a is the side of the conventional FCC unit cell. Finally we have a ′ 1 = radicalbig 3 / 2 a , a ′ 2 = a / √ 2, and a ′ 3 = 2 a / √ 3. We still need the basis, i.e. locations of the atoms in the unit cell. In the units of the othorhombic unit cell given above they are (000) , ( 1 2 1 2 0) , (0 1 3 1 2 ) , ( 1 2 5 6 1 2 ). Now, from the solutions of the previous exercise, for the FCC lattice with the xyplane in the (111) plane we have b 1 = a / √ 2 , b 2 = radicalbig 3 / 2 a , and b 3 = √ 3 a . The basis here is (000) , ( 1 2 1 2 0) , (0 1 3 1 3 )...
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 Winter '12
 Kotakoski
 Energy, Kinetic Energy, Potential Energy, .........

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