sol1 - ECE 440 HW1 Solutions Summer 2009 Thu, Jun 18, 2009...

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ECE 440 HW1 Solutions Summer 2009 Thu, Jun 18, 2009 1. Beginning with a sketch of a fcc lattice, add atoms at (1/4, 1/4, 1/4) from each fcc atom to obtain the diamond lattice. Show that only four added atoms in Fig. 1- 8a appear in the diamond unit cell. How many nearest-neighbor atoms does each atom have in a diamond structure? 2. Assuming that the lattice constant varies linearly with the composition, determine the unit cell side length for a silicon-germanium alloy (in diamond structure) which has an atomic density of 4.6x10 22 atoms/cm 3 . Also, find the composition of the alloy and its mass density. Diamond solids have eight atoms per unit cell. From the atomic density, one can find the lattice constant, a, of the alloy. (1) 4.6x10 22 atoms/cm 3 = 3 8 a [atoms/cm 3 ] (2) a = 5.583 x 10 -8 [cm] The lattice constants of both silicon and germanium are known quantities (see Appendix III in Streetman and Banerjee). The assumption made in this problem is that the relationship between 1
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sol1 - ECE 440 HW1 Solutions Summer 2009 Thu, Jun 18, 2009...

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