EE 338_HW2_Solutions

EE 338_HW2_Solutions - EE 338 Fall 2007 HW#2 Solutions 1)...

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EE 338 Fall 2007 HW#2 Solutions 1) Zincblende structure is an FCC with 2 atoms per site, where the second atom is displaced by a vector (x, y, z) = (¼, ¼, ¼). Since the molecular structure is GaAs, we should have equal number of Ga and As atoms in the crystal. # of atoms in FCC unit cell: n = x8 + ½x6 = 4 Volume of unit cell: v = a 0 3 = (0.565x10 -9 ) 3 = 1.8x10 -28 m 3 # of atoms per cm 3 (Ga or As): 22 6 10 22 . 2 10 × = × = v n N cm -3 Diamond structure is very similar to zincblend, but instead of having 2 different atoms at the same site it has 2 identical Ge atoms. In which case, the number of atoms per cm 3 is doubled. # of atoms per cm 3 (Ge): 22 6 10 44 . 4 10 2 × = × = v n N cm -3 2) Figure 1 - Tetrahedral bond structure and its location in a diamond lattice Hence, a 0 /2 a 0 2/2 8 3 0 a r = . Then nearest atom distance becomes 2r = 2.35Å. Mass density: d = N x m Ge =5.38gr/cm 3 1
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3) Here, L = 2x(r A + r B ) = 0.78nm and a 0 nm a L a 45 . 0 3 0 0 = = Also the number density for both A and B
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This note was uploaded on 02/27/2008 for the course EE 338 taught by Professor Dapkus during the Fall '07 term at USC.

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EE 338_HW2_Solutions - EE 338 Fall 2007 HW#2 Solutions 1)...

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