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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 nearestneighbor 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 silicongermanium 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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