Problem I
Figure (a):
The equation of the plane is
1
3
(111)
333
xyz
Figure (b):
The equation of the plane is
1
2
3
3
6
(233)
322
Problem II
There are twelve equivalent directions:
[110],[110],[110],[110],[101],[101],[101],[101],[011],[011],[011],[011]
Problem III
As shown in Figure 19, there are
11
224
24
4
atoms if the cube is cleaved
along a (110) plane. (Four on the corners, two on the edges, and two inside the
rectangle.)
It is easy to calculate the area of the rectangle:
2
2
Sa
Therefore, the area density is:
14
2
2
2
8
44
9.59 10
2
25
.
4
3
1
0
nc
a
cm
m
Problem IV
As shown in Figure 19.
(a) On (100) plane, there are 4 atoms on the corners of the square, and one on the
center.
The area density on the (100) surface:
14
2
2
2
8
1
14
2
4
6.78 10
5.43 10
a
cm
m
(b) The crystal structure of InP is zincblende. Therefore, given the lattice constant a,
the nearest atoms of the same kind have the distance:
8
8
.
8
71
0
2/
2
4
.
1
51
0
2
In In
cm
da
c
m
P.S: For nearest neighboring P, the answer should be the same.
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 Fall '08
 Banjeree
 Gallium arsenide, Aluminium arsenide, area density

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