Assignment # 1
1) Consider planes with Miller indices of (203) and (235) in a cubic crystal
structures. Where are the intercepts of each plane with the crystal axes?
Ans:
For Miller Indices (203):
Reciprocal (
, ∞,
)
(LCD=6 ) (
)
Intercepts (3,
)
For Miller Indices (235):
Reciprocal (
(LCD=30)
, ,
)
(
Intercepts (15, 10, 6)
2
)
Ans:
Area of a equilateral triangle with length
Number of atoms in lattice plane =
=
=2
Atomic surface density =
=
=
=
atoms/ cm2
3) Find Miller indices for the plane in the following figure. Pay attention that the
plane is not intercepting the b-axis.
Ans:
Plane intercepts are (2,
Reciprocal
(
)
,
,
)
(L.C.D=2)
( ,0 , )
Therefore Miller indices (1, 0, 1)
4) a) Determine the number of electrons per unit of area for silicon on the (100) and
(110) planes.
b) The number of Gallium valence electrons in a GaAs unit cell. (Assume lattice
constant = 5.4 angstroms for both unit cells).
Ans:
For (100) plane:
# of atoms in unit cell = 2
# of electrons in Si atom = 14
Unit area = a2 = (5.4 x 10-8 )2 cm2
# of electrons per unit area for Si =
=
ecm-2
= (0.96x 1016) e cm-2
For (110) plane:
# of atoms in unit cell = 4
# of electrons in Si atom = 14
Unit area = a2
=[ ( 5.4 x 10-8 )2
] cm2
# of electrons per unit area for Si =
=
ecm-2
= (1.358x 1016) e cm-2
b)
In GaAs a unite cell has 8 atoms where 4 atoms of Ga and 4 atoms of As.
Ga has 3 valence electrons
Therefore the total number of gallium valence electrons in a GaAs unit cell = 3x4
=12 valence electrons
5) Consider a three-dimensional cubic lattice with a lattice constant equal to a. (a)
Sketch the following planes: (i) (100), (ii) (110), (ii) (310), (iv) (230). (b) Sketch the
following directions: (i) [100], (ii) [110],
(iii) [310], and (iv) [230].
Ans:
a)
i)
ii)
iii)
(310)
b) i) (100)
ii) (110)
x
iii) (310)
iv) (230)
16
-3
6) (a) Phosphorus atoms, at a concentration of 5x10 cm , are added to a pure
sample of silicon. Assume the phosphorus atoms are distributed homogeneously
throughout the silicon. What is the fraction by weight of phosphorus? (b) If boron
18
-3
atoms, at a concentration of 10 cm , are added to the material in part (a),
determine the fraction by weight of boron.
Ans:
Concentration of Si is ( 5x1022) cm-3 and atomic weight is 28.09
Atomic weight of Phosphorous is 30.98
Thus fraction by weight of Phosphorous =
= 1.1x 10-6
Atomic weight of Boron is 10.82
Thus fraction by weight of Boron =
= 7.704 x 10-6