ENE 311
Homework # 2
Due: 12 Dec 2008
1.
An electron confined by an infinite one-dimensional potential well may be anywhere within the interval 2a. So the uncertainty in its position is x 2 a . There must be a corresponding uncertainty in the momentum of the electron and hence it must have a certain kinetic energy. Calculate this energy from the uncertainty relationship and compare it with the value obtained from the last example in lecture 2 for the ground state.
Ans: E h 2

32 ma 2
2.
An electron being trapped at the ground state in a 1-D infinite potential well of width L =
1 nm can be excited to higher energy states by absorbing energy from incident photons.
Calculate the wavelengths of the two lowest-energy photons capable of exciting electrons from the ground state. Ans: 1.1 μm, 0.414 μm
3.
Calculate the atom surface density for the (001) surface of GaAs semiconductor. The lattice constant for GaAs is 5.65 Å. Ans: 6.26 x 10
14
atoms/cm
2
4.
Calculate the density of GaAs (the lattice constant of GaAs is 5.65 Å and the atomic weights of Ga and As are 69.72 and 74.92 g/mole, respectively). Ans: 5.33 g/cm 3
5.
Find the number of atoms per square centimeter (surface density) in Si in the (100) and
(110) planes. Ans: 6.78 x 10 14 atoms/cm 2 , 9.6 x 10 15 atoms/cm 2
6.
Find the packing factor (P.F.) of the unit cell volume which can be filled by identical hard spheres in diamond lattices. Ans: 34%

Solution
Homework # 2
1.
Uncertainty principle:
 p
 x

h
 p

h/
 x = h/2a
That means p varies from zero to h/2a.
P avg
= h/(2.
 x) = h/(4a)
2.
Kinetic energy =
1
2 mv 2 avg

1
2
2
P avg m h 2
Kinetic energy =
32ma
2

2 m
2 h

12 h

13
E
4
E
3
E
2
E
1
E n

2 2 n h
8 mL
2
E
2

E
1
 h

12
 hc

12
4 h
2 h
2

8 mL
2
8 mL
2

12

 hc

12
 8    
31  

34

9 2

E

3
13


E
1

9 h
2 h
2 h
2
 
8
2
8
2 mL mL mL
2
 8   
31  

34

9 2


3.
Number of atoms for GaAs (001) surface = (4 x ¼) + 1 = 2 atoms.
Surface density = number of atoms a
2

2

8 2
(5.65 10 )
  14
6.265 10 atoms/cm
2
4.
The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and As are 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and four arsenic atoms per unit cell, therefore
4/a
3
= 4/ (5.65 × 10
-8
)
3
= 2.22 × 10
22
Ga or As atoms/cm
2
,
Density = (no. of atoms/cm
3 × atomic weight) / Avogadro constant
= 2.22 × 10
22 (69.72 + 74.92) / 6.02 × 10 23
= 5.33 g / cm
3
.
5.
For the (100) plane, there are two atoms (one central atom and 4 corner atoms each contributing 1/4 of an atom for a total of two atoms) for an area of a
2
, therefore we have
2/ a
2
= 2/ (5.43 × 10
-8
)
2
= 6.78 × 10
14
atoms / cm
2
Similarly we have for (110) plane
(2 + 2 ×1/2 + 4 ×1/4) /
2 a
2 = 9.6 × 10 15 atoms / cm 2 ,
6.
For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a total of three spheres, and

4 spheres inside the cell. The diagonal distance between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in the figure above is
D =
1
2

 a
2


2


 a
2


2


 a
2


2 a
= 3
4 a
The radius of the sphere is D/2 = 3
8
∴ Maximum fraction of cell filled
= (1 + 3 + 4)



4

3 

 a
8
3





 3
/ a
3
= π 3 / 16 = 34 %.
This is a relatively low percentage compared to other lattice structures.