EEE436/591 Fundamentals of Solid State Devices, Spring 2023
Homework Assignment 1 Solutions, 80 pts
Problems 1.11, 3.12, 4.45, Special Problem
Problem 1.11 (20). The crystal structure of sodium chloride (NaCl) is a simple cubic
with the Na and Cl atoms alternating positions. Each Na atom is then surrounded by six
Cl atoms and likewise each Cl atom is surrounded by six Na atoms. (a) Sketch the atoms
in a (100) plane. (b) Assume the atoms are hard spheres with nearest neighbors touching.
The effective radius of Na is 1.0 Å and the effective radius of Cl is 1.8 Å. Determine the
lattice constant. (c) Calculate the volume density of Na and Cl atoms. (d) Calculate the
mass density of NaCl.
Solution:
(a)
(b) In terms of the lattice constant of the periodic Rocksalt structure, it is an FCC
structure (not simple cubic) with a two atom basis as can be seen by the location of the
Cl atoms in the figure above. If one just considers the distance between the center of
Na atoms and center of the Cl atoms, the interatomic spacing is 𝑑 = 1 + 1.8 = 2.8 Å .
The lattice constant of the FCC cubic unit cell is the minimum distance between the
center of the Cl atoms along the horizontal axis above. The radius of the Na atoms is
sufficiently large that they bridge the Cl atoms along the horizontal, so that the lattice
constant of the cubic unit cell is 𝑎 = 1.8 + 2 × 1.0 + 1.8 = 5.6 Å.
Alternately, one can look at a smaller cell shown in red above, with lattice constant of
half this, 2.8 Å, although this is not a true unit cell as the basis vector does not go from
the same lattice point to identical lattice points. However, it does capture the density
of atoms in the NaCl lattice, and gives the correct volume and mass density.
1
(c) Na: Density =
(1 / 2)
(2.8 ´10 )
-8 3
(just looking at the cube formed by Na and Cl atoms)
= 2.28 ´10 22 cm -3
Cl: Density = 2.28 ´10 22 cm -3
!
!
If you consider the FCC lattice formed by Na ions, there are 6 × " + 8 × # = 4 atoms
$
per unit cubic FCC cell, and Density=(&.(×!*!" )# = 2.28 × 10"" 𝑐𝑚,- , the same.
(d) Na: At. Wt. = 22.99
Cl: At. Wt. = 35.45
So, mass per unit cell
æ1ö
æ1ö
ç ÷(22.99) + ç ÷(35.45)
è2ø
è2ø
=
= 4.85 ´10 - 23
6.02 ´10 23
Then mass density
r=
4.85 ´10 -23
(2.8 ´10 )
-8 3
= 2.21 grams/cm 3
2
Problem 3.12 (20). The bandgap energy in a semiconductor is usually a slight function of
temperature. In some cases, the bandgap energy versus temperature can be modeled by
𝛼𝑇 "
𝐸. = 𝐸. (0) −
(𝛽 + 𝑇)
where 𝐸. (0) is the value of the bandgap energy at T=0 K. For silicon, the parameter values
are 𝐸. (0) = 1.170 𝑒𝑉, 𝛼 = 4.73 × 10,$ eV/K, and 𝛽 = 636 K. Plot 𝐸. versus T over the
range 0 ≤ 𝑇 ≤ 600 K. In particular, note the value of T=300 K.
Solution:
For T = 100 K,
E g = 1.170 -
(4.73´10 )(100)
-4
636 + 100
E g = 1.164 eV
T = 200 K,
E g = 1.147 eV
T = 300 K,
E g = 1.125 eV
T = 400 K,
E g = 1.097 eV
T = 500 K,
E g = 1.066 eV
T = 600 K,
E g = 1.032 eV
2
Þ
1.2
1.15
1.1
Bandgap (eV)
1.05
1
0.95
100
200
300
400
Temperature K
3
500
600
Problem 4.45 (20). A particular semiconductor material is doped at 𝑁/ = 2 × 10!$ 𝑐𝑚,and 𝑁0 = 1.2 × 10!$ 𝑐𝑚,- . The thermal equilibrium electron concentration is found to
be 𝑛1 = 1.1 × 10!$ 𝑐𝑚,- . Assuming complete ionization, determine the intrinsic carrier
concentration and the thermal equilibrium hole concentration.
Solution:
no =
Nd - Na
æ N - Na
+ çç d
2
2
è
1.1´1014 =
(1.1´10
2
ö
÷÷ + n i2
ø
2 ´1014 - 1.2 ´1014
2
æ 2 ´ 1014 - 1.2 ´ 1014
+ çç
2
è
14
- 4 ´ 1013
) = (4 ´10 )
2
13 2
2
ö
÷ + ni2
÷
ø
+ n i2
4.9 ´10 27 = 1.6 ´10 27 + ni2
so ni = 5.74 ´1013 cm -3
po =
n i2 3.3 ´ 10 27
=
= 3 ´ 1013 cm -3
n o 1.1´ 1014
4
Special problem (20). The lattice constant of Si is 5.43 Å. Calculate the density of
surface atoms on the (a) Si (100) surface; (b) the Si (110) surface.
Solution:
(a) The arrangements of atoms on the (100) surface
are as shown to the right. In terms of the two
dimensional lattice of surface atoms, the four corner
atoms are shared with four other unit cells, and so the
total number of atoms per unit cell is 2, one for the 4
corners and 1 in the center by itself.
The area of the unit cell is 𝐴 = (5.43 × 10,# )" =
2.95 × 10,!& 𝑐𝑚"
"
Therefore, the density of surface atoms is 2 = 6.78 × 10!$ 𝑎𝑡𝑜𝑚𝑠/𝑐𝑚"
(b) For the Si (110), the arrangement is slightly
more complicated as shown to the right. The
corner atoms are shared with 4 other unit cells
contributing 1 atom to the unit cell. The side
atoms are shared with another adjacent unit cell,
and so contribute a total of 1 atom to the unit cell.
The two atoms in the interior contribute 2 more
for a total of 4 atoms per unit surface cell.
The area of the unit cell is 𝐴 = √2(5.43 ×
10,# )" = 4.17 × 10,!& 𝑐𝑚"
$
Therefore, the density of surface atoms is 2 = 9.59 × 10!$ 𝑎𝑡𝑜𝑚𝑠/𝑐𝑚"
(c) For the Si (111), the smallest arrangement of atoms is an equilateral triangle as shown
by the bisection of the unit cell shown in the figure below. The sides of the triangle are
along [110] directions with length √2(5.43 × 10,# ) = 7.68 × 10,# 𝑐𝑚. The area of an
equilateral triangle is √3/4𝑎" , where 𝑎 = 7.68 × 10,# 𝑐𝑚.
The area of the triangle is then 𝐴 = √3/4(7.68 × 10,# )" = 2.55 × 10,!& 𝑐𝑚"
5
Each corner atom is shared with 6 other cells leading to ½ atom per corner atom. The
three side atoms are shared with one other cell so 3/2 atoms per side atom. This leads to
at total of 2 atoms per triangular cell
"
Therefore, the density of surface Ga atoms is 2 = 7.83 × 10!$ 𝑎𝑡𝑜𝑚𝑠/𝑐𝑚"
6