Table Of ContentSemiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1
By D. A. Neamen Exercise Solutions
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Chapter 1
Exercise Solutions TYU 1.2
(a) Number of atoms per (100) lattice plane
Ex 1.1
(a) Number of atoms per unit cell
Surface Density
(b) Volume Density = cm
(b) Number of atoms per (110) lattice plane
cm
_______________________________________
Surface Density
Ex 1.2
Intercepts of plane; p=1, q=2, s=2
Inverse;
cm
(c) Number of atoms per (111) lattice plane
Multiply by lowest common denominator,
plane
_______________________________________
Lattice plane area
Ex 1.3
(a) Number of atoms per (100) plane
where
Surface Density
cm
(b) Number of atoms per (110) plane Then lattice plane area
Surface Density
Surface Density
cm
cm
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Test Your Understanding Solutions
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TYU 1.1
TYU 1.3
Number of atoms per unit cell
(a) For (100) planes, distance
(b) For (110) planes, distance
Volume Density
cm _______________________________________
Radius
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1
By D. A. Neamen Exercise Solutions
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TYU 1.4
(a) 8 corner atoms
(b) 6 face-centered atoms
(c) 4 atoms totally enclosed
_______________________________________
TYU 1.5
Number of atoms in the unit cell
Volume Density
cm
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
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Chapter 1
Problem Solutions
Then
1.1
(a) fcc: 8 corner atoms atom
6 face atoms atoms
Ratio
Total of 4 atoms per unit cell
(b) bcc: 8 corner atoms atom
1 enclosed atom =1 atom
(d) Diamond lattice
Total of 2 atoms per unit cell
(c) Diamond: 8 corner atoms atom Body diagonal
6 face atoms atoms
4 enclosed atoms = 4 atoms
Unit cell vol
Total of 8 atoms per unit cell
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8 atoms per cell, so atom vol
1.2
(a) Simple cubic lattice:
Then
Unit cell vol
1 atom per cell, so atom vol
Ratio
Then
_______________________________________
Ratio
1.3
(b) Face-centered cubic lattice
(a) ; From Problem 1.2d,
Unit cell vol
Then
4 atoms per cell, so atom vol
Center of one silicon atom to center of
Then
nearest neighbor
(b) Number density
Ratio
cm
(c) Body-centered cubic lattice
(c) Mass density
Unit cell vol grams/cm
_______________________________________
2 atoms per cell, so atom vol
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
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(a)
1.4
(a) 4 Ga atoms per unit cell (b)
Number density
(c) A-atoms: # of atoms
Density of Ga atoms cm
Density
4 As atoms per unit cell
Density of As atoms cm
cm
(b) 8 Ge atoms per unit cell
B-atoms: # of atoms
Number density
Density
Density of Ge atoms cm
_______________________________________
cm
1.5 _______________________________________
From Figure 1.15
1.9
(a)
(a)
# of atoms
(b) Number density
cm
_______________________________________
Mass density
1.6
gm/cm
(b)
_______________________________________
# of atoms
1.7
(a) Simple cubic:
Number density
(b) fcc:
cm
(c) bcc:
Mass density
(d) diamond: gm/cm
_______________________________________
_______________________________________
1.8
1.10
From Problem 1.2, percent volume of fcc
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
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atoms is 74%; Therefore after coffee is
Surface density cm
ground,
Volume = 0.74 cm For 1.12(a) and (b), Same material
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1.11
(b)
(b) For 1.12(a), A-atoms;
(c) Na: Density
Surface density
cm cm
Cl: Density cm
B-atoms;
(d) Na: At. Wt. = 22.99
Surface density
Cl: At. Wt. = 35.45
So, mass per unit cell
cm
For 1.12(b), A-atoms;
Surface density
Then mass density
cm
grams/cm
B-atoms;
_______________________________________ Surface density
cm
1.12
(a) For 1.12(a) and (b), Same material
_______________________________________
Then
Density of A: 1.14
(a) Vol. Density
cm
Density of B: Surface Density
cm (b) Same as (a)
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(b) Same as (a)
1.15
(c) Same material
(i) (110) plane
_______________________________________ (see Figure 1.10(b))
(ii) (111) plane
1.13 (see Figure 1.10(c))
(iii) (220) plane
(a) For 1.12(a), A-atoms
Same as (110) plane and [110] direction
Surface density
(iv) (321) plane
cm
Intercepts of plane at
For 1.12(b), B-atoms:
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________
[321] direction is perpendicular to
(321) plane
_______________________________________ So
1.16
(a)
Area of plane
(b)
cm
Surface density
_______________________________________
cm
1.17
(b) bcc
Intercepts: 2, 4, 3 (i) (100) plane:
Surface density cm
(634) plane
_______________________________________
(ii) (110) plane:
1.18 Surface density
(a)
cm
(b) (iii) (111) plane:
(c) Surface density
_______________________________________ cm
(c) fcc
1.19 (i) (100) plane:
(a) Simple cubic
Surface density cm
(i) (100) plane:
(ii) (110) plane:
Surface density
Surface density
cm
(ii) (110) plane: cm
(iii) (111) plane:
Surface density
cm Surface density
(iii) (111) plane:
cm
Area of plane _______________________________________
where 1.20
(a) (100) plane: - similar to a fcc:
Now
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
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= cm
Surface density
cm
(b) (110) plane: _______________________________________
Surface density 1.22
Density of silicon atoms cm and
cm 4 valence electrons per atom, so
Density of valence electrons cm
_______________________________________
(c) (111) plane:
1.23
Density of GaAs atoms
Surface density
cm
cm
_______________________________________ An average of 4 valence electrons per atom,
So
1.21 Density of valence electrons
cm
_______________________________________
1.24
(a) #/cm
(a)
cm
(b)
(b) #/cm _______________________________________
1.25
cm (a) Fraction by weight
(c)
(b) Fraction by weight
(d) # of atoms
Area of plane: (see Problem 1.19)
_______________________________________
1.26
Volume density cm
Area
So cm
cm We have
Then
#/cm _______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
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1.27
Volume density cm
So cm
We have
Then
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Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2
By D. A. Neamen Exercise Solutions
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Chapter 2
Exercise Solutions
or eV
Ex. 2.1
Then
(a)
eV, eV, eV
J
(b)
or eV
(b)
J
J
or
or eV
eV
_______________________________________
Then eV
Ex 2.2 eV
(a)
eV
_______________________________________
kg-m/s
m
Ex 2.4
or
(c)
J
kg-m/s Now
Set
Then
= J
or eV
_______________________________________
Ex 2.3 or
m
(a)
(a) m
J or
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2
By D. A. Neamen Exercise Solutions
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% kg-m/s
(b) m
(b)
or
%
_______________________________________
J or eV
_______________________________________
Ex 2.5
(a)
m
Then
TYU 2.2
(a) eV
s
(b)
(b) Same as part (a), s
_______________________________________
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TYU 2.3
(a)
Ex 2.6
From Example 2.6, we have
eV
= m
meV, meV,
meV
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Test Your Understanding
TYU 2.1
(b)
(a)
