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Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
Second Edition ( 2001 McGraw-Hill)
Chapter 1
1.2 Ionic bonding and NaCl
The interaction energy between Na+ and Cl- ions in the NaCl crystal can be written as
4.03 × 10
E(r) = −
r
−28
6.97 × 10
+
r8
−96
where the energy is given in joules per ion pair, and the interionic separation r is in meters. Calculate
the binding energy and the equilibrium ionic separation in the crystal; include the energy involved in
electron transfer from Cl- to Na+. Also estimate the elastic modulus Y of NaCl given that
Y≈
2
1 d E
6ro  dr2 
r= ro
Solution
The PE curve for NaCl is given by
E(r) = −
4.03 ×10−28 6.97 × 10−96
+
r
r8
where E is the potential energy and r represents interionic separation.
We can differentiate this and set it to zero to find the minimum PE, and consequently the
minimum (equilibrium) separation (ro).
dE(ro )
=0
dro
1
− 55.76
isolating ro:
ro = 2.81 × 10-10 m or 2.81 Å
96
10 ro
9
+ 4.03
1
∴
28
10 ro 2
=0
Let q = 1.602×10-19 J/eV. Then the apparent “ionic binding energy” (Eb) for the ions alone, in
eV, is
E (ro )  − 4.03 × 10−28 6.97 × 10−96  1 
E b (eV) = −
=−
+
q
ro
ro 8

 q 
∴

−4.03× 10−28
6.97 × 10−96 
1

E b (eV) = −
+

8
−19
−10
−10

J/eV 
 (2.81× 10 m) (2.81×10 m)  1.602 ×10
∴
Eb(eV) = 7.83 eV
Note however that this is the energy required to separate the Na+ and Cl- ions in the crystal and
then to take the ions to infinity, that is to break up the crystal into its ions. The actual bond energy
involves taking the NaCl crystal into its constituent neutral Na and Cl atoms. We have to transfer the
1.1
Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
electron from Cl- to Na+. The energy for this transfer, according to Figure 1Q2-1, is -1.5 eV (negative
represents energy release). Thus the bond energy is 7.83 - 1.5 = 6.33 eV as in Figure 1Q2-1.
Potential energy E(r), eV/(ion-pair)
6
–
Cl
1.5 eV
0.28 nm
0
r=
Cl
r=
–6
–6.3
Cl
–
Na
+
r o = 0.28 nm
Na
+
Separation, r
Na
Figure 1Q2-1
Sketch of the potential energy
per ion pair in solid NaCl.
Zero energy corresponds to
neutral Na and Cl atoms
infinitely separated.
If r is defined as a variable representing interionic separation, then Young’s modulus is given
by:
Y=
∴
1  d  dE(r)  
6r  dr  dr  

1
1 
1  −8.06 10 28 r3 + 501.84 10 96 r 10 
1
1
= 8.364 × 10−95 11 − 1.3433× 10−28 4
Y=

6
r
r
r


Substituting the value for equilibrium separation (ro) into this equation (2.81 × 10-10 m),
Y = 7.54 × 1010 Pa = 75 GPa
This value is somewhat larger than about 40 GPa in Table 1.2 (in the textbook), but not too far
out.
Author’s Note: As explained in the textbook’s CD, in “Bonding in Molecules and Solids” , in general,
the bulk modulus K can be found from the PE curve E = E(r) using
1  d2 E 
K=
2
9cro  dr  r= r
o
where c is a constant of the order of unity that depends on the crystal structure. (For example, for the
CsCl, c = 1.5 and NaCl, c = 2.) The elastic modulus Y and the bulk modulus K are related through the
Poisson’s ratio υ,
K = Y/[3(1 − 2υ)]
For many ceramics υ ≈ 1/4 so that Y ≈ (3/2)K and hence
Y=
1  d 2E 
2
6cro  dr  r= r
o
A detailed analysis would make c = 2 for the NaCl structure, and hence would cut the calculate Y = 75
GPa to Y = 37.5 GPa close to the experimental value.
1.2
Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
*1.3 van der Waals bonding
Below 24.5 K, Ne is a crystalline solid with an FCC structure. The interatomic interaction energy per
atom can be written as
6
12

σ 
σ 
E(r) = −2ε 14.45  − 12.13  
r
r


(eV/atom)
where ε and σ are constants that depend on the polarizability, the mean dipole moment, and the
extent of overlap of core electrons. For crystalline Ne, ε = 3.121 × 10-3 eV and σ = 0.274 nm.
a. Show that the equilibrium separation between the atoms in an inert gas crystal is given by ro =
(1.090)σ. What is the equilibrium interatomic separation in the Ne crystal?
b. Find the bonding energy per atom in solid Ne.
c. Calculate the density of solid Ne (atomic mass = 20.18 g/mol).
Solution
a
Let E = potential energy and x = distance variable between the atoms. The energy E is given by
6
12

σ
 σ 
E(x) = −2 ε 14.45  − 12.13  
x
x 

The force F on each atom is given by
11
5

σ
σ 
σ  
σ  

dE(x)
x 
= 2ε 145.56 x2 − 86.7
F(x) = −

x
x2 
dx


∴

σ 12
σ6 
F(x) = 2ε 145.56 13 − 86.7 7 
x
x 

When the atoms are in equilibrium, this net force must be zero. Using ro to denote equilibrium
separation,
F(ro ) = 0
∴

σ 12
σ6 
2ε 145.56 13 − 86.7 7  = 0
ro
ro 

∴
145.56
∴
ro13  145.56  σ 12
=
ro 7  86.7  σ 6
∴
ro = 1.090σ
σ 12
ro13
= 86.7
σ6
ro 7
For the Ne crystal, σ = 2.74 × 10-10 m and ε = 0.003121 eV. Therefore,
ro = 1.090(2.74 × 10-10 m) = 2.99 × 10-10 m for Ne.
1.3
Homework 1 Solution
EE145 Spring 2002
b
Prof. Ali Shakouri
Calculate energy per atom at equilibrium:
6
12

σ 
σ  




E(ro ) = −2ε 14.45
− 12.13


 ro  
r
o

∴
E(ro ) = −2(0.003121 eV)(1.602 ×10
∴
E(ro) = -4.30 × 10-21 J or -0.0269 eV
−19
6


 2.74 ×10 -10 m 


14.45
-10


 2.99 × 10 m 
J/eV)

12
-10

2.74
×10
m


 
−12.13
-10
 
m

2.99
×
10

Therefore the bonding energy in solid Ne is 0.027 eV per atom.
c
To calculate the density, remember that the unit cell is FCC, and density = (mass of atoms in the
unit cell) / (volume of unit cell). There are 4 atoms per FCC unit cell, and the atomic mass of Ne is
20.18 g/mol.
a
2R
a
a
a
Figure 1Q3-1
Left: An FCC unit cell with close-packed
spheres.
Right: Reduced-sphere representation of the
FCC unit cell.
Examples: Ag, Al, Au, Ca, Cu, γ-Fe (>912
°C), Ni, Pd, Pt, Rh.
Since it is an FCC crystal structure, let a = lattice parameter (side of cubic cell) and R = radius of
atom. The shortest interatomic separation is ro = 2R (atoms in contact means nucleus to nucleus
separation is 2R (see Figure 1Q3-1).
R = ro/2
and
2a2 = (4R)2
∴
r
a = 2 2R = 2 2  o  = 2 (2.99 × 10−10 m )
2
∴
a = 4.228 × 10-10 m
Therefore, the volume (V) of the unit cell is:
V = a3 = (4.228 × 10-10 m)3 = 7.558 × 10-29 m3
The mass (m) of 1 Ne atom in grams is the atomic mass (Mat) divided by NA, because NA number
of atoms have a mass of Mat.
m = Mat / NA
∴
m=
(20.18 g/mol)(0.001 kg/g)
23
6.022 × 10 mol
-1
= 3.351 × 10−26 kg
1.4
Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
There are 4 atoms per unit cell in the FCC cell. The density (ρ) can then be found by:
ρ = (4m) / V = [4 × (3.351 × 10-26 kg)] / (7.558 × 10-29 m3)
ρ = 1774 kg/m3
∴
In g/cm3 this density is:
ρ=
1774 kg/m3
3
3 × (1000 g/kg ) = 1.77 g/cm
(100 cm/m )
The density of solid Ne is 1.77 g cm-3.
*1.14 BCC and FCC crystals
a. Consider iron below 912 °C, where its structure is BCC. Given the density of iron as 7.86 g cm-3
and its atomic mass as 55.85 g/mol, calculate the lattice parameter of the unit cell and the radius of
the Fe atom.
b. At 912 °C, iron changes from the BCC (α-Fe) to the FCC (γ-Fe) structure. The radius of the Fe
atom correspondingly changes from 0.1258 nm to 0.1291 nm. Calculate the density of γ-Fe and
explain whether there is a volume expansion or contraction during this phase change.
Solution
Given:
Density of iron at room temperature: ρ = 7.86 × 103 kg/m3
Atomic mass of iron: Mat = 55.85 g/mole
a
For the BCC structure, the density is given by:
Mat × (10 −3 kg/g)
2
NA
ρ=
a3
Thus the lattice parameter a is:
1

1
Mat  3
a=
 (500 g/kg) N A ρ 
1
∴

3
1
(55.85 g/mol)
 = 2.87 × 10-10 m
a=
23
−1
3
3
(
)
6.022
×
10
mol
7.86
×
10
kg/m
500
g/kg
(
)
(
)


The radius of the Fe atom, R, and the lattice parameter, a, are related.
a 3 = 4R
∴
R=
3
3
a=
2.87 ×10−10 m) = 1.24 × 10-10 m
(
4
4
b
Fe has a BCC structure just below 912 °C (α-Fe). An Fe atom in the α-Fe state has a radius of
RBCC = 0.1258 × 10-9 m. The density of α-Fe is therefore:
1.5
EE145 Spring 2002
Homework 1 Solution
Prof. Ali Shakouri
(55.85 g/mol)(10−3 kg/g)
Mat × (10−3 kg/g)
2
2
6.022 ×10 23 mol−1 )
NA
(
ρ BCC =
=
3
3
 4RBCC 
 4 (0.1258 ×10−9 m)


 3 
3


= 7564 kg/m3 (less than at room temperature)
Fe has a FCC structure just above 912 °C (γ-Fe). An Fe atom in the γ-Fe state has a radius of
RFCC = 0.1291 × 10-9 m (Remember that for a FCC structure, a 2 = 4RFCC ). The density of γ-Fe is
therefore:
(55.85 g/mol)(10−3 kg/g)
M at × (10−3 kg/g )
4
4
6.022 ×10 23 mol−1 )
NA
(
ρ FCC =
=
= ρFCC = 7620 kg/m3
3
3
 4RFCC 
 4(0.1291× 10−9 m)


 2 
2


As the density increases, the volume must contract (the Fe retains the same mass).
1.16
Diamond and zinc blende
Si has the diamond and GaAs has the zinc blende crystal structure. Given the lattice parameters of Si
and GaAs, a = 0.543 nm and a = 0.565 nm, respectively, and the atomic masses of Si, Ga, and As as
28.08 g/mol, 69.73 g/mol, and 74.92 g/mol, respectively, calculate the density of Si and GaAs. What
is the atomic concentration (atoms per unit volume) in each crystal?
Solution
Referring to the diamond crystal structure in Figure 1Q16-1, we can identify the following types
of atoms
8 corner atoms labeled C,
6 face center atoms (labeled FC) and
4 inside atoms labeled 1,2,3,4.
The effective number of atoms within the unit cell is:
(8 Corners) × (1/8 C-atom) + (6 Faces) × (1/2 FC-atom) + 4 atoms within the cell (1,2,3,4) = 8
C
C
FC
C
FC
4
1
FC
a
2
FC
FC
3
C
a
FC
C
a
C
1.6
Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
Figure 1Q16-1 The diamond crystal structure.
The lattice parameter (length of a cube side) of the unit cell is a. Thus the atomic concentration
in the Si crystal (nSi) is
8
8
28
nSi = 3 =
atoms per m-3
−9
3 = 5.0 × 10
a
(0.543 × 10 m)
If Mat is the atomic mass in the Periodic Table then the mass of the atom (mat) in kg is
mat = (10-3 kg/g)Mat/NA
(1)
where NA is Avogadro’s number. For Si, Mat = MSi = 28.09 g/mol, so then the density of Si is
ρ = (number of atoms per unit volume) × (mass per atom) = nSi mat
 8   (10−3 kg/g)MSi 

NA
a3  
or
ρ =
i.e.
ρ =

8
 (0.543 × 10
−9
  (10 −3 kg/g)(28.09 g mol-1 )
= 2.33 × 103 kg m-3 or 2.33 g cm3 
23
−1

m)   (6.022 × 10 mol ) 
3
In the case of GaAs, it is apparent that there are 4 Ga and 4 As atoms in the unit cell. The
concentration of Ga (or As) atoms per unit volume (nGa) is
nGa =
4
4
= 2.22 × 1028 m-3
3 =
a
(0.565 ×10−9 m)3
Total atomic concentration (counting both Ga and As atoms) is twice nGa.
nTotal = 2nGa = 4.44 × 1028 m-3
There are 2.22 × 1028 Ga-As pairs per m3. We can calculate the mass of the Ga and As atoms
from their relative atomic masses in the Periodic Table using Equation (1) with Mat = MGa = 69.72 g/mol
for Ga and Mat = MAs = 74.92 g/mol for As. Thus,
 4   (10−3 kg/g)(MGa + MAs ) 
a3  
NA

ρ =
or

4
  (10 −3 kg/g)(69.72 g/mol + 74.92 g/mol) 
ρ =
−9
3

6.022 × 10 23 mol −1
 (0.565 × 10 m)  
i.e.
ρ = 5.33 × 103 kg m-3 or 5.33 g cm-3
1.18 Crystallographic directions and planes
Consider the cubic crystal system.
a. Show that the line [hkl] is perpendicular to the (hkl) plane.
b. Show that the spacing between adjacent (hkl) planes is given by
a
d=
2
h + k2 + l2
1.7
Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
Solution
This problem assumes that students are familiar with three dimensional geometry and vector
products.
Figure 1Q18-1(a) shows a typical [hkl] line, labeled as ON, and a (hkl) plane in a cubic crystal.
ux, uy and uz are the unit vectors along the x, y, z coordinates. This is a cubic lattice so we have
Cartesian coordinates and ux⋅ux = 1 and ux⋅uy = 0 etc.
(a)
uz
(b)
az1 C
az 1 C
uy
N
al
ak
B
ay1
γ
O
B
ay1
D
γ
α
β
O
ah
ax1
A
α
β
ax1
ux
A
Figure 1Q18-1 Crystallographic directions and planes
a
Given a = lattice parameter, then from the definition of Miller indices (h = 1/x1, k = 1/y1 and l =
1/z1) , the plane has intercepts: xo = ax1 =a/h; yo = ay1 = a/k; zo = az1 = a/l.
The vector ON = ahux + akuy + aluz
If ON is perpendicular to the (hkl) plane then the product of this vector with any vector in the
(hkl) plane will be zero. We only have to choose 2 non-parallel vectors (such as AB and BC) in the
plane and show that the dot product of these with ON is zero.
AB = OB − OA = (a/k)uy − (a/h)ux
ON•AB = (ahux + akuy + aluz) • ( (a/k)uy − (a/h)ux) = a2 − a2 = 0
Recall that
uxux = uyuy =1 and uxuy = uxuz = uyuz = 0
Similarly,
ON•BC = (ahux + akuy + aluz) • ( (a/l)uz − (a/k)uy) = 0
Therefore ON or [hkl] is normal to the (hkl) plane.
b
Suppose that OD is the normal from the plane to the origin as shown in Figure 1Q18-1(b).
Shifting a plane by multiples of lattice parameters does not change the miller indices. We can therefore
assume the adjacent plane passes through O. The separation between the adjacent planes is then simply
the distance OD in Figure 1Q18-1(b).
Let α, β and γ be the angles of OD with the x, y and z axes. Consider the direction cosines of the
line OD: cosα = d/(ax1) = dh/a; cosβ = d/(ay1) = dk/a; cosγ = d/(az1) = dl/a
1.8
Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
But in 3 dimensions, (cosα)2 + (cosβ)2 + (cosγ)2 = 1
(d2h2/a2) + (d2k2/a2) + (d2l2/a2) = 1
Thus,
Rearranging, d2 = a2 / [h2 + k2 + l2]
d = a / [h2 + k2 + l2]1/2
or,
1.19 Si and SiO2
a. Given the Si lattice parameter a = 0.543 nm, calculate the number of Si atoms per unit volume, in
nm-3.
b. Calculate the number of atoms per m2 and per nm2 on the (100), (110) and (111) planes in the Si
crystal as shown on Figure 1Q19-1. Which plane has the most number of atoms per unit area?
c. The density of SiO2 is 2.27 g cm-3. Given that its structure is amorphous, calculate the number of
molecules per unit volume, in nm-3. Compare your result with (a) and comment on what happens
when the surface of an Si crystal oxidizes. The atomic masses of Si and O are 28.09 g/mol and 16
g/mol, respectively.
a
a
(100) plane
a
(110) plane
(111) plane
Figure 1Q19-1 Diamond cubic crystal structure and planes. Determine what
portion of a black-colored atom belongs to the plane that is hatched.
Solution
a
Si has the diamond crystal structure with 8 atoms in the unit cell, and we are given the lattice
parameter a = 0.543 × 10-9 m and atomic mass Mat = 28.09 × 10-3 kg/mol. The concentration of atoms
per unit volume (n) in nm-3 is therefore:
8
1
1
8
3
n= 3
3 =
3
3 = 50.0 atoms/nm
9
−9
9
a (10 nm/m) (0.543 × 10 m) (10 nm/m)
If desired, the density ρ can be found as follows:
−3
M
8 at 8 28.09 × 10 kg/mol
23
−1
N
ρ = 3 A = 6.022 × 10 −9 mol3 = 2331 kg m-3 or 2.33 g cm-3
a
(0.543 × 10 m )
b
The (100) plane has 4 shared atoms at the corners and 1 unshared atom at the center. The corner
atom is shared by 4 (100) type planes. Number of atoms per square nm of (100) plane area (n) is shown
in Fig. 1Q19-2:
1.9
Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
B
A
Ge
B
A
a
(100)
E
a
a
D
E
C
a
a
D
C
Figure 1Q19-2 The (100) plane of the diamond crystal structure.
The number of atoms per nm2, n100, is therefore:
4
n100 =
∴
 1
+1
 4
a
2
(10
4
1
9
nm/m )
2
=
 1
+1
 4
(0.543 × 10
−9
1
m) (10 nm/m )
2
9
2
n100 = 6.78 atoms/nm2 or 6.78 × 1018 atoms/m2
The (110) plane is shown below in Fig. 1Q19-3. There are 4 atoms at the corners and shared
with neighboring planes (hence each contributing a quarter), 2 atoms on upper and lower sides shared
with upper and lower planes (hence each atom contributing 1/2) and 2 atoms wholly within the plane.
B
A
a¦2
A
B
(110)
a
D
(110)
C
C
D
Figure 1Q19-3 The (110) plane of the diamond crystal structure.
The number of atoms per nm2, n110, is therefore:
n110
1
1
4  + 2   + 2 

1
2
= 4
 9
2
a(a 2 )
 (10 nm/m ) 
[
]
4
 1
1
+2  +2
 4
 2


1
 9
2
2  (10 nm/m) 
∴
n110 =
∴
n110 = 9.59 atoms/nm2 or 9.59 × 1018 atoms/m2
[(0.543 × 10
−9
(
m) (0.543 × 10
−9
m)
)]
This is the most crowded plane with the most number of atoms per unit area.
The (111) plane is shown below in Fig. 1Q19-4:
1.10
Homework 1 Solution
EE145 Spring 2002
Prof. Ali Shakouri
A
A
a¦3
¦2
a¦2
30° 30°
60°
D
a¦2
C
60°
C
B
a¦2
2
a¦2
2
B
D
Figure 1Q19-4 The (111) plane of the diamond crystal structure
The number of atoms per nm2, n111, is therefore:
n111
 60 
1
+ 3 
3



1
2
360
=
 9
2
  1   2  
3    (10 nm/m) 
2  a  a
 2  2  
2  
∴
n111
 60 
 1
+
3
3



 2
1
360
=
 9
2
  1  
2 
3   10 nm/m) 
2  (0.543 × 10−9 m )   (0.543 × 10−9 m )   (
 2 
2 
2  
∴
n111 = 7.83 atoms/nm2 or 7.83 × 1018 atoms/m2
c
Given:
Molar mass of SiO2: Mat = 28.09 × 10-3 kg/mol + 2 × 16 × 10-3 kg/mol = 60.09 × 10-3 kg/mol
Density of SiO2: ρ = 2.27 × 103 kg m-3
Let n be the number of SiO2 molecules per unit volume, then:
M
ρ = n at
NA
∴
23
−1
3
−3
NA ρ (6.022 × 10 mol )(2.27 × 10 kg m )
= 2.27 × 1028 molecules per m3
n=
=
−3
Mat
60.09
×
10
kg/mol
(
)
Or, converting to molecules per nm3:
n=
2.27 × 1028 molecules/m 3
(10
9
nm/m )
3
= 22.7 molecules per nm3
Oxide has less dense packing so it has a more open structure. For every 1 micron of oxide
formed on the crystal surface, only about 0.5 micron of Si crystal is consumed.
1.11