Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Third Edition (© 2001 McGraw-Hill)

Chapter 4
Note: The first printing has a few odd typos, which are indicated in blue below. These will be
corrected in the reprint.

4.1 Phase of an atomic orbital

a. What is the functional form of a 1s wavefunction, ψ(r)? Sketch schematically the atomic
wavefunction ψ1s(r) as a function of distance from the nucleus.
b. What is the total wavefunction Ψ1s(r, t)?
c. What is meant by two wavefunctions Ψ1s(A) and Ψ1s(B) that are out of phase?
d. Sketch schematically the two wavefunctions Ψ1s(A) and Ψ1s(B) at one instant.

Solution
a. ψ1s(r) decays exponentially as exp (-r/a0), where ao is the Bohr radius (Table 3.2).
ψ1s(r)

Figure 4Q1-1: Atomic wavefunction as a function of distance from the nucleus.

b. Ψ1s(r, t) = ψ1s(r) × exp(−jEt/ ) = ψ1s(r) × exp(−jωt) where ω = E/ is an angular frequency (Section
3.2.2). The total wavefunction is harmonic in time. Recall that exp(jθ) = cosθ + jsinθ so that exp(jωt)
= cosωt + jsinωt.
c. Two sine waves of the same frequency will have a certain phase difference which represents the
time delay between the time oscillations of the two waves. Two waves will be in phase if their maxima
coincide and out of phase if the maximum of one coincides with the minimum of the other. Two 1s
wavefunctions, Ψ1s(A) and Ψ1s(B) will have the same frequency. Like two sine waves of the same
frequency, the waves can be in phase or out of phase. When they are out of phase, if Ψ1s(A) = 1 then
Ψ1s(B) = -1 and vice versa.

4.1
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

d.
Ψ1s(A) Ψ1s(B)

B
r r
A

Figure 4Q1-2: Two 1s wavefunctions which are out of phase (sketches at the same instant)

4.2 Molecular orbitals and atomic orbitals Consider a linear chain of four identical atoms
representing a hypothetical molecule. Suppose that each atomic wavefunction is 1s wavefunction.
This system of identical atoms has a center of symmetry C with respect to the center of the molecule
(midway between the second and the third atom), and all molecular wavefunctions must be either
symmetric or antisymmetric about C.
a. Using LCAO principle, sketch the possible molecular orbitals.
2
b. Sketch the probability distribution ψ
c. If more nodes in the wavefunction lead to greater energies, order the energies of the molecular
orbitals.
Note: The electron wavefunctions, and the related probability distributions, in a simple potential
energy well that are shown in Figure 3.15 can be used as a rough guide towards finding the appropriate
molecular wavefunctions in the four-atom symmetric molecule. For example, if we were to smooth the
electron potential energy in the four-atom molecule into a constant potential energy, that is, generate a
potential energy well, we should be able to modify or distort, without flipping, the molecular orbitals
to somewhat resemble ψ 1 to ψ 4 sketched in Figure 3.15. Consider also that the number of nodes
increases from none for ψ 1 to three for ψ 4 in Figure 3.15.
Author's Note to the Instructor: This has to changed to four atoms to simplify the problem.

4.2
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Solution
Center of symmetry for the PE
O

Combinations of
1,2,3 and 4
S ψa ++++
1 2 3 4

A ψb ++--

S ψc +--+
(- + + -)

A ψd +-+-

Left: Molecular orbitals based on LCAO for a linear array of 4 atoms. S is symmetric and A is
antisymmetric. Right: Probability distributions. All sketches are rough schematic illustrations. Energy
increases with the number of nodes. In the above figures, energy increases downwards from ψa to ψd.
Lowest for the top MO (molecular orbital) and highest for the bottom MO.

O
S ψa
1 2 3 4
Wavefunctions in a PE well

O
A ψb

ψc O
S

O
A ψd

Comparison of molecular orbitals based on LCAO for a linear array of 4 atoms and the first four
wavefunctions in a potential energy well. Notice the resemblance.

4.3
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Author's Note: Generally, the actual molecular oribitals are not a simple linear combination but may
be a linear combination in which each individual atomic wavefunction is scaled or weighted by a
certain amount. The similarity between the molecular oribitals and the wavefunctions in a PE well is
more apparent in the above figure from waves, Atoms and Solids, D.A. Davies, Longman (England),
1978, Figure 7.5, p214

4.3 Diamond and Tin Germanium, silicon, and diamond have the same crystal structure, that of
diamond. Bonding in each case involves sp3 hybridization. The bonding energy decreases as we go
from C to Si to Ge, as noted in Table 4.7.
a. What would you expect for the band gap of diamond? How does it compare with the experimental
value of 5.5 eV?
b. Tin has a tetragonal crystal structure, which makes it different than its group members, diamond,
silicon, and germanium.
1. Is it a metal or a semiconductor?
2. What experiments do you think would expose its semiconductor properties?

4.4
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Solution
Given the properties in Table 4.7, we have the following plots:

4.5
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Energy Gap, eV
Energy Gap, eV 8
6 7 Diamond

5 6
5
4 Diamond 4
3 3

2 2
Si
Ge 1 Si
1 Ge
Tin 0
0 Tin
-1
200 1200 2200 3200
Melting Temperature °C -2
(a) Energy gap versus melting temperature. 1 2 3 4
From the plot, it seems that diamond has an Eg Bond Energy, eV
= 3.4 eV and Sn has Eg ≈ 0 or is a metal (b) Energy gap versus bond energy. From the
plot, it seems that diamond has an Eg = 6.8 eV
and Sn has Eg ≈ −0.9 or is a metal

Energy Gap, eV
Energy Gap, eV 7
6 Diamond
6
5 Diamond
4 5

3 4
2 3
1 Si
Ge 2
0
1 Si
-1 Ge
Tin
-2 0 Tin
0.05 0.1 0.15 -1
Covalent Radius, nm 6 7 8 9 10 11 12
(c) Energy gap versus covalent radius. From the First Ionization Energy, eV
plot, it seems that diamond has an Eg = 4.7 eV
(d) Energy gap versus first ionization energy.
and Sn has Eg ≈ −1.5 eV or is a metal. From the plot, it seems that diamond has an Eg
= 6.3 eV and Sn has Eg ≈ −0.3 eV or is a metal.
Figure 4Q3-1

Each is a plot of the band gap Eg (or energy gap) vs. some property. The straight line in each is drawn
to pass through Si and Ge. Diamond and tin points are then located on this straight line at the
intersections with the vertical lines representing the corresponding properties of diamond and tin.

4.6
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

a. Diamond has an Eg greater than Si and Ge. Averaging the four Eg for diamond we find 5.3 eV which
is close to the experimental value of 5.5 eV.
b.
(1) All the four properties indicate that tin has Eg ≤ 0 or is a metal.
(2) Tin’s semiconductivity can be tested by examining its electrical conductivity and optical
absorption (see Chapter 5). For example, for metals the conductivity should NOT be thermally
activated over a wide temperature range, whereas for semiconductors there will be an Arrhenius
temperature dependence over at least some temperature range. Further, semiconductors have an
absorption edge that corresponds to hυ > Eg (Chapter 5).

4.4 Compound III-V semiconductors Indium as an element is a metal. It has a valency of III.
Sb as an element is a metal and has a valency of V. InSb is a semiconductor, with each atom bonding
to four neighbors, just like in silicon. Explain how this is possible and why InSb is a semiconductor
and not a metal alloy. (Consider the electronic structure and sp3 hybridization for each atom.)

Solution
The one s and three p orbitals hybridize to form 4 ψhyb orbitals. In Sb there are 5 valence electrons.
One ψhyb has two paired electrons and 3 ψhyb has 1 electron as shown in Figure 4Q4-1. In In there are 3
electrons so one ψhyb is empty. This empty ψhyb of In can overlap the full ψhyb of Sb. The overlapped
orbital, the bonding orbital, has two paired electrons. This is a bond between In and Sb even though
the electrons come from Sb (this type of bonding is called dative bonding). It is a bond because the
electrons in the overlapped orbital are shared by both Sb and In. The other 3 ψhyb of Sb can overlap 3
ψhyb of neighboring In to form "normal bonds". Repeating this in three dimensions generates the InSb
crystal where each atom bonds to four neighboring atoms as shown. As all the bonding orbitals are
full, the valence band formed from these orbitals is also full. The crystal structure is reminiscent of
that of Si, as all the valence electrons are in bonds. Since it is similar to Si, InSb is a semiconductor.

Sb In

Sb atom (Valency V) In atom (Valency III)

ψhyb orbitals ψhyb orbitals

Valence
Valence
electron
electron
In Sb In Sb

Sb In Sb In
Sb ion core (+5e) In ion core (+3e)

In Sb In Sb

Sb In Sb In

Figure 4Q4-1: Bonding structure of InSb.

4.7
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

4.5 Compound II-VI semiconductors CdTe is a semiconductor, with each atom bonding to
four neighbors, just like in silicon. In terms of covalent bonding and the positions of Cd and Te in the
Periodic Table, explain how this is possible. Would you expect the bonding in CdTe to have more
ionic character than that in III-V semiconductors?

Solution

Te Cd

Te atom (Valency VI) Cd atom (Valency II)

ψhyb orbitals ψhyb orbitals

Valence
Valence
electron
electron
Cd Te Cd Te

Te Cd Te Cd
Te ion core (+6e) Cd ion core (+2e)

Cd Te Cd Te

Te Cd Te Cd

Dative bonding

Figure 4Q5-1: Bonding structure of CdTe.

In CdTe one would expect a mixture of covalent and ionic bonding. Transferring 2 Cd
electrons to Te would generate Te2- and Cd2+ which then bond ionically. In covalent bonding we
expect hybridization of s and p orbitals. The one s and three p orbitals hybridize to form 4 ψhyb
orbitals. In Te there are 6 valence electrons. Two ψhyb have two paired electrons each and two ψhyb
have 1 electron each as shown. In Cd there are 2 electrons so two ψhyb are empty. An empty ψhyb of Cd
can overlap a full ψhyb of Te. The overlapped orbital, the bonding orbital, then has two paired
electrons. This is a bond between Cd and Te even though the electrons come from Te (this type of
bonding is called dative bonding). It is a bond because the electrons in the overlapped orbital are
shared by both Te and Cd. The other full ψhyb of Te can similarly overlap another empty ψhyb of a
different neighboring Cd to form another dative bond. The half occupied orbitals (two on Te and two
on Cd) overlap and form "normal bonds". Repeating two dative bonds and two normal bonds in three
dimensions generates the CdTe crystal where each atom bonds to four neighboring atoms as shown.
Since all the bonding orbitals are full, the valence band formed from these orbitals is also full. Strictly
the bonding is neither fully covalent nor fully ionic but a mixture.

*4.6 Density of states for a two-dimensional electron gas Consider a two-dimensional

electron gas in which the electrons are restricted to move freely within a square area a2 in the xy plane.
Following the procedure in Section 4.5, show that the density of states g(E) is constant (independent of
energy).

4.8
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Solution
For a two dimensional electron gas confined within a square region of sides a we have:

E=
h2
8me a 2
( 2
n1 + n2
2
)
Only positive n1 and n2 are allowed. Each n1 and n2 combination is an orbital state. Define a new
variable n as:
n2 = n12 + n22
h2
substitute: E= 2
n2
8me a
Let us consider how many states there are with energies less than E′. E′ corresponds to n ≤ n′.
h2
E′ = n′2
8me a 2

8a 2 me E ′
∴ n′ =
h2
n2
n1 2 + n2 2 = n'2
5
4
3
n1 = 1
n2 = 3 2
1
–n1 n1
0 1 2 3 4 5 6
n1 = 2, n2 = 2
–n2

Figure 4Q6-1: Each state, or electron wavefunction in the crystal, can be

represented by a box at n1, n2.
Consider Figure 4Q6-1. All states within the quarter arc defined by n′ have E < E′. The area of this
quarter arc is the total number of orbital states. The total number of states, S, including spin is twice as
many,
⎛ 2
⎞
⎛ 1 2 ⎞ ⎜ 1 ⎡ 8a me E ′ ⎤ ⎟
2
S = 2⎜ πn′ ⎟ = 2⎜ π ⎢ ⎥
⎝4 ⎠ ⎜ 4 ⎢⎣ h 2 ⎥⎦ ⎟⎟
⎝ ⎠
4πa 2 me E ′
∴ S=
h2
The density of states g is defined as the number of states per unit area per unit energy.

4.9
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

1 dS 1 4πa 2 me 4πme
∴ g= = = 2
a 2 dE ′ a 2 h 2 h
Thus, for a two dimensional gas, the density of states is constant.

4.7 Fermi energy of Cu The Fermi energy of electrons in copper at room temperature is 7.0 eV.
The electron drift mobility in copper, from Hall effect measurements, is 33 cm2 V-1 s-1.
a. What is the speed vF of conduction electrons with energies around EF in copper? By how many
times is this larger than the average thermal speed vthermal of electrons, if they behaved like an
ideal gas (Maxwell-Boltzmann statistics)? Why is vF much larger than vthermal?
b. What is the De Broglie wavelength of these electrons? Will the electrons get diffracted by the
lattice planes in copper, given that interplanar separation in Cu = 2.09 Å? (Solution guide:
Diffraction of waves occurs when 2dsinθ =λ, which is the Bragg condition. Find the relationship
between λ and d that results in sinθ > 1 and hence no diffraction.)
c. Calculate the mean free path of electrons at EF and comment.

Solution
a. The Fermi speed vF is given by:
1 2
EF = me vF
2

∴ vF = 2
EF
= 2
(7 eV )(1.602 ×10−19 J/eV )
me (9.109 ×10−31 kg )
∴ vF = 1.57 × 106 m/s
Maxwell - Boltzmann statistics predicts an effective velocity (rms velocity) which is called “thermal
velocity” given by (assume room temperature T = 20 °C = 293 K):
1 2 3
me vthermal = kT
2 2

∴ vthermal =
3Tk
=
(
3(293 K ) 1.381× 10 − 23 J/K )
me (
9.109 × 10 −31 kg )
∴ vthermal = 1.15 × 105 m/s
Comparing the two values:
Ratio = vF/vthermal = 13.7
vF is about 14 times greater than vthermal. This is because vthermal assumes that electrons do not
interact and obey Maxwell-Boltzmann statistics (Eav = 3/2kT). However, in a metal there are many
conduction electrons. They interact with the metal ions and obey the Pauli exclusion principle, i.e.
Fermi-Dirac statistics. They extend to higher energies to avoid each other and thereby fulfill the Pauli
exclusion principle.

4.10
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

b. The De Broglie wavelength is λ = h/p where p = mevF is the momentum of the electrons.
h 6.626 × 10 −34 J s
λ= =
me vF ( )(
9.109 × 10 −31 kg 1.57 ×106 m/s )
∴ λ = 4.63 × 10-10 m or 4.63 Å
The interplanar separation, d, is given as 2.09 Å. The diffraction condition is:
λ = 2dsinθ
1 λ 1 (4.63 Å )
∴ sin θ = = = 1.11
2 d 2 (2.09 Å )
Since this is greater than 1, and sinθ cannot be greater than 1, the electrons will not be diffracted.
c. The drift mobility is related to the mean scattering time τ by:

τ=
µme
=
(33 ×10 −4
)(
m 2 V −1 s −1 9.109 × 10 −31 kg )
= 1.876 × 10-14 s
e 1.602 × 10 C -19

The mean free path, lF, of electrons with speed, vF is:

lF = vFτ = (1.57 × 106 m/s)(1.876 × 10-14 s) = 2.95 × 10-8 m or 295 Å
The mean free path of those electrons with effective speeds ve (close to mean speed) can be found as
follows (EF has little change with temperature, therefore EF ≈ EFO):
1 2 3 3
me ve = EFO = EF
2 5 5

∴ ve =
6 EF
=
(
6 (7.0 eV ) 1.602 × 10 −19 J/eV
= 1.215 × 106 m/s
)
5 me 5 9.109 ×10 kg(
−19
)
∴ le = veτ = (1.215 × 106 m/s)(1.876 × 10-14 s) = 2.28 × 10-8 m or 228 Å

4.8 Free electron model, Fermi energy, and density of states Na and Au both are valency
I metals; that is, each atom donates one electron to the sea of conduction electrons. Calculate the Fermi
energy (in eV) of each at 300 K and 0 K. Calculate the mean speed of all the conduction electrons and
also the speed of electrons at EF for each metal. Calculate the density of states as states per eV cm−3 at
the Fermi energy and also at the center of the band, to be taken at (EF +Φ)/2. (See Table 4.1 for Φ)

Solution
Since Na and Au are valency I metal, their electron concentrations, n are then the atomic
concentrations multiplied by the group number, or:

nNa = (Valency)
N A d Na
= (1)
(
6.022 × 10 23 mol−1 968 kgm −3 )( )
= 2.53 × 10 28 m −3
−3
M Na 23 × 10 kg/mol

nAu = (Valency)
N A d Au
= (1)
(
6.022 ×10 23 mol−1 19300 kgm −3 )( )
= 5.9 ×10 28 m −3
−3
M Au 197 ×10 kg/mol
4.11
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

At 0 K,

( )
2 2/3
2
h 2 ⎛ 3n ⎞ 3 6.626 ×10 −34 Js ⎛ 3 × 2.53 × 10 28 m -3 ⎞
E F 0 ( Na ) = ⎜ ⎟ = ⎜⎜ ⎟⎟
8me ⎝ π ⎠ 8 × 9.1× 10 −31 Kg ⎝ π ⎠
= 5.04×10-19 J or 3.15 eV

( )
2 2/3
2
h 2 ⎛ 3n ⎞ 3 6.626 × 10 −34 Js ⎛ 3 × 5.9 ×10 28 m -3 ⎞
E F 0 (Au) = ⎜ ⎟ = ⎜⎜ ⎟⎟
8me ⎝ π ⎠ 8 × 9.1×10 −31 Kg ⎝ π ⎠
= 8.863×10-19 J or 5.54 eV
At 300 K,
⎡ π 2 ⎛ kT ⎞ 2 ⎤ ⎡ π 2 ⎛ 0.02585eV ⎞ 2 ⎤
E F ( Na ) = EF 0 ( Na ) ⎢1 − ⎜⎜ ⎟⎟ ⎥ = 3.15eV ⎢1 − ⎜ ⎟ ⎥
⎢⎣ 12 ⎝ EFO ( Na ) ⎠ ⎥⎦ ⎢⎣ 12 ⎝ 3.15eV ⎠ ⎥⎦

∴ EF(Na) = 3.15 eV
⎡ π 2 ⎛ kT ⎞ 2 ⎤ ⎡ π 2 ⎛ 0.02585eV ⎞ 2 ⎤
E F (Au) = EF 0 (Au) ⎢1 − ⎜⎜ ⎟⎟ ⎥ = 5.54eV ⎢1 − ⎜ ⎟ ⎥
⎢⎣ 12 ⎝ EFO (Au) ⎠ ⎥⎦ ⎢⎣ 12 ⎝ 5.54eV ⎠ ⎥⎦

∴ EF(Au) = 5.54 eV
Mean speed of conduction electrons:
1 3
me ve2 = Eav = E 0
2 5
ve = (6 E F 0 / 5me )
1/ 2
∴

∴ ve(Na) = (6EF0(Na)/5me)1/2 = ((6×3.15×1.6×10-19 J)/(5×9.1×10-31kg))1/2

∴ ve(Na) = 8.15×105 ms-1
and ve(Au) = (6EF0(Au)/5me)1/2 = ((6×5.54×1.6×10-19 J)/(5×9.1×10-31kg))1/2
∴ ve(Au) = 1.08×106 ms-1
Speed at EF:
1
me ve2 = EF0
2
ve = (2 E F0 / me )
1/ 2
∴

∴ ve(Na) = (2EF0(Na)/me)1/2 = ((2×3.15×1.6×10-19 J)/(9.1×10-31kg))1/2

∴ ve(Na) = 1.05×106 ms-1
and ve(Au) = (2EF0(Au)/me)1/2 = ((2×5.54×1.6×10-19 J)/(9.1×10-31kg))1/2
∴ ve(Na) = 1.4×106 ms-1
The density of states is given by

4.12
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

3/ 2

( ⎛m ⎞
)
g ( E ) = 8π 21/ 2 ⎜ 2e ⎟ E 1/ 2
⎝h ⎠
3/ 2
⎡ 9.1× 10 −31 kg ⎤
For Na, (
g EF ( Na ) = 8π 2 1/ 2
)
⎢ ⎥ (3.15 ×1.6 ×10 J )
−19 1/ 2

⎣ 6.626 × 10 Js ⎦
-34

∴ g EF ( Na ) = 7.54 × 10 46 m −3 J −1 = (7.54 × 10 46 m −3 J −1 )(10 −6 m 3cm −3 )(1.6 × 10 −19 JeV −1 )

= 1.2×1022 cm-3 eV-1

(EF + Φ)/2 = (3.15 eV + 2.75 eV)/2 = 2.95 eV
3/ 2
⎡ 9.1×10 −31 kg ⎤
∴ (
g center ( Na ) = 8π 2 1/ 2
)
⎢ ⎥ (2.95 ×1.6 ×10 J )
−19 1/ 2

⎣ 6.626 × 10 Js ⎦
-34

∴ g center ( Na ) = 7.3 ×10 46 m −3 J −1 = (7.3 × 10 46 m −3J −1 )(10 −6 m 3cm −3 )(1.6 × 10 −19 JeV −1 )
= 1.17×1022 cm-3 eV-1
3/ 2
⎡ 9.1× 10 −31 kg ⎤
For Au, (
g EF (Au) = 8π 2 1/ 2
)
⎢ ⎥ (5.54 ×1.6 ×10 J )
−19 1/ 2

⎣ 6.626 × 10 Js ⎦
-34

∴ g EF (Au) = 10 × 10 46 m −3 J −1 = (10 × 10 46 m −3 J −1 )(10 −6 m 3cm −3 )(1.6 ×10 −19 JeV −1 )

= 1.6×1022 cm-3 eV-1

(EF + Φ)/2 = (5.54 eV + 5.1 eV)/2 = 5.32 eV
3/ 2
⎡ 9.1×10 −31 kg ⎤
∴ (
g center (Au) = 8π 2 1/ 2
)
⎢ ⎥ (5.32 ×1.6 ×10 J )
−19 1/ 2

⎣ 6.626 × 10 Js ⎦
-34

∴ g center (Au) = 9.8 ×10 46 m −3 J −1 = (9.8 × 10 46 m −3J −1 )(10 −6 m 3cm −3 )(1.6 × 10 −19 JeV −1 )
= 1.568×1022 cm-3 eV-1

4.9 Fermi energy and electron concentration Consider the metals in Table 4.8 from groups
I, II and III in the Periodic Table. Calculate the Fermi energies at absolute zero, and compare the
values with the experimental values. What is your conclusion?

Solution

4.13
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Since Cu is in group I, its valency is also 1. The electron concentration n is then the atomic
concentration multiplied by the group number, or:

n = (Valency)
N AD
= (1)
(6.022 ×1023 mol−1 )(8.96 ×103 kg/m3 ) = 8.490 ×1028 m −3
M at 63.55 ×10 −3 kg/mol
Using Equation 4.22:
2
h 2 ⎛ 3n ⎞ 3 ⎛ 1 ⎞
E FO = ⎜ ⎟ ⎜ ⎟
8me ⎝ π ⎠ ⎜⎝ q ⎟⎠
2

∴ E FO =
(6.626 ×10 −34
Js) ⎛⎜ 3(8.490 ×10
2 28
)
m −3 ⎞ 3 ⎛
⎟⎟ ⎜
1 ⎞
⎟
8(9.109 ×10 −31
kg) ⎜⎝ π −19
⎠ ⎝ 1.602 × 10 J/eV ⎠
∴ EFO = 7.04 eV
Comparing with the experimental value:
7.04 eV − 6.5 eV
% difference = × 100% = 8.31%
6.5 eV
EFO can be calculated for Zn and Al in the same way (remember to take into account the different
valencies). The values are summarized in the following table and it can be seen that calculated values
are close to experimental values:
Metal n (m-3) (× 1028) EFO (eV) EFO (eV) % Difference
(calculated) (experimental)
Cu 8.490 7.04 6.5 8.31
Zn 13.15 9.43 11.0 14.3
Al 18.07 11.7 11.8 0.847

Table 4Q9-1: Summarized values for Fermi energy at absolute zero temperature.

4.10 Temperature dependence of the Fermi energy

a. Given that the Fermi energy for Cu is 7.0 eV at absolute zero, calculate the EF at 300 K. What is
the percentage change in EF and what is your conclusion?
b. Given the Fermi energy for Cu at absolute zero, calculate the average energy and mean speed per
conduction electron at absolute zero and 300 K, and comment.

Solution
a. The Fermi energy in eV at 0 K is given as 7.0 eV. The temperature dependence of EF is given by
Equation 4.23. Remember that EFO is given in eV.

4.14
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

⎛ π2 ⎡ kT ⎤
2
⎞
E F = EFO ⎜1 − ⎢ ⎥
⎟
⎜ 12 ⎣ E FO ⎦ ⎟
⎝ ⎠

∴ EF = (7.0 eV )⎜1 −
( )
⎛ π 2 ⎡ 1.381× 10 − 23 J/K (300 K ) ⎤ 2 ⎞
⎟ = 6.999921 eV
(
⎜ 12 ⎢⎣ (7.0 eV ) 1.602 × 10 −19 J/eV ⎥⎦ ⎟
⎝ ⎠
)
6.999921 eV − 7.0 eV
∴ % difference = ×100% = 0.00129%
7.0 eV
This is a very small change. The Fermi energy appears to be almost unaffected by temperature.
b. The average energy per electron at 0 K is:
Eav(0 K) = 3/5 (EFO) = 4.2 eV
The average energy at 300 K can be calculated from Equation 4.26:

3 ⎛ 5π 2 ⎡ kT ⎤ 2 ⎞
Eav (T ) = E FO ⎜1 + ⎢
⎟
⎥ ⎟
5 ⎜ 12 ⎣ E FO ⎦
⎝ ⎠

∴
3
Eav (300 K ) = (7.0 eV )⎜1 +
( )
⎛ 5π 2 ⎡ 1.381× 10 − 23 J/K (300 K ) ⎤ 2 ⎞
⎟
5 ⎜
⎝
⎢
( ⎥
12 ⎣ (7.0 eV ) 1.602 × 10 −19 J/eV ⎦ ⎟
⎠
)
∴ Eav(300 K) = 4.200236 eV
This is a very small change.
Assume that the mean speed will be close to the effective speed ve. Effective speed at absolute zero is
denoted as veo, and is given by:
1 2
Eav (0 K ) × q = me veo
2

∴ veo = 2
qEav (0 K )
= 2
(1.602 ×10−19 J/eV )(4.2 eV) = 1215446 m/s
me (9.109 ×10-31 kg )
At 300 K, the effective speed is ve:

ve = 2
qEav (300 K )
= 2
(1.602 ×10−19 J/eV )(4.200236 eV ) =1215480 m/s
me (9.109 ×10-31 kg )
Comparing the values:
1215480 m/s − 1215446 m/s
% difference = × 100% = 0.002797%
1215446 m/s
The mean speed has increased by a negligible amount (0.003%) from 0 K to 300 K.
Note: For thermal conduction this tiny increase in the velocity is sufficient to transport energy from
hot regions to cold regions. This very small increase in the velocity also allows the electrons to
diffuse from hot to cold regions giving rise to the Seebeck effect.

4.15
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

4.11 X-ray emission spectrum from sodium Structure of Na atom is [Ne]3s1. Figure 4.59a
shows formation of the 3s and 3p energy bands in Na as a function of internuclear separation. Figure
4.59b shows the x-ray emission spectrum (called the L-band) from crystalline sodium in the soft x-ray
range as explained in Example 4.6.
a. From Figure 4.59a, estimate the nearest neighbor equilibrium separation between Na atoms in the
crystal if some electrons in the 3s band spill over into the states in the 3p band.
b. Explain the origin of X-ray emission band in Figure 4.59b and the reason for calling it the L-band.
c. What is the Fermi energy of the electrons in Na from Figure 4.59b?
d. Taking the valency of Na to be I, what is the expected Fermi energy and how does it compare with
that in (c)?

Solution
a. As represented in Figure 4.59a, the estimated nearest interatomic separation is near the minimum of
the 3s band, or slightly above it, 0.36 - 0.37 nm.
b. When an electron, for some reason, is leaving the closed inner L-shell of an atom, an empty state is
created there. An electron from the energy band of the metal drops into the L-shell to fill the vacancy
and emits a soft X-ray photon in this process. The spectrum of this X-ray emission from metal
involves a range of energies, corresponding to transitions from the bottom of the band and from the
Fermi level to the L-shell. So all the X-ray photons emitted from the electrons during their transitions
to the L-shell will have energies lying in that range (band) and because all of them are emitted due to a
transition to the L-shell, this X-ray band is called L-band.
c. As explained in part b, the width of the X-ray band corresponds to the distance from the bottom of
the energy band to the Fermi level. As shown in Figure 4.59b, the position of the Fermi level with
respect to the bottom of the energy band is approximately 3.2 eV.

4.16
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

d. Theoretically, the position of the Fermi level is given by Equation 4.23. Since the temperature
dependence of the Fermi level is really very weak, we can neglect it. Then EF(T) = EF0 and using
Equation 4.22, we receive
2
h 2 ⎛ 3n ⎞ 3
EF = ⎜ ⎟
8me ⎝ π ⎠
The electron concentration in Na can be calculated, assuming that each Na atom donates exactly one
electron to the crystal. Taking from Appendix B the density of Na d (0.97 g cm-3) and its atomic mass
Mat (22.99 g mol-1), we receive
d N A (0.97 g cm −3 )(6.022 ×10 23 mol−1 )
n= = = 2.54 × 1022 cm-3 = 2.54 × 1028 m-3
M at (22.99 g mol )−1

Thus the Fermi level is

2
h 2 ⎛ 3n ⎞ 3 (6.626 ×10 −34 J s ) ⎡ 3(2.54 ×10 28 m 3 )⎤ 3
2
2

EF = ⎜ ⎟ = ⎥ = 5.05 × 10 J
-19
8(9.1× 10 kg ) ⎣
⎢
8me ⎝ π ⎠ −31
π ⎦
or EF = 3.15 eV,
which is very close to the value obtained from the X-ray spectrum.

4.12 Conductivity of metals in the free electron model Consider the general expression for
the conductivity of metals in terms of the density of states g(EF) at EF given by
1
σ = e 2 vF2τg ( EF )
3
Show that within the free electron theory, this reduces to σ = e2nτ/me, the Drude expression.

Solution
At Fermi energy
1
me vF2 = EF
2
ve = (2 E F / me )
1/ 2
∴
The density of states
3/ 2

(
g ( EF ) = 8π 2 1/ 2
) ⎛ me ⎞
⎜ 2⎟ EF1/ 2
⎝h ⎠
And the Fermi energy is
2
h 2 ⎛ 3n ⎞ 3
EF = ⎜ ⎟
8me ⎝ π ⎠

4.17
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

1
Now, σ = e 2 vF2τg ( EF )
3
Putting the value of vF and g(EF) we get
3/ 2
1 2E ⎛m ⎞
(
σ = e 2 F τ × 8π 21/ 2 ⎜ 2e ⎟
3 me
) EF1/ 2
⎝h ⎠
3/ 2
2 e 2τ
=
3 me
( ⎛m ⎞
× 8π 21/ 2 ⎜ 2e ⎟ ) EF3 / 2
⎝h ⎠
3/ 2 3/ 2
2 e 2τ ⎛ h2 ⎞
= (
× 8π 21/ 2 ) ⎛ me ⎞
⎜ 2⎟ ⎜⎜ ⎟⎟ ⎛ 3n ⎞
⎜ ⎟
3 me ⎝h ⎠ ⎝ 8me ⎠ ⎝π ⎠

e 2 nτ
∴ σ=
me

4.13 Mean free path of conduction electrons in a metal Show that within the free electron
theory, the mean free path ℓ and conductivity σ are related by
e2
σ = 1/ 3 2 / 3 n 2 / 3 = 7.87 ×10 −5 n 2 / 3
3 π
Calculate ℓ for Cu and Au, given each metal’s resistivity of 17 nΩ m and 22 nΩ m, respectively, and
that each has a valency of I. We are used to seeing σ ∝ n. Can you explain why σ ∝ n2/3?

Solution
Mean free path, ℓ = vFτ
The Fermi energy is,
1
EF = me vF2
2
ve = (2 EF / me )
1/ 2
∴
The density of states
3/ 2

(
g ( EF ) = 8π 2 1/ 2
)
⎛ me ⎞
⎜ 2⎟ EF1/ 2
⎝h ⎠
And the Fermi energy
2
h 2 ⎛ 3n ⎞ 3
EF = ⎜ ⎟
8me ⎝ π ⎠

4.18
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

1
Now, σ = e 2 vF2τg ( EF )
3
1
= e 2 ( v Fτ ) g ( E F ) v F
3
3/ 2
1 ⎛m ⎞
= e 2 × (8π 21/ 2 )⎜ 2e ⎟ E F1/ 2 × (2 EF / me )1/ 2
3 ⎝h ⎠
3/ 2
1 ⎛m ⎞ 21/ 2
= e 2 × (8π 21/ 2 )⎜ 2e ⎟ × 1/ 2 × EF
3 ⎝h ⎠ me

1 m h2 n2/ 3
= e 2 × (8 × 2π ) 3e ×
3 h 8me π 2 / 3

∴ σ=
e2
n 2/3
=
(1.602 ×10 ) −19 2
n 2 / 3 = 7.87 ×10 −5 n 2 / 3
31/ 3 π 2 / 3 31/ 3 π 2 / 3 × 1.055 × 10 −34
σ
∴ =
7.87 × 10 −5 n 2 / 3
1
∴ Cu = −9 −5
= 3.88×10-8 m or 38.8 nm
(17 × 10 )7.87 × 10 (8.45 × 10 )
28 2 / 3

1
and Au = −9 −5
= 3.81×10-8 m or 38.1 nm
(22 × 10 )7.87 × 10 (5.9 × 10 )
28 2 / 3

The statement σ ∝ n would be true if the drift mobility and hence the mean free path ℓ did not
change with n at all. ℓ depends on vF and EF and hence vF depends on n.

*
4.14 Low-temperature heat capacity of metals The heat capacity of conduction electrons
in a metal is proportional to the temperature. The overall heat capacity of a metal is determined by the
lattice heat capacity, except at the lowest temperatures. If δEt is the increase in the total energy of the
conduction electrons (per unit volume) and δT is the increase in the temperature of the metal as a result
of heat addition, Et has been calculated as follows:
∞
⎛ π 2 ⎞ n(kT ) 2
Et = ∫ Eg ( E ) f ( E )dE = Et (0) + ⎜⎜ ⎟⎟
0 ⎝ 4 ⎠ EFO
where Et(0) is the total energy per unit volume at 0 K, n is the concentration of conduction electrons,
and EFO is the Fermi energy at 0 K. Show that the heat capacity per unit volume due to conduction
electrons in the free electron model of metals is
π 2 ⎛ nk 2 ⎞
Ce = ⎜ ⎟T = γT
2 ⎜⎝ EFO ⎟⎠
where γ = (π2/2)(nk2/EFO). Calculate Ce for Cu, and then using the Debye equation for the lattice heat
capacity, find Cv for Cu at 10 K. Compare the two values and comment. What is the comparison at

4.19
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

room temperature? (Note: Cvolume = Cmolar(ρ/Mat), where ρ is the density in g cm−3, Cvolume is in J
K−1cm−3, and Mat is the atomic mass in g mol−1.)

Solution
dEt d ⎛ ⎛ π 2 ⎞ n(kT ) 2 ⎞ π 2 ⎛ nk 2 ⎞
⎜ ⎟ ⎜ ⎟
⎟ = 0 + 2 ⎜ E ⎟T = γT
Ce = = ⎜
Et (0) + ⎜ ⎟ ⎟
dT dT ⎜⎝ 4
⎝ ⎠ FO ⎠ E ⎝ FO ⎠
For Cu, n = 8.45×1028 m-3 and EFO = 7×1.6×10-19 J
π 2 ⎛ nk 2 ⎞ π 2 (8.45 ×10 28 m -3 )(1.38 × 10-23 JK −1 ) 2
∴ Ce = ⎜ ⎟T = (10 K ) = 709 JK-1m-3
2 ⎜⎝ EFO ⎟⎠ 2 7 × 1.6 ×10 −19
J

∴ Ce = 7.09×10-4 J K-1cm-3
For Cu, Debye temperature, TD = 315 K
∴ T/TD=0.0317
x 4 e x dx
Cm (Phonons) = 9 R(0.0317 )
3 31.5
∴ ∫ (e
0 x
−1 ) 2

Solving this equation using math software, we get

Cm(Phonons) = 0.0622 JK-1 mol-1
∴ Cv(Phonons) = Cm(Phonons)×(ρ/Mat)
= (0.0622 J K-1 mol-1)(8.96 g cm-3/63.5 g mol-1) = 8.77 ×10-3 J K-1cm-3
Comment: At 10 K, the heat capacity Ce due to electrons is roughly 8% of Cv(Phonons) due to lattice
vibrations (phonons) only. Ce is not quite negligible. If we decrease the temperature further, Ce will
eventually be larger than Cv(Phonons).
At room temperature, T =300 K
π 2 ⎛ nk 2 ⎞ π 2 (8.45 ×10 28 m -3 )(1.38 × 10-23 JK −1 ) 2
∴ Ce = ⎜ ⎟T = (300 K ) = 21.3×103 JK-1m-3
2 ⎜⎝ EFO ⎟⎠ 2 7 ×1.6 × 10 −19
J

∴ Ce = 21.3×10-3 JK-1cm-3
And T/TD = 0.95
From the Fig. 4.45, we get
Cm(Phonons) = 23.5 JK-1mol-1
∴ Cv(Phonons) = Cm(Phonons)×(ρ/Mat) = (23.5 JK-1 mol-1)(8.96 g cm-3/63.5 g mol-1)
= 3.316 JK-1cm-3
Comment: At 300 K, the heat capacity Ce due to electrons is negligible (0.6 %) compared to
Cv(Phonons) due to lattice vibrations (phonons) only.
4.20
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

*
4.15 Secondary emission and photomultiplier tubes When an energetic (high velocity)
projectile electron collides with a material with a low work function, it can cause electron emission
from the surface. This phenomenon is called secondary emission. It is fruitfully utilized in
photomultiplier tubes as illustrated in Figure 4.60. The tube is evacuated and has a photocathode for
receiving photons as a signal. An incoming photon causes photoemission of an electron from the
photocathode material. The electron is then accelerated by a positive voltage applied to an electrode
called a dynode which has a work function that easily allows secondary emission. When the
accelerated electron strikes dynode D1, it can release several electrons. All these electrons, the original
and the secondary electrons, are then accelerated by the more positive voltage applied to dynode D2.
On impact with D2, further electrons are released by secondary emission. The secondary emission
process continues at each dynode stage until the final electrode, called the anode, is reached
whereupon all the electrons are collected which results in a signal. Typical applications for
photomultiplier tubes are in X-ray and nuclear medical instruments

(X-ray CT scanner, positron CT scanner, gamma camera, etc.), radiation measuring instruments (e.g.,
radon counter), X-ray diffractometers, and radiation measurement in high-energy physics research.
A particular photomultiplier tube has the following properties. The photocathode is made of a
semiconductor-type material with Eg ≈ 1 eV, an electron affinity χ of 0.4 eV, and a quantum efficiency
of 20 percent at 400 nm. Quantum efficiency is defined as the number of photoemitted electrons per
absorbed photon. The diameter of the photocathode is 18 mm. There are 10 dynode electrodes and an
applied voltage of 1250 V between the photocathode and anode. Assume that this voltage is equally
distributed among all the electrodes.
a. What is the longest threshold wavelength for the phototube?
b. What is the maximum kinetic energy of the emitted electron if the photocathode is illuminated
with a 400 nm radiation?

4.21
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

c. What is the emission current from the photocathode at 400 nm illumination per unit intensity of
radiation?
d. What is the KE of the electron as it strikes the first dynode electrode? It has been found that the
tube has a gain of 106 electrons per incident photon. What is the average number of secondary
electrons released at each dynode?

Solution
a. For longest threshold wavelength,
hc / λth = E g + χ

∴ λth = (6.626×10-34 J s)(3×108 ms-1)/(1.4×1.6×10-19 J) = 8.87×10-7 m or 887 nm

With 400 nm radiation,
Eph = hc/λ = (4.13×10-15 eV s)(3×108 ms-1)/(400×10-9 m) = 3.1 eV
b. The excess energy over (Eg + χ) goes as kinetic energy
∴ KEm = Eph – (Eg + χ) = 1.7 eV
c. The intensity, I = 1 W m-2.
The number of photons arriving per unit area per unit time at the photocathode is

Γph =
I
=
( )
1 J s −1 m 2
= 2.02 × 1018 s-1 m-2
E ph ( −19
3.1eV ×1.6 × 10 J eV -1
)
The current density is then simply
J = eΓ ph QE = (1.6 × 10 −19 C )(2.02 × 1018 s −1 m −2 )(0.2 ) = 0.065 A m-2

∴ The emission current, I = J×A = 0.065 A m-2 × π(9×10-3 m)2 = 1.654×10-5 A or 16.54 µA
d. The potential difference between cathode and first anode is, Vd = 1250V/11 = 113.64 V
∴ KED1 = KEm + eVd = 1.7 eV + 113.64 eV = 115.34 eV
e. Quantum efficiency is 20 percent. So, ejected electron is 0.2 per photon
∴ The gain of the tube = 106/0.2 = 5×106 per ejected electron from the cathode.
and Average number of secondary electron released = (5×106)1/10 = 4.68

4.16 Thermoelectric effects and EF Consider a thermocouple pair that consists of gold and
aluminum. One junction is at 100 °C and the other is at 0 °C. A voltmeter (with a very large input
resistance) is inserted into the aluminum wire. Use the properties of Au and Al in Table 4.3 to estimate
the emf registered by the voltmeter and identify the positive end.

Solution

4.22
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Au

Hot 100 °C 0 °C Cold

0
Al Al
µV

Figure 4Q16-1: The Al-Au thermocouple. The cold end is maintained at 0 °C which is the reference
temperature. The other junction is used to sense the temperature. In this example it is heated to 100 °C.
We essentially have the arrangement shown above. For each metal there will be a voltage across it
given by integrating the Seebeck coefficient. From the Mott-Jones equation:
xπ 2 k 2T xπ 2 k 2 2
T T
∆V = ∫ SdT = ∫ − dT = − (T − T02 )
T0 T0
3eEFO 6eEFO

The emf (VAB) available is the difference in ∆V for the two metals labeled A (= Al) and B (= Au) so
that
V AB = ∆V A − ∆VB
where in this example, T = 373 K and T0 = 273 K. We can calculate EFAO and EFBO for each metal as in
Example 4.9 by using
2/3
h 2 ⎛ 3n ⎞
E FO = ⎜ ⎟
8me ⎝ π ⎠
where n is the electron concentration.
n = atomic concentration (nat) × number of conduction electrons per atom
From the density d and atomic mass Mat, the atomic concentration of Al is:
N A d (6.022 ×10 23 mol-1 )(2700 kg/m 3 )
nAl = = = 6.022 ×10 28 m -3
M at (0.027 kg/mol)
so that n = 3nAl = 1.807 × 1029 m-3
which leads to

E FAO =
h 2 ⎛ 3n ⎞
⎜ ⎟
2/3

=
(6.626 ×10 ) ⎛⎜ 3(1.807 ×10 ) ⎞⎟
−34 2 29 2/3

8me ⎝ π ⎠ 8(9.109 ×10 ) ⎜⎝ π

−31 ⎟
⎠
i.e. EFAO = 1.867 × 10-18 J or 11.66 eV
Similarly for Au, we find EFBO = 5.527 eV.
Substituting x and EF values for A (Al) and B (Au) we find,
π 2k 2 xA
∆VA = −
6eEFAO
(T 2
)
− T02 = -188.4 µV

π 2 k 2 xB
and ∆VB = −
6eEFBO
(T 2
)
− T02 = 211.3 µV

4.23
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

so that the magnitude of the voltage difference is

|VAB | = |-188.3 µV - 1.3 µV| = 399.7 µV
Au

157 µV
Hot Cold
Al I

Meter
188 µV

Figure 4Q16-2
To find which end is positive, we put in the resistance of the voltmeter and replace each metal
by its emf and determine the direction of current flow as in the figure. For the particular circuit shown,
the cold connected side of the voltmeter is positive.

4.17 The thermocouple equation Although inputting the measured emf for V in the
thermocouple equation V = a∆T + b(∆T)2 leads to a quadratic equation, which in principle can be
solved for ∆T, in general ∆T is related to the measured emf via
∆T = a1V + a2V2 + a3V3 + ...
with the coefficients a1, a2 etc., determined for each pair of TCs. By carrying out a Taylor's expansion
of TC equation, find the first two coefficients a1 and a2. Using an emf table for the K-type
thermocouple or Figure 4.33, evaluate a1 and a2.

Solution

4.24
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Figure 4.33: Output emf versus temperature (ºC) for various thermocouples between 0 to 1000 ºC.
From Example 4.11, the emf voltage (V) can be expressed:

V=
π 2k 2 ⎛ 1
⎜⎜ −
1 ⎞ 2
(
⎟ T − To 2
4e ⎝ E FAO E FBO ⎟⎠
)
Since we know that ∆T = T - To, and therefore T = ∆T + To, we can make the following substitution:

V=
π 2k 2 ⎛ 1
⎜⎜ −
1 ⎞
(
⎟⎟ [∆T + To ]2 − To 2
4e ⎝ E FAO E FBO ⎠
)
π 2 k 2To ∆T π 2 k 2To ∆T π 2 k 2 (∆T )2 π 2 k 2 (∆T )2
expanding, V= − + −
2eEFAO 2eEFBO 4eEFAO 4eEFBO

π 2 k 2To ⎛ 1 1 ⎞ π 2k 2 ⎛ 1 1 ⎞
factoring, V= ⎜⎜ − ⎟⎟∆T + ⎜⎜ − ⎟⎟(∆T )2
2e ⎝ E FAO E FBO ⎠ 4e ⎝ E FAO E FBO ⎠

Upon inspection, it can be seen that this equation is in the form of the thermocouple equation, V = a∆T
+ b∆T2, and therefore we know that the coefficients a and b are equal to:
π 2 k 2To ⎛ 1 1 ⎞ π 2k 2 ⎛ 1 1 ⎞
a= ⎜⎜ − ⎟⎟ b= ⎜⎜ − ⎟
2e ⎝ E FAO E FBO ⎠ 4e ⎝ E FAO E FBO ⎟⎠
Continuing with the thermocouple equation, we can rearrange it as follows to obtain a quadratic
equation:
-b(∆T)2 - a∆T + V = 0
4.25
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

This is a quadratic equation in ∆T. The solution is

a − a 2 + 4bV a a 1 + 4bV / a 2
∆T = =− +
− 2b 2b 2b

( )
1
a a
∴ ∆T = − + 1 + 4bV / a 2 2
2b 2b
n
a a ∞ ⎛ 1 / 2 ⎞⎛ 4bV ⎞
−
∴ ∆T = − + ∑n =0 ⎜ ⎟⎜1 + 2 ⎟
2b 2b ⎝ n ⎠⎝ a ⎠
Taylor expansion:

a ⎡ 1 ⎛ 4bV ⎞ ( 1 )(− 12 ) ⎛ 4bV ⎞ ( 12 )(− 12 )(− 32 ) ⎛ 4bV ⎞ ⎤

2 3
a
∆T = − + ⎢1 + ⎜ 2 ⎟ + 2 ⎜ 2 ⎟ + ⎜ 2 ⎟ + ...⎥
2b 2b ⎢⎣ 2 ⎝ a ⎠ 2! ⎝ a ⎠ 3! ⎝ a ⎠ ⎥⎦

V bV 2 b 2V 3
∴ ∆T = − 3 + 2 5 + ...
a a a
The positive root is not used because V = 0 must give ∆T = 0, and with the positive root the a/2b terms
will not cancel out. This equation is of the form given in the question,
∆T = a1V + a2V 2 +a3V 3 +...
1 −b
such that a1 = and a2 =
a a3
1 1 2eE E
∴ a1 = = 2 2 = 2 2 FAO FBO
a π k To ⎛ 1 1 ⎞ π k To (EFBO − EFAO )
⎜⎜ − ⎟
2e ⎝ EFAO EFBO ⎟⎠

π 2k 2 ⎛ 1 1 ⎞
⎜⎜ − ⎟
b 4e ⎝ EFAO EFBO ⎟⎠ 2e 2 EFAO EFBO
2 2
and a2 = − 3 = − = −
a ⎡ π 2 k 2To ⎛ 1 1 ⎞⎤
3
π 4 k 4To 3 (EFBO − EFAO )2
⎢ ⎜⎜ − ⎟⎟⎥
⎣ 2e ⎝ EFAO EFBO ⎠⎦
From Figure 4.33, at ∆T = 200 °C, V = 8 mV, and at ∆T = 800 °C, V = 33 mV. Using these values and
neglecting the effect of a3, we have two simultaneous equations we can solve:
∆T = a1V + a2V2
∆T = 200 °C,
200 °C = a1(8 mV) + a2(8 mV)2
∴ a1 = (-(64 mV2)a2 + 200 °C) / (8 mV)
∆T = 800 °C,
800 °C = a1(33 mV) + a2(33 mV)2
substitute for a1,
800 °C = [(-(64 mV2)a2 + 200 °C) / (8 mV)](33 mV) + a2(33 mV)2
4.26
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

isolate a2: a2 = -0.03030 °C/mV2

∴ 200 °C = a1(8 mV) + (-0.03030 °C/mV2)(8 mV)2
∴ a1 = 25.24 °C/mV
We can check our calculation by calculating ∆T when V = 20 mV and comparing the value we obtain
from examining Figure 4.33

4.18 Thermionic emission A vacuum tube is required to have a cathode operating at 800 °C and
providing an emission (saturation) current of 10 A. What should be the surface area of the cathode for
the two materials in Table 4.9? What should be the operating temperature for the Th on W cathode, if
it is to have the same surface area as the oxide-coated cathode?

Solution
Operating temperature T is given as 800 °C = 1073 K and emission current I is given as 10 A. The
temperature and current of the tube are related to its area by Equation 4.44:

J=
I
= BeT 2 exp ⎢
(
⎡ − Φ − βs E ⎤
⎥
)
A ⎣ kT ⎦
I
∴ A=
(
⎡ − Φ − βs E ⎤
BeT 2 exp ⎢ ⎥
)
⎣ kT ⎦
Assuming there is no assisting field emission, the area needed for Th on W is:
10 A
A=
(3 ×10 4
A m −2 K −2 ) (
(1073 K )2 exp⎢ − (2.6 eV ) 1−23.602 ×−10
⎡ −19
)
J/eV ⎤
( )
⎣ 1.381×10 J K (1073 K ) ⎦
1 ⎥

∴ A = 467 m2 (large tube)

For the oxide coating:
10 A
A=
(100 A m −2
K −2 ) (
(1073 K )2 exp⎢ − (1 eV ) 1.−602
⎡ × 10 −19 J/eV ⎤)
( )
⎣ 1.381×10 J K (1073 K ) ⎦
23 −1 ⎥

∴ A = 0.00431 m2 (small and practical tube)

To find the temperature that the Th on W cathode would have to work at to have the same surface area
as the oxide coated cathode, the area of the oxide cathode can be used in the current equation and the

4.27
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

temperature can be solved for. It is a more difficult equation, but can be solved through graphical
methods.
⎡− Φ⎤
I = ABeT 2 exp ⎢
⎣ kT ⎥⎦
The plot of thermionic emission current I versus temperature T is shown below, with A = 0.00431 m2,
Be = 3 × 10-4 A m-2 K-2, and Φ = (2.6 eV)(1.602 × 10-19 J/eV).
15

10
I (A)

1.6x10 3 1.7x103 1.8x103

T (K)
Figure 4Q18-1: Behavior of current versus temperature for the Th on W cathode.
From the graph, it appears that at 10 A of current the cathode will be operating at a temperature
of T = 1725 K or 1452 °C.

4.19 Field-assisted emission in MOS device Metal-oxide-semiconductor (MOS) transistors

in microelectronics have metal gate on an SiO2 insulating layer on the surface of doped Si crystal.
Consider this as a parallel plate capacitor. Suppose the gate is an Al electrode of area 50 µm × 50 µm
and has a voltage of 10 V with respect of the Si crystal. Consider two thicknesses for the SiO2, (a) 100
Å and (b) 40 Å, where (1 Å = 10-10 m). The work function of Al is 4.2 eV, but this refers to electron
emission into vacuum, whereas in this case, the electron is emitted into the oxide. The potential energy
barrier ΦB between Al and SiO2 is about 3.1 eV, and the field emission current density is given by
Equation 4.46a and b. Calculate the field emission current for the two cases. For simplicity take me to
be the electron mass in free space. What is your conclusion?
Solution
We can begin the calculation of field emission current finding the values of the field independent
e3
constants Ec, B = and the area A of the Al electrode.
8π hΦ B
Thus,

4.28
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

[ ]
1

8π (2me Φ 3B ) 8π 2(9.1×10 −31 kg )(3.1× 1.602 ×10 −19 J )

1/ 2 3 2

Ec = = = 3.726 × 1010 V m-1

3eh 3(1.602 ×10 −19 C )(6.626 ×10 −34 J s )

B=
e3
=
(
1.602 × 10 −19 C )
3

= 4.971 × 10-7 A V-2

( )(
8π hΦ B 8π 6.626 ×10 −34 J s 3.1× 1.602 × 10 −19 J )
A = (50 × 10 −6 m )× (50 × 10 −6 m) = 2.5 × 10-9 m2
When the thickness of SiO2 layer d is 100 Å, the field in the MOS device is
10 V
E= −10
= 1 × 109 V m-1
100 × 10 m
and the field emission current is
⎛ E ⎞
I = AJ field −emissiom = AB E 2 exp⎜ − c ⎟
⎝ E⎠

( )( )( 2
= 2.5 × 10 −9 m 2 4.971× 10 −7 A V − 2 1× 109 V m −1 exp ⎢−) (
⎡ 3.726 ×1010 V m −1 ⎤ )
⎣ 1× 109 V m −1 (⎥
⎦ )
= 8.18 × 10-14 A.
In the second case the SiO2 layer is 2.5 times thinner (40 Å) and the field in the device is 2.5 times
stronger.
10 V
E= −10
= 2.5 × 109 V m-1
40 × 10 m
The current in this case is
⎛ E ⎞
I = AJ field −emissiom = AB E 2 exp⎜ − c ⎟ =
⎝ E⎠
⎡ (3.726 ×1010 V m −1 )⎤
= (2.5 × 10 −9 m 2 )(4.971× 10 −7 A V − 2 )(2.5 ×109 V m −1 ) exp ⎢−
2

⎣ (2.5 × 10 V m ) ⎦
9 −1 ⎥

= 2.62 × 10-3 A.
So, as predicted by equation 4.47, the field-assisted emission current is a very strong function
of the electric field.

4.20 CNTs and field emission The electric field at the tip of a sharp emitter is much greater
than the “applied field,” Eo. The applied field is simply defined as VG/d where d is the distance from
the cathode tip to the gate or the grid; it represents the average nearly uniform field that would exist if
the tip were replaced by a flat surface so that the cathode and the gate would almost constitute a
parallel plate capacitor. The tip experiences an effective field E that is much greater than Eo, which is
expressed by a field enhancement factor β that depends on the geometry of the cathode–gate emitter,
and the shape of the emitter; E = βEo. Further, we can take Φ e2ff Φ ≈ Φ 3 / 2 in Equation 4.46. The final
expression for the field-emission current density then becomes
4.29
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

1.5 ×10 −6 2 2 ⎛ 10.4 ⎞ ⎛ 6.44 ×10 7 Φ 3 / 2 ⎞

J= β Eo exp⎜ 1/ 2 ⎟ exp⎜⎜ − ⎟⎟ [4.85]
Φ ⎝Φ ⎠ ⎝ β Eo ⎠
Where J is in A cm-2, Eo is in V cm-1, and Φ is in eV. For a particular CNT emitter, Φ = 4.9 eV.
Estimate the applied field required to achieve a field-emission current density of 100 mA cm-2 in the
absence of field enhancement (β = 1) and with a field enhancement of β = 800 (typical value for a
CNT emitter).

Solution
In the absence of field enhancement, β = 1 and current density, J = 100 mA cm-2 =0.1 A cm-2 and Φ =
4.9 eV
Substituting in Equation 4.85, we get
1.5 × 106 2 ⎛ 10.4 ⎞ ⎛ 6.44 ×10 7 × 4.93 / 2 ⎞
0 .1 = Eo exp⎜ 1/ 2 ⎟ exp⎜⎜ − ⎟⎟
4 .9 ⎝ 4 .9 ⎠ ⎝ Eo ⎠

⎛ 69.85 × 10 7 ⎞
or, 2.976 × 10 −9 = Eo2 exp⎜⎜ − ⎟⎟
⎝ Eo ⎠
Solving this equation we get, Eo = 13.32×106 V cm-1 (or 13.32 V/µm)
In the presence of field enhancement, β = 800
1.5 × 10 6 ⎛ 10.4 ⎞ ⎛ 6.44 × 10 7 × 4.93 / 2 ⎞
∴ 0 .1 = × 800 2 × Eo2 exp⎜ 1/ 2 ⎟ exp⎜⎜ − ⎟⎟
4 .9 ⎝ 4 .9 ⎠ ⎝ 800Eo ⎠
⎛ 69.85 × 10 7 ⎞
or, 4.65 × 10 −15 = Eo2 exp⎜⎜ − ⎟
⎝ 800Eo ⎟⎠

Solving this equation we get, Eo = 16.65×103 V cm-1 (or 0.016 V/µm)

Author's Note to the Instructor: The term exp (10.4/F) is usually ignored.

4.21 Nordhein-Fowler field emission in a FED Table 4.10 shows the results of I-V
measurements on a Motorola FED microemitter. By a suitable plot show that the I-V follows the
Nordheim-Fowler emission characteristics. Can you estimate Φ?

Solution

4.30
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

⎛ b ⎞
The Nordheim-Fowler emission characteristics is I A = aVG2 exp⎜⎜ − ⎟⎟ , where a and b are constants.
⎝ VG ⎠
So a plot of ln( I A / VG2 ) versus 1/VG must be a straight line.
VG 40.0 42 44 46 48 50 52 53.8 56.2 58.2 60.4
Iemission 0.40 2.14 9.40 20.4 34.1 61 93.8 142.5 202 279 367
(
ln I / VG2 ) -8.3 -6.7 -5.3 -4.6 -4.2 -3.7 -3.36 -3.0 -2.75 -2.5 -2.3

1/VG 0.025 0.024 0.023 0.022 0.021 0.02 0.019 0.0185 0.018 0.017 0.0165

Table 4Q21-1

2
Plot of ln(I /Vg ) versus 1/Vg

0
0.016 0.018 0.02 0.022 0.024 0.026
-1

-2

-3
ln(I/Vg )
2

-4

-5

-6

-7

-8

-9

1/Vg

Figure 4Q21-1: Plot of ln(I/Vg2) versus 1/Vg

Note: Nordhein-Fowler field emission is applicable only at high fields so we have to neglect the three
points in the low field region.
Φ cannot be calculated. Since we do not have the distance or the electrode separation to calculate
electric field.

4.22 Lattice waves and heat capacity

a. Consider an aluminum sample. The nearest separation 2R (2 × atomic radius) between the Al-Al
atoms in the crystal is 0.286 nm. Taking a to be 2R, and given the sound velocity in Al as 5100 m
s-1, calculate the force constant β in Equation 4.66. Use the group velocity νg from the actual

4.31
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

dispersion relation, Equation 4.55, to calculate the “sound velocity” at wavelengths of Λ = 1 mm, 1
µm and 1 nm. What is your conclusion?
b. Aluminum has a Debye temperature of 394 K. Calculate its specific heat capacity at 30 °C
(Darwin, Australia) and at -30 °C (January, Resolute Nunavut, Canada).
c. Calculate the specific heat capacity of a germanium crystal at 25 °C and compare it with the
experimental value in table 4.5

Solution
a. The group velocity of lattice waves is given by Equation 4.55. For sufficiently small K, or long
wavelengths, such that 1/2Ka →0, the expression for the group velocity can be simplified like in
Equation 4.6 to
β
νg = a
M
From here we cam calculate the force constant β
2
⎛ν ⎞
β = M ⎜⎜ g ⎟⎟
⎝ a ⎠
The mass of one Al atom is
M at
M=
NA
and finally for the force constant we receive

M ⎛ν
β = at ⎜⎜ g
⎞
2

⎟⎟ =
(
27 × 10 −3 kg mol−1 ) ⎛⎜ 5100 m s ⎞
−1 2

⎟⎟ = 14.26 kg s-2
NA ⎝ a ⎠ (
6.022 × 10 23 mol−1 ) ⎜⎝ 0.286 ×10
−9
m⎠
2π
Now considering the dispersion relation K = and Equation 4.55 we receive
Λ
1
⎛ β NA ⎞2 ⎛ π a ⎞
ν g (Λ ) = a⎜⎜ ⎟⎟ cos⎜ ⎟
⎝ M at ⎠ ⎝ Λ ⎠
Performing the calculations for the given wavelengths, we receive the following results:
ν g (10 −3 m ) = 5100 m s-1

ν g (10 −6 m ) = 5099.998 m s-1

ν g (10 −9 m ) = 3176.22 m s-1

It is evident that for the first two wavelengths, 1/2Ka →0 and we can use the approximation in
Equation 4.66. For the third wavelength, this is not true and we have to use the exact dispersion
relation when calculating the group velocity.

4.32
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

b. In summer, the temperature is given to be T = 30 °C = 303 K and T/TD is 303/394 = 0.769. The
molar heat capacity of Al at 30 °C is
Cm = 0.92 × (3R) = 22.95 J K-1 mol-1
The corresponding specific heat capacity is
Cm (22.95 J K −1 mol−1 )
cs = = = 0.85 J K-1 g-1
M at (27 g mol )
−1

At -30 °C, T = 243 K and T/TD is 243/394 = 0.62.

Cm (22.40 J K −1 mol−1 )
Cm = 21.94 J K mol and cs =
-1
=
-1
= 0.81 J K-1 g-1
M at (27 g mol )
−1

c. We can find the heat capacity of Ge in the way described in part b. Alternatively, we can find Cm
performing the integration in Equation 4.64 numerically
3
TD
⎡ 360
3 298 ⎤
⎛T ⎞ T 4
x e x
⎢ ⎛ 298 ⎞ x4ex ⎥
Cm = 9 R⎜⎜ ⎟⎟ ∫( dx = 3R⎢3⎜ ⎟ ∫ dx ⎥ = 3R(0.931)
⎝ TD ⎠ 0 ex −1
2
) ⎢⎣ ⎝ (
360 ⎠ 0 e x − 1) 2
⎥⎦

= 23.22 J K-1 mol-1

Thus the specific heat capacity is:

cs =
Cm
=
(23.22 J K −1 mol−1 ) = 319.9 J K-1 kg-1
M at (72.59 × 10 −3 kg mol−1 )
From Table 4.5, the specific heat capacity is 23.38 J K-1 mol-1.

4.23 Specific heat capacity of GaAs and InSb

a. The Debye temperature TD of GaAs is 344 K. Calculate its specific heat capacity at 300 K and at -
30° C.
b. For InSb, TD = 203 K. Calculate the room temperature specific heat capacity of InSb and compare
it with the value expected from the Dulong-Petit rule (T > TD).

Solution
a. T = 300 K, TD = 344 K
∴ (T / TD) = 0.87
From Figure 4.45, the molar heat capacity, Cm = 23.1 J K-1 mol-1
The specific heat capacity cs from the Debye curve is
Cm 23.1 JK −1mol-1
cs = ≈ −1
= 0.32 J K-1 g-1
M at 72.3 gmol
At -30° C, T = 243 K,

4.33
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

∴ (T / TD) = 0.706
From Figure 4.45, at -30° C the molar heat capacity, Cm = 22.7 J K-1 mol-1
The specific heat capacity cs from the Debye curve is
Cm 22.7 JK −1mol-1
cs = ≈ = 0.31 J K-1 g-1
M at 72.3 gmol−1
b. T = 300 K, TD = 203 K
∴ (T / TD) = 1.478
TD
3
⎛T ⎞ T
x4ex
0.677
x4ex
Cm = 9 R⎜⎜ ⎟⎟ ∫ (e dx = 9 R(1.478) ∫ (e
3
∴ dx
⎝ TD ⎠ 0
x
)
−1
2
0
x
−1 ) 2

= 24.37 J K-1 mol-1 = 0.977 (3R)

The calculated value is close to 3R (from Dulong-Petit rule).

4.24 Thermal conductivity

a. Given that silicon has Young’s modulus of about 110 GPa and a density of 2.3 g cm-3, calculate the
mean free path of phonons in Si at room temperature.
b. Diamond has the same crystal structure as Si but has a very large thermal conductivity, about 1000
W m-1 K-1 at room temperature. Given that diamond has a specific heat capacity cs of 0.50 J K-1 g-1,
Young’s modulus of 830 GPa, and density ρ of 0.35 g cm-3, calculate the mean free path of
phonons in diamond.
c. GaAs has a thermal conductivity of 200 W m-1 K-1 at 100 K and 80 W m-1 K-1 at 200 K. Calculate
its thermal conductivity at 25 °C and compare with the experimental value of 44 W m-1 K-1. (Hint:
Take κ ˜ T-n in the temperature region of interest; see Figure 4.48)

Solution

4.34
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Figure 4.48: Thermal conductivity of sapphire and MgO as a function of temperature.

a. Assume room temperature of 25 °C (298 K). For this temperature from Table 4.5, we can find the
thermal conductivity κ (κ = 148 W m-1 K-1) for silicon and its specific heat capacity Cs (Cs = 0.703 J
K-1 g-1). We can calculate the phonon mean free path at this temperature from Equation 4.68,
3κ
ph =
CVυ ph
where CV is the heat capacity per unit volume. CV can be found from the specific heat capacity CV =
Y
ρCs and the phonon velocity can be obtained from Equation 4.67, υ ph ≈ .
ρ
Thus the phonon mean free path in Si at 25 °C is
3κ ρ 3κ
= = =
ρCs
ph
Y Cs ρY

3(148 W m −1 K −1 )
= = 3.971 × 10-8 m
(0.703 ×10 3 −1
J K kg −1
) (2.3 ×10 3
kg m −3
)(110 ×10 9
Pa )
b. The mean free path of phonons in diamond is
3κ 3(1000 W m −1 K −1 )
= =
ph
Cs ρY (0.5 ×10 3
J K −1 kg −1 ) (3.5 × 103 kg m −3 )(830 × 109 Pa )

= 1.113 × 10-7 m
c. The temperatures at which the thermal conductivity κ is given can be considered as relatively high.
For this temperature range, we can assume that CV is almost constant and since the phonon velocity is
approximately independent from temperature according to Equation 4.68 the thermal conductivity is
proportional to the mean free path of phonons ph . Since the phonon concentration increases with

4.35
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

1
temperature, nph ∝ T, the mean free path decreases as ph ∝ . Thus, κ decreases in the same manner
T
with temperature as in Figure 4.48.
We can assume that the temperature dependence of the thermal conductivity is given by:
A
κ (T ) = +B
T
Then we have two equations and two unknowns
A
200 = +B
100
B
80 = +B
200
and for the coefficients A and B we receive: A = 2.4 × 104 W m-1 and B = -40 W m-1 K-1
The thermal conductivity at 25 °C (298 K) is
2.4 ×10 4 W m −1
κ= − 40 W m −1 K −1 = 40.5 W m-1 K-1
298 K
which underestimates the experimental value of 44 W m-1 K-1 but, nonetheless, still close.
Alternatively, we can take κ = AT−n. Then
200 = A(100)−n and 80 = A(200)−n
solving we find, A = 8.8 ×104 and n = 1.32. Thus at T = 25 + 273 K,
κ = AT−n = (8.8 ×10 4)(298)−1.32 = 47.2 W m-1 K-1
which overestimates the experimental value of 44 W m-1 K-1 but, nonetheless, still close.

*4.25 Overlapping bands Consider Cu and Ni with their density of states as schematically
sketched in Figure 4.61. Both have overlapping 3d and 4s bands, but the 3d band is very narrow
compared to the 4s band. In the case of Cu the band is full, whereas in Ni, it is only partially filled.
a. In Cu, do the electrons in the 3d band contribute to electrical conduction? Explain.
b. In Ni, do electrons in both bands contribute to conduction? Explain.
c. Do electrons have the same effective mass in the two bands? Explain.
d. Can an electron in the 4s band with energy around EF become scattered into the 3d band as a result
of a scattering process? Consider both metals.
e. Scattering of electrons from the 4s band to the 3d band and vice versa can be viewed as an
additional scattering process. How would you expect the resistivity of Ni to compare with that of
Cu, even though Ni has two valence electrons and nearly the same density as Cu? In which case
would you expect a stronger temperature dependence for the resistivity?

4.36
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

Solution
a. In Cu the 3d band is full, so the electrons in this band do not contribute to conduction.
b. In Ni both the 3d and 4s bands are partially filled so electrons in both bands can gain energy from
the field and move to higher energy levels. Thus both contribute to electrical conductivity.
c. No, because the effective mass depends on how easily the electron can gain energy from the field
and accelerate or move to higher energy levels. The energy distributions in the two bands are different.
In the 4s band, the concentration of states is increasing with energy whereas in the 3d band, it is
decreasing with energy. One would therefore expect different inertial resistances to acceleration,
different effective mass and hence different drift mobility for electrons in these bands.
d. Not in copper because the 3d band is full and cannot take electrons. In Ni the electrons can indeed
be scattered from one band to the other, e.g. an electron in the 4s band can be scattered into the 3d
band. Its mobility will then change. Electrons in the 3d band are very sluggish (low drift mobility) and
contribute less to the conductivity.
e. Ni should be more resistive because of the additional scattering mechanism from the 4s to the 3d
band (Matthiessen's rule). This scattering is called s-d scattering. One may at first think that this s-d
scattering de-emphasizes the importance of scattering from lattice vibrations and hence, overall, the
resistivity should be less temperature dependent. In reality, electrons in Ni also get scattered by
magnetic interactions with Ni ion magnetic moments (Nickel is ferromagnetic; Ch. 8 in the textbook)
which has a stronger temperature dependence than ρ ∼ T.

*4.26 Overlapping bands at EF and higher resistivity Figure 4.61 shows the density of
states for Cu (or Ag) and Ni (or Pd). The d-band in Cu is filled and only electrons at EF in the s band
make a contribution to the conductivity. In Ni, on the other hand, there are electrons at EF both in the s
and d bands. The d band is narrow compared with the s band, and the electron's effective mass in this d
band is large; for simplicity, we will assume me* is "infinite" in this band. Consequently, the d-band
electrons cannot be accelerated by the field (infinite me*), have a negligible drift mobility and make no
contribution to the conductivity. Electrons in the s band can become scattered by phonons into the d
band, and hence become relatively immobile until they are scattered back into the s-band when they
can drift again. Consider Ni, and one particular conduction electron at EF starting in the s band. Sketch
schematically the magnitude of the velocity gained |vx – ux| from the field Ex as a function of time for
10 scattering events; vx and ux are the instantaneous and initial velocities, and |vx – ux| increases
linearly with time, as the electron accelerates in the s band, and then drops to zero upon scattering. If
4.37
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

τss is the mean time for s to s-band scattering, τsd is for s-band to d-band scattering, τds is for d-band to
s-band scattering, assume the following sequence of 10 events in your sketch: τss, τss, τsd, τds, τss, τsd,
τds, τss, τsd, τds. What would a similar sketch look like for Cu? Suppose that we wish to apply Equation
4.27. What does g(EF) and τ represent? What is the most important factor that makes Ni more resistive
than Cu? Consider Matthiessen's rule. (Note: There are also electron spin related effects on the
resistivity of Ni, but for simplicity these have been neglected.)

Solution
|vx - ux| s and d band overlap and s-d scattering

(a)

τss τss τsd τds τss τsd τds τss τsd τds Time

|vx - ux|
(b)

Expected for Cu

τss τss τss τss τss τss τss τss τss τss Time

Figure 4Q26-1: (a) Upper for Ni. (b) Lower for Cu

τ in Equation 4.27 represents the scattering time averaged for all possible scattering processes, and
g(EF) is the total density of states at EF. In the presence of multiple types of scattering, we have to
apply the Matthiessen rule to find τ. For example in the presence of two bands, s and d bands,
1 1 1 1 1
= + +
τ τ ss τ sd τ dd τ ds
where τss is the scattering time in the s-band, τisd is the scattering time from the s-band to the d-band,
τdd is the scattering time in the d-band, τds is the scattering time from the d-band to the s-band. Thus,
the overall τ will be shorter than the usual mean free time τss in a single s-band conduction. On the
other hand, for Cu, there is only one s-band for conduction, and the mean free time for scattering
events is for a single type of scattering process within the s-band (in Figure 4Q26-1b). An intuitive
way to look at conduction in Ni is that, time to time, a conduction electron in the s-band gets scattered
into the d-band where it is immobile. During this time it does not gain velocity from the field (the
effective mass in the d-band is very large), and hence, its overall average velocity is less than it would
have been had it stayed in the s-band all the time.
The most important factor is the fact that the conduction electrons in Ni can be scattered into a
band where their drift mobility is very low, and this additional scattering mechanism makes Ni more
resistive than Cu.

4.38
Solutions to Principles of Electronic Materials and Devices: 3rd Edition (06 June 2005) Chapter 4

4.27 Grüneisen's law Al and Cu both have metallic bonding and the same crystal structure.
Assuming that the Gruneisen's parameter γ for Al is the same as that for Cu, γ = 0.23, estimate the
linear expansion coefficient λ of Al, given that its bulk modulus K = 75 GPa, cs = 900 J K-1 kg-1, and
ρ= 2.7 g cm-3. Compare your estimate with the experimental value of 23.5×10-6 K-1.

Solution
Given that, K = 75 Gpa = 75×109 J m-3
cs = 900 J K-1 kg-1
and ρ = 2.7 g cm-3 = 2.7×103 kg m-3
ρcs (2.7 ×103 kgm −3 )(900JK −1kg −1 )
Now λ = 3γ = 3(0.23) −3
=22.36×10-6 K-1
K (75 × 10 Jm )
9

So the estimated value is close to the experimental value.

"After a year's research, one realises that it could have been done in a week."
Sir William Henry Bragg

4.39