EEE 3394
Electronic Materials
Chris Ferekides
Fall 2014
Week 6
Electrical Conductivity
Insulators
Semiconductors
Conductors
Many ceramics
Superconductors
Alumina
Diamond Inorganic Glasses
Mica
Polypropylene
PVDF Soda silica glass
Borosilicate Pure SnO2
PET
SiO2
10-18
Metals
Degenerately
Doped Si
Alloys
Intrinsic Si
Amorphous
Intrinsic GaAs
As2Se3
10-15 10-12
10-9
10-6
10-3
100
Conductivity(Wm)-1
Te Graphite NiCr Ag
103
106
109
1012
Charges “under the influence”
• Everyone must know what happens to a
charge in the presence of an electric field
E

Charges “under the influence”
• Everyone must know what happens to a
charge in the presence of a magnetic field
B

Right vs. Left-Handed Oriented x y z
System … and the Cross Product
In order ....
i.e. x  y  z
xˆ  yˆ  zˆ
yˆ  zˆ  xˆ
In reverse order
zˆ  xˆ  yˆ
i.e. z  y  x
yˆ  xˆ   zˆ
zˆ  yˆ   xˆ
xˆ  zˆ   yˆ
The Hall Effect
Example with p-type semiconductor; i.e. holes are
the majority charge carriers;
 apply voltage in x direction i.e. current Ix.
 apply a B-field in the z-direction.
 Total Force on the charge carriers due to E and B
fields is

  
F  q(E  v  B)
 The y-component of the force is
Fy  q(Ey  v x  Bz )
 As the holes flow in the x-direction they
experience a force in the y-direction due to the Bfield.
 Holes will accumulate in the -y end of the bar
setting up an electric field, i.e. a voltage VAB in the
y direction.
 The net force in the y-direction becomes zero when the two components of the force i.e. due
to the electric and due to the magnetic field are equal.
The Hall Effect
 The “setting up” of the E-field in the y-direction is
known as the Hall Effect.
 The voltage VAB is known as the Hall Voltage.
 This experiment is used to measure the mobility
of the charge carriers as explained below:
VAB  E y w
J X  qpν X  ν X 
JX
qp
Fy  q(E y  v x  Bz )
Jx
Fy  0  E y  v x Bz 
Bz
qp
J x B z (I x wt)Bz
I x Bz
p


qE y
q(VAB /w) qtVAB
The Hall Effect
Note: the current and magnetic fields are known
quantities since they are externally applied; the
hall voltage can be measured.
 The resistivity ρ of the sample can also be
calculated by measuring the resistance R of the
bar.
R
ρL
wt
σ  qpμ p  μ p 
σ
qp
B
Bz
Jy = 0
+
Jx
Ey
eE y
+
eE y
+
+
A
z
Jx
vex
evhxB z
VAB>0
>0
+
Ex
v hx
y
evexB z
+
A
Bz
V
VAB<0
>0
x
Chapter 3
•
Wave-particle Duality … ??
•
•
Particle – electron has wave-like properties &
… wave – light has particle-like properties.
•
First let’s look at a wave …
E y (x,t) = E osin(kx - w t) where k =
2π
l
and w = 2π ×n =
Ey
Velocity = c
x
Bz
2π ×c
l
Waves
• What are some wave characteristics (behavior) ?
… interference
Constructive interference
P
S1
Destructive interference
S2
difference in distance
1
travelled proportional to (n + )l
2
Waves
• What are some wave characteristics (behavior) ?
Detector
X-rays
… diffraction
1
1
Photographic film
2

d
A
Scattered
Scattered
dsinX-rays
X-rays
Single
crystal
Powderedcrystalor
polycrystallinematerial
X-rayswith
withall
single
d X-rays
wavelength
wavelengths
Bragg's Law
2d sinq = nl
where n = 1, 2, 3, 4 ...
2
B

dsin
Atomic planes
Crystal
(c)
http://www.eserc.stonybrook.edu/ProjectJava/Brag
g/
Waves
• What are some wave characteristics (behavior) ?
Detector
X-rays
… diffraction
1
1
2
2
700
ZTO_1.5
d
d
Bragg's Law
2d sinq = nl
where n = 1, 2, 3, 4 ...
Intensity [cps]
600
AZTO_2.1 
ZTO_1.9
ZTO_2.0
ZTO_2.5
500
dsin
dsin
400
300
Atomic planes
B
Crystal
200
100
(c)
0
20
30
40
50
2θ [°]
60
70
Photoelectric Effect
• Light frees electrons which
can be collected at the anode
by applying a voltage – or
reach there if they have
enough KE
Light
CATHODE
ANODE
Electrons
Evacuated quartz tube
• Increasing the light intensity
leads to higher current
• VO – is the “stopping voltage”,
i.e. the voltage required to
“stop” electrons with enough
KE from reaching the anode
A
V
+
-
• Therefore VO is proportional
to the max KE energy of the
emitted electrons
I
I
Saturation
I2
I1
V0
0
V
Photoelectric Effect
• When the voltage is negative,
i.e. acts against the electrons,
the electrons must do work
against the potential equal to
eVO which is equal to the max KE
at the cathode
Light
CATHODE
ANODE
Electrons
Evacuated quartz tube
• If the light frequency
(wavelength) is the same, VO
remains the same regardless of
the light intensity …
-
Saturation
• Therefore, the light intensity is
independent of the electron
energy
eVO = KE max
A
V
+
I
1 2
= mv
2
I
I2
I1
V0
0
V
Photoelectric Effect
• If the light frequency is varied
then the stopping potential
changes
Light
CATHODE
ANODE
Electrons
• Therefore the frequency of the
light is proportional to the
energy of the emitted electrons
• Let’s look at the relationship
between the electron KE and the
light frequency: the KE
decreases with the frequency …
• The behavior is similar for
several metals
A
Evacuated quartz tube
V
+
KEm
Cs
1
eVO = KE max = mv 2
2
K
W
I
u03
0
slope=h
u >u
1
2
u
u01u 3 < u 2
02
• … BUT the slope is the same
I
F3 u
1
F2
V 01
F1
u2
V 02
V 03
u
u3
0
V
Photoelectric Effect
Light
• Below a certain frequency NO
electrons are emitted – i.e.
the KE of the electrons is
zero!
CATHODE
ANODE
Electrons
E
l
V
+
-
= hn - F
KE
l (nm)
A
Evacuated quartz tube
• The Photoelectric Effect
results tell us that:
– The number of electrons is
proportional to the light
intensity but independent of
max
M
the light frequency
– The energy of the electrons
is proportional to the
frequency of the light but
the frequency does not
hc the total number of 1240
affect
= hn =electrons
...emitted
also E(eV) =
I
KEm
Cs
K
u03
0
F3
F2
F1
W
slope=h
u02
u01
u
Photoelectric Effect
KEmax = hn - FM
Light
CATHODE
ANODE
Electrons
hc
1240
E = hn =
... also E(eV) =
l
l (nm)
Evacuatedquartztube
-
• CONCLUSION:
– light has particle like behavior!
• As the “intensity” of the light
increases, more light “particles”
are striking the metal target,
therefore more electrons being
emitted.
• As the frequency increases the
energy of each particle increases
KEm
V
+
Cs
K
u03
0
F3
F2
F1
• Light Particle: PHOTON!
I
A
W
slope=h
u02
u01
u
Example – (a)
• Green Light with wavelength 522 nm is the longest wavelength that
can cause electron photoemission from a sodium surface
– What is the work function of sodium?
KEmax = hn - FM
hc
1240
0 = hn - F M Þ F M = .....F M [eV] =
= 2.375eV
l
l[nm]
KEm
Cs
K
u03
0
W
slope=h
u02
u01
u
Light
CATHODE
ANODE
F3
F2
Electrons
Evacuatedquartztube
F1
-
V
+
I
A
Example – (b)
• If UV light of 250 nm is used what will be the KE of the
photo-emitted electrons?
KEmax = hn - FM
KE max
KEm
Cs
K
u03
0
1240
= hn - F M =
- 2.375 = 2.585eV
250
W
slope=h
u02
u01
u
Light
CATHODE
ANODE
F3
F2
Electrons
Evacuatedquartztube
F1
-
V
+
I
A
Example – (b)
• If the intensity of the 250 nm UV is 20 mW/cm2 and all electrons
are collected using a positive bias on the anode what will be the
current density?
2
20[mJ/(s×cm
)]
15
2
20mW/cm -2 = 20mJ/(s× cm 2 ) =
=
12.5x10
eV/(s×cm
)
-19
1.602x10 (J/eV)
12.5x1015eV/(s ×cm 2 )
= 2.520x1015 photons/(s× cm 2 )
4.96eV/photon
é electrons ù
-19
2
2.520x10 ê
x1.602x10
C
=
40.4mA/cm
2
ë s × cm úû
15
Light
CATHODE
ANODE
Electrons
Evacuatedquartztube
-
V
+
I
A
Compton Scattering
•
•
What happens when an x-ray “strikes”
an electron ?
The scattered x-ray has a different
frequency
Recoiling electron
X-ray photon
… and the electron “moves” …
•
The electron’s KE is related to the
characteristics of the incoming and
scattered x-rays
KE = hv - hv¢
c
Electron


u 
y
Scattered photon
u '  '
c
x
Scattering of an x-ray photon by a "free" electron in a conductor.
• Since the electron also gains
momentum … due to the
principle of conservation of
momentum … then the photon
(x-ray) also has momentum
p=
h
l
Therefore light … exhibits particle-like behavior
…comes in “packets”
… quanta
… is quantized
… photons!
… energy of a photon
hc
h
… has momentum
E  hu 
 ω where  
λ
h
p   k where k 
λ
2π
λ
2π
Electrons
Fluorescent screen
50kV
Two slits
h
l=
p
A 1 1 7 6 0 0 (1 9 2 8 )).
Filament
Electrons
Vacuum
(c ) E le c tro n d iffr a c tio n
p a tte r n o b ta in e d b y G .
P . T h o m s o n u s in g a
g o ld fo il ta rg e t.
Electron diffraction fringes on the
screen
(d ) C o m p o s ite p h o to g r a p h s h o w in g
d iffr a c tio n p a tte r n s p ro d u c e d w ith
a n a lu m in u m fo il b y x - ra y s a n d
e le c tro n s o f s im ila r w a v e le n g th .
L e ft: X -ra y s o f  = 0 .0 7 1 n m .
R ig h t: E le c tro n s o f e n e rg y 6 0 0 e V .
(e )
(e ) D iffr a c tio n p a tte rn p r o d u c e d b y 4 0 k e V e le c tr o n s p a s s in g th ro u g h z in c o x id e p o w d e r . T h e
d is trib u tio n o f th e p a tte rn w a s p r o d u c e d b y a s m a ll m a g n e t w h ic h w a s p la c e d b e tw e e n th e s a m p le
a n d th e p h o to g ra p h ic p la te . A n X -r a y d iffr a c tio n p a tte r n w o u ld n o t b e a ffe c te d b y a m a g n e tic fie ld .
Example
• 2000 Kg car running at 70 mph (~30m/s).
What is the wavelength !
Wavelength = 1.1 x 10-38 m
h
p
λ
Electrons Behave Like Waves
E(x,t) = EOsin(kx - w t) is a wave equation...
where E(x,t)
2
is the intensity of the wave
Y(x,t) is the wavefunction of an electron
Y(x,t) represents the probability of finding an electron ... !
2
Y(x, y,z,t)
2
is the probability of finding an electron at xyz time t
æ Et ö
Y(x,t) = y (x)exp ç - j ÷
è
ø
Electrons can behave like waves
… a behavior that can be described by
… Schrodinger’s Equation!
d Y 2m
+
(E
V)Y
=
0
2
2
dx
2
• Schrodinger’s time independent equation;
• V is only a function of space! … (otherwise we would
be in trouble!)
• Its solution results in information on the probability of
finding an electron “somewhere” and on its energy
“characteristics”
Properties of Ψ
(x) (x) not continuous
(x)
d not continuous
dx
x
(x)
x
(x) not single valued
x
Unacceptable forms of  (x)
From Principles of Electronic Materials and
Devices, Third Edition, S.O. Kasap (©
McGraw-Hill, 2005)
Fig 3.14
Infinite Potential Well
• We will solve Schrodinger’s time independent equation;
• We will confine an electron in a specified location … How?
• To complete the solution we need to apply BC’s!
d 2 Y 2m
+ 2 (E - V)Y = 0
2
dx
ψ(x)  Ae
jkx
 Be
 jkx
V(x)
Electron

 2k 2
E
2m
apply BC's ... Y(0) = 0 ... B = -A
Y(x) = A(e jkx - e- jkx ) = 2Ajsin(kx)
V= 
0
V=
V=0
0
a
x
Infinite Potential Well
Y(x) = A(e jkx - e- jkx ) = 2Ajsin(kx)
Y(a) = 2Ajsin(ka) = 0 ... therefore sin(ka) = 0
ka = 0 ...
therefore ka = nπ with n being an integer (n ¹ 0)
kn =
2m
2
k 2n 2 n 2 h 2
E ... E n =
=
2m 8ma
• THEREFORE the Energy of the Electron can only have
“quantized” values corresponding to n=1,2,3,4….
Eigen-energies!
Infinite Potential Well
y (x) = 2Ajsin(kx)
kn =
ò
a
ò
a
0
0
2m
2
k 2n 2 n 2 h 2
E ... E n =
=
2m 8ma
y (x) = ??
2
Y(x) dx = 1 = ò
2
a
0
æ 1ö
A=ç ÷
è 2a ø
1/2
nπ × x
2Ajsin(
) dx
a
2
æ 2ö
æ nπ × x ö
... therefore ... Y n (x) = jç ÷ sin ç
è aø
è a ÷ø
1/2
8
Infinite Potential Well … bottom line
V=
8
V=
8
V(x)
Electron
V=0
0
Energy of electron
E4
0
a
0
x
4
n=4
3
E3
n=3
2
E2
n=2
E1
n=1
x=0
x=a
Energy levels in the well
1
0
x
(x)  sin(nx/a)
a0
a
Probability density  |(x)|2
Electron in a one-dimensional infinite PE well. The energy of the
electron is quantized. Possible wavefunctions and the probability
distributions for the electron are shown.
Heisenberg’s Uncertainty Principle
• We cannot exactly and simultaneously know the position
and momentum of a particle along a certain coordinate …
• … mathematically
Dx × Dpx ³
• Essentially what the above means is that due to the wave nature of
quantum mechanics we cannot simultaneously know exactly the
position and momentum of a particle … this is a theoretical
limitation …
• i.e. the quantum nature of the universe limits us to only knowing the
product of position and momentum within ħ
DE × Dt ³
“Electron” in an isolated Atom …
Electron confined in a well size of 0.1nm. Calculate the ground energy. Calculate the
energy required to move it to the 3rd level. How can this energy be provided?
k 2n  2 n 2 h 2
En 

2m 8ma 2
Ground level, E1= 6.025x10-18 J or 37.6 eV
3rd level, E3 = E1 n2 = (37.6 eV)(3)2 = 338.4 eV
Energy required = 300.8 eV
This exact amount of energy can come from a photon with wavelength,
λ = hc/E = 4.12nm