Chapter2 introduction to quantum mechanics
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About This Presentation
solid states physics
Slide Content
Microelectronics I
Chapter 2: Introduction to
Quantum Mechanics
2.1 Principles of Quantum Mechanics
2.2 Schrodinger’s Wave Equation
2.3 Applications of Schrodinger’s Wave equation
2.4 Extensions of the Wave Theory to Atoms
Chapter 2: Introduction to
Quantum Mechanics
2.1 Principles of Quantum Mechanics
2.2 Schrodinger’s Wave Equation
2.3 Applications of Schrodinger’s Wave equation
2.4 Extensions of the Wave Theory to Atoms
Microelectronics I : Chapter 2
Control electron in the solid (crystal)
MPosition
M
Velocity
i
device’s speed
I
∝
n
x q x
v
Current,
Introduction 1
Need to know
electron behavior
in the crystal and
the material (energy band, etc)
M
Velocity
i
device’s speed
MNo. of electron
Density of electron velocity
Introduction of quantum mechanics (Defines electron with wave function)
iSchrodinger equation
Control electron in the solid (crystal)
MPosition
M
Velocity
i
device’s speed
I
∝
n
x q x
v
Current,
Introduction 1
Need to know
electron behavior
in the crystal and
the material (energy band, etc)
M
Velocity
i
device’s speed
MNo. of electron
Density of electron velocity
Introduction of quantum mechanics (Defines electron with wave function)
iSchrodinger equation
Microelectronics I : Chapter 2
cQuantum mechanic becomes more significant as electr onic device becomes smaller
Appearance of “
quantum effect
”
Introduction 2
Current, I
Voltage, V
Current, I
Width, W
I
Ohm’s law
“classical”
Smaller W
I
?
Electron channel
V
“classical”
V
MHere, classical Physics no longer applicable !!
cQuantum mechanic becomes more significant as electr onic device becomes smaller
Appearance of “
quantum effect
”
Introduction 2
Current, I
Voltage, V
Current, I
Width, W
I
Ohm’s law
“classical”
Smaller W
I
?
Electron channel
V
“classical”
V
MHere, classical Physics no longer applicable !!
Microelectronics I : Chapter 2
Objective:
To understand the basic of quantum mechanics
MWave-particle duality
MSchrodinger equation
-equation
-physical meaning
MApplication:
-quantized energy
-tunneling effect
Chapter 3: energy band theory of solids
Objective:
To understand the basic of quantum mechanics
MWave-particle duality
MSchrodinger equation
-equation
-physical meaning
MApplication:
-quantized energy
-tunneling effect
Chapter 3: energy band theory of solids
Microelectronics I : Chapter 2
propagates as wave (frequency, ν)
Particle having energy, hν
*h, Planck's constant = 6.625 x 10
-34
Js
light
1905, Einstein
M
Interference
1905, Einstein “Photon (discrete packet)”
Explains the
photoelectric effect
M
Interference
MRefraction
Mdiffraction
light
photoelectron
Max kinetic energy
of photoelectron
Wave-Particle duality
frequency
ν
o
propagates as wave (frequency, ν)
Particle having energy, hν
*h, Planck's constant = 6.625 x 10
-34
Js
light
1905, Einstein
M
Interference
1905, Einstein “Photon (discrete packet)”
Explains the
photoelectric effect
M
Interference
MRefraction
Mdiffraction
light
photoelectron
Max kinetic energy
of photoelectron
Wave-Particle duality
frequency
ν
o
Microelectronics I : Chapter 2 Wave-Particle duality of electron
MElectron: charged
particle
(q=1.6 x 10
-19
C)
M
De Broglie (1924)
I
∝
n
x q x v
ex:
M
De Broglie (1924)
Particle with momentum, p has
wavelength
, λ
λ= h/p
P: Planck’s constant
Ex:
e
Velocity, v=10
5
m/s
Wave nature
e
λ= h/p =h/m.v
=6.625 x 10
-34
/(10
5
x 9.11 x 10
-31
)
= 7.27 nm
MElectron: charged
particle
(q=1.6 x 10
-19
C)
M
De Broglie (1924)
I
∝
n
x q x v
ex:
M
De Broglie (1924)
Particle with momentum, p has
wavelength
, λ
λ= h/p
P: Planck’s constant
Ex:
e
Velocity, v=10
5
m/s
Wave nature
e
λ= h/p =h/m.v
=6.625 x 10
-34
/(10
5
x 9.11 x 10
-31
)
= 7.27 nm
Microelectronics I : Chapter 2 Electron gun
Double slitscreen
e
Wave-particle duality : experiment
When electron hits the screen,
a dot
will appears
Particle nature
e
①Shoot electron 4 times
Double slitscreen
Particle nature
1st
M
Electron could go through the slit
Electron gun
e
2nd
3rd
4th
M
Electron could go through the slit
MPosition of electron was random
Double slitscreen
e
Wave-particle duality : experiment
When electron hits the screen,
a dot
will appears
Particle nature
e
①Shoot electron 4 times
Double slitscreen
Particle nature
1st
M
Electron could go through the slit
Electron gun
e
2nd
3rd
4th
M
Electron could go through the slit
MPosition of electron was random
Microelectronics I : Chapter 2
②Shoot electron many time
Electron gun
slitscreen
e
http://www.hitachi.com/rd/research/em/doubleslit.ht ml
e
slitscreen
a: 8 electrons, b: 270 electrons, c: 2000 electrons , d: 160,000 electrons
MInterference pattern
Wave nature
d: 160,000 electrons
②Shoot electron many time
Electron gun
slitscreen
e
http://www.hitachi.com/rd/research/em/doubleslit.ht ml
e
slitscreen
a: 8 electrons, b: 270 electrons, c: 2000 electrons , d: 160,000 electrons
MInterference pattern
Wave nature
d: 160,000 electrons
Interpretations Microelectronics I : Chapter 2
1. Electron propagates in space like wave
2. Each electron pass through both two open slit at the same time
3. Electron interfere with itself
Electron gun
Double slitscreen
e
e
The experimental results confirm the wave-particle duality of electron
1. Electron propagates in space like wave
2. Each electron pass through both two open slit at the same time
3. Electron interfere with itself
Electron gun
Double slitscreen
e
e
The experimental results confirm the wave-particle duality of electron
Microelectronics I : Chapter 2 The Uncertainty principle (Heisenberg 1927)
i
p
i
x ≥
ħ
①
Impossible to simultaneously describe with absolute accuracy the position and
momentum of a particle
i
p
i
x ≥
ħ
i
E
i
t ≥
ħ
②Impossible to simultaneously describe with absolute accuracy the energy of
particle and the instant of time the particle has t his energy
ip: uncertainty in momentum
ix : uncertainty in position
i
E
i
t ≥
ħ
iE: uncertainty in energy
it : uncertainty in time
MCannot determine exact position of electron
use “probability”
i
p
i
x ≥
ħ
①
Impossible to simultaneously describe with absolute accuracy the position and
momentum of a particle
i
p
i
x ≥
ħ
i
E
i
t ≥
ħ
②Impossible to simultaneously describe with absolute accuracy the energy of
particle and the instant of time the particle has t his energy
ip: uncertainty in momentum
ix : uncertainty in position
i
E
i
t ≥
ħ
iE: uncertainty in energy
it : uncertainty in time
MCannot determine exact position of electron
use “probability”
Microelectronics I : Chapter 2 Schrodinger Wave Equation
Describe and discuss electron behavior
Wave functionisin, cos
k : wave number
Ψ
(x,t)= a.cos(kx
-
ω
t+
ε
))
Ψ(x,t)= exp ( j(kx- ωt))
= exp( jkx)exp(
-
jω
t)k : wave number ω: angular momentum
ε: initial phase
Ψ
(x,t)= a.cos(kx
-
ω
t+
ε
))
Use “exp”
exp(jθ)=cosθ+jsinθ
ε=0
= exp( jkx)exp(
-
jω
t)
Ψ(x,t)= Φ(t)φ(x)
position-dependent time-dependent
Describe and discuss electron behavior
Wave functionisin, cos
k : wave number
Ψ
(x,t)= a.cos(kx
-
ω
t+
ε
))
Ψ(x,t)= exp ( j(kx- ωt))
= exp( jkx)exp(
-
jω
t)k : wave number ω: angular momentum
ε: initial phase
Ψ
(x,t)= a.cos(kx
-
ω
t+
ε
))
Use “exp”
exp(jθ)=cosθ+jsinθ
ε=0
= exp( jkx)exp(
-
jω
t)
Ψ(x,t)= Φ(t)φ(x)
position-dependent time-dependent
Hφ(x)=Eφ(x)
Time-independent Schrodinger equation
Hamiltonian
Schrodinger Wave Equation2 Microelectronics I : Chapter 2
Energy of electron
Hamiltonian
Hamiltonian: total energy operator
H = kinetic energy + potential energy
)(
2
2
2 2
xV
xm
H+
∂
∂
−=
h
Energy of electron
)( )()(
2
2
2 2
x E x xV
xm
ϕ ϕ
=
+
∂
∂
−
h
Time-independent Schrodinger equation
Hamiltonian
Schrodinger Wave Equation2 Microelectronics I : Chapter 2
Energy of electron
Hamiltonian
Hamiltonian: total energy operator
H = kinetic energy + potential energy
)(
2
2
2 2
xV
xm
H+
∂
∂
−=
h
Energy of electron
)( )()(
2
2
2 2
x E x xV
xm
ϕ ϕ
=
+
∂
∂
−
h
Microelectronics I : Chapter 2
Physical meaning of wave equation
φ(x): wave function
|φ(x)|
2
:
probability
of existence of electron at x
position of electron cannot be determined precisely
∫ ∫
+∞
∞−
+∞
∞−
= =1 )()( |) (|
* 2
dxx x dx x
ϕ ϕ ϕ
total probability=1
φ*(x): complex conjugate function
total probability=1
Physical meaning of wave equation
φ(x): wave function
|φ(x)|
2
:
probability
of existence of electron at x
position of electron cannot be determined precisely
∫ ∫
+∞
∞−
+∞
∞−
= =1 )()( |) (|
* 2
dxx x dx x
ϕ ϕ ϕ
total probability=1
φ*(x): complex conjugate function
total probability=1
Microelectronics I : Chapter 2 Major quantities: energy, momentum, position of electron
)
(
)
(
x
x
i
x
H
ϕ
ϕ ∂∂h
Energy,
Momentum,
Result :equation
)(
)
(
x x
x
x
iϕ
ϕ
∂
Momentum,
Position,
Value of major quantities given by
expected value
in probability theory
∫
+∞
∂
*
ϕ
ϕ
h
∫
+∞
ϕ
ϕ
∫
+∞
dx
x
x
x
)
(
)
(
*
ϕ
ϕ
energymomentumposition
∫
∫
∞+
∞−
∞−
∂∂
dxx x
dxx
xi
x
)()(
)( )(
*
*
ϕ ϕ
ϕ
ϕ
h
∫
∫
∞+
∞−
∞−
dxx x
dxx Hx
)()(
)( )(
*
*
ϕ ϕ
ϕ
ϕ
∫∫
∞+
∞−
∞−
dxx x
dx
x
x
x
)()()
(
)
( *
*
ϕ ϕϕ
ϕ
Result :real number
)
(
)
(
x
x
i
x
H
ϕ
ϕ ∂∂h
Energy,
Momentum,
Result :equation
)(
)
(
x x
x
x
iϕ
ϕ
∂
Momentum,
Position,
Value of major quantities given by
expected value
in probability theory
∫
+∞
∂
*
ϕ
ϕ
h
∫
+∞
ϕ
ϕ
∫
+∞
dx
x
x
x
)
(
)
(
*
ϕ
ϕ
energymomentumposition
∫
∫
∞+
∞−
∞−
∂∂
dxx x
dxx
xi
x
)()(
)( )(
*
*
ϕ ϕ
ϕ
ϕ
h
∫
∫
∞+
∞−
∞−
dxx x
dxx Hx
)()(
)( )(
*
*
ϕ ϕ
ϕ
ϕ
∫∫
∞+
∞−
∞−
dxx x
dx
x
x
x
)()()
(
)
( *
*
ϕ ϕϕ
ϕ
Result :real number
Microelectronics I : Chapter 2
Region I
Region II
Boundary Condition
x=a
Condition 1:
φ(x) must be finite, single
-
valued, and
continuous
Condition 1:
φ(x) must be finite, single
-
valued, and
continuous
Condition 2: ∂φ(x)/∂x must be finite, single-valued and
continuous
ax
II
ax
I
II I
x x
a a
= =
∂
∂
=
∂
∂
=
ϕ ϕ
ϕ
ϕ
)( )(
Condition 1
Condition 2
Region I
Region II
Boundary Condition
x=a
Condition 1:
φ(x) must be finite, single
-
valued, and
continuous
Condition 1:
φ(x) must be finite, single
-
valued, and
continuous
Condition 2: ∂φ(x)/∂x must be finite, single-valued and
continuous
ax
II
ax
I
II I
x x
a a
= =
∂
∂
=
∂
∂
=
ϕ ϕ
ϕ
ϕ
)( )(
Condition 1
Condition 2
Microelectronics I : Chapter 2 Basic solution of Schrodinger equation
Consider V: constant
)( )(
2
2
2 2
x E x V
xm
ϕ ϕ
=
+
∂
∂
−
h
0)(
) (2)(
2 2
2
=
−
−
∂
∂
x
E Vm
x
x
ϕ
ϕ
h
--eq.1
Second order differential equation
∂
x
h
constant
Consider V: constant
)( )(
2
2
2 2
x E x V
xm
ϕ ϕ
=
+
∂
∂
−
h
0)(
) (2)(
2 2
2
=
−
−
∂
∂
x
E Vm
x
x
ϕ
ϕ
h
--eq.1
Second order differential equation
∂
x
h
constant
Microelectronics I : Chapter 2
1. if, E < V
0
) (2
2
2
> =
−
α
h
E Vm
eq.1
ϕ
α
ϕ
2 "
0
=
−
Solution:
x x
Be Ae
α α
ϕ
−
+ =
A,B: Coefficient
ϕα ϕϕ
α
ϕ
2 "
0
=
=
−
∞∞
Condition 1(
φmust be finite
)
x
Be
α
ϕ
−
=
1. if, E < V
0
) (2
2
2
> =
−
α
h
E Vm
eq.1
ϕ
α
ϕ
2 "
0
=
−
Solution:
x x
Be Ae
α α
ϕ
−
+ =
A,B: Coefficient
ϕα ϕϕ
α
ϕ
2 "
0
=
=
−
∞∞
Condition 1(
φmust be finite
)
x
Be
α
ϕ
−
=
1. if, E > V
0
) (2
2
2
> −=
−
−
β
h
E Vm
eq.1
ϕ
β
ϕ
=
+
Microelectronics I : Chapter 2
Solution:
xi xi
De Ce
β β
ϕ
−
+ =
C,D: Coefficient
eq.1
ϕβ ϕ
ϕ
β
ϕ
2 "
2 "
0
−=
=
+
MWave function is given by the combination of the tw o type solution
0
) (2
2
2
> −=
−
−
β
h
E Vm
eq.1
ϕ
β
ϕ
=
+
Microelectronics I : Chapter 2
Solution:
xi xi
De Ce
β β
ϕ
−
+ =
C,D: Coefficient
eq.1
ϕβ ϕ
ϕ
β
ϕ
2 "
2 "
0
−=
=
+
MWave function is given by the combination of the tw o type solution
Microelectronics I : Chapter 2 Application 1: Potential well
∞∞
region I
region II
region III
region I and III
V(x)=∞
x=L
x=0
region I
region II
region III
V(x)=∞
Electron cannot exist in the regions
region II (0<x<L)
−
∂
ϕ
φ=0
……Time-independent equation
V=0
0)(
2)(
2 2
2
= +
∂
∂
x
mE
x
x
ϕ
ϕ
h
---eq. 1
0)(
) (2)(
2 2
2
=
−
−
∂
∂
x
E Vm
x
x
ϕ
ϕ
h
∞∞
region I
region II
region III
region I and III
V(x)=∞
x=L
x=0
region I
region II
region III
V(x)=∞
Electron cannot exist in the regions
region II (0<x<L)
−
∂
ϕ
φ=0
……Time-independent equation
V=0
0)(
2)(
2 2
2
= +
∂
∂
x
mE
x
x
ϕ
ϕ
h
---eq. 1
0)(
) (2)(
2 2
2
=
−
−
∂
∂
x
E Vm
x
x
ϕ
ϕ
h
eq. 1
Microelectronics I : Chapter 2
)(
2 )(
2 2
2
x
mE
x
x
ϕ
ϕ
h
−=
∂
∂
ϕ
β
ϕ
2 "
−
=
2
2
2
hmE
=
β
ϕ
β
ϕ
−
=
2
h
Solution ;
xi xi
Be Ae
β β
ϕ
−
+ =
Boundary condition;
0 )0( )0(
=
=
II I
ϕ
ϕ
0 )( )(
=
=
L L
III II
ϕ
ϕ
0
=
+
B A0= +
−Li Li
Be Ae
β β
...eq. 2...eq. 3
Microelectronics I : Chapter 2
)(
2 )(
2 2
2
x
mE
x
x
ϕ
ϕ
h
−=
∂
∂
ϕ
β
ϕ
2 "
−
=
2
2
2
hmE
=
β
ϕ
β
ϕ
−
=
2
h
Solution ;
xi xi
Be Ae
β β
ϕ
−
+ =
Boundary condition;
0 )0( )0(
=
=
II I
ϕ
ϕ
0 )( )(
=
=
L L
III II
ϕ
ϕ
0
=
+
B A0= +
−Li Li
Be Ae
β β
...eq. 2...eq. 3
Microelectronics I : Chapter 2
From eq. 2 & 3
0 ) (
= −
−
Li Li
e eA
β β
A≠0
0
=
−
−
e
e
L
i
L
i
β
β
region II
0 ) sin( 20
==
−
−
L i
e
e
L
i
L
i
β
β
β
n; integer
2
n
L
mE
n L
=
=
π
π
β
x=L
x=0
E
n=1
E
n=2
E
n=3
2
2
2
2
n
L m
E
n
L
=
= π
π
h
h
x=L
x=0
The energy of particle is quantized
Particular discrete values
Classical; continuous values
From eq. 2 & 3
0 ) (
= −
−
Li Li
e eA
β β
A≠0
0
=
−
−
e
e
L
i
L
i
β
β
region II
0 ) sin( 20
==
−
−
L i
e
e
L
i
L
i
β
β
β
n; integer
2
n
L
mE
n L
=
=
π
π
β
x=L
x=0
E
n=1
E
n=2
E
n=3
2
2
2
2
n
L m
E
n
L
=
= π
π
h
h
x=L
x=0
The energy of particle is quantized
Particular discrete values
Classical; continuous values
Microelectronics I : Chapter 2
Wave function
xi xi
Be Ae
β β
ϕ
−
+ =
=x
L
n
iA
π
sin 2
=
x
L
n
C
L
π
sin
normalization
1 sin
0
2 2
=
∫
dxx
L
n
C
L
π
Total probability=1
2L
C
2
=
=x
L
n
L
π
ϕ
sin
2
n=1,2,3,4……
Wave function
xi xi
Be Ae
β β
ϕ
−
+ =
=x
L
n
iA
π
sin 2
=
x
L
n
C
L
π
sin
normalization
1 sin
0
2 2
=
∫
dxx
L
n
C
L
π
Total probability=1
2L
C
2
=
=x
L
n
L
π
ϕ
sin
2
n=1,2,3,4……
Microelectronics I : Chapter 2
Corresponding probability functions
Wave functions
x=L
x=L
Corresponding probability functions
Wave functions
x=L
x=L
Microelectronics I : Chapter 2 Application 2: Potential well
region Iregion II
region III
V
0
x=L
x=0
e
Region I
0)(
2)(
2 2
2
= +
∂
∂
x
mE
x
x
I
I
ϕ
ϕ
h
Region II
Region III
0)(
2)(
2 2
2
= +
∂
∂
x
mE
x
x
III
III
ϕ
ϕ
h
0)(
) (2)(
2
0
2
2
=
−
−
∂
∂
x
E Vm
x
x
II
II
ϕ
ϕ
h
region Iregion II
region III
V
0
x=L
x=0
e
Region I
0)(
2)(
2 2
2
= +
∂
∂
x
mE
x
x
I
I
ϕ
ϕ
h
Region II
Region III
0)(
2)(
2 2
2
= +
∂
∂
x
mE
x
x
III
III
ϕ
ϕ
h
0)(
) (2)(
2
0
2
2
=
−
−
∂
∂
x
E Vm
x
x
II
II
ϕ
ϕ
h
Microelectronics I : Chapter 2
Consider
E<V
0
,
,
) (2
2
0 2
h
E Vm
−
=
α
2
2
2
h
mE
=
β
Solution; wave function Solution; wave function
xi xi
I
Be Ae
β β
ϕ
−
+ =
xi xi
III
Fe Ee
β β
ϕ
−
+ =
x x
II
De Ce
α α
ϕ
−
+ =
region Iregion II
region III
V
0 A
B
C
D
E
F
x=L
x=0 F=0
xi
III
Ee
β
ϕ
=
Consider
E<V
0
,
,
) (2
2
0 2
h
E Vm
−
=
α
2
2
2
h
mE
=
β
Solution; wave function Solution; wave function
xi xi
I
Be Ae
β β
ϕ
−
+ =
xi xi
III
Fe Ee
β β
ϕ
−
+ =
x x
II
De Ce
α α
ϕ
−
+ =
region Iregion II
region III
V
0 A
B
C
D
E
F
x=L
x=0 F=0
xi
III
Ee
β
ϕ
=
Microelectronics I : Chapter 2
Boundary condition;
)
(
)
(
)0( )0(L
L
III
II
II I
ϕ
ϕ
ϕ
ϕ
=
=
Continuous wave function
)
(
)
(
L
L
III
II
ϕ
ϕ
=
)( )(
)0( )0(
' '
' '
L L
III II
II I
ϕ ϕ
ϕ ϕ
=
=
Continuous first derivative4 equation
Can solve for the
4 coefficients
B, C, D, E in term of A
Boundary condition;
)
(
)
(
)0( )0(L
L
III
II
II I
ϕ
ϕ
ϕ
ϕ
=
=
Continuous wave function
)
(
)
(
L
L
III
II
ϕ
ϕ
=
)( )(
)0( )0(
' '
' '
L L
III II
II I
ϕ ϕ
ϕ ϕ
=
=
Continuous first derivative4 equation
Can solve for the
4 coefficients
B, C, D, E in term of A
Microelectronics I : Chapter 2
Parameter of interest;
Transmission coefficient, T
region Iregion II
region III
V
0 AE
*
*
A
A
EE
T⋅⋅
=
x=L
x=0
A
A
⋅ )
2
exp(
1
16
L
E
E
T
β
−
−
≈
)
2
exp(
1
16
0 0
L
VE
VE
T
β
−
−
≈
MT is not zero
Electron penetrate the barrier
“tunneling”
Parameter of interest;
Transmission coefficient, T
region Iregion II
region III
V
0 AE
*
*
A
A
EE
T⋅⋅
=
x=L
x=0
A
A
⋅ )
2
exp(
1
16
L
E
E
T
β
−
−
≈
)
2
exp(
1
16
0 0
L
VE
VE
T
β
−
−
≈
MT is not zero
Electron penetrate the barrier
“tunneling”
Microelectronics I : Chapter 2
L
Wave function through the potential barrier
L
Wave function through the potential barrier
Microelectronics I : Chapter 2 Extensions of the wave Theory to Atoms
Potential function (coulomb attraction)
e
2
−
=
+
Nucleus; positively charged
r
e
rV
02
4
)(
πε
−
=
“Quantized
approximation
+
Quantum well
Expected results
+
Quantum well
“Quantized energy”
=x
L
n
L
π
ϕ
sin
2
n=1,2,3,…
Potential function (coulomb attraction)
e
2
−
=
+
Nucleus; positively charged
r
e
rV
02
4
)(
πε
−
=
“Quantized
approximation
+
Quantum well
Expected results
+
Quantum well
“Quantized energy”
=x
L
n
L
π
ϕ
sin
2
n=1,2,3,…
Microelectronics I : Chapter 2
Solving Schrodinger equation
0),,())( (
2
),,(
2
0 2
= − + ∇
φθ ψ φθ ψ
r rVE
m
r
h
Wave function & energy of electron in the atom Wave function & energy of electron in the atom
result
2 2 2
0
4
0
1
2) 4(n
em
E
n
h
πε−
=
n; 1, 2, 3,…
(principal quantum number)
1. Energy of atom is quantized. The value is determi ned by a quantum number,n
2. Wave function of electron also determined by quan tum numbers (n, l, m)
Solving Schrodinger equation
0),,())( (
2
),,(
2
0 2
= − + ∇
φθ ψ φθ ψ
r rVE
m
r
h
Wave function & energy of electron in the atom Wave function & energy of electron in the atom
result
2 2 2
0
4
0
1
2) 4(n
em
E
n
h
πε−
=
n; 1, 2, 3,…
(principal quantum number)
1. Energy of atom is quantized. The value is determi ned by a quantum number,n
2. Wave function of electron also determined by quan tum numbers (n, l, m)
Microelectronics I : Chapter 2
Mn: principal quantum number (determine total electron energy)
Quantum number Quantum states of electron
(determine total electron energy) N=1, 2, 3,….
Ml: azimuthal quantum number
(specifies the shape of atomic orbital)
l= n-1, n-2,……,0 (s,p,d,..)
Mm: magnetic quantum number
( direction)
|m|=l,l
-
1,…0
S
-
orbital
z
y
x
z
y
x
z
z
y
x
|m|=l,l
-
1,…0
Ms: spin quantum number
( spin of electron)
S=1/2, -1/2
S
-
orbital
y
x
p- orbital
Mn: principal quantum number (determine total electron energy)
Quantum number Quantum states of electron
(determine total electron energy) N=1, 2, 3,….
Ml: azimuthal quantum number
(specifies the shape of atomic orbital)
l= n-1, n-2,……,0 (s,p,d,..)
Mm: magnetic quantum number
( direction)
|m|=l,l
-
1,…0
S
-
orbital
z
y
x
z
y
x
z
z
y
x
|m|=l,l
-
1,…0
Ms: spin quantum number
( spin of electron)
S=1/2, -1/2
S
-
orbital
y
x
p- orbital
Microelectronics I : Chapter 2 electron
energy
ex: C (no. of electron: 14)
l=2p
+
n=1
n=2
l=1s
l=2s
l=2p
n=1
n=2
As n increases, energy of
quantum state increases
m
++
n=1
Pauli Exclusion Principle
No two electrons may occupy the same quantum state
quantum state increases
energy
ex: C (no. of electron: 14)
l=2p
+
n=1
n=2
l=1s
l=2s
l=2p
n=1
n=2
As n increases, energy of
quantum state increases
m
++
n=1
Pauli Exclusion Principle
No two electrons may occupy the same quantum state
quantum state increases
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