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ELEKTRONIKOS PAGRINDAI
1
2008
1. PROPERTIES OF ELECTRONS
1. Electrons are microparticles.
2. Quantum mechanics allows to reveal properties of microparticles.
3. Quantum theory, the branch of physics which is based on quantization,
began in 1900 when Karl Ernst Ludwig Planck (1858 – 1947) published
his theory explaining the emission spectrum of black bodies. K. Planck
stated that … electromagnetic energy could be emitted only in quantized
form, … the energy could only be a multiple of an elementary unit W = hν.
4. Quantum mechanics was developed by N. Bohr, W. Heisenberg, P. Dirac,
E. Schrődinger, V. Pauli and other famous physicists.
Objectives:
Gain knowledge about properties of electrons (using principles of
quantum mechanics), that is necessary for understanding of
semiconductor physics and studies of semiconductor and electronic
devices.
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
2
2008
Wave-particle duality of microparticles. De Brouglie waves
Light has dual nature:
λ = c /ν
W = hν = hω
W = mc 2
p = mc = W / c = hν / c = h / λ
1924: L. de Brouglie postulated:
The wave-particle duality is the common property of microparticles.
The length and the frequency of the wave, that correspond to the
microparticle with energy W and momentum p, are given by
ν =W /h
λ =h/ p
Equations are called de Broglie’s equations.
Experiments made by K. Davison and L. Gemer in 1927 approved the theory
about the wave-particle duality .
De Broglie’s waves are not electromagnetic waves. They are probability
waves.
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
3
2008
The uncertainty principle
From classical mechanics we know that six parameters define the state of a
particle: three co-ordinates and three projections of momentum.
If we know the momentum of a microparticle, we can find the length of the wave
that corresponds to this microparticle. It means that a monochromatic harmonic
wave corresponds to the microparticle. This wave exists in the space from –∞ to
+∞. So, if we know the momentum p, we cannot find the co-ordinate of the particle:
the lack of precision of the co-ordinate of the particle is infinite.
In 1927 Heisenberg published the uncertainty principle. He showed that, if an
attempt is made to measure certain pairs of variables concerned with a physical
system, then the lack of precision with which the two variables can be specified
simultaneously becomes apparent.
∆x∆p ≥ h
∆x∆px ≥ h,
∆y∆p y ≥ h,
∆z∆p z ≥ h.
...we cannot measure co-ordinate and momentum simultaneously with unlimited
accuracy. The inequalities represent fundamental limitation imposed by
nature.
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
4
2008
The Schrodinger wave equation
According to Heizenberg’s inequalities we cannot find the co-ordinates and
momentum of a microparticle simultaneously. So we cannot apply Newton’s laws
to a microparticle.
According to de Broglie microparticles behave in a wave-like manner.
Waves are defined by wavefunctions. In 1926 Schrodinger proposed an equation
for predicting the wavefunction of a microparticle. Now the theory based on
Schrodinger equation is called wave or quantum mechanics.
The wavefunction can be found from the equation:
∂Ψ (x, y, z , t ) h 2
− jh
=
∆Ψ (x, y, z , t ) − Wp (x, y, z , t ) Ψ (x, y, z , t )
∂t
2m
Ψ ( x, y, z, t ) = ψ ( x, y, z )ϕ (t )
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
5
2008
The Schrodinger wave equation and its solution
∂Ψ (x, t ) h 2
− jh
=
∆Ψ (x, t ) − Wp (x ) Ψ (x, t )
∂t
2m
Ψ ( x, t ) = ψ ( x)ϕ (t )
2
h 2 1 d ψ ( x)
1 d ϕ (t )
− Wp = − jh
.
2
2m ψ ( x ) d x
ϕ (t ) d t
The equation can be simplified by separating out the time-depending and
position-depending parts:
d 2 ψ ( x)
dx
2
d ϕ (t )
dt
+
2m
h
= −j
2
[W − Wp ( x)]ψ ( x) = 0.
W
ϕ (t ).
h


ϕ (t ) = C exp − j
W 
t  = C e − jωt
h 
... if the energy is constant, the wavefunction is a harmonic function of
time t.
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
6
2008
Properties of a free microparticle
A microparticle is free, if no force acts it and its potential energy Wp = 0.
Then W = Wk, and the Schrodinger equation can be simplified:
d 2 ψ ( x)
dx
2
d 2 ψ ( x)
dx
2
+
2m
h
2
Wk ψ ( x ) = 0
+ k ψ ( x) = 0
2
mv 2
p2
h2
Wk =
=
=
2
2m 2mλ2
k = 2π / λ = p / h = ω / c
ψ ( x) = ψ 1 ( x ) + ψ 2 ( x) = A1 e jkx + A 2 e − jkx
k is the propagation
constant or wave number
of the de Broglie wave,
corresponding to the
microparticle.
The solution describes the superposition of two waves. The first term represents
the incident probability wave and the second term represents the reflected
wave travelling in the opposite direction.
h 2 4 π 2 h 2 2 ... the kinetic energy of a free microparticle may have
Wk =
=
k any value.
2
2m λ
2m
…the energy spectrum of the free microparticle is
continuous.
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
7
2008
Conclusions
1. The wave-particle duality is the common property of microparticles.
2. The length and the frequency of the wave, that correspond to the microparticle
with energy W and momentum p, are given by
ν =W /h
λ =h/ p
3. De Broglie’s waves are not electromagnetic waves. They are probability
waves.
4. It is not possible to measure the co-ordinate and momentum of a
microparticle simultaneously with unlimited accuracy. According to the
Heisenberg’s uncertainty principle
∆x∆p ≥ h
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
8
2008
Conclusions
5. According to de Broglie, microparticles behave in a wave-like manner. Waves
are defined by wavefunctions. The wavefunction, describing a micropartcle,
can be found as a result of solving Schrodinger equation.
6. When potencial energy is constant (does not depend on time), the
Schrodinger equation can be simplified by separating out the time-depending
and position-depending parts.
7. A microparticle is free, if no force acts it and its potential energy is 0. The
energy spectrum of the free microparticle is continuous.
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
9
2008
Properties of a free microparticle. Problems
1. The length of de Broglie’s wave that corresponds to the electron is 0.1 nm.
Find the frequency of the wave.
2. Prove that Ψ·Ψ*=|Ψ|2.
3. How are the phase velocity, and the group velocity related to the velocity of a
free microparticle?
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
10
2008
Microparticles and potential barriers
Let us consider what happens when a microparticle impinges on a potential
barrier.
Steps of solution of the problem:
1. Creation of the model.
2. Solution of the Schrodinger equation. General solution is found.
3. Application of boundary conditions.
4. Finding of reflection and transmission factors, describing probabilities of
reflection and passing the barrier by the particle.
5. Analysis of derived expressions in order to find:
•
what happens when a microparticle meets a low barrier;
•
what happens when a microparticle meets a high barrier.
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
11
2008
Microparticles and potential barriers
In order to reveal what happens when a microparticle meets the barrier, we
must solve the Schrodinger equation.
1 region
d 2 ψ 1 ( x)
2 region
d x2
+
2m
Wψ 1 ( x ) = 0
2
h
ψ 1 ( x) = A1 e jk1 x + B1 e − jk1 x
k1 =
d 2 ψ 2 ( x)
d x2
k2 =
2π
λ2
VGTU EF ESK
+
=
2m
(W − Wb )ψ 2 ( x) = 0
2
h
2m(W − Wb )
h
2π
λ1
2mW
h
=
ψ 2 ( x ) = A 2 e jk 2 x + B 2 e − jk 2 x
B1
A1
2
2
2
=R
[email protected]
λ1 A2
=D
2
λ 2 A1
R + D =1
ELEKTRONIKOS PAGRINDAI
12
2008
Microparticles and potential barriers
... Boundary conditions must be satisfied:
ψ 1 (0) = ψ 2 (0),
Reflection coefficient
ψ 1 ' (0) = ψ 2 ' (0).
 k1 − k 2 

R = 
 k1 + k 2 
2
•... a microparticle can be reflected by the low potential barrier.
• If the height of the low potential barrier increases, the reflection coefficient
also increases and becomes 1 when the height of the low barrier becomes equal
to the energy of the microparticle.
If the barrier becomes
high, a microparticle
is reflected.
k2 =
jk 2∗
=j
2m(Wp − W )
h
2
2
ψ 2 = A2 e
− 2 k 2• x
•... microparticles penetrate into the high potential barrier before reflection from
the barrier.
• if the barrier is sufficiently thin, there is a small but finite probability that a
microparticle can penetrate the barrier.
• This is the essence of the tunnel effect.
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
13
2008
A microparticle in a one dimensional potential well
Let us reveal properties of a microparticle being in a potential well.
Steps of solution of the problem:
1. Creation of the model taking into account practically important cases.
2. Solution of the Schrodinger equation. General solution is found.
3. Application of boundary conditions.
4. Analysis of derived expressions in order to find:
•
energy spectrum of a microparticle;
•
probability to find microparticle.
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
14
2008
A microparticle in a one dimensional potential well
In order to reveal properties of a microparticle, we must solve the
Schrodinger equation.
d 2 ψ ( x)
d x2
k=
2π
λ
2m
+ 2 Wkψ ( x) = 0
h
=
2mW
h
ψ ( x) = A e jkx + B e − jkx
ψ (0) = 0,
ψ (a) = 0.
ψ ( x) = 2 j A sin kx = C sin kx
ψ (a) = C sin ka = 0
nπ
k=
a
h2 2
h2 2
W=
k =
n
2
2m
8ma
n is integer (n =1, 2, 3,…). It is called the quantum
number.
... The energy spectrum of a microparticle in a
potential well is discrete. The total energy of a
microparticle has particular allowed values that are
dependent upon quantum number n.
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
15
2008
Conclusions
1. A microparticle can be reflected by the low potential barrier. If the
height of the low potential barrier increases, the reflection
coefficient also increases and becomes 1 when the height of the
low barrier becomes equal to the energy of the microparticle.
2. Microparticles penetrate into the high potential barrier before
reflection from the barrier. If the barrier is sufficiently thin, there is
a small but finite probability that a microparticle can penetrate
through the barrier. This is the essence of the tunnel effect.
3. The energy spectrum of a microparticle in a potential well is
discrete. The total energy of a microparticle has particular allowed
values that are dependent upon quantum number n.
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
16
2008
Properties of microparticles. Problems
1. An electron having kinetic energy of 10 eV meets the potential barrier that
is 6 eV high. Find probability that electron will be reflected by the barrier.
2. Compute the three lowest energy levels for an electron trapped in a onedimensional potential well of length 0.1 nm.
3. An electron is confined in a one-dimensional potential energy well of
length 0.3 nm. Find the kinetic energy of the electron when in the ground
state and the spectral frequency resulting from a transition from the next
state (n = 2) to the ground level.
4. An electron is in the infinitively deep one-dimensional potential well
having rectangular form. The wave function ψ(x)=Csin(nπx/a) corresponds
to the electron. Using normalization condition find constant C at n = 1.
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
17
2008
Electrons in atoms
Taking into account properties of a microparticle in a potential well, let us
consider states of an electron in a hydrogen atom and states of electrons in
many-electron atoms.
A hydrogen-like atom consists of a central nucleus surrounded by an electron.
The potential energy of the electron is given by
Zq
Zq 2
Wp = −
4πε 0 r
According to the graph the electron is trapped in the potential well. The well
is three-dimensional.
VGTU EF ESK
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ELEKTRONIKOS PAGRINDAI
18
2008
Electrons in
atoms
Elektronų
savybės
In order to reveal properties of the electron it is necessary to solve the threedimensional Schrodinger equation using spherical coordinate system.
Because the well is three dimensional, the solution is more complicate.
Let us skip the solution and turn our attention to the main results.
Because the electron in a hydrogenic atom moves in three-dimensional space
and the Schrodinger equation is also three-dimensional, three quantum
numbers are necessary to characterise the state of the electron in the atom.
The first quantum number is the main or principal quantum number. It may
have values : n =1; 2; …
It defines the total energy of the electron:
Z 2 mq 4 1
Rch
Wn = − 2 2 2 = − Z 2 2
8ε 0 h n
n
... The electron can occupy
allowed energy levels. At n = 1,
the atom and the electron are in
the ground state.
VGTU EF ESK
Z = 1,
W1 = –13,6 eV
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ELEKTRONIKOS PAGRINDAI
19
2008
Electrons in atoms
The second quantum number is called the orbital or azimuthal quantum
number: l = 0; 1; ... ; (n-1).
l = 0 – s-state; l = 1 – p- state; l = 2 – d- state; l = 3 – f- state, …
The second quantum number is associated with the angular momentum of
the electron which is itself quantized.
L = mr × v
L = h l (l + 1 )
The third quantum number is called the magnetic quantum number. It can
take values: –l, ..., –1, 0; 1;..., l .
The values of the magnetic quantum number are associated with the
projection of the angular momentum on the direction of the external
magnetic field.
L =mh
H
l
In 1922 O. Stern and W. Gerlach found that … if
the beam from the electron gun is directed into
inhomogeneous magnetic field, the beam is split
into two parts.
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SternGerlach/SternGerlach.html
http://en.wikipedia.org/wiki/Stern-Gerlach_experiment
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
20
2008
Electrons in atoms
In 1925 R. Kronig, G. Uhlenbeck and S. Goudsmit proposed the concept of
elementary particle spin.
... a further quantum number is necessary to define completely a particular
quantum state which is not apparent from the solution of the three-dimensional
Schrodinger equation.
It is called the spin quantum number. Its possible values 1/2 and –1/2 correspond
to the spin of the electron in an either clockwise or anticlockwise direction.:
LsH = ms h
… four quantum numbers are necessary to define the quantum state of
the electron in the hydrogen atom or hydrogenic atom.
Solving of the Schrodinger equation in the case of a many-electron atom gives
that the state of an electron is also defined by four quantum numbers. The total
energy of an electron in the many-electron atom depends on the values of the
first and second quantum numbers, n and l .
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
21
2008
Electrons in many-electron atoms
Besides nucleus other electrons of a manyelectron atom act on an electron. As a result there
are more allowed energy levels in the manyelectron atom, corresponding to quantum
numbers n and l.
Electrons that have the same principal quantum number are said to be in the
same shell.
A shell may contain subshells that correspond to values of the azimuthal
quantum number.
The states that correspond to subshells usually are denoted using designation
letters s, p, d, f that correspond to values 0, 1, 2, 3 of the orbital quantum
number.
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
22
2008
Electrons in many-electron atoms
Electrons tend to fill the lowest allowed energy levels first.
According to Pauli’s exclusion principle, no more than one electron
can exist in any one quantum state.
Electronic structures of various chemical elements may be found and
the structure of the periodic table may be constructed on the basis of
these two principles and the quantum mechanical theory of atoms.
VGTU EF ESK
[email protected]
ELEKTRONIKOS PAGRINDAI
23
2008
Properties of electrons. Problems
1. How many electrons may be in the states corresponding to the various
sets of quantum numbers?
2. Find the electronic structures of Si and Ge atoms.
3. A hydrogen atom is excited from a state with n =1 to one with n = 4.
Calculate and display on the energy-level diagram the different photon
energies that may be emitted if the atom returns to its ground state.
4. The helium ion He+ has emitted energy quantum of 7.5 eV. What
happened in the ion?
5. A photon with a wavelength of 75 nm knocks out an electron from the
hydrogen atom. Find the length of the de Broglie wave corresponding to
the electron that becomes free.
VGTU EF ESK
[email protected]