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FE1002 Ph i IIFE1002 Physics II
An introduction toAn introduction to
Quantum Physics Q yAssociate Professor Zhang Qingg g
Microelectronics CentreS h l f EEESchool of EEE
Nanyang Technological Univeristy
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 1
Prescribed Text Book
“Physics for Scientists and Engineers with Modern Physics” (6th Edition)
R A S & J W J tt JR. A. Serway & J. W. Jewett, Jr. (QC23.S492P 2004)
Main ReferencesMain References
“University Physics” by Hugh D.Young & Roger A Freedman (12th
Edition) (QC21.2.Y72U 2008)
“Fundamentals of Physics” (Extended) (7th Edition) by D. Halliday, R. Resnick & J. Walker
(QC21.2H188 2005)
“Physics: principles with applications” (5th Edition)
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 2
by Douglas C. C. Giancoli (QC23G433 2005)
Topic One: IntroductionpIsaac Newtonian (1642-1727) developed the basic concepts and laws of
17th
1600
the basic concepts and laws of mechanics and proposed the law of universal gravitation, etc.
cs
18th
1700
ical
phy
sic
19thJames Clark Maxwell (1831 1879)
Dynamic characteristics of an object: position, velocity, momentum, etc
1700
clas
si 19thJames Clark Maxwell (1831-1879) developed the electromagnetic theory of light and kinetic theory of gases, etc.
1800
20th
Characteristics of electromagnetic waves: wavelength, velocity, amplitude, phase angle, etc.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 3
1900
20th
1900A major revolution in physics took place at the beginning of 20th century.
phys
ics
In 1900, Max Plank (1858-1974) introduced the concept of quantum
21st
mod
ern
p
2000
action.
In 1905 Albert Einstein (1879 1955)In 1905, Albert Einstein (1879-1955) published his special theory of relativity.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 4
1.1 BackgroundTopic One: Introduction
g
A few experimental observations could not be i d i h f k f l i h i
Blackbody radiation
interpreted in the framework of classic physics.
Topic 2
Photoelectric effect Emission of sharp spectrum lines from atoms
Topic 3
Topic 5lines from atoms
New concepts and theories must be introduced.
p
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 5
1.2 Wave & particle behaviorTopic One: Introduction
In quantum physics, wave and particle are not contradictorycontradictory.
Electromagnetic wave does show particle characteristics, i.e.,
Electrons and other particles are of
particle characteristics, i.e., momentum etc.
Topic 4
wave properties, i.e., wave package and wave length etc. Topic 6
The Schrödinger equation with certain boundary conditions links the wave and particle properties
Topics 7 & 8
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 6
the wave and particle properties together.
& 8
1.3 Applications of quantum physics
Topic One: Introduction
1.3 Applications of quantum physics
Classically, a particle will never overcome aClassically, a particle will never overcome a potential barrier whose barrier height is larger than the total energy of the particle. What about in quantum physics?
Topic 9
about in quantum physics?
Why do we have the periodic table for different elements? Topic 10different elements?
What is the difference between laser light and ordinary light? How is laser light Topic 11and ordinary light? How is laser light generated?
Topic 11
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 7
Topic Two: Blackbody Radiationp yAll objects at temperature T > 0 K emit thermal radiation.
The thermal radiation consists of a continuous distribution of wavelength from the infrared, visible and ultraviolet portions of the spectrumvisible and ultraviolet portions of the spectrum.
nsity
Mono
Spec
tral
Inte
n
Mono-chromator
Detector
As the temperature of an object increases, the major emitting portion of the spectrum shifts to a shorter
SFE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 8
emitting portion of the spectrum shifts to a shorter wavelength.
Topic Two: Blackbody Radiation
2.1 Spectral radiancySpectral radiancy or spectral intensity R(,T) tells us the radiation energy emitted by per unit area of an
p y
object, per unit time, per unit wavelength at a given temperature T and wavelength . (unit: Js-1m-3)
Intensity R(,T)d gives the radiated intensity or power per unit area for the wavelength from to +d. (unit: Js-1m-2))
ensi
ty R (,T)
pect
ral i
nte
I(T) +d
R(,T)dFigure 2-1: Spectral intensity of an emitter radiation versus wavelength at a temperature.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 9
Sp
( )
+d
I t it f λ t λ b d t i d b i t ti
Topic Two: Blackbody Radiation
Intensity from λ1 to λ2 can be determined by integrating in the spectral radiancy with respect to λ. (unit: Js-1m-2)
2
1 21
,I T R T d
(2.1a)
Radiant intensity or power density I(T) is the rediation energy emitted from per unit surface area of an object in
dTRTI
, (2.1)
gy p jper unit time for entire wavelength. (unit: Js-1m-2)
dTRTI 0 , ( )
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 10
2.2 Blackbody
Topic Two: Blackbody Radiation
2.2 Blackbody
A blackbody is defined as an object that absorbs all the i id t it
Good approximation of a blackbody is a hole leading to the inside of a hollow object.
energy incident on it.
A blackbody is also a perfect emitter since an object in thermal equilibrium emits as much energy as it absorbs.
Figure 2-2 : The opening to the cavity inside a hollow object is a good approximation of a blackbody.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 11
blackbody.
The radiation of a blackbody depends only on the
Topic Two: Blackbody Radiation
The radiation of a blackbody depends only on the temperature of the blackbody and not on the material of which the blackbody is made.The radiation in the visible region is only a small portion of the entire
Fi 2 3 S t l i t it f th
small portion of the entire radiation.
inte
nsity
Figure 2-3: Spectral intensity of the blackbody radiation versus wavelength (spectral radiancy) at three temperatures. Note that the 3500 KSp
ectra
l i
pamount of radiation emitted (area under the curve) increases with increasing temperature.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 12
visible region
2.3 The Stefan-Boltzmann lawTopic Two: Blackbody Radiation
The Stefan-Boltzmann Law: total radiated intensity or power per unit surface area of a blackbody isor power per unit surface area of a blackbody is only determined by the temperature T of the blackbody as
4TTI (2.2)
where =5.670 10-8 Wm-2K-4 is the Stefan-Boltzmann constantBoltzmann constant.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 13
Topic Two: Blackbody Radiation
Ordinary objects radiate less efficiently than a blackbody and the corresponding radiant intensity becomesbecomes
4TTI (2.3)
where 0< <1 is the emmisivity of the surface of the ordinary objects.
For a blackbody, = 1.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 14
2.4 Wien’s displacement law
Topic Two: Blackbody Radiation
2.4 Wien s displacement lawmax means the wavelength where the intensity or radiancy is maximum
• Wien’s Displacement Law: For a blackbody, the product of max and temperature T is a universal
radiancy is maximum.
p max pconstant, i.e,
Km 10898.2 3max T (2.4)
max
y
Figure 2-4: Wavelength max of the
ctra
l int
ensi
ty
peak of the curve shifts to shorter wavelengths at higher temperatures 3500 KSp
ecFE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 15
2.5 Two classical models
Topic Two: Blackbody Radiation
Rayleigh & Jeans Model: assuming that atoms are h i ill t th t it l t ti t
2.5 Two classical models
harmonic oscillators that emit electromagnetic waves at all wavelengths, Lord Rayleigh and James Jeans developed the following formula in 1900;
42,
TckTR BRJ (2.5)
where kB= 1.38×10-23 J/K is Boltzmann’s constant and cis the speed of light.
Eq.(2.5) fits the blackbody’s radiation curves well at long wavelengths (> 50 m), but completely fails at short wavelengths, because of RRJ(,T) as 0.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 16
g , RJ( , )
Wien Model: Wilheim Wien also gave an expression
Topic Two: Blackbody Radiation
Tb
W eaTR
5, (2.6)
based on a conjecture (“guess”)
W 5, ( )
where a and b are empirical constants chosen to give the best fit to the experimental data.
Wein’s formula fits the curves well at short wavelengths but
best fit to the experimental data.
gdeparts noticeably at longer wavelengths.
Figure 2-5: Comparison of the Wein and the Rayleigh-Jeans theories to that of Planck, which closely follows the experimental data
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 17
the experimental data.
2.6 Planck’s modelTopic Two: Blackbody Radiation
In 1900, Max Planck, seeking to reconcile these two laws, made an inspired interpolation which fits the data at all a elengthsall wavelengths.
12, 5
2
hhcTR (2 7)
1, 5
Tkhc
Be (2.7)
h = ?
Tk
hc
B is dimensionless quantity. TkB
1
BB k Tk T mJh Jsc c ms
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 18
Topic Two: Blackbody Radiation
Plank adjusted the constant h so that his simulated curve could perfectly match the observed blackbody’s radiation curves. y
h is Planck’s constant: h = 6.626 10-34 J·s
h l f h i i d d f h bl kb d ’The value of h is independent of the blackbody’s materials and temperatures.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 19
2 7 Planck’s two assumptions
Topic Two: Blackbody Radiation
Assumption 1: Oscillating atoms/molecules can have
2.7 Planck s two assumptions
only discrete values of energy En, given by
nhfEn (2.8). . . ,3,2,1,0n
where n is called a quantum number and f is the natural frequency of oscillation of the atoms/molecules.Radiation energy is quantized.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 20
Topic Two: Blackbody Radiation
Each value of n represents a specific quantum state.
molecule energy
Figure 2-6: Allowed energy levels for a molecule that oscillates at its natural frequency f.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 21
natural frequency f.
Assumption 2: The atoms/molecules emit or absorb
Topic Two: Blackbody Radiation
Assumption 2: The atoms/molecules emit or absorb energy in discrete packets (called photons) by jumping from one quantum state to another. D d t iti it
Energy of one photon is the product of h and f:Upward transition absorbs energy.Downward transition emits energy.
hchfE (2.9)
Energy of one photon is the product of h and f:
where c is the speed of light in vacuum.
Figure 2-7: A representation of photons. Each photon has a discrete energy hf
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 22
has a discrete energy hf.
Example 2-1*Topic Two: Blackbody Radiation
Find the peak wavelength of the blackbody radiation emitted by each of the following objects (as blackbodies):blackbodies):
(a) The human body when the skin temperature is 35oC(b) The tungsten filament of a light bulb operating at
2000 K(c) The Sun, which has a surface temperature of about
5800 K Solution: From Wien’s Displacement Law, Eq.(2.4), we can easily determine max as follows
For human body,
μm4.9Km10898.2 3
max
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 23
μm4.9K35273max
This radiation of 9.4 m is in the infrared region and
Topic Two: Blackbody Radiation
ginvisible to the human eye.
For tungsten filament,
μm4.12000K
mK10898.2 3
max
It is also in the infrared region so that most of emitted energy is not visible to us. For the sun,
50mK10898.2 3
μm5.05800Kmax
This is near the center of the visible spectrum.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 24
p
2.8 Max Planck (1858 –1947)
Topic Two: Blackbody Radiation
Max Karl Ernst Ludwig Planck (April 23, 1858 October 4 1947) was a German
2.8 Max Planck (1858 1947)
1858 – October 4, 1947) was a German physicist. At first Planck considered that thequantisation was only as “a purely formal assumption ... actually I did not think much about it ”; nowadays this assumption incompatibleabout it… ; nowadays this assumption, incompatible with classical physics, is regarded as the birth of quantum physics and the greatest intellectual
li h t f Pl k'Planck was awarded the Nobel prize for physics in 1918.accomplishment of Planck's career.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 25
Data Source: http://en.wikipedia.org/wiki/Max_Planck
Topic Three: The Photoelectric
The photoelectric effect: light incident on certain
Effectp g
metallic surfaces can cause electrons to be ejected from the surfaces. Th j d l ll d h lThe ejected electrons are called photoelectrons.
The induced current is called photoelectric current.
hf photoelectronsFigure 3-1: Photoelectrons are generated from the
metal
are generated from the surface of a metal under impact of an incident light beam.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 26
Topic Three: The Photoelectric Effect
metallic target Electron collection electrode
li d i l photoelectric currentapplied potential difference
Figure 3-2: An apparatus used to study the photoelectric effect. When a beam of monochromatic light with appropriate wavelength shines on the metallic target T photoelectrons will be collected by electrode C and the
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 27
metallic target T, photoelectrons will be collected by electrode C and the photoelectric current i is detected by the ammeter.
Topic Three: The Photoelectric Effect
3.1 Stopping potential For a large positive V, the current reaches a maximum value.
pp g p
The current increases as incident light intensity increases.
Saturated photo-electric current
Figure 3-3: Photoelectric current versus applied potential differences–
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 28
versus applied potential differences for two light intensities.
When a negative potential is applied to electrode C
Topic Three: The Photoelectric Effect
When a negative potential is applied to electrode C, photoelectrons are repelled, leading to a decrease in photocurrent. Stopping potential, VS: Any negative bias V < -|VS|, no photocurrent is detectable.VS is independent of radiation intensity, but dependent ofVS is independent of radiation intensity, but dependent of the frequency of the incident light. The maximum kinetic energy of the photoelectrons is
max | |sK q V (3.1)
where q=1 610-19 C is the elementary charge
Classically Kmax should depend on intensity.
where q 1.610 C is the elementary charge.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 29
max
3.2 Cutoff frequencyTopic Three: The Photoelectric Effect
|Vs| linearly increases with fCutoff frequency, f0: any frequency of the incident light b l f ill h t l t j tibelow f0 will cause no photoelectron ejection. f0 is characteristic of the target materials, independent of radiation intensity.
Classically, effect should occur at
f
sodium target
any frequency provided intensity is high enough.
Figure 3-4: The stopping potential Vstop as a function of the frequency f of the
g g
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 30
q y fincident light.
Topic Three: The Photoelectric Effect
Kmax increases with increasing light frequency.
Cl i ll K h ld b i d d t f f
Electrons ejected from surface almost instantaneously, l li h i i i
Classically Kmax should be independent of frequency.
even at low light intensities.
Classically, electrons were expected to require some time to absorb radiationtime to absorb radiation.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 31
Topic Three: The Photoelectric Effect
Simulation & Illustration
Photoelectric Effect
Key points: This cartoon illustrates the influences of light intensity, frequency and bi l h l ibias voltage, etc on photoelectric current.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 32
3.3 Einstein’s modelTopic Three: The Photoelectric Effect
In 1905, Einstein assumed that light of frequency f can be considered as a stream of photons with energy Ebe considered as a stream of photons with energy Egiven by E=hf [i,e, Eq.(2.9)].Photon is so localized that it gives all its energy to a single electron in the metal target.
The maximum kinetic energy of photoelectrons can be determined by the difference between photon energy and the work function (or the minimum energy with which an electron is bound to the metal), see Fig.3-5. ), g
hfKmax (3.2)
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 33
Einstein’s model explains the previously mentioned
Topic Three: The Photoelectric Effect
Einstein s model explains the previously mentioned features which the classical theory cannot explain.(1) The effect is not observed below f0 .
(2) K is independent of the incident light intensity.
® The photon energy of incoming photon must be .
(2) Kmax is independent of the incident light intensity.® Increasing intensity only increases the number of
photons.(3) Kmax increases with increasing f. ® Kinetic energy of photoelectrons depends only on f.(4) Th h t l t j t d i t t l(4) The photoelectrons are ejected instantaneously.® Incident energy appears in small packets and there
is a one-to-one interaction between photons and
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 34
is a one to one interaction between photons andelectrons.
Topic Three: The Photoelectric Effect
Electronic energy states in a metal.Case 1: hf1<, no photoelectrons are ejected.Case 2: hf2=, the critical case.Case 3: hf3>, photoelectronsCase 3: hf3 , photoelectrons with certain kinetic energy are generated. Kmax
E=0
The vacuum level
hf3
hfEF
level
Work function: the minimum energy
hf1hf2
The Fermi energy level: the highest
energy level
minimum energy required to move an electron from
the metal to
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 35
occupied by electrons at 0 K
vacuum. Figure 3-5: Electronic energy states in a metal.
I i l h l i i iExample 3-1*
Topic Three: The Photoelectric Effect
In a particular photoelectric experiment, a stopping potential of 2.1 V is measured when ultraviolet light with a wavelength of 290 nm is incident on a metal. Using the g gsame setup and metal, determine the stopping potential if blue light with a wavelength of 440 nm is used, instead of the ultraviolet lightthe ultraviolet light.
Solution
34 8
1 19 9
6.66 10 3.0 10 2.1 4.3 2.1 2.2 (eV)1 6 10 290 10
S
S
hf qV
hf qV
34 8
2 19 9
1.6 10 290 106.66 10 3.0 10 2.2 2.8 2.2 0.6 (eV)1.6 10 440 10SqV hf
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 36
2So, 0.6 (V)SV
Example 3-2*Topic Three: The Photoelectric Effect
The human eye can respond to as little as 10-18 J of light energy. Determine the number of photons that will lead to an observable flash at a wavelength of 550 nmto an observable flash at a wavelength of 550 nm.
Solution:
18 9
34 8
10 550 10 36 6 10 3 10
E ENhf hc
6.6 10 3 10hf hc
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 37
3.4 Albert Einstein (1879 –1955)Topic Three: The Photoelectric Effect
Albert Einstein (March 14, 1879 – April 18, 1955) was a German-born theoretical physicist, one of the greatest physicists of all timephysicists of all time. He played a leading role in formulating the special and general theories of relativity, quantum theory and statistical mechanics. He was awarded the 1921 Nobel Prize for Physics for his e o ys cs o sexplanation of the photoelectric effect in 1905 (his "wonderful year" or "miraculous year") andyear or miraculous year ) and "for his services to Theoretical Physics".
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 38
Data Source: http://en.wikipedia.org/wiki/Albert_Einstein
3.5 X-ray photoelectron Topic Three: The Photoelectric Effect
X Ph t l t S t (XPS) i
spectroscopy*X-ray Photoelectron Spectroscopy (XPS) is a quantitative spectroscopic technique that measures the empirical formula, chemical state and electronic state of the elements that exist within a material.
XPS spectra are obtained by irradiating a testing material with a beam of x-rays while simultaneously measuring the kinetic energy and number of electrons that escape from the top 1~10 nm of the material beingthat escape from the top 1 10 nm of the material being analyzed.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 39
Topic Three: The Photoelectric Effect
Figure 3-6: Schematic of an X-ray photoelectron spectroscopy.
Figure 3-7: An XPS spectrum for a testing sample which contains Si O C F N Sn etc
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 40
Si, O, C, F, N, Sn, etc.
3.6 Photomultiplier tubes*Topic Three: The Photoelectric Effect
Photomultiplier tubes (photomultipliers or PMTs) are extremely sensitive detectors of light in the ultraviolet, visible and near infrared regionsvisible and near infrared regions.
0V
500V500V
1500V
2500V
1000V2000V
Figure 3-8: Schematic of the structure of a typical PMT
4500V3500V3000V
4000V
5000V
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 41
structure of a typical PMT and a photo of PMTs.
Topic Three: The Photoelectric Effect
These detectors multiply the signal (number of electrons) produced by incident light by many times, from which a single photon could be resolvedfrom which a single photon could be resolved.
The combination of high gain, low noise, high frequency response and large area of collectionfrequency response and large area of collection makes PMTs suitable for applications in nuclear and particle physics, astronomy, medical imaging and motion picture film scanning etcand motion picture film scanning, etc.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 42
Topic Four: The Compton EffectIn 1923, A. H. Compton did an experiment that confirmed the particle-like aspect of electromagnetic
p p
p p gradiation.The Compton effect: the scattering phenomenon associated with incident x ray colliding with electronsassociated with incident x-ray colliding with electrons.
Figure 4-1: Compton’s apparatus. A beam of x-rays of wavelength = 71 1 i di t d t71.1 pm is directed onto a carbon target. The rotating crystal plays a role of a spectrometer.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 43
p
4.1 Scattering of x-raysTopic Four: The Compton Effect
g yThe scattered x-ray only y ypeak at 0 and ’ which is a function of
graphite
0o
90o
function of the angle.
Classically, the
45o
graphite
135o
ywavelength of the scattered x-ray is a broad distribution incident
x-ray 0
Figure 4-2: Scattering of x-ray as the function of
135obroad distribution at a given angle.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 44
Figure 4-2: Scattering of x-ray as the function of the angle with respect to the incident direction.
Topic Four: The Compton Effect
4.2 Scattering of electronsThe electron is scattered through an angle with respect to the incident photon as if this was a billiard-ball t pe collisionball type collision.
Figure 4-3: Illustration of the scattered x-ray and electron in the Compton effect.
Classically, the electron would be pushed along the incident direction due to radiation pressure and set into oscillatory
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 45
direction due to radiation pressure and set into oscillatory motion driven by the wave’s oscillating electric field.
4.3 Compton’s modelTopic Four: The Compton Effect
To fully explain the effect, Compton and his co-workers had to treat photons, not as waves, but rather as point-
p
had to treat photons, not as waves, but rather as pointlike particles assuming that the energy and momentum of any colliding photon-electron pair are conserved, respectivelyrespectively.
The energy of photons E = hf [Eq.(2.9)] and the t
(4.1)hhfp
momentum
(4.1)c
p
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 46
Using these ass mptions and the la s of energ
Topic Four: The Compton Effect
Using these assumptions and the laws of energy conservation and momentum conservation in both x and y directions, he derived the Compton Shift Equation;
cos1' c0 (4.2)
where ’ and 0 are the scattered and incident light wavelength, respectively, as indicated in Fig.4-3. = h/m c=0 00243 nm is called the Comptonc= h/mec=0.00243 nm is called the Compton wavelength, me=9.10910-31 kg is the rest mass for electron and c is light speed in vacuum.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 47
Topic Four: The Compton Effect
Compton’s measurements were in excellent agreement with his predictions and these really convinced physicists of the fundamental validity of
Photon transferred energy to the electrons during
convinced physicists of the fundamental validity of the quantum theory.
collision and hence the scattered x-ray has a longer wavelength.The unshifted peak (0) is due to scattering from theThe unshifted peak (0) is due to scattering from the atoms’ tightly bound inner electrons which do not gain significant energy from incident photons.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 48
Example 4-1*Topic Four: The Compton Effect
pA 1.0×10-12 m wavelength photon collides with a free electron initially at rest. After the collision, the photon recoils directly backward. (i) Determine the wavelength, momentum and energy of the scattered photon and (ii) determine the momentum and kinetic energy of the de e e e o e u a d e c e e gy o escattered electron.Solution:
θ =180o
λ f λ′ f ′
Figure 4 4: Illustration of momentum of the scattered photon and
λ0, f0 λ′, f ′
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 49
Figure 4-4: Illustration of momentum of the scattered photon and electron in the Compton effect.
Topic Four: The Compton Effect
Solution: From Eq.(4.2), we have the wavelength of the scattered photon
0 c
3412 o
' 1 cos
6.626 10 J s1.0 10 1 cos180
31 8 1
12 12 12
1.0 10 1 cos1809.11 10 kg 3.00 10 m s
1.0 10 4.8 10 =5.8 10 (m)
The momentum and energy of the scattered photon:34
226.6 10 1 1 10 (k / )h
22 8 141 1 10 3 0 10 3 3 10 (J)hcE p c
2212
6.6 0 1.1 10 (kgm/s)5.8 10ph
hp
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 50
1.1 10 3.0 10 3.3 10 (J)ph phE p c
Topic Four: The Compton Effect
From momentum and energy conservation, we have:h h
0e
h h p
epheph EEEE 00
Therefore, the momentum and kinetic energy of the
and
3412 12
0
1 16.6 101.0 10 5.8 10e
h hp
, gyscattered electron are:
022
1.0 10 5.8 10
7.7 10 (kgm/s)
0 01 1
k h hE E E E E hc
0 00
34 8 1312 12
1 16.6 10 3.0 10 1.6 10 (J)1 0 10 5 8 10
k e e ph phE E E E E hc
12 121.0 10 5.8 10
Note: according to classic physics
22 227 7 10p
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 51
13classic 31
7.7 103.3 10 (J)
2 2 9.1 10e
ke
pE
m
Wrong!
Topic Four: The Compton Effect
4.4 A. H. Compton (1892-1962) p ( )Arthur Holly Compton (September 10th, 1892 - March 15th, 1962) discovered the Compton effect, which clearly illustrates the particle concept of electromagnetic radiation, wasof electromagnetic radiation, was afterwards substantiated by C. T. R. Wilson who, in his cloud chamber, could show the presence of the trackscould show the presence of the tracks of the recoil electrons. For this discovery, Compton was awarded the Nobel y pPrize in Physics for 1927 (sharing this with C. T. R. Wilson).D t S
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 52
Data Source:http://nobelprize.org/nobel_prizes/physics/laureates/1927/compton-bio.html
Topic Five: Atomic SpectraLow-pressure gas subject to electric discharge emits discrete line spectrum (emission spectra)
p p
discrete line spectrum (emission spectra)
Mono-chromator
Detector
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 53
Figure 5-1: Emission line spectra for hydrogen, mercury and neon.
Absorption spectrum is obtained by passing light from
Topic Five: Atomic Spectra
Absorption spectrum is obtained by passing light from a continuous source through the gas or dilute solution.Absorption spectra of low pressure gases show a series of discrete dark lines superimposed on a continuous spectrum of light source.
Mono-chromator
Detector White light
Figure 5-2: Absorption spectrum for hydrogen gas.
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Topic Five: Atomic Spectra
5.1 Hydrogen emission line series In 1884, Johann Balmer realized that the wavelengths of the first 4 lines in the visible spectrum of hydrogen
y g
p y gwere related by
543111
R ...5,4,3, 1
211
22H
n
nR
(5.1)
h R 1 097 107 1 i ll d h R dbwhere RH= 1.097107 m-1, is called the Rydberg constant.
1/λ is called wavenumber.
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Other series of lines in hydrogen were soon found and
Topic Five: Atomic Spectra
Other series of lines in hydrogen were soon found and Balmer’s equation was generalized.
111 . . . 2, 1, ,111
22121
22
H
nnn
nnR
(5.2)
B l ’ i f 2 ( i ibl )Balmer’s series for n2 = 2 (visible)Lyman’s series for n2 = 1 (UV)Paschen’s series for n2 = 3 (IR)
All of these equations were purely empirical. No
Brackett’s series for n2 = 4 (IR)
theoretical basis existed for them. Classic electromagnetic theory could not explain these.
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these.
Topic Five: Atomic Spectra
5.2 The Bohr atomIn 1913, Niels Bohr proposed an atomic theory that accounted for the spectra lines of hydrogen. The model contains: (1) The electron moves in circular orbits around the proton under the influence of the Coulomb force ofproton under the influence of the Coulomb force of attraction.
Figure 5-3: Diagram represent-g g ping Bohr’s model of a hydrogen atom, in which the orbiting electron is allowed to be only in specific orbits
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in specific orbits.
Topic Five: Atomic Spectra
(2) Only certain orbits are stable i e electrons in these(2) Only certain orbits are stable, i.e, electrons in these orbits do not emit or radiate energy.
In classical physics, orbiting electrons should In classical physics, orbiting electrons should continuously radiate, losing energy and causing it to spiral into the nucleus.
(3) Radiation is emitted when electrons jump from a more energetic initial orbit to a lower energy orbit. Frequency of the emitted radiation depends on theFrequency of the emitted radiation depends on the difference between the energy of the initial state, Ei, and the energy of the final state, Ef and Ei > Ef.
hfEE fi (5.3)
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(4) The allowed orbits are those for which the electron’s bi l l b h l i
Topic Five: Atomic Spectra
orbital angular momentum about the nucleus is an integral multiple of ħ = h/2.
nvrm (5 4)nvrme (5.4)
Using these four assumptions, the allowed energy levels f h d b l l dof a hydrogen atom can be calculated:
32112
nekE e (5 5)...3,2, 1, ,
2 20
nna
En (5.5)
where ke=1/(40)=8.988109 Nm2/C2 is the Coulomb’s constant and a0= h2/[meke(2e)2]= 0.0529 nm is the Bohr radius, which corresponds to the orbit with the smallest radius.
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radius.
R di f ll d bit i
Topic Five: Atomic Spectra
02anrn (5.6)
Radius of any allowed orbit is
Fi 5 4 Th fi hFigure 5-4: The first three circular orbits predicted by the Bohr model of the hydrogen atom.
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y g
Topic Five: Atomic Spectra
With Eq.(5.5), we have the energy of hydrogen:
eV 606.132n
En (5.7)...3, 2, 1,n
Only energies satisfying Eq. (5.7) are permitted.n = 1 is the ground state with E1 = -13.606 eV.g 1 Vn = 2 is the first excited state with E2 = -3.401 eV.n = is the free electron state with E = 0 eV.
The energy needed to completely remove an electron from the atom in its ground electronic state is called the ionization energythe ionization energy.The ionization energy for hydrogen is E-E1=13.6 eV
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Frequency of the photon emitted when electron jumps
Topic Five: Atomic Spectra
Frequency of the photon emitted when electron jumps from an outer orbit i to an inner orbit f can be determined from Eqs.(5.3) and (5.5):
22
0
2 112 if
efi
nnhaek
hEE
f (5.8) f
The wavenumber or 1/ is:
2 111 ekf
22
0
112
1if
e
nnhcaek
cf
(5.9)
RH
Comparing Eq.(5.9) with Eq.(5.2), the Rydberg constant is
1-72
H m10097.1 ekR e (5 10)
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0H 2 hca (5.10)
The theoretical
Topic Five: Atomic Spectra
The theoretical Rydberg constant is accurate to within 1% of the experimentally determined value.The spectra series forThe spectra series for hydrogen can be interpreted as
IR
transitions between the energy levels.
visible
Figure 5-5: An energy level di f h d
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diagram for hydrogen
UV
5.3 Niels Bohr (1885-1962)
Topic Five: Atomic Spectra
( )Niels (Henrik David) Bohr (October 7, 1885 – November 18, 1962) was a , )Danish chemist who made fundamental contributions to understanding atomic structure andunderstanding atomic structure and quantum mechanics. Bohr is widely considered one of the greatest h i i f h i hphysicists of the twentieth century
(even though he self-identified as a chemist).)He received the Nobel Prize for Physics for his atomic theory in 1922.
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Data Source: http://en.wikipedia.org/wiki/Niels_Bohr
Topic Six: Wave-Particle DualityTopic Six: Wave-Particle DualityBlackbody experiments, photoelectric effect, Compton effect and atomic spectra offer ironclad evidence that when light and matter interact, they behave like particles.particles.
A photon has a speed of c, energy of hf=hc/ and momentum of hf/c=h/ and no mass.f
On the other hand, light and other electromagnetic waves exhibit interference and diffraction effects which could be interpreted only with wave characteristics.
Electromagnetic wave has its amplitude, frequency and
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phase angle.
Q Which model (particle or wave) is more appropriate?
Topic Six: Wave-Particle Duality
A. Depends on the phenomenon observed – some experiments are better or solely explained using
Q. Which model (particle or wave) is more appropriate?
experiments are better or solely explained using particle model while others are better or solely explained using wave model.
Light has a dual nature. It exhibits both wave and particle characteristicsand particle characteristics.
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Topic Six: Wave-Particle Duality
To understand why photons are compatible with electro-magnetic waves, let us consider a radio wave and x-ray.
Consider a 2.5 MHz radio wave (=120 m).
Its energy E =hf 10–8 eV – too small to be detected.
Requires a lot of photons to produce a detectable signal
At higher frequencies energy of photon is higher and
q p p g– graininess is lost.
At higher frequencies, energy of photon is higher and graininess can be easily detected.For very high frequencies such as x-ray and -ray, the photons are easily detected as a single event, but wave effects are difficult to observe.
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Topic Six: Wave-Particle Duality
6 1 The Wave properties of particles6.1 The Wave properties of particlesIn 1923, L. V. de Broglie, in his doctoral dissertation, postulated that because photons have wave andpostulated that because photons have wave and particle characteristics, perhaps all forms of matter have wave as well as particle properties.
Electrons have particle-wave nature electrons in motion exhibit wave properties.motion exhibit wave properties.
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From the momentum wavelength relationship of p=h/
Topic Six: Wave-Particle Duality
From the momentum-wavelength relationship of p=h/for photons, de Broglie applied this relation to other particles. de Broglie wavelength: de Broglie suggested that material particles of momentum p are associated with their characteristic wavelength :
mvh
ph (6.1)
their characteristic wavelength :
The frequencies of the matter waves obey the Planck relationship
mvp
relationship.
hEf (6.2)
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h
Topic Six: Wave-Particle Duality
Example 6-1*If a proton and an electron have the same de Broglie wavelength, will they have the same kinetic energy? Account for your answer If not determine the ratio of theAccount for your answer. If not, determine the ratio of the proton’s kinetic energy to that of the electron.
S l i Th h diff ki i i22 1
hpE ek
Solution: They have different kinetic energies
22
mmk
31
2
101921
h
mE 427
31
2 105.5107.1101.9
21
2
p
e
e
p
electronk
protonk
mm
hm
mEE
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2 em
6.2 Louis de Broglie (1892-1987)
Topic Six: Wave-Particle Duality
6.2 Louis de Broglie (1892 1987)Louis-Victor-Pierre-Raymond, 7th duc de Broglie, generally knownduc de Broglie, generally known as Louis de Broglie (August 15, 1892–March 19, 1987), was a French ph sicistFrench physicist. He introduced his theory of electron waves. This included theelectron waves. This included the wave-particle duality theory of matter. For this he won the Nobel Prize in Physics in 1929Prize in Physics in 1929
Data Source: http://en.wikipedia.org/wiki/Louis, 7th duc de Broglie
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Data Source: http://en.wikipedia.org/wiki/Louis,_7th_duc_de_Broglie
6.3 Double-slit experimentTopic Six: Wave-Particle Duality
pIf de Broglie’s hypothesis is right, we should observe the interference pattern caused by a beam of electrons incident to double slits.
The experimental requirements: the slit width<< the p qseparation between the two slits; the detector is far behind the slits.
The experimental procedures: (1) Close one slit and only leave the other open for a given time and observe the pattern. (2) Open the both slits at theobserve the pattern. (2) Open the both slits at the same time for the given time.
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Topic Six: Wave-Particle Duality
Figure 6-1: The two blue curves in the middle represent the patterns of individual slits with the upper or lower slit closed.
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Topic Six: Wave-Particle Duality
Figure 6-2: The single blue curve on the right represents the accumulated from the two blue curves. The brown curve represents the interference pattern with both slits open at the same time.
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p
Topic Six: Wave-Particle Duality
Simulation & Illustration
Particles through double slits
Key points: This simulation shows interference patterns for a beam of particles
i th h d bl lit Th i flpassing through double slits. The influences of slit width and particle mass on the patterns can be seen.
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Example 6-2*Topic Six: Wave-Particle Duality
A particle of charge q and mass m has been accelerated from rest through a potential difference V. Show its d B li l th h/(2 V)1/2de Broglie wavelength =h/(2mqV)1/2.
Solution: when a charged particle is accelerated from rest through a potential difference V the gain in kineticthrough a potential difference V, the gain in kinetic energy mv2/2 must equal the loss in potential energy qV
p1 22
V
mpmvVq22
1 2
where the momentum p=mv=(2mq V)1/2
v=0
vwhere the momentum p=mv=(2mq V)1/2, so that
Vhh
2
Shown!
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Vmqp 2
Topic Six: Wave-Particle Duality
6.4 Determination of the wavelength
I 1927 C J D i d L H G d d i
6.4 Determination of the wavelength of electron
In 1927, C. J. Davisson and L. H. Germer succeeded in measuring the wavelength of electrons.
I th i tIn their apparatus, electrons are accelerated from a
hot filament
heated filament F by an adjustable potential difference V.
Figure 6-3: The apparatus used in the Davisson-Germer
difference V.
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experiment.
The electron beam with kinetic energy of eV, is i id i k l l C
Topic Six: Wave-Particle Duality
incident to a nickel crystal C.A detector at D at an angle reads the current I of electrons entering the detector for various potentialelectrons entering the detector for various potential difference V.
There is a strong diffracted beam at = 50 and V = 54
Figure 6-4: The results obtained by Davisson and Germer for five different accelerating voltages, shown as polar plots of current I as a function of the angle .
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There is a strong diffracted beam at 50 and V 54 V but not otherwise.
Topic Six: Wave-Particle Duality
Figure 6-5: A simplified representation of a nickelrepresentation of a nickel crystal.
Dsin
The crystal surface acts like a diffraction grating with i Dspacing D.
Because of the low energy of the electrons, they cannot penetrate very far into the crystal and diffraction takes
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 79
penetrate very far into the crystal and diffraction takes place in the plane of atoms on the surface.
This situation is very similar to light reflected from
Topic Six: Wave-Particle Duality
This situation is very similar to light reflected from diffraction grating where the maxima must satisfy the following equation.
...3, 2,1, sin mDm (6.3)
For nickel crystal, D = 2.15 Å and m = 1, we have = D sin50o =1.65Å.
Since the kinetic energy of the electrons is gained from electric potential acceleration Eke=p2/2m=eV, we can have the classic momentum p=(2eVm)1/2 and the de Brogliethe classic momentum, p (2eVm) and the de Broglie wavelength is =h/p=h (2eVm)-1/2=1.6710-12 m=1.67 Å.
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Topic Six: Wave-Particle Duality
In 1928, G. P. Thomson of Scotland also observed electron diffraction patterns by passing electrons through
thi ld f ilvery thin gold foils. These results show conclusively the wave nature of electrons and confirmed de Broglie’s hypothesis.electrons and confirmed de Broglie s hypothesis.
Q Would the wavelengths of 10 eV photon, electron and t b th ?proton be the same?
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Topic Six: Wave-Particle Duality
6.5 Principle of complementarityp p yThe dual nature of matter and radiation is conceptually difficult to understand because the two models seem to contradict each other. Neils Bohr’s principle of complementarity states that th d ti l d l f ith tt
Q How do they complement each other?
the wave and particle models of either matter or radiation complement each other.
Q. How do they complement each other? Position of particle can be localized in both space and time but a wave cannot, being spread out in both oftime but a wave cannot, being spread out in both of these dimensions.
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6.6 Uncertainty principleTopic Six: Wave-Particle Duality
Localizing a wave in space and considering wave at arbitrary instant t = 0.
0
2cosxby
2b 02b
0 20 0
2 4, 0, 0k k
Δx =
→0
Δk =0b
Figure 6-6: (a) A harmonic wave viewed at t = 0. (b) The distribution f b h l t f th lit d f th h i
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of wave numbers, shown as a plot of the amplitude of the harmonic component as a function of its wave number.
The wave can be expresses as:
Topic Six: Wave-Particle Duality
The wave can be expresses as:
0
2cos, xbtx t
(6.4) 0
where b is a constant.
This wave has a sharply defined wavelength and a corresponding sharply defined wave number k0= 2/0kor k=0.
If this wave is to represent a particle, the uncertainty xi th i i i fi it (i h l th i )in the x axis is infinite (i.e. anywhere along the x axis).
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Add sine or cosine waves with properly chosen wave numbers amplitudes and phases see Fig 6 7(b) and
Topic Six: Wave-Particle Duality
numbers, amplitudes and phases, see Fig.6-7(b), and sum up all the waves over, we have a wave packet only over a certain region x and zero everywhere else, Fig.6-7(a).
Δx ≠
2b
0 20 0
2 4, 0, 0k k
Δx ≠
b
Figure 6-7: (a) A wave packet of length x, viewed at t = 0. (b) The
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relative amplitudes of the various harmonic components that combine to make up the packet. The central peak has a width k.
Topic Six: Wave-Particle Duality
Simulation & Illustration
Length of waves x & width of wave number k
Key points: Through the Fourier transform, this illustration shows clearly that wave number (k=2/) distribution becomes broad when the length of the wave reduces.wave reduces.
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Topic Six: Wave-Particle Duality
Wave is now localized, but purity of the wave is sacrificed since the packet contains a spread of wave numbers (k) centered about k0. u be s ( k) ce te ed about k0.
Hence the sharper the wave packet (or smaller x), the broader the range of wave numbers (k) we must use to
In general, as x increases, k decreases. The f ll i l ti hi h ld
build up the wave packet and vice versa.
following relationship holds
1 xk (6.5)
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Topic Six: Wave-Particle Duality
Form de Broglie wavelength Eq.(6.1), we have
x ppk
222 (6.6)xp
hhk
(6.6)
A l i E (6 5) thApplying Eq.(6.5), then
12 xp
hxk x
(6.7)h
2hxp (6 8)
or
2xpx (6.8)
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Taking into account that momentum is a vector and for id l i
Topic Six: Wave-Particle Duality
non-ideal instruments,
4hxpx
4
4
h
hypy
x
(6.9)
4hzpz
These are the mathematical formulations of the Heisenberg Uncertainty Principle for position-momentum relation.
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Though an individual measurement of momentum of a
Topic Six: Wave-Particle Duality
Though an individual measurement of momentum of a particle can yield an arbitrarily precise value, that value can be anywhere in a range px about the “true” px.
Repeated measurements on identically prepared systems generate results clustered about px with statistical di ib i f id h
It is not possible to determine both the position and th t f ti l ith li it d i i
distribution of width px.
the momentum of a particle with unlimited precision.
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Observing a wave at x = 0.
Topic Six: Wave-Particle Duality
g
00
2cos,T
tbtx x (6.10)
where T0 is the period and the angular frequency isdefined as 0 (= 2/T0) and b’ is a constant.
If this wave has a sharp angular frequency of 0= 2f0, it must be a continuous harmonic for infinite time t, see Fi 6 8Fig. 6-8.
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Topic Six: Wave-Particle Duality
t
2b’
T0Δt =
→0
= 0b’
→0
0000
b
Figure 6-8: (a) A harmonic wave viewed at x = 0. (b) The distribution of angular frequencies, shown as a plot of the amplitude of the harmonic component as a function of its angular frequencies.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 92
component as a function of its angular frequencies.
Similarly, add sine or cosine waves with properly
Topic Six: Wave-Particle Duality
chosen angular frequency , amplitudes and phases and sum up all the waves over, we have an impulse only over a certain duration t and zero beyond it.over a certain duration t and zero beyond it.
If the duration t will decrease, the spread of angular frequencies will increase. Thus we have:
1 t (6.11)
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As the energy of the particle is E = hf the uncertainty in
Topic Six: Wave-Particle Duality
As the energy of the particle is E hf , the uncertainty in frequency is related to the uncertainty in energy.
Ef (6 12)
hf (6.12)
Substitute Eq.(6.12) into Eq.(6.11) with = 2f= 2E/h,2E/h,
2htE (6.13)
The following
is the Energy-Time Uncertainty Relationship
g
4htE (6.14)
It is not possible to determine both the energy and the time coordinate of the particle with unlimited precision
is the Energy-Time Uncertainty Relationship.
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time coordinate of the particle with unlimited precision.
ll i i h i
Topic Six: Wave-Particle Duality
All energy measurements carry in inherent uncertainty. The lowest state of an atom (ground state) has well defined energy because the atom normally exists gy yindefinitely in that state.
All other states at higher energies are less precise because the atom (sooner or later) will move spontaneously to a lower energy state.
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Topic Six: Wave-Particle Duality
Example 6-3*pThe speed of an electron is measured to be 5.00×103 m/s to an accuracy of 0.00300%. Find the minimum uncertainty in determining the position of this electron.
Solution: The momentum of the electron is
m/skg1056.41000.51011.9 27331 mvpx
The uncertainty in px is 0.00300% of this value, so px=0.0000300×4.56×10-27 kg·m/s. From Eq.(6.9),
38000038010626.6 34
h mm38.0m00038.010368.114.34
10626.64 31
xphx
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6.7 Werner Heisenberg (1901 –1976)Topic Six: Wave-Particle Duality
Werner Karl Heisenberg (December 5, 1901 – February 1,
g ( )
1976) was a celebrated German physicist. He invented matrix mechanics theHe invented matrix mechanics, the first formalization of quantum mechanics in 1925. His
i i i l d l d iuncertainty principle, developed in 1927, states that the simultaneous determination of two pairedp
He received the Nobel Prize in physics in 1932quantities, has an unavoidable uncertainty.
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Data Source: http://en.wikipedia.org/wiki/Heisenberg
Topic Seven: A Particle in a 1-DBox
7.1 Wave function and probabilityWave function is a mathematical expression (usually a complex quantity) to describe the waves
i i l
p y
representing particles.If we know the wave function (x, y, z, t) for every point in space and every instant of time we know all about the
Probability density is determined by | |2 (probability
in space and every instant of time, we know all about the behavior of the particle.
per unit volume, or area, or length), defined as the probability of experimentally finding the particle described by the wave function at x, y, z at time t.
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y , y,
Probability that the particle will be somewhere must be
Topic Seven: A Particle in a 1-D Box
Probability that the particle will be somewhere must be equal to unity (100% chance of finding it) in the space.
12 dV (7 1)(normalization condition)1 dV (7.1)(normalization condition)
To normalize a wave function is to multiply it by a constant such that Eq (7 1) is satisfied
For a one-dimension system, where the particle must be located along the x axis and its wave function is only a
constant such that Eq.(7.1) is satisfied.
located along the x-axis and its wave function is only a function of x, we replace dV with dx and
2
12 dxx (7.2)
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Topic Seven: A Particle in a 1-D Box
Figure 7-1: The probability of a particle being in the interval a≤x≤ b is the area under the curve from a to b.
Probability of finding the particle in the interval a xbis
dxPb
aab 2(7.3)
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Expectation value of a particle’s position x (for 1 D
Topic Seven: A Particle in a 1-D Box
Expectation value of a particle s position x (for 1-D case) is determined by
dxxx
2 (7 4)dxxx (7.4)
The expectation value a particle’s position x is the p p paverage value of x. Once the wave function is known, it is possible to calculate the average position x or the expectation value of the particle
In general, the expectation value of a function f(x) is
expectation value of the particle.
dxxfxf
2 (7.5)
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7.2 A particle in a 1-D boxTopic Seven: A Particle in a 1-D Box
Classically, if a particle is confined to moving along the x-axis and bouncing back and forth between two impenetrable walls with a speed of v, then its momentum (mv) and its kinetic energy are constants.
Figure 7-2: A particle of mass m and velocity v confined tom and velocity v confined to moving parallel to the x axis and bouncing between two impenetrable walls.
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7.2.1 The wave function of the particle
Topic Seven: A Particle in a 1-D Box
Quantum mechanics’ approach to this problem is very different and requires that the appropriate wave function consistent with the conditions of the situation be foundconsistent with the conditions of the situation be found.If a string of length L is fixed at each end, the standing waves set up in the string must have nodes at two ends.
y(x)=0
y(x)=0
Figure 7-3: Standing waves set up in a stretched string of length L.
y(x)=0
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g g
x=0 x=L
R hi d l h l th i i t l
Topic Seven: A Particle in a 1-D Box
Resonance achieved only when length is some integral multiple of half-wavelengths.
2orn LL n (7 6)n = 1 2 3 or 2
nnL n
n (7.6)n = 1, 2, 3, . . .
The standing waves are only the form of stable waves whose wavelengths are quantizedwhose wavelengths are quantized.
2 x For the standing waves, it can be shown that
2sin sinnn
xy x A k x A
(7.7)
where A is the amplitude. Substitute Eq.(7.6), we have
LxnAxy sin (7.8)
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For a particle in a 1 D box of length L the de Broglie
Topic Seven: A Particle in a 1-D Box
For a particle in a 1-D box of length L, the de Broglie waves of the particle, in analogy with standing waves on a string, must form standing wave.
xnAx sin (7.9)n = 1 2 3
The allowed wave functions for the particle is
Allowed de Broglie wavelengths are those of standing
L
Ax sin ( )n = 1, 2, 3, . . .
waves n=2L/n.
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Topic Seven: A Particle in a 1-D Box
( ) 0
(x)=0|(x)|2=0
(x)=0
(x)=0
|(x)|2=0
|(x)|2=0
|(x)|2=0
Figure 7-4: The first three allowed states for a particle confined to a one-dimensional box. (a) The wave functions for n = 1, 2, and 3. (b) The probability distributions for n = 1 2 and 3
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probability distributions for n 1, 2, and 3.
7.2.2 The probability density of the particle
Topic Seven: A Particle in a 1-D Box
The probability density: 2
can be either positive or negative, but 2 is always p g ypositive.2 is zero at the boundaries, in other words, it is i ibl t fi d th ti l t th i timpossible to find the particle at these points. Where2 is the maximum and zero depends on n.
Th b bili d i di ib i i h iThe probability density distribution is shown in
Fig.7-4(b).
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7.2.3 The energy of the particle
Topic Seven: A Particle in a 1-D Box
The kinetic energies of the allowed (stable) states are
2 222 Lh
gy p
2
222
2
222
hm
Lnhm
pmvEn (7.10)
3,.... 2, 1,n 8
22
n
mLh
E i ti d ith thEnergy is quantized with the lowest energy corresponding to n = 1 (zero-point energy).
Figure 7-5: Energy level diagram for a particle confined to a one-dimensional
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particle confined to a one dimensional box of width L. E1=h2/8mL2.
Cl i ll th i i f th ti l
Topic Seven: A Particle in a 1-D Box
Classically, the minimum energy of the particle can be zero.
When a particle drops from E3 to E2, it emits a photon of energy hf = E3 – E2.It l b b h t d j f E t E ifIt can also absorb a photon and jumps from E1 to E2 if the incident photon has energy hf = E2 - E1.
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Topic Eight: Schrödinger EquationTopic Eight: Schrödinger Equation
8 1 1 D tiGeneral form of the wave equation for waves traveling along the x axis
8.1 1-D wave equation
along the x axis
t
txvx
tx
,1 2
22
2(8.1)
tvx where and v are the amplitude and propagation speed of the wave. In quantum mechanics, particles’ behaviors can described using de Broglie waves, which should satisfy Eq.(8.1).
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8.2 A system with constant energy Topic Eight: Schrödinger Equation
We consider an isolated bound system and the paticles’ interaction inside it is ignorable, the total energy E = hfg gy fof the particle will remain constant.The wave function of the de Broglie waves is
l t t di t i b
txΨtx cos (8.2)
analogous to standing waves on a string can be expressed as
txtx cos ( )
Substituting into Eq.(8.1),
tΨxΨdt
22
coscos
xΨd
txΨvxd
t
22
2 cos cos
(8 3)
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xΨvxd
xΨd
2
(8.3)or
Recall that = 2f = 2 v/ and p = h/, then
Topic Eight: Schrödinger Equation
2
22
2
22
2
2 42
pphv
(8.4)
E th t t l f ki ti (K) d
where ħ=h/2π is the reduced Planck constant.
Express the total energy as sum of kinetic (K) and potential energy (U).
2pE K U U
2
2 2
pE K U Um
p m E U
(8.5)
(8.6) 2
2
2
2 2
2v
p m E U
and
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 112
v
Substituting Eq (8.6) into Eq (8.3), we have
Topic Eight: Schrödinger Equation
(8.7) xEΨxΨxUxd
xΨdm
2
22
2
kinetic energy
potential total energy
This is 1-D time-independent Schrödinger Equation.For 3-D time-independent Schrödinger Equation;
energy energy
z
zyxΨy
zyxΨx
zyxΨ ,,,,,,2
2
2
2
2
2
zyxΨzyxUEmy
,,,,22
(8.8)
In principle, if U(x,y,z) and two appropriate boundary conditions are known, we can solve the Schrödinger equations to obtain the wave functions and energies
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 113
equations to obtain the wave functions and energies for the allowed states.
Example 8-1*Topic Eight: Schrödinger Equation
A free electron has a wave function ψ(x)=Aexp(ikx), where x is the position and k is a constant. (i) Show that its non relativistic kinetic energy may be expressed asits non-relativistic kinetic energy may be expressed as
mkE
2
2
em2
where ħ is the reduced Planck constant and me is the rest mass of electron.Solution:For free electron, U=0. The time-independent Schrödinger equation:equation:
22
22d Ψ x
EΨ xm d x
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2m d x
d ik Since
Topic Eight: Schrödinger Equation
22
2
ikdx
d kdx
Since
dx
Substituting the above into the time time-independent Schrödinger equation, we have
2
2
2k x E x
m
g q ,
2mSo
2kh
2 e
Em
shown
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8.3 Erwin Schrödinger (1887-1961)
Topic Eight: Schrödinger Equation
8.3 Erwin Schrödinger (1887 1961)Erwin Rudolf Josef Alexander Schrödinger (August 12, 1887 –Schrödinger (August 12, 1887 January 4, 1961), an Austrian physicist, achieved fame for his contributions to quantumcontributions to quantum mechanics, especially the Schrödinger equation.He received the Nobel Prize in 1933 for his great contributions.
Data Source: http://en wikipedia org/wiki/Schrodinger
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Data Source: http://en.wikipedia.org/wiki/Schrodinger
8.4 Determination of
Topic Eight: Schrödinger Equation
8.4 Determination of The solutions to the Schrödinger Equation should satisfy the following conditions:satisfy the following conditions: Since it is second order differentiation equation, one must need two boundary conditions to determine the wave function. At the boundary of two contiguous regions, the wave function must join smoothly at the boundary betweenfunction must join smoothly at the boundary between regions. For 1-D at the boundary x=x0,
xxΨxxΨ (8 9) xxdΨxxdΨ
xxΨxxΨ
00
00 (8.9)
(8.10)
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dxdx
(x) must obey normalization condition :
Topic Eight: Schrödinger Equation
(x) must obey normalization condition :
12
dxxΨ (8.11)
0xΨ (8.12)
(x) must be single-valued and must also be
Note : Steps leading to Eq (8.7) do not represent a
(x) must be single-valued and must also be continuous for finite values of U(x).
derivation of the Schrödinger Equation. Rather, the procedure represents a plausibility argument based upon an analogy with other wave phenomena that are alreadyan analogy with other wave phenomena that are already familiar to us.
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8.5 An example-a particle in a 1-D
Topic Eight: Schrödinger Equation
8.5 An example a particle in a 1 D box revisited
Figure 8-1: Diagram of a one-dimensional box of width L and
infinitely high walls.
The1-D impenetrable box is a 1-D
infinite infinitely high walls.
Potential energy distribution:U(x) = for x 0 and x L
potential well
U(x) for x 0 and x LU(x) = 0 for 0 < x <L (8.13)
The particle does not feel any external forces exerted to it.
Thus, the particle can never escape from the box and
(x) = 0 for x 0 and x L (8.14)
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(x) 0 for x 0 and x L (8.14)
I id th b 0 < < L th S h ödi E ti
Topic Eight: Schrödinger Equation
Inside the box 0 < x < L, the Schrödinger Equation Eq.(8.7) becomes
mEΨd 22
(8.15)
mE
ΨkΨmEdxΨd
2
2 222
(8 16)h
mEk 2 (8.16)where
The general solution to Eq.(8.15) isg q ( )
kxBkxAx cossin (8.17)
h A d B t i t ti t t t bwhere A and B are two integration constants to be determined as follows
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Using the boundary condition, Eq.(8.14),(0)=0, we
Topic Eight: Schrödinger Equation
Applying the other boundary condition, Eq.(8.14),
Using the boundary condition, Eq.(8.14),(0) 0, we have B=0
0sin kLA (8.18)
(L)=0, we have
We have to require sin(kL)=0 This requirement leads to
We can not require A=0. Otherwise, =0 everywhere.
We have to require sin(kL) 0. This requirement leads to
nkL n=1, 2,3…. (8.19)
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Substituting k in Eq (8 19) into Eq (8 17) we have the
Topic Eight: Schrödinger Equation
Substituting k in Eq.(8.19) into Eq.(8.17), we have the final wave functions, see Fig.8-2:
xnAxn sin n=1, 2, 3…. (8.20)
Ln , , ( )
Applying the normalization condition Eq.(8.11): 2
1sin0
2
dx
LxnA
L
1
(8.21)2
1
2
LA
Th b bili d i Fi 8 2(b) iThe probability density, see Fig.8-2(b), is
Lxn
Lxn
22 sin2 n=1, 2,3…. (8.22)
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LL
With Eq.(8.16) and Eq.(8.19), we have obtained the
Topic Eight: Schrödinger Equation
(8 23)nLmEkL 2 n = 1 2 3
energy of the particle, see Fig.8-3:
(8 24)22hE
(8.23)nLkL
n = 1, 2, 3, . . .
(8.24)228
nmL
En
These results agree with those obtained in the previousThese results agree with those obtained in the previous section.
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Topic Eight: Schrödinger Equation
(x)=0|(x)|2=0
(x)=0
(x)=0|(x)|2=0
Fi 8 2 Th fi t th ll d t ti t t
|(x)|2=0
Figure 8-2: The first three allowed stationary states for a particle confined to a one-dimensional box. (a) The wave functions for n = 1, 2, and 3. (b) the probability densities 2 for n = 1, 2, and 3. p y , ,
Figure 8-3: Energy level diagram for a particle confined to a 1-D box of width L. The lowest
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 124
allowed energy is E = h2/8mL.
Example 8-2*Topic Eight: Schrödinger Equation
pA 1.00 mg object is confined to moving between two rigid walls separated by 1.00 cm. Calculate the
i i d f h bjminimum speed of the object.Solution: The minimum speed corresponds to the state for which n=1 From Eq (8 24)for which n 1. From Eq.(8.24),
J1049.501.01018
1063.68
5826
234
2
2
1
mLhE
01.010188mLThis energy corresponds to kinetic energy of the object,
/103131049.52/2 262/158
E m/s1031.3
101/2 26
61
mEv
The speed is so small that it can be considered to be at
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 125
rest, which is what one would expect for the minimum speed of a macroscopic object!
Example 8-3*Topic Eight: Schrödinger Equation
A particle confined to a 1-D infinite potential well from x=0 to x=+L has a wave function given by
2 sin n xL L
determine the probability that it will be found between(1) x=0 and x=L/4, (2) x=L/4 and x=L/2, (3) x=0 and x=L/2
22 1 2sin 1 cosn x n xdx dxL L L L
Solution:Since
1 2sin2
L L L L
x n xL L
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 126
2L n L
Yielding:
Topic Eight: Schrödinger Equation
2 / 4/ 4
0 / 4 00
2 1 2sin sin2
LL
Ln x x n xP dx
L L L n L 0
1 1 sin4 2 2
nn
2 / 2/ 2
/ 4 / 2 / 4
2 1 2sin sin2
LL
L L L
n x x n xP dxL L L n L / 42
1 1 sin4 2 2
LL L L n L
nn
21
2/4/4/02/0 LLLL PPP
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2
Topic Eight: Schrödinger Equation
8.6 Characteristics of the particle in the box
Classically the particle would be expected to stay stillClassically, the particle would be expected to stay still with zero kinetic energy at very low temperatures. In quantum mechanics, the particle cannot be at rest in the box.The stable wave functions (standing waves) are caused by multiple reflections of the particle’s de y p pBroglie wave at the two walls of the box. Zero-point energy is the lowest energy (ground state)
d 1 dcorresponds to n = 1 and
E h1
2
28 (8.25)
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mL1 28
Depending on the quantum state n, the electron spends
Topic Eight: Schrödinger Equation
p g q , pmore time at certain position of the box than in others.In the ground state, the electron is more likely to be f d th t th it ll Thi fi difound near the center than near its walls. This finding contradicts classical theory. As n increases, the distribution of electron probability , p ydensity becomes more uniform and approaches classical theory.
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Topic Eight: Schrödinger Equation
8.7 Finite potential wellV0= p0.5 n=3
0.3
0.4
V0=0.3 eV
0.2
E/e
V
n=2
0.1
n=1
5-50.0
x/nm x/nmFi 8 4 T i l ll f ( ) i fi i d h d (b) fi i d h (0 3
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Figure 8-4: Two potential wells of (a) infinite depth and (b) finite depth (0.3 eV), respectively, are compared. The wells have the same width of 10 nm.
The following characteristics can be found when we deal
Topic Eight: Schrödinger Equation
gwith the more realistic case of a finite quantum well:1) Note that0 at the two walls of the well. It
should satisfy boundary conditions Eqs. (8.9) and (8.10). As a result, there is a spilling over of the exponential tail of the probability curve.p p y
2) The energy levels En are lower than the corresponding ones in the infinite deep quantum
ll ith th ll idth3) If the particle energy is higher than the barrier height
V0, it will not be trapped. If lower, it may be trapped
well with the same well width.
0, pp , y ppwith a probability of escaping from the trap. It is also possible to find the particle beyond the well.
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Topic Eight: Schrödinger Equation
Simulation & Illustration
Quantum well of finite height
Key points: Influences of the well height, width and particle energy on the wave p gyfunctions, energy of the states, etc are illustrated.
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Topic Nine: Tunneling p g9.1 Tunneling through a rectangular
b ibarrier Consider a particle of energy E incident on a rectangular barrier of finite height U and width L where E<Ubarrier of finite height U and width L, where E<U.
Classically, the particle must be reflected as it does not havereflected as it does not have sufficient energy to overcome the barrier. U
EIn quantum mechanics, the amplitude of the de Broglie wave associated with the particle is L
E
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pnonzero everywhere.
Tunneling or barrier penetration: the possibility of
Topic Nine: Tunneling
finding the particle on the far right side of the barrier.
Figure 9-1: for a particle incident from the left on the b i f h i ht U ibarrier of height U is sinusoidal in regions I and III, but exponentially decaying in region II. y g g
The probability of tunneling can be described with a transmission coefficient T and a reflection coefficient R.
1RT
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1 RT (9.1)
9.2 Transmission coefficient of Topic Nine: Tunneling
tunnelingT i i ffi i t f t li T i thTransmission coefficient of tunneling T is the probability that the particle penetrates the energy barrier.
An approximate expression for T (when T <<1) is
EUmL 24
hEUmL
T24
exp
(9.2)
h h i Pl k’ t t d E th dwhere h is Planck’s constant, m and E are the mass and energy of the incident particle, L and U are the width and barrier potential height, respectively.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 135
Topic Nine: Tunneling
Simulation & Illustration
Quantum Tunneling
Key points: We pay our attention to the waveforms of incident, penetrated and reflected de Broglie waves and the influences of U, L and E on them.
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9.3 Low contact resistance th h t li *
Topic Nine: Tunneling
through tunneling*For a metal in contact with highly doped semicon-ductor ND1019 cm-3, the barrier width becomes very narrow, the tunneling current becomes dominant and contact resistance Rand contact resistance RCdecreases rapidly with increased doping.
Figure 9-2: Calculated and measured values of specific contact resistance. Upper insert shows the tunneling
L i h h i i
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process. Lower insert shows thermionic emission over the low barrier.
9.4 Scanning tunneling microscope* Topic Nine: Tunneling
g g pAn electrically conducting probe with a very sharp tip p obe w t a ve y s a p t pis brought near the surface to be studied.
The empty space between the tip and the surface represents the “barrier”.
Figure 9-3: Schematic view of a scanning tunneling microscope (STM). The tip is mounted on a piezoelectric xyz p p yscanner. A scan of the tip over the sample can reveal contours of the surface down to the atomic level. An STM image is
d f i f di l d
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 138
composed of a series of scans displaced laterally from one another.
The tip and the surface are two walls of the “potential
Topic Nine: Tunneling
well”.The STM allows highly detailed images of surfaces with resolutions comparable to the size of a single atom.gAn STM image of the surface of graphite is shown in Fig 9-4shown in Fig.9 4.A conductive surface is needed.
Figure 9-4: The surface of graphite as “viewed” with an STM. This type of image is capable of a lateral resolution of about 0.2 nm and a vertical resolution of 0 001 nm
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 139
vertical resolution of 0.001 nm.
Example 9-1*Topic Nine: Tunneling
A 30-eV electron is incident on a barrier whose cross-session is a rectangle of height 40 eV. What is the probability that the electron will tunnel through the p ob b y e e ec o w u e oug ebarrier if its thickness is 1.0 nm? and 0.1 nm?Solution: In this situation, U-E=40-30=10 eV=1.6×10-18 J. F E (9 2)From Eq.(9.2),
24exp
hEUmL
T
1831 106.11011.9214.34exp L
h
34106.6exp
When L=1 nm T8 510-15 If L=0 1 nm T0 039
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 140
When L 1 nm, T8.510 . If L 0.1 nm, T0.039.
Topic Ten: Quantum States ofTopic Ten: Quantum States of an Atom
In a hydrogen atom a single electron is bound to a
10.1 Hydrogen atom revisitedIn a hydrogen atom, a single electron is bound to a single proton by the attractive Coulomb’s force.Atomic system can be viewed as an electron trapper in
Electron can exist only in a discrete set of quantum t t h ith t i
y ppwhich an electron is confined to a region of space.
states each with a certain energy.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 141
Potential energy of the hydrogen atom is
Topic Ten: Quantum States of an Atom
rekrU e
2
(10.1)
h k 8 99 109 N 2/C2 i h C l bwhere ke=8.99 109 N·m2/C2 is the Coulomb constant and r is the radial distance between the proton and electron.Substitute U(r) into a 3-D Schrödinger equation and find the appropriate solution.Solving the equation and the energies of the allowed states for the hydrogen atom are,
2 eV 6.1312 22
0
2
nnaekrE e
(10.2)n = 1, 2, 3, . . .
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 142
ll d d d h b
Topic Ten: Quantum States of an Atom
Allowed energy states depend on the quantum number n.
This result is exactly the same as that obtained in the B h thBohr theory. For circular electron orbit, one quantum number n is sufficient to characterize a stationary statesufficient to characterize a stationary state. n is called the principal quantum number, an integer, ranging from 1 to .
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10.2 Quantum numbers of atomic Topic Ten: Quantum States of an Atom
For a more general electronic orbit, three quantum system
numbers are needed to define it.They correspond to three independent degrees of freedom.
Quantum Number
Name Allowed Values Number of Allowed States
n Principal quantum 1, 2, 3, . . . Any positive number integer
l Orbital quantum number
0, 1, 2, . . . , n - 1 n
mlOrbital magnetic quantum number
- l, - l + 1, . . . , 0, . . . , l - 1, l
2l + 1
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Any orbits that violate the above rules cannot exist.
For historical reasons all orbits having the same
Topic Ten: Quantum States of an Atom
For historical reasons, all orbits having the same principal quantum number are said to form a shell. Those orbits having the same value of n and l are said to
n Shell Symbol l Subshell symbol
gform a subshell.
y y
1 K 0 s
2 L 1 p
3 M 2 d
4 N 3 f
5 O 4 g5 O 4 g
6 P 5 h
. . . . . .
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 145
10 3 Spin magnetic quantum number
Topic Ten: Quantum States of an Atom
10.3 Spin magnetic quantum number The spectra of certain gases, such as sodium vapor, show two very closely spaced lines called a doubletshow two very closely spaced lines called a doublet.To explain this, Goudsmidt and Uhlenbeck proposed a new quantum number mS, called the p p q S,spin magnetic quantum number. Convenient (but incorrect) to think of mS as describing
l t i i it i it bit than electron spinning on its axis as it orbits the nucleus.
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There are only two values mS = ½, corresponding to
Topic Ten: Quantum States of an Atom
two directions, i,e, the electron can “spin up” (mS = ½) or “spin down” (mS = -½).
I t l tiIn external magnetic field, the energy of the electrons differs slightly for the two spin directions. This accounts for the doublet.for the doublet. Electron is a point particle, without
Figure 10-1: The spin of an electron can be either (a) up or (b) down relative to an
spatial extent.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 147
( ) p ( )external magnetic field.
O h bi l i h “ i ” ( ½)On each orbit, an electron can either “spin up” (mS = ½) or “spin down” (mS = -½).
A f b d fi d b fA quantum state of an atom must be defined by a set of four quantum states, i.e., n, l, ml and ms.
Q H ibl ldQ. How many possible quantum states could an atom have?
A Determined by a combination of all possible values ofA. Determined by a combination of all possible values of n, l, ml and ms.
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Topic Ten: Quantum States of an Atom
Example 10-1*For a hydrogen atom, determine the quantum numbers associated with the possible states that correspond to h i i l bthe principal quantum number n=2.
n l ml ms subshell shell Number of states in subshell
1/2112
-1/2002
1/20022s L 2
1/2012
-1/2112
1/2112
1/2-112
-1/2012
1/2012 2p L 6
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 149
-1/2-112
10.4 The exclusion principleTopic Ten: Quantum States of an Atom
p p
It turns out that n, l, ml, and ms can be used to describe all the electronic states of an atom regardless of theall the electronic states of an atom regardless of the number of electrons in its structure.
Q. How many electrons can have a particular set of quantum numbers?
orHow many electrons can a particular set of quantum numbers accommodate? qua tu u be s acco odate?
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Pauli, in 1925, answered this question in the
Topic Ten: Quantum States of an Atom
Pauli, in 1925, answered this question in the following statement:
No two electrons in the same atom can ever be inNo two electrons in the same atom can ever be in the same quantum state. Therefore no two electrons can have the same set of quantum numbersof quantum numbers. If this principle were not valid, an atom could radiate energy until every electron in the atom is in the lowestenergy until every electron in the atom is in the lowest possible energy state and the chemical behavior of the elements would be grossly modified. Nature as we know it would not existknow it would not exist.
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10.5 Wolfgang Pauli (1900-1958)Topic Ten: Quantum States of an Atom
g g ( )Wolfgang Ernst Pauli (April 25, 1900 – December 15, 1958) was
i h i i d fan Austrian physicist noted for his work on the theory of spin, and in particular the discovery of p ythe exclusion principle, which underpins the whole of chemistrychemistry.In 1945, he received the Nobel Prize in Physics for his “decisive contribution through his discovery in 1925 of g ya new law of Nature, the exclusion principle or Pauli principle.” He had been nominated for the prize by Einstein
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Einstein.Data Source: http://en.wikipedia.org/wiki/Wolfgang_Pauli
Q. How are the quantum states for an atom filled with l ?
Topic Ten: Quantum States of an Atom
electrons?Electrons intend to fill up the state (subshell) with the lowest energy state firstlowest energy state first.
To a first approximation, energy depends only on the principal quantum number n. It increases with
Once a subshell is filled, the next electron goes to the t l t t b h ll
p p qincreasing n.
next lowest energy vacant subshell.If an atom is not at its ground state (or the lowest energy state available), it will radiate energy until itenergy state available), it will radiate energy until it reaches this state.
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10.6 The periodic tableTopic Ten: Quantum States of an Atom
pAn orbit for an electron in an atom refers to the quantum state of an electron characterized by the quantum
The exclusion principle suggests that there can be only t l t i bit l ith +½ () d
y qnumbers n, l and ml.
two electrons in any orbital with ms = +½ () and ms=-½ ().
Allowed Quantum States for an Atom having n = 3
210100l
3M
2L
1K
n
Allowed Quantum States for an Atom having n 3.
-2-1012-1010-10100ml
2d
1p
0s
1p
0s
0s
l
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 154
ms
2 8 18 Total=28
n = 1 K shell can only accommodate 2 electrons
Topic Ten: Quantum States of an Atom
n = 1 K shell can only accommodate 2 electrons. n = 2 L shell has 2 subshells and 4 orbitals and is capable of accepting a total of 8 electrons. p gn = 3 M shell has 3 subshells and 9 orbitals and is capable of accepting a total of 18 electrons.
In general, each shell can have up to 2n2 electrons.
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Hydrogen has 1 electron with the ground state
Topic Ten: Quantum States of an Atom
Hydrogen has 1 electron with the ground state described by 2 sets of quantum numbers : 1, 0, 0, ½ or 1, 0, 0, -½. The electronic configuration is written as 1 11s1.
Electronic configuration: 1s1
numbers of electronson the subshell
n l2p2s1s
or
Helium has 2 electrons with ground state quantum numbers of 1, 0, 0, ½ and 1, 0, 0, - ½. The electronic configuration is written as 1s2 - K shell filled.
2p2s1s
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Lithium has 3 electrons with 2 in the 1s subshell and
Topic Ten: Quantum States of an Atom
1 in the 2s subshell (2s subshell has slightly lower energy than 2p subshell). The electronic configuration is written as 1s22s1is written as 1s 2s .
2p2s1s
or
Beryllium has electronic configuration of 1s22s2.
2p2s1s
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Boron has electronic configuration of 1s22s22p1. The
Topic Ten: Quantum States of an Atom
2p electron may be described by 6 sets of quantum numbers corresponding to 6 states of equal energy.
221 2p2s1s
b h l d h l h hCarbon has 6 electrons and the 2p electrons are such that they occupy different orbits with unpaired spins ().
2p2s1s
2p2s1s
or
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H d’ R l Wh t h bit f l
Topic Ten: Quantum States of an Atom
Hund’s Rule: When an atom has orbits of equal energy, the order in which they are filled by electrons is such that a maximum number of electrons have unpaired spins. Exceptions to the rule occur in an atom having subshells close to being filled or half filledclose to being filled or half-filled.
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The Exclusion Principle can be illustrated as follows :
Topic Ten: Quantum States of an Atom
3
Atomic number
K L
3
4
5
6
Figure 10-2: The filling of electronic states
7
8must obey both the exclusion principle and the H d’ l
9
10
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Hund’s rule. 10
Atomic number Z
Symbol Ground state configuration
1 H (Hydrogen) 1s1
K1 H (Hydrogen) 1s1
2 He(Helium) [inert] 1s2
3 Li (Lithium) 1s2 2s1
4 Be (Beryllium) 1s2 2s2
L
4 Be (Beryllium) 1s 2s
5 B (Boron) 1s2 2s22p1
6 C (Carbon) 1s2 2s22p2
7 N (Nitrogen) 1s2 2s22p3
8 O (Oxygen) 1s2 2s22p4
9 F (Fluorine) 1s2 2s22p5
10 Ne (Neon) [inert] 1s2 2s22p6 M11 Na (Sodium) 1s2 2s22p6 3s1
12 Mg (Magnesium) 1s2 2s22p6 3s2
13 Al (Aluminum) 1s2 2s22p6 3s23p1
14 Si (Sili ) 1 2 2 22 6 3 23 214 Si (Silicon) 1s2 2s22p6 3s23p2
15 P (Phosphorus) 1s2 2s22p6 3s23p3
16 S (Sulfur) 1s2 2s22p6 3s23p4
17 Cl (Chlorine) 1s2 2s22p6 3s23p5
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17 Cl (Chlorine) 1s 2s 2p6 3s 3p5
18 Ar (Argon) [inert] 1s2 2s22p6 3s23p6
19 K (Potassium) 1s2 2s22p6 3s23p6 4s1
20 Ca (Calcium) 1s2 2s22p6 3s23p6 4s2
K L M
M
N
( ) p p
21 Sc (Scandium) 1s2 2s22p6 3s23p6 4s2 3d1
22 Ti (Titanium) 1s2 2s22p6 3s23p6 4s2 3d2
23 V (Vanadium) 1s2 2s22p6 3s23p6 4s2 3d3
24 Cr (Chromium) 1s2 2s22p6 3s23p6 4s1 3d5
25 Mn (Manganese) 1s2 2s22p6 3s23p6 4s2 3d5
26 Fe (Iron) 1s2 2s22p6 3s23p6 4s2 3d6
27 Co (Cobalt) 1s2 2s22p6 3s23p6 4s2 3d7
28 Ni (Nickel) 1s2 2s22p6 3s23p6 4s2 3d8
29 Cu (Copper) 1s2 2s22p6 3s23p6 4s1 3d10
30 Z (Zi ) 1 2 2 22 6 3 23 6 4 2 3d10 N30 Zn (Zinc) 1s2 2s22p6 3s23p6 4s2 3d10
31 Ga (Gallium) 1s2 2s22p6 3s23p6 4s2 3d10 4p1
32 Ge (Germanium) 1s2 2s22p6 3s23p6 4s2 3d10 4p2
33 As (Arsenic) 1s2 2s22p6 3s23p6 4s2 3d10 4p3
N
33 As (Arsenic) 1s 2s 2p 3s 3p 4s 3d 4p
34 Se (Selenium) 1s2 2s22p6 3s23p6 4s2 3d10 4p4
35 Br (Bromine) 1s2 2s22p6 3s23p6 4s2 3d10 4p5
36 Kr (Krypton) [inert] 1s2 2s22p6 3s23p6 4s2 3d10 4p6 O
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( yp ) [ ] p p p
37 Rb (Rubidium) 1s2 2s22p6 3s23p6 4s2 3d10 4p6 5s1
38 Sr (Strontium) 1s2 2s22p6 3s23p6 4s2 3d10 4p6 5s2 (4d5p6s4f5d)
s1 s2 p1 p2 p3 p4 p5 s2p6
K (n=1)
L ( 2)L (n=2)
M (n=3) d3d2d1 d10 d10
N (n=4)?
O (n=5)?
P (n=6)?( )
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Topic Ten: Quantum States of an Atom
Noble gases formed when either shell or subshell are
Elements in periodic table are arranged so that those in a column have similar chemical properties. Noble gases formed when either shell or subshell are filled or there is a large gap in energy before the next possible level is encountered.
He (helium, Z=2): 1s2 first (K) shell filled
Ne (neon, Z=10): 1s22s22p6 second (L) shell filled( , ) p
Ar (argon, Z=18): 1s22s22p63s23p6 p subshell in (M) shell filled
Kr (krypton Z=36):1s22s22p63s23p63d104s24p6 p subshell in (N) Kr (krypton, Z=36):1s22s22p63s23p63d104s24p6shell filled
Xe (xenon, Z=54): 1s22s22p63s23p63d104s24p64d105s25p6
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p subshell in (O) shall filled
Topic Eleven: Lasers andTopic Eleven: Lasers and Laser Light
LASER means Light Amplification for Simulated Emission of Radiation.
Laser light is an intense, concentrated, and highly parallel beam of coherent light.
L i h h f MASER i il d iLaser is the outgrowth of MASER, a similar device used in microwaves instead of visible light.
In 1960 the first laser was built by T H Maiman of theIn 1960, the first laser was built by T. H. Maiman of the Hughes Aircraft Company Laboratories.
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11.1 Absorption
Topic Eleven: Lasers and Laser Light
To understand the operation of a laser, we must be familiar with the processes describing the emission and
11.1 Absorption
absorption of radiation by atoms. Consider atomic system with twosystem with two lower states, of energies E1 and E2.
Figure 11-1: Stimulated absorption of a photon. The
dots represent electrons. One electron is transferred from the ground state to the excited state
when the atom absorbs a
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 166
when the atom absorbs a photon of energy hf = E2 – E1.
Stimulated absorption: when a photon of frequency f
Topic Eleven: Lasers and Laser Light
Stimulated absorption: when a photon of frequency f(such that the photon energy hf = E2 - E1) is incident and acting with the atom, the photon vanishes and the atomic s stem is e cited to make the p ard transitionsystem is excited to make the upward transition.
Excited states: atoms are raised to allowed higher energy levelsenergy levels.
Excited states are metastable. Instability of the excited states depends on the excited atoms or systemsstates depends on the excited atoms or systems.
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11.2 Spontaneous emissionTopic Eleven: Lasers and Laser Light
Spontaneous emission: in the instable excited states (typically ~10-8 s), the atom will jump back to a lower
l l d it h t i th d denergy level and emit a photon in the downward transition process.
Figure 11-2: Spontaneous emission of a photon by an p yatom that is initially in the excited state E2. When the
atom relaxes it to the ground state it emits a photon ofstate, it emits a photon of
energy hf=E2 – E1.
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In spontaneous emission process, a photon of energy hf
Topic Eleven: Lasers and Laser Light
(= E2 - E1) is emitted under no external influence.
Spontaneous emission happens naturally. Emitted h d l diff f h h iphotons are randomly different from each other in
propagation direction, phase angle, etc.
Phosphorescent materials glow because of a similar process, but the excited atoms may remain in an excited state for periods ranging from a few seconds to several hours. Because of this reason phosphorescent materials emitBecause of this reason, phosphorescent materials emit light after being placed in the dark.
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11.3 Stimulated emissionTopic Eleven: Lasers and Laser Light
Consider an atomic system in its excited state in the presence of a radiation field of frequency f, such that hf E Ehf = E2 - E1.
Figure 11-3: Stimulated emission of a photon by anemission of a photon by an
incoming photon of energy hf. Initially, the atom is in the
excited state. The incoming photon stimulates the atom to
emit a second photon of energy hf = E2 – E1.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 170
Sti l t d i i th h t i t t ith th
Topic Eleven: Lasers and Laser Light
Stimulated emission: the photon interacts with the excited system to drive it downward transition to its lower energy by emitting an additional photon. The emitted photon is identical with the ‘triggering’ or ‘stimulating’ photon - same energy, direction, phase, and state of polarizationThese two photons can cause other stimulated emissions, leading to a chain reaction of similar
state of polarization.
gprocesses - amplification.
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11.4 Conditions for laser action
Topic Eleven: Lasers and Laser Light
11.4 Conditions for laser action1. Population inversion must be realized.
For an atomic system at thermal equilibrium, the number of atoms occupying a state at an energy Eis determined by the exponential factor of s de e ed by e e po e a ac o oexp(E/kBT), governed by the Maxwell-Boltzmann distribution.
The ratio of the number of atoms in the upper excited level E2 to the number in the lower level E1isis
TkEE
EnEn
B
12
1
2 exp (11.1)
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 172
TkEn B1
Topic Eleven: Lasers and Laser Light
Since E2 > E1, the ratio of n(E2)/n(E1) will always be less than unity, meaning that there are fewer atoms in the higher energy state than in the lower one in thermalthe higher energy state than in the lower one in thermal equilibrium.
5
6
7
TFigure 11-4: The
2
3
4
k BT
(E2-
E1)
/kB
gMaxwell-Boltzmann
distribution. 0
1
0 0.2 0.4 0.6 0.8 1 1.2
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 173
exp(-E /k BT )n(E2)/n(E1)e-1=0.368e-3=0.050
Topic Eleven: Lasers and Laser Light
Example 11-1*Estimate relative populations at room temperature (300 K) of two energy levels such that a transition from the hi h t th l l l i i ibl di ti fhigher to the lower levels gives a visible radiation of 550 nm.
Energy difference between the two levels,
105501031063.6
9
834
12
hcEE
gy ,
From Eq.(11.1), eV 25.2J106.3
1055019
q ( ),
192 37
231
3.6 10exp exp 87 101.38 10 300
n En E
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 174
1
Population inversion: there are more atoms in a higher
Topic Eleven: Lasers and Laser Light
Population inversion: there are more atoms in a higher energy state than in a lower state at non-thermal equilibrium. P i i th f ti l ti
This can be done optically by creating an intense
Pumping is the process of creating a population inversion. This can be done optically by creating an intense, continuous light source around the lasing material, or electrically by gas discharge.
2. The excited state of the system must be a metastable state so that when population inversion is met, stimulated emission occurs before spontaneous emission.
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3. The emitted photons must be confined in the system
Topic Eleven: Lasers and Laser Light
p ylong enough to stimulate further emission from other excited atoms.
Fi 11 5 S h i di f l d i Th b i hFigure 11-5: Schematic diagram of a laser design. The tube contains the atoms that are the active medium. An external source of energy “pumps” the atoms to the excited state. The parallel end mirrors confine the photons to the tube. One mirror is made totally reflecting and the other is slightly transparent to
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 176
y g g y pallow part of the laser beam to escape.
Simulation & Illustration
Spontaneous emission, stimulated emission, laser light
Key points: This illustration shows clearly spontaneous emission, stimulated emission, importance of an optical resonator formed by two mirrors for lasing, etc.lasing, etc.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 177
11.5 Helium-neon lasers*
Topic Eleven: Lasers and Laser Light
The three principal elements of a laser are (1) an energy pump (2) an optical gain medium and (3) an
11.5 Helium neon lasers
energy pump, (2) an optical gain medium, and (3) an optical resonator.
(1)Energy pump:A 1400 V high voltage, DC power supply maintains a glowmaintains a glow discharge or plasma in a glass tube containing
ti l i tan optimal mixture (typically 5:1 to 7:1) of helium and neon gas.
Fig re 11 6: A photo of a t pical HeNe laser
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 178
Figure 11-6: A photo of a typical HeNe laser.laser=632.8 nm
Topic Eleven: Lasers and Laser Light
Figure 11-7: Diagram of optical and electrical components in a typical HeNe
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 179
g g p p yplaser.
(2) Optical gain medium: To achieve laser action, it is
Topic Eleven: Lasers and Laser Light
necessary to realize population inversion.
20.61 eV 20.66 eVHe* Ne*632 8nm
18.70 eV
Spontaneous
Through the collision, Ne atoms are excited into a
632.8nm
Figure 11-8: Simplified atomic energy level diagram showing li i
emissions deplete the lower level to maintain the population
excited into a metastable state.
diagram showing excited states of atomic He and Ne relevant to the
ti f th
Helium is “pumped” up to excited
population inversion.
operation of the laser at 632.8 nm.
excited states by electrical discharge
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 180
Helium Neon
(a) An energetic electron collisionally excites a He atom
Topic Eleven: Lasers and Laser Light
to the state He* with an energy 20.61eV. (b) The excited He* atom collides with an unexcited Ne
atom and the atoms exchange their internal energyatom and the atoms exchange their internal energy, causing the unexcited Ne atom to be excited, i.e., Ne* with 20.66 eV. This energy exchange process occurs
ith hi h b bilit l b f th id t lwith high probability only because of the accidental near equality of the two excitation energies of the two levels in these atoms.
(c) The Ne* is metastable and it deexcites to a lower excited energy level of 18.70eV by emitting a photon of
l th 6328 Åwavelength 6328 Å.
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 181
(d) The excited neon rapidly deexcites to its ground state by
Topic Eleven: Lasers and Laser Light
emitting additional photons or by collisions with the plasma tube walls.
(d) Because of the extreme quickness of the deexcitation(d) Because of the extreme quickness of the deexcitation process, at any moment in the HeNe plasma, there are more Ne atoms in Ne* state (20.66 eV) than in the l it d t t (18 70 V) d l tilower exited state (18.70 eV) and a population inversion is said to be established between these two levels.
(3) Optical resonator or cavity is formed by two highly reflecting mirrors along the axis of the discharge. Thus, the photons of 632 8 nm along the axis of the cavitythe photons of 632.8 nm along the axis of the cavity can be reflected hundreds of times. These reflecting photons can interact with other excited Ne* atoms and
h i 632 8 li h i k
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 182
cause them to emit 632.8 nm light in a process known as stimulated emission.
11.6 Laser Applications*Topic Eleven: Lasers and Laser Light
ppOptical CommunicationHolographyLaser cutting Laser marking Laser weldingLaser welding
Figure 11-9: Components of an optoelectronican optoelectronic
communication system and a bundle of optic febres .
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Quantum Physics & Classic
Quantum theory must agree with classical theory in the
Physics Quantum theory must agree with classical theory in the limit in which classical theory is known to agree with experiments.I h d h i h l i lIn other words, quantum theory must agree with classical theory in the limit of large quantum numbers.
This is because En-En-10 when n . The energy becomes continuous, not discrete. *For a particle in a1-D box (or infinite deep quantum p ( p qwell), the probability density becomes more uniform when n .
FE1002 Physics II – Quantum Physics – A/P Zhang Qing (EEE, NTU) 184
For electromagnetic waves, when their hf is muchFor electromagnetic waves, when their hf is much larger than kBT (classical measure of the mean translation energy of a particle at temperature T), a single photon can cause a detectable measurement
For a particle which is trapped to a small space whose
single photon can cause a detectable measurement. Thus, the waves could show particle properties.
For a particle which is trapped to a small space whose dimension is comparable to the particle’s de Broglie wavelength, the wave character of the particle becomes
ti l d th b ti dessential and the energy becomes quantized. Electron behaviors in nanoelectronic devices could only be interpreted through quantum theorybe interpreted through quantum theory.
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Acknowledgments
I thank Professor Tjin Swee Chuan (EEE) for hi t h l ith i th this great help with preparing the notes.
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