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General Physics (PHY 2140)
Lecture 27
Modern Physics
Quantum Physics
Blackbody radiation
Planks hypothesis
Chapter 27
http://www.physics.wayne.edu/~apetrov/PHY2140/
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Quantum Physics
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Introduction: Need for Quantum Physics
Problems remained from classical mechanics that relativitydidnt explain:
Blackbody Radiation The electromagnetic radiation emitted by a heated object
Photoelectric Effect Emission of electrons by an illuminated metal
Spectral Lines Emission of sharp spectral lines by gas atoms in an electric discharge tube
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Development of Quantum Physics
1900 to 1930
Development of ideas of quantum mechanics
Also called wave mechanics
Highly successful in explaining the behavior of atoms, molecules, and nuclei
Quantum Mechanics reduces to classical mechanics when applied
to macroscopic systemsInvolved a large number of physicists
Planck introduced basic ideas
Mathematical developments and interpretations involved such people asEinstein, Bohr, Schrdinger, de Broglie, Heisenberg, Born and Dirac
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26.1 Blackbody Radiation
An object at any temperature is known to emit
electromagnetic radiation
Sometimes called thermal radiation
Stefans Law describes the total power radiated
The spectrum of the radiation depends on the temperature andproperties of the object
4P AeT
emissivityStefans constant
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Blackbody
Blackbody is an idealized system that absorbs incident
radiation of all wavelengths
If it is heated to a certain temperature, it starts radiate
electromagnetic waves of all wavelengths
Cavity is a good real-life approximation to a blackbody
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Blackbody Radiation Graph
Experimental data fordistribution of energy inblackbody radiation
As the temperature increases,
the total amount of energyincreases
Shown by the area underthe curve
As the temperature increases,the peak of the distributionshifts to shorter wavelengths
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Wiens Displacement Law
The wavelength of the peak of the blackbody distribution
was found to follow Weins Displacement Law
max T = 0.2898 x 10-2
m K
maxis the wavelength at the curves peak
T is the absolute temperature of the object emitting the radiation
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The Ultraviolet Catastrophe
Classical theory did not matchthe experimental data
At long wavelengths, thematch is good
Rayleigh-Jeans law
At short wavelengths, classical
theory predicted infinite energyAt short wavelengths,experiment showed no energy
This contradiction is called theultraviolet catastrophe
4
2 ckTP
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Plancks Resolution
Planck hypothesized that the blackbody radiation was produced byresonators
Resonators were submicroscopic charged oscillators
The resonators could only have discrete energies
En= n h
n is called the quantum number
is the frequency of vibrationh is Plancks constant, h=6.626 x 10-34 J s
Key point is quantized energy states
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QUICK QUIZ
A photon (quantum of light) is reflected from a mirror. True or false:
(a) Because a photon has a zero mass, it does not exert a force on
the mirror.
(b) Although the photon has energy, it cannot transfer any energy to
the surface because it has zero mass.
(c) The photon carries momentum, and when it reflects off the mirror,it undergoes a change in momentum and exerts a force on
the mirror.
(d) Although the photon carries momentum, its change in momentum
is zero when it reflects from the mirror, so it cannot exert a
force on the mirror.
(a) False
(b) False
(c) True
(d) Falsep Ft
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27.2 Photoelectric Effect
When light is incident on certain metallic surfaces, electrons areemitted from the surface
This is called thephotoelectric effect
The emitted electrons are calledphotoelectrons
The effect was first discovered by HertzThe successful explanation of the effect was given by Einstein in1905
Received Nobel Prize in 1921 for paper on electromagnetic radiation, ofwhich the photoelectric effect was a part
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Photoelectric Effect Schematic
When light strikes E,
photoelectrons are emitted
Electrons collected at C and
passing through the ammeterare a current in the circuit
C is maintained at a positive
potential by the power supply
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Photoelectric Current/Voltage Graph
The current increases with
intensity, but reaches a
saturation level for largeVs
No current flows for voltagesless than or equal toVs, the
stopping potential
The stopping potential is
independent of the radiationintensity
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Features Not Explained by Classical
Physics/Wave Theory
No electrons are emitted if the incident light frequency is
below some cutoff frequencythat is characteristic of the
material being illuminated
The maximum kinetic energy of the photoelectrons is
independent of the light intensity
The maximum kinetic energy of the photoelectrons
increases with increasing light frequency
Electrons are emitted from the surface almost
instantaneously, even at low intensities
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Einsteins Explanation
A tiny packet of light energy, called aphoton, would be emittedwhen a quantized oscillator jumped from one energy level to thenext lower one
Extended Plancks idea of quantization to electromagneticradiation
The photons energy would be E = h
Each photon can give all its energy to an electron in the metal
The maximum kinetic energy of the liberated photoelectron is
KE = h
is called the work function of the metal
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Explanation of Classical Problems
The effect is not observed below a certain cutofffrequency since the photon energy must be greater thanor equal to the work function Without this, electrons are not emitted, regardless of the intensity
of the light
The maximum KE depends only on the frequency andthe work function, not on the intensity
The maximum KE increases with increasing frequency
The effect is instantaneous since there is a one-to-oneinteraction between the photon and the electron
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Verification of Einsteins Theory
Experimental observations of a
linear relationship between KE
and frequency confirm
Einsteins theory
The x-intercept is the cutoff
frequency
cf
h
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27.3 Application: Photocells
Photocells are an application of the photoelectric effect
When light of sufficiently high frequency falls on the cell,a current is produced
Examples Streetlights, garage door openers, elevators
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27.4 The Compton Effect
Compton directed a beam of x-rays toward a block of graphite
He found that the scattered x-rays had a slightly longer wavelengththat the incident x-rays
This means they also had less energy
The amount of energy reduction depended on the angle at which the
x-rays were scatteredThe change in wavelength is called the Compton shift
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Compton Scattering
Compton assumed the
photons acted like other
particles in collisions
Energy and momentum
were conserved
The shift in wavelength is)cos1(
cm
h
e
o
Compton wavelength
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Compton Scattering
The quantity h/mec is called the Compton wavelength
Compton wavelength = 0.00243 nm
Very small compared to visible light
The Compton shift depends on the scattering angle and not onthe wavelength
Experiments confirm the results of Compton scattering and
strongly support the photon concept
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Problem: Compton scattering
A beam of 0.68-nm photons undergoes Compton scattering from free
electrons. What are the energy and momentum of the photons that
emerge at a 45 angle with respect to the incident beam?
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QUICK QUIZ 1
An x-ray photon is scattered by an electron. The frequency of the
scattered photon relative to that of the incident photon (a)
increases, (b) decreases, or (c) remains the same.
(b). Some energy is transferred to the electron in the scattering
process. Therefore, the scattered photon must have less energy
(and hence, lower frequency) than the incident photon.
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QUICK QUIZ 2
A photon of energy E0 strikes a free electron, with the scattered photon
of energy Emoving in the direction opposite that of the incident
photon. In this Compton effect interaction, the resulting kinetic energy
of the electron is (a) E0 , (b) E, (c) E0 E , (d) E0 + E , (e) none of the
above.
(c). Conservation of energy requires the kinetic energy given to
the electron be equal to the difference between the energy of the
incident photon and that of the scattered photon.
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QUICK QUIZ 2
A photon of energy E0 strikes a free electron, with the scattered photon
of energy Emoving in the direction opposite that of the incident
photon. In this Compton effect interaction, the resulting kinetic energy
of the electron is (a) E0 , (b) E, (c) E0 E , (d) E0 + E , (e) none of the
above.
(c). Conservation of energy requires the kinetic energy given to
the electron be equal to the difference between the energy of the
incident photon and that of the scattered photon.
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27.8 Photons and Electromagnetic Waves
Light has a dual nature. It exhibits both wave and particlecharacteristics
Applies to all electromagnetic radiation
The photoelectric effect and Compton scattering offer evidence for
the particle nature of light When light and matter interact, light behaves as if it were composed of
particles
Interference and diffraction offer evidence of the wave nature of light
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28.9 Wave Properties of Particles
In 1924, Louis de Broglie postulated that because
photons have wave and particle characteristics, perhaps
all forms of matter have both properties
Furthermore, the frequency and wavelength of matter
waves can be determined
The de Broglie wavelength of a particle is
The frequency of matter waves is mv
h
h
E
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The Davisson-Germer Experiment
They scattered low-energy electrons from a nickel target
They followed this with extensive diffraction measurements fromvarious materials
The wavelength of the electrons calculated from the diffraction dataagreed with the expected de Broglie wavelength
This confirmed the wave nature of electrons
Other experimenters have confirmed the wave nature of otherparticles
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Review problem: the wavelength of a proton
Calculate the de Broglie wavelength for a proton (mp=1.67x10-27 kg )
moving with a speed of 1.00 x 107 m/s.
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Calculate the de Broglie wavelength for a proton (mp=1.67x10-27 kg ) moving with a
speed of 1.00 x 107 m/s.
Given:
v = 1.0 x 107m/s
Find:
p = ?
Given the velocity and a mass of the proton we can
compute its wavelength
p
p
h
m v
Or numerically,
34
14
31 7
6.63 10
3.97 10
1.67 10 1.00 10ps
J sm
kg m s
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QUICK QUIZ 3
A non-relativistic electron and a non-relativistic proton are movingand have the same de Broglie wavelength. Which of the
following are also the same for the two particles: (a) speed, (b)
kinetic energy, (c) momentum, (d) frequency?
(c). Two particles with the same de Broglie wavelength will have the same
momentump = mv. If the electron and proton have the same momentum, they
cannot have the same speed because of the difference in their masses. For the
same reason, remembering that KE = p2
/2m, they cannot have the same kineticenergy. Because the kinetic energy is the only type of energy an isolated particle can
have, and we have argued that the particles have different energies, Equation 27.15
tells us that the particles do not have the same frequency.
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QUICK QUIZ 2
A non-relativistic electron and a non-relativistic proton are movingand have the same de Broglie wavelength. Which of the
following are also the same for the two particles: (a) speed, (b)
kinetic energy, (c) momentum, (d) frequency?
(c). Two particles with the same de Broglie wavelength will have the same
momentump = mv. If the electron and proton have the same momentum, they
cannot have the same speed because of the difference in their masses. For the
same reason, remembering that KE = p2
/2m, they cannot have the same kineticenergy. Because the kinetic energy is the only type of energy an isolated particle can
have, and we have argued that the particles have different energies, Equation 27.15
tells us that the particles do not have the same frequency.
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The Electron Microscope
The electron microscope dependson the wave characteristics ofelectrons
Microscopes can only resolve details
that are slightly smaller than thewavelength of the radiation used toilluminate the object
The electrons can be accelerated tohigh energies and have smallwavelengths
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27.10 The Wave Function
In 1926 Schrdinger proposed a wave equation that
describes the manner in which matter waves change in
space and time
Schrdingers wave equation is a key element in
quantum mechanics
Schrdingers wave equation is generally solved for the
wave function,
i Ht
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The Wave Function
The wave function depends on the particles position and
the time
The value of ||2 at some location at a given time is
proportional to the probability of finding the particle at
that location at that time
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27.11 The Uncertainty Principle
When measurements are made, the experimenter isalways faced with experimental uncertainties in themeasurements
Classical mechanics offers no fundamental barrier toultimate refinements in measurements
Classical mechanics would allow for measurements witharbitrarily small uncertainties
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The Uncertainty Principle
Quantum mechanics predicts that a barrier to measurementswith ultimately small uncertainties does exist
In 1927 Heisenberg introduced the uncertainty principle
If a measurement of position of a particle is made with precisionxand a simultaneous measurement of linear momentum is made withprecision p, then the product of the two uncertainties can never besmaller than h/4
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The Uncertainty Principle
Mathematically,
It is physically impossible to measure simultaneously the
exact position and the exact linear momentum of aparticle
Another form of the principle deals with energy and time:
4
hpx x
4
htE
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Thought Experiment the Uncertainty
Principle
A thought experiment for viewing an electron with a powerfulmicroscope
In order to see the electron, at least one photon must bounce off it
During this interaction, momentum is transferred from the photon tothe electron
Therefore, the light that allows you to accurately locate the electronchanges the momentum of the electron
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Problem: macroscopic uncertainty
A 50.0-g ball moves at 30.0 m/s. If its speed is measured to an
accuracy of 0.10%, what is the minimum uncertainty in its
position?
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A 50.0-g ball moves at 30.0 m/s. If its speed is measured to an accuracy of 0.10%,
what is the minimum uncertainty in its position?
Given:
v = 30 m/s
dv = 0.10%
m = 50.0 g
Find:
dx = ?
Notice that the ball is non-relativistic. Thus, p = mv,
and uncertainty in measuring momentum is
2 3 2
50.0 10 1.0 10 30 1.5 10
p m v m v v
kg m s kg m s
d
Thus, uncertainty relation implies
24
32
3
6.63 103.5 10
4 4 1.5 10
h J sx mp k g m s
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Problem: Macroscopic measurement
A 0.50-kg block rests on the icy surface of a frozen pond, which we
can assume to be frictionless. If the location of the block is measured
to a precision of 0.50 cm, what speed must the block acquire becauseof the measurement process?
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