Download Lecture 27: Quantum Physics

Chapter 27:Quantum Physics
Blackbody Radiation and Planck’s Hypothesis
Homework : Read and understand the lecture note.
Sample homework problems : 1,18,24,30,37,46
 Thermal radiation
• An object at any temperature emits electromagnetic radiation called
thermal radiation.
• The spectrum of the radiation depends on the temperature and
properties of the object.
• From a classical point of view, thermal radiation originates from
accelerated charged particles near the surface of an object.
 Blackbody
• Is an ideal system that absorbs all radiation
incident on it.
• An opening in the cavity of a body is a good
approximation of a blackbody.
Blackbody Radiation and Planck’s Hypothesis
 Blackbody radiation
• The nature of the blackbody radiation depends
only on the temperature of the body, not on the
material composition of the object.
• The distribution of energy in blackbody radiation
varies with wavelength and temperature.
- The total amount of energy (area under the
curve) it emits increase with the temperature.
- The peak of the distribution shifts to shorter
wavelengths. This shift obeys Wien’s displacement law: max T  0.2898 102 m  K
• The classical theory of thermal radiation at the
end of 19th century failed to explain the
distribution of energy of the blackbody radiation.
Blackbody Radiation and Planck’s Hypothesis
 Planck’s hypothesis
• To solve the discrepancy between the classical physics prediction
and the observation of the blackbody radiation spectrum, in 1900
Planck developed a formula for the spectrum that explains the
observed spectrum behavior.
• Planck’s hypothesis:
- Blackbody radiation is produced by submicroscopic charged
oscillation (resonators).
- The resonators are allowed to have only certain discrete energies
given by: En  nhf n= quantum number (positive integer)
f = frequency of vibration of the resonators
h= Planck’s constant 6.626 x 10-34 J s
• Energy is quantized.
• each discrete energy value represents a different quantum state,
where the quantum number n specifies the quantum state.
Photoelectric Effect and Particle Theory of Light
 Photoelectric effect
• Light incident on certain metallic surfaces
causes the emission of electrons from the
surfaces.
• This phenomenon is called photoelectric
effect and the emitted electrons are called
photoelectrons.
• For an electron to reach Plate C when
DV<0, its kinetic energy must be at least
eDV.
• When DV is equal to or more negative
than –DVs, the stopping potential, no
electrons reach C and the current is zero.
• The maximum kinetic energy of the photoelectrons is : KEmax  eDVs
• The stopping potential is independent of
the radiation intensity.
DV=VC-VE
Photoelectric Effect and Particle Theory of Light
 Photoelectric effect (cont’d)
Observations
Predictions by wave theory
No electrons are emitted if the
Wave theory predicts that this effect
incident light frequency falls below should occur at any frequency,
a cutoff freq. fc.
provided the intensity is enough.
The max. kinetic energy of the
photoelectrons is independent of
light intensity.
Light of higher intensity carries more
energy into the metal per unit time
and eject photoelectrons with higher
energies.
The max. kinetic energy of the
photoelectrons increases with
light frequency.
No relation between photoelectron
energy and incident light frequency
is predicted.
Electrons are emitted from the
surface almost instantaneously
even at low intensities (10-9 s) .
It is expected that the photoelectrons
need some time to absorb the incident
radiation before they acquire enough
kinetic energy to escape.
Photoelectric Effect and Particle Theory of Light
 Einstein’s particle theory of light
• Einstein successfully resolve the mystery in 1905 by extending Planck’s
idea of quantization to electromagnetic waves.
• Einstein’s theory:
- A localized packet of light energy (photon) would be emitted when a
quantized oscillator made a jump from an energy state En=nhf to the
next lower state En-1=(n-1)hf.
- From conservation of energy, the photon’s energy is E=hf.
- A well localized photon can give all its energy hf to a single electron
in the metal.
- An electron in the metal is bound by
electromagnetic force and it needs to
gain a certain energy (work function f )
to be liberated:
KEmax  hf  f
Photoelectric Effect and Particle Theory of Light
 Einstein’s particle theory of light
• Predictions of Einstein’s theory:
Cutoff frequency
Photoelectrons are created by absorption of a single photon that
has enough energy to overcome the work function.
Independence of KEmax of light intensity
KEmax depends on only the frequency of light and the work function.
Linear dependence of KEmax on light frequency
KEmax=hf-f explains it.
Instantaneous production of photoelectrons
The light energy is concentrated in packets.
If the light has enough energy (frequency),
no time is need to knock-off a photoelectron.
KEmax  hc  f  0  h
c 
hc
f
c
c
f  0
X-Rays
 Roentgen’s discovery of X-rays
• Roentgen noticed that a fluorescent
screen glowed even when placed several
meters from the gas discharge tube and
when black cardboard was placed between
the tube and the screen. Discovery of x-rays
• It was found that x-rays were beams of
uncharged particles.
• von Laue suggested that if x-rays were
electromagnetic waves with very short
wavelengths (~0.1 nm), it should be possible
to diffract them using the regular atomic
spacings of crystal lattice as a diffraction
grating.
• Experiments demonstrated that his
suggestion was indeed valid.
X-Rays
 Mechanism of x-ray production
• X-rays are produced when high-speed
electrons are suddenly slowed down.
This happens when a metal target is struck
by electrons that are accelerated by a
potential difference.
• The spectrum of x-rays produced by an x-ray
tube shows two components in the intensity
vs. wavelength curve – a continuous broad
spectrum and a series of sharp intense lines.
- The continuous spectrum is produced by
bremsstrahlung, radiation by electrons
when they undergo acceleration under
influence of electromagnetic field of
nuclei of atoms.
- The line spectrum is produced by
transitions from one quantum state to
another and it depends on the target
material.
X-Rays
 Bremsstrahlung
• When an electron passes close to
positively charged nucleus in the target
material, it is deflected and accelerated
and at the same time it radiates a photon.
• If the electron loses all its energy to the
emitted photon (x-ray in this case),
eDV  hf max 
min 
hc
min
hc
eDV
• The reason for the continuous spectrum is that many of electrons
do not lose all their energy at once in a single collision.
 Application of x-rays
Crystallography, imaging of organs, radiation therapy, security screening…
Diffraction of X-Rays by Crystals
Red spheres : Na+
2d sin   m (m  1,2,3,...)
NaCl crystal
• If atoms are aligned in a regular
manner, they act as a diffraction
grating device. This is realized
in a crystal.
• When the incident beam of x-ray
are reflected at the upper and
lower plane, the path difference
produces interference pattern.
• If Bragg’s law is satisfied,
constructed interference is produced:
Blue spheres : Cl-
 Principle
Diffraction of X-Rays by Crystals
 Principle (cont’d)
• If well-collimated beam of x-ray is
used, each configuration of layers
made by atoms in a crystal produces
a point on the screen.
• A pattern of spots (Laue pattern)
created on the screen including
the positions and intensities of the
spots gives information about the
structure of the crystal.
Double-helix structure
of DNA was determined by
x-ray crystallography.
Compton Effect
 Photon nature of light
• Compton’s experiment:
- X-ray beam of wavelength 0
directed to a lock of graphite.
- Compton observed that the
scattered x-rays had a slightly
longer wavelength (lower energy).
Compton shift
D  '  0
- The amount of reduction in energy depends on the amount of angle
x-rays are deflected.
• Compton’s explanation :
- If a photon behaves like a particle, its collision with other particles is
similar to a collision between two billiard balls.
- A photon collides with an electron at rest and transfer part of its energy
and momentum to the electron. Then from conservation of energy and
momentum he derived a formula :
h
D 
(1  cos  ) me:electron mass
h/mec = 0.00243 nm Compton wavelength
me c
 : scattering angle
Compton Effect
 Example 27.6 : Scattering x-rays
• X-rays of wavelength 0 =0.200000 nm are scattered from a block of
material. The scattered x-rays are observed at an angle of 45.0o to the
incident beam.
(a) Calculate the wavelength of the scattered x-rays at this angle.
h
6.63 1034 J  s
D 
(1  cos  ) 
(1  cos 45.0)  7.111013 m
31
8
me c
(9.1110 kg)(3.00 10 m/s)
  D  0  0.200711 nm
(b) Compute the fractional change in the energy of a photon in the
collision.
E  hf  h
c

DE E f  Ei hc /  f  hc / i
D
0.000711 nm




 3.54 103
E
Ei
hc / i
i
0.200711 nm
Dual Nature of Light and Matter
 Light and electromagnetic radiation
• The photoelectric effect and the Compton scattering suggest particle
nature of light with energy hf and momentum h/.
• At the same time light behaves like wave.
Light has a dual nature, exhibiting both wave and particle characteristics.
 Wave properties of particles
• De Broglie’s idea :
All forms of matter have both properties - wave and particle
characteristics
• According to de Broglie’s idea, electrons like light have a dual
particle-wave nature with the following relation among the momentum,
energy and wavelength (de Broglie wavelength of a particle) :
  h / p  h /( mv)
An application: electron microscope
f  E/h
• His postulate was verified by diffraction pattern produced by electrons.
Wave Function
 Light and electromagnetic radiation
• In 1926, Schroedinger proposed a wave equation that described how
wave change in space and time : The Schroedinger wave equation
• Solving Schroedinger’s equation determines a quantity Y called the
wave function.
Y
2
i

DY  U ( x, y, z )Y
t
2m
Don’t worry. Just to
be cool.
• Putting aside historical controversies over the interpretation of a wave
function Y , this wave function describes a single particle, the value of
|Y|2 at some location at a given time is proportional to the probabilities
per unit volume of finding the particle at that location at that time.
• Adding up all the values of |Y|2 in a given region gives the probability
of finding the particle in that region.
Uncertainty Principle
 Uncertainty principle
• How accurately can we measure the position and speed of a particle
at any instance? Is there any limit?
• In 1927, Heisenberg answered this question and introduced a surprising
principle:
Heisenberg’s uncertainty principle:
If a measurement of the position of a particle is made with precision
Dx and a simultaneous measurement of linear momentum is made
with precision Dpx, then the product of the two uncertainties can never
be smaller than h/4p.
DxDp x 
h

4p
It is physically impossible to measure simultaneously the exact
position and exact linear momentum of a particle.
Uncertainty Principle
 Interpretation of uncertainty principle
• Consider a thought experiment:
You are trying to measure the position
and momentum of an electron as
accurate as possible using a powerful
microscope.
As the momentum of incoming photon
is h /  , the maximum uncertainty in
the electron’s momentum after collision
is Dpx  h /  .
It is a reasonable guess from the wave
property of photon, you can determine
the position of the electron with an
accuracy of Dx   . Then DxDt  h .
• Another form of the uncertainty principle:
h
DEDt 

4p
Uncertainty Principle
 Example 27.8 : Locating an electron
• The speed of an electron is measured to be 5x103 m/s to an accuracy
of 0.00300%. Find the minimum uncertainty in determining the position
of this electron.
p x  me v  (9.1110 31 kg)(5.00 103 m/s)  4.56 10-27 kg  m/s
Dp x  0.0000300 p  1.37 10 31 kg  m/s
h
h
DxDp x 
 Dx 
 0.384 10 3 m  0.384 mm
4p
4pDp x
 Example 27.9 : Excited states of atoms
• Electrons in atoms can be found in certain high states of energy called
excited states for short periods of time. If the average time that an
electron exists in one of these states is 1.00x10-8 s, what is the minimum
uncertainty in energy of the excited state?
h
6.63 1034 J  s
 27
-8
DE 


5
.
28

10
J

3.30

10
eV.
8
4pDt 4p (1.00 10 s)