Transcript Chapter 27

Chapter 27
Quantum Physics
Need for Quantum Physics
• Problems remained that classical mechanics couldn’t
explain:
• Blackbody radiation – 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
Blackbody Radiation
• An object at any temperature emits
electromagnetic radiation, sometimes
called thermal radiation
• Stefan’s Law describes the total power
radiated
• The spectrum of the radiation depends
on the temperature and properties of
the object
• As the temperature increases, the total
amount of energy increases and the
peak of the distribution shifts to shorter
wavelengths
Josef Stefan
1835 – 1893
Wien’s Displacement Law
• The wavelength of the peak of the blackbody
distribution was found to follow Wein’s displacement
law
λmax T = 0.2898 x 10-2 m • K
• λmax – wavelength at which the curve’s peak
• T – absolute temperature of the object emitting the
radiation
Wilhelm Carl Werner Otto Fritz Franz Wien
1864 – 1928
The Ultraviolet Catastrophe
• Classical theory did not match the
experimental data
• At long wavelengths, the match is
good
• At short wavelengths, classical
theory predicted infinite energy
• At short wavelengths, experiment
showed no or little energy
• This contradiction is called the
ultraviolet catastrophe
Planck’s Resolution
• Planck hypothesized that the blackbody radiation was
produced by resonators, submicroscopic charged
oscillators
• The resonators could only have discrete energies
En = n h ƒ
• n – quantum number, ƒ – frequency of vibration, h –
Planck’s constant, 6.626 x 10-34 J s
• Key point is quantized energy states
Max Karl Ernst Ludwig Planck
1858 – 1947
Photoelectric Effect
• Photoelectric effect (first
discovered by Hertz): when
light is incident on certain
metallic surfaces, electrons are
emitted from the surface –
photoelectrons
• The successful explanation of
the effect was given by Einstein
• Electrons collected at C
(maintained at a positive
potential) and passing through
the ammeter are a current in the
circuit
Albert Einstein
1879 – 1955
Photoelectric Effect
• The current increases with intensity, but reaches a
saturation level for large DV’s
• No current flows for voltages less than or equal to –
DVs, the stopping potential, independent of the
radiation intensity
• The maximum kinetic energy of the photoelectrons is
related to the stopping potential: KEmax = e DVs
Features Not Explained by Classical
Physics
• No electrons are emitted if the incident light
frequency is below some cutoff frequency that 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
KEmax = hƒ – f
Einstein’s Explanation
• Einstein extended Planck’s idea of quantization to
electromagnetic radiation
• A tiny packet of light energy – a photon – is emitted
when a quantized oscillator jumps from one energy
level to the next lower one
• The photon’s energy is E = hƒ
• Each photon can give all its energy to an electron in
the metal
• The maximum kinetic energy of the liberated
photoelectron is KEmax = hƒ – f
•
f is called the work function of the metal
KEmax = hƒ – f
Einstein’s Explanation
• The effect is not observed below a certain cutoff
frequency since the photon energy must be greater
than or 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 and
the work function, not on the intensity
• The maximum KE increases with increasing frequency
• The effect is instantaneous since there is a one-to-one
interaction between the photon and the electron
KEmax = hƒ – f
Verification of Einstein’s Theory
• Experimental observations of a
linear relationship between KE
and frequency confirm
Einstein’s theory
c = λƒ
• The x-intercept is the cutoff
frequency
• The cutoff wavelength is
related to the work function
• Wavelengths greater than lC
incident on a material with a
work function f don’t result in
the emission of photoelectrons
lc 
hc
f
Chapter 27
Problem 16
An isolated copper sphere of radius 5.00 cm, initially
uncharged, is illuminated by ultraviolet light of wavelength
200 nm. What charge will the photoelectric effect induce
on the sphere? The work function for copper is 4.70 eV.
X-Rays
• X-rays (discovered and named by Roentgen):
electromagnetic radiation with short – typically about
0.1 nm – wavelengths
• X-rays have the ability to penetrate most materials with
relative ease
• X-rays are produced when high-speed electrons are
suddenly slowed down
Wilhelm Conrad Röntgen
1845 – 1923
Production of X-rays
• X-rays can be produced by
electrons striking a metal
target
• A current in the filament
causes electrons to be
emitted
• These freed electrons are
accelerated toward a dense
metal target (the target is
held at a higher potential than
the filament)
X-ray Spectrum
• The x-ray spectrum has two
distinct components
• 1) Bremsstrahlung: a
continuous broad spectrum,
which depends on voltage
applied to the tube
• 2) The sharp, intense lines,
which depend on the nature of
the target material
Bremsstrahlung
• An electron passes near a target nucleus and is
deflected from its path by its attraction to the nucleus
• This produces an acceleration of the electron and
hence emission of electromagnetic radiation
• If the electron loses all of its energy in the collision,
the initial energy of the electron is completely
transformed into a photon
• The wavelength then is
eDV  hƒmax 
hc
lmin
Bremsstrahlung
• Not all radiation produced is at this wavelength
• Many electrons undergo more than one collision
before being stopped
• This results in the continuous spectrum produced
Diffraction of X-rays by Crystals
• For diffraction to occur, the spacing between the lines
must be approximately equal to the wavelength of the
radiation to be measured
• The regular array of atoms in a crystal can act as a
three-dimensional grating for diffracting X-rays
Diffraction of X-rays by Crystals
• A beam of X-rays is incident on the crystal
• The diffracted radiation is very intense in the
directions that correspond to constructive interference
from waves reflected from the layers of the crystal
Diffraction of X-rays by Crystals
• The diffraction pattern is detected by photographic film
• The array of spots is called a Laue pattern
• The crystal structure is determined by analyzing the
positions and intensities of the various spots
Max Theodore Felix
von Laue
1879 – 1960
Bragg’s Law
• The beam reflected from the lower surface travels
farther than the one reflected from the upper surface
• If the path difference equals some integral multiple of
the wavelength, constructive interference occurs
• Bragg’s Law gives the conditions for constructive
interference
2 d sin θ = m λ
m = 1, 2, 3…
Sir William Lawrence
Bragg
1890 – 1971
The Compton Effect
• Compton directed a beam of x-rays toward a block of
graphite and found that the scattered x-rays had a
slightly longer wavelength (lower energy) that the
incident x-rays
• The change in wavelength (energy) – the Compton
shift – depends on the angle at which the x-rays are
scattered
Arthur Holly Compton
1892 – 1962
The Compton Effect
• Compton assumed the photons acted like other
particles in collisions with electrons
• Energy and momentum were conserved
• The shift in wavelength is given by
h
Dl  l  lo 
(1  cos  )
mec
Arthur Holly Compton
1892 – 1962
The Compton Effect
• The Compton shift depends on the scattering angle
and not on the wavelength
• h/mec = 0.002 43 nm (very small compared to visible
light) is called the Compton wavelength
h
Dl  l  lo 
(1  cos  )
mec
Arthur Holly Compton
1892 – 1962
Chapter 27
Problem 33
A 0.45-nm x-ray photon is deflected through a 23° angle
after scattering from a free electron. (a) What is the kinetic
energy of the recoiling electron? (b) What is its speed?
Photons and Electromagnetic Waves
• Light (as well as all other electromagnetic radiation)
has a dual nature. It exhibits both wave and particle
characteristics
• 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
• On the other hand, interference and diffraction offer
evidence of the wave nature of light
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
h
h
l 
p mv
• The frequency of matter waves is
E
ƒ
h
Louis de Broglie
1892 – 1987
Wave Properties of Particles
• The de Broglie equations show the dual nature of
matter
• Each contains matter concepts (energy and
momentum) and wave concepts (wavelength and
frequency)
• The de Broglie wavelength of a particle is
h
h
l 
p mv
• The frequency of matter waves is
E
ƒ
h
Louis de Broglie
1892 – 1987
The Davisson-Germer Experiment
• Davisson and Germer scattered low-energy electrons
from a nickel target and followed this with extensive
diffraction measurements from various materials
• The wavelength of the electrons calculated from the
diffraction data agreed with the expected de Broglie
wavelength
• This confirmed the wave nature of
electrons
• Other experimenters confirmed the
wave nature of other particles
Clinton Joseph Davisson (1881 – 1958) and Lester Halbert Germer (1896 – 1971)
Chapter 27
Problem 40
A monoenergetic beam of electrons is incident on a single
slit of width 0.500 nm. A diffraction pattern is formed on a
screen 20.0 cm from the slit. If the distance between
successive minima of the diffraction pattern is 2.10 cm,
what is the energy of the incident electrons?
The Wave Function
• In 1926 Schrödinger proposed a wave equation that
describes the manner in which matter waves change in
space and time
• Schrödinger’s wave equation is a key element in
quantum mechanics
• Schrödinger’s wave equation is generally solved for the
wave function, Ψ, which depends on the
particle’s position and the time
• The value of Ψ2 at some location at
a given time is proportional to the
probability of finding the particle Erwin Rudolf Josef Alexander
Schrödinger
at that location at that time
1892 – 1987
The Uncertainty Principle
• When measurements are made, the experimenter is
always faced with experimental uncertainties in the
measurements
• Classical mechanics offers no fundamental barrier to
ultimate refinements in measurements and would allow
for measurements with arbitrarily small uncertainties
• Quantum mechanics predicts that a barrier to
measurements with ultimately small uncertainties does
exist
The Uncertainty Principle
• In 1927 Heisenberg introduced the uncertainty
principle: If a measurement of position of a particle is
made with precision Δx and a simultaneous
measurement of linear momentum is made with
precision Δpx, then the product of the two
uncertainties can never be smaller than h/4
• Mathematically,
h
DxDp x 
4
• It is physically impossible to measure
simultaneously the exact position and the
Werner Karl Heisenberg
exact linear momentum of a particle
1901 – 1976
The Uncertainty Principle
• Another form of the principle deals with energy and
time:
h
DEDt 
4
• Energy of a particle can not be measured with
complete precision in a short interval of time Dt
Werner Karl Heisenberg
1901 – 1976
The Uncertainty Principle
• A thought experiment for viewing an electron with a
powerful microscope
• In order to see the electron, at least one photon must
bounce off it
• During this interaction, momentum is transferred from
the photon to the electron
• Therefore, the light
that allows you to
accurately locate
the electron changes
the momentum of the electron
The Uncertainty Principle
Dx ≈l
Dp ≈ p 
DxDp ≈l 
h
l
h
l
h
Chapter 27
Problem 46
(a) Show that the kinetic energy of a nonrelativistic
particle can be written in terms of its momentum as KE =
p2/2m. (b) Use the results of (a) to find the minimum
kinetic energy of a proton confined within a nucleus
having a diameter of 1.0 × 10−15 m.
Answers to Even Numbered Problems
Chapter 27:
Problem 12
5.4 eV
Answers to Even Numbered Problems
Chapter 27:
Problem 20
(a) 8.29 × 10−11 m
(b) 1.24 × 10−11 m
Answers to Even Numbered Problems
Chapter 27:
Problem 24
6.7°
Answers to Even Numbered Problems
Chapter 27:
Problem 28
1.8 keV, 9.7 × 10−25 kg⋅m/s
Answers to Even Numbered Problems
Chapter 27:
Problem 34
(a) 1.98 × 10−11 m
(b) 1.98 × 10−14 m
Answers to Even Numbered Problems
Chapter 27:
Problem 52
191 MeV