General Physics (PHY 2140)
Lecture 14
 Modern Physics
1. Relativity
 Einstein’s General Relativity
2. Quantum Physics
 Blackbody Radiation
 Photoelectric Effect
 X-Rays
 Diffraction by Crystals
 The Compton Effect
Chapter 26
Chapter 27
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Lightning Review
Last lecture:
Einstein’s Postulates:
For all inertial frames of reference:
 Velocity of light in vacuo, c, is the same (~2.998 × 108 m/s).
 The form of the laws of physics (E/M, Mechanics, thermodynamics, etc.) are
identical.
Relativity – consequences
• Time dilation, length contraction
• Relativistic energy, momentum
• Relativistic addition of velocities
vab 
vad  vdb
v v
1  ad 2 db
c
p
mv
1 v c
2
2
t 
  mv
t p
1 v c
2
2
L  Lp 1  v 2 c 2
KE = mc2 – mc2
Problem: relativistic proton
A proton in a high-energy accelerator is given a kinetic energy of
50.0 GeV. Determine
(a) the momentum and
(b) the speed of the proton.
A proton in a high-energy accelerator is given a kinetic energy of 50.0 GeV.
Determine (a) the momentum and (b) the speed of the proton.
Given:
Recall that E2 = p2c2 + (mc2)2. This can be used to
solve for p:
2
2
p
E = 50.0 GeV
Thus,



 mc
2
 
 KE  mc
2

2
c
 KE   2 mc 2
2
p
2
c
mc2
=
946.3 MeV

E  mc
2
  KE 
c
 50.9 GeV
c
Similarly with velocity:
Find:
p=?
v =?


E   mc 2   
1
1  v2 c2

E
mc 2
2
 E 
v  1  1  2   0.9998c
 mc 
Problem: relativistic pion
The average lifetime of a p meson in its own frame of reference (i.e., the
proper lifetime) is 2.6 × 10–8 s. If the meson moves with a speed of
0.98c, what is
(a) its mean lifetime as measured by an observer on Earth and
(b) the average distance it travels before decaying as measured by an
observer on Earth?
(c) What distance would it travel if time dilation did not occur?
The average lifetime of a p meson in its own frame of reference (i.e., the proper
lifetime) is 2.6 × 10–8 s. If the meson moves with a speed of 0.98c, what is
(a) its mean lifetime as measured by an observer on Earth and
(b) the average distance it travels before decaying as measured by an observer
on Earth?
(c) What distance would it travel if time dilation did not occur?
Given:
v = 0.98 c
tp = 2.6 × 10–8 s
Recall that the time measured by observer on Earth
will be longer then the proper time. Thus for the
lifetime
t
tp
1  v2 c2
 1.3 107 s
Thus, at this speed it will travel
Find:
t=?
d=?
dn =?



d  vt   0.98  3 108 m s 1.3 10 7 s  38m
If special relativity were wrong, it would only fly about



d  vt p   0.98  3 108 m s 2.6 108 s  7.6m
Problem: more spaceships…
A spaceship travels at 0.750c relative to Earth. If the spaceship fires a
small rocket in the forward direction, how fast (relative to the ship) must it
be fired for it to travel at 0.950c relative to Earth?
A spaceship travels at 0.750c relative to Earth. If the spaceship fires a small rocket
in the forward direction, how fast (relative to the ship) must it be fired for it to travel
at 0.950c relative to Earth?
Given:
vSE = 0.750 c
vRE = 0.950 c
Find:
vRS = ?
Since vES = -vSE = velocity of Earth relative to ship, the
relativistic velocity addition equation gives
vRS
vRE  vES

v v
1  RE 2 ES
c
0.950 c   0.750 c 

 0.696c
0.950 c  0.750 c 

1
c2
Mass – Inertial vs. Gravitational
Mass has a gravitational attraction for other
masses
m m'
Fg  G
g
g
r2
Mass has an inertial property that resists
acceleration
 Fi = mi a
The value of G was chosen to make the values
of mg and mi equal
Einstein’s Reasoning
Concerning Mass
That mg and mi were directly proportional
was evidence for a basic connection
between them
No mechanical experiment could
distinguish between the two masses
He extended the idea: no experiment of
any type can distinguish the two masses
Postulates of General Relativity
All laws of nature must have the same
form for observers in any frame of
reference, whether accelerated or not
In the vicinity of any given point, a
gravitational field is equivalent to an
accelerated frame of reference without a
gravitational field

This is the principle of equivalence
Implications of General
Relativity
Gravitational mass and inertial mass are not just
proportional, but completely equivalent
A clock in the presence of gravity runs more
slowly than one where gravity is negligible
The frequencies of radiation emitted by atoms in
a strong gravitational field are shifted to lower
frequencies

This has been detected in the spectral lines emitted
by atoms in massive stars
QUICK QUIZ 26.5
Two identical clocks are in the same house, one
upstairs in a bedroom and the other downstairs in
the kitchen. Which statement is correct? (a) The
clock in the kitchen runs more slowly than the
clock in the bedroom. (b) The clock in the
bedroom runs more slowly than the clock in the
kitchen. (c) Both clocks keep the same time.
QUICK QUIZ 26.5 ANSWER
(a). The downstairs clock runs more
slowly because it is closer to the Earth
and hence experiences a stronger
gravitational field than the upstairs clock
does.
The time difference between the two clocks is about
4x10-16 s for each second that passes on the clock.
Too small to observe without expensive equipment!
More Implications of General
Relativity
A gravitational field may be “transformed
away” at any point if we choose an
appropriate accelerated frame of reference
– a freely falling frame
Einstein specified a certain quantity, the
curvature of time-space, that describes the
gravitational effect at every point
Curvature of Space-Time
There is no such thing as a gravitational
field

According to Einstein
Instead, the presence of a mass causes a
curvature of time-space in the vicinity of
the mass

This curvature dictates the path that all freely
moving objects must follow
Testing General Relativity
General Relativity predicts that a light ray passing near
the Sun should be deflected by the curved space-time
created by the Sun’s mass
The prediction was confirmed by astronomers during a
total solar eclipse
Black Holes
If the concentration of mass becomes
great enough, a black hole is believed to
be formed
In a black hole, the curvature of spacetime is so great that, within a certain
distance from its center, all light and
matter become trapped
Quantum Physics
Chapter 27
Objectives
27.1 Explain how blackbody radiation implies the quantization of electromagnetic
energy.
27.2 Describe how Einstein's photoelectric equation implies the existence of the
photon.
27.3 Define bremsstrahlung radiation.
27.4 Explain how X-ray diffraction can be used to determine crystal structure.
27.5 Solve sample problems that demonstrate the particle-like aspects of radiation
as predicted by the Compton effect.
27.6 Identify the particle and wave-like aspects of electromagnetic radiation.
27.7 Identify the wave-like aspects of particles, citing examples.
27.8 Define the wave function for particles.
27.9 Summarize the key aspects of the uncertainty principle.
27.10 Describe the operation of the scanning tunneling microscope.
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Need for Quantum Physics
Problems remained from classical mechanics
that relativity didn’t 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
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 systems
Involved a large number of physicists


Planck introduced basic ideas
Mathematical developments and interpretations involved
such people as Einstein, Bohr, Schrödinger, de Broglie,
Heisenberg, Born and Dirac
Blackbody Radiation
An object at any temperature is
known to emit electromagnetic
radiation


Sometimes called thermal radiation
Stefan’s Law describes the total power
radiated
P   AeT 4
Stefan’s constant

emissivity
The spectrum of the radiation depends
on the temperature and properties of
the object
Black Body Radiation
Blackbody Radiation Graph
Experimental data for
distribution of energy in
blackbody radiation
As the temperature
increases, the total
amount of energy
increases

Shown by the area under
the curve
As the temperature
increases, the peak of the
distribution shifts to
shorter wavelengths
Wien’s Displacement Law
The wavelength of the peak of the blackbody
distribution was found to follow Wein’s
Displacement Law
lmaxT = 0.2898 x 10-2 m • K
lmax is the wavelength at the curve’s peak
T is the absolute temperature of the object emitting
the radiation
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
energy
This contradiction is called
the ultraviolet catastrophe
Planck’s Resolution
Planck hypothesized that the blackbody
radiation was produced by resonators

Resonators were submicroscopic charged oscillators
The resonators could only have discrete
energies

En = n h ƒ
n is called the quantum number
ƒ is the frequency of vibration
h is Planck’s constant, 6.626 x 10-34 J s
Key point is quantized energy states
Problem: a lightbulb
Assuming that the tungsten filament of a lightbulb is a blackbody,
determine its peak wavelength if its temperature is 2 900 K.
Assuming that the tungsten filament of a lightbulb is a blackbody, determine its
peak wavelength if its temperature is 2 900 K.
Given:
Recall Wein;s law: λmax T = 0.2898 x 10-2 m • K. It can
be used to solve for lmax:
T = 2900 K
lmax
0.2898 102 m  K

T
Thus,
Find:
lmax = ?
0.2898 10


2
lmax
2900K
m K
  9.99 10
7
m  999nm
This is infrared, so most of the electric energy
goes into heat!
Photoelectric Effect
When light is incident on certain metallic
surfaces, electrons are emitted from the surface


This is called the photoelectric effect
The emitted electrons are called photoelectrons
The effect was first discovered by Hertz
The successful explanation of the effect was
given by Einstein in 1905

Received Nobel Prize in 1921 for paper on
electromagnetic radiation, of which the photoelectric
effect was a part
Photoelectric Effect Schematic
When light strikes E,
photoelectrons are
emitted
Electrons collected at C
and passing through the
ammeter are a current in
the circuit
C is maintained at a
positive potential by the
power supply
Photoelectric Current/Voltage
Graph
The current increases
with intensity, but reaches
a saturation level for
large V’s
No current flows for
voltages less than or
equal to –Vs, the
stopping potential

The stopping potential is
independent of the
radiation intensity
Features Not Explained by Classical
Physics/Wave Theory
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
Einstein’s Explanation
A tiny packet of light energy, called a photon, would be emitted when
a quantized oscillator jumped from one energy level to the next
lower one

Extended Planck’s idea of quantization to electromagnetic
radiation
The photon’s energy would be E = hƒ
Each photon can give all its energy to an electron in the metal
The maximum kinetic energy of the liberated photoelectron is
KE = hƒ – Φ
Φ is called the work function of the metal
Explanation of Classical
“Problems”
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
More Explanations
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
Verification of Einstein’s Theory
Experimental
observations of a
linear relationship
between KE and
frequency confirm
Einstein’s theory
The x-intercept is the
cutoff frequency

fc 
h
Problem: photoelectric effect
Electrons are ejected from a metallic surface with speeds ranging
up to 4.6 × 105 m/s when light with a wavelength of λ = 625 nm
is used.
(a) What is the work function of the surface?
(b) What is the cutoff frequency for this surface?
Electrons are ejected from a metallic surface with speeds ranging up to 4.6 × 105 m/s
when light with a wavelength of λ = 625 nm is used.
(a) What is the work function of the surface?
(b) What is the cutoff frequency for this surface?
Given:
v = 4.6x105 m/s
l = 625 nm
Recall that KEmax=hf - f. This can be used to solve
for f. First find the kinetic energy
KEmax
2
mvmax
1

 9.111031 kg 4.6 105 m s  9.6 1020 J
2
2

Thus,   hf  KEmax 
Find:
f=?
n =?

hc
l

 KEmax  2.2 1019 J
Which equals 1.4 eV
Cutoff frequency is

2.2 1019 J
14
fc  

3.3

10
Hz
34
h 6.63 10 J  s
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
X-Rays
Electromagnetic radiation with short
wavelengths



Wavelengths less than for ultraviolet
Wavelengths are typically about 0.1 nm
X-rays have the ability to penetrate most
materials with relative ease
Discovered and named by Roentgen in
1895
Production of X-rays, 1
X-rays are produced when
high-speed electrons are
suddenly slowed down

Can be caused by the electron
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
Production of X-rays, 2
An electron passes near
a target nucleus
The electron is deflected
from its path by its
attraction to the nucleus

This produces an
acceleration
It will emit
electromagnetic radiation
when it is accelerated
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
For X-rays, the regular array of atoms in a
crystal can act as a three-dimensional
grating for diffracting X-rays
Schematic for X-ray Diffraction
A continuous beam of Xrays is incident on the
crystal
The diffracted radiation is
very intense in certain
directions

These directions correspond
to constructive interference
from waves reflected from the
layers of the crystal
The diffraction pattern is
detected by photographic
film
Photo of X-ray Diffraction
Pattern
The array of spots is called
a Laue pattern
The crystal structure is
determined by analyzing the
positions and intensities of
the various spots
This is for NaCl
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…
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 wavelength that 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 scattered
The change in wavelength is called the Compton
shift
Compton Scattering
Compton assumed the
photons acted like
other particles in
collisions
Energy and momentum
were conserved
The shift in wavelength
is
h
l  l  lo 
(1  cos )
me c
Compton Scattering, final
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 on the wavelength
Experiments confirm the results of Compton
scattering and strongly support the photon
concept
QUICK QUIZ 27.1
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.
QUICK QUIZ 27.1 ANSWER
(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.
QUICK QUIZ 27.2
A photon of energy E0 strikes a free electron,
with the scattered photon of energy E moving
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.
QUICK QUIZ 27.2 ANSWER
(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.
QUICK QUIZ 27.3
A photon of energy E0 strikes a free electron
with the scattered photon of energy E moving
in the direction opposite that of the incident
photon. In this Compton effect interaction, the
resulting momentum of the electron is:
(a) E0/c
(c) > E0/c
(e) (E  Eo)/c
(b) < E0/c
(d) (E0  E)/c
Ephoton  pc
QUICK QUIZ 27.3 ANSWER
(c). Conservation of momentum requires the
momentum of the incident photon equal the
vector sum of the momenta of the electron
and the scattered photon. Since the scattered
photon moves in the direction opposite that of
the electron, the magnitude of the electron’s
momentum must exceed that of the incident
photon.
pelectron
E  E0

c