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Infrared radiation
Chapter 26 Lecture
Quantum
Optics
 Hot star
 Hot coal
Hot objects radiate energy
Black body radiation
 A black body absorbs all
incident light and converts the
light's energy into thermal
energy (no light is reflected).
 The black body then radiates
electromagnetic waves based
solely on its temperature.
SPECTRAL CURVE
Hotter objects higher
intensity (power)
Cooler objects (higher
wavelength)
Model of a black body
Hot objects radiate
energy.
Infrared –
Electromagnetic
spectrum
What characterizes a black body?
 If we raise the temperature of the box and again
measure the spectrum of the EM radiation being
emitted from the hole, we find that:
 The total power output from the hole is now
greater.
 The spectral curve rises at all wavelengths.
 The peak of the power per small wavelength
interval shifts to a shorter wavelength.
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Studies of black body radiation
 The total radiation output (power per unit area of
the emitting object) increases dramatically as the
temperature of the black body increases.
Wavelength at which maximum intensity
occurs
 In 1893, the German physicist Wilhelm Wien
(1864–1928) quantified a second aspect of black
body radiation.
 He defined a relationship for the maximum
wavelength at which a black body emits
radiation of maximum power per wavelength:
 Stefan's law:
WHITEBOARD
 What is the max for:
The electromagnetic spectrum
 The range of frequencies and wavelengths of
electromagnetic
waves
is
called
the
electromagnetic spectrum.
 Human T = 310 K
 Molten iron T = 1810 K
WHITEBOARD
 The maximum power per wavelength of light
from the Sun is at a wavelength of about
510 nm, which corresponds to yellow light.
What is the surface temperature of the Sun?
The color of stars
 Very hot stars (around 10,000 K) look blue.
 Even hotter stars are generally invisible to our
eyes.
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The ultraviolet catastrophe
 No models built on the physics of the late 19th
century made predictions that were consistent
with the drop in intensity at high frequencies.
 This problem became known as the
ultraviolet catastrophe.
Energy Quantization
 In classical physics, standing wave frequencies
are quantized.
Planck's hypothesis
 Planck hypothesized that charged particles
could radiate energy only in discrete portions
called quanta.
 He proposed that “a” microscopic oscillating
charged particle had some kind of
fundamental portion of energy, which was
proportional to the frequency of its
oscillation.
 According to Planck, the particle could emit
amounts of energy equal only to multiples of
this fundamental portion.
Planck's hypothesis
 Using his energy quantization model, Planck
was able to derive a mathematical function
describing the black body spectral curve:
 The energy of these standing waves can take
on any positive value, because the energy of
the wave depends on the amplitude of the
wave and not on its frequency.
 In quantum physics, the energy of the oscillator
is quantized, rather than its frequency.
 The proportionality constant h became known as
Planck's constant.
PHOTOELECTRIC EFFECT
 The photoelectric effect was first observed in 1887 by Heinrich Hertz
The Photoelectric Effect
 Hertz helped establish the photoelectric effect (which was later explained
by Albert Einstein) when he noticed that a charged object loses its charge
more readily when illuminated by ultraviolet light. In 1887, he made
observations of the photoelectric effect and of the production and reception
of electromagnetic (EM) waves, published in the journal Annalen der
Physik. His receiver consisted of a coil with a spark gap, whereby a spark
would be seen upon detection of EM waves. He placed the apparatus in a
darkened box to see the spark better. He observed that the maximum
spark length was reduced when in the box. A glass panel placed between
the source of EM waves and the receiver absorbed ultraviolet
radiation (UV) that assisted the electrons in jumping across the gap. When
removed, the spark length would increase. He observed no decrease in
spark length when he substituted quartz for glass, as quartz does not
absorb UV radiation. Hertz concluded his months of investigation and
reported the results obtained. He did not further pursue investigation of this
effect, nor did he make any attempt at explaining how the observed
phenomenon was brought about
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Photoelectric current produced immediately
The Experimental Setup
The intensity and
wavelength/frequency of the
incident light can be changed
 Absorbing one quantum of light of sufficient
energy could instantly eject an electron from the
metal.
Vacuum tube
for a free path
from emitter to
collector plate
The voltage of the battery can
be adjusted to change the
electric field between the
emitting and collecting plates.
 Energy does not need to accumulate in the
metal before we see an effect.
The amount of EPE that the electrons gain
(KE that they lose) if they cross the full gap is
KE = qΔV
Wave Model of Light – Light is a wave
(of crossed electric and magnetic fields.)
Predictions made by the Wave Model
of Light
(Caution: These are the predictions made by a disproved model of light)
“If light is a wave, its energy depends on its amplitude (intensity).”
“Therefore the electrons ejected by high-intensity light should have
more kinetic energy than the electrons ejected by low-intensity light.”
Prediction
The energy of a wave depends on…
the amplitude of the wave!
Low-Intensity
Light
High-Intensity
Light
The wave model of light states that the energy of
a beam of light depends on its intensity
(amplitude of the E and B fields)
What actually happens
Low-Intensity
Light
High-Intensity
Light
Conclusion
The wave model of light cannot be correct!
The energy of light does not depend on its intensity.
Another model of light is needed to explain the
way that light interacts with matter.
The intensity of the incident light does not affect
the energy of the ejected electrons!
However, high-intensity light will eject more
electrons per second from the metal.
This discovery, made in the year 1900,
sparked the birth of quantum physics.
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Evolution of ideas concerning the nature of
light
Photoelectric Whiteboard I
Ultraviolet light is incident upon a sodium surface,
ejecting electrons into the surroundings. The intensity
of the light is then steadily increased.
a) Sketch a graph the maximum kinetic energy of the
ejected electrons as a function of the intensity of
the light (KE vs. Intensity)
b) Sketch a graph the current through the circuit as a
function of the intensity of the light. (I vs. Intensity)
Photon
Kmax
The intensity of light
does not affect the
energy that it transfers
to each electron.
Ephoton = hf
Intensity
I
A photon is a discrete portion of electromagnetic
radiation that has energy
The intensity of light
affects the number of
photons that are
incident on the metal.
f = Frequency of the electromagnetic radiation
h = Planck’s constant 6.63E-34 Js
Intensity
What does affect the energy of light?
Frequency.
Higher frequency photons have more energy
Ephoton = hf
h = 6.6 ×
J*s
 Estimate the energy of a quantum of radio waves
(frequency of 1 x 106 Hz), infrared radiation (3 x
1013 Hz), visible light (5 x 1014 Hz), and UV
radiation (1 x 1015 Hz).
 Radio Waves
E = 6.63x10-28 J
 Infrared Radiation E = 1.99x10-20 J
h is called Planck’s constant.
10-34
WHITEBOARD
An extremely
tiny number!
 Visible Light
E = 3.32x10-19 J
 UV Radiation
E = 6.63x10-19 J
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Photoelectric Whiteboard II
An electron at rest absorbs a photon of
wavelength 450 nm.
An electron at rest absorbs a photon of
wavelength 450 nm.
v
eEnergy Conservation!
Ephotoninitial = Kelectronfinal
What is its recoil velocity?
me = 9.1 × 10-31 kg
qe = 1.6 × 10-19 C
c = 3 × 108 m/s
h = 6.6 × 10-34 J*s
Hint: You will need to use another relationship that
relates the frequency and wavelength of light!
hf = ½ mv2
Using the wave equation (c = λf), we can calculate the frequency of the photon.
fphoton = 6.7 × 1014 Hz
Work function Φ
Work function (Φ)
The work function of a metal is
the minimum amount of energy
that is needed to eject electrons
from the surface of the metal.
Electrons that are deeper below
the surface will require more
energy than Φ to be ejected.
The electrons on the surface will require the least amount of energy to be
ejected, and will have the greatest kinetic energy when they leave the metal.
The work function is the bare minimum needed to eject a surface electron.
 The work function is the minimum energy
needed to remove a free electron from a metal.
 The work function has units of energy but is
measured in electron volts because it is
typically very small.
 The greater the work function of a metal, the
more tightly the free electrons are bound to
the metal and the more energy must be
added to separate them.
When an electron in the metal absorbs a
photon,
Stopping potential
Leftover kinetic energy = Photon energy – Energy needed to break it free
K = hf – (Energy binding electron to the metal)
If the electron is a surface electron, then the photon only had
to spend Φ to break it free. The electron will be ejected with
the maximum possible kinetic energy!
KEmax = hf - Φ
This relationship can be used to calculate the
KE of the most energetic ejected electrons.
vfinal ≈ 1 × 106 m/s
Light has extremely high frequencies!
v0
By applying a voltage across the plates,
we can determine exactly how much
kinetic energy those surface electrons
were ejected with!
UE = qΔV
ΔV
The voltage that is able to just barely stop the most
energetic electrons is called the stopping potential.
Energy conservation as
they cross the gap
gives:
qeΔVstop = KEmax
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The net result:
Photoelectric Whiteboard III
When violet light of wavelength 400 nm is incident
upon a Calcium surface, a voltage of -0.2 Volts is
required to completely stop the photoelectric current.
qeΔVstop = hf - Φ
In order for the current to drop to zero, you must stop the
most energetic electrons (surface electrons).
What would be the stopping potential if
ultraviolet light of wavelength 250 nm were
incident on the surface?
If you stop those, then none of the electrons will
be able to complete the circuit.
qe = -1.6 × 10-19 C
c = 3 × 108 m/s
qeΔVstop = hf - Φ
c = λf
1. Calculate the work function of Calcium
fviolet = 7.5 × 1014 Hz
Φca = 2.9 eV
2. Use the work function to determine the new
stopping potential.
fUV = 1.2 × 1015 Hz
ΔVstop = -2.1 Volts
A greater voltage is needed to stop the electrons now,
since UV light has more energy (higher frequency)!
Ultraviolet light is incident upon a sodium surface, ejecting
electrons into the surroundings. The frequency of the light
is then steadily increased.
h = 6.6 × 10-34 J*s
Challenging Conceptual Whiteboard!
Ultraviolet light is incident upon a sodium surface,
ejecting electrons into the surroundings. The
frequency of the light is then steadily increased.
Which of the following happens as a result?
a) Electrons are ejected at the same rate, and have
the same maximum kinetic energy.
b) Electrons are ejected at the same rate, and have a
greater maximum kinetic energy.
c) Electrons are ejected at a greater rate, and have
the same maximum kinetic energy.
d) Electrons are ejected at a greater rate, and have a
greater maximum kinetic energy.
f = fthreshold
f > fthreshold
f >> fthreshold
The light has just enough
energy to just barely
eject the surface
electrons from the metal.
The light has enough
energy to eject electrons
from deeper within the
metal, and the ones from
the surface are ejected
with some spare KE.
The light has enough
energy to eject electrons
from very deep within
the metal, surface
electrons are ejected with
lots of spare KE.
More electrons, and some
leftover energy for KE
More electrons, and lots of
leftover energy for KE
Electrons are ejected at a greater rate, and have a greater
maximum kinetic energy.
Light of higher frequency has more energy per photon, and is able to
eject electrons that are deeper within the metal. This results in more
ejected electrons, and a greater current through the circuit!
Also, electrons from the surface of the metal that absorb those photons
are ejected with greater KE.
Very few electrons, and no
leftover energy for KE
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Threshold frequency
Threshold Frequency
Kmax = hf - F
With low-frequency light, no electrons at all
will be ejected from the metal!
This is because each photon does not have enough energy to
overcome the work function of the metal.
Kmax
When light with the threshold frequency is incident on the
metal, electrons at the very surface are just barely able to
absorb enough energy to escape.
But they have no
leftover energy
once they do!
fthreshold
f
No photocurrent below cutoff frequency
0 = hfthreshold - F
fthreshold =
F
h
Cutoff frequencies for selected metals
 We can express the cutoff frequency in terms of
the work function of the metal and Planck's
constant:
if the energy of one quantum is less than the
work function of the metal, and no photocurrent
is produced.
An Amazing Discovery!
The wave model of light predicted that if you shine really
bright red light on a metal, the combined energy of all of
that light should be able to give at least some of the
electrons enough energy to be ejected.
How does light interact with matter?
When light interacts with matter, it’s either all or nothing.
Light transfers its energy to matter in individual
packets (quanta) of amount hf.
However, light doesn’t work that way!
Light can only transfer its energy in
distinct amounts of quantity hf
Since hf for red light is not enough to overcome the
work function of the metal, the electrons will never
be able to absorb any of the red light’s energy!
These packets cannot be combined or split, and will
either be absorbed completely or not at all.
Therefore, increasing the intensity of the light will do
nothing if none of the individual hf packets have
enough energy to eject an electron!
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What is “Quantum”?
A quantum is the smallest possible amount of something
that can exist in the physical world.
If some property is quantized, it means that it is made up of
indivisible “building blocks”, or quanta.
Electric charge is quantized, because there cannot exist an
amount of electric charge that is smaller than 1.6 × 10-19 C.
Mass is quantized, because there cannot exist an amount of
mass that is smaller than 9.1 × 10-31 kg (mass of an electron).
And now we know that energy is quantized, because it
travels in indivisible “packets” (photons) with energies hf.
The Photon Model of Light
Photons
Light is made up of individual packets of energy, called photons.
 Physicists started to think of light as being
composed of particle-like photons (the quanta of
light).
Photons are neither waves nor particles, but rather
have properties of each in different scenarios.
 To
explain
interference
and
diffraction
phenomena, the photons also had to have
wave-like properties.
When a photon travels through empty space, it behaves like a wave.
It it spreads out in all directions, obeys the superposition principle,
and it can create an interference pattern.
When a photon interacts with matter, it behaves like a particle.
It completely transmits all its energy in an amount hf.
The Wave-Particle Duality
 This is called the dual particle-wave property of
photons.
Wave-like and particle-like properties of
photons
This marks the beginning of a new era in physics, where
the distinction between waves and particles vanishes.
A photon behaves like a wave when traveling through
empty space, but collapses and transfers all of its energy
when it interacts with a proton or electron.
The Photon Model is the currently accepted and
exhaustively-tested model of light.
Don’t worry. It gets much, much crazier.
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DOUBLE SLIT EXPERIMENT
DOUBLE SLIT EXPERIMENT
(PARTICLE MODEL)
 Only two bright bands should appear (images of
the slits themselves).
 As intensity decreases, individual flashes are
expected.
DOUBLE SLIT EXPERIMENT
(WAVE MODEL)
 Many alternating bright and dark bands appear.
 As intensity decreases, bright bands disappear.
DOUBLE SLIT EXPERIMENT
(DUAL WAVE-PARTICLE (PHOTON) MODEL)
 Many alternating bright and dark bands appear.
 As intensity decreases, individual flashes are
expected.
SUMMARY OF MATH MODELS
t  t0  
L
p   mv
WHITEBOARD
 Write an expression for the momentum of a
photon in terms of h, and 
L0

E   mc
p   mv
E    m  c2
E photon  h  f
2
E photon  h  f
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SUMMARY OF MATH MODELS
p   mv
E    m  c2
p
mv
E

m  c2

p
E

m  v m  c2
p
Photon momentum
 Photons participate in collisions with electrons
inside metals (the photoelectric effect).
E
c
c  f
E photon  h  f
p
vc
h f
p
Cathode ray tubes
 Physicists discovered
that although a physical
gap was present
between the cathode
and the anode in a
cathode ray tube, when
the cathode was
heated, a current
appeared in the circuit
and the tube would
glow.
The accidental discovery of X-rays
 The first X-ray image of
a human was made by
Roentgen of his wife
Anna Bertha's hand on
December 22, 1895.
must have
 f
p  c2  E  c
pc  E
 This suggests that photons
momentum.
h

 In 1922, Arthur Compton performed a testing
experiment to determine whether this
expression was reasonable.
Conceptual Exercise 26.8
 Imagine that you can see the beam
of electrons shooting from the
cathode to the anode of a cathode
ray tube.
 A. Draw the direction of the electric
force exerted on the electron.
 B. Draw the direction of the
magnetic force exerted on the
electron.
 What would happen if you placed
the tube in a region with both an
electric field and a magnetic field?
Explanation of X-rays
 Not deflected by either electric or magnetic fields
 Do not cause the viewing screen to become
electrically charged
 Cause photographic paper to darken
 Produce a diffraction pattern of dark and bright
bands on the screen after passing through a
single narrow slit
 Can be polarized
 Can go through many materials that are opaque
to visible light
 Can ionize gases
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Producing X-ray photons
WHITEBOARD
 In a cathode ray tube used to produce X-rays for
imaging in a hospital, the potential difference
between the cathode and the anode is 90.0 kV
(90,000 V). If the X-rays are generated by
electrons in the tube that stop abruptly when
they collide with the anode and emit photons,
what are the energy and the wavelength of each
photon?
 Conservation of energy. KE
becomes radiating energy (x-rays)
The Compton effect and X-ray interference




Ephoton = KE
Ephoton = qV
Ephoton = 1.44x10-14 J
Ephoton = 90,000 eV




Ephoton = hf
Ephoton = hc/
 =hc/Ephoton
 =0.14 nm
Compton effect
 If a photon of wavelength i scatters off a
charged particle of mass m and moves at an
angle  relative to its initial direction of motion,
the scattered photon will then have a longer
wavelength f given by the following equation:
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