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4/9/2017
Infrared radiation
AP PHYSICS 2
UNIT 7
Quantum Physics,
atomic, and nuclear
physics
 Hot star
CHAPTER 26
Quantum Optics
 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
Joseph Stefan
 The total radiation output (power per unit area of
the emitting object) increases dramatically as the
temperature of the black body increases.
 Stefan's law:
𝑃 = 𝜀∙𝐴
Maximum Wavelength
Wilhem Wien
 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:
𝑃 = 𝜎 ∙ 𝑇4 ∙ 𝐴
𝜎 = 5.67𝑥10−8
𝑊
𝑚2 ∙ 𝐾 4
WHITEBOARD
 What is the max for:
 Human T = 310 K
 Molten iron T = 1810 K
 The range of frequencies and wavelengths of
electromagnetic
waves
is
called
the
electromagnetic spectrum.
𝜆𝑀𝐴𝑋 = 9354 𝑛𝑚
𝜆𝑀𝐴𝑋 = 1602 𝑛𝑚
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?
𝜆𝑀𝐴𝑋 =
𝑇=
2.90 ∗ 10−3
𝑇
2.90 ∗ 10−3
𝜆𝑀𝐴𝑋
The electromagnetic spectrum
𝑇=
The color of stars
 Very hot stars (around 10,000 K) look blue.
 Even hotter stars are generally invisible to our
eyes.
2.90 ∗ 10−3
510 ∗ 10−9
𝑇 = 5686 𝐾
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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.
 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.
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:
𝐸0 = ℎ𝑓
 The proportionality constant h became known as
Planck's constant.
Photon
Mathematical model that describes the black body
spectral curve
E0 = hf
f = Frequency of the electromagnetic
radiation
h = Planck’s constant
−𝟑𝟒
𝒉 = 𝟔. 𝟔𝟑 ∗ 𝟏𝟎
𝑱∙𝒔
The Photoelectric Effect
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PHOTOELECTRIC EFFECT
The Experimental Setup
 The photoelectric effect was first observed in 1887 by Heinrich Hertz
 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
Photoelectric current produced immediately
The intensity and
wavelength/frequency of the
incident light can be changed
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.
The amount of EPE that the electrons gain
(KE that they lose) if they cross the full gap is
𝑼𝑬 = 𝒒 ∙ ∆𝑽
Wave Model of Light – Light is a wave
(of crossed electric and magnetic fields.)
 Absorbing one quantum of light of sufficient
energy could instantly eject an electron from the
metal.
𝑼𝑬 = 𝒒 ∙ ∆𝑽
𝑼𝑬 = ∆𝑲𝑬
 Energy does not need to accumulate in the
metal before we see an effect.
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
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).”
Low-Intensity
Light
High-Intensity
Light
“Therefore the electrons ejected by high-intensity light should have
more kinetic energy than the electrons ejected by low-intensity light.”
Prediction
Low-Intensity
Light
High-Intensity
Light
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.
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Conclusion
Evolution of ideas concerning the nature of
light
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.
This discovery, made in the year
1900, sparked the birth of
quantum physics.
Photoelectric Effect: 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)
Kmax
The intensity of light
does not affect the
energy that it transfers
to each electron.
The electrons gain Kinetic energy.
Where is the energy coming from?
(energy must be conserved)
Intensity
I
ENERGY
The intensity of light
affects the number of
photons that are
incident on the metal.
Intensity
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It wasn’t Amplitude . . . . It is the frequency !
Photon
A photon is a discrete portion of
electromagnetic radiation that has
energy
Ephoton = hf
 f = Frequency of the
electromagnetic radiation
 h = Planck’s constant
𝒉 = 𝟔. 𝟔𝟑 ∗ 𝟏𝟎−𝟑𝟒 𝑱 ∙ 𝒔
An extremely
tiny number!
Higher frequency photons have more energy
WHITEBOARD
Photoelectric Effect: Whiteboard II
An electron at rest absorbs a photon of
wavelength 450 nm.
Estimate the energy of a quantum of:
e Radio waves (frequency of 1 x 106 Hz)
E = 6.63x10-28 J
 Infrared radiation (3 x 1013 Hz)
E = 1.99x10-20 J
 Visible light (5 x 1014 Hz)
 UV radiation (1 x 1015 Hz).
E = 3.32x10-19 J
E = 6.63x10-19 J
An electron at rest absorbs a photon of
wavelength 450 nm.
What is its recoil velocity?
me = 9.1 × 10-31 kg
qe = 1.6 × 10-19 C
c = 3 × 108 m/s
h = 6.63 × 10-34 J*s
Hint: You will need to use another relationship that
relates the frequency and wavelength of light!
An electron at rest absorbs a photon of
wavelength 450 nm.
v
v
Energy Conservation!
𝑬𝒑𝒉𝒐𝒕𝒐𝒏 = 𝑲𝑬𝒆𝒍𝒆𝒄𝒕𝒓𝒐𝒏
𝒉𝒇 =
𝒄 = 𝝀𝒇
𝒎 ∙ 𝒗𝟐
𝟐
𝒇=
𝒄
𝝀
𝒉𝒄 𝒎 ∙ 𝒗𝟐
=
𝝀
𝟐
𝒗=
𝟐𝒉𝒄
𝝀𝒎
vfinal = 985,610 m/s
𝒇=
𝒄
𝝀
fphoton = 6.7 × 1014 Hz
Light has extremely high
frequencies!
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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 (remove a free
electron from a 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.
Work function Φ
 The work function has units of energy but is
measured in electron volts (eV) because it is
typically very small.
The work function has units
of energy but is measured in
electron volts because it is
typically very small.
Work function Φ
 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.
 Electron Volt: Energy to move one electron
through a potential difference V = 1 V.
𝑼𝑬 = 𝒒 ∙ ∆𝑽
𝒒=𝒆
∆𝑽 = 𝟏 𝑽
𝑼𝑬 = 𝟏. 𝟔 ∗ 𝟏𝟎−𝟏𝟗 𝑪 ∙ 𝟏 𝑽
𝟏𝒆𝑽 = 𝟏. 𝟔 ∗ 𝟏𝟎−𝟏𝟗 𝑱
When an electron in the metal absorbs a
photon,
Stopping potential
Leftover kinetic energy = Photon energy – Energy needed to break it free
KE = 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.
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 Effect: Whiteboard III
qeΔVstop = hf - Φ
When violet light of wavelength 400 nm is
incident upon a Calcium surface, a voltage
of -0.21 Volts is required to completely
stop the photoelectric current.
In order for the current to drop to zero, you must stop the
most energetic electrons (surface electrons).
What is the work function for Calcium?
(in Joules and eV)
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
h = 6.63 × 10-34 J*s
𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑 = 𝒉𝒇 − 𝝓
Photoelectric Effect: Whiteboard IV
𝝓 = 𝒉𝒇 −𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑
What would be the stopping potential if
ultraviolet light of wavelength 250 nm
were incident on the calcium surface?
𝝓=𝒉
𝒄
− 𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑
𝝀
𝑼𝒔𝒆 𝒒 = −𝒆
𝝓 = 𝟒. 𝟔𝟒 ∗ 𝟏𝟎−𝟏𝟗 𝑱
𝝓 = 𝟐. 𝟗 𝒆𝑽
qe = -1.6 × 10-19 C
c = 3 × 108 m/s
𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑 = 𝒉𝒇 − 𝝓
∆𝑽𝒔𝒕𝒐𝒑 =
𝒉𝒇 − 𝝓
𝒒
𝒉𝒄 𝝓
∆𝑽𝒔𝒕𝒐𝒑 =
+
𝝀𝒒 𝒒
∆𝑽𝒔𝒕𝒐𝒑 = −𝟐. 𝟎𝟕 𝑽
h = 6.63 × 10-34 J*s
𝒒 ∙ ∆𝑽𝒔𝒕𝒐𝒑 = 𝒉𝒇 − 𝝓
 in Joules
-0.21 V for violet light
-2.07 V for ultraviolet light
When  divided by q,
it will be the same
magnitude as eV
(-q then +) in V
A greater voltage is needed to
stop the electrons now, since
UV light has more energy
(higher frequency)!
A greater voltage is
needed to stop the
electrons, since UV light
has more energy (higher
frequency)!
 Constant for
Calcium (2.9 eV)
𝒄
𝒇=
𝝀
7.50x1014 Hz for violet light
1.20x1015 Hz for ultraviolet light
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Photoelectric Effect: Whiteboard V
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.
Threshold or cutoff frequency
Ultraviolet light is incident upon a sodium
surface,
ejecting
electrons
into
the
surroundings. The frequency of the light is
then steadily increased.
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.
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
𝑲𝑬𝒎𝒂𝒙 = 𝒉𝒇 − 𝝓
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!
𝟎 = 𝒉𝒇 − 𝝓
𝒇𝒄𝒖𝒕𝒐𝒇𝒇
𝝓
=
𝒉
Very few electrons, and no
leftover energy for KE
Cutoff frequency
HAPPY SPRING BREAK
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.
KEmax
fcutoff
f
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No photocurrent below cutoff frequency
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!
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.
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!
The Photon Model of Light
Light is made up of individual packets of energy, called photons.
Photons are neither waves nor particles, but rather
have properties of each in different scenarios.
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.
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Photons
The Wave-Particle Duality
 Physicists started to think of light as being
composed of particle-like photons (the quanta of
light).
This marks the beginning of a new era in
physics, where the distinction between
waves and particles vanishes.
 To
explain
interference
and
diffraction
phenomena, the photons also had to have
wave-like properties.
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.
 This is called the dual particle-wave property of
photons.
The Photon Model is the currently accepted and
exhaustively-tested model of light.
The Wave-Particle Duality
A photon behaves like a wave
when traveling through empty
space, but collapses and
transfers all of its energy when it
interacts (collides) with a proton
or electron.
Energy
vv==cc
momentum
m = 0 kg
RELATIVISTIC ENERGY & MOMENTUM
(vphoton = c)
Relativistic Energy
𝑬=
𝟏−
𝒎𝒄𝟐
𝒗
𝟏−
𝒄
𝒗
𝒄
𝟐
=
𝟐
𝒎𝒄𝟐
𝑬
Relativistic Momentum
𝒑=
𝟏−
𝒎𝒗
𝒗
𝟏− 𝒄
𝒗
𝒄
𝟐
=
𝟐
𝒎𝒗
𝒑
RELATIVISTIC ENERGY & MOMENTUM
(vphoton = c)
𝒎𝒄𝟐 𝒎𝒗
=
𝑬
𝒑
𝒄𝟐 𝒗
=
𝑬 𝒑
𝒄𝟐 𝒄
=
𝑬 𝒑
𝒄 𝟏
=
𝑬 𝒑
𝐩=
𝐩=
𝑬
𝒄
𝒉𝒇
𝝀𝒇
𝐩=
𝒉
𝝀
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Photon momentum
Whiteboard
 Photons participate in collisions with electrons
inside metals (the photoelectric effect).
 This suggests that photons
momentum.
must have
𝒉
𝐩=
𝝀
What is the momentum of one of these
photons?
 In 1922, Arthur Compton performed a testing
experiment to determine whether this
expression was reasonable.
𝑵=
𝑬𝒕𝒐𝒕𝒂𝒍
𝑬𝟏
𝟏 ∗ 𝟏𝟎−𝟑 𝑱
𝟕. 𝟗𝟓𝟔 ∗ 𝟏𝟎−𝟏𝟓 𝑱
𝐩=
𝑵 = 𝟏. 𝟐𝟔 ∗ 𝟏𝟎𝟏𝟏 𝒑𝒉𝒐𝒕𝒐𝒏𝒔
𝐩=
𝑬 = 𝒉𝒇
𝑬=
𝒉𝒄
𝝀
𝑬 = 𝟕. 𝟗𝟓𝟔 ∗ 𝟏𝟎−𝟏𝟓 𝑱
Wave-like and particle-like properties of
photons
Whiteboard
𝑵=
Determine the number of photons (N)
absorbed during a single chest X-ray if
the body absorbs a total of 1x10-3 J of
0.025 nm wavelength X-ray photons.
𝒉
𝝀
𝟔. 𝟔𝟑 ∗ 𝟏𝟎−𝟑𝟒
𝟎. 𝟎𝟐𝟓 ∗ 𝟏𝟎−𝟗
𝒑 = 𝟐. 𝟔𝟓𝟐 ∗ 𝟏𝟎−𝟐𝟑
DOUBLE SLIT EXPERIMENT
𝒌𝒈 ∙ 𝒎
𝒔
DOUBLE SLIT EXPERIMENT
(PARTICLE MODEL)
 Only two bright bands should appear (images of
the slits themselves).
 As intensity decreases, individual flashes are
expected.
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DOUBLE SLIT EXPERIMENT
(WAVE MODEL)
 Many alternating bright and dark bands appear.
 As intensity decreases, bright bands disappear.
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.
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.
 C. What would happen if you placed
the tube in a region with both an
electric field and a magnetic field?
DOUBLE SLIT EXPERIMENT
(DUAL WAVE-PARTICLE (PHOTON) MODEL)
 Many alternating bright and dark bands appear.
 As intensity decreases, individual flashes are
expected.
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.
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
 =0.014 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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