Chapter 38
Photons: Light Waves
Behaving as Particles
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
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https://www.youtube.com/watch
?v=LJtLrfKdG3A
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Poisson’s Spot
Spot of Arago
Goals for Chapter 38
• To consider the fundamental constituent of light, the photon
• To study the removal of an electron by an incident photon,
the photoelectric effect
• To understand how the photon concept explains x-ray
production, x-ray scattering, and pair production
• To interpret light diffraction and interference in the photon
picture
• To introduce the Heisenberg uncertainty principle
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Introduction
• Until the late 19th and early 20th centuries, light was well
understood as an electromagnetic wave.
• When Einstein and others published work on the photoelectric
effect, scientists began to understand light also as a discrete unit,
the photon.
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The photoelectric effect
•
•
•
Blue light striking cesium causes the cesium to emit electrons. Red light
does not.
Einstein’s explanation: Light comes in photons. To emit an electron, the
cesium atom must absorb a single photon whose energy exceeds the
ionization energy of the outermost electron in cesium. A blue photon has
enough energy;
a red photon does not.
Refer to Figure 38.3
at right.
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Einstein’s explanation of the photoelectric effect
• A photon contains a discrete
amount of energy. For light of
frequency f and wavelength ,
this energy is E = hf or
E = (hc)/ , where h is Planck’s
constant 6.626 × 10−34 J • s.
• This explains how the energy
of an emitted electron in the
photoelectric effect depends on
the frequency of light used (see
Figure 38.6 to the right).
• The momentum of a photon of
wavelength  is p = h/.
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Q38.1
In an experiment to demonstrate the photoelectric effect, you
shine a beam of monochromatic blue light on a metal plate.
As a result, electrons are emitted by the plate.
If you increase the intensity of the light but keep the color of
the light the same, what happens?
A. More electrons are emitted per second.
B. The maximum kinetic energy of the emitted electrons
increases.
C. both A. and B.
D. neither A. nor B.
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A38.1
In an experiment to demonstrate the photoelectric effect, you
shine a beam of monochromatic blue light on a metal plate.
As a result, electrons are emitted by the plate.
If you increase the intensity of the light but keep the color of
the light the same, what happens?
A. More electrons are emitted per second.
B. The maximum kinetic energy of the emitted electrons
increases.
C. both A. and B.
D. neither A. nor B.
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Q38.2
This graph shows the
stopping potential as a
function of the frequency
of light falling on a metal
surface. If a different type
of metal is used,
A. the graph could have a different slope.
B. the graph could intercept the horizontal axis at a
different value.
C. both A. and B.
D. neither A. nor B.
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A38.2
This graph shows the
stopping potential as a
function of the frequency
of light falling on a metal
surface. If a different type
of metal is used,
A. the graph could have a different slope.
B. the graph could intercept the horizontal axis at a
different value.
C. both A. and B.
D. neither A. nor B.
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The photoelectric effect—examples
•
•
•
•
•
Read the Problem-Solving Strategy 38.1.
Follow Example 38.1—Laser-pointer photons.
Follow Example 38.2—A photoelectric-effect experiment.
Follow Example 38.3—Determining  and h experimentally.
Refer to Table 38.1 below.
Insert Table 38.1
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X-ray production
•
An experimental arrangement for making x rays is shown in Figure
38.7 at lower left. The greater the kinetic energy of the electrons that
strike the anode, the shorter the minimum wavelength of the x rays
emitted by the anode (see Figure 38.8 at lower right).
•
The photon model explains this behavior: Higher-energy electrons can
convert their energy into higher-energy photons, which have a shorter
wavelength (see Example 38.4).
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Q38.3
A beam of electrons is accelerated to high speed and aimed at a
metal target. The electrons brake to a halt when they strike the
target, and x-ray photons are produced. How do the photon energy
and wavelength change if we increase the voltage used to
accelerate the electrons?
A. photon energy increases and photon wavelength increases
B. photon energy increases and photon wavelength decreases
C. photon energy decreases and photon wavelength increases
D. photon energy decreases and photon wavelength decreases
E. it won’t deflect at all
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A38.3
A beam of electrons is accelerated to high speed and aimed at a
metal target. The electrons brake to a halt when they strike the
target, and x-ray photons are produced. How do the photon energy
and wavelength change if we increase the voltage used to
accelerate the electrons?
A. photon energy increases and photon wavelength increases
B. photon energy increases and photon wavelength decreases
C. photon energy decreases and photon wavelength increases
D. photon energy decreases and photon wavelength decreases
E. it won’t deflect at all
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X-ray scattering: The Compton experiment
•
In the Compton experiment,
x rays are scattered from
electrons. The scattered x rays
have a longer wavelength than
the incident x rays, and the
scattered wavelength depends on
the scattering angle .
•
Explanation: When an incident
photon collides with an electron,
it transfers some of its energy to
the electron. The scattered photon
has less energy and a longer
wavelength than the incident
photon (see Figure 38.10 right).
Follow Example 38.5.
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Pair production
• When gamma rays of
sufficiently short wavelength
are fired into a metal plate,
they can convert into an
electron and a positron
(positively-charged electron),
each of mass m and rest
energy mc2.
• The photon model explains
this: The photon wavelength
must be so short that the
photon energy is at least
2mc2. Follow Example 38.6.
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Diffraction and uncertainty
• When a photon passes through a narrow slit, its momentum
becomes uncertain and the photon can deflect to either side
(see Figure 38.17 below). A diffraction pattern is the result
of many photons hitting the screen. The pattern appears
even if only one photon is present at a time in the
experiment.
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Q38.5
A photon of wavelength 500 nm passes through a narrow slit of
width 250 nm. At which of these angles is there zero probability of
detecting the photon after it passes through the slit?
A. 0°
B. 30°
C. 45°
D. 60°
E. none of these
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A38.5
A photon of wavelength 500 nm passes through a narrow slit of
width 250 nm. At which of these angles is there zero probability of
detecting the photon after it passes through the slit?
A. 0°
B. 30°
C. 45°
D. 60°
E. none of these
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The Heisenberg Uncertainty Principle
• You cannot simultaneously know
the position and momentum of a
photon with arbitrarily great
precision. The better you know
the value of one quantity, the less
well you know the value of the
other (see Figure 38.18).
• In addition, the better you know
the energy of a photon, the less
well you know when you will
observe it.
• Follow Example 38.7.
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Q38.6
A beam of photons passes through a narrow slit. The photons land
on a distant screen, forming a diffraction pattern.
In order for a particular photon to land at the center of the
diffraction pattern, it must pass
A. through the center of the slit.
B. through the upper half of the slit.
C. through the lower half of the slit.
D. impossible to decide
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A38.6
A beam of photons passes through a narrow slit. The photons land
on a distant screen, forming a diffraction pattern.
In order for a particular photon to land at the center of the
diffraction pattern, it must pass
A. through the center of the slit.
B. through the upper half of the slit.
C. through the lower half of the slit.
D. impossible to decide
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Q38.7
A photon has a position uncertainty of 2.00 mm. If you decrease
the position uncertainty to 1.00 mm, how does this change the
momentum uncertainty of the photon?
A. the momentum uncertainty becomes 1/4 as large
B. the momentum uncertainty becomes 1/2 as large
C. the momentum uncertainty is unchanged
D. the momentum uncertainty becomes twice as large
E. the momentum uncertainty becomes 4 times larger
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A38.7
A photon has a position uncertainty of 2.00 mm. If you decrease
the position uncertainty to 1.00 mm, how does this change the
momentum uncertainty of the photon?
A. the momentum uncertainty becomes 1/4 as large
B. the momentum uncertainty becomes 1/2 as large
C. the momentum uncertainty is unchanged
D. the momentum uncertainty becomes twice as large
E. the momentum uncertainty becomes 4 times larger
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Q38.8
A photon has a momentum uncertainty of 2.00  10–28 kg • m/s. If
you decrease the momentum uncertainty to 1.00  10–28 kg • m/s,
how does this change the position uncertainty of the photon?
A. the position uncertainty becomes 1/4 as large
B. the position uncertainty becomes 1/2 as large
C. the position uncertainty is unchanged
D. the position uncertainty becomes twice as large
E. the position uncertainty becomes 4 times larger
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A38.8
A photon has a momentum uncertainty of 2.00  10–28 kg • m/s. If
you decrease the momentum uncertainty to 1.00  10–28 kg • m/s,
how does this change the position uncertainty of the photon?
A. the position uncertainty becomes 1/4 as large
B. the position uncertainty becomes 1/2 as large
C. the position uncertainty is unchanged
D. the position uncertainty becomes twice as large
E. the position uncertainty becomes 4 times larger
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