PEGS Unit 4 Physics:
Interactions of light and matter
Name:
INTERACTIONS OF LIGHT
AND MATTER
Solar cells at the Vic Market
produce electricity by using an
interaction between
light and matter
These images are taken with an
electron microscope using electrons rather
than photons to obtain very high resolution
images.
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IDEAS ABOUT LIGHT AND MATTER
Throughout the ages, people have wondered about the nature of the world in which they lived. What were
things made of? If an object could be chopped into smaller and smaller pieces, what would we end up with?
The idea of atoms has actually been around for thousands of years. It has only been in the past hundred
years, however, that physicists have obtained experimental data that has allowed them to establish a model
for the structure of the atom.
What is your understanding of atoms, the building blocks of matter?
In much the same way, people have also thought about light. What were the Sun and other stars producing?
How was light able to travel around? Did it need a medium to travel through? How fast did it move? Was
light a particle or a wave? It was very difficult to determine the nature of light because it travels so fast.
Again, physicists have had to perform experiments and establish models that can be used to explain the many
observed properties of light such as reflection, refraction, dispersion and diffraction.
What is your understanding of the nature of visible light
and the other parts of the electromagnetic spectrum?
We now use a variety of light sources such as light globes, LEDs, lasers, etc. These light sources basically
function in by one of two different processes:-by making materials very hot or :- by making electrons in
excited atoms jump between energy levels. In this topic we will be looking at both of these processes.
We now understand a great deal about the structure of atoms and subatomic particles – the stuff that matter is
made of. Our understanding of light has also developed greatly over the past 150 years and we can use this
knowledge to figure out what stars billions of light years away are made of! Devices such as lasers, electron
microscopes, television picture tubes, solar cells, fluorescent tubes and X-ray imaging machines make use of
the interaction between light and matter and our understanding of this process.
PEGS Un it 4 Physics:
What is light???
Interactions of light and matter
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As we work through the topic, use this table to keep tabs on the relative success of each of the competing
models for light.
Property of light
Demonstration
Particle model
Wave model
Travels in straight lines
Regular reflection
Diffuse reflection
Snell’s Law
sin i
=n
sin r
Refracts towards normal
when slowing down
Light travels slower in a
denser medium
Dispersion – splitting up
into the colours of the
rainbow
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Property of light
Demonstration
Particle model
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Wave model
Higher frequencies are
refracted more
Light beams pass
through each other
unaffected
Partial transmission /
partial reflection at an
interface
Intensity ∝ 1/d2
Diffraction
Interference
Polarization
Light can travel through
a vacuum
Photoelectric effect
Pressure of light
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11.1 Light−waves or particles?
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From the seventeenth through to the early twentieth century, physicists debated the nature of light. Sir Isaac
Newton, the most influential physicist of his time, favoured the idea that light consisted of particles. He was
supported by the scientific community in general and his ideas survived for many years after his death in
1727.
Newton's particle model stated that:
• light consists of a stream of particles or corpuscles; and
• light particles travel in straight lines at extremely high speeds
Over time, however, experimental evidence supporting an opposing model mounted. This model was called
the wave model and its proponents included physicists such as Huygens, Young, Fresnel and Maxwell.
Read from page 405 to page 408 of the text, then describe how each model explains the following properties
of light.
Reflection
Partial reflection/partial transmission (not in text)
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Refraction:
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the direction that light bends as it passes into another medium.
the speed at which light travels as it passes into another medium.
The evidence supporting the wave model of light seemed to be so conclusive that by 1870 it was the
accepted theory throughout the scientific community.
However; towards the end of the nineteenth century, further experimental evidence in the form of the
photoelectric effect again threw the debate wide open.
By the time this debate was settled in the early twentieth century, neither model was found to provide a
satisfactory explanation for the behaviour of light. The resultant theory changed forever the way scientists
viewed light and matter.
 Homework 11.1 1-9
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11.2 Evidence supporting the Wave Model of Light
Both the wave and the particle model could successfully explain properties of light such as reflection and
propagation in a straight line, but these models were able to explain the other properties of light with varying
degrees of success.
Diffraction of light waves
You have studied the diffraction of water waves in year 11 and diffraction of sound waves in year 12.
Diffraction is the bending that takes place as a wave passes through an opening or past a barrier.
The amount of diffraction that is present depends on the relationship between the wavelength λ and the
opening or barrier size w.
Amount of diffraction ≈
!
w
Redraw these 3 scenarios showing the amount of
diffraction present if a higher frequency water wave
was used in each case.
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Diffraction of light was observed around 1660 by Jesuit priest, Francesco Grimaldi. He noticed
that when sunlight shone through a small hole, it did not make a clearly defined spot of light. The spot of
light was surrounded by a fuzzy ring of light. The light was diffracting as it passed through the hole.
Diffraction of light is not an effect that you would usually notice in everyday life. Special conditions such as
a small, monochromatic light source and very narrow slits are required.
If you try to create a diffraction pattern for light by shining laser light through a very narrow slit, you might
expect to observe just a single fuzzy band of light as the light diffracts past the opening. In fact, a very
distinctive interference pattern, consisting of a series of bright and dark bands, is formed. This is known as a
single slit diffraction pattern.
Red laser light
Draw a pattern for blue laser light
Homework questions
Question 1: Diffraction is not evident when light passes through an opening 1 mm wide. What does this
suggest about the wavelength of visible light?
Question 2: Explain why coloured fringes can sometimes be seen at the edges of diffraction patterns that are
formed when white light is shone through a narrow slit.
Question 3: A diffraction pattern is formed on a screen when green light is incident upon a narrow single
slit. What happens (if anything) to the pattern if:
(a) the width of the slit is increased?
(b) a more intense green light source is used?
(c) violet light is used instead of green light?
(d) the screen is moved further from the slit?
(e) the green source is moved closer to the slits?
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11.2 More evidence supporting the Wave Model of Light
Interference of waves
Interference of water waves was covered in year 11. Label these points in the ripple tank diagram as nodes
or antinodes. Indicate whether the interference at each point is constructive or destructive.
Complete the “How interference patterns are formed” worksheet on the next page. Draw the nodal lines and
antinodal lines in different colours. Label the central antinode n = 0, and the other antinodes n = 1, 2, 3, …
Label the innermost nodal line n = 1 and the remaining nodal lines n = 2, 3, ..
Mark a point A0 on the central (n = 0) antinode. Determine the path difference at A0 in terms of λ.
Repeat for a point A1 on the next (n = 1) antinode.
Repeat again for a point A2 on the next antinode after these.
Now work out the path difference for a point N1 on the first nodal line in terms of λ.
Repeat for a point N2 on the second nodal line.
and finally for a point N3 on the third nodal line.
Can you see any patterns forming here?
Rules for interference of waves
•
On a nodal line: path difference PD= (n−½)λ
•
On an antinode: path difference PD = nλ
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11.2 Interference of light waves
Perhaps the most conclusive experiment supporting the wave model of light was performed by an English
physicist, Thomas Young, in 1801. He shone light through two narrow parallel slits and observed the pattern
that formed on a screen behind the slits. If the particle model was correct, then a simple pattern consisting of
two bright bands of light should have formed. However, this is not what Young found!
The pattern on the screen in fact consisted of a series of bright and dark bands. This is a most interesting
phenomenon. The pattern is formed by the light from both slits hitting the screen. Considering the dark bands
then, it follows that these are the resultant of light from both slits adding together. In other words, when light
from one slit combines with light from the other slit, the result is a dark region; i.e. light + light = dark!
Again, the particle model cannot explain this phenomenon. You would not expect light particles to add
together to produce darkness.
Constructive and destructive interference of light
The light that Thomas Young passed through the two slits produced an interference pattern similar to that
produced by water waves from two sources in phase.
The patterns that Thomas Young observed were interference patterns for light. They consisted of a series of
bright and dark bands. These patterns of fringes can only be explained using the wave model for light.
•
The bright bands are the result of _____________ interference.
•
The dark regions are the result of ______________ interference.
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The diagram below shows the interference pattern formed when light is shone through a double
slit.
Figure 1
At each of the bright regions on the screen (A, C, E, G and I), the light from each of the sources (S1 and S2)
is in phase and constructive interference takes place.
At point E on the central bright band, light has travelled equal distances from S1 and S2. The path
difference at this point is zero. Crests from S1 will meet crests from S2, and troughs from S1 will meet
troughs from S2. A double amplitude light wave (i.e. brighter light) is formed here due to this constructive
interference.
Explain in detail what is happening at point C on the interference pattern and indicating why this point is an
antinode.
Repeat again for point A.
The general rule for constructive interference is:
XPath difference = nλx
where n = 0, 1, 2, 3,
Now let us consider the regions in the pattern where there is darkness. On Figure 1, this refers to points B, D,
F and H. At each of these points, light from S1 is again meeting light from S2, but the light is cancelling out
to produce darkness! This is destructive interference.
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Let us consider point D in the fringe pattern. Waves from S2 have to travel half a wavelength
!
further than waves from S1 to reach this point i.e. the path difference is . This means that the waves are in
2
opposite phase as they meet and so destructive interference occurs here. A similar process applies at point
F.
Explain in detail why there are dark regions at points B and H.
The general rule for destructive interference is:
xPath difference = (n- ½ )λx
where n = 1, 2, 3, …
Problem: Red light of wavelength 620 nm is shone through a pair of narrow slits. An interference pattern as
shown below is formed on a screen 2.5 m away from the slits.
What is the path difference (in nanometres) at:
(a)
point A?
(b)
point B?
(c)
point C?
Answers: (a) 310 nm (b) 1.24×103 nm (c) 1.55×103 nm
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Factors that affect double-slit interference patterns
You may recall from your study of Waves in year eleven that an interference pattern for water waves is
affected by the wavelength of the waves and also by the distance between the sources of the waves.
Sketch the nodal pattern for higher frequency water waves.
Sketch the nodal pattern if the sources are further apart.
It is much the same for light. A double-slit interference pattern for light will have fringes that are more
bunched up (i.e. narrower bands) when:
•
•
light with a higher frequency (shorter wavelength) is used, or
the slits are further apart.
Activity: observe double slit patterns formed by red and blue light. Sketch the patterns that you observe.
Red:
Blue:
v Check out the 2-D interference (interactive) website.
 Homework: 11.2 1-9
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11.3 Evidence Supporting the Particle Model of Light
As the nineteenth century drew to a close, the wave model seemed to successfully explain the observed
behaviour of light. The particle model was well and truly out of favour, especially since Young's double slit
interference experiments.
However, in 1887 Heinrich Hertz observed an interaction between light and matter that came to be known
as the photoelectric effect. Further experimental work on the photoelectric effect by Phillip Lenard between
1899 and 1902 led Albert Einstein to conclude that light does in fact behave like a particle. The wave model
could not explain the photoelectric effect.
The photoelectric effect
The experiments performed by Hertz and Lenard showed that electrons could be made to escape from the
surface of a metal just by shining light onto the metal. This phenomenon became known as the photoelectric
effect and it seemed to indicate that the energy in the light was somehow making the electrons in the metal
escape.
Lenard started off by shining monochromatic light of different intensities onto a metal surface in a vacuum.
Bright light
Dim light
Photocell
+
+
-
Photocell
e
-
e
e
Metal plate
-
Small I
A flows
Metal plate
A
Larger I
flows
When brighter light of the same colour was shone onto the metal surface, more electrons were released from the metal.
As the light struck the metal surface of the photocell (see above), the electrons that had escaped were
attracted across to the positive terminal thus causing a small current flow to register on the ammeter. Lenard
found that as the light intensity increased, so too did the size of the current. More electrons were escaping
from the metal when the light was brighter.
Lenard also varied the frequency (i.e. colour) of the light shining onto the metal. He found that there was a
frequency at which the electrons began to be emitted from the metal. This was called the threshold or
cut-off frequency fo. Below this frequency, no emission occurred; even for very intense light. (See below).
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High intensity
red light
Dim red light
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Photocell
Photocell
-
+
-
+
A I=0
Metal plate
A I=0
Metal plate
(ii) No photoelectric emission
(i) No photoelectric emission
Violet light
Green light
Photocell
Photocell
+
-
-
-
+
e
Metal plate
e
Small I
A flows
-
Metal plate
Larger I
A flows
(iv) Photoelectric emission continues
(iii) Photoelectric emission starts
(threshold frequency)
Lenard investigated different metals and found that each metal had its own particular
threshold frequency. He then modified the circuit so that he could work out the energy of the escaping
electrons. Lenard reversed the battery so that it was now providing a retarding potential on the electrons.
At small retarding voltages, many of the emitted electrons still had enough energy to get across to the
opposite terminal.
Photocell
-
-
e
-
+
e
Metal plate
A
Small I
flows
Small retarding voltage
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Lenard increased the size of this retarding
voltage until all of the electrons were stopped
from reaching the opposite terminal. This
voltage is known as the stopping (or cut-off)
potential Vo .
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Photocell
+
-
e
e
-
A I=0
Metal plate
VO
This enabled the maximum kinetic energy of the photoelectrons to be calculated:
Maximum kinetic energy of emitted electrons : EK
MAX
= e.Vo
Graphically, these results can be shown as:
Light of the
same colour
Current
Bright
Medium
Dim
Voltage
VO
0
Stopping
voltage
When only the intensity of the incident light is altered, the size of
the photocurrent changes but the stopping voltage is unaffected.
Higher frequency
light
Light of the
same intensity
Current
Lower frequency
light
Stopping
voltages
0
Voltage
When only the frequency of the incident light is altered, the
stopping voltage changes but the size of the photocurrent is unaffected.
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Lenard found that:
Interactions of light and matter
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o the kinetic energy of the emitted electrons was equal for both dim and bright light of a particular
frequency;
o the kinetic energy of the emitted electrons was dependent on the frequency of the light. For example,
violet light ejects faster-moving electrons than green light; and
o above the threshold frequency, photoemission begins immediately; even for extremely dim light.
The wave model could not explain these results. Read page 424-5 of your text and explain what the
shortcomings of the wave model were.
Einstein's photon model
It was Albert Einstein in 1905 who explained what was going on. Einstein suggested that light was not a
continuous wave, but instead travels in discrete packets or quanta. All light of a certain frequency comes in
packets that have the same amount of energy. These packets of light energy came to be known as photons.
Electronvolts are often used in these calculations. 1 electronvolt =
The energy of a photon depends on its frequency and is given by:
Photon energy E = hf =
where
hc
ë
h = Planck's constant 6.6×10-34 J s = 4.1!10-15 eV s
f = frequency (Hz )
! = wavelength (m)
c = speed of light = 3.0×108 m s-1
 Homework: 11.3 1, 2, 4, 6 plus Photon Energy questions 1-6 (PTO)
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Photon Energy
1. How much energy (in joules) does a red light photon of frequency 3.85×1014 Hz have?
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2. A photon has a frequency of 2.0×104 Hz.
(a) What is its wavelength?
(b) What type of photon is it? (Refer to page 418 of your text)
(c) What is the energy of the photon (in electronvolts)?
3. (a) What is the wavelength of a photon with energy 3.5×10-14 J?
(b) What type of photon is this?
4. It is found that a neutron travelling at 1.98×104 m s-1 has the same energy as a photon of frequency
5.0×1014 Hz. What is the mass of the neutron?
5. The energy required to ionise an aluminium atom is 4.3 eV. Will light of wavelength 3.5×10-7 m be able to
eject an electron from an aluminium atom?
6. How fast would a speck of iron of mass 1.75×10-6 g have to travel to have the same amount of energy as a
photon with a wavelength of 4.62×10-13 m?
Answers: 1. 2.55×10-19 J 2. (a) 1.5×104 m (b) radio (c) 8.2× 10-11 eV 3. (a) 5.7×10-12 m (b) gamma 4. 1.7× 10-27 kg 5. No, the
photons have only 3.5 eV of energy. 6. 0.022 m s-1
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11.3 Einstein and the photoelectric effect
Einstein's explanation of the photoelectric effect was that each photon of light gave up its energy completely
when it collided with an electron in the metal. The energised electron used up some of this energy in
overcoming the binding force of the atoms in the metal and escaped with the remaining energy, EK.
Photon with
energy E
Metal
-
EK
The electron absorbs the energy of the photon, uses some of this energy to
break free from the metal and escapes with remaining energy EK.
The energy that the electron uses up to escape from the metal is called the binding energy or work function
W of the metal. This quantity varies from metal to metal.
Thus the maximum kinetic energy EK MAX of the escaped electron is given by:
Kinetic energy of photoelectron EK
MAX
= e.Vo = hf - W
Problem:
Light of frequency 1.3×1015 Hz strikes a metal surface with a work function of 3.0×10-19 J .
(a)
Calculate the photon energy in joules and eV.
(b)
Calculate the binding energy of the metal in eV.
(c)
Work out the maximum energy of the escaping electrons in joules and eV.
Solution:
(a)
Photon energy
(b)
(c)
Binding energy
Photon
E = 5.3 eV
Metal
-
EK = 3.4 eV
1.9 eV
Electron in
metal
Answers: (a) 8.6×10-19 J, 5.3 eV (b) 3.0× 10-19 J, 1.9 eV (c) 5.6× 10-19 J, 3.4 eV
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At frequencies below the threshold, the photons do not have enough energy to enable the
electrons to escape from the metal.
f < fO
Metal
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The energy of the photon hfo is less than the binding energy W
of the metal, so the electron does not escape from the metal.
No photoemission occurs.
At the threshold frequency, the energy of the photon is equal to the work function of the metal.
f = fO
Metal
Work Function: W = hf0
Photoemission begins at frequencies greater than f o .
The maximum energy of the photoelectrons can therefore be found by using:
Ek (max) = hf – W = hf - hf0
Graphically, the relationship between the energy of the emitted electrons and the frequency of the photons is:
Maximum EK of
Threshold
photoelectrons (J)
frequency
Binding
energy W
0
Gradient = Planck’s
constant, h
No emission
below fO
fO
Above threshold
frequency
Frequency of
photons (Hz)
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Questions:
Light of frequency 1.2×1015 Hz is shone onto a zinc metal plate. The energy of the photoelectrons emitted
from the zinc plate is shown in the graph below (h= 6.6×10-34 J s).
EK (max.)
-19
(! 10 J )
6
4
2
0
-2
-4
-6
2
4
6
8
10 12 14 16
18
20
Frequency
( 1014 Hz)
(a)
Calculate the energy of these photons.
(b)
Use the graph to find the binding energy of the zinc.
(c)
Use the graph to determine the maximum energy of the emitted photoelectrons.
(d)
Use the graph to find the threshold frequency for zinc.
Answers: (a) 7.9× 10-19 J; (b) 5.9× 10-19 J; (c) 2.0× 10-19 J; (d) 9.0×1014 Hz.
 Homework 11.3 3,5,7,9
 11.4 1- 8
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Photoelectric Effect Problems
1. A certain metal has a work function of 6.4×10-19 J. What is the longest wavelength of light that result in
photoelectrons being released from this metal?
2. The threshold frequency of rubidium is 4.87×1014 Hz. Calculate the binding energy of this metal in
electronvolts.
3. Green light of wavelength 5.46×10-7 m illuminates a clean surface of calcium metal. The work function of
calcium is 4.2×10-19 J. Are photoelectrons emitted in this situation?
4. Radiation of wavelength 2.86×10-7 m strikes the surface of a material that has a work function of 2.2 eV .
Calculate the:
(a) energy of the incident photons (in eV). (b) maximum energy of the photoelectrons(in eV).
5. Electromagnetic radiation of frequency 1.15×1015 Hz strikes the surface of a metal for which the binding
energy is 4.99×10-19 J. Calculate the:
(a) maximum kinetic energy of the emitted photoelectrons.
(b) threshold frequency of the metal.
(c) cut-off wavelength for the metal.
6. Light of wavelength 3.0×10-7 m illuminates the surface of a material having a threshold frequency of
6.0×1014 Hz. Calculate the maximum speed of the photoelectrons emitted from the material (in km s-1).
7. UV light with a wavelength of 3.55×10-7 m from a mercury lamp strikes a clean metal surface causing
some electrons to be emitted. The work function of the metal is 1.65 eV.
(a) What is the frequency of the UV radiation?
(b) How much energy (in eV) does each UV photon have?
(c) What is the maximum kinetic energy (in eV) of the emitted photoelectrons?
(d) What is the longest wavelength of radiation that will eject electrons from this metal?
Answers: 1. 3.1×10-7 m 2. 2.0 eV 3 No 4 (a) 4.3 eV (b) 2.1 eV 5. (a) 2.62×10-19 J (b) 7.54×1014 Hz (c) 3.97×10-7 m
6. 760 km s-1 7 (a) 8.45×1014 Hz (b) 3.49 eV (c) 1.84 eV (d) 7.52×10-7 m
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11.3 The momentum of photons – the ‘Compton effect’
In 1920, Einstein was still one of the few believers in the photon theory of light. Then an American physicist,
Arthur Compton, conducted a series of experiments in the early 1920s that provided overwhelming evidence
that Einstein was correct.
Arthur Compton and Albert Einstein
Compton found that when high energy photons (he used X-rays) collided with electrons, the electrons moved
off with energy and momentum and the photons scattered or recoiled having lost energy and momentum. In
other words, the collision could be treated like a simple mechanics problem involving two billiard balls. It
was like an elastic collision between particles.
Before collision
After collision
Photon
8
3 10 m s
Electron at
rest
-1
-
3 108 m s-1
-
v
1. Why doesn’t the speed of the photon change??
2. The electron clearly gains some energy here. Where does this energy come from?
3. Which photon has more energy, the incident or scattered photon? What must be different about the
incident and scattered photons?
The collision with the photon has caused the electron to recoil and so we can consider the photon has having
momentum (a very very small amount!).
The momentum of a photon can be found by using:
Momentum of photon
p=
hf h
=
c
"
CANNOT USE eV HERE!
This interaction is known as the Compton Effect. Its discovery led to light being treated as a particle that
had wave properties
such as frequency and wavelength.
!
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The momentum of a photon depends on its frequency and wavelength.
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Question: Calculate the momentum of a photon of:
red light (λ = 7.0×10-7 m)
violet light (λ = 4.0×10-7 m)
a gamma ray (λ = 3.0×10-12 m)
A photon of violet light has a ________ frequency and a ________ wavelength than a photon of red light;
therefore the violet photon has _____ momentum than the red photon. A photon that has more energy will
also have more momentum when comparing two photons.
Questions:
In an experiment used to verify the Compton Effect, two beams of high energy photons were aimed at a
crystal. Beam 2 was of higher frequency than beam 1.
Beam 1: f = 2.0 1019 Hz
Beam 2: f = 2.0 1020 Hz
(a)
Calculate the energy of a single photon in each of the beams.
(b)
How does the momentum of a photon in beam 2 compare with that of a single photon in beam 1?
(c)
Calculate the momentum of a photon in beam 1. If you were hit by this photon, would it knock you
over?
(d)
If a photon in each beam experienced an identical elastic collision with an electron in the crystal, how
would the recoil speeds of the respective electrons compare?
(e)
How will the recoil speeds of the different photons compare?
Crystal
-14
-13
Answers: (a) Beam 1: 1.3×10 J; beam 2: 1.3 ×10 J (b) Photons in beam 2 have ten times the momentum of photons in beam
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-1
1 (c) 4.4×10
kg m s ; no (d) Electrons hit by a beam 2 photon will recoil faster (e) Both will recoil at 3 0 × 108 m s-1.
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Wave-Particle Duality of Light
The argument about whether light was a wave or a particle was eventually settled in the 1920s. The wave
model explained refraction, diffraction and interference of light. The particle model explained the
photoelectric effect and the Compton effect. So which model is correct?
Light is neither a wave nor a particle. Light is light! Photons exhibit both wave and particle properties. This
is called wave-particle duality.
The behaviour of photons varies across the electromagnetic spectrum. Low frequency photons such as radio
waves and microwaves exhibit distinctly wave-like behaviours such as diffraction and interference, but have
no particle-like properties. Around the middle of the spectrum in the visible light region, photons have both
wave and particle properties. They interfere and diffract like waves, and also interact with electrons in the
photoelectric effect as particles do. At the high frequency end of the spectrum, X-ray and gamma ray photons
behave much more like particles than waves.
Indicate the regions of the spectrum where the electromagnetic radiation exhibits:
predominantly wave characteristics
predominantly particle characteristics
both characteristics
 Homework: Chapter Review Q7, 8, 11, 12
ü Interference of Light and Matter website
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Investigating the Nature of Matter
Line emission spectra
During the second half of the nineteenth century, scientists discovered that gases produce their own
characteristic emission spectra. Many experiments were conducted where a gas such as neon or oxygen was
excited using an electric current in a discharge tube. When light from the excited gas was analysed using a
spectroscope, a distinctive series of lines was observed. Each element was found to have its own unique
emission spectrum. It was many years before this well-known interaction between matter and light could be
fully explained and understood.
R
V
The emission spectrum for hydrogen has 5 lines in the visible region.
Activity: use a spectroscope to observe and sketch the emission spectrum for:
sodium, Na
mercury, Hg
Line absorption spectra
In 1862, a different form of spectroscopy was used to show that hydrogen was present on the Sun. It was
observed that when white light was shone through a particular gas, a distinctive pattern of dark bands would
be seen when viewed through a spectroscope. Each particular gas seemed to be absorbing certain discrete
bands of light. Again, each element had its own unique pattern of absorption spectral lines.
R
V
The absorption spectrum for hydrogen has 5 lines in the visible region.
In 1911, Ernest Rutherford, a New Zealand
physicist, proposed a model of the atom that
consisted of a small, dense, positively charged
nucleus surrounded by a large cloud of
electrons. A problem with this model was that
the orbiting electrons, because of their
accelerated motion, should radiate energy and
spiral into the nucleus. Clearly, this does not
happen!!
.
Orbiting
electrons
Nucleus
The Rutherford model of the atom.
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Quantised energy levels in atoms−the Bohr model
In 1913, Niels Bohr, a Danish physicist, suggested a solution to the problem.
Bohr studied the emission spectra of hydrogen in great detail. He said that the electrons in hydrogen atoms
should not be considered to be orbiting like planets. Bohr said that they simply existed outside the nucleus
with certain amounts of energy. According to Bohr, the electrons in a hydrogen atom existed in certain
discrete energy levels or quantum states. He said that:
•
electrons can only exist in one of these allowable energy levels, not in between. In
other words, energy levels are quantised.
•
if an atom is given extra energy, an electron can move up to a higher energy level by
absorbing an amount of energy equal to the difference between the energy levels. The
atom is then said to be in an excited state.
•
when an electron in a higher energy level returns to its normal (ground state) energy
level, it emits energy in the form of a photon. The energy of the photon (E = hf) is
equal to the difference in energy levels the electron moves between, so:
Photon energy Ephoton = hf = EM - EN
where EM, EN = energy levels (eV or joules)
h = Planck’s constant = 6.63×10-34 J s = 4.14×10-15 eV s
f = frequency of emitted photon (Hz)
Ionisation
5
4
etc
Excited States
3
2
1
Ground State
The Bohr model of the energy levels in an atom.
The Bohr model helped to explain the spectral lines seen in emission and absorption spectra. These lines
form when the atoms are irradiated with light in which the photons do not possess enough energy to ionise
the atom as happened in the photoelectric effect.
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According to the Bohr model:
o the electrons in the atom can only accept particular amounts of energy.
o the excited electrons will then move to a higher energy level (for less than a microsecond!) before
returning to ground state.
o in moving back to ground state, the electrons could drop back in one step or could drop back via any
combination of the intermediate energy levels.
o as the electron moves back, dropping from one energy level to another, a photon is emitted each
time. The energy of the emitted photon is given by the energy difference between energy levels
and accounts for one of the bands on the line emission spectrum.
Metal vapour lamps, such as sodium lamps used for street lighting, work on this principle.
They are known as narrow-spectrum discrete-line sources. The sodium gas in the lamps has
energy levels that cause photons with wavelengths of 589.0 nm and 589.6 nm to be emitted
when the atoms are excited. Sodium lamps therefore emit photons with just two discrete
wavelengths in the visible part of the electromagnetic spectrum. These photons are from the
yellow/orange region of the spectrum and give sodium lamps their characteristic colour.
Bohr’s model could also explain the existence of line absorption spectra. When light from a continuous
source, such as white light, is passed through a sample of a gas, the photons from the light source interact
with valence electrons in the gas. In this case, the electrons only accept particular amounts of energy as
determined by their energy levels. For example, if the element has an energy level of 5 eV, then photons with
5 eV of energy will interact with and give all their energy to the atoms. The evidence suggesting that these
photons have been absorbed is a dark band on the spectrum.
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Question 1:
Several of the energy levels of an electron in a
hydrogen atom are shown on the right.
Ionisation
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4
3
13.6 eV
12.7 eV
12.1 eV
2
10.2 eV
1
0 eV
Ground State
(a)
What could happen if this hydrogen gas was irradiated with electrons with 14 eV of energy?
(b)
What could happen if the hydrogen was irradiated with a beam of electrons with 11 eV of energy?
(c)
What could happen if the hydrogen was irradiated with a beam of photons with 11 eV of energy?
(d)
(i)
An electron in the hydrogen is excited to the n = 3 energy level.
What are the energies that the emitted photons could have as the electron returns to ground state?
(ii)
Calculate the possible wavelengths of the photons emitted as this electron returns to the ground state.
(e) Use the diagram above to determine the transition that will result in a photon with the lowest frequency
on the emission spectrum.
 Homework: 12.2 1-9
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12.1 The wave-like nature of matter
In 1923 a French physicist, Louis de Broglie, made a radical suggestion about matter. Up until that time,
scientists had considered electrons and the other particles of matter to be just that – particles. However de
Broglie, for reasons of symmetry, speculated that since light waves could behave like particles, then particles
of matter should behave like waves.
Experiments that have been performed with electrons clearly show that they can diffract and interfere with
each other. Protons and neutrons also have been shown to exhibit this wave-like behaviour.
De Broglie suggested that the wavelength of a particle of matter could be found by using the same
relationship that applies to photons; i.e. p =
MATTER
h
.
!
de Broglie matter waves " =
h
h
=
p mv
CAN’T USE eV!!
The wavelength of any particle is inversely proportional to its momentum. A fast-moving electron therefore
has a shorter wavelength than a slow-moving electron.
8
-1
!
1 10 m s
5
-1
1 10 m s
A proton, due to its greater mass, will have a shorter wavelength than an electron travelling at the same
speed.
5
-1
5
-1
1 10 m s
1 10 m s
-+
Example:
Calculate the de Broglie wavelength of these objects:
(a)
An electron (m = 9.1×10-31 kg) moving at 1.5× 106 m s-1 .
(b)
A marble of mass 0.050 kg moving at 6.0 m s-1.
Solution:
(4.8×10-10 m)
(2.2×10-33 m)
Question:
Why doesn’t a marble diffract as it rolls through a 2cm wide opening???
Electrons do diffract as they pass through layers in crystals and atoms. Why???
Read about electron microscopes on page 442 of your text.
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Electron diffraction patterns
Louis de Broglie’s theory of matter waves was verified in experiments performed in 1927. G.P.Thomson
fired a beam of electrons at tiny crystals. The electrons passed through the crystals and diffracted as they
passed through the spaces between the atoms in the crystals. Distinctive circular diffraction patterns were
formed.
Electrons diffract as they pass through a crystal
resulting in a circular diffraction pattern.
Thomson went on and calculated the wavelength of the electrons in the beam. He then fired a beam of high
energy photons (in the form of X-rays) with the same wavelength as the electrons at the same crystals. The
diffraction pattern formed by the X-rays was strikingly similar to the diffraction pattern that was formed
using the electrons.
X-rays diffract as they pass through the gaps in a
crystal. The similarity to the electron diffraction pattern
suggests that electrons have wave properties.
Since these early experiments, other particles such as protons, neutrons, hydrogen atoms and helium atoms
have been used in similar experiments. These particles have also produced diffraction patterns providing
further evidence in support of de Broglie and his theory of the wave-like nature of matter.
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Questions:
An electron is accelerated from rest through a potential difference of 1000 V. It then passes through a crystal
in which the gaps between the atoms in the crystal lattice measure 2.0 "10!10 m. (me = 9.1×10-31 kg)
-
v
Crystal
(a)
Calculate the final energy of the electron (in eV and J) after it has passed through this potential
difference.
(b)
What is the speed and momentum of the electron as it reaches the crystal?
(c)
Calculate the de Broglie wavelength of the electron.
(d)
Will the electron diffract as it passes through the crystal? Explain.
Answers: (a) 1000 eV , 1.6×10-16 J; (b) 1.9×107 m s-1, 1.7×10-23 kg m s-1 (c) 3.9×10-11 m (d) Yes, its wavelength is about the same
as the gap size.
Read Worked Example 12.1A on page 438 of your text.
 Homework: 12.1 1-8
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12.3 Electron orbitals in atoms – standing waves!!
De Broglie hypothesised that a particle such as an electron would have a wavelength.
MATTER
de Broglie matter wavelength " =
h
h
=
p mv
Can’t use eV!
If we assume that electrons move in circular paths around the nucleus, then it follows that an electron wave
must be a circular wave. Also, for the electron wave to exist in an orbit, it must constructively interfere
! this to occur, the circumference of the orbit would have to be a whole number of
with itself! For
wavelengths of the electron wave i.e. nλ = 2πr.
Look at Figure 12.19 on page 456 of your text.
If the orbit’s circumference was not exactly equal to an integral number of wavelengths, then the wave
would be out of phase with itself and the energy of the electron would not be one of the atoms energy levels.
Only particular values of wavelength would fit the circumference of the orbit and so only particular energy
values would exist. These are the energy levels of the atom and so atoms are quantised.
Many years prior to this, Bohr had determined the energy levels of hydrogen. The theories developed by de
Broglie were finally able to explain why these energy levels existed.
De Broglie’s theory that electrons have a wave-like nature explains why electrons can only exist in specific
orbits / energy levels. Atomic emission and absorption spectra are also evidence of quantum behaviour of
atoms. We live in a quantum universe!
 Homework 12.3 1-6
REVISION: Chapter 12 Review and Exam-style questions
Louis de Broglie
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