Vu4 light&matter2009

VCE Physics
Unit 4
Topic 2

Interactions of
Light & Matter

Unit Outline
To achieve the outcome the student should demonstrate the knowledge and skills to :
• Explain the production of incoherent light from wide spectrum light sources including the
Sun, light bulbs and candles (descriptive) in terms of thermal motion of electrons.
• Explain the results of Young’s double slit experiment as evidence for the wave like nature
of light including:
– constructive and destructive interference in terms of path difference
– qualitative effect of wavelength on interference patterns
• Interpret the pattern produced by light when it passes through a gap or past an obstacle in
terms of the diffraction of waves and the significance of the magnitude of the λ/w ratio
• interpret the photoelectric effect as evidence of the particle like nature of light including the
KE of emitted photoelectrons in terms of the energy of incident photons Ek max = hf – W, using
energy units of both joules and electron-volts, effects of intensity of incident irradiation on
the emission of photoelectrons.
• interpret electron diffraction patterns as evidence of the wavelike nature of matter
expressed as the De Broglie wavelength λ = h/p
• compare momentum of photons and of particles of the same wavelength including
calculations using p = h/λ
• interpret atomic absorption and emission spectra, including those from metal vapour lamps
in terms of the quantised energy level model of the atom, including calculations of the
energy of photons emitted or absorbed. ∆E = hf
• explain a model of quantised energy levels of the atom in which electrons are found in
standing wave states
• use safe and responsible practises when working with light sources, lasers and related
equipment.

Chapter 1
Topics covered:
• The Nature of Light.
• Interference.
• Incoherent Light.
• Coherent Light
• Young’s Experiment.
• Path Difference
• Single Slit Diffraction
• Diffraction around Objects

1.0 The Nature of Light
EMR is a self propagating wave
Light is a form of ENERGY. consisting of mutually
It is described as ELECTRO - perpendicular, varying
MAGNETIC RADIATION (EMR). ELECTRIC and MAGNETIC
FIELDS.
EMR travels through a vacuum
at 300,000 kms-1, (3.0 x 108 ms-1)
Changing
Magnetic Field
Direction of
Electromagnetic Wave
Movement
Changing
Electric Field

1.1 Superposition
Single waves, called Once the
pulses. have the ability superposition is
to pass through one complete the pulses
another and, while continue their
occupying the same journey unaffected.
space, add together in a
process called A series of pulses
SUPERPOSITION. together form a
Trough Crest wave train with
alternating crests
When two wave trains interact and troughs
with one another they also Constructive interference occurs when
undergo SUPERPOSITION and the two wave trains are in phase
will either:
1. Add together to produce a
larger wave - a process called
CONSTRUCTIVE INTERFERENCE
or
2. Subtract from one another to
Destructive interference occurs when
produce no wave – a process called the two wave trains are 1800 out of
DESTRUCTIVE INTERFERENCE phase

1.2 Interference
Light behaves in a similar manner.
When light (with certain properties)
Destructive is passed through two narrow slits,
Interference an “interference pattern” is
Constructive produced, showing constructive
Interference (light bands) and destructive (dark
bands) interference.
Bright bands occur
where a crest and a
Waves, in this case water crest (or a trough and a
waves, when passed trough) arrive at the
through two narrow slits, screen at the same time
“interfere” or interact with
one another to produce
areas of large disturbances Dark bands
Crest Bright
(Constructive Interference) Band occur when a
Trough
or areas of no disturbance crest and a
(Destructive Interference) trough arrive at
Dark the screen at
Band the same time

1.3 Incoherent Light
Light is generated by luminous bodies, eg,
The Sun, light globes, burning candles.
Light is produced when atoms of the
filaments or source become electrically
excited and produce an electromagnetic
or light wave. Since the excitations occur in an
unpredictably random fashion, the
Typical light sources such as those
light waves are NOT produced in
mentioned above have an inbuilt
regular repeating manner and so do
irregularity in the way they produce light.
not maintain a constant “phase
relationship” with one other.
About once every 10-8 sec, a source
will randomly alter its phase.
Incoherent light, when combined,
This leads to the sources giving off a produces rapidly moving areas of
broad spectrum of white light composed constructive and destructive
of all colours in the rainbow. interference and therefore do not
Each of the millions of colours have produce a stable, visible interference
waves that are random to each other. pattern.
This is called INCOHERENT LIGHT.

Light & Matter Revision
Question Type: Incandescent Light

Q1: The light from a candle can best be described as
A. coherent, arising from the vibrations of electrons.
B. incoherent, arising only from the transition of electrons in excited
energy levels falling to lower energy levels.
C. coherent, arising only from the transition of electrons in excited energy
levels falling to lower energy levels.
D. incoherent, arising from the vibrations of electrons.

Question Type: Incandescent Light
The spectrum of wavelengths produced by a particular
incandescent light globe is shown in Figure 1 below.

Q2: The light produced by an
incandescent light globe can
best be described as
A. coherent.
B. incoherent.
C. monochromatic.
D. in phase.

Q3: Describe the mechanism by which light is
produced in an incandescent light globe.

The thermal/random excitation of
electrons in the filament leads to the
emission of a broad/continuous spectrum.

1.4 Coherent Light
Coherence is a property of waves that
measures their ability to interfere with
Coherent waves (zero phase each other.
difference)
Shown are monochromatic (single
colour) light waves of the same
frequency. Coherent waves (constant
They are coherent and in phase, and will phase difference)
combine constructively to produce bright
Two waves that are coherent can
white light
be combined to produce an
unmoving distribution of
constructive and destructive
interference (a visible
interference pattern) depending
on the relative phase of the
waves at their meeting point.
Lasers generate light at a single
wave length and frequency and all
of the waves (and PHOTONS) are in
phase.
This is called COHERENT LIGHT.

Question Type: Coherent Light

Q4: Which of the following light sources will
produce coherent, monochromatic light
A. Sunlight
B. LED
C. Light globe with filter
D. LASER

1.5 Young’s Experiment
First performed by This pattern has a series of equally
Thomas Young Thomas Young in the spaced coloured and black bands
1773 –
1829 early 1800’s this spread across the screen onto
experiment proved which it is projected.
light was a wave. The width of the coloured bands
It has been voted the and their spacing depends on
most elegant
Incident Light
the wavelength of the light used.
experiment ever Short wavelength, BLUE light
devised. produces a pattern with narrow blue
When light of a single frequency bands which are closely spaced.
(colour) is passed through a pair
Long wavelength, RED light
of closely spaced, narrow slits
produces a pattern with wider
an “interference pattern” is
red bands which are spread
produced.
farther apart.

BLUE Light
using the same slits

Screen
Double Slits
RED Light

Question Type: Young’s Experiment
Thomas Young’s double slit experiment The slits are now moved further
has been replicated in the experimental from the screen.
arrangement shown in Figure 4. Q6: What effect would this have
on the pattern observed on the
screen?
The pattern will spread further
across the screen.

Q5: Explain using the wave theory of
light why a series of bright and dark
bars are observed on the screen.
Slits provide 2 coherent sources
leading to superposition which
produces constructive and
destructive interference leading to
bright and dark bars on screen

Question Type: Young’s Experiment
A physics teacher has
apparatus to show
Young’s double slit
experiment. The
apparatus is shown in
Figure 4.
The pattern of bright and
dark bands is observed
on the screen.

Q7: Which one of the following actions will increase the distance, Δx,
between dark bands in this double slit interference pattern?
A. decrease the slit width
B. decrease the slit separation
C. decrease the slit screen distance
D. decrease the wavelength of the light

1.6 Path Difference
Light travels the same distance from S1 and S2
to reach the central bright band, BC
The path difference The dark band (D1) on
S1 S2 S1BC – S2BC = 0. either side of AC occurs
So a crest and a crest
because a crest and a
(or trough and trough)
trough are arriving at the
arrive at AC at the same
same time leading to
time, leading to destructive interference.
constructive Path difference:
interference or a bright S1D1 – S2D1 = ½ λ
band The next dark band (D2)
Slits For the next bright
has path difference = 1½λ
S1 S2 band path difference
S1B1 – S2B1 = 1 λ
For the next bright So dark bands occur
band path difference when
would be: path difference = (n + ½)λ
S1B2 – S2B2 = 2λ where n =
So bright bands occur 0,1,2,3, etc
when path difference = nλ
where n = 0,1,2,3,etc
D2 B1 D1 B C D1 B1 D2

Question Type: Interference – Path Difference
In the following diagram, laser light of
wavelength 600 nm is shone onto a pair of Q9: Estimate the difference in
parallel slits and a pattern of alternating length between P1 and P2.
light/dark bands is projected onto a wall.
The indicated position is on
the sixth antinode from the
centre.
The path difference is
6λ = 6 × 600nm = 3.6 ×10−6 m

Q8: Explain how the observed pattern
on the wall supports a wave model for
light.
The alternating pattern represents an interference pattern which is a wave
phenomenon. The bright bands represent regions of constructive interference
where the difference in path length from each source is a multiple of
wavelengths. Waves therefore arrive from both sources in phase. The dark
regions represent bands of destructive interference with the waves being half
a wavelength out of phase due to path differences.

Question Type: Interference – Path Difference
Q10: What is the difference in length
(S2P2 - S1P2) where P2 is the second
maximum away from the central axis.

Path difference to 2nd max = 2λ = 5.6 cm

Students have set up an experiment similar
to that of English physicist Thomas Young.
The students’ experiment uses microwaves
of wavelength λ = 2.8 cm instead of light.
The beam of microwaves passes through
two narrow slits shown as S1 and S2 in
Figure 3.
The students measure the intensity of the
resulting beam at points along the line
shown and determine the positions of
maximum intensity. These are shown as
filled circles and marked P0, P1 . . .

1.7 Single Slit Diffraction
Appreciable diffraction will occur if
When light of a single THE EXTENT OF
the ratio λ/w is between about 0.1
wavelength shines DIFFRACTION
and 50. Outside this range,
through a narrow gap or DEPENDS UPON
diffraction is not observed.
single slit, a “diffraction THE RATIO λ /w,
pattern” is produced. WHERE λ = Diffraction patterns for blue and
WAVELENGTH red light show that the shorter
The pattern consists of AND w = SLIT wavelength blue light produces the
a rather wide, coloured WIDTH. more “compact” pattern, while

Screen
central maximum with the longer wavelength red
a series of thinner produces a more spread out
coloured and dark pattern
bands spreading out In other words, the spacing
from the centre. between the lines is
Width = w wavelength dependent and
Single Slit patterns with the same line
Incident Light spacing were made by light
Wavelength = λ of the same wavelength
BLUE light using same slit

Note: The wider the slit the narrower central maximum RED light

Question Type: Single Slit Diffraction
In an experiment,
monochromatic laser light of
wavelength 600 nm shines
through a narrow slit, and the
intensity
of the transmitted light is
recorded on the screen some
distance away as shown in
Figure 2a. The intensity
pattern seen on the screen is
shown in Figure 2b.
Q11: Which one of the
intensity patterns (A-D) below
best indicates the pattern that
would be seen if a wider slit
was used?

1.8 Diffraction around Objects
In addition to diffraction Sunlight diffracting around
occurring when light passes a car tail light lens
through gaps, it can also
occur when light passes
around objects

The shadow of a
Red Light hand holding a coin
Diffracting around a illuminated by a He –
You can see a
pinhead Ne Laser.
diffraction
Point your hand pattern
toward a light through your
source and view fingers
through this gap

Chapter 2
Topics covered:
• Models for Light Behaviour.
• Particle like Properties of Light.
• Wave like Properties of Light.

2.0 Models for Light Behaviour
• Light is/has energy and thus Two separate models have
can do work. We see this in been proposed:
things like plants growing or 1. The Particle Model:
the chemical reactions on This says that all properties of
photographic films. light can be explained by
• It is quite easy to list or show assuming light is made up of
what light DOES but more a stream of individual
difficult to say what light IS ! particles.
• Much time and effort has
been put into trying to find a 2. The Wave Model:
THEORY or MODEL which
This says that all properties of
will explain ALL the
light can be explained by
properties and behaviours
assuming light is made up of
exhibited by light.
a stream of waves.

2.1 Particle like Properties of
Light
• Most properties of light can
be explained using the “Wave
Model”.
• However, some properties
can only be explained by
assuming light is made up of
a “stream of particles”.
4. Certain colours of light can cause
• These particle like properties electrons to be emitted from some
are: metal surfaces. This is called the
1. Light exerts pressure. Photoelectric Effect.

2. Light can travel through a 5. Light reflects from shiny
vacuum as well as through surfaces (both a wave and particle like
transparent media. property).

6. Different colours correspond to
3. Beams of light are bent by different energies of light (both a wave
gravity. and particle property).

2.2 Wave Like Properties of Light
• Light exhibits many properties which
can be demonstrated by other forms
of waves eg. Water waves or waves on
springs or strings.
• These wavelike properties of light are: 4. Light undergoes Diffraction when
passing through narrow gaps.
1. Light travels very fast, but its speed
does depend upon the medium 5. Light beams can pass through one
through which it moves. another unaffected.

2. Light undergoes Reflection according 6. Different colours of light correspond
to the Laws of Reflection. to different energies.

3. Light undergoes Refraction according 7. Light forms interference patterns
to the Laws of Refraction. when passed through a pair of
closely spaced, narrow slits.

Question Type: Light Properties

Q12. Light sometimes behaves as a particle and
sometimes as a wave. Which one or more of the
following properties does light sometimes show?
A. mass
B. momentum
C. charge
D. energy

Question Type: Models for Light
Q 13: The table below contains some predictions for the
behaviour of light incident on a shiny metal sheet. Complete the
table by placing a .Y. (Yes) or .N. (No) in the appropriate boxes if
the prediction is supported by the wave and/or particle model of
light. Some answers have already been provided. It is possible for
predictions to be supported by both models

Y
Y
N
N

Chapter 3
Topics covered:
• The Photoelectric Effect.
• Electron Energies
• Energy Units
• Practical Applications

3.0 The Photoelectric Effect
Albert Einstein won his Nobel
The most important of the Prize for his explanation of this
particle like properties of phenomena not for his work on
light is the Photoelectric Relativity
Effect.
Simply put, the photoelectric
effect occurs when light
shines on a metal surface
causing electrons to be
ejected from the metal.
There are, however, certain restrictions:
(a) Only certain metals respond.
(b) Light must be above a certain
frequency - the Cut Off Frequency.

Einstein called these individual
light particles PHOTONS

Louis De Broglie working in the 1920’s
suggested a photon looked like this.

3.1 The Photoelectric Effect (2)
The photoelectric effect is best studied Incident Photon
Evacuated
using the demonstration circuit shown. Glass Tube e
An incident photon of the right frequency +
(that is above the Cut Off Frequency)
Ejected Electron
striking a susceptible metal (such as
aluminium, copper, lead, zinc and many Galvanometer G
others), will free an electron.
Rheostat
The electron will cross the gap in the
evacuated tube, due to electrostatic
attraction.
+

So a stream of incident photons striking
the metal will release a stream of electrons,
setting up a current in the circuit which will
be detected by the galvanometer (a
sensitive ammeter able to detect currents
in the μA range).

The current (of ejected photoelectrons)
does NOT depend upon the VOLTAGE
between the plates.
Incident Photon
The size of the current DOES depend Evacuated
upon the NUMBER of photons striking Glass Tube e
the plate. +
The NUMBER of photons depends Ejected Electron
upon the INTENSITY of the incident
light.
Low Intensity or Dim Light of a
Current (I)
particular colour or frequency will
produce a small current. Bright Light
High Intensity or Bright Light of the
same frequency will produce a large Dim Light
current.
Voltage (V)

To study the energy possessed by the
electrons ejected from the metal the Evacuated
Evacuated
Evacuated
Glass Tube +
Glass Tube
Glass Tube +
circuit is changed. +
+
+
+
The power supply has been reversed, +
reversing the polarity of the plates. Some e-e-electrons crossenough
No morenow don’t have the gap
All still have enough
thus energy to crossgap gap
Current = 0
energy to cross the the
Galvanometer G
Galvanometer
Moving the slider on the rheostat
increases the voltage between the
plates.

This increasing voltage stops the least Rheostat
Rheostat
Rheostat
energetic electrons crossing the gap,
reducing the current in the circuit. +
+

The slider is again moved until no Current (I)
electrons have enough energy to cross
the gap and the current drops to zero.

At this point the “Cut Off Voltage”
(VC) has been reached.
Voltage (V)

VC

The fact that the current falls to zero
only slowly as the reverse voltage
increases indicates the photoelectrons
are ejected from the metal surface with
VARYING AMOUNTS OF ENERGY.

Different Intensities of the same Current (I)
colour (frequency) have the
same cut off voltages

Different frequencies and/or -Vc Voltage (V)
different metals will have
differing cut off voltages

Question Type: Photoelectric effect
A student carries out an experiment to
Q14: What is the maximum kinetic
investigate the photoelectric effect. She
energy (in eV) of the electrons
shines a monochromatic light onto a
ejected from the plate P?
metal plate (P) inside a sealed glass
1.7 eV
chamber as shown in Figure 2.
Q15: What is the subsequent
maximum speed of these electrons
ejected from plate P?
KEMAX = 1.7 eV = (1.7)(1.6 x 10-19) J
= 2.72 x 10-19 J
KE = ½ mv2
v = 7.73 x 105 ms-1

The current in the circuit changes as the voltage is varied as shown in Figure 3.

Blue light of wavelength 360 nm was Q16: Determine the threshold
incident on a potassium metal plate. The frequency of potassium.
photo electrons emitted from the W = hfo
potassium plate were collected by a fo = W/h
cathode and anode at varying voltages to = 2.3/4.14 x 10-15
obtain the curve in Figure 3. Potassium = 5.56 x 1014 Hz
has a work function of 2.3 eV.
Q18: Which of the
following (A-D) would
occur if the frequency
was decreased to less
than the threshold
frequency?
A Increased
photocurrent
B Decreased
photocurrent
Q17: UV light of 200 nm was now shone onto the C Lower stopping
potassium plate at the same intensity striking the potential
cathode. Sketch the resulting curve on the graph in D No signal.
Figure 3 above.

3.5 Electron Energy
When initially placed into the field, the
Before proceeding with more on electron will possess Electrical Potential
the Photoelectric Effect a short Energy given by the product of the
detour into the world of electron charge on the electron (q) and the size of
energy is required. the voltage difference or accelerating
When an electron is placed in a voltage (V) So, E.P.E. = qV
region where a voltage difference At this point the electron is stationary
exists (ie. an electric field), it will so its KE =0
undergo an energy conversion
in passing through that voltage
difference.

The electron arrives at Plate 2 KE
KE = 0 KE KE KE
where all its initial E.P.E. has e- e- e- e- e-
been converted to K.E. Thus EPE EPE EPE EPE
E.P.E.at start = K.E.at finish EPE
=0
Mathematically:
qV = ½mv2
Plate 1, Plate 2,
V=0 V = +V

3.6 Energy Units
• Electrons are tiny and the amount of
energy they carry, even when exposed
to accelerating voltages of tens of K
e- E
millions of volts, is also tiny. e-
• In order to deal with this situation the
normal energy unit of Joules (a very
large unit) can be replaced by a EPE
special energy unit called the
“electron-Volt” (eV).
• Plate 1, V = 0 Plate 2, V = +1V
The electron-Volt is defined as the
energy change experienced by an
electron in passing through a Voltage
On Plate 1 at Plate 2 thethe e-of
At arrival the E.P.E. of K.E.
of 1 Volt.
= qVe- hasx 10-19 J = 1 eV eV
the = 1.6 increased by 1
• Mathematically:
E.P.E. = qV Thus an electron passing through
= (1.6 x 10 )(1)
-19
1 thousand volts will have
= 1.6 x 10-19 J increased its energy by 1keV and
Thus 1 eV = 1.6 x 10-19 J an electron passing through 10
million volts will have increased its
energy by 10 MeV

• The ENERGY of the incoming
photons depends upon the The Energy carried by the incoming
FREQUENCY OF THE LIGHT. So, Photon is TOTALLY ABSORBED by
blue photons are more energetic the electron with which it collides.
than red ones.
• Mathematically: Some of the photon energy is used free
E = hf the electron from the metal lattice and the
where E = Energy (J or eV) rest is converted into the Kinetic Energy
h = Planck’s Constant of the ejected electron. This can be
f = Frequency (Hz)
summarised mathematically as:
• Planck’s constant can have 2 values
depending upon the energy units
KEMAX = hf - W
used:
If Joules: h = 6.63 x 10-34 J s; where:
if eV: h = 4.14 x 10-15 eVs KEMAX = Maximum KE of ejected
electron (J or eV) hf =
Photon Energy (J or eV) W=
Work Function (J or eV)

The term, W, called the Work Function, is
the energy needed to bring the electron
from within the metal lattice to the surface
before it can be ejected.

A plot of the KEMAX 1. Remember KE MAX = hf - W, 3. Different metals have
of the ejected thus a graph of KE MAX vs different cut off
electron versus the frequency is a straight line frequencies.
frequency of the graph of the form y = mx + c
4. Different metals have
incoming photons is with m = h and c = W
different work functions.
shown below. This 2. When KEMAX = 0, hf = W
graph has a number and this frequency is called 5. All metals will produce
of important the “cut off frequency” (fO) graphs with the same
features: slope = h (Planck’s
for that particular metal.
constant)
Light below this frequency,
no matter how intense, will never
liberate electrons from this
K.E.MAX metal.
(eV or J )
Slope = h Slope = h

fo fo
Frequency (Hz)

W

W

Susan and Peter conducted a photo-
electric experiment in which they used a
light source and various filters
to allow light of different frequencies to fall
on the metal plate of a photo-electric cell.
The maximum kinetic energy of any emitted
photo-electrons was determined by
Figure 3 shows the stopping
measuring the voltage required, VS
voltage, VS, as a function of the
(stopping voltage), to just stop them
reaching the collector electrode. The frequency (f) of the light falling on
apparatus is shown in Figure 2. the plate.

Table 1 shows the work
functions for a series of metals.

Q19: Use the information above to
identify the metal surface used in
Susan and Peter’s experiment.
The work function corresponding to
sodium could be found by drawing a
line of best fit through the points on the
graph to determine the y-intercept.

Q20: Use the results in Figure 3 to
calculate the value for Planck’s constant
that Susan and Peter would have
obtained from the data.
You must show your working.

The gradient of the line of best fit gave Planck’s
constant as approximately 4.5 x 10-15 eV s.

Measurements of the kinetic energy of
electrons emitted from potassium
metal were made at a number of
frequencies. The results are shown in
Figure 1.
Q21: What is the minimum energy
of a light photon that can eject an
electron from potassium metal?
1.8 eV
The work function for silver
metal is higher than the work
function for potassium metal.

Q22: Which one of the graphs (A-
D) would best describe the result
if the experiment was repeated
with silver metal instead of
potassium metal?

Some students are
investigating the
photoelectric effect. They
shine light of different
wavelengths onto a
rubidium plate. They
measure the maximum
kinetic energy of
photoelectrons emitted
from the plate. Their data
of maximum kinetic energy Q23: From the data on the graph, what is the
of ejected photoelectrons minimum energy, W, required to remove
as a function of the photoelectrons from the rubidium plate?
frequency of incident light 2.0 eV
is shown in Figure 1. The students shine light of wavelength λ = 400 nm
In answering the following onto the rubidium plate.
questions, you must use Q24: From the graph, with what maximum kinetic
the data from the graph. energy would the photoelectrons be emitted?
Take the speed of light to
1.0 eV
be 3.0 × 108 m s-1

PHOTOELECTRIC EFFECTS OBSERVED WAVE MODEL PREDICTIONS

1. Below a certain frequency, 1. The energy of the light beam
different for different metals, NO arrives uniformly and continuously at
photoelectrons are observed, no the metal surface. If the intensity
matter how INTENSE the light. increases, the energy available to the
electrons increases, so KEMAX of the
ejected electrons should also
increase.
2. The KEmax of the ejected 2. The energy of the light beam is
photoelectrons is INDEPENDENT completely described by its INTENSITY,
of the INTENSITY of the incident so the KEMAX of the ejected
light, but DEPENDENT on the photoelectrons should be INDEPENDENT
FREQUENCY of that light. of Frequency.

3. Because of the nature of the energy
3. The photoelectrons are emitted
delivery process, electrons should take
IMMEDIATELY the metal is exposed to
SOME TIME to build enough energy to
light above the cut off frequency.
be ejected. A TIME DELAY should exist.

These anomalies between the observed and predicted results meant the death of the wave
theory as a complete explanation for light behaviour. A new theory was required.


The students use a light source that Q25: Comment whether this
emits a large range of frequencies. experimental evidence supports
They use filters which allow only the wave like or the particle-like
certain frequencies from the source theory of light.
to shine onto the plate. Most of the
students filters produce frequencies The experimental evidence supported
below the cut-off frequency. the particle theory.
Alice says that if they increase the
The wave theory predicts that
intensity of light, these frequencies
photoelectrons would be emitted at
below the cut-off
any frequency if the intensity was
frequency will be able to produce
sufficient.
emitted photoelectrons.
They experiment and find Alice is the particle theory, the energy of the
incorrect. photons is related to the frequency,
not the intensity, so there would be
no electrons emitted below the
threshold frequency.

As mentioned previously it Photons are neither Photons can be
was in 1905 that Einstein waves nor particles, visualised as a series of
proposed that light travels having properties individual particles each
as a series of discrete units similar to particles when of which displays some
or particles or “quanta” travelling through a wave like properties.
which he called “photons” vacuum and when in a
and each carried a discrete gravitational field, while
amount of energy also having properties
dependent upon the light’s similar to waves when
Individual Photon
frequency. (ie E = hf). refracting and
interfering.

Light beam

Light Beams can be
regarded as a series of
photons radiating out
from the source.

3.11 Practical Applications
The photoelectric effect has
practical applications in many
areas:
Solar Cells

Spacecraft
The photoelectric effect will cause
spacecraft exposed to sunlight to
develop a positive charge.
This can get up to the tens of volts.
This can be a major problem, as
Night Vision Goggles other parts of the spacecraft in
shadow develop a negative charge
(up to several kilovolts) from
nearby plasma, and the imbalance
can discharge through delicate
electrical components.

Chapter 4
Topics covered:
• Investigating the Electron
• Electrons and Matter as Waves.
• Electron Interference
• Electron Diffraction
• Electrons and X Rays
• Photon Momentum.
• Photons as Waves.

4.0 Investigating the Electron
• Electrons are stripped from atoms
and then used as individual
particles in many applications.
• One important application is in
Cathode Ray Tubes - the basis for
Cathode Ray Oscilloscopes
• The electron is the smallest of the 3 (CRO’s), T.V. and non LCD video
fundamental particles which make display units.
up all atoms. • These devices rely on a stream of
• It carries the basic unit of electric “energetic” electrons, which are
charge (1.6 x 10-19 C) “boiled off” a hot wire filament.
• It has a mass of 9.1 x 10-31 kg • Their energy is then increased by
• When in an atom, the electron being accelerated through a large
circulates around the nucleus BUT Voltage.
only in certain allowed positions or • Their trajectory can be manipulated
orbits. by using magnetic or electric fields.
• This restriction on position means
that electrons are behaving more like
waves than the particles we know
them to be.
• It was not until the development of
“Quantum Mechanics” that the
problem of restricted position was
satisfactorily explained.

4.1 Electrons and Matter as Waves
In 1924 French Physicist
Louis DeBroglie suggested
that “the universe is
symmetrical” so, if radiation (light) A racing car, mass 800 kg, travelling at
displayed both particle and wave 200 kmh-1 (55.6 ms-1) has a DeBroglie
like properties simultaneously, wavelength of:
matter should also display similar λ = h/p = h/mv
= 6.63 x 10-34/(800)(55.6)
behaviours.
= 1.5 x 10-38 m
• DeBroglie suggested that matter (with An indescribably small number. Too
mass, m, moving with velocity, v), will small to measure or even notice.
have a wavelength (called the DeBroglie
wavelength) associated with its motion.
v
• This wavelength is calculated from the
momentum equation: An electron of mass 9.1 x 10-31 kg travelling
λ = h/p = h/mv at 1.4 x 104 ms-1 has a DeBroglie wavelength
where λ = DeBroglie wavelength (m) of:
h = Planck’s Constant λ = h/p = h/mv
p = Particle Momentum = 6.63 x 10-34/(9.1 x 10-31)(1.4 x 104)
(kgms )
-1
m = Mass (kg) = 5.2 x 10-8 m
v = Velocity (ms )
-1
A measurable quantity, so electrons are
able to display wave like properties like
interference and diffraction.

Question Type: De Broglie Wavelength
A sketch of a cathode ray tube (CRT) is Q26: Calculate the de Broglie
shown in Figure 5. In this device, electrons wavelength of the electrons. You
of mass 9.10 × 10-31 kg are must show your working.
accelerated to a velocity of 2.0 × 10 m s . A
7 -1
λ= h/mv
fine wire mesh in which the gap between the
wires is w = 0.50 mm the de Broglie wavelength was
has been placed in the path of the electrons, found to be 3.64 x 10-11 m.
and the pattern produced is observed on the Q27: Explain, with reasons,
fluorescent screen. whether or not the students
would observe an electron
diffraction pattern on the
fluorescent screen due to the
presence of the mesh.
For diffraction, the gap width
must be the same order of
magnitude as the wavelength.
Since the wavelength was much
smaller than the gap, no
diffraction pattern would be
observed.

Question Type: De Broglie Wavelength
Neutrons are subatomic particles and, like electrons, can
exhibit both particle-like and wave-like behaviour.
A nuclear reactor can be used to produce a beam of
neutrons, which can then be used in experiments.
The neutron has a mass of 1.67 × 10-27 kg.
The neutrons have a de Broglie wavelength of 2.0 × 10-10 m.

Q28: Calculate the speed of the neutrons.
λ= h/mv gives
2.0 × 103 ms–1 (or 1985 ms–1)

Q29: The neutron beam is projected onto a
metal crystal with interatomic spacing of 3.0 ×
10-10 m.
Would you expect to observe a diffraction
pattern? Explain your answer.
There would be a diffraction pattern because
the wavelength is of the same order of
magnitude as the interatomic spacing.

4.2 Electron Interference
The now familiar Young’s Initially, with only small
.
. .. . double slit experiment is numbers, electrons arrive
performed with the light at the screen in a random
source replaced by an fashion.
“electron gun” which fires Even as the numbers reach
a single electron at a time. the thousands still nothing
unusual appears on the
screen.
However as the numbers
climb past 5000 the familiar
light and dark bands begin
to appear
The total exposure time
8 electrons 270 electrons from picture (a) to the
picture (d) was 20 min.

This result shows
particles (in this case
electrons) can display
wave like properties.
2000 electrons 6000. electrons

4.3 Electron Diffraction
Just as waves diffract
when hitting a solid
object so electrons
can diffract from a
well-ordered
arrangement of atoms
on the surface of a
sample.

Electrons which have been
accelerated through a potential
of 30 to 500 volts (i.e., have
The arrangement of the K.E.’s of 30 to 500 eV), have a
spots is interpreted to Platinum Sample de Broglie wavelength between
provide information about 2.2 x 10-10 m and 0.5 x 10-10 m.
the ordered arrangement of This fits nicely into the range of
atoms on the surface and distances between atoms in
the distances between the solids and can therefore
spots gives information on strongly diffract from them.
the distance between the
atoms. e- energy = 65 eV

Question Type: Electron Diffraction
A beam of electrons, pass through two
very thin slits (approximately 10-10 m)
Q30: Explain whether the
and are detected on a electron detector
particle model or the wave
screen as shown in Figure 5.
model best explains the
expected observations from
this experiment.

The wave model best explains
this.
The wave model is better:
Interference pattern (a wave
phenomenon) will be
observed.
The Particle model would not
predict the interference
pattern, rather two zones
where the electrons would
strike the screen.

4.4 Electrons and X Rays
Diffraction
Pattern

Al foil

High energy
X Ray Beam
electron beam

A high energy electron
beam is fired at an The distance between the bright lines in
Aluminium target both patterns is the same, meaning the
The diffraction pattern wavelengths of both beams was the same.
shown is produced.
Hence if λX Ray is known, then the De
When the electron beam is Broglie wavelength of the electrons is
replaced with an X Ray also known.
beam a somewhat similar
diffraction is produced

Question Type: X ray Energy
A beam of X-rays, wavelength λ = 250
pm (250 × 10-12 m), is directed onto a
thin aluminium foil as shown in Figure
4a. The X-rays scatter from the foil onto
the photographic film.

Q31 : Calculate the energy, in
keV, of these X-rays.

E=hc/λ
= (4.14 x 10-15)(3.0 x 108)/ 250 x 10-12)
= 4.97 x 103 ev
= 4.97 keV or 5.0 keV

Question Type: X ray Diffraction
After the X-rays pass through the foil, Q32: Explain why the electrons
a diffraction pattern is formed as produce a diffraction pattern similar
shown in Figure 4b. to that of the X-rays.
In a later experiment, the X-rays are Electrons have a de Broglie
replaced with a beam of energetic wavelength which was the same
electrons. Again, a diffraction pattern (or very similar) to the wavelength
is observed which is very similar to of the X-rays.
the X-ray diffraction pattern. This is Q33: Assuming the two diffraction
shown in Figure 4c. patterns are identical, estimate the
momentum of the electrons. Include
the unit.
p=h/λ
= (6.63 x 10-34)/(250 x 10-12)
= 2.7 x 10-24 kgms-1
The diffraction pattern were the
same, so the wavelength of the
electrons must equal that of the X-
rays.

Question Type: X ray & electron Diffraction
In 1927, G.I. Thomson fired a beam of
electrons through a very thin metal foil
and the emerging electrons formed
circular patterns, similar to the pattern
obtained when the exercise was repeated
with a beam of X-rays. The following
diagram shows a comparison of the
patterns obtained.
Q35: If the X-rays have frequency of
Q34: How does this evidence support the 1.5 x 1019 Hz, what is the wavelength
concept of matter waves? of the electrons?
The circular pattern is a diffraction
pattern. With X-rays, the bright lines λ = c/f = 3.0 x 108/1.5 x 1019
represent regions of constructive = 2.0 x 10-11 m
interference alternating with regions of Q36: What is the momentum of an X-
destructive interference. ray photon?
The corresponding electron pattern
shows similar diffraction properties p = h/λ = 6.63 x 10-34/2.0 x 10-11
which suggests electrons have wave = 3.3 x 10-23 kgms-1
properties, as diffraction is a wave
phenomenon.

4.5 Evidence for Photons as Waves
It appears trying to predict the fate of
individual photons is not possible,
Young’s double slit experiment is
but as their numbers build the wave
well known as evidence for the wave
model allows the prediction of their
like nature of light.
average behaviour with great
This experiment creates difficulties
accuracy.
for the particle (photon) model.
Thus photons (en masse) can display
How can individual photons, which
wave like properties although they
must pass through one or other of
are generally regarded as discrete,
the slits, interact with themselves to
individual particles.
produce an interference pattern ?
When the experiment is carried out
at extremely low light intensities, it
is found that, for a while, no
interference pattern is noticed.
However, as time passes the
number of photons arriving at the
screen builds and an interference
pattern does emerge with more
photons going to the areas of
bright bands and less to areas of
dark bands.

4.6 Photon Momentum
• When a light beam strikes a surface, the
individual photons will transfer MOMENTUM
as they are absorbed or reflected and so will
exert a pressure on that surface.
• The size of the MOMENTUM can be
calculated from
p = E/c …………(1)
where, p = Photon Momentum (kgms-1)
E = Photon Energy (J or eV)
c = Speed of Light (ms-1)

• Using Photon Energy, E = hf, Equation 1 can
be rewritten as:
p = hf/c …………(2)
• Using the Wave Equation, c = fλ
• Equation 2 can be rewritten as:
p = hf/f λ
p = h/λ

Question Type: Photon energy
A monochromatic violet light of wavelength
390 nm is emitted by a light source with a
power of 200 Watt.

Q37: How many photons leave this
light source in a 10 second interval?

P = W/t = E/t  E = Pt = (200)(10)
= 2000 J
Each photon has Energy = hc/λ =
= (6.63 x 10-34)(3.0 x 108)/(390 x 10-9)
= 5.1 x 10-19 J
No of photons in 2000 J = 2000/(5.1 x 10-19)
= 3.92 x 1021

Q38: What is the momentum of
each of these photons?

p = h/λ = 1.7 x 10-27 kgms-1

Chapter 5
• Topics covered:
• Quantised Energy Levels for Atoms.
• Emission Spectra
• Absorption Spectra
• Solar Spectrum.
• Quantised Energy Levels
• Wave Particle Duality.

5.0 Quantised Energy
Levels for Atoms
• The Bohr model for the atom,
restricting electrons to The 1st five
certain regions or orbits with electron shells
fixed energy levels, worked with their standing
well in describing atomic
structure and behaviour. waves
• But it was not until DeBroglie
suggested electrons had a
wavelength, was there a
satisfactory explanation of
why fixed orbits and energies
existed.
• DeBroglie suggested that for Disallowed orbit
an electron to survive in an
orbit, it must form a standing
wave which represents a
stable state for the electron.
• Only certain radius orbits
allow the standing wave to be
set up, so electrons can only
exist certain regions.

Allowed Orbit

5.1 Emission Line Spectra
When atoms of substances are exposed to The energy absorbed corresponds
certain kinds of energy sources, the to jumps between the fixed energy
electrons surrounding the nucleus can levels of the electron shells within
gain energy and move up to higher energy the atom
levels, called “excited states”.
The excited electrons
return to their ground
state by emitting a
photon of a certain
frequency (colour)

The emitted photons from all electrons transitions
form “a picture” called an “Emission Spectrum”.
These Spectra are unique to each element and
are used to identify unknowns in a field called
spectroscopy.
Helium Emission Spectra Sodium Emission Spectra

Question Type: Emission Spectra
A class looks at spectra from two
sources. Q40: State the electron mechanism,
i. an incandescent light globe in each of the sources below, that
ii. a mercury vapour lamp produces each spectrum.
They observe that the spectra are i. Incandescent light globe
of different types. ii. Mercury vapour lamp

Q39: State the type of spectrum i Thermal motion of free electrons
seen from each source.
i. Incandescent light globe ii Energy transition of bound electrons
ii. Mercury vapour lamp

i Continuous (or broad)
ii Discrete (or individual lines)

5.2 Absorption Line Spectra
If you look at the spectrum
from a common incandescent
light bulb, you'll see a
relatively smooth spectrum
with no part being brighter or
darker than any other part.
This type of a spectrum is
called a continuous spectrum.
Electrons in an atom can jump between
The pattern of absorption lines in
discreet energy levels by absorbing a
a spectrum will be unique to a
photon.
chemical element so we can use
So certain atoms under certain
absorption lines to detect the
conditions will absorb very specific
presence of specific elements in
wavelengths of light dependent on the
astronomical objects.
configuration of their electrons.
This will remove energy and leave blank An emission spectrum is simply the
spaces (dark lines) in the spectrum. reverse of an absorption spectrum.

Question Type: Absorption Spectra
Figure 5a shows part of the emission
spectrum of hydrogen in more detail.

With a spectroscope, Val examines the
spectrum of light from the sun. The
spectrum is continuous, with colours
ranging from red to violet. However
there were black lines in the spectrum,
as shown in Figure 5b.

Q41: Explain why these dark lines are
present in the spectrum from the sun.

The dark lines exist because photons are
absorbed which correspond to the energy
levels in hydrogen. This indicates the
presence of hydrogen.

5.3 The Solar Spectrum
Solar Spectrum When the “White Light” arriving at
the Earth’s surface is passed
through a prism, it is broken up into
its constituent colours.
This produces the “solar spectrum”

It was later discovered that each
In 1802 English chemist William Wollaston was chemical element was
the first to note the appearance of a number of associated with a set of spectral
dark lines in the solar spectrum. lines, and that the dark lines in
the solar spectrum were caused
In 1814, Joseph Von Fraunhofer by absorption by those elements
independently rediscovered the in the upper layers of the sun.
lines and began a systematic Some of the observed features
study and careful measurement of are also caused by absorption in
the wavelength of these features. oxygen molecules in Earth’s
In all, he mapped over 570 lines. atmosphere.

5.4 Quantised Energy Levels
The energy is associated with
Electrons surrounding the nucleus these “jumps” is given by E = hf
can “jump” up and down between where
allowed energy levels by absorbing E = Energy of the emitted photon
or emitting a photon. for a downward jump OR the
The ABSORPTION and EMISSION energy absorbed from an incident
SPECTRA for atoms show the photon for an upward jump. (eV)
energy values associated with the h = Planck’s Constant
“jumps”. f = Frequency of Photon
(Hz)
Mercury Energy Levels

Ionisation
10.4 eV n=5
8.8 eV n=4
6.7 eV n=3
Electronsabsorbing State5(n = 3 can of
States in = 2 to n = represent
The n excited state n 10.4
Electrons Ground more than = 1) eV
4.9 eV n=2 energy from a photon1.8 eV by: be
represents thea state electron
2. By emitting lowest photon
return to the groundcollision willfor
allowed energy levels
stripped completely4.9 eV photon
Emitting a single 6. 7the Mercury
electrons a from eV atoms.
1. followed byin Mercuryphoton
energy level
atom leaving it ionised
OR
0 eV n=1
Ground State

Question Type: Energy Levels
The spectrum of photons emitted by Q42: What is the lowest energy
excited atoms is being investigated. photon that could be emitted
Shown in Figure 6 is the atomic energy from the excited atoms?
level diagram of the particular atom The smallest energy gap was
being studied. Although most of the that from n = 3 to n = 2, an
atoms are in the ground state, some energy difference of 1.8 eV.
atoms are known to be in n = 2 and n =
Q43: Calculate the wavelength of
3 excited states.
the photon emitted when the atom
changes from the n = 2 state to the
ground state (n = 1).
Data: h = 4.14 × 10-15 eV s ,
c = 3.0 × 108 m s-1
The energy of the photon was 3.4 eV.
By substituting this into the equation
E =hc/λ,
the wavelength was 3.65 x 10-7 m.

Question Type: Energy Levels
Figure 2 shows the energy levels
of a sodium atom. A sodium atom is initially in an n = 4
excited state.
Q44: Calculate the highest frequency
of light that this sodium atom could
emit.
ΔE = 3.61 = h f
Hence f = 8.7 x 1014 Hz

Figure 2 shows that electrons in a Electrons have an equivalent or de
sodium atom can only occupy specific Broglie wavelength.
energy levels.
Q45: Describe how the wave nature of Only orbits of an integral number of
electrons can explain this. wavelengths are allowed as these
will form a standing wave

Question Type: Energy Level Diagrams
Part of the visible region of the
spectrum of light emitted from excited
hydrogen gas has three lines as shown
in Figure 3.
The energy level diagram for the
hydrogen atom is shown in Figure 4.
The binding energy is 13.6 eV.
Q46: What is the energy of the
photons with a wavelength 434.1
nm in Figure 3?
E = hc/λ
= 2.86 eV
Q47: A different photon has an energy
of 3.0 eV.
On Figure 4 indicate with an arrow the
electron transition that leads to
emission of a photon of light with this
energy.

5.5 The Wave - Particle Duality
• As you can see from what has been
presented, attempting to
characterise light as only wave like
or only particle like in nature has
failed.
• In fact, as of today, light (and matter)
are regarded as some part particle
like and some part wave like in
nature.
• This conflict is summarised in the
use of the term “wave-particle
duality” to describe light’s known BUT THIS IS WHAT PHYSICS IS ALL
behaviour. ABOUT –
• This is an unsatisfactory situation, PUSHING THE ENVELOPE –
as it defies our need for simple all EXPLORING THE UNKNOWN –
encompassing explanations for the TRYING TO FIT RATIONAL AND LOGICAL
behaviours we observe around us
and in the universe. EXPLANATIONS TO THAT WHICH HAS
BEEN OBSERVED.
• It would appear, at this stage, we
have not yet unravelled all the details
of the behaviour of light and matter.

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