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CfE Higher Physics Unit 2
Particles and Waves
Higher Physics
Particles and Waves
H
H
He
n
+
Energy
Pupil Notes
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page 1
CfE Higher Physics Unit 2
Particles and Waves
The Standard Model
The Standard Model is a model for classifying sub-nuclear particles and
their interactions.
The Greeks, a philosopher Democritus, came up with the idea of a
fundamental particle, the atom. This word comes from the Greek for
indivisible ‘atmos’. He wondered if he could keep breaking an object in half
indefinitely. He came to the conclusion that eventually you would get
something that couldn’t be split.
Next came John Dalton, we only had to wait 2000 years after the ancient
Greeks. He used his data from chemistry experiments to propose that not
only were substances made of atoms, the atoms of an element were
identical and different elements had different types of atom.
By the end of the 1800’s J.J. Thomson had discovered the electron, so now
physicists knew atoms were actually made of smaller component parts. This
started a race to find out what the smaller parts were.
Orders of Magnitude
The following table is used to determine the size of sub-atomic particles and
universally large objects compared to a human. Humans are taken as the
“standard size” particle.
Particle or Object
Neutrino
Proton
Hydrogen Atom
Dust
Human
Earth
Sun
Doon Academy
Order of Magnitude
≈ 10-24 m
10-15 m
10-10 m
10-4 m
100 m
107 m
109 m
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CfE Higher Physics Unit 2
Particles and Waves
Solar System
Nearest Star
Galaxy
Distance to a Quasar
1013 m
1017 m
1021 m
1026 m
Fundamental Particles
The standard model was developed in the early 1970’s in an attempt to tidy up
the number of particles being discovered and the phenomena that physicists
were observing.
At present physicists believe that there are 12 fundamental mass particles split
into two groups
 Leptons
 Quarks
There are also 4 force mediating particles.
 Photon
 Gauge Boson
 Gluon
 Graviton
Hadrons can be made up of a number of different particles, each containing a
certain number of quarks. The hadrons are called Mesons and Baryons. The best
way to remember how many are in each is:
 Meson  Me –son  Two syllables  Two Quarks
 Baryon  Bar-y-on  Three syllables  Three Quarks
Doon Academy
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CfE Higher Physics Unit 2
Particles and Waves
fermions
u
c
bosons
t

quarks
charm
top
photon
d
s
b
Z
leptons
down
strange
bottom
Z boson
e


W
electron
neutrino
muon neutrino
tau neutrino
W boson
e


muon
tau
g
electron
force carriers
up
gluon
Charge and Antimatter
Every particle/quark has an associated antimatter particle. The purpose of the
antimatter particle is to cancel out the original particle. The antimatter
particles have the same mass, but opposite charge to their original particles.
 Up  Anti-Up
 Electron  Positron
 Anti-Down  Down
Each quark has an associated charge with it also. When quarks combine to form
particle, the overall charge of the particle is 1.
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CfE Higher Physics Unit 2
Particles and Waves
Example: A particle consists of an up quark, an anti-down quark and a charm
1
quark. The up quark has a charge of
and the anti-down quark has a charge of
3
2
 . What is the charge of the charm quark?
3
1  2
     CQ  1
3  3
1
  CQ  1
3
1 3 1
CQ  1   
3 3 3
CQ 
4
3
Fundamental Forces
There are four fundamental forces that dictate how particles interact with
each other whether in close proximity or a massive distance away. The
fundamental forces are:




Strong Nuclear Force
Weak Nuclear Force
Gravitational Force
Electromagnetic Force
The strong and weak nuclear forces only act over a very small range – falling
away to zero just outside the nucleus of an atom. The electromagnetic and
gravitational forces can act over an infinite range.
In the nucleus of an atom where there is more than one proton, the
electromagnetic force tries to repel the protons from the nucleus. The strong
nuclear force is able to overcome this and keep the nucleus together. Weak
nuclear force is felt during beta decay – when an electron is released from the
atom.
Doon Academy
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CfE Higher Physics Unit 2
Doon Academy
Particles and Waves
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CfE Higher Physics Unit 2
Particles and Waves
Electric Fields
Links to National 5 – Energy Conservation
In Physics, when we talk about a field we mean a region within space in which an
object experiences a force without being touched. We do not mean a place
where cows graze!!!
In a gravitational field it is a mass that experiences a force. In an electric field
it is a charge that experiences a force.
Electric field lines show the direction and strength of force that a charged
particle will experience. Electric field lines will always go from positive to
negative. If two particles of the same charge are brought together, they will
repel each other’s electric field.
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Particles and Waves
An electric field can be used to accelerate a charge particle from one end to
the other. It is accelerated because it experiences a force between the two
electric plates.
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Particles and Waves
Definition of a Volt
If 1 joule of work is done in moving 1 coulomb of charge between two points,
there is a potential difference of 1 volt between the points.
Ew  QV
Ew = Work Done (J)
Q = Charge (C)
V = Voltage (V)
The work done is the minimum amount of energy required to move the particle
from the lower plate to the upper plate and keeping it there. When the particle
is released it will accelerate back to the lower plate. Using the conservation of
energy, the final speed of the particle as it reaches the plate can be calculated.
EW  Ek
1
QV  mv 2
2
v
2QV
m
Magnetic Fields
Charged particles can also experience a force when placed into a magnetic field.
This force will cause a charged particle to move in a different direction –
usually in a circular motion. The speed of the particle does not change this time
The direction in which the charged particle will experience the force once in
the magnetic field can be determined using the “right hand rule”:
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Particles and Waves
Current – original direction of travel for charge particle.
Field – the direction of the magnetic field lines
Motion – the direction in which the particle will be forced.
N.B. – The right hand rule only applies for NEGATIVELY charged particles. For
positively charged particles, the force will in the opposite direction.
Current – Right
Field – Down
Motion – “Out of the page”
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Current - Down
Field – “Into the page”
Motion - Left
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Particles and Waves
Radioactive Decay
Links to National 5 – Alpha, Beta & Gamma; Ionising Radiation; Half-Life; Fission
and Fusion
Radioactive Particles – Atomic and Mass Numbers
All stable nuclei have a set number of protons and neutrons. The atoms then
consist of set numbers of electrons orbiting at different levels. If an atom
becomes unstable, it will emit a particle or energy in the attempt to stabilise
itself.
The number of particles that a nucleus has can be determined using information
from the periodic table. The information is always given in the following format:
X
Z
A
A = the element
X = Mass Number (protons and neutrons)
Z = Atomic Number (protons only)
For example
45
21
Sc
Element is Scandium
It has 21 protons
It has an atomic number of 45  24 neutrons
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Particles and Waves
Radioactive Decay
An unstable atom can emit either particles or energy in an attempt to stabilise
itself. There are three types of ionisation radiation that are emitted:
 Alpha (α)– the nucleus of a helium atom with the atomic arrangement of
4
He
2
 Beta (β) – a fast moving electron. These are emitted when neutrons decay
into protons and electrons. The atomic arrangement is 10 e
 Gamma (γ) – pure energy. This radiation has no mass or charge associated
with it.
Examples – Alpha and Beta Decay
Ac  219
Fr 24 He
87
223
89
On either side of the arrow, the numbers on the top
and bottom lines must equal each other i.e. for alpha
decay total of 223 on the top and 89 on the bottom
Po 215
At  01 e
85
215
84
Nuclear Fission and Fusion
Nuclear fission is the splitting of an atomic nucleus into two smaller nuclei. This
process releases a neutron and some energy. There are two types of nuclear
fission:
 Induced – a neutron is fired into an atomic nucleus causing it to split.
 Spontaneous – an atom will split on its own.
Nuclear fusion is the merger of two atomic nuclei to form a new substance. In
both processes, fission and fusion, energy is given out in order to stabilise the
newly formed substances. This energy is released because a small amount of
mass is lost during the reaction.
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Particles and Waves
The amount of energy being released can be determined using, probably, the
most famous Physics equation:
E  mc2
E = Energy released
m = Mass lost in reaction
c = Speed of light
Example
Plutonium undergoes the following fission reaction:
Pu 01 n137
Te 100
Mo 301 n
52
42
239
94
The masses of the nuclei and particles involved are shown below
Particle
Mass (kg)
n
1.675 x 10-27
Pu
396.741 x 10-27
Te
Mo
227.420 x 10-27 165.809 x 10-27
Calculate the energy released by this reaction.
N.B. This is the only question in which you do not round the masses when added
together. Only your final answer can be rounded to an appropriate decimal
place.
Mass Before = 1.675 x 10-27 + 396.741 x 10-27 = 398.416 x 10-27 kg
Mass After = 227.420 x 10-27 + 165.809 x 10-27 + (3 x 1.675 x10-27) = 398.254 x
10-27 kg
Mass Lost = (398.416 – 398.254) x 10-27 = 0.162 x 10-27 kg
E  mc2
2
E  0.162x1027  3x108 
E  1.458x1011 J
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Particles and Waves
Interference
Links to National 5 – Wave Characteristics
Interference occurs when two waves meet. Depending on how these waves meet
will determine what is observed.
Constructive Interference
Constructive interference occurs when two waves meet in phase. If two waves
are in phase this means that a peak will meet a peak/ a trough will meet a
trough. The two waves will add together to produce a larger wave. This results
in a sound wave appearing louder in certain places for example.
Destructive Interference
Destructive interference occurs when two waves meet when they are not in
phase. This means that a peak will line up with a trough. The causes the wave to
cancel itself out and nothing will be heard for example.
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Particles and Waves
Coherent sources
Sources are coherenet if they have the same frequency and are in phase with
each other. It is possible to create to coherent waves using only one source. If
they source wave is passed through two small gaps spaced, roughly, one
wavelength apart, it can create two coherent sources.
Path Difference
When using two different output sources, this can produce either coherent or
incoherent waves, in turn creating constructive or destructive interference. An
example of this is walking between two loudspeakers that are playing the same
note from the same source. As you move from one speaker to the other, the
note will apear louder and quieter in places.
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This change in interference is caused by the path difference. The path
difference is created as each sound wave has a different distance to travel to
reach your ear. As you move between the speakers, the distance change for
each sound wave, creating points of constructive and destructive interference.
C
On the diagram above, the waves are meeting at point P. The distance the wave
from source 1 travels can be written as S1P. The distance the waves from
source 2 travels can be written as S2P.
Path Difference = S2P – S1P
Point C on the diagram indicates the first area of constructive interference.
This is known as the Central maximum. The path difference will be zero as point
C is the same distance from both sources. As an observer moves from point C
towards point P, they will hear louder and quieter areas as the path difference
is changing. It can be determined whether an area will produce constructive or
destructive interference using the following.
(Constructive) path difference = mλ
(Destructive) path difference = (m + ½)λ
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m = 0,1,2,3… (m = 0 is central)
m = 0,1,2,3… (m = 0 is central)
CfE Higher Physics Unit 2
Particles and Waves
Example
Calculate the wavelength of the following source when P is the 3rd order
maximum.
Path Difference = S2P – S1P
Path Difference = 550 – 520
Path Difference = 30mm
Path Difference = mλ
30 = 3λ
λ = 10mm
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Example
State, with calculation, whether point P is a point of constructive or destructive
interference when the wavelength of the source is 30 mm.
Path Difference = S2P – S1P
Path Difference = 300 – 225
Path Difference = 75 mm
Path Difference = mλ
(always assume to be constructive unless calculated otherwise)
75 = 30m
m = 2.5
m=2+½
Point P is destructive
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Particles and Waves
Diffraction Gratings
A diffraction grating is a thin film of plastic or glass that can be used to cause
an interference pattern using a laser or other monochromatic light source. The
film has thousands of narrow slits etched onto it.
The distance between each slit on the diffraction grating is incredibly small as
there is often thousands of slits per millimetre of film. To calculate the
distance between each slit the following formula can be used.
Slit Separation 
Size of film
Number of slits
Example – There are 15,000 slits per millimeter on a diffraction grating. What
is the slit separation?
1x10-3
Slit Separation 
15000
Slit Separation  6x10-8 m
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Similar interference patterns are created and there will be areas of
constructive and destructive interference (areas or light and dark). On this
occasion a screen has to be placed a certain distance from the source in order
to observe the interference pattern.
The wavelength of light being used can be determined using the angle between
the central maximum and area of interference, as well as the slit separation on
the diffraction grating.
d sin   m m  0,1,2,3... (constructive)
1

d sin    m   m  0,1,2,3...(destructi ve)
2

d = slit separation (m)
Θ = Angle between central maximum and point of interference (°)
λ = Wavelength (m)
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Particles and Waves
Example
Light from a monochromatic source strikes a diffraction grating of 400 lines
per mm. The first order maximum is produced at an angle of 140.
Find a) the slit separation b) the wavelength of the source.
Length
1x10-3
(a) Slit Separation 

 2.5x10-6 m
Number of Lines
400
(b)
m  d sin 
1  2.5 x10  6 sin 14
  6 x10  7 m
Example
Light of wavelength 6.5 x10 -7 m is shone onto a grating. The angle between the
zero and third order maxima is 31.5°. Calculate:
(a)
(b)
The spacing between the slits on the grating.
The number of lines per mm on the grating.
m  d sin 
3  6.5 x10  7  d sin 31.5
(a)
0.52 d  1.95 x10  6
d  3.75 x10  6 m
Slit Separation 
(b)
Length
Number of Slits
Length
1x10-3
Number of Slits 

 266.6 lines per mm
Slit Separation 3.75x10-6
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Refraction of Light
Links to National 5 – Refraction; Total Internal Reflection; Critical Angle
Refraction of light occurs when a ray of light passes from one medium to
another – from air into glass for example. Refraction causes the speed of light
to change (slow down). Even though it causes a change in direction this is not
the meaning of refraction!!!!!!
As the angle of incidence increases, so does the angle of refraction. Each angle
is measured between the ray of light and the normal outside and inside the
glass respectively. At this stage there is no clear relationship between how
much each angle increases by.
However, if the sine of each angle is taken, the is a direct correlation between
each angle.
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The above graph shows that
sin 1
 constant
sin 2
This constant is called the refractive index (n). In air, the refractive index is 1.
sin 1 n2

sin 2 n1
N.B. Be careful not to use Θi and Θr for angle of incidence and angle of
refraction and na and ng for the refractive indices. Some questions can begin
inside a material other than air so n1 may not always be 1.
Example
A ray of red light passes from air into water
The refractive index of water for this light is 1.33. Calculate the angle of
refraction in water.
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n 2 sin  1

n1 sin  2
1.33 sin 70

1
sin  2
sin 70
sin  2 
 0.7065
1.33
 2  45
When light passes from air into glass, the speed decreases and the wavelength
also decreases. Frequency will only decrease if the source of light is changed.
By how much the speed and wavelength decrease by is also linked to the
refractive index a material.
n2 sin 1 1 v1



n1 sin 2 2 v2
The Critical Angle
Whether a ray of light will refract or totally internally reflect whilst inside a
material is all depends on the critical angle.
If the angle of incidence is less than the critical angle it will refract.
If the angle of incidence is equal to the critical angle it will refract at 90°
If the angle of incidence is greater than the critical angle it will totally
internally reflect.
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At the critical angle, the value for refraction in air is 90°
sin 1 n2

sin 2 n1
sin  c 1

sin90 n
sin90
n
sin  c
Since sin 90  1
1
n
sin  c
Example
A ray of red light passes from air into glass. The refractive index is 1.5 for this
light in glass. Show by calculation whether the ray is totally internally reflected
or not.
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1
sin  c
1
1.5 
sin  c
1
sin  c 
1.5
 c  42
n
The angle of incidence (63°) is greater than the critical angle (42°) so the light
will totally internally reflect.
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Irradiance
The irradiance of radiation (light) at a surface is the power per unit area of
surface the light falls on.
I
P
A
I = Irradiance (Wm-2)
P = Power (W)
A = Area (m2)
This can be used for light rays that spread out the further they get from the
source or for point sources i.e. a laser.
Example
A lamp shines onto a surface of area 4 m2. The irradiance at the surface is 0.02
Wm-2. Calculate the power of the incident light.
I
P
A
0.02 
P
4
P  0.08W
The irradiance of a light source will decrease over as the distance from the
source increases. The diagram below shows how the power is spread out thus
reducing the irradiance.
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The relationship between irradiance and the distance from the source can be
given as:
I1d12 = I2d22
The overall combination of irradiance and distance must be the same before and
after the light source has been moved.
Example
A light source produces an irradiance of 0.2 Wm-2 at a distance of 2m from the
light source. What will be the irradiance at a distance of 4m from the source?
I1d1  I2d2
0.2  22  I2  42
0.8
I2 
16
I2  0.05Wm2
2
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CfE Higher Physics Unit 2
Particles and Waves
Photoelectric Effect and Spectra
A beam of radiation can be regarded as a stream of individual energy bundles
called photons.
Each photon will carry a certain amount of energy. The amount of energy it has
is related to the frequency of that photon.
E  hf
E = Energy of photon (J)
h = Planck ’s constant (6.63 x 10-34 Js)
f = Frequency (Hz)
Electromagnetic radiation above a certain frequency can eject electrons from
the surface of some metals. When an electron is ejected from a piece of metal,
this is known as photoelectric emission.
Below a certain frequency, known as the Threshold Frequency, there will be no
photoelectric emission. This is due the photons not carrying enough energy to
release an electron from the piece of metal.
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When the frequency increases, along with an increase of irradiance, the
photoelectric emission increases causing a larger flow in current.
Work Function
Each electron requires a minimum amount of energy needed from a photon
before it will release itself from a piece of metal. This minimum amount of
energy is known as the Work Function.
Example
The work function of a piece of metal is 5 x 10-19J. A photon of frequency 4.48
x 1014 Hz is incident on the metal. Show, by calculation, whether an electron will
be released or not.
E  hf
E  6.63x1034  4.48x1014
E  4.95x1019 J
An electron will not be released as the energy of the photon is less than the
work function.
If a photon is incident on a piece of metal that has an energy which is higher
than the work function, the electron will have a certain amount of kinetic
energy.
Ek  hf  hfo
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Emission Spectra
In the Bohr model of the atom electrons are confined to certain orbits (shells).
Electrons can move between energy levels. Light is emitted when an electron
falls from a high energy orbit to a low energy orbit.
Electrons can also jump from low energy levels to higher ones in a photon is
incident on the atom. Just like with the photoelectric effect, the photon must
have the right amount of energy in order for an electron to jump.
The ground state is the lowest energy level in the atom. The ionistation level is
the point at which an electron is ready to be released if given enough energy.
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All energy levels are in negative values. The ground state will have a larger
numerical value than W1 but is still a lower energy level.
When an electron drops from a higher energy level to a lower one, it can emit
visible light photons. The colour of these photons can be determined using the
frequency produced. However, all forms of electromagnetic radiation can be
emitted from falling electrons.
Example
(a)
(b)
Which transistion produces radiation with the longest wavelength?
Calculate the frequency of the photon produced when an electron falls
from E3 to E2.
(a) Longest wavelength  Shortest Frequency  Smallest energy  E4  E3
(b)
E3  E2   2.4  ( 5.6)x1019  3.2x1019 J
E  hf
3.2x1019  6.63x1034  f
f  4.8x1014Hz
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