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Modern Physics
By
Neil Bronks
Atoms
Mass Number Number of protons +
Neutrons.
12
6
C
Atomic Number Number of protons
In a neutral atom the number of electrons and protons are the
same. In Carbon it is……… 6
Hydrogen
The simplest
atom has one
negative
electron orbiting
one positive
proton.
The electron is
very light
compared to the
proton.
Electron
Proton
Helium
In this atom we
see two neutrons
and two protons
forming the
nucleus.
The Neutron has
no charge but is
the same mass
as the proton.
Electron
Strong Nuclear –
Holds nucleus together
Electromagnetic –
Positive and negative
- Very Short Range
4 Forces of Nature
(Order of strength)
Weak NuclearAssociated with beta
decay
Gravitational
- Only Positive
- Very long range
Radiation
J’ai fait ça !
 Decay of nucleus by the emission of a
particle or a ray.
 Discovered by Henri Becquerel
 Units 1 Bq is one decay per second
 Natural happens without outside
bombardment
 Artificial happens due to bombardment
Safety
Wear Gloves or
Apron of lead
Don’t point at
anyone
Don’t eat!!!
Nuclear Equations
Top and bottom must add up
Top is mass number
Bottom is atomic number
Proton HH Neutron
1
1
And Alpha
1
0
n
electron
U  Th He
238
92
234
90
4
2
0
1
e
Alpha Particles 
U  Th He
238
92
234
90
Helium Nuclei
Positive Charge
Heavy so not very penetrating
Very Ionizing
14 N
7
+42He 178O + 11H
4
2
Beta Particle ß




Th Pa 
234
90
234
91
Fast electron from the nucleus
Negative charge
Moderately Penetrating
Moderately ionizing
14
14 O + 0 ß
N

7
8
-1
0
1
Gamma Ray 
 High energy e-m wave (A Photon)
 No charge - not deflected by field
 Very penetrating – Need lead to stop most
of them
 Not very ionizing
 Release energy after reaction
Penetrating Power



Paper
Al foil
Concrete

H/W
LC Ord 2007
Q11
Charged particles move in a
circular path as the force is
always at right angles to the
direction of motion-
Particles in Fields


Fleming's Left Hand Rule

Radioactive
Source
Cloud
Chamber
Click here for internet demo
Ionisation
We can prove that
radiation creates ions
as we bring a source
close to a charged
electroscope

Ionising Power
Alpha is heaviest and so does most
damage – poison with Polonium
Beta is only moderately ionising
Gamma is only slightly ionising but
difficult to stop
Solid State Detector
This a P-N junction in reverse bias.
This creates a huge depletion layer.
-
+
P
N
A piece of radiation passes through the depletion
layer and creates enough electron -hole pairs to
allow one pulse of current.
Geiger
Muller
Tubeincident
particle
ionises
gas
H/W
LC Ord 2004
Q10
Experiments
All experiments the same stick a
DETECTOR in front of a source and
count the decays.
Move it away for distance and plot
Time for half life and plot
Put things in front for penetration
Penetration
A Gieger Muller
Tube and Counter.
Plot the activity against the thickness or
the type of barrier
Distance
r
A Gieger Muller
Tube and Counter.
Plot the activity against the distance r.
Half Life
Time it takes for half
the atoms to decay
A Gieger Muller
Tube and Counter.
Plot the activity against the time
Half-Life – time it takes for half the
radioactive particles to decay
Atoms
Not
Decayed
Time
1
2
3
4
Half life demo from internet click
here
Half Life Calculations
4000 particles
time=0
2000 particles
time=3s
1 half-life
1000 particles
500 particles
time=6s
time=9s
2 half-life
3 half-life
250 particles
time=12s
4 half-life
125 particles
time=15s
5 half-life
Calculations – we use the decay
constant λ in our calculations.
=0.693/T½
=0.693/3s
-1
=0.231s
Activity Calculations
Rate of Decay =
 x number you started with
dN/dt = -
 xN
Start with 4000 particles and =0.231
Activity = 4000 x 0.231=924 Bq
Calculations
1) You start with 100 grams of sulfur-35,
which has a half life of 87.51 days. How
much time will it take until only 12.5
grams remain?
How many half lives?
100>50>25>12.5 so 3 half lives
Time = 3 x 87.51 = 262.53 days
Calculations
2) You measure the radioactivity of a substance,
then when measuring it 120 days later, you find
that it only has 25 % of the radioactivity it had
when you first measured it. What is the half life of
that substance?
How many half lives
100%>50%>25%
2 half lives =120 days
1 half life = 60 days
Calculations
 3) Your professor gives you 64g of phosphorus32 (half life = 14.263 days).
 (a) What is its decay constant ?
 (b) What is its activity (Rate of Decay)?
(a) Using the formula =0.693/T½
 =0.693/(14.263x24x60x60)
 = 6.62 x 10-7 s-1
Calculations
 3) Your professor gives you 64g of phosphorus32 (half life = 14.263 days).
 (a) What is its decay constant ?
 (b) What is its activity (Rate of Decay)?
= 6.62 x 10-7 s-1
(b) Using Activity =dN/dt = -N
N= Moles x 6x1023 = 2 x 6x1023
Activity = 6.62 x 10-7 x 12x1023 =
= 7.3 x1017 Bq
Isotopes
Same
atomic
number
different
mass
number
Isotopes
Same
Atomic
number
different
Mass
number
Carbon-14 Dating
 At death all animals contain the same ratio
of C-14 to C-12
 The rate of decay of C-14 is fixed
 The C-14 left tells us how long ago it died
%C-14
time
Isotope pp
Uses of Radioactive Isotopes
Medicine – treatment and imaging
Smoke detectors
Food Irradiation
Carbon-14 Dating
H/W
LC Ord 2005 Q12(d)
LC Higher 2003 Q11
LC H 2007 12(d)
Do I look like
Freddie?
Rutherford Scattering
Rutherford on internet
Rutherford Scattering – alpha
particles fired at gold foil.
Most pass unaffected - So the nucleus is very small
Rutherford Scattering – alpha
particles fired at gold foil.
Nucleus
A small number of high energy alphas are Deflected
Some reflected completely back - Nucleus totally positive.
Rutherford Scattering – alpha
particles fired at gold foil.
Nucleus
A small number of high energy alphas are Deflected
More pass unaffected - So the nucleus is very small
Some reflected completely back - Nucleus totally positive.
Cockcroft and Walton
Hydrogen
discharge tube
Proton
Accelerated by
An huge electric
Field (700000v)
Alpha strikes the
screen
Producing a flash
that
Is seen with the
microscope
Alpha 
Alpha

Lithium
Target
Alpha
Alpha 
Internet explanation
Nobel Prize for Physics
 Proton + Lithium  2xAlpha + Energy
 Proves Einstein’s Law E=mc2
 First Transmutation by artificial
Bombardment of an element
7
3
Li  p     Energy
1
1
4
2
4
2
Ernest Walton
Binding Energy
 The total nucleus weighs less than all its
parts
 Difference is Mass Defect
 Converted to energy to hold the nucleus
together
2
E=mc
 As Iron is the most stable as you go towards it
you release energy
 So Carbon-12 is lighter than 12 protons
 The difference is the binding energy
Binding Energy of a Deuteron
 A deuteron is the nucleus of a deuterium atom, and
consists of one proton and one neutron. The masses of
the constituents are:
 mproton = 1.007276 u (u is Atomic mass unit)
 mneutron= 1.008665 u
 mproton + mneutron = 1.007276 + 1.008665 = 2.015941 u
 The mass of the deuteron is:
 Atomic mass 2H = 2.013553 u
 The mass difference = 2.015941 - 2.013553
= 0.002388 u
Convert to Kg
Multiply by conversion factor
1u = 1.66x10-27 Kg
Mass = m = (0.002388) x 1.66x10-27
Mass = m = 3.96x10-30 Kg
Use Famous Formula
E=mc2
E=
3.96x10-30 Kg
x
(3x108 m/s)
E = 3.56x10-13 Joules
2
Fusion – The sun and the stars
 Fusion is the joining together of 2 light nuclei
to make one nucleus with release of energy.
 Caused by a super fast collision at high
temperature in a magnetic bottle.
2
1H
2
1H
Fission
 The breaking apart of a heavy nucleus to form
smaller nucleus with release of energy.
 Caused artificially by the bombardment of
the right speed of neutron.
 In both fusion and fission the products are
lighter than the reactants and the MASS
DEFECT is turned into Energy E=mc2
Also produced 3 fast neutrons that can
cause another fission and so a chain
reaction
1. Subtract mass
in a.m.u.
2. Convert to kg
3. Use E=mc2
Uranium-235
Nuclear Equation
U  n Kr  Ba 3 n  Energy
238
92
1
0
97
36
139
56
1
0
In the isotope U-238 the neutrons must be
slowed down by a moderator - Graphite
Moderators
(Graphite) slow
down the neutrons
to the right speed
Fuel rods contain the
Uranium-235 (Enriched to
ensure chain reaction)
Control Rods (Boron
Steel) absorb
neutrons to stop the
reaction and prevent
meltdown
Heat to heat
exchanger
prevents
Radiation
escaping
steam to
turbine
H/W
LC Ord 2006
Q 9
Leptons
Fundamental particles
1/1846 of an a.m.u.
Does not feel the strong nuclear force
Matter – Electron , Muon, Tau, ……
Anti-matter – Positron, Anti-Tau
Anti-matter first suggested by Paul Dirac
Annihilation
Matter combining with anti-matter to
form energy in the form of e-m
radiation
e+
ee+ + e-
2hf (2)
2 photons conserve momentum ?
Annihilation
2 photons conserve
momentum
Matter combining with anti-matter to
form energy in the form of e-m
radiation
e+
e-
An electron and a positron collide to make energy.
All the mass of the electrons gets turned into
gamma waves
Matter turns
2
So Energy E=mc
Into energy
To find frequency of wave E = 2h.f
Wave made by Anihilation
 A proton and a Anti proton. The masses of the
constituents are:
 mproton = 1.007276 u (u is Atomic mass unit)
 mproton + manti = 1.007276 + 1.007276 =
2.014552 u
 The mass difference = 2.014552 u
 To use this in a calculation we covert to kg
Convert to Kg
Multiply by convertion factor
1u = 1.66x10-27
Mass = m = (2.014552) x 1.66x10-27
Mass = m = 3.34x10-27 Kg
Use Famous Formula
E=mc2
E=
3.34x10-27 Kg
x
(3x108 m/s)
E = 3.01x10-10 Joules
2
Use Planks Equation
E=hf
h= planks constant
3.01x10-10 Joules = (6.6x10-34 js)x(f)
f= 3.01x10-10 Joules / 6.6x10-34 js
= 4.56x1023 Hz
In practice this is low as KE from particles
increases this.
Pair Production
A matter and anti-matter
pair being created by energy
from an e-m wave
e+
eAn electron and a positron are created from a gamma
ray. (We can also get a proton and an anti-proton)
We do the calculation in reverse
To find energy of wave E = h.f
As we get 2 electrons E = 2mc2
Annihilation and Production
p+
p-
p+
p-
+
0
-
New particles are produced from the KE of the colliding
protons They must conserve charge
If we carry in 4Gev (1.6x10-19 . 4x109= 6x10-10 J)
As Energy to make 3 Pions is E=mc2 =(3x 2.4842x10-28xcxc)
=6.7x10-11 J
Subtracting we find the KE after collision.
H/W
LC Higher 2009 10(a)
Quarks - Inside the Hadrons
6 Quarks
6 Anti-Quarks – Opposite Signs
UP
+2/3
DOWN
-1/3
STRANGE
-1/3
TOP
CHARMED
+2/3
BOTTOM
-1/3
+2/3
Hadrons
Baryons
Mesons
3 quarks
Quark+anti-quark
Proton
Pion
uud
ud
Feels strong nuclear
force
Feels strong nuclear
force
Baryon
Meson
Hadrons on internet
Particle Zoo
Hadrons
Subject
to all
forces
Baryons
Mesons
3 Quarks
Quark +
Anti-quark
Proton
uud
Pion
ud
Leptons
Fundamental
particles
Do not feel
Strong Nuclear
Force
Ghost Particle Mystery
 By 1930 most of the particle physics world was
understood
 However the decay of the neutron producing a
beta particle did not obey conservation of
mass-energy or conservation of momentum
n0 → p+ + e-
Pauli said there must be a new particle called a
neutrino
Beta decay
 The weak force converts a neutron into a proton
while emitting an electron and an antineutrino so
weak force most important to neutrino
n0 → p+ + e- + νe
This explains loss in energy and momentum.
Pauli proposed it’s existence in 1930 but was
not discovered until 1956 as it is so weakly
interacting with other particles. (No charge and
tiny mass compared to electron)
Nuclear Formula
1
0
n p    
1
1
0
1
0
0
Particle Accelerators-Linear
Very high Voltage
electric fields
Electro-magnetic
attraction pulls
particles down.
Circular Accelerators
Particles spiral in fields
(Flemings Left hand
rule)
Cyclotron- We put the
field at right angles e
more power with
oscillating field
CERN
Particles can
travel in
opposite
directions and
double the
collision energy
Magnets force
particles in
circular path so
stay in tube
More velocity
more KE so
more new
particles made
Vacuum to
avoid collisions
increase mean
free path
Circular more
compact
High velocity
needed to
overcome
repulsion
Detectors
H/W
LC Ord 2012 Q11
LC Higher 2010 10(a)
H/W Roundup
2007 q11
2004 q10
2003 q11
2005 q12(d)
2006 q 9
2009 10(a)
(Yeh har)