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7/28/2019 Section VI 27 Nuclear Physics
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Nuclear Physics 27. Nuclear Physics
Content
27.1 The nucleus 27.2 Isotopes
27.3 Nuclear processes
27.4 Mass excess and nuclear binding energy
27.5 Radioactive decay
Learning Outcomes
(k) show an appreciation of the association between energy and mass asrepresented by E = mc2 and recall and solve problems using this relationship.
(l) sketch the variation of binding energy per nucleon with nucleon number.
(m) explain the relevance of binding energy per nucleon to nuclear fusion andto nuclear fission.
(n) define the terms activity and decay constant and recall and solve problemsusing A = N.
* (o) infer and sketch the exponential nature of radioactive decay and solveproblems using the relationship x = xoexp(t) where x could represent activity,number of un-decayed particles or received count rate.
(p) define half-life.
(q) solve problems using the relation = 0.693/t
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Mass defect At the nuclear level, since the masses are very small, instead of measuring
them in kilogram, we measure the masses of nuclei and nucleons in atomic
mass unit One atomic mass unit (1 u) is defined as being equal to one-twelfth of the
mass ofcarbon-12 atom. 1 u is equal to 1.66 x 10-27 kg
Using this scale we have:
proton mass mp = 1.007276 u
neutron mass mn = 1.008665 u
electron mass me = 0.000549 u
e.g. the mass of a helium-4 nucleus which consists of 2 protons and 2
neutrons should be (2 x 1.007276) + (2 x 1.008665) = 4.031882 u
However the actual mass of a helium nucleus is 4.001508 u, less by0.030374 u
This difference between the expected mass and the actual mass of a nucleus
is called the mass defectof the nucleus(i.e. mass of the product < the sum ofmasses of reactants)
Hence mass defect of a nucleus is the difference between the total mass ofthe separate nucleons and the combined mass of the nucleus
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Example
Calculate the mass defect for a carbon-14 nucleus. The measured mass is
14.003240 u.
Solution
The carbon-14 nucleus contains 6 protons and 8 neutrons,
Hence, total mass of separate nucleons is,(6 x 1.007276) + (8 x 1.008665) = 14.112967 u
Given, combined mass is = 14.003240 u
Therefore the difference is the mass defect = 0.109736 u
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Mass-energy equivalence principle
In 1905 Albert Einstein proposed that there is a equivalence between mass
and energy, the relationship being
E = mc2
where c is the speed of light in metres per second , E is measured in joules
and m in kilograms
In the example of the helium nucleus earlier, since the mass defect is0.030374 u, converting u into kg and using the above Einsteins equation,the energy equivalent is 4.54 x 10-12 J
This energy in Joules can be converted into eV which is a more convenient
energy unit by dividing by 1.6 x 10-19 J giving 28.4 MeV
1 u change in mass is the equivalent of 931 MeV The mass of object when it is at rest is called its rest mass
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Binding energy
Within the nucleus there are strong forces which bind the protons and
neutrons together, and to separate all these nucleons requires energy i.e.work must be done, referred to as the binding energyof the nucleus(bindingmeans held together)(actually more appropriate to call it unbindingenergy)
Stable nuclei which have little or no tendency to disintegrate, have largebinding energies whereas less stable nuclei have smaller binding energies
Similarly, to join together protons and neutrons together to form a nucleus,
this binding energy must be released.The binding energy is the equivalentof the mass defect
Referring again to the helium example earlier, since the energy equivalent of
the mass defect is 28.4 MeV which is the binding energy, 28.4 MeV of
energy is required to separate to infinity the 2 protons and 2 neutrons ofthis nucleus
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Example
Calculate the binding energy in MeV of a carbon-14 nucleus with a mass defect
of 0.109736 u.
Solution
Using the equivalence 1 u = 931 MeV,
0.109736 u = 102 MeV
Since the binding energy is the energy equivalent of the mass defect,the binding energy = 102 MeV
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Stability of nuclei Within the nucleus of an atom the nucleons experience 2 major forces of attraction
and repulsion
The attractive forcewhich is called the strong nuclear forceacts like a glue to holdthe nucleons together and it is a short range force The repulsive forces are the electric forces between the positively charged protons
(Coulombic force), which is a long range force The strong force is strong enough to overcome the Coulombic repulsion between
protons otherwise the protons would fly apart
Stability of a nucleus is determined by the equilibrium between these forces
Stable nuclei have much larger attractive forces than repulsive forces Gravitational force of attraction also exist, but are negligible in comparison to the
other 2 forces
A stable nucleus is one which has a very low probability of decay
Stable nuclides generally have approximately the same number of protons to theneutrons in the nucleus i.e. neutron-to-proton ratio is close to 1
A useful measure of stability is the binding energy per nucleon which is defined asthe total energy needed to completely separate all the nucleons in a nucleusdivided by the number of nucleons in the nucleus
The most stable nuclides are those with the highest binding energy per nucleon e.g.iron
Typically very stable nuclides have binding energies per nucleon of about 8 MeV
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Neutron vs proton number curve
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Binding Energy/nucleon curve
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Example
The binding energy of a helium-4 nucleus is 28.4 MeV. Calculate the binding
energy per nucleon
Solution
The helium-4 nucleus has 4 nucleons.
The binding energy per nucleon is 28.4/4 = 7.1 MeV
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Fusion and fission
For a system to become stable, it tends to give out energy so that it has less
energy content.
e.g. chemical reaction : acid + base,
mechanical : horizontal cylinder is more stable than standing
cylinder
Hence, a system also tends to be more stable if it has less mass. This is
demonstrated in nuclear reactions. Light nuclei may combine or fuse to form larger nuclei with larger binding
energies per nucleon in a process called nuclear fusion
For this process to take place conditions of very high temperature andpressure are required, as in stars e.g. the Sun
Heavy nuclei when bombarded with neutrons may break into 2 smallernuclei, again with larger binding energy per nucleon values in a process
called nuclear fission. Fission rarely happens spontaneously
Whenever nuclear fusion or fission takes place, the nucleon numbers of the
nuclei involved change, a higher binding energy is achieved and this is
accompanied by an enormous release of energy 11
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Nuclear fission
Being uncharged, neutrons can penetrate the nucleus more easily, but the
problem is that they cannot be directed and controlled by electric andmagnetic fields
In heavy nuclei such as uranium and plutonium there are far more neutrons
than protons, giving a neutron/proton ratio of more than 1
e.g. uranium-235 has 92 protons and 143 neutrons giving a neutron/proton
ratio of 1.55
This leads to a much lower binding energy per nucleon compared with iron
Any further increase in the number of neutrons in such nuclei is likely to
cause the nucleus to undergo nuclear fission
Nuclear fission is the splitting of a heavy nucleus into 2 lighter nuclei ofapproximately the same mass
One element changing into another is called transmutation
The energy released per atom by fission(about 200 MeV) is about 50 million
times greater than that per atom from a chemical reaction such as burning
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Induced nuclear fission
When nuclear fission is started by the capture of a neutron by say a uranium
nucleus, it is known as induced nuclear fission When a uranium-235 nucleus absorbs a neutron, it becomes unstable and
splits into 2 lighter more stable nuclei. Although other fission reactions arepossible, the most likely reaction is
235U92 +1n0 141Ba56 +92Kr36 +31n0 + energy
If these released neutrons are absorbed by other uranium-235 nuclei, thesetoo may become unstable and undergo fission, thereby releasing even moreneutrons causing a chain reaction
In such a case, the reaction continues uncontrolled and a great deal ofenergy is released in a short time
If the reaction is controlled so that the number of fissions per unit isconstant, the rate of release of energy can be controlled
This is precisely what is done in a modern nuclear power station wheresome of the neutrons released in the reactions are absorbed by control rods
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Nuclear fission reactor There are several possible fission reactions which may occur in a nuclear
reactor represented by the general equation235
U92 +1
n0 236
U92 2 new nuclides + 2 or 3 neutrons + energy The nuclides formed in this reaction are called fission fragments Main type of reactor is the Pressurised water-cooled reactor(PWR)pg 372
fig 13.15 Physics by Chris Mee
Inside a reactor, uranium-235 is placed in long fuel rods surrounded bypressurised (150 atm) heavy water (deuterium) called a moderator whose
role is to slow down the neutrons releasedduring fission Slow moving neutrons are more readily absorbed by uranium-235 The number of neutrons available to trigger further fission reactions is
controlled by using control rodsmade of boron or cadmium Pushing the control rods further into the reactor core causes more neutrons
to be absorbed and hence the rate of fission reactions decreases
The heat energy that is produced is removed from the reactor by a coolant(water) which is then used to produce high temperature steam in a heatexchanger to drive turbine generators
The other main type of commercial nuclear reactor for generation ofelectricity is the Advanced Gas-Cooled reactor(AGR) where helium gas isthe coolant
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Disadvantages of nuclear fission
If the fission material is less than a certain critical mass, too many neutrons
escape without hitting the nuclei
The fission of uranium-235 produces medium-speed neutrons, while slow
neutrons are better at causing fission
1 neutron from each fission must cause further fission to maintain a chain
reaction
Less than 1% of natural uranium is uranium-235 while over 99% is
uranium-238 which absorbs medium-speed neutrons without fiffion taking
place
Uncontrollable chain reactions
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Nuclear fusion
Most of the energy on Earth comes from the Sun i.e. solar energy through
thermonuclear reactions It is produced by nuclear fusion reactions whereby light nuclei such as
isotopes of hydrogen join together to produce heavier more stable nuclei andin doing so release energy
One of the fusion reactions is 2H1 +2H1 3He2 +1n0 + energy
The binding energy per nucleon for light nuclei such as hydrogen is low, but
if 2 light nuclei are made to fuse together, they may form a stable newheavier nucleus which has a higher binding energy per nucleon thusreleasing energy
Fusion is much more difficult to achieve than fission
In order to produce these thermonuclear reactions requires extremely hightemperatures(108 or more K) and pressures like that in the Sun because the
hydrogen nuclei have to be brought very close together against thecoulombic repulsive forces which repel each other
We are at present unable to duplicate this reaction in a controlled manneralthough work is being carried out by Joint European Torus (JET) andInternational Tomahawk Engineering Research (ITER) an internationalconcern
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Advantages of nuclear fusion
Advantages are:
Fuels will be readily obtainable for example deuterium can be extracted
from sea-water
The main waste product is helium which is not radioactive
Fusion reactors have built in safety features, as, if the system fails, fusion
stops
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Radioactive decay series
The daughter nuclide of a radioactive decay may itself be unstable and so
may emit radiation to give another different nuclide
This sequence is called a radioactive decay series
(to get picture of decay series)
Alpha decay tends to occur in heavy nuclides which are below the stabilitylineof the NZ curve
- - Beta decay(actually beta minus decay) tends to occur in heavy nuclides
which are above the stability lineof the NZ curve
During beta minus decay, a neutron is converted into a proton, an electron
and an almost undetectable particle with no charge and near-zero mass
called an antineutrino + - Beta plus decay is an emission of a positron with the same mass as an
electron but a charge of +1 e. It is the antiparticleof an electron
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Neutron vs proton number curve
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Detecting/measuring radioactivity
Some of the methods of detecting/measuring radioactivity are based on the
ionising properties of the particles or radiation
The Geiger Countera tube with argon gas at low pressure and a thin micawindow to allow radiation in. Presence of radiation causes ions pairs in the
gas which are accelerated through a potential difference and counted
Photographic plates when radioactive emission strikes a photographicfilm, the film reacts as if it had been exposed to a small amount of visible
light which when developed, a fogging or blackening is seen. Used in film
badge dosimeter where the radiation passes through different filters
consequently the type of radiation as well as the quantity can be assessed
The Scintillation counter this is a device that uses the principle ofphotoelectric emission. When radiation is incident on zinc sulphide it emits
tiny pulses of light called a scintillation which causes emission ofphotoelectrons from the negative electrode of a photomultiplier tube which
amplifies the current and is measured or counted
All measurements should take into consideration the background radiationdue to natural radioactivity and man-made sources which should be
subtracted to obtain the correct count or reading 20
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Nature of radioactive decay
The emission of radiation is due to radioactivity is both spontaneous and
random Spontaneous means it is not affected by any external factors such as
temperature or pressure
Randommeans that it is not possible to predict which nucleus in a samplewill decay next
However there is a constant probability or chance that a nucleus willdecay in any fixed period of time
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Random nature of radioactive decay-simulation
A dice has 6 faces and if 6 dice are thrown simultaneously it is likely that
one of them will show a 6
If 12 dice are thrown, it is likely that 2 of them will show a 6 and so on
While it is possible to predict the likely number of sixes, it is impossible to
say which of the thrown dice will actually show a 6
If a large number of dice say 6000 is thrown, and every time a 6 shows that
die is removed, below is a likely scenario
No of throws No of dice remaining No of dice removed
0 6000
1 5000 1000
2 4173 827
3 3477 696
4 2897 580
5 2414 483
6 2012 402
7 1677 335
8 1397 280 22
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cont
If a graph ofno of dice remaining vs no of throws is plotted, a decay curve
will be obtained which is not linear but has a pattern similar to the decay ofcurrent in a circuit containing a capacitor and a resistor
This experiment can be applied to model radioactive decay where the dice
represent the radioactive nuclei, and the 6 on the dice representing the
radioactive emission.
Once a nucleus has undergone radioactive decay it is no longer available for
further decay
A graph of the number of un-decayed nuclei in a sample against time has a
typical decay curve as above
The half-life of a radioactive nuclide is the time taken for the number ofun-decayed nuclei to be reduced to half its original number
Half-life may also be expressed in terms of the activityof the material
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Example
The half-life of francium-221 is 4.8 minutes. Calculate the fraction of a sample
of francium-221 remaining un-decayed after a time of 14.4 minutes.
Solution:
14.4 mins is 3 half-lives since 14.4/4.8 = 3
therefore after 1st
half-life only remainsafter 2nd half-life only remains
after 3rd half-life only remains
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Activity and decay constant
On investigation, we find that the greater the number of radioactive nuclei in
the sample, the greater the rate of decay i.e. the more undecayed nuclei thereare, the more frequently disintegrations are likely to occur
Mathematically this is described as dN/dt N which gives
dN/dt = - N
where N is the number of un-decayed atoms in the sample, and dN/dt is therate at which the number in the nuclei is changing, so dN/dt represents therate of decay and is called the activity A of the source. A is measured inbecquerels (1 Bq = 1 s-1)
Combining A = - dN/dtand dN/dt = - N we have A = N
where
is a constant of proportionality known as the decay constant, of units-1, yr-1 etc
Decay constant , is defined as the probability per unit time that a nucleuswill undergo decay
This is an important equation because it relates a quantitydN/dt which we
can measure to a quantity which we cannot i.e. number of un-decayed nuclei Also we will see that is directl related to the half-life of the nuclei
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Example
Calculate the number of phosphorus-32 nuclei in a sample which has an
activity of 5.0 x 106 Bq. (given decay constant of phosphorus-32 is 5.6 x 10-7s-1)
Solution
from dN/dt = - N, N = (-dN/dt)/ = (-5.0 x 106)/(5.6 x 10-7) = -8.9 x 1012
The minus sign indicates a decay
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Mathematical descriptions of radioactive decay
To solve the equation dN/dt = - N requires higher level mathematics;
however it is important to know the solution in order to find the variationwith time of the number of nuclei remaining in the sample
The solution is N = Noe-t or N = Noexp(-t)
where Nois the initial number of un-decayed nuclei in the sample, and Nis the number of un-decayed nuclei at time t
This equation represents an exponential decay
Since the activity A is proportional to N, the curve ofA against t is of thesame shape and hence we can write
A = Aoe-t or A = Aoexp(-t)
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Example
A sample of phosphorus-32 contains 8.6 x 1012 nuclei at time t = 0. The decay
constant of phosphorus-32 is 4.8 x 10-2 day-1. Calculate the number of un-decayed phosphorus-32 nuclei in the sample after 10 days.
Solution
using N = No
e-t , we have N = 8.6 x 1012 x e-0.048x10 = 5.3 x 1012
(always ensure that and t are consistent in their units i.e. days mathced to
days, seconds matched to seconds etc)
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Decay constant and half-life
Using N = Noe-t we can derive an equation which relates the half-life to the
decay constant
For any radioactive nuclide, the number of un-decayed nuclei after 1 half-
life is by the definition, equal to No/2 where No is the original un-decayednuclei.
Using N = Noe-t we have at time t = t
N = Noe-t
and dividing each side by No
N/No = 1/2 = e-t
or 2 = etand taking natural logs of both sides
lne 2 = t so that t= ln 2/ t= 0.693/
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Example
Calculate the half-life of radium-226 which has a decay constant of 1.42 x 10-11
s-1.
Solution
Using t= 0.693/ =4.88 x 1010 s
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Matter particles
Matter particles
Hadrons Leptons
(feel the strong nuclear force) (do not feel the strong nuclear force)
(have no size, low or no mass)
Baryons Mesons(includes protons (particles lighter
neutrons & heavier than protons)
particles)
Proton - Pion - Electron
Neutron - Kaon - Electron-neutrino
Lambda - etc - Muon
Sigma - Tau
Omega
etc32
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Quantum numbers & quarks
Particles have various quantum numbersassigned to them to represent
other quantities seen during interactions
Examples: Charge, Lepton number, Baryon number, Strangeness, Charm,
Spin, Topness, Bottomness
The properties and quantum numbers of hadrons can be accounted for by
assuming that each particle is a combination of others called quarkswhichhave fractional charge
Example: Baryons are each made up of 3 quarks, Mesons are made up of
a quark & antiquark
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Force carriers
Nucleons need not be in contact to exert forces on each other
To explain how the strong force is carried from one nucleon to another, the
idea ofexchange particles is used
Particles that carry the fundamental forces are known as gauge bosons
Quarks are bound together by gluons
Force Gauge bosons
strong gluon
electromagnetic photon
weak W plus, W minus
gravitational graviton
Existence of a graviton is speculation only
Grand unified theories(GUTs) seeks to link all the above forces
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