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 as represented 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 and to nuclear fission.

• (n) define the terms activity and decay constant and recall and solve problems using A = λN.

• * (o) infer and sketch the exponential nature of radioactive decay and solve problems using the relationship x = xoexp(–λt) where x could represent activity, number of un-decayed particles or received count rate.

• (p) define half-life.

• (q) solve problems using the relation λ = 0.693/t½

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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 of carbon-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 by

0.030374 u

• This difference between the expected mass and the actual mass of a nucleus

is called the mass defect of the nucleus(i.e. mass of the product < the sum of

masses of reactants)

• Hence mass defect of a nucleus is the difference between the total mass of

the separate nucleons and the combined mass of the nucleus2

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 is

0.030374 u, converting u into kg and using the above Einstein’s 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

4

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 energy of the nucleus(binding

means held together)(actually more appropriate to call it unbinding energy)

• Stable nuclei which have little or no tendency to disintegrate, have large

binding 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 equivalent

of 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 of

this 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 force which is called the strong nuclear force acts 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 betweenprotons 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 theother 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

Binding Energy/nucleon curve

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 and

pressure are required, as in stars e.g. the Sun

• Heavy nuclei when bombarded with neutrons may break into 2 smaller

nuclei, 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

Nuclear fission

• Being uncharged, neutrons can penetrate the nucleus more easily, but the

problem is that they cannot be directed and controlled by electric and

magnetic 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 of

approximately 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 uraniumnucleus, it is known as induced nuclear fission

• When a uranium-235 nucleus absorbs a neutron, it becomes unstable andsplits 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 equation235U92 + 1n0

236U92 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 372fig 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 whoserole is to slow down the neutrons released during fission

• Slow moving neutrons are more readily absorbed by uranium-235

• The number of neutrons available to trigger further fission reactions iscontrolled by using control rods made of boron or cadmium

• Pushing the control rods further into the reactor core causes more neutronsto 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 throughthermonuclear reactions

• It is produced by nuclear fusion reactions whereby light nuclei such asisotopes 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, butif 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 thehydrogen 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 stability

line of the N – Z curve

• β- - Beta decay(actually beta minus decay) tends to occur in heavy nuclides

which are above the stability line of the N – Z 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

• β+ - 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 antiparticle of an electron

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Neutron vs proton number curve

Detecting/measuring radioactivity

• Some of the methods of detecting/measuring radioactivity are based on the

ionising properties of the particles or radiation

• The Geiger Counter – a tube with argon gas at low pressure and a thin mica

window 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 photographic

film, 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 of

photoelectric emission. When radiation is incident on zinc sulphide it emits

tiny pulses of light called a scintillation which causes emission of

photoelectrons 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 radiation

due to natural radioactivity and man-made sources which should be

subtracted to obtain the correct count or reading20

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

• Random means that it is not possible to predict which nucleus in a sample

will decay next

• However there is a constant probability or chance that a nucleus will

decay 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

cont…

• If a graph of no 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 of

current 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 of

un-decayed nuclei to be reduced to half its original number

• Half-life may also be expressed in terms of the activity of 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 ½ remains

after 2nd half-life only ¼ remains

after 3rd half-life only ⅛ 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 there

are, the more frequently disintegrations are likely to occur

• Mathematically this is described as dN/dt α N which gives

dN/dt = - λN

where N is the number of un-decayed atoms in the sample, and dN/dt is the

rate at which the number in the nuclei is changing, so – dN/dt represents the

rate of decay and is called the activity A of the source. A is measured in

becquerels (1 Bq = 1 s-1)

• Combining A = - dN/dt and dN/dt = - λN we have A = λN

where λ is a constant of proportionality known as the decay constant, of unit

s-1, yr-1 etc

• Decay constant λ, is defined as the probability per unit time that a nucleus

will undergo decay

• This is an important equation because it relates a quantity – dN/dt which we

can measure to a quantity which we cannot i.e. number of un-decayed nuclei

• Also we will see that λ is directly 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 λ of phosphorus-32 is 5.6 x 10-7

s-1)

Solution

from dN/dt = - λN, N = (-dN/dt)/λ = (-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 = - λN requires higher level mathematics;

however it is important to know the solution in order to find the variation

with time of the number of nuclei remaining in the sample

• The solution is N = Noe-λt or N = Noexp(-λt)

where No is the initial number of un-decayed nuclei in the sample, and N

is 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 of A against t is of the

same shape and hence we can write

A = Aoe-λt or A = Aoexp(-λt)

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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 = Noe-λt , we have N = 8.6 x 1012 x e-0.048x10 = 5.3 x 1012

(always ensure that λ 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-λt 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-decayed

nuclei.

• Using N = Noe-λt we have at time t = t½

N = Noe-λt½ and dividing each side by No

N/No = 1/2 = e-λt½

or 2 = eλt½

and taking natural logs of both sides

lne 2 = λt½ so that t½ = ln 2/λ t½ = 0.693/λ

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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 t½ = 0.693/λ = 4.88 x 1010 s

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Particle Zoo

• By the mid 1930s, the understanding of the fundamental structure of matter

seemed almost complete

• Ordinary matter is made up of protons, neutrons and electrons

• The 30s’ and 40s’ saw the discovery of a number of particles not found in

normal matter through high energy collisions in particle accelerators, followed

by a whole host of unstable particles after 1960, most of which are short-lived

• These matter particles are classified into 2 main groups: hadrons and leptons

• There are now 12 known leptons, but more than 100 hadrons!

• It is clear that these do NOT represent the fundamental structure of matter!

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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

Quantum numbers & quarks

•Particles have various quantum numbers assigned 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 quarks which

have 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 of exchange 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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