7/10/2006 nuclear (© f.robilliard) 1flai/theory/lectures/nuclear.pdf · 7/10/2006 nuclear (©...

7/10/2006 Nuclear (© F.Robilliard) 1


Introduction:

In this introduction to nuclear physics we will firstly summarise some basic information about the atom, and its nucleus, and present some standard notation.

We will then discuss some fundamental and necessary concepts of mass and energy, and develop the idea of binding energy.

We will then look at the statistics, and mechanisms, for radioactivity.

This will lead us to considerations of nuclear reactions, and in particular, nuclear fusion and fission.

Finally, we will see the application of these ideas in nuclear reactors.


The Nucleus:The Atom:

Consists of a small, compact, positively-charged nucleus, surroundedby a distribution of negatively-charged electrons. The negative electrons are kept in the vicinity of the nucleus by their Coulomb attraction to the positive nucleus.

The Nucleus:is composed of two types

of particle – positively-charged protonsand un-charged neutrons. +nucleus

-electrons

+

neutron

proton+

Nomenclature:

Atomic Number = Z = Number of protons in nucleus

Neutron Number = N = Number of neutrons in nucleus

Mass Number = A = Number of neutrons + protonsin nucleus

A = Z + N


Significance of Z & A:

Protons and electrons have equal, but opposite, charges. Thus, in a neutral atom, there must be one electron for each proton. The Z of an atom, therefore determinesits total number of electrons.

Protons and neutrons have similar masses, which is about 2000 times greater than that of an electron. Thus A is a measure of the total mass of an atom.It does not determine chemical properties.

The chemistry of an atom is determined by how its energy levels are filled.Thus Z determines the chemistry of an atom. Each chemical element is characterised by a unique value of Z.

Electrons are able to exist only in certain discrete energy levels about the nucleus. As the number of electrons increases, these levels fill up sequentially starting from the lowest. As a level fills, the next electron must go into the level of next highest energy.

Significance of Z:

Significance of A:


Notation:

XAZ

n) 143 p 92 nucleus (Uranium U

n) 30 p 26 nucleus(Iron Fe

n) 3 p 3 nucleus (Lithium Li

n) 2 p 2 nucleus (Helium He

p) 1 nucleus(Hydrogen H

23592

5626

63

42

11

+=

+=

+=

+=

=

The type of nucleus is represented by the number of protons, Z, and the total number of particles, A, from which it is composed, and is written:

where Z = atomic numberA = mass number

and X = the chemical symbol for the element

Examples:p = protonn = neutron

(Note: sometimes the Z is omitted – example –4He (Helium-4), 56Fe (Iron-56))


Nomenclature:

nuclide"" aor nucleus, particular arepresent can XAZ

Nucleus:

This is the central part of a particular atom, consisting of protons and neutrons.

Nuclide:

This is a type of nucleus.

Nucleon:

A nucleon is a component particle of a nucleus. Protons and neutrons are nucleons.


Isotopes:Nucleii with a given number of protons occur with a range of numbers of neutrons.

Nuclides with – the same Zbut different Aare called different isotopes of an element

Examples:

Uranium.of isotopes :UU, U, U, U, U,

Carbon. of isotopes :CC, C, , C

Hydrogen. of isotopes :(Tritium) H), (Deuterium H, (Hydrogen) H

23992

23892

23692

23592

23392

23292

146

136

126

116

31

21

11

= unstable & radioactive)(where

Many isotopes occur naturally, and naturally-occurring elements are generally a mixtureof different isotopes. The more radioactive isotopes may have occurred naturally in the past,but have long since decayed to other species. These isotopes can be produced artificially.


Units:In nuclear physics special non-SI units are used, which are more suited to the microscopic scale of the nucleus.

Mass: unified mass unit, u:

Mass of a single Carbon-12 nucleus is defined to have a mass of exactly 12 u.

1 u = 1.660559 x 10-27 kgCharge: electronic charge, e:

1 electronic charge = e = charge on one electron

1 e = 1.602 x 10-19 coulombEnergy: electron volt, eV:

1 eV = the work done to move an electron through a potential of 1 Volt.= charge x potential= (1.602 x 10-19 ) x 1 = 1.602 x 10-19 joule

1 keV = 103 eV 1 MeV = 106 eV


Mass & Charge for Proton, Neutron, Electron:

Particle mass charge (u) (kg) (e)

Proton (p) 1.007 276 470 1.672 621 71 x 10-27 +1

Neutron (n) 1.008 664 904 1.674 92 x 10-27 0

Electron (e) 5.485 799 03 x 10-4 9.109 56 x 10-31 -1

Note: masses of p and n are both about 1 u. Mass of p ~ 2000 times mass of e.Charges of p and e are equal but opposite.

The masses above are for an isolated particle.


Mass-Energy Equivalence:

Mass, m0 , is equivalent to a proportional amount of energy, E . The constant of proportionality is the square of the speed of light.

( )N2............ cmE 20=

Because c is a large number, a small amount of mass, is equivalent to a very large amount of energy.

As was discussed in the notes on Quantum Physics, the total energy, E, the rest mass, m0 , and the momentum, p, of a particle are related by the following equation from the theory of relativity -

( ) ( ) ( )N1............ cm pc E22

022 +=

where c = speed of light = (2.9979 x 108) m/s ~ (3.0 x 108) m/s.

If the particle is at rest, p = 0, and equation (N1) becomes -

Mass-energy equivalence is fundamental to the understanding nuclear processes.


Direct Evidence:Mass can be converted into energy, and energy can be converted into mass.

Mass into Energy:

If an electron and a positron collide, they can mutually annihilate and be replaced by two photons (pure electromagnetic energy). This is called “pair annihilation”. The particles disappear, and their masses are converted into electromagnetic energy.

Energy into Mass :If a photon of sufficiently high energy interacts with a nucleus, the photon can disappear, and be replaced by two particles, an electron and an antielectron(an antielectron is also called a positron). This is called pair production. Energy of the photon has been converted into the mass of the two particles.

There are many other situations where such mass-energy conversions occur, as we shall see.

A direct example of such conversions involves the electron.

(For any particle, there is an antiparticle, whose quantum properties are all exactly opposite to those of the particle.)


Example: Carbon-12:Find the energy equivalent of 1 atom of Carbon-12 ( mass = 12.00 u).

E = m0 c2 = [12.00x(1.66x10-27)] . (3.00x108)2 = 1.79x10-9 J = 1.79 nJ

In SI Units:

In Nuclear Units (MeV & u):Since:E [Joule] = m [kg] (c [m/s]) 2

E [in MeV] (1.6022x10-19)x106 ={ m [in u] (1.6605x10-27 )} x (2.9979x108)2

Giving: E [MeV] = 931.5 m [u] when the constants are combined

In practice we use this form:E [in MeV] = 931.5 m [in u]

c = 2.9979x108 m/s1 u = 1.660559 x 10-27 kg

1 eV = 1.6022 x 10-19 joule

Here: E [MeV] = 931.5 m [u] = 931.5 x 12.00 = 1.12x104 MeV= 11.2 GeV


Strong Nuclear Force:

Consider a nucleus of Deuterium 12H, which is composed of1 proton plus 1 neutron, held together by the “strong nuclear force”.

Let’s pull this nucleus apart. We will need to apply equal, but opposite forces, F , to the two nucleons.

FF

p nIn this process we do work on the nucleus.

But the nucleons are stationary after being separated, and thus have no kinetic energy. There is no potential energy, since the strong force field is negligible outside the nucleus. No energy has escaped from the system. So where did the work go that we did on the system?

The equivalence between mass and energy is essential to the understanding of the physics of the nucleus.

The protons in a nucleus are positive, and repel each other strongly by Coulomb repulsion. They are held together by another much stronger attractive nuclear force, called the “strong nuclear force”. This force attracts any proton or neutron, to any other proton or neutron. It is a short range force, and therefore its action is essentially confined to the nucleus.


Answer:The work we did has turned into mass!

The total mass of the separated proton (p) and neutron (n) is greater thanThe mass of the deuterium nucleus.

E = δm. c2Numerically:

For deuterium nucleus: measured mass of 12H = 2.014102 u(p mass) + (n mass) = 1.007825 u +1.008665 u = 2.016490 u

Difference = δδδδm = 0.002388 u

Energy equivalent of δm = W = (0.002388 u) x 931.5 = 2.22 MeV

If the p & n were recombined to form a deuterium nucleus again, this mass increase would convert back into energy.

The mass difference, δm, and its energy equivalent, E, are related by -

This is the work we did to strip the nucleus.


Binding Energy:This work that we must do to split up the nucleus, is called the “binding energy ”(BE) of the nucleus.

(binding energy) + (input energy)= 0

binding energy = -(input energy)

It is energy we must input to the nucleus to separate it.

By convention, we take the separated state to be the zero of nuclear potential energy for the nucleus.

If we must add energy to the nucleus to bring it to a zero energy separated state, then the binding energy of the nucleus must be negative.

negative energy


BE/nucleonAs the total number of nucleons in a nucleus increases, the total BE of the nucleus also increases. However, due to quantum effects within the nucleus, the total binding energy does not increase in proportion to the number of nucleons (= the mass number A).

We usually consider the average BE per nucleon = BE/A

The BE per nucleon is a measure of the average energy required to extract a nucleon from a given nucleus. It is therefore a measure of the stability of the nucleus.

The greater the BE/nucleon, the more stable the nuclide.

By measuring isotopic masses, we can plot a curve that represents the nuclear stability of a range of nuclides.


BE/nucleon Curve:region of max stability

The most stable nuclei occur for maximum BE/nucleon

Iron-56 is one of the most stable.


How Did We Calculate the BE/nucleon Curve?:

X; of mass is n; of mass is p; of mass is where

))(.(

nucleusof mass

neutronsof mass

protons

of mass

for

differenceMass

AZ

Xnp

Xnp

mmm

mmZAmZ

m

X

−−+=

��

��

��

�

�−��

�

�+��

�

�≡≡

��

�

�

�

δ

2AZ

.nucleus

of BEcm

Xδ=��

�

�

�

( )( ) A

cmX

2

AZ

.nucleons ofNumber BE Total

nucleus for

nucleonper BE δ=≡��

�

�

(SI units)

We calculate the BE/nucleon curve from experimentally measured masses of the nuclei of the various isotopes. These can be measured to high precision, using mass spectrometry.


Comments on the BE/nucleon Curve:From about 2.2 MeV/A (=MeV per nucleon) for deuterium (H-2), the curve rises steeply to a maximum of about 8.5 MeV/A near iron (Fe-56).

On the rising edge are peaks at He-4, C-12, and O-16, which are particularly stable nuclides.

Beyond iron, the curve drops slowly to uranium (U-235). As uranium is approached, the stability of nuclides decreases. Uranium itself is unstable and radioactive. Elements with higher atomic number, because of their increasing instability, do not occur naturally.


Nuclear Processes:

Nuclear processes generally involve changes in the structure of a nucleus. Changes in mass and energy are involved.We will consider these under three headings.

Radioactivity -This where unstable nuclei spontaneously decay, with the emission of small particles, at high energy.

Nuclear Reactions -Here changes in nuclear structure are initiated when particles are fired at nuclei. New nuclei are produced.

Fission and Fusion reactions -These are special nuclear reactions, where either smaller nuclei fuse into a larger nucleus, or a large nucleus splits into two smaller fragments.

We are now in a position to consider some important processes involving nuclei.


Radioactivity:Some nuclei are unstable, and spontaneously decay to other nuclei by emitting some of their component particles. Such nuclides are called radioisotopes, and are said to be radioactive.The emitted particles exit the nucleus with significantly large kinetic energies.

There are 4 mechanisms for natural radioactivity.

αααα-emission: an αααα-particle is emitted from the nucleus.

ββββ-emission: a ββββ -particles is emitted.

γγγγ-emission: a γγγγ-particle is emitted.

electron capture: an inner orbital electron is captured, and a neutrino (symbol νννν) emitted.

Let’s look firstly, at the nature of each of these emitted particles.


αααα & ββββ Particles:An αααα-particle consists of a group of 2 p + 2 n, combined as a helium nucleus(

24He). We saw earlier, that the binding energy/nucleon for this grouping of

protons and neutrons is high, which is why they tend to hold together as a particle.It is a relatively heavy particle, which interacts strongly with any matter in its path.

A β β β β -particle is an electron. There are two types of electron:

ββββ++++ is a positive electron (called a positron). It is the antiparticleof the negative electron. Each of its quantum properties is the oppositeof those of the negative electron. It has an equal, but opposite charge.It has the same mass but is made of antimatter. If a negative electron and apositron collide, they annihilate each other and their two masses convertcompletely into energy.

ββββ−−−− is a negative electron (sometimes called a negatron).... These are identical to the orbital electrons of an atom.

(In fact, each type of elementary particle has its own corresponding antiparticle.)


γγγγ’s & νννν’s:

A γ γ γ γ -particle is a lump of pure high-frequency electromagnetic energy.It has zero charge and zero rest mass.

A neutrino (νννν) is lump of non-electromagnetic energy, with zero charge and very small rest mass.

Note: The α-particle is a relatively large particle, and α-decay is a common decay mechanism for larger nuclides ( say A >144 ), but not for smaller nuclides. β-decay occurs over a wide range of A values.

γ-decay is generally associated with either α- or β-decay.


Statistics of Radioactive Decay:

It is not possible to predict precisely when a particular nucleus will decay, as the process is probabilistic. However we can specify the probability of a decay.This probability will depend on the particular nuclide & decay mechanism.

We deal firstly with the statistics of the decay process, for any of the radioactivity mechanisms (α, β, γ, or EC).

All nucleii of a given nuclide have the same decay probability, for a given mechanism.

Thus, for a large population of a given nuclide, the number of decays that occur,per unit time, is proportional to the total number, N, of nucleii present.

Let dN be the number of decays that occur in a time interval dt.


Decay-Rate Equation:

negative.) thusis and decay, afor decrease a is dN since required, issign negative the(where

unit timeper nucleusper decay a ofy probabilitconstant.decay thecalledality proportion ofconstant � where

�NdtdN

therefore

N dtdN

remaining) nucleii active of (nr. second)per decays of (nr.decay) of (rate

≡≡

−=

∝

∝≡

(This is a simple differential equation, which can be solved for N.)


Solution of Equation:

�t-0

0

0

t

0

N

N

eN and

t-�) NN

ln( thus

0at t NN where

with t)sidesboth ng(Integratidt �..dtdtdN

N1

and

�dtdN

N1

gtransposin

�NdtdN

0

N=

=

=≡

−=

−=

−=

Radioactive samples decay exponentially


�

693.0�

)2ln(T

�T)2ln(

eNN21

gives T at t N21

N Puting

eNN

1/2

1/2

�T00

2/10

�t0

1/2

==

−=−

=

==

=

−

−

Decay Curve - Half Life:

T1/2

2T1/23T1/2

�t0eNN −=

Half-life = T1/2

= time taken for half the original nuclei to decay.

We can plot the solution:

(Half-life in terms of Decay const.)


After n Half Lives:

1/2n0

1/230

1/220

1/20

0

nTat t 2N

N

T3at t 2N

N

T2at t 2N

N

Tat t 2

N Nthen

0at t NN If

==

==

==

==

==

�

During any time interval of T1/2 ,the number of active nuclei halves.It doesn’t matter when you start the time interval.

2N

lives-halfn after remaining nucleii active

ofNumber The

n0=

��

��

�

��

��

�

Summary:


Some Half-lives:Half-lives vary greatly.


Activity Initial0)at t R(R where

�NR cesineRR thus

eNN sincee�N

negative) is dN since used are || (where �NdtdN

R therefore

dtdN

ondDecays/sec ateDecay RActivity

0

�t0

�t0

�t0

≡=≡≡=

==

=≡

≡≡≡≡

−

−−

R

Activity:

Activity decreases exponentiallyin proportion to the number, N, of active nucleii remaining.

Activity is a measure of the level of radioactivity of a sample. This is the average number of decays occurring per second.


Units for Activity:

Curie (Ci):

This is an older non-SI unit for activity, that is still sometimes used.

1 Ci is the activity of 1 gram of pure Radium ( 88226Ra )

1 Ci = 3.7 x 1010 Bq1 mCi = 10-3 Ci1 µCi = 10-6 Ci

This is the SI unit for the activity, R, of a radioactive sample.

1 Bq = 1 decay per second 1 kBq = 103 Bq1 MBq = 106 Bq

Bequerel (Bq)


Example:7

13N has a half-life of 10 minutes. If the initial number of 713N nucleiin a sample is 1024x106 nuclei, how many will be left after 50 minutes?

By Table:

t (min) 0 10 20 30 40 50

N (M nucleii) 1024 512 256 128 64 32

Using the fact that the number of active nuclei halves after every half-life:

Answer:32x106 nuclei remain.

( ) ( ) nuclei 1032 e 101024 e N N

min 50 t , 101024 N min, 10 T Here

T2ln

� whereeN N

62.ln -6

tT

2ln

0

60

t� -0

1050

21

21

21

×=×==∴

=×==

==

��

�

�

�

−

By Formula:

Same answer!


Example:

88236Ra has a half-life of (1.6 x 103 ) years.

What is the activity of a sample containing (3 x 1012) nuclei ?

Answer: 41 Bq


Decay Mechanisms:We deal next with the detailed mechanisms for α-, β-, γ-decay and for electron capture (EC).

This parent nucleus is a system to which conservation laws apply.

These processes involve changes in the nuclear structure of the unstable parent nucleus.

The total charge, the total mass-energy, and the total momentumof this system (including any emitted particles) must be the same, before and after the decay.


αααα-Decay:

HeYX AZ

AZ

42

)4()2( +→ −

−

HeRnRa 42

22286

22688 +→

HeThU 42

23490

23892 +→

αααα-Decay: In this decay, 2 protons + 2 neutrons are ejected from the nucleus as an α-particle, which is a Helium-4 nucleus.

General α-particle

Examples:

α-decay of Radium226:

α-decay of Uranium-238:

Rn (radon) is itself unstable:HePoRn 4

221884

22286 +→


Conservation of Z & A:

� �=

��

�

�=��

�

�

�

LHS RHS

ZZ

RHS on the protonsofnumber totalThe

LHS on the protonsofnumber totalThe

Conservation of charge:

Conservation of nucleons:

�� =

��

�

�=��

�

�

�

RHSLHS

AA

RHSon nucleons ofnumber Total

LHSon nucleons ofnumber Total

The following two rules apply to nuclear changes:

atomic number, Z, is conserved

mass number, A, is conserved


Q for a Nuclear Process:

Definition of Q:

This is the amount of energy released by the process.

This energy can only come from the conversion of mass into energy during the process.

( ) ( )�� −=

≡=

RHSLHS

mm

products of mass - reactants theof mass m wherem.cQthen

process, theduringlost mass theis m If2

δδ

δ


Q for αααα-Decay:

Q for an αααα-decay:

2

2

).(

.

YXFor

cmmm

cmQ

YX α

δα

−−==

+→ where mX = mass of XmY = mass of Ymα = mass of α

In α-decay, the Q goes mainly into the KE of the α. (Α small amount will go into the recoil energy of nucleus Y.)

Ιf δm = (mx – my – mα) is not greater than zero, there will be no energy for the emission of the α, and consequently the decay will not happen.


Example:

HePbPo 42

20682

21084 +→

Polonium-210 α-decays to Lead with a half-life of 138.38 days. Find the Q for this process.

The atomic masses of nuclides are given in tables (see Serway Table A-3)

MeV 5.400005797.050.931

)(50.931)(get we. From

005797.0002603.497446.20598286.209

2

=×=

×===

−−=−−=

umMeVQcmQ

u

mmmm HePbPo

δδ

δ

Most of this energy (5.400MeV) goes into the KE of the ejected α. Some goes into the recoil KE of the Pb-206 nucleus.


ββββ-Decay:

General:

Examples: Note:conservation of A and Z apply.

Instabilities within the nucleus can result in the ejection of either an electron (β-), or its antiparticle the positron (β+), from the nucleus. Another particle, called a neutrino (ν), or its antiparticle, the antineutrino ( ), is always emitted with the β. A neutrino is emitted with the β+, and an antineutrino, with the β-. Both these particles carry energy and momentum.

�

neutrino with positron

antineutrino with electron

The properties of the positron are precisely opposite to those of the electron – equal but opposite charge, equal antimass, spin ½ .


Properties of the Neutrino ( 00ν ν ν ν ):

A neutrino (νννν) is an elementary particle which -

interacts very weakly with matter and is consequently difficult to detect (a νννν can pass transparently right through planet Earth). Neutrinos interact only by the weak or gravitational forces – not by strong or electromagnetic forces.

has angular momentum (spin = 1/2), directed in the opposite direction to its momentum, p.

has zero charge. has a small rest mass, m0 ( 0.05 eV < m0 < 0.3 eV )

has energy E given by E2 = (cp)2 + ( m0c2 )2, where m0 is small.

has momentum p

travels near the speed of light c

The ν was first postulated in 1936 to explain anomalies in the conservation of energy and momentum, in β decay. However, because of its weak interaction with matter, was not directly experimentally confirmed until 1956.


More Neutrino Properties:

The anti-neutrino, because it is the anti-particle of the neutrino, must have properties, that are precisely opposite to those of the neutrino. This means that the charge will also be zero, the rest mass will be equal, but antimatter, and the direction of its spin, must be in the same direction as its momentum, p, since the spin vector for the neutrino is oppositely directed to its momentum.

Recent experiments show that there are three types (or flavours) of neutrino – electron neutrino, muon neutrino, and tau neutrino. As a neutrino travels through space, it can oscillate between these three types. This is called neutrino flavour oscillation.

The neutrinos in ββββ decay, are electron neutrinos.

Although the neutron is stable inside the nucleus, outside the nucleus it is unstable, and β-decays to a proton, with a half life of 10.6 minutes -

ββββ-decay of the neutron:


Atomic Mass Values & Q:The masses listed in tables, are traditionally atomic mass values, which include the masses of the orbital electrons of the neutral atom. For example, the atomic mass for (2

4He) = 4.002603 u is the mass of the nucleus plus the mass of the two orbital electrons.

In the case of α-decay, the number of orbital electrons on the LHS of the decay reaction, is equal to the total number on the RHS.

Thus the orbital electron masses cancel out in the calculation of Q, and the atomic mass values will give the same result as using the nuclear mass values. Thus the atomic mass values are generally used.

The masses we need for Q calculations are the masses of the nuclear species taking part in the nuclear change – the nuclear masses.

(atomic mass) = (nuclear mass) + (mass of orbital electrons)

HeRnRa 42

22286

22688 +→For example: has 88 on the LHS,

and (86+2)=86 on the RHS.


In ββββ-decay:

Let mX = neutral atomic mass of XmY = neutral atomic mass of Yme = mass of electron

Using nuclear masses:nuclear mass = mass of nucleus

= (mass of neutral atom) – (total mass of orbital electrons)

For a neutral atom:The number of orbital electrons = the number of protons = Z

Note: we assume the rest mass of neutrino to be negligible.

In β-decay and EC, electrons are directly involved in the nuclear change, and we must either use nuclear masses, or use atomic masses, but take careful account of the electrons.

Let’s see what happens in the calculation of Q for β+ and β- decay.


Q For ββββ-decay

To get the Q: Q = δm.c2

or Q(MeV) = 931.50 δm(u)For β-decay to occur, Q must be > 0

orbital electrons & β-

cancel out!

electrons don’t cancel out


γγγγ-decay

γ00

AZ

AZ :General +→∗ XX

If a nucleus is left in an excited state, it can emit its excitation energy in the form of a quantum of electromagnetic energy - the γ-photon.

γ-decay is often associated with other decay mechanisms, which leave the nucleus in an excited state.

]decay -[ by followed

decay]-[ Example126

126

126

125

γγ

βυβ

+→

++→∗

−∗

CC

CB

(The superscript * represents the fact that the nucleus X is in an excited state.)

Wave-particle duality allows us to treat the photon as a particle, that travels at the speed of light, c, and has zero charge, zero rest mass, energy = E = hf (Planck’s equation), momentum = E/c, spin = 1. [see notes on Quantum Physics]


Electron Capture (EC):

ν001

01 :EC General +→+ −− YeX A

ZAZ

ν00

73

01

74 :EC of Example +→+− LieBe

YX

eYeeX

mm

mZmmZmm

m

−=−−−+−=

+=])1([][][

Y) mass(nuclear - mass)(electron X) massnuclear (δ

In this process, an (inner) orbital electron is captured by the nucleus, and a neutrino is emitted.

EC is closely connected with β+-decay

m(u) 931.50 Q(Mev)Or m.c Q :And 2

δδ

==


Nuclear Reactions:

YbaX +→+

As we have seen, unstable nuclei can decay spontaneously, producing a change in nuclear structure – radioactivity!.

It is also possible to trigger changes in nuclear structure by firing a high energy particle at a nucleus.

This is a nuclear reaction, in which a new nucleus is formed, with the emission of second particle.

Target nucleus X is bombarded by high energy particle a. Structural changes result in the emission of a second particle b, leaving a product nucleus Y.

This is often written: YbaX ),(Let’s look at some examples.


Examples:

HepLi

HeHeHLi42

73

42

42

11

73

),( α+→+

Lithium-7 bombarded by a proton:

Berylium-9 bombarded by an alpha particle:

CnBe

CnHeBe126

94

126

10

42

94

),(α+→+

This reaction was used to experimentally confirm the existence of the neutron.

This was the first nuclear reaction produced by artificially accelerated particles.


More Examples:

LinB

LiHenB73

105

73

42

10

105

),( α+→+

NapMg

NaHMg2411

2512

2411

11

2512

),(γγ +→+

Neutron bombardment of Boron-10:

Absorption of a gamma photon, leading to the photodisintegration of the absorbing nucleus:

where the 1124Na is unstable and undergoes β-decay -

ννννββββ ++→ −MgNa 2412

2411


Q For a Nuclear Reaction:

)-m-mm(m�m

cmQ

bYaX +== where,. 2δ

YbaX ),(For:

Note: as we have noted earlier, nuclear masses should be used to calculate δm. In practice, as we have seen, neutral atomic masses ( those usually found in tables, and which include the masses of orbital electrons) are used, since the orbital electron masses are the same on both sides of the reaction, and consequently cancel out.


Example:HepLi 4

273 ),( α

u 018623.0)002603.4(2007825.1016004.7

2

=−+=

−+=

−−+=

HeHLi

HepLi

mmm

mmmmm αδ

MeV 35.17018623.050.931

)(50.931)(

=×=

×= umMeVQ δ

This energy will take the form of the kinetic energy of the two α-particles.

Note: that we need to use the atomic masses of H, and He, rather than the masses of isolated protons and alpha particles, so that the orbital electron masses will cancel out.


Exothermic & Endothermic Reactions:

Exothermic: Q >0 :energy is produced.

Endothermic: Q<0 : energy is absorbed.

For endothermic reactions to proceed, energy greater than the Q must be supplied to the reaction, generally in the form of the KE of the reacting particles. This energy is a threshold energy for the reaction.

In exothermic reactions, the evolved energy usually takes the form of the KE of the product particles.

There are two types of nuclear reaction: those where energy is produced, and those where is is absorbed.


Nuclear Fusion and Fission :

Fission is the inverse operation, where a larger nucleus splits into two smaller nuclei.

Fusion is where two smaller nuclei join together to form a single larger nucleus

P

Q

RS Fusion: could involve fusing two nuclei

near P to form a single larger nucleus near Q, or fusing two nuclei near S to form a single larger nucleus near R.

Fission: could involve splitting a single nucleus near R into two smaller nuclei near S, or splitting a single nucleus near Q into two smaller nuclei near P.

These are two important classes of nuclear reaction.

To understand the important energy implications of these two processes, we need to revisit the Binding Energy/Nucleon curve.


Increase in BE :

P

Q

RS

Of the possibilities shown by the arrows in the figure, two are exothermic, and consequently of interest in power generation –

Fusion P �� Q Fission R �� S

The other two are endothermic - & not of interest.

Both these reactions involve going up the BE/nucleon curve – that is, both involve an increase in BE/nucleon.

In fusion and fission, the number of nucleons composing the reactants, and the number composing the products, is the same. If the BE/nucleon has increased during the reaction, the total overall BE of the products is greater than that of the reactants. The total BE has increased.But BE is the work we need to put into a nucleus to strip it into into its constituent nucleons.

BE can only increase, if the original nuclei have lost the extra energy during the reaction, that we now need to add, to strip them. That is, the reaction must be exothermic (Q > 1)


Fusion and Fission:

)(

:nucleus He)( a form tonucleus H)( a with collides nucleus H)(A 32

11

21

32

11

21

energyHeHH +→+

Example of fusion P Q:

Example of fission R S: ( Note: most fissions are not spontaneous like this example, but are triggered by the absorption of neutrons.)

( )energy )(3

:neutrons 3 ofemission with theand

fragments intofission sly spontaneoucan

10

14355

9037

23692

14355

9037

23692

+++→ nCsRbU

CsRb

U

Mass:In both the above examples, the total mass of the reactants is greater

than the total mass of the products. This difference in mass has been converted into the energy that is released.

We will conclude by looking, in more detail, at two important cases -nuclear fusion in stars, and fission in nuclear reactors.


Nuclear Fusion In Stars:

.......(3).......... )(2

.......(2)..........

.......(1)..........

11

42

32

32

00

32

21

11

00

01

21

11

11

HHeHeHe

HeHH

HHH

+→+

+→+

++→+ +

γνβ

The energy of stars, including the sun, is produced by exothermic nuclear fusion processes in stellar cores.

One mechanism for this is called the p-p cycle, and consists of three steps:

The overall effect is to fuse hydrogen into helium with the evolution of energy. Positrons, neutrinos and gammas are also produced. The overall Q is about 25 MeV.

)(2)(2)(2HeH)(4

:(3) 2(2)2(1) :reaction Overall00

00

01

42

11 γνβ +++→

+++

Research is ongoing to achieve nuclear fusion efficiently on Earth.


Nuclear Fission in Reactors:

Uranium-235 is the only naturally-occurring nuclide that will undergo n-induced fission. There are artificially produced nuclides such as Plutonium-239, and Uranium-233 that will also fission in this way.

Fission of heavy nuclei can be triggered by bombardment with particles other than neutrons, such as protons, deuterons, alphas, or gamma photons, but such fissions are not self sustaining.

Nuclear fission can be exothermic, and therefore offers possibilities for energy generation. However, there are a number of problems that need to be overcome, before this is possible on a large scale.

If a neutron is fired at a large nucleus, the neutron can be absorbed, producing an unstable new nucleus. This unstable nucleus can fission into two fragments, with the release of excess neutrons, plus energy. This is a n-induced fission.

Because the excess neutrons produced by the fission, can go on to produce further fissions, this process has the potential to be self sustaining – that is, we don’t have to keep providing the neutrons.


Fission of Uranium-235:

QnSrXeUUn

QnKrBaUUn

+++→→+

+++→→+

)(2

)(310

9438

14054

*23692

23592

10

10

9236

14156

*23692

23592

10

QneutronsYXUUn +++→→+ *23692

23592

10

X and Y are the “fission fragments”. There are many different X, Y pairs possible, for U-235 fission, for example:

The number of neutrons will vary, depending on the X and Y. The average number of neutrons over all possible X, Y pairs is ~2.5.

Q is the energy released

The fission fragments are generally radioactive, and will subsequently decay.


Q for:

QnKrBaUn +++→+ )(3 10

9236

14156

23592

10

δm = mn + mU – mBa – mKr – 3mn= 235.0439231 – 140.9144069 – 91.92615313 -2(1.008664915)= 0.1860332 u

Q(MeV) = 931.50.δm(u)= 931.50 x 0.1860332= 173.290 MeV

(Q is the energy generated by 1 fission.)Compare:(energy released by 1 molecule of octane in petrol combustion)

= 200 eV

Fission is a highly energetic process!


Sustained Nuclear Fission:

Since each fission of uranium-235 produces, on the average, 2.5 neutrons, and it takes only one neutron to induce another fission, a sustained nuclear reaction is possible, assuming that we lose no more that an average of 1.5 neutrons per fission.

Because of its high Q in fission, and because it occurs naturally, uranium-235 can be used for large scale energy generation, provided that the reaction can be made self-sustaining.

2. absorption by neutron radiative capture (n, γγγγ)

1. elastic collisions (no absorption).

Management of the neutrons is important. We need to minimise losses, that can occur through absorption by non-fissionable nuclei, and maximise absorption by fissionable nuclei. Apart from being absorbed and causing fission, the following are ways in which neutrons can interact with nuclei.

Both of these are important to achieve a sustainable nuclear fission reaction.


1. Elastic Collisions:

A fast neutron will lose some of its KE to the nucleus, in such a collision. After many collisions, with many nuclei, the neutron approaches the same average thermal KE as the nuclei themselves. We say that the neutrons have been thermalised, or moderated.The fraction of its KE lost by a neutron in a collision, will be greatest for nuclei whose masses are comparable to that of the neutron. Unfortunately some of these nuclei, such as H-1, have a high probability for neutron absorption.

Rather than be absorbed, neutrons can collide elastically with some nuclei.

Materials that efficiently transfer neutron KE to their nuclei by elastic collisions, are called moderators.

Moderators slow down neutrons.

Common moderator materials are heavy water ( D2O ,where D=Deuterium), beryllium (Be), and graphite (12C).

It turns out, that the probability of absorption by fissionable nuclei increases as the energy of the neutron decreases, making moderation useful.


2. Neutron Radiative Capture:Some nuclei can absorb neutrons, in a collision, and not fission.

The nucleus can get rid of this excess energy by emitting it as a gamma photon.

Such an absorption leaves the resultant compound nucleus in an excited state.

γ+→→+ ++ XXXn AZ

AZ

AZ

1*110 XnX A

ZAZ

1),( +γor:General:

This process removes neutrons from a nuclear reaction.To achieve sustained fission, such losses need to be minimised. Once the reaction has been started, however, controlling the neutron flux is critically important, and can be achieved through such absorption.

CdnCd 11448

11348 ),( γExample:


Reproduction Constant, K:For a given mass of fissionable material, neutrons can be lost to fission by

1. (n, γγγγ ) absorption

2. escape from the surface of the material.

��

�

�≡

fissionanother causeat fission th each from sn' ofnumber average the

K

Taking these factors into account, we can define the reproduction constant, K -

For a self-sustained nuclear reaction, K=1If K<1 the reaction will die out.If K>1 the reaction will increase in neutron flux, and energy output, and could result in a nuclear meltdown, or, in the extreme, a nuclear explosion.


Critical MassFor neutrons produced by fissions in a mass of fissionable material of dimension d:

- the net production rate (production rate – absorption loss rate) is ~proportional to the volume of the material, which is ~proportional to d3

- the loss rate is ~proportional to the surface area of the material, which is ~ proportional to d2

d

fissionablematerial

Thus as d increases, d3 increases faster than d2. More n’s are produced by the increasing volume, but a smaller fraction of them is lost by leakage through the surface.

The critical mass is the mass for which K = 1, and a self-sustained nuclear reaction is achieved in the mass of fissionable material.

A nuclear explosion can be caused when the mass is suddenly increased beyond critical – super-critical mass! In this case, we get a runaway chain reaction.


Chain Reaction For Super-critical Mass (K=2):The n’s released in a n-induced fission can collide with other fissionable nuclei to produce further fissions, which produce further n’s, which cause yet more fissions.

n

n

n

n

n

n

n

n

n

n

n

n

nn

n

In this case, the number of fissions increases as 2n

where n is the number of the fission = 1,2,3,4…

This is a positive feedback, avalanche process called a chain reaction.

We have assumed that an average of 2 n’s from each fission induce another fission (K=2).


Controlled Nuclear Reaction:

If the chain reaction can be controlled, the energy released by fission, can be utilised for energy generation.

This involves control of the neutron flux producing new fissions.

We need to design the reactor for an excess of neutron flux (K>1), which can be diminished to K = 1 by the controllable absorption of excess neutrons.

These conditions have been achieved in the nuclear reactor.


Nuclear ReactorsThe following diagram shows a simplified nuclear reactor cross-section:

Base

Fuel rodsControl rodsModerator

Shielding

There are several design and operational issues that need to be taken into account for the reactor to go critical, and then be controllable in operations.

These factors relate to neutron flux.


Fuel Rods:The essential component in the fuel rods is enriched uranium.

Naturally occurring uranium consists of a mixture of two isotopes -238U (99.3%) and 235U (0.7%). However only the 235U is fissionable by thermal neutrons.Furthermore, the 238U absorbs neutrons, thereby removing neutrons from the fuel rods, and stifling the chain reaction.

For the reactor to go critical, the fraction of 235U must be increased (enriched) up to a few %, which requires special technology.

Plutonium (94239Pu) is also fissionable, and can be used in fuel rods.

Pu-239 does not occur naturally, but can be produced from U-238 by n-absorption, followed by two steps of β-decay.:

( )( )2.3daysT � �Pu Np

23minT � � NpU

Un U

21

21

00

-01-

23994

23993

00

-01-

23993

23992

23992

10

23892

=++→

=++→

→+∗

∗


Moderator:

Neutrons produced by fissions in one fuel rod, can produce fissions in other fuel rods.

The fuel rod array is immersed in a moderator, which slows theseneutrons to thermal energies in this transit between rods.

The probability of n-absorption by 235U increases as the neutrons slow, whereas, for the 238U, the opposite is the case - absorption decreases for slower n’s.

Thus, the probability of fission in uranium-235 is significantly enhanced by the action of the moderator.

Graphite and heavy water (deuterium dioxide) are used as moderator materials.


Control Rods:If we make the fuel rods sufficiently enriched, and immerse the fuel rod array in a moderator, the reactor core will go super critical. That is, we will have a meltdown situation!!

To prevent this, we must reduce the neutron flux in the reactor core.

This is achieved by inserting rods composed of a material that will readily absorb neutrons.

These control rods are inserted into the reactor core from the top. The level of n-absorption depends on how far into the core the rods are inserted,

Thus we are able to control the reproduction factor, K, of the reactor.

K = 1 is the operational condition.

K < 1 will shut down the reactor.

Cadmium is an efficient neutron absorber used in control rods.


Power Generation:The heat of the reactor core is used for electrical power generation.

This is commonly achieved by using heavy water as the moderator. This is cycled through the core and then through a heat exchanger, where its heat energy is used to generate steam, before returning to the core.

The steam so produced, is then used to drive a steam turbine, which is connected to a generator for electric power generation.

So that the moderator material does not vaporise, it is operatedat high pressure.

The steam is then condensed and recycled to the heat exchanger.

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