nuclear and particle physics lectures sumner
TRANSCRIPT
Nuclear & Particle Physics Prof. T.J. Sumner 2001/2
Nuclear Physics
[Last Update 14/11/2001]
1. Introduction ...........................................................................................................31.1 Course Information...........................................................................................3
1.1.1 Book list .....................................................................................................31.1.2 Office hours and contact details.................................................................31.1.3 Problem sheets ...........................................................................................3
1.2 What is nuclear physics? ..................................................................................31.3 Review of atomic and nuclear parameters........................................................4
2. Nuclear Properties.................................................................................................42.1 Size ...................................................................................................................42.2 Composition .....................................................................................................42.3 Mass and charge density...................................................................................52.4 Binding energy..................................................................................................52.5 Internal Structure ..............................................................................................7
3. Semi-Empirical Mass Formula ............................................................................8(a) Energy Equation ...............................................................................................8
(i) Volume energy, Ev............................................................................................8(ii) Surface tension ..............................................................................................9(iii) Electrostatic.................................................................................................10(iv) Asymmetry term..........................................................................................11(v) Pairing term.................................................................................................13
(b) SEMF summary..............................................................................................14(c) Magic numbers....................................................................................................14
4. Radio-Active Decay .............................................................................................15(a) Beta-decay – Q value......................................................................................15(b) Types of decay................................................................................................17
(i) Beta-decay ......................................................................................................17(ii) Positron-decay.............................................................................................18(iii) Electron capture...........................................................................................18(iv) Nuclear fission.............................................................................................18
(c) Weak decays ...................................................................................................18(i) Mass parabolae from SEMF...........................................................................18
(d) Alpha decay ....................................................................................................20(i) SEMF..............................................................................................................20(ii) Theory .........................................................................................................21(iii) Observation .................................................................................................24(iv) Decay series.................................................................................................25
5. Shell Model ..........................................................................................................27(a) Magic numbers ...............................................................................................28
(i) Atoms..............................................................................................................28(ii) Nuclei ..........................................................................................................28
(b) Energy levels and potentials ...........................................................................28(c) Spin-orbit coupling .........................................................................................31(d) Nuclear spin....................................................................................................31(e) Parity...............................................................................................................32
(f) Magnetic Dipole Moments ................................................................................326. Formation of Nuclei ............................................................................................33
(a) Big-Bang nucleosynthesis ..............................................................................33(b) Stellar nucleosynthesis ...................................................................................35
(i) Main Seqeunce ...............................................................................................35(ii) Red Giants ...................................................................................................36
(c) Supernova explosions .....................................................................................36(i) s-process .........................................................................................................37(ii) r-process ......................................................................................................38
7. Nuclear Reactors .................................................................................................38(a) Fission reactors ...............................................................................................38(b) Fusion reactors................................................................................................40
1. Introduction
1.1 Course InformationThese notes on Nuclear Physics constitute about half of the ‘Nuclear and ParticlePhysics’ lecture course.
1.1.1 Book listThe main recommended book, which contains most of the required material at theright level, is Nuclear & Particle Physics , by Williams. Other useful books, whichalso cover the material, and more, are Introductory Nuclear Physics , by Krane,Nuclei and Particles , by E. Segré.
1.1.2 Office hours and contact detailsDuring term time two hours will be kept as formal office hours on Thursdays between15.00 and 17.00. Other times are possible but need to be agreed individually, subjectto availability.The office room number is 1111 in the Blackett Laboratory within the AstrophysicsGroup. The internal phone number is 47552, and the email address [email protected]
Other staff associated with this course are Dr. Jordan Nash, who delivers the ParticlePhysics section of this course, and Dr. Paul Dauncey and Prof. Peter Dornan who arecourse associates for the whole course. Help and advice can also be sought fromthese if required. Their contact details areDr. Jordan Nash: room 524, telephone 47804, and email [email protected]. Paul Dauncey: room 524, telephone 47803, and email [email protected]. Peter Dornan: room 538, telephone 47882, and email [email protected]
1.1.3 Problem sheetsThere will be three problem sheets with answers for this part of the course. Theproblem sheets are designed to consolidate some of the lecture material and, in places,introduce extensions to it. The questions on the problem sheets should not be taken tobe indicative of examination questions in style or complexity. Previous examinationpapers are available from the Undergraduate Office.
1.2 What is nuclear physics?Nuclear Physics is the study of how physics needs to be applied within the atomicnucleus in order to understand how nuclei exist, what their compositions are, whattheir properties are and how they behave. The existence of nuclei depends on thepresence of a very short range force, called the strong force. A large range of nucleiis known to exist with differing compositions. However only certain compositionsseem to be allowed. Properties of nuclei that need to be understood are their size,charge, mass, angular momentum and magnetic dipole moments. Nuclei are notnecessarily passive objects and ‘behave’ with differing levels of stability, can exhibita number of decay processes and can take part in nuclear interactions.As well as allowing us to understand atomic nuclei, Nuclear Physics shows us how toexploit the behaviour of nuclei in practical uses, such as power generation in nuclear
reactors, and how to understand the formation of elements within the Universe, withclose connections to cosmology and astrophysics.
1.3 Review of atomic and nuclear parametersAtoms are composed of light negatively charged electrons in orbitals around amassive central tiny positively charged nucleus. Typically the mass of the nucleus is
enucleus mZm *2000**2≈ , where Z is the number of electrons and me is the mass of
the electron. The electrons are held in place by electrostatic forces and quantummechanical effects. As the nucleus is so small the electrostatic force seen by eachelectron is more or less central. Quantum mechanics dictates that the electronenergies fall into discrete ‘levels’. Atomic radii are characterised by the so-called
Bohr radius, m 103.5 112
2 −×≈=cm
hae
o . Typical energy level values can be
estimated from the uncertainty relationship between position and momentum,
h≈∆∆ xp . Putting oax ≈∆ gives ( ) eV few 2 2
2≈≈
eoe ma
hε [1eV = 1.6x10-19 J].
The particle/wave connection applied to the electron gives a rest mass electron scale
size (wavelength) of m 103.1 12−×≈≈ cmh
eeλ .
Nuclei are composed of a collection of quasi-equal mass particles (nucleons) heldtogether in a spherical distribution by mutual attraction. Nucleons come in twovarieties, neutrons and protons. The protons provide the positive charge of thenucleus and the mutual attraction must be strong enough to overcome the Coulombrepulsion between the protons. The mutual attraction is a two body interactionbetween pairs of nucleons and is so short range it does not necessarily extend acrossthe whole nucleus. Hence the attractive force seen by any particular nucleon is notcentral. Thinking about the nucleus as ‘particles in a box’ suggests quantummechanical effects will be present. Nuclear radii are characterised by,
m 103.1 15−×≈or . Typical nucleon ‘energy level’ values can again be estimated from
the uncertainty relationship between position and momentum, this time by putting
orx ≈∆ giving ( ) MeV few 2 2
2≈≈
eon mr
hε .
2. Nuclear Properties
2.1 SizeNuclear radii, R, are seen to increase with the total number of nucleons, A, as
31
ArR o= (1) [R]
This implies that the nuclear volume simply scales with the total number of nucleons.Each nucleon occupies a similar volume independent of its ‘position’ within thenucleus.
2.2 CompositionA nucleus with A nucleons, comprised of N neutrons and Z=(A-N) protons has a masswhich will be denoted by M(A,Z). A is the atomic mass number, not to be confusedwith Z which is the atomic number, i.e. the number of protons and electrons. Theatomic mass number is simply the sum of the numbers of neutrons and protons,
ZNA += (2) [R]
2.3 Mass and charge densityThe mass density, ρ, of a nucleus is its mass divided by its volume. As neutrons andprotons have equal masses to within 0.14% the mass density is
317
33
kg/m 1082.1constant
4
3
4
3
×==
=≈=o
n
o
n
r
m
Ar
Am
V
M
ππρ
(3) [R/D]
where mn is the mass of the neutron. This implies the nuclear density does notdepend on the atomic mass number.Similarly a charge density, ρc, can be defined as
Ar
Ze
V
Q
o
c 34
3
πρ ≈= (4) [R]
where e is the charge on the electron. It will be seen later that stable nuclei haveN~Z~A/2 (figure 1). This means the charge density is also approximately constant.
3
33C/m
8
3
4
3
oo
cr
e
Ar
Ze
V
Q
ππρ ≈≈= (5) [D]
2.4 Binding energyProperties of nuclei, such as stability and decay schemes are determined by energyconsiderations. For a given value of A the most stable isotopes will have the lowestoverall energy, which will manifest itself as the lowest rest mass energy. Thedifference between the rest mass energies of the individual nucleons and the overallnucleus is accounted for by the interaction energy between the nucleons, which isholding the nucleus together. This ‘interaction’ energy is referred to as the bindingenergy, BE, and is given by
( ) 222 ,)( cZAMcmZAcZmB npE −−+= (6) [R]
where mp is the mass of the proton. As noted earlier the proton and neutron restmasses are very similar, but not equal, and can be expressed in terms of the atomicmass unit as:-
amu 00728.1
amu 00866.1
==
p
n
m
m| kg 106605.1amu 1 27−×= (7) [G]
Figure 2 shows the binding energy per nucleon, which is the more usual way in whichthe binding energy is shown.
( )A
cZAMcmA
ZA
cZm
A
Bn
pE2
22
,)1( −−+= (8) [R]
In figure 2 the open circles denote actual measured values of the binding energy pernucleon.
Points to note from figure 2 are:-• The scale coverage on the vertical axis is an expanded range which only spans
from 7.4<BE/A<8.8 MeV/nucleon for 10<A<240. BE/A is almost a constant!• The observation data follow a smoothish curve but with some features at
particular values of A i.e. there are obvious bumps at A ~ 95, 140, 205. These‘features’ point to internal structure ⇒ shell orbitals ⇒ quantum effects ⇒ magicnumbers (see later)
• The behaviour of BE/A shows how the nuclear energy varies with A and hence willdetermine which nuclei are stable and which are likely to exhibit radioactivedecay
• The solid dots show predictions of a semi-empirical mass formula for M(A,Z).This is not rigorously founded physical theory, but is based on a simple startingmodel with successively finer corrections. This will be the subject of the nextchapter.
2.5 Internal StructureMany experiments show clear evidence for regular internal structure with nuclei.Figures 3a and 3b show two such observations. The first is for proton scattering froma nucleus (in this case 10B) as a function of proton energy. The rate of scatteringevents shows resonant enhancements at specific proton energies. The second showsγ-ray spectroscopy measurements of 238U nuclei after bombardment with 136Xenuclei. The γ-rays are only emitted at certain energies and there is a regularity to thesequence.These types of experiment provide clear evidence for detailed energy level structuresfor both the neutrons and protons within the nuclei. This suggests the system isprobably quantum mechanical with discrete energy levels. In chapter 5 we shall seehow this is likely to work. The nucleus is actually more complex to deal withquantum mechanically then atoms and we will not seek rigorous solutions but identifythe main features needed to confront the observational data.
3. Semi-Empirical Mass FormulaThe aim is to produce a predictive formula which reproduces the measured values ofBE to better than 5%.(a) Energy EquationIn general the total nuclear energy, ∑∑ +=
jj
ii cmEE 2 (9) [R]
( ) ( )
fermions close
between
mass coupling rm teenergy energy energy
rest termpairing assymetry ticelectrosta surface volume
, ...
31
32
123
222
−
−
∝∝∝
∝∝∝
↑↑↑↑↑↑
+−=++++++
AAA
RRR
cZmcmZAcZAMEEEEE pnpaeSV
The first two terms are from the so-called liquid drop model, which appeals to someproperties of liquids to characterise the main terms in the interaction energy. To theseare then added successive approximations that add in additional effects.
(i) Volume energy, EvThis is the main term in BE. As noted previously in equation 3 the nuclear densityappears to be independent of size, which implies incompressibility. This isreminiscent of a liquid. Another similarity with a liquid is that the main force holdingthe nucleus together is a very short-range force, which only acts between nearestneighbours. To a first approximation there is the same amount of energy associatedwith each nucleon added, i.e.
MeV 56.15==
+=∝
vE
vE
aA
B
AaAB(10) [R]
A term like this can also be justified by appealing to a fermi gas model for theneutrons and protons contained in the nucleus. Each of these fermion types can betreated as separate free gases in a box and their fermi energy, fε , is given by
( )
n
nfn m
n
2
3 32
22 πε h= for the neutrons (11a)[G]
and ( )
p
pfp m
n
2
3 32
22 πε
h= for the protons (11b)[G]
with VZnV
Nn pn == and being the number densities of neutrons and protons
respectively. To a first approximation it will be seen later than the number of neutrons
is roughly equal to the number of protons, i.e. 2AZN ≈≈ . Then total energy
associated with the two nucleon fermi gases assuming they are both degenerate (i.e.
fkT ε<< ) is then
( ) ( )n
fpfn mV
AAZNU
223
5
3
5
33
222 π
εεh
=+= (12)[G]
From equation (1) it can be seen that AV ∝ which then implies AU ∝ as required.Although this is a useful exercise to see how an energy level structure can give a maintotal energy term which scales as volume if the effective value of av implied byequation (12) is evaluated it turns out to be too small. This is partly because a fermigas model for the nucleon energy levels is not appropriate, as we shall see later, andalso because the assumption that the effective temperature is such that fkT ε<< is
also not valid..Figure 4 shows how the various energy terms combine to predict a form for BE/Awhich eventually approximates the data shown in figure 2 to within 5%. The volumeterm provides the main component and appears on this figure as a positive constant of15.56 MeV/nucleon.
(ii) Surface tensionThis term is a correction, which allows for the fact that the outer nucleons will be lesstightly bound than those inside the nucleus, whilst the previous volume term assumedall nucleons were equally tightly bound. As the strong binding force is very short-range only nearest neighbour interactions are important and the outer nucleons havefewer nearest neighbours. This term is negative as the overall binding energy islowered and the size of the correction scales as the number of nucleons affected,which is proportional to the surface area. Hence for this term in the binding energywe have:-
MeV 23.17
31
32
=
−=
−=−
s
sE
sE
a
AaA
B
AaB
(13) [R]
In figure 4 the effect of this negative correction can be seen.
(iii) ElectrostaticSo far the positive charge on the protons has been ignored. These will obviously repeleach other, which lowers the binding energy. Assuming the protons are distributeduniformly throughout the nucleus the electrostatic energy associated with a small shellwithin the nucleus interacting with those protons at smaller radii is
drr
r
rdrrdE
o
c
o
cc
e
ερπ
πε
ρπρπ
3
4
43
44
24
32
=
= (14) [R]
The total electrostatic energy is then
25
0
24
53
4
3
4c
o
R
co
e
Rdr
rE ρ
επρ
επ == ∫
Substituting for cρ from equation 4
r
R
dr
gives ( )
R
eZ
R
eZRE
o
oe
πε
πεπ
45
3
3453
4
22
622
225
=
=
(15)[D]
As charge is quantised the integration over radius should rigorously only start once asingle charge has been enclosed. If this is done the total electrostatic energy is
( )R
eZZE
oe πε4
1
5
3 2−= (16)[D]
Functionally this means 311 −∝∝ A
REe and a negative term must be added to the
binding energy to account for the electrostatic repulsion which lessens the overallbinding strength. Usually this is written in as:-
MeV 7.0
34
2
31
2
=
−=
−=
c
cE
cE
aA
Za
A
B
A
ZaB
(17)[R]
Figure 4 illustrates how this correction affects the overall binding energy per nucleoncurve.
(iv) Asymmetry termThis term relates to the imbalance, or asymmetry, between the numbers of neutronsand protons in a nucleus. It arises because neutrons and protons are both fermionsand each particle must have a different wave function to all other particles of the sametype. This means that as more neutrons (or protons) are added to a nucleus thoseneutrons (or protons) must occupy successively higher energy levels. This contributesto the total energy of the nucleus and it turns out that with two entirely separatespecies (neutrons and protons), with their own separate energy level states, that aminimum overall energy, for a given total number of nucleons, occurs when there areequal numbers of each species. This can be demonstrated by using the fermi gasmodel again. The total energy, U, (assuming fkT ε<< ) is given by
( ) ( )
pnnp m
VZ
Zm
VN
NUUU2
3
5
3
2
3
5
33
22
2
32
2
2ππ
hh +=+= (18)[D]
( )
+
=
32
32
32
22
2
3
5
3
V
ZZ
V
NN
mn
πh
( )
+= 3
53
5
32
32
22
2
3
5
3ZN
Vmn
πh
However at this point substitute equation (1) and the fact that A=N+Z to get
( )
−+
= 3
53
5
32
32
3
2 1
4
9
10
3NAN
ArmU
on
πh(19)[D]
If N is allowed to vary this is a minimum when
( ) 03
5
3
51
4
9
10
33
23
2
32
32
3
2
=
−−
= NAN
ArmdN
dU
on
πh(20)[D]
which requires 2
i.e. )(A
NNAN =−= . The total energy corresponding to the
minimum configuration of N=Z=A/2 is already contained within the main volumeenergy term in the binding energy. If N ≠ Z then the internal energy will be higherthan the minimum and this will weaken the binding by an amount
22 Anp UUUU −+=∆ (21a)[R]
2
22
5
3
5
3 Af
Nf
Zf
ANZ εεε −+= (21b)[G/D]
From equation (20) it can be seen that
( )
( )
−−+∝
−−+
∝∆
35
235
31
35
31
35
35
353
2
2
12
1
22
1
AZAAZAA
AZAZ
AU
(22)[D]
Now substitute δδ −=+=2
and 2
AN
AZ to get
( ) ( ){ }
−
−+
+∝
−−++∝
−
−+
+∝∆
23
5
23
5
2
235
31
35
31
35
23
5
313
5
31
22
12
11
2221
2
12
22
1
AA
AA
AA
AAAAAA
AA
AA
AA
U
δδ
δδ
δδ
(23)[D]
Assuming 1<<A
δand expanding to second order in δ gives
A
AAA
AAA
AA
U
2
2
2
22
2
22 2
!2
4
3
2
3
52
3
51
!2
4
3
2
3
52
3
51
1
δ
δδδδ
∝
−
+−+
++∝∆
(24)[D]
From the basic definition of δ this becomes
( )A
ZN
AU
22 −∝∝∆ δ(25)[D]
This now shows the basic form for a correction to the binding energy resulting froman imbalance or asymmetry between the numbers of neutrons and protons in anucleus. Writing this in the usual way gives
( )
( )
MeV 3.23
2
2
2
=
−−=
−−=
c
aE
aE
aA
ZNa
A
B
A
ZNaB
(26)[R]
Figure 4 illustrates the size of this correction as a function of A for the stable nuclei.This implies the number of neutrons is, in general, not quite equal to the number ofprotons for the stable nuclei and the reason is that a nucleus will attempt to find a stateof lowest overall energy and so there is an interplay between the previous Coulombterm and this asymmetry term. This is dealt with in detail in the next chapter.
(v) Pairing termThe main binding energy term accounts for the short-range strong force interactionbetween pairs of nucleons. It assumes all nearest neighbour pairs are equally stronglybound, such that the overall binding energy scales as the number of nucleons.However the actual nearest neighbour separation between pairs of nucleons may vary.Classically the closest a pair can get is to touch each other, which means their centre-to-centre separation is nucleon diameter. Quantum-mechanically the situation is verydifferent and nucleon wavefunctions can actually overlap. The spatial overlap caneven be complete, as long as the spin part of the wave function is different –remember both neutrons and protons are fermions, with S=½, and obey the PauliExclusion principle. This means there can be pairs of nucleons with complete spatialoverlap but with spins opposing. These pairs of nucleons will be very tightly boundby the strong force. As neutrons and protons have their own separate, and different,energy levels, this ‘pairing’ effect is most pronounced for pairs of nucleons of thesame type. Pairing tends to be complete for even numbers of neutrons/protons.Nuclei for which Z and Nare both even will show the strongest binding as all nucleons tend to be in boundpairs. Those with Z and N both odd will always have at least 1 unpaired neutron and1 unpaired proton and benefit the least from the pairing effect. Nuclei with A odd willbe intermediate to these other two cases. The main binding energy term includes acontribution for the intermediate case and the binding energy correction is put in as
MeV 12
oddboth &
odd 0
evenboth &
oddboth &
odd 0
evenboth &
23
23
21
21
=
−=
=
+=
−==
==
+==
−
−
−
−
p
pE
E
pE
pE
E
pE
a
NZAaA
B
AA
B
NZAaA
B
NZAaB
AB
NZAaB
δ
δδ
(27)[R/G]
Note that the actual form used for this term does vary from one text to another.
One effect of this correction is that, for even values of A, even-even nuclei are moretightly bound and have the lowest overall energy. This is reflected in the fact thatthere are 177 stable even-even nuclei but only 6 stable odd-odd nuclei. There are 121stable nuclei for which A is odd.Further evidence for nucleons coupling in opposing spin pairs comes frommeasurements of the magnetic moments of nuclei. The overall magnetic moment willbe the net results of adding together the magnetic moments of all the individualnucleons. There is zero contribution from pairs with equal and opposite spins. Henceif most nucleons are paired the overall nuclear magnetic moment should be close tothat expected from just a few unpaired nucleons. The neutron and proton havedifferent magnetic moments with
Np
Nn
µµµµ
793.2
913.1
=−=
where 1-27 T J 10051.5
magnetonNuclear −×=
=Nµ(28)[G]
Most measured magnetic moments are very small or zero. Even for the heavier nucleithe maximum nuclear magnetic moments to be found are below Nµ6 which implies
most nucleons are paired together.
(b) SEMF summaryThis chapter has dealt with a number of terms which are required for the nuclearbinding energy to produce a reasonable agreement with measured data. The bindingenergy is a measure of how tightly the nucleus is bound. Nuclei will tend to find astate corresponding to the tightest binding possible, and this can also be thought of asfinding a lowest energy configuration, which in turn implies a lowest mass as theoverall energy can ultimately be expressed as the rest mass energy.Recall that
( ) ( ) 2/,, cZABZmNmZAM Epn −+=Collecting all the terms together for the binding energy produces
( ) ( ) δ+−−−−−= −− 1231
32
)1(, AZNaAZZaAaAaZAB acsvE (29a)[R/G]
( ) ( ) 12234
31
)1(, −−−− +−−−−−= AAZNaAZZaAaa
A
ZABacsv
E δ (29b)[R/G/D]
with lyrespective odd-odd odd,-A even,-evenfor ,0, 21
21 −− −= AaAa ppδ .
Figure 2 shows that this expression agrees with the measured values to within 5% forall the stable nuclei with A>20. There are some values of A for which systematicvariations do occur and these are the subject of the next section.
(c) Magic numbersThe pairing effect discussed in section 3(a)(v) leads to some nuclei being more stablethan others. If the outer nucleon energy levels are s-shells and are paired this givesthe strongest coupling to the central potential well. This also happens in atoms wherethe ‘magic’ atoms are those with closed s-shells which enhances the electrostaticcoupling due to penetration of the electron wavefunctions into the nucleus. Thesemagic atoms are the hardest to ionise – they are the inert gases – see figure 5.For nuclei the effect is more complex as there are two separate species (neutrons andprotons) and each of these can have a ‘magic’ configuration. There can be ‘double-
magic’ nuclei. The most significant deviations of the measured binding energy fromthe predictions of the SEMF seen in figure 2 occur near these values. This will becovered more fully later.
4. Radio-Active DecayA nucleus might be unstable to decay if the net energy of the final products is lessthan that of the starting nucleus. The energy difference, Ei – Ef, is defined as theenergy deficit, or Q value. Decay is possible if Q>0 and the decay (or transition) isnot disallowed by any other selection rule (or conservation law). The energy releasedin the decay will come out in the kinetic energy of the fragments or in nuclearexcitation followed by relaxation and γ-ray emission.As an example the case of beta-decay, which changes a neutron into a proton insidethe nucleus will be studied.
(a) Beta-decay – Q valueIn the beta decay process a neutron inside the nucleus is converted into a proton. Anelectron is ejected from the nucleus together with a neutrino. The Q value is definedas
( ) ( ) 222 1,, cmcZAMcZAMQ e−+−= (30)[R]
Recasting this in terms of binding energies gives
( ) ( ) ( ) ( ) 22222 1,11, cmZABcmZcmNZABcZmcNmQ eEpnEpn −+++−−−−+=
( ) ( )1,,222 ++−−−= ZABZABcmcmcm EEepn (31)[D]
Using the SEMF the binding energies are
( ) ( ) ZAacsvE AZNaAZZaAaAaZAB ,123
13
2)1(, δ+−−−−−= −−
(32)[D]
( ) ( )( ) 1,123
13
211)1(1, +
−− ++−−−+−−=+ ZAacsvE AZNaZAZaAaAaZAB δThe difference between these two is
( ) ( ) ( )( ) ( ) ( )( )( ) ( )( ) ( )( ) ZAZAac
ZAZAacEE
AZNZNZNaZAa
AZNZNaAZZZaZABZAB
,1,1223
1
,1,1223
1
442
211,1,
δδ
δδ
−+−−−−+−−−=
−+−−−−−−−+−=−+
+−−
+−−
( ) ZAZAaac AaZNAaZAa ,1,113
1442 δδ −+−+−−= +
−−− (33)[D]
The pairing energy difference will depend on the starting nucleus. Consider the caseof an odd A nucleus, say odd-even, then it remains odd A after decay but becomeseven-odd. In this case 0,1, ==+ ZAZA δδ and
( ) 1131
222 442 −−− −+−−−−−= AaZNAaZAacmcmcmQ aacepn (34)[D]
Substituting values in units of MeV yields
( )( )( ) 13
1
131
131
122.934.194
122.934.18.0
12.934.1511.03.9386.939
−−
−−
−−
+−−=
−−+−=
−−+−−−=
AZZA
AZAZA
AZNZAQ
(35)[D]
As long as Q>0 beta decay can occur. As successive decays occur the value of Zincreases until stability is reached. Stability happens when
( )
+−−=
+−−=−−−
−−
131
1
131
4.1864.12.9394
122.934.1940
AAZA
AZZA
and hence ( )
+
−=−−
−
131
1
4.1864.1
2.9394
AA
AZ (36)[D]
Evaluating for A=101 and A=202 predicts stable nuclei with Z=43 and Z= 80respectively. This shows that the values of Z expected for stable nuclei are somewhatless that A/2. This can be seen clearly in figure 1. Figure 6 illustrates this moreclearly with data for nuclei around A~101. Note this figure does not keep A constant.Those with a ‘β-’ in the last column are unstable to beta decay. The nucleus withA=101 and Z=44 is stable. Nuclei with a ‘ε’ in the final column are also unstable, butthis time it is to positron decay which decreases Z and so the stability criterion can beapproached from either side.
Radioactive decay is a means of achieving the lowest energy state for a nucleus.
The higher the value of Q the more energetically favourable the decay is and ingeneral the more quickly it happens. The penultimate column in figure 6 shows the
decay time constants and illustrated this effect clearly.
Figure 6 Nuclear wall chart for Ruthenium (Z=44)
(b) Types of decayThere are four types of common decay. The first three are weak decays.(i) Beta-decayAt the quark level this is eeduuddu ν++→ −
For a specific neutron this is eepn ν++→ − (37a)[R]
For a nucleus this is eA
ZAZ eYX ν++→ −
+1 (37b)[R]
In general ( ) ( )( ) 21,, cmZAMZAMQ e−+−= (37c)[R]
Beta-decay happens for free neutrons.
(ii) Positron-decayAt the quark level this is eedduduu ν++→ +
For a specific proton this is eenp ν++→ + (38a)[R]
For a nucleus this is eA
ZAZ eYX ν++→ +
−1 (38b)[R]
In general ( ) ( )( ) 21,, cmZAMZAMQ e−−−= (38c)[R]
Beta-decay does not happen for free protons.
(iii) Electron captureIn this mode an atomic electron is captured by the nucleus. Quantum mechanicallythis is enhanced by the penetration of some atomic electron wavefunctions into thenucleus.
For a nucleus this is eA
ZAZ YXe ν+→+ −
−1 (39a)[R]
In general ( ) ( )( ) 21,, cZAMmZAMQ e −−+= (39b)[R]
(iv) Nuclear fissionIn this mode the nucleus splits into two nuclear fragments.For a nucleus this is WYX B
CBACZ
AZ +→ −
− (40a)[R]
Specific cases are:- B=1 and C=0 - neutron dripB=1 and C=1 - proton dripB=4 and C=2 - alpha decay
In general ( ) ( ) ( )( ) 2,,, cCBMCZBAMZAMQ −−−−= (40b)[R]
(c) Weak decaysIn this section a more general procedure will be used to predict which nuclei arestable. The approach is based on looking for a minimum in the rest mass for anucleus of fixed A value. Nuclei will decay towards this minimum from both sides.Nuclear masses will be derived using the SEMF and so-called ‘mass parabolae’ willbe used.
(i) Mass parabolae from SEMFFor a fixed value of A there is only 1 free parameter for a nucleus; either N or Z.Using Z the mass can be written ( ) ( ) ( ) 2/,, cZABZmmZAZAM Epn −+−= (41)[R]
Substituting for the binding energy and expanding as a power series in Z leaves( ) 2
210, ZCZCCZAM ++= (42)[R/D]
where ( )13
1
2
31
21
32
20
4
4
−−
−
+=
−−−=
+++−=
AaAaC
aAacmmC
AaAaAacAmC
ac
acnp
asvn δ
(43)[D/G]
C1 is < 0 whilst C2 >0 which means there will be a minimum with respect to Z and aplot of the mass against the value of Z will define a parabola, called the ‘massparabola’. The result of plotting this for A=101 is shown in figure 7. The smallpoints joined by the solid curve are the predicted points from the SEMF. The larger
dots are the measured values. The location of the minimum in the parabola can befound by differentiating the equation.
Figure 7: Mass parabola for an A odd Figure 8: Mass parabolae for A evennucleus (δδδδ=0, A=101) nucleus (δδδδ=±±±±∆∆∆∆, A=100)
ZCCdZ
dM21 2+=
The minimum is then when 2
1
2C
CZ = . For A=101 the minimum is at Zmin=44, as can
be seen from figure 7 and the discussion in the previous section. Also shown onfigure 7 are the decay process showing beta-decays if Z < Zmin and positron decay orelectron capture if Z>Zmin. These decays are allowed energetically as long as the massdifference is greater than the electron rest mass - this is half a division on the verticalscale of figure 7. If A is even the situation is more complicated because the pairingcorrection, δ, is no longer zero. For an even A nucleus successive decays, whetherbeta decays or positron decays, changes the type of nucleus as
→→→→→→→→
odd even odd even
even odd even odd
Z
N . The value of δ changes in sign each
time a decay occurs. This means every other nucleus has the same sign for δ andthere will effectively be two mass parabolae defined by alternate nuclei. This can beseen in figure 8, which is plotted for A=100. Nuclei on the sides of the parabolaework their way done towards the valley moving from one parabola to the other. In thecase of A=100, it can be seen from figure 8 that there are actually two end points.Nuclei decaying from low Z will end up at Z=42 as a decay to Z=43 is notenergetically allowed an this is directly due to the effect of δ changing sign. Nucleidecaying from high Z will end up at Z=44. A nucleus at Z=43 can decay either wayand which way a specific nucleus goes is purely statistical with probabilitiesdetermined by so-called branching ratios.The two stable nuclei for A=100 are both even-even as this parabola is lowest inenergy/mass. For an odd-odd nucleus to be stable the mass parabolae must be verysteep. This does happen for a few nuclei. 14N is an example of a stable odd-oddnucleus.
(d) Alpha decayThe alpha decay reaction is
HeYX AZ
AZ
42
42 +→ −
− (44)[R]The energetics of alpha decay will first be examined using the SEMF beforedeveloping a deeper theoretical foundation, which will allow other properties of thisradioactive decay mode to be studied.(i) SEMFThe Q value for the decay is given by
( ) ( ) ( )[ ]( ) ( ) ( )ZABBZAB
cMZAMZAMQ
EEE ,2,42,4
2,42,4, 2
−+−−=−−−−=
(45)[D]
For alpha decay to be energetically allowed the value of Q must be positive. Q > 0 if( ) ( ) ( )2,4,2,4 −−−> ZABZABB EEE (46)[R]
This can obviously be worked out explicitly for any particular nucleus. In general it isknow that it is only the heavier nuclei which tend to show alpha decay. For large Athe form of the binding energy per nucleon is linear with A. This allows asimplification to the calculation using
( ) ( ) AA
BBZABZAB EE ∆
∂∂≈∆=−−− 2,4, (47)[D]
The rate of change of B with A is related to the form of B/A using
( )2
1
A
B
A
B
AAA
B−
∂∂=
∂
∂ which can be rearranged to give
( )A
B
AA
BA
A
B +∂
∂=
∂∂
.
Substituting this gives
( )( )
+
∂
∂>
A
B
AA
BABE 42,4 (48)[D]
From figure 2 the rate of change of B/A with A for large A is
( )23 MeV/A 107.7 −×−≈
∂
∂
AA
B (49)[G/C]
and
( )
×−> − A
A
BBE
3107.742,4 (50)[D]
The binding energy of an alpha particle (helium nucleus) is 28.3 MeV [G]. Using thisvalue
AA
B
AA
B
3
3
107.7075.7
or 107.743.28
−
−
×+≤
×−>
(51)[D]
is then the criterion for alpha decay to be energetically possible. This criteriondefines a region of parameter space on the binding energy per nucleon plot of figure2. This shows that decay becomes energetically possible for A>151.Other points which can be seen by doing an explicit calculation of Q using the SEMFare that:• For fixed A, Q increases for large Z• For fixed Z, Q decreases with increasing A
(ii) TheoryAlthough the SEMF is very useful in showing what is possible it does not explain‘how’ the reaction occurs or give any means to calculate other observable properties,such as life-time (or half-life). These two questions will be used to encourage atheoretical approach.
The Decay MechanismTo understand how a composite particle can be emitted from a nucleus it is necessaryto think about the potential wells seen by nucleons ‘trying to escape’.
~6 MeV ~6 MeV
The neutrons, on the left, see a potential well created by the strong force. Thetopmost occupied energy level, shown here with two neutrons in it, will typically beabout 6 MeV below zero. For the protons, on the right, there is also the Coulombinteraction. Paradoxically, once inside the nucleus, this acts as an additionalimpediment to escape as the immediate potential barrier is higher. Of course once aproton is moved far enough away to get onto the downward slope it then is repelledaway. For any of these nucleons to escape they would need to tunnel through thebarrier. For an alpha particle to escape two neutrons and two protons would need tosimultaneously tunnel out. This is highly unlikely. However, what is thought tohappen is that there is a finite chance of two protons and two neutrons occasionallygetting close enough together within the nucleus to form a sub nucleus within thelarger nucleus. This can release a certain amount of energy equivalent to the effectivebinding energy of this sub-nucleus and the newly formed alpha particle finds itself ina shallower energy level as shown below. The alpha is in a level with energy Qα
~6 MeVQα
r = b
and is still classically forbidden to escape. However quantum tunnelling of this sub-nucleus as an entity is much more likely. Once the particle wavefunction hasextended out to r = b Coulomb repulsion will help repel the particle away. Theprobability of quantum mechanical tunnelling out to r = b can be estimated using arectangular potential barrier approximation.
freebarriernucleus
E
ψ~Ceikrψ~αeKr
+βe-Krψ~eikr
+Be-ikr
U
In the regions where U = 0, (i.e. inside the nucleus and beyond the barrier) the
wavenumber k is given by MEk 2=h (52a)[R] where M is the mass of the particle.
Inside the barrier ( )EUMK −= 2h (52b)[R]. A standard exercise in quantum
mechanics is then to match the wavefunction and its derivative at all boundaries to get
( )222
222
22
2
22
22
1
2
21
2
222
4
4
114
Kk
Kke
Kk
k
kK
Ke
k
K
K
keC
Kt
Kt
Kt
+=
+
+
=
+
+=
−
−
−−−
(53)[D]
where t is the barrier thickness. If KteCKkEU 22 and 2 −≈≈⇒≈ (54)[R/D/G].
A closer approximation to the 1/r Coulomb potential barrier shape can be obtained byusing a sequence of rectangular barriers of equal width, dr, but successively smallerheight.
E
rr = bR
If the thin rectangular barrier sections are numbered sequentially out from the nucleusthe overall transmission through the potential barrier is simply the product of thetransmission probabilities through each of the thin slabs, i.e.
( )
( )( )G
drErVM
drrKdrK
drKdrKdrK
ee
ee
eeeCT
b
R
b
Rii
−
−−
−−
−−−
≡∫=
∫⇒∑
=
==
22
22
2222...321
h
(54)[D]
where G is the Gamov factor.
For a Coulomb barrier the potential is
( )r
ZerV
oπε4
2 2
≈ (55)[R]
The integration ends when r = b at which point the Coulomb barrier potential is equalto the alpha particle kinetic energy:-
( )b
ZeEbV
oπε4
2 2
== (56)[D]
Evaluating the integral eventually gives (see problem sheet):-
RZE
ZG 97.296.3 −≈ (57)[G]
where E is the alpha particle energy in MeVR is the nuclear radius in fmZ is the atomic number
The value of G depends sensitively on E. This means the transmission factor whichvaries as GeT −= is incredibly sensitive to the energy, E.For example, consider the radioactive nucleus U238
92 . It has a radius of 10fm and emits
alpha particles of energy ~4.2MeV.
Hence 3785 108.1
and 8595180−− ×==
=−=eT
G(58)[C]
If the alpha energy were a factor of ~2 higher at 9MeV, say, the numbers would be13103.1 30 −×=⇒= TG . (59)[C]
In this example the transmission factor, T, has changed by a factor of 1024 for afactor of 2 change in alpha particle energy! This is a startling prediction of thistheory and it will be seen how this stands up to observational scrutiny in the nextsection.
Decay Half-lifeIf there are a certain number of radioactive nuclei to start with the rate at which theydecay is given by
Ndt
dN
eNN to
λ
λ
−=
= −
(60)[R]
The half-life, 2
1τ , is defined as the time over which half the nuclei decay, i.e.
( )( )λ
τ
λτ
λτ
5.0ln
5.0ln
5.0
21
21
21
−=
==
−e
(61)[R/D]
A theoretical estimate of λ can be obtained from the preceding theory. λ is the rate ofdecay and this will depend on the probability of quasi-alpha particles existing withinthe nucleus, the rate at which these quasi-alpha particles then collide with thepotential barrier walls, and the probability of a barrier tunnelling at each collision.
The probability of alpha particles existing within the nucleus can be characterised byusing a number giving the mean number of alpha particles averaged over a long timeperiod, Nα. Hence
fTNαλ ≈ (62)[R]
where f is the frequency of wall collision and T is the barrier transmission factor dueto tunnelling. In general Nα will be <<1. The frequency of wall collision will beroughly the distance between wall collisions divided by the alpha particle speed. Thebarrier transmission is given from the Gamov factor in the preceding theory. Puttingthese together and taking logarithms gives
RZZEERm
N
GRm
EN
GR
vN
97.296.312
lnln
12lnln
lnlnln
21
21
+−+=
−+=
−+=
−αα
αα
α
αα
ααλ
(63)[D]
The largest term in this is the third one and hence the main trend should be
21
ln−∝ ZEλ (64)[R/D].
How well this agrees with observation will be seen in the next section.
(iii) ObservationMeasurements of radioactive alpha particle decay rates exist for a number of heavyelements with values of A between 210 and 240. The logarithm of these is plotted in
figure 9 as a function of αE
Z . According to the theory this plot should show a
linear relationship. The solid symbols are the measurements and the dashed lineshows the best linear fit to the data. The fit is a good match over a range of half-lives ranging from 10µµµµs up to the age of the Universe!! The theory works over afactor of 1023 in half-life.
Figure 9: Logλλλλ against Z/E1/2 for Figure 10: Log ττττ1/2 against Qαααα for the isotopes ofvarious nuclei with a linear trend various elements. The form of these curves is wellline superimposed. matched to the Geiger-Nuttall rule.
The understanding of alpha particle decay using quantum theory and tunnelling cameabout through the work of Gamov, Condon and Gurnley. Before quantum mechanicswas available an empirical relationship between the decay rate and the alpha particlerange, Rα, in air had been established. This was known as the Geiger-Nuttall rule andtook the form:-
CRB −= ατ 10
21
10 log1
log (65)[G]
If Rα is in cm and τ in seconds then B = 57.5. The value of C depends on the elementand equals 41 for uranium. The range in air, Rα, is proportional to the alpha particleenergy, Eα, and the energy of the alpha is close to the Q value as all the excess energyis released as kinetic energy of the lightest product. Figure 10 shows plots for variouselements with radioactive isotopes and the systematic variation of half-life with Qvalue is clearly seen. The curves connecting isotopes of the same element are of theform given by the Geiger-Nuttall rule.
Figure 11: Qαααα against A over a more extended range than figure 10.
Figure 11 shows a more extensive plot covering a wider range of A. The generaltrend of decreasing Q as A increases for fixed Z is clearly seen, although there is alsosome change in this behaviour around A=210. The reason for this will becomeapparent in the next chapter.
(iv) Decay seriesThe decay of the heavy transuranic elements tends to produce chains of successivedecays as the nucleus works its way to the top of the stable nucleus locus ( see figure1). For example 238U decays to 234Th by alpha decay. 234Th is unstable to beta decayand then decays to 234Pa. Likewise 234Pa is unstable to beta decay and decays to 234U.
The decays, of various types, continue until a stable nucleus is reached, which in thiscase is 206Pb. The overall sequence is called a 'decay series' or 'chain'. Figure 12shows the whole series for 238U.
Figure 12: Decay series for 238U
The main features of the 238U decay series are:-• The decay half-life of 238U is extremely long. It is 6.5×109 years, which is a
significant fraction of the age of the Universe. Uranium is produced in supernovaexplosions (see later) and its long lifetime ensures that it is still around long afterthe supernova ejecta has dispersed and mixed into its surroundings. Uranium isfound naturally on Earth and this will have come from ancient supernovae.
• The decay series involves 15 steps with successive alpha and beta decays endingup with Pb206
82 (see figure 12)
• 238U has the longest life-time of all the nuclei in the series (apart from 206Pb,which does not decay of course). All the ‘daughter’ nuclei have short enough life-times that once a parent nucleus has decayed the whole series proceeds quicklyenough that there is no bottle-neck in the chain. This means that for a samplecontaining a large number of uranium nuclei, which has reached equilibrium:-! the ‘activity’, or number of decays per second, of each daughter type is the
same.! the amount of each daughter nucleus is proportional to its lifetime.
• Some decays leave an excited nucleus (see next chapter) and this leads to γ-rayemission as the nucleus relaxes to its ground state. Figure 13 shows a detailedexample of this for the decay of 226Ra. About 5.4% of the alpha decays leave theresulting 222Rn nucleus in an excited state. This decays in about 0.3ns emitting aγ-ray of energy 186keV.
• The end of the chain is a particular stable nucleus.• For uranium there are four possible decay series depending which isotope the
chain starts from. These are PbUPbU
BiUPbU20882
23692
20782
23992
20983
23792
20682
23892
→→
→→.
It is intriguing that all the decay series for U end up on a stable nucleus with A~208and Z~82.
Q. Why is this?A. The stable nuclei must be so tightly bound that it is not energetically favourable foranother decay to occur.
Q. Why are these nuclei so tightly bound?A. Because they are ‘magic’ or even ‘double-magic’
Q. Why are they magic nuclei?A. Because the nucleus is a quantum system, which has separate energy levelstructures for the neutrons and protons. Magic behaviour occurs when there are filledshells. This is the topic of the next chapter.
Figure 13: Detail for the alpha decay step involving 226Ra showing possible branchesfor alpha decay leaving excited nuclear states with subsequent gamma-ray emission.
5. Shell ModelThe Shell Model is an attempt to understand the existence of features in nuclearproperties which show systematic divergence from the SEMF predictions at particularvalues of N and/or Z.The assumption is that there must be quantum mechanical energy levels within thenucleus for both neutrons and protons and that the ‘magic’ numbers occur where thebinding energy is enhanced through detail in the energy level structure andoccupancy. If this is the correct reason then it should be possible to recover the magicnumbers from a predicted energy level diagram.
(a) Magic numbersMagic numbers are also displayed by quantum system with energy level structures. Ingeneral there are special characteristics associated with systems in which completeenergy level shells are occupied. A brief reminder about atomic systems will be givenfirst.(i) AtomsFor atoms we know that closed shell effects can give tighter binding of atomicelectrons, leading to increased ionisation potentials at Z = 2, 10, 18, 36, 54, 80, 86….(see figure 5). These are noble gases with filled outer p-shells causing the outerelectrons to ‘see’ more clearly the Coulomb potential.(ii) NucleiThe are a number of effects which are manifestations of the magic number behaviourof nuclei. These are:-• BE varies smoothly from the SEMF prediction near the magic numbers• It is more difficult to remove single neutrons from nuclei with magic numbers of
neutrons.• It is more difficult to remove single protons from nuclei with magic numbers of
protons.• Elements with magic numbers of protons tend to have more isotopes• Nuclei with N magic tend to have more isotomes (range of Z values)• The abundances of the elements are enhanced for magic number nuclei - see
figure 14a• Magic nuclei interact less readily with other particles - e.g. it is more difficult to
induce neutron capture as more energy is needed to change to nuclear state awayfrom a preferred magic on - see figure 14b
The magic numbers at which these magic nuclei occur are
N and/or Z = 2, 8, 20, 28, 50, 82, 126, … (66)[G]
Note that Pb20882 is actually double magic !!!
In the next section various energy level schemes will be studied to try to understandthe magic number sequence.
(b) Energy levels and potentialsPreviously the neutrons and protons were considered as separate gases of fermionscontained in a box - the nucleus. In this section the energy level structure will belooked at more closely together with the effect of using different forms for theretaining potential.
For a general quantum mechanical approach a solution to the relevant Shrodingerequation is needed.
( ) Ψ=Ψ+Ψ∇− ErVm
22
2
h(67)[R]
where it is assumed the wavefunction can be separated into( ) ( )φθ ,m
lrR Υ=Ψ (68)[R]
l is the orbital angular momentum, m is the magnetic quantum number ( lml ≤≤− )
Also ( ) hh mLllL Z =+= and 1 22 (69)[R]
Figure 14a: Abundances of elements measured Figure 14b: Cross-section for neutronin solar system. Peaks are seen around magic capture by nuclei shows lower ease ofnumber nuclei capture by magic nuclei.
The radial equation can be written using ( ) ( )rUr
rR1= (70)[R] as
( ) ( ) ( ) ( ) ( )rEUrUmr
llrV
r
rU
m=
+++∂
∂−2
2
2
22
2
1
2
hh(71)[D]
The solutions ( )rU nl are eigenfunctions which depend on n the principal quantum
number, and l the orbital angular momentum.
A similar notation will be used for nuclear levels as that used for atomic levels, i.e.
3 2 1 0
====↑↑↑↑
llll
nfndnpns
Each level can accommodate ( )122 +l neutrons/protons (72)[R].
The big question now is how to represent ( )rV .A number of options have been tried including:-
• A Coulomb potential with ( )r
VrV o−= (73a)[R]
⇒ atomic type energy levels with nl <
• A Simple Harmonic Oscillator with ( ) rrV 2
2
1 µω= (74b)[R]
• An Infinite Square Well with ( ) −∞=rV for Rr <
• A Finite Square Well with ( ) oVrV −= for Rr <
• A rounded square well (Saxon-Woods) with ( )
−+
−=
a
Rr
VrV o
exp1
(74c)[G]
None of these reproduce the magic numbers. Figure 15 shows the energy levelstructures produced by each of the potential forms. On the far right of the diagramcan be seen the numbers of fermions required to fill each of the levels and the largesteffects characteristic of magic numbers would be seen where there are the biggestgaps above the levels. If these are worked out for the first four level schemes (putting
( )122 +l fermions into each level) in figure 15 it will be seen that the predicted magicnumber sequence is 2, 8, 20, 34, 40, 58, 92, 112, 138, 168 … Numbers up to 20 arecorrect, but beyond that there is no correspondence. A more radical change to theform of the potential is needed. Looking beyond the form of the main attractivepotential, the next level modification to the Hamiltonian, prompted by experience
with atomic systems, would be to add a spin-orbit coupling term involving SLrr
⋅ .
Figure 15: Energy level diagrams for various assumed central potentials. The magicnumbers are only successfully predicted when a spin-orbit term is included.
(c) Spin-orbit couplingFor atomic systems a significant addition to the Hamiltonian beyond the simpleelectrostatic potential is a spin-orbit coupling term. As the nuclear magic numbers donot seem to be predicted correctly with a simple nuclear potential the next levelmodification to the nuclear Hamiltonian worth trying is a nucleon spin-orbit couplingterm. This is often added in to the potential using:-
( ) ( ){ }SLWrVrV ls
rr⋅+= 1 (75)[R]
The mechanics of this is dealt with in the companion ‘Particle Physics’ half of thecourse currently given by Dr. Jordan Nash.
Spin-orbit coupling creates a new angular momentum vector SLJrrr
+= (76)[R].According to the rules of quantum mechanics the coupling can produce a new vectorwith a range of possible values of j, from l+s to l-s in integer steps. For neutrons andprotons s=1/2, and so there are only two possible new angular momentum options ofl+½ and l-½. This means that each of the energy levels in the first 4 potential welloptions in figure 15 splits into two levels. This can be seen in the last ‘column’ infigure 15. A level of degeneracy in l-states has been lifted. The new levels arelabelled jnl and each state has an possible occupancy of ( )12 +j corresponding to the
allowed values of jm . From figure 15 it can be seen that the l+½ level is always
lower than the l-½ level; i.e. states with 0>⋅ SLrr
are more tightly bound in deeperenergy levels.
The predicted magic numbers after adding in the spin-orbit interaction are shown onthe far right of figure 15. There is a good match between numbers just below thelargest gaps between the levels and the observed magic number sequence. Armedwith this new energy level structure it is now possible to predict other nuclearproperties, such as nuclear spin, parity, and magnetic moment.
(d) Nuclear spinFor a given number of neutrons/protons the level structure can be used to predict thevalues of n, l, and mj for each particle. The overall properties of the are then obtainedby summing over all the particles. In the case of nuclear spin, this is simplified bynoting that in general the particles tend to form pairs with opposing spins, and so theoverall spin is caused by nucleons which are not paired; this usually means only theoutermost nucleons are relevant.
Neutrons Protons
even, even nuclei all have spin zero
n or p p or n
A - odd nuclei spin = j of add
nucleon, which isalways half-integer
n or p p or n
odd, odd nuclei↑↑ or ↑↓ or ↓↓ or ↓↑
Jtotal = Jn + Jp
e.g. 14N has Z=N=7Jn = ½h & Jp = ½h
⇒ Jtotal = 0(✖) or 1(✔)
(e) Parity[This topic is covered in more depth in the ‘Particle Physics’ half of the course by DrJordan Nash].The overall wavefunction for the nucleus, ψ, involves all the nucleons, and the parityof the nucleus is then the eignevalue, λ, of the parity operator, P, which inverts thecartesian coordinate system such that zzyyxx −→−→−→ and , , .
( ) ( ) 1 and such that ±=−== λλψψλψψ rrP (77)[R]For a single nucleon with orbital angular momentum, l
( ) ( ) ( )rr lψψ 1−=− (78)[R]For a nucleus with A nucleons the overall parity of the nucleus, P, is the product ofthe parities of all the individual nucleons.
( )∏=
−=A
i
liP1
1 (79)[R]
(f) Magnetic Dipole MomentsThe only contribution to the overall magnetic moment of a nucleus is from unpairednucleons. This means even-even nuclei all have zero nuclear magnetic moment.
For the unpaired nucleons there are two contributions to the individual magneticmoment, µ, of that nucleon. These are from its spin, s, and its orbital angularmomentum, l. These combine to give:-
lgsg ls +=µ (80)[R]
Nuclear magnetic moments are quoted in units of the nuclear magneton
≡
pm
e
2
h.
Neutrons and protons have different intrinsic magnetic moments and in units of thenuclear magneton:-
)charge! (no 0 and 8261.3 have neutrons
1 and 5856.5 have protons
=−===
ls
ls
gg
gg(81)[G]
For unpaired nucleons their value of j must be used to work out their magneticmoment using
jg j=µ (82)[R]
with ( ) ( ) ( )
( )( ) ( ) ( )
( )
+
+−++++
+
+++−+=12
111
12
111
jj
sslljjg
jj
sslljjgg lsj (83)[G]
6. Formation of NucleiVirtually all the nuclei in existence today have been created in an astrophysicalcontext. There three main scenarios:-• The lightest nuclei were mainly created during the very early Universe• Nuclei up to the most stable ones, Ni & Fe, were made by fusion processes inside
stars during their normal lifetimes, during which the energy released is responsiblefor producing their starlight. Some of these nuclei are then released into theenvironment when stars explode or eject their envelopes.
• The heaviest nuclei are made in catastrophic events creating high density neutron-rich conditions.
(a) Big-Bang nucleosynthesisBig-Bang cosmology assumes the Universe started out as a very high density and hightemperature event. The Universe then expanded adiabatically, cooling as it did soaccording to
K 105.1 2110 −×≈ tTUNIV (84)[R]
During the very early Universe the density is so high that radiation and matter are inthermal equilibrium and the effective thermal energy of the material in the Universe is
MeV 29.1 21−≈≈ tkTE UNIVUNIV (85)[R]
Whilst the density is high enough for a thermal equilibrium between the radiationfield and matter to exist there will be on-going particle-antiparticle formation goingon as long as the temperature/energy is high enough to allow pairs to be producedfrom the radiation field. This is true when:-
s 67.1K 102.1 GeV 86.1 production-pairelectron For
s 5.0K 101.2 GeV 86.1 production-pairnucleon For 10
13
<⇒×≥⇒≥
<⇒×≥⇒≥
tTE
tTE
UNIV
UNIV µ
Once nucleon production has come to an end at ~0.5 µs after the start of the Big-Bangthe Universe then contains mainly neutrons and protons in the ratio
( )
−−≈
UNIV
pn
p
n
kT
cmm
N
N2
exp (86)[D/G]
This is very close to 1, i.e. pn NN ≈After this there is very little further nucleon production and a series of predictableinteractions will now take place.
To start with there will be a number of immediate reversible processes, which directlyaffect the neutron to proton ratio. These are:-
e
e
e
epn
enp
nep
υυ
υ
++↔
+↔+
+↔+
−
+
−
The first more complex nucleus (deuterium) to form come from γ+→+ dnp , whichreleases 2.225 MeV of energy for each deuterium nucleus formed. However in thevery early stages this process is negated by a dissociation induced by γ-ray absorptiongiving npd +→+γ . Until the ambient γ-ray flux or energy becomes low enough,as the Universe expands and becomes cooler and more dilute, the deuteriumabundance will be kept down. This continues to be the case until about 250s when
K 105.9 8×≈UNIVT and the black-body γ-rays no longer have enough energy to
destroy the deuterium. However what happens now is that, as soon as the deuteriumis allowed to survive long enough, a whole chain of processes takes place leading the4He production. These processes actually suppress the deuterium build up even moreeffectively but open the way for heavier nuclei to be produced. Figure 16 showsgraphically how the mass fraction of various nuclei are though to evolve with time.
Figure 16: Formation of light nuclei during the first three minutes after the Big-Bang.
The reactions which have been included in the calculations leading to figure 16 are:-
γ
γγγ
+→+
→+
+→+
+→++→+
+→+
+→+
HenHe
HepH
nHedd
pHdd
Hepd
Hnd
dnp
43
43
3
3
3
3
There is very little production of nuclei beyond A=4 partly because there are no stablenuclei with A=5 or A=8 to act as stepping stones. However there are still a fewheavier nuclei produced via:-
pLinBe
LiHeH
BeHeHe
+→+
+→+
+→+
77
743
743
γγ
The main result of Big-Bang nucleosynthesis, which lasts for about the first 3 minutesof the Universe, is that the surviving matter content of the Universe is
107
53
4
4
102~
102~
10~
%77~
%23~
−
−
−
×
×Li
He
D
H
He
(b) Stellar nucleosynthesisThe first stars to form did so from a gas containing 77% hydrogen and 23% helium.(i) Main SeqeunceDuring the main part of their lifetime stars derive their energy by converting hydrogeninto helium. Our Sun is in this phase at the moment. This requires temperaturesaround 107 K and the starting interaction is
eedpp υ++→+ + (87)[R]
This reaction is enabled by quantum mechanical tunnelling through the Coulombbarrier. The probability of this happening is given by a Gamov factor in a similar wayto that seen earlier, with
GeT −= (88)[R]
E
mcZZG r
BA
2 and πα= (89)[G]
In this case ZA=ZB=1 for collisions between two protons, and mr is the reduced massfor the collision. Typical timescales for the p-p starting interaction is 6×109 years,which ensures the Sun burns ‘gently’, but the large numbers of protons availableenable enough interactions to provide the energy output of the Sun. Once the firstinteraction has taken place a number of further interactions take place leading tohelium production:-
HeHepLi
LieBe
BeHeHe
ppHeHeHe
Hepd
e
447
77
743
433
3
or
+→+
+→+
+→+++→+
+→+
− υγ
γ
The overall result is MeV 7.24224 4 +++→ +eeHep υ (90)[R]
For heavier stars with core temperatures ~5×107 K there are other interactions whichcan make use of existing heavier nuclei as catalysts. These heavier nuclei exist in
second generation stars in which ashes from previous stars have been incorporated.Carbon, nitrogen and oxygen are the most effective and give rise to
HeCpN
eNO
OpN
NpC
eCN
NpC
e
e
41215
1515
1514
1413
1313
1312
+→+
++→+→+
+→+
++→
+→+
+
+
υγγυ
γ
The overall result is again eeHep υ224 4 ++→ + and this has used 12C as a catalyst.
(ii) Red GiantsEventually main sequence stars will use up all the available hydrogen in the core andwhen this happen they undergo gravitational collapse and the core becomes hotter. Atthis stage the main material in the core is helium. If the temperature gets high enough(~108 K) then quantum mechanical tunnelling can happen in helium-helium collisionsleading to:-
MeV 31.9
MeV 73.4
MeV 16.7
MeV 37.7
MeV 091.0
24204
20164
16124
1284
844
=+→+
=+→+
=+→+=→+
−=→+
QMgNeHe
QNeOHe
QOCHe
QCBeHe
QBeHeHe
γγ
γ
When the helium in the core is exhausted and energy production stops, the core willcollapse again and if the temperature subsequently rises to ~109 K fusion can start inthe next generation of nuclei, such as:-
γ+→+++→+
+→+
+→+
SOO
HeHeMgOO
pNaCC
HeNeCC
321616
44241616
231212
4201212
For the most massive stars this succession of fusion, core collapse, more fusion, morecollapse can continue until the most stable nuclei have been produced. The moststable nuclei are those with A=56, i.e. CoNiFe 56
275628
5626 ,, . Beyond this there is no
further energy release and collapse would continue with no further fusion.
(c) Supernova explosionsAfter the red-giant phase is completed and all fusion processes have run their course astar will collapse until it becomes one of three types of dead star:-• A white dwarf - this is the fate of stars with masses < 1.5M
". The force of
gravity is balanced by repulsion between electrons arising out of the Pauliexclusion principle - a degeneracy pressure.
• A neutron star - these are slightly more massive than white dwarfs and electrondegeneracy pressure is not sufficient to combat gravity. The star collapses further,electrons are ‘squashed’ out of existence through recombination with protons toform neutrons, and then neutron degeneracy pressure stops the collapse.
• A black hole - these have masses so high that neutron degeneracy pressure can notstop gravitational collapse, which continues unopposed until the object is so smalland compact that even photons can not escape from the surface.
The process of collapse results in very high densities being reached and neutron richenvironments forming. Sometimes the collapse is catastrophic and results in asubsequent explosion probably caused by radiation pressure from intense neutrinoproduction. From the point of view of production of nuclei beyond A=56 the criticalpoint is that neutron rich environments can develop in dense regions. If this happensthere are two possible routes to heavy nuclei being formed. Both of these involveneutron capture by existing nuclei. If an explosion does happen this is a mechanismfor then dispersing the elements into the outside environment. The Earth andeverything on it, including us, are formed from the remnants of stellar processing andexplosion.
(i) s-processIn moderate density situations the time between successive neutron captures for aspecific nucleus is long enough that if an unstable nucleus is formed at any point ithas time to undergo beta-decay before the next capture. This effectively allows thenucleus to siddle along the line of stable nuclei, gradually increasing both A and Z.This is called the ‘slow’ or s-process. Figure 17 shows which nuclei are likely to beformed during the s-process and this can be compared to the line of stable nuclei infigure 1.
Figure 17: Build up of heavier nuclei during supernova explosions. Two processes areshown – the s-process (slow enough that beta decay happens between successive neutroncaptures) and the r-process (too rapid for beta decay to keep up with neutron capture).
(ii) r-processMuch more rapid neutron capture is likely to occur in very high density environments,such as during the collapse leading to explosion and formation of a neutron star orblack hole. In this case it may be possible to capture several neutrons before beta-decay has a chance to happen and restore stability.
e.g. minutes 6
56161
6156
υ++→→+
−eCoFe
FenFe
Typical combinations of nuclei formed by the ‘rapid’ or r-process are shown in figure17. At positions corresponding to magic numbers it becomes more difficult to capturethe next neutron and this causes a pause during which decay can occur. This explainsthe discontinuities at the magic numbers.
7. Nuclear ReactorsA useful power reactor needs to be able to produce large amounts of power, over longperiods of time, in a controllable fashion. Energy can be released in nuclearinteractions which result in nuclear masses moving closer to the peak, at A=56, in thebinding energy per nucleon curve of figure 2. This can be done in two; either bycombining light nuclei into larger ones or by splitting very large nuclei into smallerones. These correspond to fusion and fission respectively.
(a) Fission reactorsFission causes large nuclei to split into two almost equal sized fragments. Fromfigure 2 it can be seen that about 1 MeV/nucleon can be released by doing this, so foruranium, for example, some 200 MeV per fission is potentially available. The fissionprocess can be thought of as a progressive deformation of the nucleus into two halves.As the halves begin to separate they first see an attractive potential pulling them backtogether. However if they can separate far enough they get over this attractive barrierand then ‘fall’ apart. The height of the barrier opposing the break up into two roughlyequal sized fragments is called the fission barrier energy, Ef, and can be approximated
by ( )MeV 49455.02
AZE f −≈ . This is about 6.1 MeV for 238U. The height of this
barrier determines how likely it is for spontaneous fission to occur. To get usefulamounts of energy release requires sufficient amounts of the material plus a highenough fission probability. The straight forward spontaneous fission rates are nothigh enough. However in some nuclei it is possible to enhance the fission rate byusing neutron irradiation. Moreover there are some of these nuclei which alsoproduce free neutrons during the fission process. This enables a chain-reaction wherea neutron causes a fission which releases more neutrons which cause more fissions …Nuclei which exhibit this process are called ‘fissile’ nuclei. The neutrons which causefission have typically thermal energies (~1/40 eV). The neutrons emitted duringfission have much higher energies, up to 1 MeV. An example is 235U. The followingbullets show the considerations needed for a sustainable reactor process using thisisotope of uranium• 235U produces 2-3 neutrons with energies of 0.1 to 1 MeV during fission.• At least 1 of these neutrons must induce another 235U nucleus to fission. However
various other things can happen to the neutrons preventing them from causingfission in another 235U nucleus. These other processes include capture by 238U and235U without fission. The relative likelihood of these other processes and the
useful fission one are shown in figure 18. Only for low energy neutrons(<200meV) does the absorption by 235U to produce fission dominate ( ( )fnU ,235
92 .
Figure 18: Neutron capture probabilities in uranium isotopes together with subsequentdecay products (f – fission, γγγγ - gamma-ray emission) as a function of incident neutron
energy
This is compounded by the fact that the natural abundance of U is 99.3% 238U andonly 0.7% 235U. The only way an efficient process can be achieved is by deliberatelyreducing the energy of the neutrons, En, released by fission of 235U quickly enough.This is done using a ‘moderator’ in which elastic neutron scattering to transfer kineticenergy to recoiling nuclei, and hence reduce En. In addition it is possible to enhancethe fraction of 235U to some extent by ‘enrichment’.
The most efficient moderator would be one in which as much energy was lost by theneutron at each scattering. This implies a moderator with nuclei of same mass as theneutron; i.e. one containing lots of single protons, such as water. However protonsactually capture neutrons quite efficiently via γ+→+ dnp . The next best is to use
deuterium ⇒ heavy water.
The conditions for stable operation can be derived by noting that this would result inan equilibrium number of thermal neutrons, N ; i.e.
( )τ1
0 NkNdt
dN −== (91)[R]
where τN
is the rate at fissions are being induced by thermal neutrons, and τ
kNis the
rate at which new thermal neutrons are being produced. There are several effectswhich contribute to the value of k. These are:-
ευ fpfPk = (92)[G]
where υ is the number of fast neutron released per fission (~2-3), p is the probabilityof thermalisation without absorption, f is the probability of subsequent interaction
with a U nucleus (~0.9), fP is the probability that a fission is induced, ε is a
correction for other fission routes (~1.03). For stable operation k = 1 and once viabledesign has been this found allowing this (figure 19) control can be maintained usingadjustable neutron absorbing material to change the thermal neutron flux by changingp .
Figure 19: Possible configurations for fission reactors showing the major components
(b) Fusion reactorsThe biggest problem for man-made nuclear fusion reactors is to overcome theCoulomb barrier and bring two light nuclei close enough together to get tunnelling tooccur with subsequent fusion. There are several possible interactions which havebeen tried, such as:-
{
{ MeV 6.17
MeV 0.4
MeV 3.3
MeV 8.23
MeV 42.0
42
31
21
31
21
21
32
21
21
42
21
21
21
=+→+
=+→+
=+→+
=+→+
=++→+ +
QnHeHHDT
QpHHH
QnHeHH
QHeHH
DD
QeHpppp
γ
γ
A popular choice is the DT process as this is enhanced by a resonant state in anintermediate He5
2 nucleus. This reduces the typical temperature required to
~ K 10200~ 6× . The critical requirement for useful power generation is acombination of temperature, density (or pressure) and time, called the fusion product.Since work began in the 1950s this ‘fusion product’ has increased by about 7 ordersof magnitude. It is currently about 1 order of magnitude below that needed. Themain techniques in use are:-• High-power laser implosion of pellets of D and T• Magnetic confinement of plasma heated by high currents - tokamak.• Z-pinch - plasma in which a high current produces a torroidal magnetic field
which compresses the plasma - ICSTM MAGPIE type device.Figure 20 shows designs for the next generation of a tokamak-type device calledSTART.
Figure 20: Cut-away diagram of proposed fusion reactor configuration using magneticconfinement in a ‘spherical tokamak’.
The energy released in fusion reactors comes out in the form of kinetic energy of theproducts. For DT this is in neutrons. One way of extracting this energy from theneutrons is to use the following reactions
nuclei) tritiumand helium tod transfere- MeV 4.8 (plus
nuclei) tritiumand helium tod transfere-energy less MeV 2.46 (has 347
347
HHeLin
nHHeLin
+→+
++→+
Once the energy has been transferred to a charged particle (alpha or tritium nucleus) itis readily transferred to other forms by electromagnetic interactions. One option forSTART is to use a lithium ‘blanket’ (see figure 20) which then gets hot.
Much more information on fusion can be found on http::/www.fusion.org.uk