“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Chapter 3 – Nuclei ..............................................................................................................5
3.1
Perspective ...........................................................................................................8
In the last chapter… .................................................................................................8
In this chapter… .......................................................................................................9
In the next chapter… ................................................................................................9
3.2
Introducing nuclides- the nuclear force ..............................................................10
some new terminology – nucleon, nuclide, nucleus, element.................................10
3.2.1
The nuclear force without pions ..................................................................10
3.2.2
the nuclear potential energy graph .............................................................11
3.2.3
Mass loss and binding energy ....................................................................12
mass-energy bank accounts ..................................................................................14
a shocking result? ..................................................................................................14
3.2.4
Pions and the nuclear force - a technical section .......................................14
3.2.5
A technical note on data sources and calculations .....................................15
sources...................................................................................................................15
Calculations............................................................................................................17
the atomic mass unit (amu) ....................................................................................17
mass-energy accounting ........................................................................................18
3.2.6
the simplest nuclide – the deuteron ............................................................18
the proton-neutron balance ....................................................................................19
why don’t diprotons and dineutrons exist? - a technical section .............................21
3.2.7
Conclusion ..................................................................................................22
3.3
Building nucleon clusters – a simple nuclear model ..........................................23
varying the proton:neutron ratio .............................................................................23
varying the cluster size ...........................................................................................25
Summary ................................................................................................................27
3.4
Real nuclei ..........................................................................................................27
3.4.1
what does a nucleus look like? ...................................................................27
the nucleus has a fuzzy boundary ..........................................................................28
some real nuclei .....................................................................................................28
how big is a nucleon? ............................................................................................30
A football as heavy as Everest ...............................................................................31
3.4.2
the nuclide plot ...........................................................................................31
3.4.3
the nuclear valley........................................................................................34
3.4.4
The nuclear binding energy and mass curves ............................................37
carbon-12 – doing the accounting for mass and binding energy ............................39
energy “invested” in matter.....................................................................................39
mass per nucleon and binding energy per nucleon – a technical section ..............40
3.5
Cluster-12 - an individual cluster seeks stability .................................................41
3.5.1
the cluster-12 family in nuclide-space.........................................................41
3.5.2
the cluster-12 family in the nuclear valley ...................................................44
3.6
An inner structure to the nuclear cluster .............................................................45
3.6.1
nucleon pairs - clusters-100 and -101 ........................................................45
cluster-101 .............................................................................................................45
cluster-100 .............................................................................................................46
clusters 100 and 101 in the nuclear valley .............................................................47
Patterns of stability .................................................................................................48
3.6.2
Magic numbers ...........................................................................................49
magic numbers 50 and 82 ......................................................................................50
3.6.3
Super-heavy clusters and the "Island of Stability" ......................................51
3.7
Fusion and fission – changing cluster size .........................................................51
D:\116105936.doc
Page 1 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
3.7.1
fusion - the energetics of cluster growth .....................................................52
small clusters are isolated by their mutual repulsion ..............................................53
3.7.2
fission - the energetics of cluster splitting ...................................................53
the spectrum of nuclear fission reactions ...............................................................53
the pattern of decay of the known nuclides ............................................................54
the three decay modes of nuclide (77p,90n) ..........................................................56
competition between nucleon configurations..........................................................57
3.7.3
Spontaneous fission ...................................................................................58
fission is energetically viable for clusters >~100 nucleons .....................................58
spontaneous fission of very heavy nuclei ...............................................................59
the nucleus as a liquid drop....................................................................................60
why there are stable nuclides >100 nucleons ........................................................61
3.7.4
alpha-decay ................................................................................................62
why is alpha-decay so common? ...........................................................................62
alpha-decay is viable for clusters >145 nucleons ...................................................62
why there are stable nuclides >145 nucleons ........................................................64
alpha particles in a potential well ............................................................................64
inside the nuclear well ............................................................................................66
outside the nucleus ................................................................................................66
tumbling and tunnelling ..........................................................................................67
a “cloud” of possibilities? ........................................................................................69
3.7.5
Why there is an arc of stable nuclides ........................................................69
motive, means and opportunity ..............................................................................69
a surprising conclusion? .........................................................................................70
3.8
Nuclear reactions ...............................................................................................71
nuclear Lego ..........................................................................................................71
nuclear and chemical reactions ..............................................................................72
3.9
Life in the nuclear valley ....................................................................................72
3.9.1
a balance of conflicting factors creates the nuclear valley ..........................72
3.9.2
The pathways towards stability ...................................................................72
the limited options for nuclide decay ......................................................................72
average mass/nucleon is absolute, but stability is relative .....................................73
decay pathways .....................................................................................................74
nuclide “ancestors” and their “descendents” ..........................................................75
3.9.3
a rain of nucleons .......................................................................................75
draining the nuclear valley......................................................................................76
3.10 The emergent nuclide ........................................................................................77
nucleosynthesis – making nuclei ............................................................................78
3.11 Nucleosynthesis 1 - The first quarter of an hour ................................................78
3.11.1 A time-line – linking temperature, time and energy......................................79
3.11.2 Before the first threshold – temperature, T > 1015 K (energy, E > 100 GeV,
time, t < 10-10 seconds) ..............................................................................................80
3.11.3 Threshold for creating W/Z particles (mass ~80GeV), T ~ 1015 K (E ~ 100
GeV, t ~ 10-10 seconds)..............................................................................................80
3.11.4 Threshold for creating protons and neutrons (mass ~940 MeV), T ~ 1013 K
(E ~ 1000 MeV, t ~ 10-6 seconds) ..............................................................................81
3.11.5 Threshold for creating electrons (mass ~ 0.5 MeV), T ~ 6 x 109 K (E ~ 0.5
MeV, t ~ 1 second).....................................................................................................82
the end of neutron creation ....................................................................................83
one tick of the clock ................................................................................................83
D:\116105936.doc
Page 2 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
3.11.6 Protons and neutrons start to combine, T ~ 109 K (E ~ 100 keV, t ~ 200
seconds) – the first nuclei ..........................................................................................84
the deuteron ...........................................................................................................84
The first clusters of protons and neutrons ..............................................................86
the creation of helium-4 ..........................................................................................87
3.11.7 The end of nucleus formation – T ~ 3 x 108K (E ~ 30 keV, t ~ 13 minutes) .89
3.11.8 The threshold for ionisation - T ~ 3,000 K (E ~ 0.3 eV, t ~ 300,000 years) the first neutral atoms ................................................................................................90
Space becomes transparent ..................................................................................91
The first atoms .......................................................................................................92
3.11.9 review ..........................................................................................................92
From radiation-dominated to matter-dominated .....................................................92
3.12 Nucleosynthesis 2 - the next 14 billion years .....................................................94
3.12.1 The cosmic microwave background ............................................................94
photons are stretched by expanding space ............................................................94
ripples in the microwave background .....................................................................96
3.12.2 Collapsing gas clouds..................................................................................96
gravity takes over the show ....................................................................................96
the first stages of collapse ......................................................................................97
pumping up a bike tyre ...........................................................................................97
3.12.3 Stars and the elements ..............................................................................100
The basic nucleosynthesis reactions ....................................................................100
the creation of the elements .................................................................................101
3.12.4 Describing nuclear reactions .....................................................................102
3.12.5 Hydrogen-burning - the proton-proton chain..............................................104
a very slow nuclear reaction .................................................................................106
the proton-proton chain ........................................................................................107
3.12.6 The carbon (CNO) cycle ............................................................................112
3.12.7 Helium burning - the " triple alpha" process ...............................................114
a trio of cosmic coincidences ...............................................................................117
3.12.8 Filling in the gaps - capturing protons and neutrons ..................................118
The p-process – making proton-rich nuclides.......................................................118
Neutron capture processes ..................................................................................119
3.12.9 The life cycles of stars ...............................................................................122
stars of different masses ......................................................................................122
Our sun - the life cycle of a star of 1 solar mass ..................................................124
matter under pressure ..........................................................................................125
3.12.10
The alpha-process - burning all the way to iron ....................................128
carbon burning .....................................................................................................128
oxygen burning.....................................................................................................131
silicon burning ......................................................................................................131
an onion-like structure ..........................................................................................134
3.12.11
running out of nuclear fuel .....................................................................135
the “death” of a star ..............................................................................................136
a technical note - why not nickel-62? ..................................................................136
3.12.12
Supernova .............................................................................................138
The Core's Tale ....................................................................................................139
The outer layers’ tale ............................................................................................141
Supernova SN1987A ...........................................................................................145
3.13 Review .............................................................................................................148
3.13.1 Seeding inter-stellar space - the "cosmic stock-pot" ..................................148
D:\116105936.doc
Page 3 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
3.13.2 Reviewing nucleosynthesis processes ......................................................149
3.13.3 Flooding the nuclear valley ........................................................................150
the array of nuclide species..................................................................................151
3.13.4 Nuclides and the stellar eco-system ..........................................................151
3.14 The emergent atom .........................................................................................151
3.14.1 From nuclides to atoms .............................................................................151
parallels and contrasts .........................................................................................152
Element abundances and ancestries ...................................................................158
3.14 The emergent atom .........................................................................................160
3.15 The next chapter ..............................................................................................163
References ..................................................................................................................164
D:\116105936.doc
Page 4 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Chapter 3 – Nuclei
Quotes
… who knows what mysteries are hidden within the nucleus of an atom, which,
although a million million times smaller than the smallest living thing, is yet a universe
in itself?
Louis de Broglie, “Matter and Light – the New Physics”, W.W. Norton, 1939, p. 10
Neutron - walking into a bar: "A pint of beer please. How much will that be?"
Barman: "For you, no charge."
Atoms, instead of being placed in the universe fully formed, had come in kit form. …
The atoms forged long ago in the fireball of the big bang, in countless stars across the
length and breadth of the galaxy, became incorporated into human beings. They
became, in short, the atoms of curiosity. Now, why should the universe be constructed
in such a way that atoms acquire the ability to be curious about themselves?
Chown, Furnace, p. 216
In all the great cities of the world we have detached ourselves from night. If you are a
city-dweller who doesn't believe this, travel at least a hundred miles into the
countryside, mount the highest hill and stare at the sky. It is not the same sky at all. In a
city, the stars overhead glitter like lights on a distant roof-top, and the sky begins
beyond the horizon. On a clear night in the mountains, you become part of the sky. The
stars reach out and touch you, and suddenly you feel the embrace of a galaxy.
Krauss, Atom, p.96
… it is in the highest degree unlikely that this earth and sky is the only one to have
been created and that all those particles of matter outside are accomplishing nothing.
This follows from the fact that our world has been made by nature through the
spontaneous and casual collision and the multifarious, accidental, random and
purposeless congregation and coalescence of atoms whose suddenly formed
combinations could serve on each occasion as the starting-point of substantial fabrics –
earth and sea and sky and the races of living creatures.
Lucretius, “On the Nature of the Universe”, trans. Ronald Latham. Penguin, 1968….p. 91
Look around. Our familiar world is built from the debris of stars. The rocks beneath our
feet, the steel and glass in our skyscrapers, the air we breathe - all are made from
D:\116105936.doc
Page 5 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
atoms in very hot places: the interiors of stars and the outrushing shock waves of a
supernova. Blown away into space, these atoms later condensed into the sun and
planets of out solar system.
Laurence Marschall, p. 197.
It is a grand scheme, no doubt about it. The destructive power of supernovae is,
paradoxically, a major agent for creation and change in the universe. Supernovae
produce and distribute the elements, develop the solar system, and shape the evolution
of life on one of its planets. Supernovae are at the root of our existence.
Laurence Marschall, p.215.
The stuff of which we are made, was “cooked” once, in a star, and spit out.
Richard Feynman, p. 61.
How I'm rushing through this! How much each sentence in this brief story contains.
"The stars are made of the same atoms as the earth." I usually pick one small topic like
this to give a lecture on. Poets say science takes away from the beauty of the starsmere globs of gas atoms. Nothing is "mere." I too can see the stars on a desert night,
and feel them. But do I see less or more? The vastness of the heavens stretches my
imagination - stuck on this carousel my little eye can catch one-million-year-old light. A
vast pattern-of which I am a part - perhaps my stuff was belched from some forgotten
star, as one is belching there. Or see them with the greater eye of Palomar, rushing all
apart from some common starting point when they were perhaps all together. What is
the pattern, or the meaning, or the why? It does not do harm to the mystery to know a
little about it. For far more marvellous is the truth than any artists of the past imagined!
Why do the poets of the present not speak of it? What men are poets who can speak of
Jupiter if he were like a man, but if he is an immense spinning sphere of methane and
ammonia must be silent?
Richard Feynman, footnote to p. 59.
I celebrate myself, and sing myself,
And what I assume you shall assume,
For every atom belonging to me as good as belongs to you.
Walt Whitman, Song of Myself, Leaves of Grass (1881-82)
D:\116105936.doc
Page 6 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
SUBTLE. Why, what have you observ'd, sir, in our art,
Seems so impossible?
SURLY. But your whole work, no more.
That you should hatch gold in a furnace, sir,
As they do eggs in Egypt!
SUBTLE. Sir, do you Believe that eggs are hatch'd so?
SURLY. If I should?
SUBTLE. Why, I think that the greater miracle.
No egg but differs from a chicken more
Than metals in themselves.
SURLY. That cannot be.
The egg's ordain'd by nature to that end,
And is a chicken in potentia.
SUBTLE. The same we say of lead and other metals,
Which would be gold, if they had time.
MAMMON. And that Our art doth further.
SUBTLE. Ay, for 'twere absurd To think that nature in the earth bred gold
Perfect in the instant: something went before.
There must be remote matter.
SUBTLE, the Alchemist, PERTINAX SURLY, a Gamester, SIR EPICURE MAMMON, a
Knight, in , “The Alchemist”, Ben Jonson, 1610.
..the universe is set up in such a way that the production of carbon, oxygen and
nitrogen… …is an inevitable consequence of the life cycles of stars, and it is inevitable
that planets like the Earth will form around stars like the sun and be laced with complex
organic molecules, originally from interstellar clouds, by the arrival of comets. We are
made of stardust because we are a natural consequence of the existence of stars…
John Gribbin, p.186
D:\116105936.doc
Page 7 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
I cannot believe that a sense requires that we be ignorant and that wonder fades in the
face of knowledge. If that is so, then it is a poor sense of wonder.
Robert Gilmore, page x
nucleons represent a level in the organisation of matter having exceptional stability
unique in the universe...the proton and the neutron...[are]...a hundred thousand times
smaller than the smallest atom...Yet as small as that might be, there is a world hidden
inside each one.
Timothy Paul Smith, p.5
Quantum mechanics tells us that if a process is not strictly forbidden, then it must
occur. Pairs of every conceivable article and antiparticle are constantly being created
and destroyed at every location across the universe.
Kaufmann and Freedman, p.733
3.1 Perspective
In the last chapter…
… we saw the materialisation of energy and the emergence of discrete particle-waves,
which exist in three dimensions of space and one of time. We encountered the quantum
vacuum, matter and anti-matter, and the uncertainties inherent in a material world of
particle-waves.
We met the Standard Model of fermions and bosons. We saw how fermions always
remain separate, becoming the matter of the physical universe, while bosons freely
mingle and merge, and mediate the interactions, or “forces”, between the fermions. Of the
fermions, the leptons stay single, while quarks are bound by gluons into groups - either in
duos (mesons) or trios (baryons). Of all the combinations of quarks, only the lightest, the
the proton (uud), is stable. The ever so slightly heavier neutron (udd) is nearly stable, and
an isolated neutron has an average lifetime of about 15 minutes before it decays to a
proton.
We met (1) the weak interaction carried by the W  / Z0 exchange particles, that changes
the flavour of quarks, and bridges the divide between quarks and leptons and matter and
anti-matter, and (2) the colour interaction, that binds quarks into clusters. We saw how the
colour interaction “leaks” out of protons and neutrons, to give the powerful but very short
range nuclear force, mediated by virtual pions.
D:\116105936.doc
Page 8 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
We saw the emergence of the proton and neutron, the quark trios bound by the colour
interaction. We also saw how the extension of this, the nuclear force, can bind protons
and neutrons into clusters – we saw the first hint of the emergent nucleus.
In this chapter…
… we will see how the nuclear force binds protons and neutrons into hundreds of stable
nuclei. We will see the delicate balance of energies that sets a limit on the maximum
stable nucleus size, and decides the precise ratio of protons and neutrons for stability. We
will see why there is only a finite number of stable nuclei, and why the mid-sized nuclei
are most stable. We will see all the ways that a combination of protons and neutrons
reduces its size and adjusts its ratio to arrive at stability.
We will see our material universe of radiation (photons) and matter (protons, neutrons and
electrons) emerge from the cosmic fireball, and within a few minutes assemble the first
few nuclei. We will see how over the next few billion years successive generations of stars
go on to create all the known nuclei.
This chapter is wholly about quarks, and how the protons and neutrons they create,
themselves combine into big nuclear clusters. Finally, we will see how these nuclei
acquire electrons, to create the atoms , and we see the emergence of the ~100 or so
chemical elements of our atomic world.
In the next chapter…
…we'll see how, on their stable nuclear foundations, the electrons bind atoms into a rich
and subtle diversity of different chemical substances. We see solids and liquids emerge
as states of matter in addition to gases.
D:\116105936.doc
Page 9 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
3.2
Introducing nuclides- the nuclear force
We now leave behind the world of quarks and gluons, and now regard protons and
neutrons as unitary particles that attract each other by the nuclear force, mediated by the
exchange of virtual pions. We will now look at how protons and neutrons gather in
clusters, bound by this short range nuclear force.
some new terminology – nucleon, nuclide, nucleus, element
We need, at this point, to introduce some important terminology, and also explain a
necessary inconsistency. I will call a cluster of nucleons a “nucleus”, and may also give it
a chemical name, such as helium-4 or lead-208. Historically, the nucleus was identified as
the centre of every atom, defining its identity as a chemical element. But we have only just
encountered the nucleons’ ability to cluster; there are as yet no atoms, and the concept of
a chemical element is meaningless. The name “nucleus” implicitly denotes something that
is at the heart of a bigger structure, and we are looking at nuclear clusters in their own
terms – we don’t even know yet what sort of clusters they can form.
But “nucleus” is the accepted term, and other terms are derived from it: so, “nuclei” is the
plural, protons and neutrons are “nucleons”, equal citizens of the cluster community, and
a “nuclide” is a cluster of a specific number of protons and neutrons, for example, (6p,6n).
This is a member of the cluster-12 family of nuclides, all of which contain 12 nucleons;
other members are (5p,7n) and (7p,5n).
Our atomic world is made of a hundred or so chemical elements - carbon, hydrogen,
oxygen, iron and so on. What defines each element is the number of protons in the nuclei
of its atoms - for example helium has 2, carbon has 6, and oxygen has 8 - while the
number of neutrons can vary. So I will refer to some nuclear clusters by their chemical
element name, for example, helium-4 (2p,2n). This is the nucleus of helium (symbol He),
which comprises 2 protons and 2 neutrons. I will refer to the more common elements by
their chemical symbols, thus He for helium. This is rather cumbersome, we will see the
foundations of our atomic world appear, and also be able to follow what the nucleons are
doing.
3.2.1
The nuclear force without pions
Let's now consider a proton and a neutron, close enough for each to be within the other’s
pion "cloud", so they experience the attraction of the nuclear force – figure 3.1.
D:\116105936.doc
Page 10 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
nucleons - a fermions
that resists merging
p+
n0
clouds of pions - bosons
that readily merge
the "evanescent web" of pions "fluttering back and
forth" binding the proton and neutron together
Figure 3.1: A proton and neutron, close enough to be within range of their pion
“clouds”, and so experiencing the attractive nuclear force. This is a deuteron – the
simplest nuclear cluster.
In the space between the nucleons, we can perhaps see in our imagination "the pions
fluttering back and forth" creating an "invisible, evanescent web" between the proton and
neutron, thus binding them together. We have our first nuclear cluster, the simplest
nuclide, known as a deuteron.
Smith sums the nuclear force like this: “it is repulsive at about half a fermi [fm], it is
attractive at a fermi, and it essentially vanishes at about 3-4 fermis. That means that two
nucleons will be oblivious to each other if they are separated by more than 4 fermis, and
they cannot get closer to each other than half a fermi. The whole regime of nuclear
physics is essentially defined by this distance scale.”
So we will go with a simple and pragmatic description of the nuclear force as (1) mediated
by pions, and (2) being very strong, but (3) with a very short range of attraction of ~3-4 fm.
We will see that we can understand nucleon clusters very well just with ideas (2) and (3).
3.2.2
the nuclear potential energy graph
We're now in a position to read and understand the nuclear potential energy graph, where
we bring a neutron and a proton together (figure 3.2).
D:\116105936.doc
Page 11 of 168
12/02/2016
Potential Energy (MeV)
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
300
(a) the attraction is
negligible beyond
about 2 fm
250
200
(e) the two
nucleons
resist merging
and repel each
other
150
100
(d) the forces of repulsion and
attraction are balanced
(b) a weak
attraction
(c) the attraction is
greatest at a bit
more than 1 fm
50
0
-50
0.0
0.5
1.0
1.5
2.0
2.5
-100
separation (fm)
repulsion
attraction
Figure 3.2: The forces acting on two nucleons close to each other. The arrows on
the graph give an idea of the relative sizes and directions of the net force. The
shaded block arrows below the graph give some idea of the ranges and varying
strength of the forces – darker shading means stronger
The graph shows the potential energy (PE) of the system of two nucleons as zero when
their centres are 2.5 fm apart (a), because they are outside the range of the nuclear force.
At closer distances (b), the nucleons attract each other, and the system's PE becomes
negative - we have to supply energy to pull them apart. The slope of the graph is steepest
- the attractive force is strongest - at a bit more than 1 fm separation (c). But now the
nucleons themselves are getting so close that we see the fermion repulsion force begin to
act. At about 0.9 fm separation (d) the attractive and repulsive forces are balanced - the
net force is zero, and the system's PE is a minimum. If we try to push the nucleons closer
than this (e), then the repulsion force very quickly dominates, and the PE steeply
increases.
3.2.3
Mass loss and binding energy
The nuclear PE curve shows how the proton-neutron pair bound into a deuteron have a
negative PE, sitting at the bottom of an energy “well”, since the system of two nucleons
has less energy when they are together, than when they are apart. We can visualise the
D:\116105936.doc
Page 12 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
separate nucleons “falling” together into this potential energy “well”, as they are attracted
by the nuclear force – figure 3.3.
p
(b)
n
the binding energy is
lost from the system
binding
energy
force is needed to
separate the
bound nucleons
PE of the system
(a)
p n
a deuteron – the
simplest nuclide
0
binding
energy
Separate nucleons…beyond
the reach of the short-range
nuclear forces…there is no
binding energy…all massenergy of the system is
wholly in the masses.
Nucleons "fall" into the binding energy "well", and
are bound by the nuclear force … the total massenergy of the system is in the masses and the
binding energy … the binding energy is released,
leaving the mass of the nucleus less than the sum
of its separate parts.
Figure 3.3: The proton and neutron “fall” into energy well created by the attractive
nuclear force.
The total mass-energy of the system of two nucleons comprises the masses of the
particles and any energy bound up in their interaction. When the nucleons are separate,
there is no interaction, and all the system’s energy is in its mass. In combining, the
nucleons lose the binding energy, so the system’s energy is the total mass of the separate
nucleons minus the binding energy. In a sense, the binding energy is "paid for" by the
nucleons' mass-energy. The equivalence of mass and energy means that a system of
nucleons has less mass when bound in a nucleus than when separate; the deuteron
weighs less than the separate proton and neutron. The "'missing mass' is emitted in the
form of gamma radiation when the proton and neutron join to form the deuteron."
The energy that is needed to separate the bound nucleons is called, not surprisingly, the
nuclear binding energy. "The binding energy of a nucleus, which is conceptually the
energy needed to separate all the nucleons in the nucleus, is easily calculated if we
remember that it should be equal to the mass loss when the nucleus is formed." Or, as a
word equation...
D:\116105936.doc
Page 13 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
binding energy = mass-energy of separate nucleons - mass-energy of bound nuclear
cluster
mass-energy bank accounts
We can liken this to someone with two bank accounts - say, a current and a deposit
account. Money can be transferred between the accounts, but the total sum always
remains the same. In the same way, when nucleons are separate all their mass-energy is
in the "mass" account, and when they are bound by the strong nuclear force into a
nucleus, some of the "mass" account is transferred to the "binding energy" account. But
the total must always remain the same.
a shocking result?
This has been known for about a century, and the explanation in terms of the equivalence
of mass and energy is universally accepted and understood. Nonetheless, it may well be
deeply unsettling, even shocking, for it undermines our everyday experience of matter as
inviolable. How can the mass decrease? How can “stuff” disappear? It takes an effort to
look at our everyday world and see that “matter-stuff” has at some time in the past been
made from “energy-stuff”. Yet our continued existence depends on the radiation from the
sun, the result of vast amounts of matter transforming every second to light energy. We
have met an enormously important principle of nuclear interactions.
3.2.4
Pions and the nuclear force - a technical section
This is a technical bit, trying to get the nature of the nuclear force clear, that the reader
can skip without losing the thread.
The basic nucleon-nucleon (NN) interaction can be modelled in terms of the exchange of
pions. At large distances, more than about 2 fm, it is the exchange of single pions (OPE –
one pion exchange): “The force field between two protons, or two neutrons, can only be
produced by the exchange of neutral pions. Between a proton and a neutron the
exchange can be done by means of charged pions.” The medium range interaction
(between 1 and 2 fm, and in the strongly attractive part of the potential/distance graph in
figure xXx) “can be described by the exchange of two pions”. The short-range interaction
“is due to the exchange of three pions or more”, where the pions resonate and combine to
make a higher energy meson, that is responsible for a strong repulsion. Finally, at very
close range, the nucleons are assumed each to have some sort of “core”, that makes
D:\116105936.doc
Page 14 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
them repel very strongly, so the NN potential rises towards infinity for distances less than
about 0.4 fm.
Bertulani reminds us that the nuclear force binding nucleons is a residual of the colour
force operating inside every nucleon - “we can demystify the [single pion interaction] in the
sense that the exchange of real particles (pions) is, in fact, not its essential element … on
a deeper level it is an effect of the color force between color-polarized composite
particles”. Though both nucleons are colour-neutral, each is a composite of three colourcharged quarks, which can individually interact when they get close enough. So when
nucleons get very close, “they start to touch and overlap”, and can exchange quarks and
gluons. At small distances the nucleon-nucleon interaction becomes very complicated.
The interaction between two nucleons is analogous to the van der Waals force between
two molecules that are electrically neutral overall, but which contain a distribution of
electrical charge within them. The force between two molecules can be described in terms
of the exchange of photons of radiation. “The pions take the place of the photons in the
case of nuclear forces…In this way, the nuclei are bound by a type of van der Waals
force.”
Pions are described as the exchange particles that mediate, or “carry”, the nuclear force.
Nucleon interactions can effectively be described in terms of pion exchange, though
“There are small details about the nuclear force that are not completely accounted for by
the pion. However, with the addition of other, less abundant particles … which have also
been observed, a complete description of the nuclear force does exist.”
3.2.5
A technical note on data sources and calculations
sources
Nuclear data has been obtained from the Atomic Mass Data Centre (AMDC), at…
http://www.nndc.bnl.gov/amdc/
From here you can get to the 2003 Atomic Mass Evaluation (AME2003) and the AMDC
mass.mas03 database, at http://www.nndc.bnl.gov/amdc/web/masseval.html, which is based
on…
G. Audi, A.H. Wapstra and C. Thibault, “The AME2003 atomic mass evaluation (II).
Tables, graphs and references”, Nuclear Physics A, volume 729 (2003), pages 337-676
(available as the file “Ame2003b.pdf” in the AMDC files, at…
D:\116105936.doc
Page 15 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
http://www.nndc.bnl.gov/amdc/masstables/Ame2003/filel.html
A prior paper describes how the data was evaluated and prepared…
A.H. Wapstra, G. Audi and C. Thibault, “The AME2003 atomic mass evaluation (I),
Evaluation of input data, adjustment procedures", Nuclear Physics A, volume 729 (2003),
pages 129-336 (file “Ame2003a.pdf” in the AMDC files).
AMDC also provide the NUBASE2003 nuclide table, which gives mass excess and decay
characteristics, and is based on…
G. Audi, O. Bersillon, J. Blachot and A.H. Wapstra, “The NUBASE evaluation of nuclear
and decay properties”, Nuclear Physics A, volume 729 (2003), pages 3-128, which can be
downloaded from…http://www.nndc.bnl.gov/amdc/web/nubase_en.html
The NUBASE2003 table can be downloaded as an ASCII file from…s
http://www.nndc.bnl.gov/amdc/nubase/nubtab03.asc
AMDC provide the Windows program NUCLEUS to display the contents of the NUBASE
table…
http://www.nndc.bnl.gov/amdc/web/nubdisp_en.html
Nuclide data and graphs given in this chapter are from the mass.mas03 database. This
gives values for mass excess, binding energy and atomic mass in ASCII format, that can
be converted to an EXCEL spreadsheet. The AMDC mass.mas03 and NUBASE tables
give different nuclide properties, but give the same nuclide mass excess values (Audi et
al., “NUBASE”, p.6). The NUCLEUS program has been used to create 2-D and 3-D views
of the the chart of nuclides.
The National Nuclear Data Center (NNDC), at the Brookhaven National Laboratory
(BNL)…http://www.nndc.bnl.gov/ provides a wealth of information, including the excellent
Interactive chart of nuclides, at…http://www.nndc.bnl.gov/chart/ and the Nudat database,
at… http://www.nndc.bnl.gov/nudat2/
“Qcalc”, for finding Q-values for nuclear reactions, is at…http://www.nndc.bnl.gov/qcalc/
Other sources of data on nuclides are…
The Particle Data Group, http://pdg.lbl.gov/ and
The National Institute of Standards and Technology http://physics.nist.gov/cuu/index.html
(All web-sites accessed on 6 April 2011)
D:\116105936.doc
Page 16 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Calculations
I've taken 1 amu = 931.494 MeV/c2, and the electron mass, me = 0.511 MeV/c2. Values
have been rounded, usually to 1 d.p., so there will be some rounding “errors”. Masses
should strictly be given in units of MeV2/c, but I shorten this to just MeV, for simplicity, and
as many others do.
The mass of a nucleus, that is without its electrons, can be calculated two ways...
1) via the binding energy…
mnucleus = mass of free protons + mass of free neutrons – nuclear binding energy.
So mdeuteron = p + n – BEdeuteron = 938.27 + 939.57 – 2.2 = 1875.6 MeV
2) or via the mass excess (or mass defect)…
matom = (number of nucleons x amu) + mass excess for that atom.
So matom = (2 x 931.494) + 13.1 = 1876.1 meV.
But this is for the deuterium atom, a deuteron nucleus plus its electron, and we need to
subtract the electron mass…
So mdeuteron = 1876.1 – 0.5 = 1875.6 MeV.
The two deuteron mass values agree.
the atomic mass unit (amu)
We’ve seen that nucleons lose mass when they combine into nuclei. So we need some
reference nucleus to provide a standard value for nuclear masses. That standard is the
carbon-12 atom, with a nucleus of 6 protons and 6 neutrons and containing also 6
electrons. The standard atomic mass unit (amu) is 1/12th of the mass of a carbon-12
atom, equal to 931.494 MeV/c2.
Unbound free nucleons have masses significantly bigger than 1 amu. We can get at these
from their mass excess values.
For a neutron, mn = 1 amu + neutron mass excess = 931.494 + 8.071 = 939.6 MeV (1d.p.)
We can find the mass of a hydrogen (H) atom, 1 proton plus 1 electron…
mH = 1 amu + hydrogen mass excess = 931.494 + 7.3 = 938.8 MeV
and then subtracting the electron mass gives us the mass of a proton…
mp = 938.8 – 0.5 = 938.3 MeV (1d.p.)
D:\116105936.doc
Page 17 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
It’s important to know if the data is for “naked” nuclei or for whole atoms (the “mas03”
table, for example, is for atoms). Nuclear properties are very accurately measured, and
the data is freely available to all – a wonderful resource. Readers are welcome to browse
the data and use it to calculate nuclear masses and other properties. In what follows, I'll
give nucleus masses and binding energies in MeV; the example of the deuteron nucleus
shows how these have been calculated.
mass-energy accounting
The particle data resources are the result of a huge amount of meticulous work by
thousands of physicists, that have been made freely available. The sheer amount of data
can be overwhelming, and clearly the full story is very complex. I will stick to the simple
picture we’ve seen with the deuteron nucleus – that the total mass-energy of a system of
nucleons remains the same whether they are separate or bound together. The mass loss
of the separate nucleons in forming the nucleus is the energy involved in their being
bound by the nuclear force. From this point of view, I will just be doing the mass-energy
accounting.
3.2.6
the simplest nuclide – the deuteron
We're now ready to try some mass-energy accounting for the proton-neutron pair, the
mass,
MeV
deuteron – figure 3.4.
no
939.6
938.3
p+
Mass of separate
nucleons
= 939.57 + 938.27
= 1877.8 MeV
Binding Energy (BE)
= mass loss of system
= 1877.8 – 1875.6
= 2.2 MeV
So BE/nucleon
= 1.1 MeV
p+ no
937.8
mass of bound deuteron cluster = 1875.6 MeV
so average mass of each bound nucleon = 937.8 MeV
Figure 3.4: The bound deuteron has less mass than its components
This simple example shows how the mass-energy accounting works (all figures in MeV)...
D:\116105936.doc
Page 18 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
mass of bound nucleus + binding energy = total mass of separate nucleons = m p + mn
1875.6
+
2.2
=
1877.8
= 938.27 + 939.57
The mass “lost” when a proton and neutron are bound together is about 2.2 MeV – the
binding energy for this nucleus. In order to compare nuclei with different numbers of
nucleons, we’ll calculate the average binding energy per nucleon. For the deuteron it is
quite simply 2.2 / 2, which is 1.1 MeV/nucleon. This quantity, the binding energy per
nucleon, will be important when we look at the full range of nuclides.changing the protonneutron balance.
the proton-neutron balance
The deuteron (pn) is only one way to combine two nucleons - we can also have a pair of
protons, a diproton (pp), or a pair of neutrons, a dineutron (nn). We'll use these three
combinations in a hypothetical example see the effects of changing the proton-neutron
balance in the cluster. Figure 3.5 shows the mass-energies of the three nucleon pairs.
proton-rich
neutron-rich
System energy, MeV
neutron-neutron
2mn = 2x939.6
= 1879.2 MeV
no
1879.2
no
proton-proton
2mp + electrical p.e.
= 2x938.3 + 1.5 = 1878.1 MeV
1878.1
p+ p+
change pn
 in electrical p.e.
is bigger than
 in mass-energy
1877.9
change np
 in mass-energy
no change in
electrical p.e. here
p+
no
proton-neutron
mp + mn = 938.3 + 939.6 = 1877.9 MeV
Figure 3.5: The proton-neutron pairing has the least mass-energy of the three
possible pairings of two nucleons
D:\116105936.doc
Page 19 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Since a neutron is heavier than a proton, the pp pair should have the least mass and the
nn pair the most. However, we also need to consider the electric repulsion between the
protons in the pp pair. We have to force the protons together, thereby storing potential
energy in the system. This is just like a compressed spring - release the protons, and they
will fly apart. When two protons are 1 fm apart this electrical repulsion adds ~1.5 MeV to
the system’s mass-energy. Protons are lighter, but bring the energy of their mutual
electrical repulsion; neutrons have no repulsion, but are heavier. Between these extremes
there is a balance of protons and neutrons with least mass-energy, lying at the bottom of
a mass-energy “valley”.
cluster stability arises from conflicting factors
This simple example illustrates the conflicting factors that decide the stability of larger
nuclei. In general, a proton-rich nucleus has lighter constituents, but has a bigger total
mass-energy due to the protons' mutual electrical repulsion. The transformation of a
proton to a neutron (I will use the shorthand pn) will reduce this mass-energy, if the loss
of electrical pe is bigger than the gain in mass. Conversely, a neutron-rich nucleus has
heavier constituents, but a smaller electrical pe contribution to the total mass-energy. The
transformation of a neutron to a proton (np) will reduce this mass-energy, if the loss in
mass is bigger than the gain in electrical pe. We can't see this with the nn pair, but we can
see it will apply to bigger clusters of nucleons, with more than one proton.
Figure 3.5 gives a hypothetical situation, arranged to show the importance of the protonneutron balance. We will see that for real nuclides, stability is found with the balance of
protons and neutrons that minimises the cluster’s mass-energy. Proton-rich clusters
reduce their mass-energy, and move towards stability by transforming protons to
neutrons; conversely, neutron-rich nuclei transform neutrons to protons.
Of the three combinations, only the pn pairing, the deuteron, is only just stable, and the
bound diproton and dineutron do not exist. This reveals the fundamental nature of the
particle-wave, that if it is confined, in effect “squeezing” its wave aspect, then it speeds up:
“a particle confined to a very small space must move very quickly, so a large force is
required to keep it from getting away”. The nuclear force, powerful as it is, can only just
confine a proton and a neutron in a deuteron. The potential energy well for the deuteron is
~100 MeV deep, but the proton and neutron rattle around so fast that “the deuteron
almost jumps out from the potential hole!” The deuteron is weakly bound system, that
needs only 2.2 MeV to overcome the small binding energy and separate the proton and
D:\116105936.doc
Page 20 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
neutron - “It is almost unstable; it has no excited state, the smallest rotation or vibration
tears it apart”. It turns out that the nuclear force is strongest between unlike nucleons, and
so is just unable to confine pairs of neutrons or protons.
The universe’s first step in building nuclei is not a very firm one. If the strong force were a
little less strong then making the deuteron, the first step in creating the elements, would
not be possible. However, if the strong force were much stronger than it is, sufficient to
bind diprotons, then this would affect the sequence of nuclide formation in the early
universe.
why don’t diprotons and dineutrons exist? - a technical section
This is a technical section, that the reader can skip, that tries to answer these questions: if
the nuclear force is so strong, then (1) why does the deuteron, in a potential well ~100
MeV deep, have a binding energy of only ~2 MeV, and (2) why are the diproton and
dineutron unstable?
The first part of the answer is to be found in the close confinement of the nucleons in the
nucleus. Quantum mechanics tells us that there are no objects but wave-packets; no
lumpen “things”, but constructs of energy, aggregates of universal action. Nothing can
ever be still, it is always on the move, and cannot be pinned down. The more you try to
confine a particle’s position, the less certain you are of how its moving. There is a
mathematical relationship between these uncertainties…
∆p x ∆x  h
This tells us that the product of the uncertainties in momentum (p) and position (x) is
approximately equal to Planck’s constant. The more confined the particle, the faster it can
be moving. This can be rearranged as
∆p  h/∆x
If the separation of the two nucleons in deuterium is ~2 fm, then that sets the uncertainty
in their positions, ∆x. The kinetic energy, 0.5mv2 can be written as 0.5m2v2/m and so as
0.5p2/m, where p is the momentum, mv. So for a particle with an uncertainty s in its
position, will have an uncertainty in its kinetic energy given by
∆Ek  0.5∆p2/m = 0.5h2/m∆x2
= 0.5 x 6.6 x 10-34 / 1.6 x 10-27 x (2.6 x 10-15)2 = 3.4 x 10-11 Joules = 212 MeV
D:\116105936.doc
Page 21 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
The nucleons in the deuteron may be confined, but they are rattling around very fast in
their confinement, with kinetic energies that may be a couple of hundred MeV. This is
comparable with the depth of the deuteron potential energy well, about 100 MeV. The
powerful nuclear force is only just able to confine the proton and neutron within the
deuteron – “the deuteron almost jumps out from the potential hole!” The deuteron is
weakly bound system, that needs only a small energy input of 2.2 MeV to separate the
proton and neutron - “It is almost unstable; it has no excited state, the smallest rotation or
vibration tears it apart”.
The two protons in the diproton repel each other, but this adds only about 0.56 MeV,
which is much less than the deuteron’s 2.2 MeV binding energy. George Marx says that
“the simplest explanation is that the mild electrical repulsion destabilises the 2He
[diproton]”. We can see that there’s more to it than this. And also, why is the dineutron,
with no repulsion, unstable?
The second part of the answer lies in the spin of the nucleons. The nuclear force is
stronger between nucleons with parallel spin than between nucleons with opposite spin.
The proton and neutron in the deuteron, being different fermions, can have parallel spin.
But the diproton and dineutron each comprise two identical particles, and so “the Pauli
exclusion principle requires that the nucleons … have opposite spin”. Consequently, the
nuclear force is weaker and not able to hold these particles together.
It is, however, a close-run thing, for an increase of only 9% in the strength of the strong
force would make the dineutron stable. A slightly bigger increase of 13% is needed for the
diproton, in order to overcome the protons’ repulsion. The diproton falls short of being
bound and stable by only ~92 keV.
So the extremely strong nuclear force, under the most favourable conditions, is only just
able to confine two differing nucleons, with parallel spins, into a stable configuration. If the
nucleons are the same, and have opposite spins, then no stable structure is possible. In
helium-3 (2p,1n), each nucleon is attracted to two others, and the binding energy per
nucleon (BE/A) is ~2.6 Mev, more than twice that of the deuteron.
3.2.7
Conclusion
We have a nuclear force that is very powerful but short range, that can bind nucleons
when they get as close as their own size, regardless of whether they are protons or
neutrons. We thus have a means of building nucleons into bigger structures. It looks fairly
D:\116105936.doc
Page 22 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
simple, but we've seen that there are conflicting factors: protons are lighter but carry an
electrical charge, neutrons carry no charge but are heavier. We can use these simple
ideas to explore model nuclei.
3.3
model
Building nucleon clusters – a simple nuclear
This section describes the factors that decide the stability of nucleon clusters, and the
ways that unstable clusters rearrange themselves to become more stable. We’ll start with
a simple model of a cluster and see how this can start to explain how each of these
factors affects its stability. We’ll see that “the competition between gluonic [strong] and
electric forces (which tend to drive the nuclei apart) creates a rich arena of nuclear
phenomena and determines which stable chemical elements can exist in nature”.
Clusters of nucleons vary in two ways: (1) by their balance of protons to neutrons – the
p:n ratio, and (2) by their size. Thus we can think of the cluster-12 “family”, which
comprises 12 nucleons in any combination, from 12 protons (12p,0n), through more
balanced ratios, like (6p,6n), to 12 neutrons (0p,12n). Each specific combination of
protons and neutrons has its own identity and characteristics, and is called a nuclide.
Thus the nuclide (6p,6n) is very different from its nearest “sibling”, (7p,5n), in the cluster12 family.
We know that nucleons are closely clustered into roughly spherical nuclei, rather like
marbles in a string bag. Imagine a simpler, two-dimensional analogue of this, where the
nucleons are represented by flat discs, closely packed together, like coins on a table. This
would be like taking a slice through the centre of a spherical cluster. We’ll use this to
explore the effects of varying the p:n ratio, in a cluster of constant size.
varying the proton:neutron ratio
Imagine a 2-D cluster of 14 nucleons, about 4 nucleons across, and with a 1p:1n ratio,
thus (7p,7n) – the middle cluster in figure 3.6. Changing one neutron into a proton gives
cluster (a), and doing the opposite gives cluster (c). We already know that protons are
lighter than neutrons and mutually repel. So the mass-energy of a cluster such as this is
the sum of the nucleon masses, as bound by the nuclear force, and the mutual repulsions
of the electrically charged protons.
D:\116105936.doc
Page 23 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
proton-rich
(a) cluster-14 (8p,6n)
7 more p-p repulsions
mass decreases by (mn-mp)
Key: proton
neutron-rich
(b) cluster-14 (7p,7n)
ratio 1p:1n
neutron
(c) cluster-14 (6p:8n)
6 fewer p-p repulsions
mass increases by (mn-mp)
electrical repulsion
Figure 3.6: (a) the arrows show the 7 extra p-p repulsions from changing one
neutron to a proton (the dotted circle), (b) cluster-14 with a 1:1 proton:neutron ratio,
(c) the arrows show the 6 p-p repulsions that disappear when a proton is changed
to a neutron (the dotted circle)
If we change one neutron into a proton, going (b) to (a), then there will be 7 more protonproton (p-p) repulsions to add to the cluster’s mass-energy, and a loss of mass (mn-mp).
If we go the other way and replace one of the middle cluster's protons by a neutron, then
there will be 6 fewer repulsions, and a gain in mass (mn-mp). Whichever way we go, there
is the same change in mass, but a different change in the number of p-p repulsions.
If we now imagine that the energy gain due to 7 extra p-p repulsions is just slightly bigger
(by less than the value of a single p-p repulsion) than the energy loss associated with the
mass difference (mn-mp), then the cluster (a) (8p,6n) will have slightly more mass-energy
than cluster (b) (7p,7n). Going the other way, the energy due to 6 p-p repulsions will be
slightly less than the energy associated with the mass difference (m n-mp), and so cluster
(c) (6p,8n) will also have slightly more mass-energy than cluster (b) (7p,7n). Thus the
mass-energy of cluster-14 is a minimum for the combination (7p,7n); shifting the p:n ratio
either way increases it.
The general argument above suggests that every cluster will have some p:n ratio that
gives it the minimum value of mass-energy. Too many protons, and the cluster mass
increases due to their repulsions, but too many of the heavier neutrons will also increase
D:\116105936.doc
Page 24 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the mass of the cluster. Since the “lost” mass is released as binding energy, the nucleons
will be most strongly bound in the cluster with the least mass. This is a general extension
of the pattern we saw with the deuteron in section 3.2.6.
We have seen how energy becomes incorporated in matter, according to Einstein’s
equation, E = mc2, and also how a particle is unstable if it can transform to another
particle with a smaller mass. Thus we have seen that only the generation I quarks, the
ones with least mass, are stable. The heavier quarks in generations II (charm and
strange) and III (top and bottom) all decay to the light up and down flavours. This fits with
the idea that each cluster family has only one stable p:n ratio, the one with least mass. It
is energetically favourable for the other heavier configurations in to “move” towards this
stable ratio, by interchanging protons and nucleons within the cluster.
varying the cluster size
Now we’ll keep the p:n ratio constant, and vary the cluster size. A cluster of protons and
neutrons is subject to two competing forces. The powerful nuclear force attracts all
nucleons, but is very short range, reaching little further than the next-neighbour nucleon.
The much weaker electrical repulsion affects only the positively charged protons, but is
long range, and can "reach" across a big cluster. Figure 3.7 shows a series of clusters of
increasing size, all with ratio close to 1p:1n. Because we're keeping the proton:neutron
ratio the same, we don’t have to consider the proton-neutron mass difference. If we
consider the forces acting on one proton at the edge, we can see that the nuclear
attractions from the adjacent nucleons compete with electrical repulsions from all other
protons in the cluster.
(a) cluster-2
1p:1n
(b) cluster-4
2p:2n
Key: proton
neutron
D:\116105936.doc
Page 25 of 168
(c) cluster-7
3p:4n
nuclear force attraction
12/02/2016
(d) cluster-14
7p:7n
electrical repulsion
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Figure 3.7: attractive and repulsive forces in small 2-D clusters with a 1:1
proton:neutron ratio
We can see that the number of next-neighbours very quickly builds up as the cluster size
increases. In this 2-D case the maximum number is three, and this is reached in cluster-4.
So while the total attractive forces quickly increase, they also quickly reach their
maximum. In contrast, the electrical repulsions build up slowly but steadily. They reach
right across the cluster, and are roughly proportional to the number of protons - doubling
the number of protons will about double the number of repulsions on the proton at the
edge. Figure 3.8 shows the shifting balance of these competing forces, as the cluster
grows.
The nuclear force
attractions are limited
by the maximum
number of adjacent
nucleons. They can't
get bigger than this
This cluster is most
strongly bound, where
the nuclear attractions
have built up the
biggest "lead" over the
electrical repulsions
Forces on the edge proton
attractivenuclear forces, very short range, limited to adjacent nucleons...
quickly build up…
...then slowly approach a limit
electrical
repulsions
are
potentially
limitless
The strong and
electrical forces
are balanced. This
is the biggest
possible stable
cluster.
Clusters larger
than this are
unstable.
0
The electrical
repulsions start
slowly, and at first
lag behind...
maximum
stability
maximum
size
cluster size
…but keep steadily increasing, and
finally "catch up" the strong attractions
Figure 3.9: how the balance of competing nuclear and electrical forces varies with
cluster size, keeping a 1:1 proton:neutron ratio
The attractive nuclear forces quickly build up a "lead" over the repulsive electrical forces,
establishing a range of stable clusters, with a maximum stability at the lower end of the
range. But after an initial fast start, the nuclear forces are soon approaching their limit,
and their curve flattens out. The total electrical force is roughly proportional to the number
of protons, and is represented here by a straight line. So there must come a cluster in
D:\116105936.doc
Page 26 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
which the forces are balanced, and beyond which no cluster is stable. It's like the race
between the hare and the tortoise, where the hare starts off fast, but the slow patient
tortoise eventually catches up.
The model suggests that there is a maximum stable cluster size, beyond which clusters
will “down-size”, as a proton at the edge is ejected by the repulsions from all the other
protons. This is in contrast to the behaviour of clusters smaller than the stable maximum,
which look like they will stay the same size, and internally adjust their p:n ratio.
Summary
The picture that emerges shows a nucleon cluster as being subject to conflicting internal
forces, and having a mass-energy that is sensitive to its p:n ratio. The cluster may be
unstable, and either fragment, or transform to a configuration with less mass-energy. We
can view a cluster as a transient configurations of nucleons, that will transform to a more
stable configuration if it gets the chance. A nuclide, a specific configuration of protons and
neutrons, is then a stage in this process.
Our brief look at a very simple nuclear model suggests that…
(1)
a cluster of a constant total number of nucleons has a minimum mass-energy, and
is therefore stable, at some particular p:n ratio,
(2)
clusters with a constant p:n ratio have an optimum size, at which they are most
tightly bound, and…
(3)
… a maximum size, beyond which they are unstable, and will down-size by ejecting
a proton.
We’ll now look at the real nuclides and see if these inferences are valid.
3.4 Real nuclei
3.4.1
what does a nucleus look like?
We're ready now to stand "outside" the nucleus, and look at the way the nucleons are
packed inside. We're now viewing the nucleons as simple spheres, like marbles gathered
together in a spherical cluster in a string bag. However, this simple picture is static, and
we will see that the nucleons in the cluster are in incessant motion: “Perceived from
outside, a stable atomic nucleus looks like a solid citizen of nature; but way inside there is
a boiling microcosm - a world of complex unstable hadron interactions”.
D:\116105936.doc
Page 27 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
We’ll now look inside a nucleus, and try to get a “feel” of the tiny world to which nucleons
are confined.
particle scattering tells us the size
We probe very small objects like nuclei, by firing even smaller particles at them, and
seeing how these are scattered. When we use negatively charged electrons as probes,
they respond to the oppositely charged protons in the nucleus, and thereby give us an
idea how closely the nucleons are packed.
the nucleus has a fuzzy boundary
Such scattering experiments show that the nucleus is not a hard sphere, like a snooker
ball, but has a “fuzzy” edge, where its density gradually falls to zero. The nucleons are
packed tightly together in the central region, giving a constant density, and thin out at the
edge – figure 3.10.
pictorial visualisation of the distribution
of nucleons in a large nucleus
central dense region of closely
packed nucleons
the nucleus "skin" where the
nucleons are thinning out and the
density is decreasing
the outermost nucleons, where
the nucleus density has fallen to
zero
a plot of the density of nucleons
in the nucleus
Figure 3.10: upper – a pictorial representation of the protons and neutrons in a
nucleus. The effective radius can be simply expressed as 1.2 times the cube root of
the total number of nucleons (A) – se the text below.
lower – a plot of the density of nucleon packing across the nucleus
some real nuclei
D:\116105936.doc
Page 28 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Figure 3.11 shows the charge density plots for several nuclei, from small helium, with only
2 protons, to massive lead, with 82. Here the density of electric charge on the protons is
used as a measure of how tightly the nucleons are packed.
the approximate plot for hydrogen, H(1p),
goes way up beyond this graph
lead (Pb) nucleus
dense central region
skin
edge
if the lead nucleus was a
hard sphere it would
have a charge
distribution like this
Figure 3.11: charge distributions for a range of nuclei: helium, He (2p,2n); calcium,
Ca (20p,20n); nickel, Ni (28p,30n); samarium, Sm (62p,90n) and lead, Pb (82p,126n).
Only part of the plot for hydrogen, H (1p,0n) can be shown, since the charge is so
concentrated.
In the lead nucleus the nucleons are tightly packed together, out to about 5 fm, beyond
which is the "skin", where they are packed more and more loosely, with the outermost
nucleons being about 9 fm from the nucleus centre. Helium has only 2 protons, but is very
compact - about 3 fm to the outer nucleons - and so has a large central charge density.
The hydrogen nucleus, a single proton, has its single charge concentrated in a very small
region.
checking the empirical radius formula
The nuclear radius depends on the number of nucleons, and approximately follows the
empirical equation…
rnucleus = r0 A1/3
where A1/3 is the cube root of the number of nucleons, and the constant r0~1.2 fm.
The "effective" nuclear size is usually taken as the radius at a point about half way up the
charge density slope, as is shown for calcium (Ca) by the red arrow (labelled R) in figure
22. Using this, we can check the empirical formula for some of the nuclei.
D:\116105936.doc
Page 29 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
For helium-4, He-4, rHe = 1.2 x 41/3 = 1.9 fm,
and for lead-208, Pb-208, rPb = 1.2 x 2081/3 = 7.1 fm.
In each case, the formula puts the effective nucleus boundary 1-2 fm inside the point
where the charge density fades to zero. The charge on a single proton is very
concentrated, so only part of its curve can be shown, and this suggests a radius a little
less than 1 fm. This agrees fairly well with the current value of the proton charge radius,
which is given as 0.877 fm.
a simple sum
We can also do a simple sum, which might help us get more of a “feel” for the physical
presence of the nucleus. The dotted blue line shows the charge distribution in a lead
nucleus, if it was truly like a snooker ball, and had a uniform charge density and a sharp
boundary. In this form it’s a sphere with a radius close to 7 fm, and a charge density about
0.06 electron charges/fm3 (e/fm3). We can use this to estimate the number of proton
charges in the nucleus.
So the lead nucleus volume = 1.33  r3 = 1.33 x  x 73 = 1437 fm3
With a uniform charge density of ~0.06 e/fm3, this will contain 1437 x 0.06 = 86 proton
charges (remembering that protons and electrons carry the same amount of charge, just
opposite in sign). This rough figure of 86 is satisfactorily close to the figure of 82 proton
charges the lead nucleus actually holds.
how big is a nucleon?
We can use the figures above to get an idea of how closely the nucleons are packed
inside a nucleus. A lead nucleus, with a volume of 1437 fm 3 contains 208 nucleons. So
each nucleon occupies a volume of 1437/208 = 6.9 fm 3. If we treat this as a cube,
enabling the nucleons to be closely packed in a 3-D cluster, then its side length is the
cube root of 6.9 = 1.9 fm. So the nucleons are about 1.9 fm apart; they have a "size" of a
bit under 2 fm. This makes the average nucleon radius a bit less than 1 fm, which fits with
the value given above for the proton charge radius, and with Bertulani - "The radius of
protons and neutrons that compose the nucleus is of the order of 1 fm."
So, we have a working grasp of the size of a nucleon – it has a radius of about 0.8-0.9 fm,
and so is bit less than 2 fm across. If we scale nucleons up so that 1 fm becomes 1 cm,
then a proton will sit on the tip of one finger. A helium nucleus is about 3-4 finger widths,
and a lead nucleus is about two hand breadths across.
D:\116105936.doc
Page 30 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Generally, protons and neutrons are described as having a size of “about a fermi”. In the
end, thankfully, precise sizes are not an issue for us here!
A football as heavy as Everest
We’ve perhaps got a “feel” for a nucleus as something like a string bag full of marbles –
something we can imagine holding in our hand. It’s worth reminding ourselves that this is
not ordinary material. Nuclei are incredibly dense – “if a solid football were made of pure
nuclear matter, it would weigh as much as Mt Everest.”
3.4.2
the nuclide plot
Figure 3.12 is a standard plot of the known nucleon clusters - the known nuclides - in
“nucleon-space”.
the biggest stable
nucleus - cluster-208
45o
this
diagonal line is for cluster-200
(Z+N=200). The p:n ratio varies along the line,
with one stable combination, at (80p,120n).
large clusters have
ratios of 1p:~1.5n
increasingly proton-rich
arc of stable and
long-lived nuclides
(black cells)
nuclides with p:n ratios further from
the stable arc are more unstable
increasingly
neutron-rich
ratio
about
1p:1n
highly unstable nuclides, with
half lives < 0.1 s (red cells)
all nuclides on a
vertical line have the
same number of
neutrons (here N=20)
all nuclides on a
horizontal line have
the same number of
protons (here Z=20)
Figure 3.12: Decay half lives for the known nuclei. The colour code follows the
visible spectrum, with black for stable and very long-lived nuclei, then from purple
through to red as half lives get shorter. Note that this makes bismuth-209, with a
very long half-life, appear as the last stable nuclide.
The neutron and proton numbers are plotted along the x- and y-axes, respectively. The
nuclides range in size from 1 proton up to massive clusters of about 180 neutrons and
120 protons. There are thus nearly 22,000 potential proton/neutron combinations, out of
which, about 3,200 are known, and of these a mere 260 or so are stable, the thin arc of
stable nuclides, shown as black cells running diagonally across the diagram. The coloured
D:\116105936.doc
Page 31 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
cells represent known but unstable nuclides, and beyond these, in the grey area of the
plot, the clusters are so unstable and short-lived that they effectively don't exist. There is a
limit to the size of a stable cluster - the biggest has 82 protons and 208 nucleons, and is
about 14 fm, roughly 7 nucleons, across. Bigger clusters than this are found naturally and
can be artificially made, but they are all unstable.
moving around in nucleon-space
We need to be able to “move around” comfortably in nucleon-space. Vertical lines mark
clusters with the same number of neutrons, while horizontal lines mark clusters with the
same number of protons. The 45o diagonals are important lines that mark clusters of the
same size, where the total of protons and neutrons (Z + N) is constant. So the nuclides on
a diagonal line all belong to the same cluster “family”, having the same number of
nucleons, but different p:n ratios. The diagram shows the diagonal for cluster-200. Of all
the possible p/n combinations in this cluster, only one is stable – the nuclide (80p,120n).
We could draw 208 diagonals for all the clusters up to the largest stable nuclide.
stability is sensitive to p:n ratio
Up to the maximum stable size of 208, all cluster families behave similarly, in that stability
is sensitive to the p:n ratio. Shifting a cluster’s p:n ratio along the diagonal away from the
stable value in either direction makes it increasingly unstable, and consequently have a
shorter half life. The outermost nuclides are very unstable, with half-lives less than 0.1
seconds. This far out, the p:n ratio is so extreme that a nuclide may eject a proton or
neutron, so quickly that “the nucleus may not have a distinct existence before the nucleon
is emitted”.
the shifting p:n ratio and the arc of stable nuclides
In small clusters the stable ratio is very close to 1p:1n, shifting to 1p:~1.5n in the biggest
stable clusters. Having worked with the simple model, we can understand this shift.
Decreasing the p:n ratio (that is, having more neutrons than protons) will tend to stabilise
larger clusters, since converting pn will reduce the number of proton repulsions within
the cluster. But this can only go so far, for there comes a point where the benefit of
removing another repulsive proton is outweighed, literally, by the gain in mass due to the
heavier neutron. We thus have the “arc of stable nuclides”, that makes a gentle curve in
nuclide space.
cluster size and stability
D:\116105936.doc
Page 32 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
A nuclide plot, with the nuclides colour-coded to show the average binding energy for
each nucleon is shown in figure 3.13.
unstable clusters - repulsive forces
are bigger than attractive forces
the biggest stable cluster, where the
attractive and repulsive forces are just
balanced. This holds 208 nucleons and
is about 14 fm across.
the spine of the
binding energy "ridge"
following a constant sized
cluster across the “ridge”
the colour coding shows binding
energy decreasing outwards from the
central black region
the clusters with the biggest binding energies hold about 60
nucleons (~9-10 fm across), with a p:n ratio close to 1:1
Figure 3.13: Binding energies per nucleon for all nuclei. B/A = binding
energy/number of nucleons, colour coding gives values in keV/nucleon
The colour coding shows the nuclide binding energies to form a long narrow ridge in
nuclide space. The maximum binding energy occurs at a cluster size of about 60, with a
p:n ratio of about 1:1. The binding energies steadily decrease as you move away from this
size and ratio.
the binding energy ridge
If we view this as a contoured geographical map, we can see a narrow ridge, with its
summit near the south-west end. If we start at the south-west end and follow the spine
along its length, we climb steeply to the summit, then have a long gentle descent to the
biggest clusters. These still have quite large binding energies, but the steady
accumulation of protons and their growing mutual repulsions eventually make the clusters
unstable, as we saw in figure 3.9.
In walking along the ridge we are changing the cluster size, and keeping the
proton:neutron ratio pretty much the same. But what if we walk across the ridge? Now we
are keeping the cluster size the same, and changing the proton:neutron ratio. Wherever
D:\116105936.doc
Page 33 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
we cross, we'll have a short climb to the spine, and then a similarly short descent; the
binding energy rapidly peaks and then decreases.
the simple model checks out
These nuclide plots confirm the inferences of the simple 2-D model of nuclei. Cluster
stability is sensitive to p:n ratio, with the stable nuclides in each cluster-family having the
greatest binding energy, and therefore the minimum mass-energy (prediction 1). The
binding energy per nucleon is greatest for clusters of a certain size, ~60 nucleons
(prediction 2). There is a maximum stable cluster size, which is 208 nucleons (prediction
3).
stable and unstable
The nuclide chart shows us that the vast majority of the known proton/neutron
combinations are unstable, either because of their p:n ratio, or because they are too big.
The simple cluster model explains this, but questions remain: what do unstable nuclides
do? And what decides that a nuclide will be stable? We will see that an unstable nuclide
has a limited number of available options, that a nuclide’s stability depends on its
neighbours, and that most stable nuclides are not all that they seem.
3.4.3
the nuclear valley
We’ve seen that the mass of a set of nucleons is less when they are bound in a cluster,
than when they are free, and that this mass loss is released as the binding energy. The
greater the mass loss, the more binding energy is released, and the more tightly the
cluster is bound. Thus the average mass of the nucleons in a cluster is a measure of how
tightly they are bound. If we take the 2-D nuclide plot of figure xXx, and incorporate the
average nucleon mass on the vertical axis, we get the 3D plot shown in figure 3.14. This
is the “nuclear valley”, in which all the known nuclides reside.
D:\116105936.doc
Page 34 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
mass per
nucleon
The most stable nuclides, with the smallest
mass per nucleon, are at the lowest point
in the valley, around cluster-60
the largest
nuclides,
and highly
unstable
neutrons
,N
the smallest nuclides
proto
ns
Z
The long shallow mass per
nucleon curve for all stable
nuclides. (black cells)
the beta-decay
curve for one
cluster family
(~115 nucleons)
add more curves
the biggest
stable nuclide –
cluster-208
nuclides further from the arc of
stability have bigger values for
mass/nucleon, and are higher up
the valley slopes
mass
per
nucleon
highly unstable nuclides
with very short half lives,
<0.1 seconds (red cells)
Figure 3.14: A panoramic view of the nuclear valley
finding our way around the nuclear valley
Plotting the average nucleon mass on the vertical axis has transformed the flat 2-D
nuclide plot into the 3-D nuclear valley. The entirety of the valley is best seen from the
large nuclide end, with the small nuclides in the far distance. The valley is long and
narrow, with steep sides at the small nuclides end, widening in the middle, and becoming
more of a broad slope for the very large nuclides at the near end. The 2-D and 3-D plots
both show the nuclides colour coded by half life, and can be directly compared. Every one
of the roughly 3,200 cells in the valley terrain represents a known nuclide, with the cell
height representing its average nucleon mass.
Of all the possible proton/neutron combinations, these coloured cells represent the few
nuclides that exist long enough to be known and named, and of these, only a small
minority are stable. The valley is bordered by the highly unstable, short-lived nuclides.,
beyond which the nuclides “survive for such a brief period, they can hardly be said to exist
D:\116105936.doc
Page 35 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
at all”. The terrain of the nuclear valley is the “habitat”, in which the nuclides co-exist and
interact.
the nuclear valley is defined by two curves
The shape of the nuclear valley can be visualised in terms of two curves - or rather, one
long curve that is intersected by a set of transverse curves. This resembles the skeleton
of a wooden boat, in which the long curving keel is intersected by a set of transverse ribs
(figure 3.15).
the ribs: individual curves
for the cluster families
mass/nucl
eon
the keel: the arc of
stable nuclides
neutro
ns
2-D nuclide plot
 diagonal for a cluster-family
 curving arc of stability
proton
s
line for ratio 1p:
1n
Figure 3.15: The curves of the nuclear valley superimposed on the skeleton of a
boat. The 2-D plot can be seen as a projection of the 3-D plot on to a flat surface.
The 2-D nuclide plot is shown as a projection of the boat skeleton, helping us to relate the
two views. Thus the arc of stable nuclides can be seen as the “keel” of the nuclear valley,
and the set of cluster-family curves then are the “ribs”. All nuclides (up to the 208 stable
maximum) lie on a rib, and the keel connects the lowest point of each transverse rib.
the arc of stable nuclides follows the valley bottom
The 2-D plot (figure 3.12) showed the stable nuclides as the black cells, lying on a long
curving arc in nuclide space. We can now see that this stable arc runs along the bottom of
D:\116105936.doc
Page 36 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the nuclear valley (figure 3.14), dropping steeply from the single proton at the far end, and
then rising slowly to end at the largest stable nuclide, cluster-208. The very lowest point
on this stable arc, at a cluster size of about 60 nucleons, represents nuclides with the
smallest average nucleon mass.
the cluster families traverse the valley
In the 2-D plot each cluster family, containing a fixed number of nucleons, lay on a straight
diagonal line, that crossed the arc of stable nuclides. In the 3-D view, we can see that
each cluster family traverses the nuclear valley in a U-shaped curve, dropping down the
slope on one side, crossing the stable arc at the bottom, and continuing up the other side.
Just about every cluster family, up to 208 nucleons, has one, maybe two, stable nuclides,
situated at or near the bottom of its U-shaped mass-energy curve. The stable nuclide(s) in
every cluster are those with the lowest average nucleon mass, that release the maximum
binding energy.
We can now see that the further a nuclide deviates from its cluster’s stable p:n ratio, the
higher it is up the valley slope, and generally its half life is shorter. There are no stable
nuclides to be found on the valley slopes.
the valley’s lowest point is around cluster-60
The arc of stable nuclides follows the valley bottom, with the lowest point at around
cluster-60, where the nuclides have the smallest average mass per nucleon. All the other
stable nuclides have more than this minimum value, and are therefore less strongly
bound, and so, strictly speaking, they are unstable.
an inconsistency?
We seem to have an inconsistency. If we look at a transverse curve for a single cluster,
the only stable configuration is the nuclide with least mass per nucleon. However, if we
look down the length of the valley, there is an almost unbroken series of nuclides carrying
excess mass that are stable. It looks as if an individual cluster can readily rearrange itself
to the stable lowest mass configuration, but there are constraints to clusters acquiring or
ejecting nucleons, and thereby changing size. In terms of the terrain of the nuclear valley,
individual clusters can internally rearrange themselves to descend transversely towards
the valley bottom, but the stable nuclides can’t gain or lose nucleons, and move towards
cluster-60 at the lowest point of the valley.
3.4.4
The nuclear binding energy and mass curves
D:\116105936.doc
Page 37 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
If we now select only the stable nuclides, and plot their average values for binding
energy/nucleon and mass/nucleon, then we get the two graphs in figure 3.16.
binding energy/nucleon (MeV)
oxygen-16 (8p,8n)
neon-20 (10p,10n)
magnesium-24 (12p,12n)
9
8
carbon-12: 6p,6n
nucleus
mass: 12 x 931.24 =
11,175
binding energy 12 x 7.68 =
92
total
11,267
Separate nucleons
6 protons: 6 x 938.27=
5630
6 neutrons: 6 x 939.57 =
5637
total
11,267
7
6
5
4
3
iron-56: 26p,30n
nucleus
mass
52,090
binding energy
492
total
52,582
Separate nucleons
26 protons
24,395
30 neutrons
28,187
total
52,582
2
deuterium: 1p,1n
mass: 2 x 937.8= 1875.6
binding energy:2x1.1= 2.2
total
1877.8
1 proton
938.27
1 neutron
939.57
total
1877.8
mass/nucleon (MeV)
1
hydrogen: 1p
mass
938.27
binding energy 0
total
938.27
0
939
938
937
lead-208: 82p,126n
nucleus
mass
193,687
binding energy
1,636
total
195,323
Separate nucleons
82 protons
76,938
126 neutrons
118,386
total
195,324
helium-4: 2p,2n
nucleus
mass: 4 x 931.85 =
3,727.4
binding energy: 4 x 7.07= 28.3
total
3,755.7
Separate nucleons
2 protons: 2 x 938.27 = 1876.5
2 neutrons: 2 x 939.57= 1879.1
total
3,755.7
936
935
934
933
932
931
930
0
20
40
60
80
100
120
140
160
180
200
220
cluster size
Figure 3.16: Graphs of binding energies (upper) and mass (lower) per nucleon for
the stable nuclides. The figures in the box for lead-208 appear inconsistent, but this
is due to rounding to the nearest whole MeV.
mirror image graphs
Plotting the average values per nucleon means that the graphs are independent of
nucleon identity and cluster size. We have recently been introduced to these curves: the
upper graph is the profile of the binding energy ridge (figure 3.13), and the lower graph is
the profile of the bottom of the nuclear valley (figure 3.14).
D:\116105936.doc
Page 38 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
In the upper graph we can see the binding energy start at zero for a single unbound
proton, rise rapidly to a maximum of nearly 9 MeV/nucleon around cluster-60, and then
slowly decrease as the cluster size increases to the biggest stable nuclide. The lower
graph, for the average mass per nucleon, is the mirror image of the upper, since the
binding energy of the cluster is "paid for" by the nucleons' loss of mass (see section 3.2.3
on the mass-energy bank accounts).
We’ll now see how the mass-energy accounting works with carbon-12.
carbon-12 – doing the accounting for mass and binding energy
In carbon-12 (6p,6n), the average nucleon mass is only 931.24 MeV, so the cluster of 12
nucleons has a mass of 11,175 MeV, much less than the mass of the separate nucleons.
The mass difference of 92 MeV is the binding energy. Thus "we see that [carbon-12] is
considerably less massive than the sum of its twelve building blocks. The disappearing
mass is caused by the negative energy of the binding of those twelve nucleons together.
Einstein's E=mc2 explains the amount exactly !"
The “negative energy of the binding” is due to the attraction of the nuclear force. If we
want to separate the nucleons in the carbon-12 nuclide, we have to supply the binding
energy in order to restore the missing mass. The nucleus of the hydrogen atom is a single
proton, so its binding energy is zero. This is the only nucleus that suffers no mass loss.
The example of deuterium, that introduced the concept of binding energy, is now shown in
context, and there are also figures for other nuclides, with the working shown for helium-4
and carbon-12.
The general pattern for the bound nuclei is...
mass of bound nucleus + binding energy = total mass of separate unbound nucleons.
The helium-4 (2p,2n) nuclide is a very stable cluster, with a large binding energy, and so it
shows as a peak on the upper graph. There are nuclides that are multiples of He-4, that
also have relatively large binding energies, and these show as smaller peaks on the rising
binding energy curve. We'll see later how alpha particles play an important part in the
build-up of small nuclides, and also the decay of large ones.
energy “invested” in matter
If we see a nuclide as the result of the conversion of energy into matter, then the lower
graph shows how the relative “efficiency” of this conversion varies with cluster size. For
D:\116105936.doc
Page 39 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
helium-4 the conversion rate is about 932 MeV/nucleon. For the slightly better rate of 931
MeV/nucleon you can assemble clusters of about 20 or about 180 nucleons. But the
nuclide iron-56 has the lowest mass per nucleon of all the nuclides – we get a cluster of
26 protons and 30 neutrons at the bargain rate of only 930.175 MeV/nucleon. No other
nuclide offers a better deal.
mass per nucleon and binding energy per nucleon – a technical section
We’ve got to be a bit careful here, with binding energies and masses.
Average
mass/nucleon
Iron-56
(26p,30n), Nickel-60 (28p,32n), Nickel-62 (28p,34n),
930.175 MeV
930.181 MeV
930.187 MeV
Binding
energy/nucleon
Nickel-62, 8.795 MeV
iron-58, 8.792 MeV
Iron-56, 8.790 MeV
So, if you want to release the maximum binding energy per nucleon, bring together 28
protons and 34 neutrons into a nickel-62 nuclide. If you want to have the least mass per
nucleon, then assemble 26 protons and 30 neutrons into an iron-56 nuclide. So nickel-62
is the most strongly bound nuclide, but iron-56 has the least mass/nucleon. Iron-56 has
slightly less binding energy, but has a smaller proton:neutron ratio, and so fewer of the
heavier neutrons, and so has the least mass per nucleon. Similarly, if you choose the
smaller apples from the available selection, then your average price per apple is less.
So, the nuclide iron-56 (26p,30n) sets the very lowest point in the nuclear valley, but it is
not the most strongly bound. However, the values of mass/nucleon for the different
nuclides define the nuclear valley, and allow us to determine whether a nuclear reaction is
energetically favourable. If a set of nucleons can be rearranged with a loss of mass, then
there will be a release of binding energy, and the process will be energetically favourable.
Wallerstein gives a nice example of this for two nuclides in the cluster-36 family: argon-36
(18p,18n) and chlorine-36 (17p,19n). The chlorine nucleus has a slightly larger binding
energy, but is slightly heavier, because of the larger proportion of heavier neutrons. Argon
should decay to the more tightly bound chlorine, but does not do so, because this would
involve a mass increase.
Iron-56 has commonly been described as the most tightly bound nuclide, but in fact it
comes (a very close) third in the binding energy competition. Yet iron-56 is more abundant
in the universe than nickel-62. We shall see that the creation of the nuclides in stars was
influenced by the ongoing nuclear reactions as much as nuclide stability.
D:\116105936.doc
Page 40 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Iron-56 is a very important nuclide, that marks a threshold in the lives of stars and the
creation of the chemical elements. I shall describe it as the nuclide with the smallest
average nucleon mass. I shall follow the example of Williams who states, “The binding
energy per nucleon of nuclei is greatest near A = 60”, where A is the physicists’ symbol
for cluster size, and this will be the basis for our understanding fusion and fission
processes.
3.5 Cluster-12 - an individual cluster seeks stability
We will now look at an individual cluster to start to learn what unstable nuclides do, and
what decides if a nuclide is stable. We’ll take the small cluster-12 as representative for all
the clusters up to 208, the biggest cluster capable of stability.
3.5.1
the cluster-12 family in nuclide-space
Cluster-12, the “family” of 12 nucleons, lies down at the bottom left corner of nuclidespace (figure 3.12). Figure 3.17 shows a detail of this region of nuclide-space, with the
cluster-12 family of nuclides lying along the diagonal line.
D:\116105936.doc
Page 41 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
proton number, Z
All nuclides in the cluster-12 family lie on this
diagonal line so that in every case Z + N = 12.
Arc of
stable
nuclides
12
(7p,5n)
decays by pn
T1/2 = 11 ms
(8p,4n)
decays by
ejecting a proton
T1/2 < 10 -15 s
11
10
9
(6p,6n)
stable
8
7
(5p,7n)
decays by np
T1/2 = 20 ms
(4p,8n)
decays by np
T1/2 = 21 ms
6
proton-rich
5
(3p,9n)
decays by ejecting
a neutron
T1/2 < 10 -8 s
4
3
2
1
the atomic
mass of each
nuclide along
the cluster12 diagonal
is plotted in
the graph
neutron-rich
0
1
2
3
atomic mass
0
4
5
6
7
8
9
10 11 12
neutron number, N
11,230
11,220
The (6p,6n)
combination has the
lowest mass of the
cluster-12 family
11,210
11,200
11,190
11,180
11,170
4
5
6
7
8
9
neutron number
proton emission
beta-plus decay and electron capture (pn)
stable nuclide
beta-minus decay (np)
neutron emission
alpha-decay (ejection of (2p,2n)
Figure 3.17: The cluster-12 mass-energy valley. The decay modes are colour coded
as shown. The inset graph plots the atomic masses of the
The nuclides in the cluster-12 family run from the very proton-rich (8p,4n) to the very
neutron-rich (3p,9n), with only the nuclide (6p,6n) being stable, and the inset graph shows
that this has the least mass of all the nuclides in the cluster-12 family.
D:\116105936.doc
Page 42 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the weak interaction
In the last chapter we encountered the weak interaction, mediated by the W  and Z0
bosons, that interchanges protons and neutrons. We can summarise these nuclear
reactions as follows…
beta-minus (np) decay: n0  p+ + e- + 
beta-plus (pn) decay: p+  n0 + e+ + 
We saw in chapter 2 that this reaction is energetically favourable if it reduces the atomic
mass by more than 2 electron masses (2 x m e = 1.02 MeV). If this is not the case, then
the transformation can be achieved if the nuclide captures an electron.
electron capture (effectively pn): p+ + e-  n0 + 
In this chapter, we will see that these beta-decay reactions are the means for an unstable
configuration of nucleons to move towards stability.
the beta-decays of the unstable nuclides
Nuclides with extreme p:n ratios eject a nucleon, to become a smaller cluster, and nearer
to the arc of stable nuclides. In less extreme cases, nuclides undergo one form of betadecay, transforming one nucleon type to the other, and moving one diagonal step towards
the stable ratio. Thus, “beta decay is the process by which complex nuclei return towards
the line of stability by emitting electrons or positrons, or by electron capture.”
the cluster-12 family traverses the nuclear valley
The cluster-12 family traverses the nuclear valley, and so a plot of the nuclide masses is a
cross-section of the valley at that point. The nuclides in the cluster-12 family bear out the
predictions of the simple model. A cluster has an optimum p:n ratio that minimises its
mass: too many protons, and the mass increases due to their electrical repulsions, but too
many of the heavier neutrons also increases the mass. Nuclides on either side of this
optimum ratio are unstable due to their excess mass, and undergo beta-decay to
interchange protons and neutrons and reduce it. Nuclides further from the stable (6p,6n)
ratio have bigger masses and shorter half lives.
a sequence of decay steps
Unstable nucleon combinations transform themselves in a sequence of decay steps to
become the stable minimum mass nuclide. Nuclide (4p,8n) becomes (5p,7n), which
becomes the stable (6p,6n), each decay “moving” the configuration one cell down the
D:\116105936.doc
Page 43 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
slope of the nuclear valley. The beta decay reactions enable an individual cluster to
reduce its mass and work towards stability.
Clusters very far from stability take a more violent course and eject a proton or neutron,
thus becoming a cluster one nucleon smaller, which then follows a similar decay
sequence.
3.5.2
the cluster-12 family in the nuclear valley
Figure 3.18 shows the view of the nuclear valley as seen from the high peaks at the end
with the smallest nuclides. The arc of stable nuclides runs along the valley bottom into the
far distance, with the proton-rich nuclides on the left, and the neutron-rich on the right.
the nuclides with the least mass per
nucleon, at the bottom of the valley
eject a neutron
(lilac)
unstable proton-rich
nuclides move towards
stability by transforming
pn (yellow cells)
unstable neutron-rich
nuclides move towards
stability by transforming
np (blue cells)
the dotted line
shows the nuclides
in the cluster-12
family
eject a
proton (pink)
(4p,8n)
(8p,4n)
proton
number
(3p,9n)
(7p,5n)
(5p,7n)
neutron
number
(6p,6n)
Figure 3.18: Looking down to the bottom of the nuclear valley from the high peaks
of the small nuclides. The decay of the cluster-12 family of nuclides can be seen in
the context of the bigger pattern. Nuclides are colour-coded by decay mode.
The dotted line shows the atomic mass curve for the cluster-12 family (see back to the
last figure). The stable configuration of this cluster has the lowest mass-energy, and sits
at the bottom of the "valley of stability". The nuclides to either side are like “boulders
D:\116105936.doc
Page 44 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
perched up the side of the valley”, and they decay so as to reduce their larger massenergies, "falling" down the energy slopes towards stability.
boulders roll, water flows, nuclides decay
Every cluster-family has its own mass-energy curve, that traverses the valley and
intersects the arc of stability. The unstable nuclides interchange protons and neutrons,
and move in a series of beta-decays downhill towards the arc of stability at the valley
bottom. Boulders roll, water flows, nucleon clusters decay – all seeking the state of lowest
energy.
energy invested in matter
We have seen that matter is created from energy, according to the equation…
E = mc2
or rather,
m = E / c2
Thus energy is converted into matter, and in a sense, energy is incorporated or invested
in matter. The valley of stability for cluster-12 shows the “efficiency” of converting energy
to 12 nucleons of matter. The most efficient configuration of matter is as the cluster
(6p,6n), anything else has a greater mass. Thus the unstable cluster-12 nuclides decay to
reduce their mass, and thereby increase efficiency of the energy to matter conversion.
3.6 An inner structure to the nuclear cluster
We’ve seen that the average mass per nucleon of a cluster is determined by its p:n ratio
and its size, and in explaining this, we have regarded nucleons in the cluster as if they are
all mixed up, like marbles in a string bag, or like molecules in a drop of liquid water. But
there is evidence that indicates that nuclei have an inner organisation and structure, that
favours configurations with pairs and “magic” numbers of nucleons.
3.6.1
nucleon pairs - clusters-100 and -101
It has been found experimentally that like nucleons “pair up” in the nucleus, so two
protons are always more strongly bound than a single proton – and the same goes for
neutrons. The greater stability of nucleon pairs affects the stability of clusters bigger than
~20. We’ll look at two examples – cluster-101 and cluster-100.
cluster-101
Figure 3.19 shows the mass of different p:n combinations in cluster-101.
D:\116105936.doc
Page 45 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
proton-rich
pn reaction or
electron capture
neutron-rich
np decay
atomic mass (MeV)
94,002
large mass losses, decay
by electron capture and
pn reaction
94,001
94,000
93,999
Decay by np
93,998
small mass loss,
decay only by
electron capture
93,997
93,996
93,995
Only one combination
is stable
93,994
93,993
93,992
54
55
56
57
58
59
60
neutron number
all proton and
neutron
numbers are
odd-even– one
Figure 3.19: The mass-energy curve for cluster-101 has only one minimum
stable p:n ratio
Proton number 47
odd or even? oe
46
eo
45
oe
44
eo
43
oe
42
eo
41
oe
The pattern is similar to what we saw for cluster-12. There is one stable p:n combination,
with the smallest mass. On either side of this the masses increase, due either to the
electrical repulsions between protons on the left, or to the heavier neutrons on the right.
Neutron-rich combinations decay by the beta-minus np reaction. Proton-rich
combinations can convert a proton to a neutron, either by the beta-plus pn reaction, or
by electron capture. We’ve seen that if the atomic mass difference is bigger than twice the
electron mass, then both of these reactions can occur, and this applies to nuclides
(47p,54n) and (46p,55n). However, nuclide (45p,56n) can only decay to (44p,57n) by
electron capture, since the atomic mass difference is only about 0.6 MeV.
Cluster-101 has an odd number of nucleons, so whatever the p:n combination, one of
them will be odd, and one even. These are shown as “oe” and “eo” under the graph. Thus
there is one unpaired nucleon in every combination.
cluster-100
Now compare this with cluster-100, whose atomic mass graph for the different p:n
combinations is quite different (figure 3.20).
D:\116105936.doc
Page 46 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
atomic mass (MeV)
proton-rich
pn reaction or
electron capture
neutron-rich
np decay
93,074
93,072
electron
capture
only
93,070
two decay routes...
 by electron capture
(~2 in 1,000 decays)
 or by np (~100%)
not completely
stable...half life 8.5 x
1018 y, decay by double
np to (44p,56n)
93,068
93,066
93,064
the only truly
stable combination
93,062
93,060
53
54
55
56
57
58
Proton number 47
Odd or even? oo
46
ee
45
oo
44
ee
43
oo
42
ee
59
60
neutron number
41
oo
40
ee
proton and neutron
numbers are either
“oo” or “ee”
Figure 3.20: The mass-energy curve for cluster-100 has two minima - but only one
truly stable p:n ratio
This cluster has an even number of nucleons, so the proton and neutron numbers are
either both odd, “oo”, or both even, “ee”. In the broad valley of stability the alternation
between “oo” and “ee” affects the cluster mass, so there are now two minima in the curve
– the main one at (44p,56n) and a second at (42p,58n). The first is truly stable - this is the
combination with the lowest mass-energy. The second must surely be stable too, after all,
the nuclides to either side have larger masses. But if this nuclide (42p,58n) could
somehow transform two neutrons to protons at the same time, then it could become the
nuclide (44p,56n). And this nuclide is in fact unstable, though it is often shown as stable,
having a very long half life of of 8.5 x 1018 years, and its decay mode is by a double np
decay.
Thus cluster-100 is only truly stable in one combination, (44p,56n), but effectively stable in
another, (42p,58n). The combination (43p,57n), in between these two, has two possible
decay reactions, one giving a much bigger reduction in mass than the other. The main
decay is by the np reaction, losing a neutron, and about 3.2 MeV of mass. But in about
about 2 in 1,000 decays it can “move” the other way and capture an electron, and thus
lose a proton, but now only lose about 0.2 MeV of mass.
clusters 100 and 101 in the nuclear valley
D:\116105936.doc
Page 47 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
If we position ourselves in the nuclear valley around cluster-90, looking in the direction of
the heavy nuclides, we see the view shown in figure 3.21.
the peak of the largest
nuclides, marking one
end of the valley
the largest stable
nuclide, lead-208
the cluster-101 curve
has only one minimum one true stable nuclide
44p,57n
the cluster-100 curve has
two minima – one truly
stable nuclide, and one
with a very long half-life
44p,56n
stable
42p,58n
very slow decay by double
np reaction to 44p,56n
43p,57n
has 2 modes of decay to a
lower mass nuclide
Figure 3.21: The curves for clusters 100 and 101 nuclide families are in the
foreground, with the mass peak for the biggest nuclides in the background.
Nuclides are colour-coded by half life.
In the middle distance the end of the black line marks the largest stable nuclide, and
beyond that is the summit of the largest known nuclides, marking the end of the valley. In
the foreground are the nuclides in clusters 100 and 101. The curve for cluster-101 has
one true minimum, and one truly stable nuclide. The curve for cluster-100 has a little
bump in it, with a minimum on either side. Thus there is one truly stable nuclide (44p,56n),
and the other has such a long half life that it is marked here as stable. Thus the slight
effects of nucleon pairings deep in the nucleus reveal themselves in the terrain of the
nuclear valley.
Patterns of stability
If we extend what we’ve seen with cluster-100, this suggests that clusters with an even
number of nucleons can have more than one stable combination. This is true, and we can
now see why there are more than 208 stable nuclides. Stable nuclei with even numbers of
D:\116105936.doc
Page 48 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
protons are more numerous than ones with odd numbers, and the same is true for
neutron numbers. Consequently, stable nuclides with even numbers of nucleons, “oo” and
“ee”, are more numerous, and the latter even more so. Conversely, it looks as if there may
be no stable odd-odd “oo” nuclides, and this is nearly true – the only exceptions are four
light nuclei, each with less than 20 nucleons: (1p,1n), (3p,3n), (5p,5n) and (7p,7n).
3.6.2
Magic numbers
Nucleon pairing is part of a bigger pattern, for it has been found that nuclides with certain
numbers of protons or neutrons, commonly called magic numbers, are more tightly bound
than neighbouring nuclides. The existence of these magic numbers can only be explained
by seeing the nucleus as having a comprehensively organised structure, where nucleons
arrange themselves in concentric shells, rather like the spherical layers of an onion, and
quite unlike a liquid water drop. The “magic numbers are a signature that the nucleons lie
in simple orbits”. The "magic" nucleon numbers which fill an inner shell and confer extra
stability on a cluster are:
2, 8, 20, 28, 50, 82 and 126
and they apply to both proton and neutron numbers separately (figure 3.22).
magic numbers:
protons…
neutrons
proton number, Z
2
8
20 28
50
82
128
the largest truly
stable nuclide, lead208 (82p,126n) is
doubly magic
82
there tend to be more stable nuclides
with magic proton or neutron numbers
50
the doubly magic nuclide (50p,82n)
28
20
8
2
magic proton numbers
helium-4 (2p,2n) is doubly magic
neutron number, N
D:\116105936.doc
Page 49 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Figure 3.22: The magic shell numbers in the chart of nuclides
A cluster with a full shell of either type of nucleon is more tightly bound, and the nuclide
chart shows that there tend to be more stable nuclides when the proton or neutron
number is magic. The tightly bound helium-4 (2p,2n) nuclide is doubly magic, as is the
largest stable nuclide lead-208 (82p,128n). Elements with magic numbers of protons or
neutrons are more abundant on Earth than elements with similar, but non-magic, nucleon
numbers.
magic numbers 50 and 82
The larger binding energies of nuclides with magic numbers of nucleons means that their
average nucleon masses are slightly smaller, and we can discern this in the 3-D nuclear
valley. Figure 3.23 shows a region of the nuclear valley where two magic numbers, 50
and 82, intersect at nuclide (50p,82n).
neutron number
50
proton
number
82
nuclide (50p,82n)
all clusters along this
line have the magic
number 82 neutrons
arc of stable
nuclides
All clusters along this line have the magic
number 50 protons. Hence, thay are more
stable, have less mass/nucleon, and so
make a small step in the local slope.
Figure 3.23: The intersection of the magic numbers 50 and 82 in the nuclear valley.
Nuclides are colour coded by decay mode. Compare this with the 2-D nuclide plot.
All the nuclides on the red dotted line have 50 protons, and those on the green line have
82 neutrons, and each of the intersecting rows of cells makes a slightly uneven step in the
D:\116105936.doc
Page 50 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
overall valley slope. As with nucleon pairings, the presence of a closed shell of nucleons
in the magic number nuclides alters the topography of the nuclear valley.
3.6.3
Super-heavy clusters and the "Island of Stability"
At the end of the arc of stability is cluster-208, which is doubly "magic", containing 82
protons and 126 neutrons. The next "magic" number is 184, and so it is believed that
there may be an "island of stability" centred on a cluster with 184 neutrons and 114 or 126
protons - there is some uncertainty which of these is the "magic" number in clusters of this
size. Various clusters containing 114 protons have been produced, which are
comparatively stable for clusters this big. Such super-heavy clusters can be produced by
colliding selected smaller clusters together. For example…
cluster-48 (20p:28n) + cluster-244 (94p:150n)  cluster-289 (114p:175n) + 3n ejected
The product of this collision had a half life of about 2.6 seconds, before it decayed by
emitting an alpha particle. This is clearly unstable, but the cluster is 9 neutrons short of
the magic 184. The doubly magic cluster-298, comprising 114p:184n, is predicted to be
more stable, with a half life of around 17 days. This kind of work takes patience: in some
experiments it's taken around 5 billion billion collisions to produce a single super-heavy
nuclide.
3.7 Fusion and fission – changing cluster size
So far, we have focussed on individual cluster-families, each lying on a transverse curve
that crosses the nuclear valley. We have seen how the nucleons are interchanged by
beta-decay reactions to minimise the cluster’s mass-energy, and arrive at a stable
configuration. Now we step back and widen our view to consider all the nuclides along the
length of the valley, and look at how clusters gain or lose nucleons in the search for
stability.
There are two basic nuclear processes that change cluster size. Fusion is cluster growth,
where two clusters merge into a single bigger cluster. Fission is the splitting of a cluster
into smaller fragments. We have seen how the nucleon mass curve shows that the most
energetically favourable cluster size is about 60 nucleons - figure 3.24.
D:\116105936.doc
Page 51 of 168
12/02/2016
mass/nucleon (MeV)
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
939
fusion of small
clusters <~60
reduces the
average
mass/nucleon
938
937
1 x helium-4(2p,2n)
3 x helium-4(2p,2n)
1 x carbon-12(6p,6n)
936
= 3727.4 MeV
= 11,182 MeV
= 11,175 MeV
saving =
7 MeV
2 x carbon-12(6p,6n)
= 22,350 MeV
1 x magnesium-24(12p,12n = 22,336 MeV
saving =
14 MeV
935
934
fragmentation
of large
clusters >~60
reduces the
average
mass/nucleon
933
932
931
930
0
20
40
60
80
100
120
140
160
180
200
220
cluster size
Figure 3.24: Clusters of nucleons can reduce their average mass/nucleon by
converging on the minimum at cluster size of 60. Small clusters grow by fusion;
large clusters fragment by fission. The box gives the energy “savings” from the
fusion of small clusters into bigger ones.
For small clusters, less than about 60 nucleons, the average nucleon mass decreases as
the cluster grows, up to about 60 nucleons. So, it is energetically favourable for a small
cluster to undergo fusion and grow, since it will reduce its average nucleon mass, thereby
releasing binding energy. For clusters larger than about 60 nucleons, past the minimum
mass, the situation is reversed. The average nucleon mass is now increasing with cluster
size, so now it is energetically favourable for large clusters to be smaller, and to downsize by fission.
We’ll now look at how the processes of fusion and fission operate, and see what are the
constraints on them.
3.7.1
fusion - the energetics of cluster growth
The inset box in figure 3.24 gives some specific examples of the energy savings in a
sequence of fusion reactions. Combining the three helium quartets into a single cluster of
a dozen “saves” 7 MeV of mass-energy, which is released as binding energy. Bringing
two such clusters of twelve together “saves” another 14 MeV. Fusion is energetically
favourable only up to a cluster size of about 60. Further enlargement beyond that
increases the average nucleon mass, thus requiring energy, rather than releasing it.
D:\116105936.doc
Page 52 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
We will see shortly how a sequence of fusion reactions such as these power the stars,
and in so doing create many of the elements of our atomic world.
small clusters are isolated by their mutual repulsion
Clusters of less than ~60 nucleons are, strictly speaking, unstable, since it is energetically
favourable for them to enlarge, at least up to a size of about 60. Yet our world is largely
made of atoms with these small nuclei - such as carbon, oxygen, magnesium - that have
been stable for billions of years. Why have these small clusters not all aggregated into
clusters of 60 nucleons?
Nucleon clusters carry large positive electrical charges, due to the protons they contain,
and so they repel each other strongly. It is only at enormously high temperatures - tens of
millions of degrees and more – that small clusters have enough kinetic energy to
overcome this repulsion, and come together so their nucleons can mingle into a single
larger cluster. At lesser temperatures, anywhere outside the interior of a star or a particle
physics experiment, clusters never come together, but are forever isolated by their mutual
repulsion.
Thus nuclear fusion is a reaction that “wants” to happen, but can’t. What about nuclear
fission?
3.7.2
fission - the energetics of cluster splitting
the spectrum of nuclear fission reactions
There is only one way for nuclear clusters to merge and undergo fusion, but for fission,
there is a “spectrum of possibilities in which a heavy nucleus breaks into two (or more)
parts…at one end of the spectrum the result is one small part and one large part, at the
other it is two almost equal parts.” - figure 3.25.
D:\116105936.doc
Page 53 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
(a)…by ejecting a
single proton or
neutron…
a nucleus
can decay
by fission…
(b)…by ejecting a He-4
nucleus (2p,2n)
alpha-decay
(c)…or by ejecting a small
nucleus like Ne-24, C-14
cluster decay
(d)…or by splitting into two smaller
nuclei, and some free neutrons
spontaneous fission
Figure 3.25: the range of nuclear fission processes
Unstable nuclides can decay by ejecting a single nucleon (a), or a small cluster (b) and
(c), or by splitting into two clusters of comparable size (d). The ejection of a quartet of
nucleons, (2p,2n), is a common mode of decay among larger nuclides. This quartet is the
nuclide helium-4, also known as an alpha particle, so this is commonly called alphadecay. A few nuclides that undergo alpha-decay also eject larger nuclides, such as
carbon-14 (6p,8n) or neon-24 (10p,14n) - this rare mode is known as cluster decay.
the pattern of decay of the known nuclides
We’re ready now to view all these decay processes in the nuclide chart, now colour-coded
by decay mode (figure 3.26).
D:\116105936.doc
Page 54 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
alpha-decay
(yellow cells)
the diagonal for
cluster-145
spontaneous
fission
(green cells)
beta-plus (pn)
decay of unstable
proton-rich nuclei
(orange cells)
largest truly stable
cluster-208
ejection of
a proton
(red cells)
arc of stable nuclides (black cells)
beta-minus (np) decay of unstable
neutron-rich nuclei (blue cells)
ejection of a neutron (purple cells)
stable nuclide
beta-plus decay and electron capture (pn)
beta-minus decay (np)
proton emission
neutron emission
alpha-decay
spontaneous fission
Figure 3.26: The major decay modes of the known nuclides, colour coded as shown
in the diagram.
We can see the arc of stable nuclides, curving through nucleon-space from a single
proton up to the maximum cluster size of 208 nucleons. On one side of this arc the
proton-rich nuclides seek stability by the beta-plus (pn) decay reaction; on the other
side, decay is by the beta-minus (np) reaction. The beta-decay reactions are important
decay modes for nearly all clusters, up to about 250 nucleons. Decay by the ejection of a
single nucleon is associated with highly unstable nuclides, with extreme proton/neutron
ratios, that are situated at the edge of the nuclear valley. Alpha-decay becomes a
significant decay mode for clusters bigger than ~145 nucleons, especially if they are
proton-rich. Spontaneous fission is confined to the massive nuclides, way beyond the
maximum stable cluster size. The one process that does not appear on this chart, of
course, is fusion.
This chart shows only the major decay modes of each nuclide. We shall see that many
unstable nuclides, especially the larger ones, decay by more than one reaction, as the
following example shows.
D:\116105936.doc
Page 55 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the three decay modes of nuclide (77p,90n)
Figure 3.27 shows nuclide (77p,90n), situated high up on the proton-rich slope of the
nuclear valley.
protons
the nuclide (77p,90n)
has 3 modes of decay…
77
(2) eject a proton
mass-energy loss
939 MeV > mass of free proton
probability: 32%
(1) eject an alpha-particle
mass-energy loss
= 3734 MeV > mass of
alpha particle
probability: 48%
(3) convert p  n
mass-energy loss
= 9 MeV
probability: 20%
76
atomic m
Ir-167 (7
Os-166
Re-163
Os-167
Re-165
alpha
Ir-167
155,5
3734
75
the nuclide (76p,90n) can only
eject an alpha-particle (72%
probability) or convert pn (28%).
It cannot eject a proton, since this
would decrease its mass-energy
by only 937 MeV, less than the
mass of a free proton.
74
73
87
88
89
90
91
92
neutrons
Figure 3.27: the decay choices open to nuclides
This highly unstable nuclide (it has a half life of only 35 ms) has three available modes of
decay: it can fission, by ejecting either (1) an alpha-particle or (2) a proton, or it can (3)
transform one proton to a neutron (beta-plus decay).
the energy accounting of fission
We have seen that fusion is energetically favourable if merging two smaller clusters into
one larger one reduces the total mass. Similarly, fission is energetically favoured if the
total mass of the fragments is less than the mass of the initial cluster - that is, the total
mass of all the nucleons is reduced by the rearrangement. The initial decaying nuclide is
D:\116105936.doc
Page 56 of 168
12/02/2016
proton
Ir-167
155,5
939.8
beta-p
Ir-167
155,5
Os-16
.72…
proton
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
commonly called the parent nuclide, and the smaller nuclide that is produced is called the
daughter.
Thus, proton ejection is energetically favourable if…
mass of parent nuclide > mass of daughter nuclide + mass of proton (938 MeV)
Similarly, for alpha-decay to be viable the condition is...
mass of parent nuclide > mass of daughter nuclide + mass of alpha particle (3,727 MeV)
So if the mass difference between the parent and daughter nuclides is bigger than the
mass of the fragment to be ejected, then that mode of decay is energetically favourable.
It is energetically favourable for nuclide (77p,90n) to undergo alpha-decay, because the
mass of the daughter nuclide is less by 3734 MeV, which exceeds the mass of an alpha
particle. Similarly, proton-decay is favourable because the the 939 MeV loss exceeds the
proton mass. And finally, the nuclide can undergo the familiar beta-plus decay,
transforming pn.
a balance of probabilities
The configuration of nucleons in the parent nuclide (77p,90n) thus has a choice of three
ways to reduce its total mass - three decay routes across the nuclear valley terrain, all
going downhill. The balance of probabilities is such that of 100 parent nuclides, about 48
will undergo alpha-decay, 32 will eject a proton, and the remaining 20 will undergo betaplus decay. If we piled one million parent nuclides on the cell (77p,90n) of nuclide space,
like casino chips, then within a second they would be all gone, and a million new daughter
nuclides would appear on the three nearby cells.
What then happens to the daughter nuclides? We’ll look at one example, the daughter
nuclide (76p,90n). The alpha and beta decay modes are available to this nuclide, but
since ejecting a proton would reduce its mass by only 937 MeV, just less than the mass of
a free proton, this mode of decay is not energetically viable (see figure 3.27).
competition between nucleon configurations
We’re discovering that a nuclide is not in itself inherently unstable. What makes a nuclide
unstable is the existence of an accessible configuration with a smaller mass. In a sense,
configurations compete for nucleons, and any configuration that can rearrange a set of
nucleons with a reduction in mass is energetically favoured. If no energetically viable
decay reaction is available, then the nuclide is, de facto, stable.
D:\116105936.doc
Page 57 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
is it enough for a decay reaction to be energetically favourable?
We’ve seen that for a decay reaction to be viable, it must be energetically favourable - it
must “pay its way”. But that does not mean that every decay mode that is energetically
viable will occur - the small stable nuclides of less then ~60 nucleons have shown us that.
It’s time for us to look at the fission of clusters larger than 60 nucleons.
3.7.3
Spontaneous fission
fission is energetically viable for clusters >~100 nucleons
Figure 3.28 is a close-up view of the minimum in the nucleon mass curve for the stable
nuclides. We can see the mass per nucleon values fall steeply from cluster size 40 to the
minimum at ~60, then begin their slow rise as the cluster gets bigger.
mass/nucleon (MeV)
930.40
930.35
no change in the
average nucleon mass
fission increases the
average nucleon mass
930.30
fission reduces the
average nucleon mass
930.25
930.20
930.15
40
50
60
70
80
90
nucleons
100
Figure 3.28: The minimum in the nuclide mass-energy curve. Fission only becomes
energetically favourable for nuclides bigger than ~100 nucleons
We can use the graph to get a rough idea when it is energetically favourable for a larger
cluster to split into two equal clusters. The average nucleon mass for a cluster of 80
nucleons is less than for a cluster of 40. Splitting a cluster of 80 nucleons into two clusters
of 40 would increase the total mass, so it’s not energetically favourable. However, the
graph shows that a cluster of around 95 or so nucleons can split into two equal clusters,
with no change in the average mass per nucleon. It is then energetically favourable for
D:\116105936.doc
Page 58 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
clusters bigger than this to split in half, since the average nucleon mass is reduced, as the
cluster-100 example shows. The fission process will become even more favourable as the
cluster size increases further.
Yet the nuclide chart shows the arc of stable nuclides to be nearly continuous up to
cluster-208, and that spontaneous fission is the major mode of decay only for massive
nuclides. So what prevents the stable nuclides larger than 100 nucleons decaying?
spontaneous fission of very heavy nuclei
It is only the very heaviest nuclides, beyond uranium, that decay by spontaneous fission,
and usually in conjunction with other decay modes. Figure 3.29 shows a portion of the
proton number, P
nuclide chart beyond the stable cluster-208 limit.
the cluster-238 family of nuclides
...beta-minus, np, ~8%
uranium-238 (92p,146n)
...alpha,
~18%
cluster-250
(96p, 154n)
has 3 modes
of decay…
...spontaneous fission into 2
similar size fragments, ~74%
cluster-208, the end of the arc of stability
neutron number, N
Figure 3.29: Very large nuclides decay by a number of processes: spontaneous
fission (green), alpha-decay (yellow) and beta-minus decay (blue). Some of the cells
have an inset box; this gives information on other nuclear processes.
We can see that spontaneous fission becomes a significant decay mode beyond cluster238. Cluster-250 decays by three processes, beta, alpha, and spontaneous fission. The
familiar beta pn reaction "moves" the cluster one cell along the cluster-250 diagonal,
while the alpha-decay reaction, which ejects an alpha-quartet of nucleons (2p,2n), moves
the cluster two cells diagonally in nucleon-space towards stability. Compared to these two
D:\116105936.doc
Page 59 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
small, precise decay reactions, spontaneous fission is like a hyper-space jump. You can't
predict what the fragments will be, though the usual outcome is that one has 90-100
nucleons, and the other has 130-140. Each decay mode has its own probability, as we
saw for nuclide (77p,90n).
the nucleus as a liquid drop
If spontaneous fission is energetically favourable for nuclides larger than ~100 nucleons,
why are there stable nuclides bigger than this? The reason is that in order for the
nucleons to get to the lower energy state, arranged in two smaller nuclei, they must go
through a higher energy configuration, that is, surmount an energy barrier.
We’re familiar with the way water drops, falling from a dripping tap, pull themselves into a
spherical shape as they fall. We might think of the surface tension, the attraction between
the water molecules in the surface, as pulling the drop into a sphere. However, it’s more
helpful to think of the surface tension giving the drop a surface energy, which is minimum
for a spherical shape. A sphere is the shape with the least surface area for its volume,
and any distortion in the drop’s shape increases its surface area, and hence its energy.
The nucleus is similarly bound by the attractive nuclear force, and behaves in some ways
like a liquid drop, and so nuclear fission then becomes analogous to a water drop splitting
(figure 3.30).
D:\116105936.doc
Page 60 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
(a) A large nucleus, within which the nucleons are all
randomly moving and colliding, making the nucleus
“wobble”, and its shape change.
(b) Nucleon collisions have disorted the nucleus
shape – its surface area and energy have increased.
(c) The nucleus shape has distorted further. Its energy
is now a maximum, and it can either go back to the
spherical shape (a), or distort further…
(d) …into two smaller nuclei, with a
neck developing between them.



(e) Two smaller nuclei, of comparable size, with free
neutrons and the conversion of the lost mass into
energy.
Figure 3.30: A large nucleus fissions like a water drop splits in two.
A large nucleus “wobbles” due to the incessant movements and collisions of its nucleons
(a), so its surface area gets bigger, and its energy increases (b). A large wobble distorts
the nucleus, so its shape is something between a large stretched nucleus and two smaller
nuclei very close together (c). At this point the nucleus energy is a maximum, and the
nucleus may relax back into the single cluster, or carry on (d) and split into two smaller
nuclei, with some free neutrons, and a release of energy from the loss of mass (e). We’ve
seen that bigger clusters need more neutrons per proton for stability, so when a big
cluster fissions into smaller fragments, the products are themselves unstable and there
are some free neutron “leftovers”.
The energy barrier opposing the splitting of a cluster of 238 nucleons is only 6-8 MeV
high, a tiny fraction of the cluster’s mass of more than 200,000 MeV, but this is enough to
make spontaneous fission nearly impossible. This energy barrier gets smaller as the
cluster size increases, and beyond a cluster size of ~238 nucleons spontaneous fission
becomes more common.
why there are stable nuclides >100 nucleons
D:\116105936.doc
Page 61 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
So it is the existence of a small but insuperable energy barrier that prevents clusters
bigger than ~100 nucleons undergoing spontaneous fission. Spontaneous fission might
be energetically favourable, but impossible in practice. It’s perhaps like having a winning
lottery ticket, but you have to post it off to claim your prize, and you can’t afford the price
of a stamp.
3.7.4
alpha-decay
We’ve calculated that alpha-decay, the ejection of a helium-4 (2p,2n) nuclide, becomes
energetically favourable for the stable clusters bigger than ~145, and this is confirmed by
the nuclide chart (see back to figure 3.26). We can see that it’s the proton-rich nuclides
that favour alpha-decay, since this moves them down the valley slope towards the arc of
stable nuclides.
why is alpha-decay so common?
The nuclide chart (figure 3.26) shows that by far the most common mode of fission is the
ejection of an alpha-particle (2p,2n). Why this particular combination of nucleons?
We have seen how for a large nuclide to fission, the mass difference between parent and
daughter nuclides must exceed the mass of the ejected fragment. Consequently, the more
tightly bound is the fragment, the smaller is its average nucleon mass, and the more likely
that its ejection will be a viable mode of decay. Hence, the tightly bound helium-4 nuclide,
with its small mass per nucleon, is a very suitable candidate for ejection. For example, we
saw in section 3.7.2 that the nuclide (76p,90n) can eject an alpha particle quartet, with the
small average nucleon mass of 931.9 MeV/nucleon, but not a proton with a large unbound
mass of 938.3 MeV.
alpha-decay is viable for clusters >145 nucleons
We’ve seen that alpha-decay becomes viable if the mass of the daughter nuclide plus
alpha particle is less than the parent, or to put this another way…
mparent - mdaughter > malpha (3,727 MeV)
The parent nuclide can undergo alpha-decay if down-sizing to the smaller daughter
reduces its mass by more than the mass of a free alpha particle, 3,727 MeV.
The nucleon mass curve tells us that as a cluster increases in size beyond about 60, at
some point alpha decay should become energetically favourable. But at what size would
D:\116105936.doc
Page 62 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
this occur? Some simple mass-energy accounting for the alpha-decay of nuclide
mass/nucleon (MeV)
(60p,85n), the stable nuclide in the cluster-145 family, will give us an idea - figure 3.31.
939
free alpha particle: He-4 (2p,2n)
mass, malpha = 4 x 931.85 = 3,727.4 MeV
938
937
cluster-145: average nucleon mass = 930.7 MeV
mass loss…
 due to ejecting (2p,2n) = 4x930.7
= 3,722.8 MeV
 of the remaining 141 nucleons = 141 x 0.032 =
4.5 MeV
 Total mass loss
= 3,727.3 MeV
936
935
934
933
alpha-decay
energetically
favourable
alpha-decay
energetically
unfavourable
932
931
930
0
20
40
60
80
100
120
straight line fit to mass/nucleon graph
140
160
180
200
220
cluster size
Above a cluster size of ~100 the graph is nearly straight.
Losing one nucleon reduces the mass per nucleon by about 0.008 MeV.
So losing 4 nucleons (2p,2n) reduces the mass per nucleon by about 0.032 MeV.
Figure 3.31: the mass-energy accounts for the alpha-decay of cluster-145
When nuclide (60p,85n) ejects the (2p,2n) quartet its mass is reduced in two ways…
1) Simply losing the 4 nucleons (2p,2n) reduces the mass of cluster-120 by 4 times the
average nucleon mass, that is, 4 x 930.7 = 3,722.8 MeV. This is less than the alphaparticle mass, so this is not enough in itself.
2) The daughter nuclide is now a little closer to the minimum in the nucleon mass curve,
and so its average nucleon mass is slightly less. The slope of this straight part of the
nucleon mass graph tells us that losing 1 nucleon reduces the average mass/nucleon by
about 0.008 MeV. So, in losing 4 nucleons, the remaining 141 nucleons each have about
0.032 MeV less mass. This reduces the cluster’s total mass by a further 141 x 0.032 = 4.5
MeV
Adding these two gives the total mass difference parent - daughter as 3,727.3 MeV,
almost exactly the same as the mass of a free alpha particle. So there is no energy
advantage in cluster-145 undergoing alpha-decay. But what about clusters smaller or
larger than this?
D:\116105936.doc
Page 63 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
The ejection of 4 nucleons reduces the mass of each nucleon remaining in the cluster by
0.032 Mev, regardless of the cluster size, so the bigger the cluster, the bigger its mass
loss. Cluster-145 is a break-even size, where the mass loss of the fragmenting cluster is
almost exactly equal to the mass of a free alpha particle. But for clusters larger than 145,
the ejection of four nucleons will reduce the cluster’s mass by more than a free alpha
mass; for smaller clusters, the reduction will be less. Thus, the nucleon mass curve tells
us that it is energetically favourable for stable clusters bigger than about 145 nucleons to
decay by ejecting an alpha particle.
The nuclide chart (figure 3.26) indeed shows that alpha-decay becomes a significant
decay mode for nuclides with more than about 145 nucleons, in agreement with the
calculation above. So, if alpha-decay is energetically favourable for nuclides of more than
~145 nucleons, then why is there a nearly continuous arc of stable nuclides all the way up
to 208 nucleons?
why there are stable nuclides >145 nucleons
One of the features of alpha-decay is that the the rate of decay is very sensitive to the
energy released with the ejected alpha particle. The bigger the alpha particle energy, the
faster the decay rate, and the shorter the nuclide’s half life. Nuclides that undergo alphadecay have an enormous range of decay rates, with half lives ranging from microseconds
to millions of years.
In the cluster size range 144-206, there are seven unstable alpha-emitting nuclides that
can be found naturally because they have very long half lives, comparable to the age of
the Earth. The very low energies of their emitted alpha particles mean that they can only
decay very slowly. Williams concludes, “it is therefore certain that although most nuclei in
this range on the line of stability may be energetically able to decay by alpha-emission,
they do not do so at a detectable level because the [decay] rate is too small.”
Thus, alpha-decay is energetically favourable for the nuclides on the stable arc in the size
range 145-208, but for most of them the decay rates are negligible, and so these nuclides
are effectively stable.
alpha particles in a potential well
We have so far seen alpha-decay simply as the ejection of an alpha particle (2p,2n) from
the nuclide, and we’ve done the energy accounting for the process. But how does an
alpha particle manage to escape from the immensely powerful grip of the nuclear force?
D:\116105936.doc
Page 64 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Alpha decay is an example of a process known as “quantum tunnelling”. We have seen
waves and particle-waves can pass through barriers that would appear to be
impenetrable, and we will see that alpha particles can behave in the same way.
Alpha-decay differs from beta-decay, in that the nucleons making the alpha particle
already exist inside the nucleus, whereas the electron is first created by the weak
interaction and then ejected from the nucleus. So our first question is how any nucleons,
let alone a quartet, elude the grasp of the nuclear force. Nuclei which undergo alphadecay have a huge range of half lives – from less than a microsecond to nearly a
thousands of years. A second question, then, is why is there such an enormous range of
rates for what is “essentially the same process”?
The nucleons are packed together, “jostling about in a very small volume”, confined in the
nucleus by the strong nuclear force. So “when a nucleon approaches the surface and tries
to fly off the nucleus, it suffers an attractive force by the nucleons that are left behind,
forcing it to return toward the interior. Inside the nucleus it feels the attraction forces of all
the nucleons that are around it, resulting in a net force approximately equal to zero. We
can imagine the nucleus as a balloon, inside of which the nucleons move freely, but
occupying states of different energy”.
The nucleons are confined in a potential energy “well”, since energy must be supplied to
remove them. Figure 3.32 shows how the potential energy of an alpha-particle (2p,2n)
varies with distance from the centre of a typical heavy nucleus of about 236 nucleons.
D:\116105936.doc
Page 65 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the energy peak is ~27 MeV
potential energy (MeV)
30
(d) an incoming alphaparticle with 9 Mev kinetic
energy only gets as close
as this to the nucleus
25
20
(c) some of these excited
nucleons can briefly come
together as a quartet.
15


10
5
0
4
MeV



 



 
 
 
 
9
MeV
-5
-10
-15
-20
-100
(e) …yet alpha-particles are
emitted with a typical kinetic
energy of ~4 MeV
-80
-60
-40
(b) some nucleons gain extra
energy from collisions
-20
0
-6
MeV
(a) Nucleons filling the
potential well from the bottom,
up to an energy of ~ - 6 MeV
20
the potential well
inside the nucleus,
radius ~10 fm
40
60
80
100
distance from centre of nucleus (fm)
outside the nucleus, beyond ~10 fm, an alpha particle is electrically repelled
Figure 3.32: For a big nuclide to undergo alpha-decay a (2p,2n) quartet must break
out of a the nuclide’s deep potential energy well.
inside the nuclear well
Nucleons fill the central potential well inside the nucleus, up to an energy of about -6 MeV,
so that it takes about 6 MeV to remove one nucleon (a). The constantly jostling nucleons
can briefly gain energy from random collisions, and so there are always excited nucleons
higher up in the potential well (b). This jostling also means that there are brief local
groupings within the nucleus, such as the alpha particle quartet (c), though all sorts of
groupings will occur. Whilst we should not think of alpha particles (2p,2n) as having a
permanent existence inside a larger nucleus, it is very likely that at any moment there is
somewhere in the nucleus an alpha-like grouping of nucleons. So, while “the simple
intuitive picture of alpha particles bouncing around in nuclei, each with its own unique set
of nucleons, is not quite accurate … alpha particle-like structures do occur in the nucleus
and sooner or later these appear ouside the nucleus as alpha particles”.
outside the nucleus
D:\116105936.doc
Page 66 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Beyond the very strong but short range attraction of the nuclear force an alpha particle,
carrying an electrical charge of +2, is electrically repelled by the protons in the nucleus. It
“falls” down the potential energy slope, towards zero potential energy, effectively out of
range of the repulsion. An alpha particle needs about 27 MeV of kinetic energy to
overcome this electrical repulsion and enter the nucleus. An alpha-particle with, say, only
9 MeV kinetic energy never gets near the nucleus, but runs part way up the potential
energy “hill”, and then rolls back down again (d). Similarly, an alpha particle would need
27 MeV to surmount the barrier and escape from the nucleus, but it would then be ejected
with 27 MeV of kinetic energy. Yet alpha particles are emitted from decaying nuclei with
only 4 MeV of kinetic energy (e). It is as if they somehow “emerge” part-way up the
potential energy hill at 4 MeV, and from there roll away from the nucleus.
tumbling and tunnelling
How do any nucleons manage to escape from the nuclear attraction? And why is it an
alpha particle particle, a quartet of nucleons, that emerges? The answer is that the alpha
particles escape the nucleus by quantum tunnelling. We know that, like any particle, an
alpha particle should be regarded as a particle-wave, and so in colliding with an
impenetrable barrier there is the probability that it will pass through.
The jostling nucleons, randomly coming together in alpha-quartets, are confined in the
nucleus by the strong nuclear attraction. We can see this as alpha particles colliding with
a potential barrier, and knowing that an alpha particle behaves as a particle-wave, we can
see that its wave function will penetrate some way into the barrier, and has a finite chance
of emerging on the other side. Figure 3.33 illustrates the sequence.
D:\116105936.doc
Page 67 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
potential energy (MeV)
30
(c) a quartet of excited nucleons (2p,2n)
collide with the walls of the energy well…
20
(d) …tunnel
through …
(e) … appear as an
alpha particle outside the
nucleus, with 4 MeV
potential energy
10
4 MeV
0
(b)…and some gain extra energy
from collisions with other nucleons.
-10
(a) nucleons “jostle”
around at the bottom of the
potential energy well…
-20
-20
0
20
40
60
80
100
distance (fm)
Figure 3.33: Alpha-particles collide with the walls of the energy well, and have a
finite probability of tunnelling through.
Nucleons jostle around in the energy well (a), and some gain extra energy from collisions
(b). A quartet of excited nucleons, corresponding to an alpha particle with about 4 MeV
energy, collides with the barrier (c), tunnels through (d), and appears on the other side (e).
The alpha particle is free of the nuclear attraction, is subject only to the electrical
repulsion, and so shoots away from the nucleus with 4 MeV kinetic energy (a speed of
~1.4 x 107 m/s, 14 million metres/second).At an energy of 4 MeV the barrier is 50 fm wide
(60 -10 = 50 fm, see the graph). But the range of the nuclear attraction is only a few fm at
the most – so why must the alpha particle tunnel the full 50 fm? The reason is that energy
must be conserved. The alpha particle goes into the barrier with 4 MeV, so it must emerge
with the same energy. If it emerged anywhere else on the potential energy hill, say at 20
MeV, then it would have somehow acquired another 16 MeV for free. So in the figure the
dotted line representing the tunnelling process must be horizontal – no forbidden energy
changes!
If the alpha particle has only a little more energy, say 6 MeV, then the barrier width is
much less, only about 30 fm (40 -10 fm). We can now qualitatively explain the huge
variation in half life for alpha-decay. An alpha particle with more energy, not only has a
better chance of penetrating a barrier of the same width, but also finds that the barrier is
D:\116105936.doc
Page 68 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
narrower. So the probability of tunnelling is extremely sensitive to the alpha particle
energy.
a “cloud” of possibilities?
We’ve seen that some nuclei emit larger nuclear clusters as well as alpha particles – for
example, radium-223 emits one C-14 nucleus for every billion alpha particles.
Presumably, bigger clusters are brought together less often by random jostlings of
nucleons. But very occasionally this happens, and if it is energetically favourable, then the
large cluster is emitted by the nucleus.
If C-14 clusters can be emitted, then we must infer that all possible nucleon groupings can
form inside the nucleus, collide with the potential barrier, and have some small but finite
probability of tunnelling through it. We can perhaps envisage a sort of “cloud” of possible
particles around the heavy nucleus. But the only permitted outcomes are those that lead
to a reduction in the mass of the system – that is, the mass of the daughter plus emitted
nucleus must be less than the parent nucleus. We’re reminded of Kaufmann and
Freedman’s aphorism: everything is possible, unless it is forbidden. There’s a host of
potential possibilities, but only those that are energetically allowed can actually occur.
3.7.5
Why there is an arc of stable nuclides
We have seen that a nuclear cluster of ~60 nucleons has the least possible mass per
nucleon, and the greatest binding energy. Thus it is energetically favourable for all other
clusters to converge on this optimum size. We can now explain why this does not occur,
and why there is a long arc of stable nuclides running through the nuclear valley.
motive, means and opportunity
As in the best murder mysteries, successful nuclide decay depends on motive (a
reduction in average nucleon mass, and therefore the release of binding energy), means
(a viable nuclear reaction), and opportunity (a significant probability of this reaction
occurring). Figure 3.34 summarises this for the stable nuclides.
D:\116105936.doc
Page 69 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
mass/nucleon (MeV)
small nuclides <~60
Fusion:  motivation  means  opportunity
939
nuclides >~100
fission…
 motivation
 means
 opportunity
938
937
nuclides ~60
 motivation
936
nuclides >~145
alpha-decay…
 motivation
 means
? opportunity
935
nuclides ~60 to ~100
fission…
 motivation
 means
934
933
932
931
930
0
20
40
60
80
100
120
140
160
180
200
220
cluster size
Figure 3.34: The decay options for the stable nuclides that are larger and smaller
than the ~60 nucleon optimum size.
Small nuclides, <~60 nucleons, would readily merge if they had enough kinetic energy to
overcome their mutual repulsions. They have the motive (a reduction in mass), and the
means (a viable nuclear reaction), but not the opportunity - they can’t get close enough to
undergo fusion.
Middle-sized nuclides, ~60 - 100 nucleons, have the motive of mass reduction, but there
is no rearrangement that will reduce it - they have no means.
Larger nuclides, >~100 nucleons, have the motive and the means, via a fission reaction,
but the insurmountable energy barrier denies them the opportunity.
Even larger nuclides >~145 nucleons have the motive and the means, via alpha particle
ejection, but the opportunities for decay are extremely rare.
a surprising conclusion?
Thus we have the rather surprising, and maybe unsettling, conclusion that our physical
world of ”stable” nuclides is not what it seems. It is energetically favourable for stable
nuclides that are smaller or larger than ~60 nucleons to decay, but they are “trapped”
because the options for decay are either unavailable, or too slow. These “stable” nuclides,
the lowest mass nuclides of each cluster-family, will in time become the nuclei of the
chemical elements, and create our atomic world.
D:\116105936.doc
Page 70 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
3.8
Nuclear reactions
The collision of two clusters can lead to a nuclear reaction, rearranging the nucleons into
new nuclides. For example, if we fire protons at an aluminium-27 (13p,14n) nucleus, there
are a number of possible reactions (figure 3.35).
silicon-28* (14p,14n) + 
p + aluminium-27
(13p,14n)
silicon-28*
(14p,14n)
- an excited
intermediate
state
silicon-27 (14p,13n) + p
sodium-24 (11p,13n) + 3p + n
magnesium-24 (12p,12n) + helium-4 (2p,2n)
Figure 3.35: A nuclear reaction can have a range of possible outcomes
The colliding particles combine to make an excited silicon-28 nucleus, indicated by the
asterisk (*), which then decays in a number of ways. It may remain as it is, and just lose
its excess energy by emitting a photon of gamma radiation. This is rather like a wet dog
shaking the water off its fur, except that to properly mimic the nucleus, the dog would
have to shake all its water off in one single amount, not a large number of small drops.
Alternatively, it may fragment into smaller nuclei and protons and neutrons.
In certain situations three nuclei can fuse…
3 helium-4 (2p,2n)  carbon-12 (6p,6n)
and we’ll soon look at this important reaction occurring in the interiors of stars.
A nuclear reaction doesn’t need two colliding nuclei, it can be initiated by a photon of
radiation…
1)
 + copper-63 (29p,34n)  nickel-62 (28p,34n) + p
2)
 + uranium-233(92p,141n)  rubidium-90(37p,53n) + caesium-141(55p,86n) + 2n
and…
In the first reaction a gamma ray photon knocks a proton off a nucleus, and in the second
it induces a giant nucleus to fission into two smaller nuclei and a pair of neutrons.
nuclear Lego
D:\116105936.doc
Page 71 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
We can take a simple view of all these reactions as analogous to rearranging Lego bricks.
The number and type of bricks are constant, and they can be rearranged in any
combinations, as long as they are energetically allowed. Thus, we see the same number
of protons and neutrons on either side of the reaction, they are just differently arranged.
We’ll see shortly how the hot dense interiors of stars force all sorts of nuclear reactions to
occur.
nuclear and chemical reactions
There is a similarity between nuclear and chemical reactions. The former rearrange
nucleons in nuclear clusters, and the latter rearrange atoms in molecules. Both types of
reaction can release energy, but nuclear reactions typically yield about a million times
more energy than chemical, because the forces between nucleons are so much greater.
3.9
Life in the nuclear valley
3.9.1
a balance of conflicting factors creates the nuclear
valley
We have seen how nucleons are bound into a cluster by the strong nuclear force, with
virtual pions “fluttering” to and fro, constantly interchanging protons and neutrons. The
mass-energy, and hence the stability, of a cluster of a fixed number of nucleons is
decided by its proton/neutron ratio. But the nucleons in the cluster form relationships,
albeit very limited ones; thus, like nucleons form pairs and magic numbers of them form
closed shells, with slight effects on the cluster’s mass and binding energy. ”The nucleons
are the bricks, and the nuclear forces provide the mortar, while everything is under the
control of rigorously enforced planning regulations provided by the quantum rules”.
Because the nuclear attraction force is stronger than the electrical repulsion, stable
nuclear clusters are possible. Because the electrical repulsion has the longer range, there
is a limit on stable cluster size. Because of the protons’ mutual repulsion and the
neutron’s greater mass, each cluster has its minimum mass-energy at only one or two
precise p:n ratios. Each stable cluster lies at the bottom of its own transverse section of
the nuclear valley. The stable nuclides in all the clusters then form the stable arc that runs
along the valley bottom.
3.9.2
The pathways towards stability
the limited options for nuclide decay
D:\116105936.doc
Page 72 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
A configuration of nucleons is unstable if there is a viable decay reaction that can
rearrange them with a reduction in total mass, and also at a reasonable rate. We have
imagined a nuclide as a boulder perched on the slope of the nuclear valley. There may be
many nuclides lower down the slope, all with lower values of average nucleon mass, and
all closer to the arc of stability, down in the valley bottom. However, this unstable parent
nuclide has only a very limited range of nuclear reactions, whereby it can decay to a
daughter nuclide, and reduce its average nucleon mass, thereby releasing more binding
energy (figure 3.36).
proton number, Z
proton-rich parent nuclide
beta-plus decay (pn)
eject p
alpha-decay
(eject (2p,2n)
arc of stable nuclides
along the nuclear
valley bottom
beta-minus decay (np)
neutron-rich parent nuclide
eject n
parent nuclides
neutron number, N
daughter nuclides
Figure 3.36: The limited decay options available to unstable nuclides
On one side of the nuclear valley, proton-rich nuclides have only three options: (1)
transform a proton to a neutron by the beta-plus decay reaction, (2) eject an alpha particle
or, in the case of an extreme proton excess, (3) eject a proton. On the opposite valley
slope, neutron-rich nuclides have the “mirror-images” of only the first and last options:
beta-minus decay or neutron ejection. Alpha-decay is not available, for it would not
“move” the nuclide closer to the arc of stability.
average mass/nucleon is absolute, but stability is relative
A nucleon configuration may be unstable in principle, but the constraints on the viable
nuclear reactions may make decay impossible, or negligibly slow. The balance of
competing reactions depends solely on the relative properties of the parent and daughter
D:\116105936.doc
Page 73 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
nuclides, and takes no account of the origins of the parent nuclide, or of the decay modes
of the daughter nuclides. Thus, while the average nucleon mass in a cluster is a fixed
quantity, the cluster’s stability, and its mode of decay are relative, and are determined
only by the parent and daughter nuclides.
decay pathways
The nuclear valley is criss-crossed by decay pathways, as the unstable nuclides “move”
downhill towards the arc of stability running along the valley bottom. Figure 3.37 shows
the final sections of three decay series.
nuclide X, (69p,84n)
thulium-153
nuclide Y, (63p,90n)
europium-153
decay series A
decay series B
decay series C
nuclide Z, (60p,85n)
neodymium-145
alpha-decay
beta-plus (pn)decay
beta-minus (np) decay
The thickness of the arrow gives an indication of the relative decay probabilities.
Figure 3.37: Decay pathways in the nuclear valley; proton-rich nuclides have alphadecay and the pn reaction, while neutron-rich nuclides have only the np
reaction. Decay pathways can branch on the proton-rich side, so an unstable
nuclide can produce two stable nuclides. The nuclides are colour-coded by their
major decay mode.
Nuclide X, part way up the proton-rich side of the valley, is the start of a cascade of
decays that run down the slope to terminate in nuclides Y and Z. A minority of nuclides
D:\116105936.doc
Page 74 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
have two modes of decay, so the pathway on the proton-rich side branches in places.
Nuclides Y and Z are the end-points of two other decay series, B and C. We have seen
that neutron-rich nuclides can only decay by the beta-minus (np) reaction, so these
decay paths do not branch.
Nuclides Y and Z share a common “ancestor” in nuclide X on the proton-rich side, and
different ancestors on the neutron-rich side. However, nuclide X is not the source of decay
series A, for it is itself a decay product of nuclides even higher up the valley slope. The
entire nuclear valley terrain is criss-crossed by decay pathways, that run down the slopes
and end at a nuclide on the stable arc.
nuclide “ancestors” and their “descendents”
Thus every unstable configuration of nucleons is on a journey to stability, following a
decay path down the slope of the nuclear valley, with each nuclide being a stage on that
journey. Every step on the path is downhill, releasing more binding energy, with the
nuclides getting closer to stability, and with increasing “efficiency” of conversion of energy
to matter. Every stable nuclide is the last “descendent” of a series of unstable nuclide
“ancestors”, that no longer exist. Its nucleons were once part of different clusters, and
what is now, say, a proton was maybe once a neutron. We can see a nuclide not so much
as a fixed thing, but rather as a transient configuration of nucleons. The nuclear valley,
then, is the environment, or habitat, in which the different nucleon configurations co-exist
and compete for existence.
decay pathways are determined by local topography
We’ve seen that the balance of the competing decay reactions is determined by the
relationships between the parent and daughter nuclides. So, while the decay pathway
must always be downhill in the nuclear valley, each step is decided solely by local factors,
and the decay series has no overall aim or preferred end-point. Similarly, water on a slope
will run downhill, always reducing its potential energy, but the precise path it takes is
determined solely by the local terrain it encounters, and not by the overall topography.
3.9.3
a rain of nucleons
We have seen all the possible combinations of protons and neutrons laid out like a chess
board, so that every possible nucleon combination has its own location in nuclide-space.
Now imagine a rain of nucleons, so that every cell is occupied by its nuclide. The great
majority of clusters break up almost as soon as they are formed; they are so unstable that
D:\116105936.doc
Page 75 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
they hardly even exist. Only the clusters within a narrow diagonal region of nuclide-space
last long enough to exist as independent entities, and these define the nuclear valley (figure 3.38).
N
Z
a rain of nucleons
the lowest
point in the
valley, at ~60
nucleons
some nuclides
spontaneously fission
into smaller nuclei
neutron-rich nuclides
decay by the beta-minus
(np) reaction towards
the arc of stability
proton-rich nuclides
decay by the betaplus (pn) reaction
and electron capture
towards the arc of
stability
very big nuclides undergo alpha and beta
decay, and ‘move’ towards the largest
stable nuclide, containing 82 protons
Figure 3.38: A rain of nucleons on the nuclear valley. Z and N are the proton and
neutron numbers, respectively. The nuclides are colour coded by their major decay
modes.
All over the terrain of the nuclear valley, the nuclides are decaying as the clusters
rearrange themselves. The decay pathways show the pattern of nuclide “migration”,
always downhill, towards the arc of stability running along the valley bottom.
draining the nuclear valley
If we watch an unstable nuclide decay, we see the cell it occupied become empty, and the
transformed nucleon configuration reappear as the daughter nuclide in a nearby cell.
Nuclides further up the slopes are further from stability and decay faster, so we see the
outer cells in the valley empty first. Almost all the moves are very short, either to an
adjacent or nearby cell by beta or alpha-decay. A few very big nuclides make great jumps
right down the valley, splitting into two smaller nuclides by spontaneous fission.
D:\116105936.doc
Page 76 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
The region of occupied cells shrinks, like water draining out of a bathtub. Finally, all that
are left are the stable nuclides, occupying the cells along the valley bottom. If the stable
nuclides could flow like water, they would run in two streams downhill and meet at the
lowest point, around cluster-60. If you wanted to drain the nuclear valley completely, that
is where you would fit the plug.
3.10
The emergent nuclide
We have seen the emergence of a new structure – the nuclide, a community of protons
and neutrons - figure 3.39.
… a nuclide – a cluster of protons
and neutrons
A stable nuclide has
emerged, with a large
mass, and carrying as
many units of positive
charge as it contains
protons
+
p
n0
0
0 n
+ n
p
p+
n0
…create and sustain…
Continuous interactions between
protons and neutrons, mediated by
pions…

+
p

n0
free protons are stable, but
are isolated by their mutual
repulsions, and unstable
neutrons cannot exist in
isolation.
We don’t consider the quarks
inside protons and neutrons.
Figure 3.39: The emergent nuclear cluster is created and sustained by continuous
activity within each nucleon.
The continual exchange of pions fluttering between protons and neutrons creates “an
invisible, evanescent web … binding them together”. Thus a nuclide emerges from the
continual interactions between all the nucleons, mediated by pions. This nuclear force that
binds nucleons together, is an extension of the colour force operating inside each
nucleon. Nuclides are massive particles, carrying as many units of positive electric charge
as they have protons.
goodbye to quarks and the strong colour force
The quarks have effectively withdrawn, gathered into protons and neutrons by the strong
colour force. The residual of that force, the nuclear force, binds protons and neutrons into
nuclei. Like a building’s foundations, the quarks and colour force are “underneath”,
holding everything up. It’s hard now even to distinguish between protons and neutrons –
they are just nucleons bound into a cluster by the nuclear force. Inside the nucleus the
D:\116105936.doc
Page 77 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
weak interaction interchanges protons and neutrons, enabling unstable clusters to “move”
towards stability.
nucleosynthesis – making nuclei
We have now surveyed the array of known nuclides, and seen which are viable and why,
and how they transform from one to another. We know quite a lot about how real nuclides
behave, but nothing as yet about how they have been made.
There have been two phases of nucleus-building – nucleosynthesis – in the evolution of
the universe. The first ended when the universe was about quarter of an hour old, and the
second has been going on for the last 14 or so billion years.
We're now just emerging into a level of the physical universe that we can recognise. It is
made of only 4 particles: two quarks - up and down, and two leptons - electrons and their
neutrinos. We'll see that anti-matter, as positrons (anti-electrons) and anti-neutrinos, plays
a crucial röle in the creation of our world of atomic matter. The u and d quarks make
protons and neutrons, but only the proton is stable. On its own a neutron has a life time of
about 15 minutes, yet with the proton it has sustained our physical universe for about 14
billion years. Out of all the potential particles, these are all that are viable. How did the
Universe cook up its rich material banquet with only these few ingredients in the larder?
Almost everything we see when we look up into the night sky is the result of nuclear
reactions.
Ray Mackintosh et al, p.94
Every atom of carbon and oxygen on Earth (and in us) was forged inside stars that died
before our solar system formed. We are stardust: or less romantically, the nuclear waste
from stars.
Martin Rees, in Craig Hogan, p.viii
3.11
Nucleosynthesis 1 - The first quarter of an hour
Here we will outline the events of the first 15 minutes or so after the big bang, when the
foundations of the material Universe were being laid. As Lawrence Krauss, writing about
oxygen, puts it, “these are the conditions when the gist of our oxygen atom came to be,
when nothing became something”.
D:\116105936.doc
Page 78 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
3.11.1
A time-line – linking temperature, time and energy
Figure 3.40 gives an outline of events, and links four important parameters – temperature,
temperature (K)
time, energy and particle mass.
1.E+15
Particle
masses
10 GeV
1.E+13
1 GeV
100 MeV
10 MeV
1.E+11
1 MeV
100 keV
1.E+09
10 keV
1 keV
1.E+07
100 eV
300,000 years
1 year
1 day
1.E+05
10 eV
1 eV
1 minute
1.E+03
1.E-10
1.E-06
1.E-02
1.E+02
1.E+06
1.E+10
1.E+14
time (seconds)
Figure 3.40:A time-line for the universe, linking temperature, time and energy. The
temperatures are given in scientific notation, so “1.E+05” is 1 with 5 zeros, 100,000
degrees, and “1.E+09” is 1 billion degrees.
The relation between the temperature of the universe and its age is well established, so
“the cosmic temperature can be used as a sort of clock, cooling instead of ticking as the
universe expands.”
We’ll now follow the series of temperature thresholds in the early “fireball” universe, with
each heading giving the temperature (T) in Kelvin (K), followed by the energy (E) in
electronVolts (eV), and then the time (t) - all values being approximate. In doing this we
will pass through levels in the universe’s hierarchy, that we have looed at in previous
chapters.
D:\116105936.doc
Page 79 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
3.11.2
Before the first threshold – temperature, T > 1015 K
(energy, E > 100 GeV, time, t < 10-10 seconds)
At these enormous temperatures, "equivalent to energies far higher than anything
achieved at an accelerator on Earth, fundamental particles of matter and antimatter
emerged and annihilated continuously. The universe was an expanding froth of quarks,
antiquarks, leptons, antileptons, photons, W particles, Z particles, gluons, and maybe
other particles as yet unknown to experiment or undreamed of by theorists." Individual
protons and neutrons can't yet exist, for "any quarks that did temporarily bind together
would be easily blasted apart again by collisions with the high energy photons".
This stage is sometimes referred to as the quark plasma, with a tiny excess of quarks
over anti-quarks. We can imagine pairs of photons and particles interchanging continually,
with particle-antiparticle pairs continually appearing out of the "sea" of background
radiation, and then disappearing back into it.
3.11.3
Threshold for creating W/Z particles (mass ~80GeV),
T ~ 1015 K (E ~ 100 GeV, t ~ 10-10 seconds)
This first threshold temperature is for the heavy W/Z particles, with mass-energies around
80GeV. As the temperature drops, fewer and fewer photons have enough energy to
create these, for example…
 +   Z0 + Z0
and the same for the W + and W -
The W/Z particles annihilate back to photons, or decay to lighter particles, for example…
W -  e- +  e
and the neutral Z0  e- + e+
After this threshold was passed, “the W and Z adopted their role of carrying the weak
interactions between particles, and had no independent existence except where they were
produced (briefly) in high-energy events involving collisions between particles, either
naturally or in particle accelerators designed for the purpose”. The quarks now "came into
their own” and started binding into hadrons – both mesons (pairs) and baryons (trios),
especially the light protons and neutrons.
The temperature is high enough for photons directly to create protons and neutrons in
particle-anti-particle pairs...
+p+p
D:\116105936.doc
and
+n+n
Page 80 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
The tiny quark excess carries over to the same excess of protons and neutrons over their
anti-matter counterparts. A slightly heavier neutron needs more energy for its creation
than a proton, but at this temperature the radiation photons easily have enough energy to
produce both in more or less equal numbers.
At the end of this phase the universe comprises photons of radiation and particles of both
both matter and antimatter: quarks that are free, or bound in protons and neutrons, and
electrons and neutrinos. Because the temperature is well above the thresholds for the
formation of all these particles, they are present in their matter and anti-matter forms in
almost equal numbers.
3.11.4
Threshold for creating protons and neutrons (mass
~940 MeV), T ~ 1013 K (E ~ 1000 MeV, t ~ 10-6 seconds)
The temperature has fallen to the point where the free quarks no longer have enough
energy to resist the strong force, and they become bound into pairs (mesons) and triplets
(baryons), mostly the light protons and neutrons. We have seen that the nature of the
strong colour force is such that it increases as quarks are pulled apart. The result is that
quarks will be confined in these particles for the next 14 billion years, up to the present
time - except for the few that find themselves at the core of a large star or in a particle
accelerator. So, by the time the universe is a millisecond old (t~10-3 s, T~3 x 1011K),
single quarks have effectively disappeared, “hiding exclusively inside protons and
neutrons”, and will never be seen free again.
Photon energies have decreased until they can only create the two lightest baryons protons and neutrons. Finally, the threshold for their creation comes at T~1013 K, and
these particles and their anti-matter counterparts annihilate each other back to radiation…
p+p + 
...and the same for neutrons.
The vast numbers of protons and neutrons and their antimatter counterparts annihilate
each other, leaving a tiny residue of protons and neutrons. Never again will anti-protons
and anti-neutrons be viable constituents of the material universe; they will be only the
ghostly, fleeting products of nuclear processes in stars and particle experiments.
The tiny matter excess now reveals itself, and we’re left with 1 proton or neutron to about
109 photons of radiation, a ratio we can measure in the universe today. We can
summarise it like this…
D:\116105936.doc
Page 81 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
(109 + 1) protons + 109 anti-protons  109  radiation photons + 1 matter proton
The residue of protons and neutrons remain vastly outnumbered in a universe awash with
radiation photons. Because the temperature is still well above the thresholds for the
creation of electrons and neutrinos, these particles and their antiparticles are still present
in almost equal numbers. The numbers of particles were decided by the balance between
the processes of creation and annihilation, and the “number and average energy of the
photons was about the same as for electrons, positrons and neutrinos.”
neutrino reactions
Neutrinos interact so little with ordinary matter that they can pass through an entire planet
like the Earth, and barely notice. However, the universe at this time is so dense, with the
particles pressed so close together, that the neutrinos readily interchange protons and
neutrons, via the weak interaction…
n0 +   e- + p +
and
p+ +   n0 + e +
When the universe is a bit less than 1 second old the protons and neutrons are changing
their identities in this way about 10 times every second. This at first keeps the numbers of
protons and neutrons about equal, but as the temperature continues to fall, and the
average neutrino energy decreases, the neutrons’ slightly greater mass becomes an
increasing barrier to their creation, and they start to be outnumbered by protons.
These protons and neutrons are not yet bound into nuclei. The energy required to break
up a nucleus is 6-8 MeV per nuclear particle, and the thermal energy at ~10 11K is more
than this, so any aggregates of protons and neutrons are destroyed as fast as they form.
3.11.5
Threshold for creating electrons (mass ~ 0.5 MeV), T
~ 6 x 109 K (E ~ 0.5 MeV, t ~ 1 second)
The temperature has now fallen below the threshold for electron creation, so the
reaction…
 +   e- + e +
can no longer occur. Electrons and positrons (anti-electrons) annihilate, to leave a
remnant of electrons.
D:\116105936.doc
Page 82 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
This is where the original “anti-universe” finally takes its leave. We’re left with the universe
of matter – baryons (protons and neutrons) and leptons (electrons and neutrinos) – awash
with radiation photons – about 1 billion photons for each baryon.
This is the last phase of the creation of matter from photons. As the universe has
expanded and cooled the photons’ “spending power” has decreased, and the electron is
the last particle species to be created. The situation can be likened to a lottery winner in a
time of rampant inflation. Initially his winnings would enable him to buy a stately home, but
as time passes their value depreciates, and he can afford less and less – a penthouse
apartment, then a detached house, then a small terrace, and finally he can only afford a
garden shed.
the end of neutron creation
The universe has now expanded to be about 1 light-year across, not far off the distance
from Earth to the nearest star. All particles are flying apart, and the average distance
between a proton and the nearest electron is now about 1000 proton diameters. The
neutrinos are becoming increasingly isolated, and effectively cease interacting with
matter, so that the production of neutrons ends at T~3 x 109 K (t~13s), when the
proton:neutron ratio has shifted to about 83p:17n. The neutron’s greater mass has led to
its being outnumbered by protons in the cooling universe. We’ve seen that free neutrons
are unstable, and now that their creation has ceased, their numbers slowly fall as they
decay to protons.
Neutrinos remain in the universe – there are about 550 neutrinos in every cubic
centimetre, about the tip of your little finger – but they hardly react with matter, and fly
through the Earth and our finger tips at close to the speed of light, almost undetectable.
one tick of the clock
The universe is about 1 second old - one tick of the digital clock on my desk. So much
seems to have happened in so little time. But to get a clearer picture we need a different
kind of clock, one that ticks each time one particle interacts with another. Lawrence
Krauss calculates that in the universe's first second the particles in a volume of 1 cm3
(roughly the tip of your little finger) experience about 10 89 interactions - a stupendous
number. He goes on…”For comparison, during its 5 billion years of burning, in each cubic
centimetre in the fiery core of the sun a total of about 1055 interactions have taken place.
This is about 10 million billion billion billion times fewer collisions than occurred in the
D:\116105936.doc
Page 83 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
same volume in the universe's first second. The number of collisions of atoms in this
volume of air during the 4 billion year history of life on Earth is about 1045, about 10 billion
times smaller still!”.
So the clock time is deceptive. The universe may appear very young, but the enormous
temperatures mean that the particles have experienced huge numbers of collisions, and
opportunities for interactions. In 1 second of clock time, two universes of matter have
been created, each capable of independent existence. One has been totally destroyed,
and the other very nearly so, such that only one billionth part of it remains, awash in a sea
of radiation.
energy budgets and lifespans
There is a link between this and the link between time and size in animals. The total
number of heartbeats in a lifetime is about the same for a hummingbird as for an elephant
or a whale. If you divide the life span by the number of heart beats you get about the
same number regardless of the size of the animal - "the total budgets for their actions are
the same". Metabolism, the rate of chemical reactions that consume biological fuel,
decreases as animal size increases; small animals "live faster". Our universe's
"metabolism", the rate of its nuclear reactions, increases with temperature and density, so
our small and hot young universe "lived faster".
3.11.6
Protons and neutrons start to combine, T ~ 109 K (E ~
100 keV, t ~ 200 seconds) – the first nuclei
As we’ve watched the universe expand and cool we’ve crossed three thresholds – and
seen the end of the creation process of three particles; first of the massive W/Z, then of
protons and neutrons, and finally of light electrons. We have seen an entire universe of
anti-matter annihilated, to leave a tiny residual universe of matter. “For every billion
particles of matter and anti-matter, one was left behind. … The little particle that was left
behind, for every billion that were annihilated, is what makes galaxies, stars, planets and
people”. It is this “small seasoning of leftover electrons and nuclear particles” which are
“the main constituents of the author and the reader”.
the deuteron
So far, the universe has contained only isolated protons and neutrons. Even the strong
nuclear force can’t withstand the disruptive energy of very hot photons, and any nucleon
clusters were immediately broken up. However, as the universe expands, the radiation
D:\116105936.doc
Page 84 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
photons are all the time being being stretched and cooled, and we now pass another
threshold. This one is marked, not by the end of production of an existing particle species,
but by the start of production of a new one. Now that photons have insufficient energy to
prevent it, we see protons and neutrons combining to make the first composite particle,
the first nuclear cluster – the deuteron,
p + n  D(p,n)
This reaction starts when the temperature has fallen below about 1 billion degrees, and
the universe is about 3 minutes old.
neutrons are saved by deuteron formation
We’ve seen that a free neutron will decay into a proton by the weak interaction…
n0  p+ + e- +  + 0.782 MeV
A free neutron's lifetime, its average "survival" time, is 886 seconds, about 15 minutes. In
the 3 minutes or so since neutron production ended they have been decaying, further
shifting the proton:neutron ratio, from 83p:17n to about 87p:13n. If things were to continue
like this all the neutrons would be lost, but instead, the remaining free neutrons are
gathered up into deuteron clusters. This is the first of a series of nuclear reactions that will
build up ever-bigger clusters of protons and neutrons. It is the first step in the long and
arduous journey towards our atomic world.
A free proton can’t decay to a neutron, because the proton is the lighter particle. But there
is another nuclear reaction that can consume protons – electron capture to create a
neutron and a neutrino
e- +
p+
0.5 + 938.3

n0
< 939.6
+

~0
MeV - the neutron is heavier by about 0.8 MeV.
The stability of protons is essential to the existence of the physical universe. Fortunately,
neither of these proton-consuming reactions can occur, because of mass difference
between the proton and neutron, but “it is a very small margin on which our existence
depends”.
We have seen that the strength of the nuclear force is a finely balanced thing: “a few
percent stronger and two protons (or two neutrons) would bind together. Nuclei consisting
of just two protons do not exist, but if they did all the hydrogen would have been
consumed in the Big Bang leaving none to power ordinary stars, including the sun….A
D:\116105936.doc
Page 85 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
few percent weaker and the deuteron would not be bound”. In both cases, the
development of the physical universe would have been quite different, and the evolution
of life would be impossible.
All the nuclear reactions
The full set of 12 nuclear reactions in hydrogen-burning…
cluster
size
1
p
p
2
n
+
H-2
p,n
4
H-3
p,2n
He-3
2p,n
5
6
7
He-4
2p,2n
Li-7
3p,4n
n  H-2
n
p
3
+
+
H-2
 H-3

H-2
2 H-2  H-3
p
2 H-2

2 H-2

+
He-3
+
p
He-3 + n
He-4

H-3
n
He-4
He-3  He-4
+
2 He-3  He-4
H-2
+
H-2
H-3
+
He-4
H-3

He-4
+
He-3  He-4
+
2p

+
+
Li-7
n
p
The first clusters of protons and neutrons
We have seen that each element is defined by the number of protons in the nuclei of its
atoms, while the number of neutrons can vary. There are three "versions" of hydrogen
atoms, and two of helium, and these are summarised in table 3.1.
element
hydrogen
hydrogen-2
or deuterium
hydrogen-3
or tritium
helium-3
helium-4
chemical symbol
H-1
H-2
H-3
He-3
He-4
1
2
3
3
4
(1p)
(1p,1n)
(1p,2n)
(2p,1n)
(2p,2n)
number of nucleons in the
nucleus
protons and neutrons in
the cluster
D:\116105936.doc
Page 86 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Table 3.1: The different atomic “versions” of hydrogen and helium.
Thus, hydrogen-3, for example, specifies 1 proton in the nucleus, and a total of 3
nucleons, so there must be 2 neutrons.
This is the point where our atomic universe starts to come into view, with names that are
familiar. It's perhaps like the return from a foreign holiday - you know the town names, you
can read the advertisements, and the radio plays pop songs you know.
the creation of helium-4
We have seen that the nucleus of deuterium (1p,1n) is loosely bound – it is broken apart
at temperatures above about one billion degrees (1 x 10 9 K). Once the temperature is low
enough for deuterium to be stable, further reactions quickly occur that build up nuclei of
helium-4 (2p,2n). Figure 3.41 shows a particle view of the main nuclear reactions in the
fireball.
p n
p
n
H-2 or
deuterium
p p
p p
n
p n
n n
He-4 or helium-4
or...
n
+
He-3 or
helium-3
+
p p
p
p n
p
n n
n n
+
+
H-3 or
tritium
p n
p and n collide and
create a pn pair,
hydrogen (H-2),
deuterium...
p
…then two pn pairs
collide and create
trios of p and n,
either He-3 or H-3…
n
…and these trios collide with
more pn pairs and create
helium-4 (He-4) quartets.
Figure 3.41: The sequence of nuclear particle reactions producing helium-4 (2p,2n)
– to save space protons and neutrons are given as p and n respectively
We can imagine particles flying at high speed through space, colliding, fusing,
fragmenting - continually rearranging themselves, but steadily building up larger clusters,
mainly helium-4. This sequence proceeds in 3 stages, with protons and neutrons
produced in the intermediate stages being fed back into the mix. We can think of these
D:\116105936.doc
Page 87 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
nucleons behaving rather like Lego bricks, "clicking" together to make a variety of larger
clusters.
Nuclear “snakes and ladders”
However, this diagram can only show a few of the many reactions, and doesn't make
clear the systematic build-up of larger clusters. An alternative is to plot the nuclear
reactions in a graphical format in what we might call "nucleon-space" (figure 3.42). Here
the proton and neutron contents of each cluster are plotted on the y- and x-axes
respectively; adding protons moves upwards, adding neutrons moves to the right. We
thus have a 2-dimensional view of nuclear reactions as moves on a sort of nucleon
chessboard.
protons, p
Li-6 (3p,3n)
Li-7 + p  2 He-4
(3p,4n) + p 2(2p,2n)
lithium-Li, 3
the proton
is fed back
into the
mix
He-3 (2p,n)
He-4 (2p,2n)
He-4 + H-3  Li-7
(2p,2n)+ (1p,2n) (3p,4n)
He-3+H-2He-4 + n
helium-He, 2
Li-7
3p,4n
+n
Hydrogen-H, 1
0
+p
+n
H-3
(p,2n)
H-2
p,n
+p
add 1 proton
free
proton, p
2H-2He-3 + p
H-1(p)
+p
+n
the neutron
is fed back
into the mix
free
neutron, n
add 1 neutron
0
1
2
3
4
neutrons, n
Figure 3.42: The nuclear reactions in the fireball plotted as a series of moves in
"nucleon-space”. Follow the arrows, add up the nucleons, see where you land. The
coloured squares represent unstable nuclides- we’re not concerned with these yet.
This figure shows the reactions given previously in figure 3.41, and a few more, including
the one producing lithium-7, where cluster (2p,2n) acquires cluster (1p,2n) and so moves
2 squares along and one up to arrive on the (3p,4n) square. If this cluster is hit by a
D:\116105936.doc
Page 88 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
proton it can fragment into two helium-4 clusters. There is thus a rich tapestry of
interactions; small clusters fusing into larger ones, larger clusters being broken down into
smaller ones, and leftovers being recycled - all can be plotted as "moves" in nucleon
space.
The nuclear reactions in the rest of this chapter will be shown as moves in nucleon-space.
This will allow us to see the details of reactions and also the inexorable progress to the
right and upwards, as bigger nuclei are formed.
3.11.7
The end of nucleus formation – T ~ 3 x 108K (E ~ 30
keV, t ~ 13 minutes)
In order for small nuclei to fuse into bigger clusters they have to approach close enough
for the short range nuclear force to bind them together. This means the background
temperature must be high enough so they have enough speed to overcome the mutual
repulsion of their positive charges. When the universe is about 13 minutes old the
temperature has dropped so far that the nuclei are moving too slowly to make contact,
and the nuclear reactions cease. We left the universe at t ~ 3 minutes, with the ratio
87p:13n. Out of every 200 particles the 26 neutrons will combine with 26 protons to make
13 helium-4 clusters, each four times the mass of a proton. This gives a mass ratio of
helium as 13 x 4/200 = 26%, which is in good agreement with measurements, and is one
of the several experimental confirmations of the big bang theory. This is an impressive
example where calculations based on the behaviour of sub-atomic particles in a particle
accelerator give a result that agrees with measurements made on the entire visible
universe. So far, all the stars in the universe have burned very little (about 4%) of their
hydrogen to helium, so the current ~24 - 28% cosmic abundance of helium is well
explained. “It is a profound fact that the universe is made almost entirely of hydrogen and
helium, and this is because of what happened early in the Big Bang”.
After 13 minutes, about the time to the first ad-break in a commercial tv programme, the
universe looks a bit like this (figure 3.43).
D:\116105936.doc
Page 89 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
p+
p+
protons or hydrogen-1
~ 93 particles in 100
(74% by mass)
neutrino hardly
interacts
with matter
hydrogen-2 (p,n)
or deuterium
about 0.01% by
mass
p
+
hydrogen-3(p,2n)
or tritium, about
0.0001% by mass
n0
0
helium-3 (2p,n)
about 0.001% by
mass
helium-4 (2p,2n)
~6-7 in every 100
particles (26% by mass)
-
n0 n0
p+ p+
n
+
n0 p
-
p+
.
-
electron
1 for every
proton
.
n
-
-
-
0
-
.
p+
n0
0
0 n
p+ n p+
n0
- lithium-7(3p,4n)
about 10-7 % by
mass, 1 particle in
about 5 billion
Figure 3.43: Nucleons and electrons in the cooling fireball; protons and neutrons
are gathered into small clusters, photons strongly interact with the charged nuclei,
so the universe shimmers with scattered light
The intense but brief heat of the early universe has fused all the surviving neutrons and
some of the protons into small clusters; pairs (H-2, deuterium) and trios (H-3 and He-3),
but mostly quartets of helium-4, and also a few septets of lithium-7, the biggest cluster
made.
The electrons and the oppositely charged protons and nucleon clusters are mutually
attracted, but as soon as they combine as neutral atoms “another photon would collide
with the atom and knock the electron free. With 1 billion energetic photons per electron,
the cards [are] stacked against the latter”.
Photons interact strongly with electrically charged particles, so they don't travel much
further than the nearest electron or nuclear cluster, "following a zig-zag path through
space like a high-speed ball in a crazy pinball machine." Thus “the universe was
completely filled with a shimmering expanse of high-energy photons colliding vigorously
with protons and electrons. This state of matter, called a plasma, is opaque, just like the
glowing gases inside … a neon advertising sign.” This would be like being in a fog, where
the light is scattered by the tiny drops of water suspended in the air, so that the whole fog
glows and you can’t see your way.
3.11.8
The threshold for ionisation - T ~ 3,000 K (E ~ 0.3 eV,
t ~ 300,000 years) - the first neutral atoms
D:\116105936.doc
Page 90 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
After this burst of activity, the “rush of creation began to subside. Physical processes
slowed … minutes turned to hours, hours to days, days to years, years to millenia.”
"Nothing much happens now for the next 300,000 years or so”. The fireball universe
expands and cools, the particles gradually slow down, the wavelengths of the photons
steadily increase. After about 300,000 years the temperature has fallen to about 3,000 K,
close to the boiling point of iron, and the average photon energy is now only ~0.26 eV.
The photons have a range of energies, however, and a tiny proportion of them have the
full 13.6 eV needed to remove an electron from the proton to which it is bound. But even
with 1 billion photons for each proton, this becomes insufficient, as the average energy
decreases. As the temperature keeps falling, the balance keeps shifting in favour of the
electrons, which settle in greater numbers on the protons and the nucleon clusters.
“Suddenly the face of matter was ready to change” and the universe now looks like this
(figure 3.44).
p+
-
-
p+
hydrogen-3,
tritium
hydrogen-2,
deuterium
n0
-
n0 n0
hydrogen-1
.
-
.
neutrino
p+
.
neutrino
-
p+ p+
p+ n0
+
n0 p
n0
-
helium-3
-
p+
n0
0
0 n
p+ n +
p
- n0
-
helium-4
lithium-7
Figure 3.44: The separation of matter and radiation after the fireball. The nuclei are
the same, and in the same places, but now they are combined with electrons, and
the photons fly through without interacting. The universe is now transparent.
Space becomes transparent
No new particles have been created; things are merely rearranged. Each nucleon cluster
has gathered a number of orbiting electrons to make an electrically neutral atom. The
photons of radiation rarely interact with these neutral atoms, and fly unimpeded through
space. The "fog" of charged particles has cleared and space becomes the transparent
D:\116105936.doc
Page 91 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
medium we're familiar with. The neutrinos continue as if nothing had changed – and they
still do!
We can imagine electrons settling around the nuclei rather like falling snow gently settles
on a landscape - on trees, fences, houses, cars. We know it’s the same solid, hard-edged
world underneath, but the shapes and patterns of things have changed and softened. So
it is with the electrons – they enfold the nuclei and enable a new level of subtle and
intricate particle interactions. All we've seen is the universal "fog" clear, and the light shine
freely, but the universe is transformed, and ready to emerge at a new level of complexity,
as we shall see in the next chapter.
Atoms of matter and photons of radiation now follow a quite different paths. We now have
separate matter and light - radiation has decoupled from matter. There are about 1 billion
photons for each proton or neutron, “a large number because matter when it was created
was only a trace contaminant born of the light”. The tidal wave of energy has subsided,
leaving small composite matter particles washed up on the shore of existence. So
"nothing" has become "something". What is this something?
The first atoms
For the first time there are atoms of substances we can recognise. There is hydrogen: the
flammable gas that lifted the first airships; the gas produced when we drop zinc into acid,
or if we over-charge a car battery. There is helium: the gas used in party balloons, that
also makes our voice go squeaky. And there is lithium: the metal used in batteries and
also to treat bipolar disorder. There the list stops - where are the familiar elements such
as carbon, oxygen, iron and copper? But if nucleons can gather in clusters of up to 7, and
containing up to 3 protons, then maybe they can go further.
3.11.9
review
From radiation-dominated to matter-dominated
We’ve gone from a universe dominated by radiation, where photons directly created
matter, to a universe that is now dominated by that matter. In our current universe the
average density of matter is tiny, equivalent to a few hydrogen atoms in each cubic metre.
In contrast, there are on average about 550 million radiation photons in every cubic metre,
with all possible wavelengths from radio through visible light to gamma-rays. “In other
words, the photons in space outnumber atoms by roughly a billion to one. In terms of total
number of particles, the universe thus consists almost entirely of microwave photons.”
D:\116105936.doc
Page 92 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
However, despite their overwhelming numbers, these photons have little influence over
matter, since their wavelengths have been stretched, and their energies reduced, by the
universe’s expansion. In the matter-dominated universe we will see matter go its own
way, more or less regardless of the huge numbers of photons passing by. Steven
Weinberg gives a perspective; “well before the contents of the universe became
transparent, the universe could be regarded as composed chiefly of radiation, with only a
small contamination of matter. The enormous energy density of radiation in the early
universe has been lost by the shift of photon wavelengths to the red as the universe
expanded, leaving the contamination of nuclear particles and electrons to grow into the
stars and rocks and living beings of the present universe”.
the first ad break
The first phase of nucleosynthesis – nucleus-building - has lasted about thirteen minutes,
roughly the time to the first ad break in a TV programme. We’ve seen how nuclear
reactions can only occur at modest temperatures, high enough to bring the nuclei
together, but not high enough to break them up. But the temperature window for this is
small – between 100 and 1,000 million Kelvin (10 8 – 109K), and the universe cooled
through this window in a few minutes, enough time for only three elements to appear.
The next phase of nucleosynthesis has been running for the last 13 billion years or so,
starting when the universe was about a billion years old. Nuclei are built up in stars, which
recreate the heat and pressure of the early universe, and yet this phase is powered by the
weakest of the universal forces – gravity. We'll look at this in section 3.12, but before this
we need to understand the principles of nucleon clustering.
The physical universe now comprises:

nuclei – that is, protons and their clusters with neutrons, up to cluster-7, each
carrying a positive electric charge

electrons, carrying negative electric charge, in numbers that equal the protons

neutrinos, with no charge, that barely acknowledge matter’s existence

photons of radiation, about 1 billion for each nucleon, that interact weakly with
matter
So there are just two “players” left on the universal stage: negatively charged electrons
and nuclei, nuggets of highly dense matter, and as far as the electrons are concerned,
differing only by their positive charge. The weak and strong interactions confine their busy
D:\116105936.doc
Page 93 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
activities in the nuclei. The much weaker electric interaction binds the oppositely charged
nuclei and electrons into neutral atoms. And the even weaker gravity can now act on the
atoms because they have neither attract nor repel, and they have mass. All the time the
neutrinos and radiation photons pass by, effectively indifferent to the doings of the neutral
atoms. Matter is now subject to gravity, by far the weakest of all the forces, but the only
one remaining.
3.12
Nucleosynthesis 2 - the next 14 billion years
The universe, having cooled to below 3,000K, now comprises almost exclusively neutral
atoms. This simple change has two enormous consequences: (1) matter and radiation
now go their separate ways, and radiation photons are free to stream without restraint
through the entire universe; (2) the electrical interaction is effectively confined into neutral
atoms, so the universe is a uniformly neutral medium, with no long range electrical
attarctions or repulsions. The weak and strong interactions are fundamentally very shortrange, and don’t extend outside the nuclei. This leaves the force of gravity, so much
weaker than the other three interactions, free to work its infinitely patient attraction on
these neutral atoms that now fill space. We’ll look now at what follows from these two
changes.
3.12.1
The cosmic microwave background
matter and radiation go separate ways
The fog of scattered light cleared, and our universe became transparent - but what was
there to see? There came a moment when each photon interacted with a charged particle
for the last time, after which it travelled freely through space, uninfluenced by neutral
atoms. When we look all around us out into space we see this radiation as the cosmic
microwave background, from the last time photons were scattered by matter at an
temperature of about 3,000 K.
photons are stretched by expanding space
What does an object at 3,000 K look like? The temperature of the tungsten filament in a
incandescent light bulb is 2,000 – 3,300 K; the visible surface of the sun is about 6,000 K.
So we can imagine the high energy photons, corresponding to the white-hot matter from
which they were scattered. But the universe has expanded to be about 1,000 times bigger
since then. Photons with energies corresponding to 3,000 K have had their wavelengths
D:\116105936.doc
Page 94 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
stretched by a factor of about 1,000, and consequently their energies reduced by the
same factor (figure 3.45).
The big
bang
singularity
separate
nuclei and
electrons
-
Radiation
decouples
from
matter…
+
-
…and
arrive at
the here
and now.
…and photons can now travel freely
through space, with their energy
reducing as the universe expands…
expanding space stretches the photons
+
- +
Time:
0
about 14x109 y
~300,000 y
the universe has expanded by a factor of ~1,000
Temperature:
3,000 K
the temperature has fallen by a factor of ~1,000
3K
Figure 3.45: The source of the cosmic microwave background radiation
We are now surrounded by photons that have been travelling unimpeded through space,
that now have a temperature of about 2.7 K. “The sky is not actually dark; it is just like the
surface of the sun, only 2,000 times cooler”. We can no longer see these photons with the
naked eye, for they are microwaves, but if we look out into space with the right detector, in
every direction we see this cosmic microwave background (CMB) – figure 3.46.
Figure 3.46: Left: The cosmic microwave background (CMB) radiation from the
universe all around us. The temperature is almost uniform, but there are tiny
D:\116105936.doc
Page 95 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
“ripples”, with cooler (blue) and warmer (red) regions differing by a few millionths
of a degree from the 2.73 K average. Even the smallest details in the inset picture
are big enough to develop into clusters of galaxies (100 μK is 100 millionths of a
degree).
Right: The colour coding gives an idea of the size of the tiny temperature variations
across the universe. (1 K = 1 millionth of a degree)
a wall of light
The microwave background, this “wall of light” from the last scattering of radiation photons
off the charged nuclei at ~3,000 K, shows us the entire universe at an age of about
300,000 years. We can’t see further back than this, the dotted line in the diagram, for the
universe was then an impenetrable fog of scattered light. The average temperature is 2.73
K, just above absolute zero, and is almost, but not completely uniform. The false colours
suggest big temperature variations, but the differences are tiny - only millionths of a
degree between the coolest blue and the warmest red regions.
ripples in the microwave background
It's thought that these temperature ripples originated in tiny quantum fluctuations in the
energy density of space in the very early universe, and were stretched by the universe's
subsequent expansion. “Tiny temperature differences are the scars left by the quantum
vacuum on our universe. These irregularities, created in the first moments of existence by
the teeming quantum vacuum, meant that the matter of the universe didn’t spread out
completely evenly. Rather, it formed vast clumps, that would evolve into the galaxies and
clusters of galaxies that make up the universe today. … It now appears as if the quantum
world, the place we once thought of as empty nothingness, has actually shaped
everything we see around us.”
The temperature "ripples" show that the density of matter in the early universe was very
slightly uneven. The scale of these ripples was huge; even the smallest features in the
inset picture would expand to be as big as a whole cluster of galaxies. These density
fluctuations were the seeds that later formed the enormous chains and clusters of
galaxies that we see today. Our own galaxy is the result of one such quantum fluctuation.
“The idea that an object with billions of stars, like the Milky Way, began life as a quantum
fluctuation … of the vacuum, an object of sub-microscopic scale, is mind-boggling.”
3.12.2
Collapsing gas clouds
gravity takes over the show
D:\116105936.doc
Page 96 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
This brings us to the second consequence of radiation decoupling from matter - the work
of gravity. Once there were no long range repulsions between charged nuclei, "gravity
completely took over the show", and started to work on the almost imperceptible clumps
of matter. Gravity can't create clumps of matter in a perfectly uniform gas, it can only
amplify pre-existing clumps. As the universe expanded, these clumps expanded very
slightly less than the surrounding regions; every time the universe doubled in size these
clumps did not quite double, and the density differences slowly grew. Once the local gas
density was more than about twice the average value, then the local effect of gravity was
strong enough to overpower the universal expansion. Huge clouds of gas atoms thus
developed, their mutual attraction brought the local expansion to a halt, and they then
began to draw in on themselves - they started to collapse. Their slight density excesses
grew at the expense of their surroundings - "the rich get richer, the poor get poorer". After
about 1-2 billion years, the first super-clusters of galaxies were well established, within
which our own Milky Way galaxy emerged.
We're now going to follow one large gas cloud, with 8-50 times the mass of our sun – that
is, 8-50 solar masses - as it collapses. Large gas clouds create large stars, which undergo
all the processes of building up the heavy nuclei, so we can use this to learn how the all
the elements in the modern universe have been created.
the first stages of collapse
We now have a cloud of gas that has started collapsing; the atoms are falling in on
themeselves. You can perhaps think of the giddy feeling at the top of a fairground ride, as
you start to fall slowly at first, then faster and faster. Our gas cloud will experience millions
of years of such falling, as its atoms of matter come together. When this starts, the cloud's
average density is about 1 atom/10 cm 3 (1 atom in your little fingertip), so the distance
between atoms is over 100 million times their individual size. The chances of collisions
between atoms is tiny, but the cloud is so huge, and there are are so many atoms, that
collisions will occur, and become more frequent, and more energetic. The gas cloud is
starting to compress under its own gravitational attraction, its own weight - this is the birth
of a star. Gravity and matter meet and compete, and this will be the constant theme
throughout the cycle of the star's life and death. We'll see that gravity will always win,
even to the extent of crushing some of the star out of its material existence, but the
universe will be immeasurably enriched in the process.
pumping up a bike tyre
D:\116105936.doc
Page 97 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
So, let's start with the process of compressing a gas, for example, in pumping up a bike
tyre. After only a few vigorous strokes, the end of the pump feels hot - where has that heat
energy come from? It's worth taking a moment to think about the processes going on
inside the bike pump - figure 3.47.
gas particle rebounds
slightly faster from the
moving piston
piston
moving in
collisions with the
walls produce the
gas pressure
to
tyre
average particle speed
is about 500 m/s.
gas particles collide
with each other
the same number of gas particles
are confined in a smaller volume so the pressure increases…
the end of the
pump feels hot
when the pressure inside the
pump is bigger than the tyre
pressure, then air will flow
into the tyre.
…and the particles are moving faster…they
are hotter - increasing the pressure still more.
Figure 3.47: How compressing a gas heats it up
A typical bike pump, with the piston drawn back will contain about 5 x 1021 particles of air five thousand billion billion. At a "normal" room temperature of about 20 oC (close to 300K)
these air particles are moving at an average speed of close to 500 metres/second - half a
kilometre each second. The gas particles also collide with and rebound from each other.
In the air we breathe, the particles have a size of about 0.1 nm, and are on average about
3 nm apart, a distance they can cover in about 7 million millionths of a second (7x10 -12 s).
The average air particle will then be experiencing something like 150 billion collisions
each second, with other particles and with the walls of its container. It's the huge numbers
of these tiny collisions against the inside walls of the bike tyre that keep it inflated. This is
a world of unceasing and furious activity. And yet at normal temperatures the energies
involved are tiny. Remembering our rule of thumb, that a temperature of ~12,000K is
equivalent to an energy of 1 eV, we can see that at our normal air temperature of 300 K,
D:\116105936.doc
Page 98 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the average particle kinetic energy is a mere 0.03 eV - the slightest pat on the back for an
atom.
compressing leads to heating
In the bike pump the air particles gain speed when they rebound from the incoming piston.
Faster means hotter; compressing the gas heats it up. For a gas cloud in space, gravity is
the piston; the gas particles fall in on each other, converting potential to kinetic energy;
the gas cloud warms as it is compressed. The gas particles strike each other with more
and more energy as time passes and the cloud shrinks. At first the particles rebound from
each other with no loss of energy, but as the temperature rises, some of the energy of
collisions is emitted as light radiation. So in our shrinking gas cloud, we see gravitational
potential energy converted to kinetic energy and radiation, in roughly equal proportions.
The loss of energy as radiation means the gas cloud does not heat up so much, and this
speeds up the contraction.
the gas cloud starts to glow
The gas cloud thus starts to emit radiation, starting in the invisible infra red, and working
towards visible light. The intensity of radiation can be enough to blow away the outer
layers of the cloud, ejecting as much as half of its original mass. When the temperature
reaches several thousand degrees the hydrogen molecules and helium atoms have their
electrons stripped from them, leaving just naked nuclei. The last time the nuclei were free
of electrons was in the early universe, about 300,000 years old. The presence of charged
nuclei means that the radiation can't escape, but is trapped inside the gas cloud. The gas
cloud dims and its contraction slows as the internal temperature starts rising rapidly,
reaching several million degrees.
the hot gas cloud becomes a true star
Our huge diffuse gas cloud has compacted and heated up, and is emitting huge amounts
of energy as radiation - but it is not yet a star, for its energy output has come from
converting its gravitational energy to heat and light. Finally, with a core temperature of
about 10 million degrees, there occurs the first of a series of nuclear reactions that
generate energy - our gas cloud has become a true star. Figure 3.48 shows the
development of galaxies and stars, enormously dense concentrations of matter, from the
slight ripples in the CMB.
D:\116105936.doc
Page 99 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
a
b
c
e
d
Figure 3.48: From ripples in the cosmic microwave background (CMB) to galaxies
and stars: (a) the view of the CMB across the whole sky; (b) a cold spot in the
microwave radiation; (c) within the forming clouds of gas the first stars light up
(about 100 million years); (d) the early universe is lit up by a variety of stars and
galaxies, and (e) the WMAP satellite looking through today's universe back in time
to the CMB.
We're ready now to look at the remarkable processes whereby our star assembles
protons into all the elements in the universe.
3.12.3
Stars and the elements
Our story of stars focusses solely on their rôle as nuclear furnaces that forge light
elements into heavier ones. We've seen how nucleosynthesis - the formation of the nuclei
of the elements - in the fireball was limited by the rapidly falling temperature and short
time available. Now we'll see stars recreate the conditions in the fireball and sustain them
for millions of years.
We'll continue the story of our gas cloud, now that it has become a large star, and this will
show us in one narrative, the main processes that have laid the foundations of our
material existence. ”Thus it is possible to say that you and your neighbor and I, each one
of us and all of us, are truly and literally a little bit of stardust”.
The basic nucleosynthesis reactions
the battle between gravity and matter
D:\116105936.doc
Page 100 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
A star reveals – in a sense, it embodies - the conflict between gravity and matter - (figure
3.49).
Gravitational
contraction
gas
cloud
?
Exhaustion
of fuel
Core
heating
Nuclear
burning
Figure 3.49: The conflict between gravity and matter: stars go round the gravitymatter circuit, undergoing a series of nuclear reactions, until all their fuel is
exhausted.
A gas cloud contracts under gravity, until its temperature and pressure are high enough to
ignite a nuclear fuel, when the cloud becomes an active star. The fuel “burns” and
releases the energy that supports the star’s weight, and halts contraction. When this fuel
is exhausted, further contraction under gravity increases the star’s core temperature until
the next nuclear fuel ignites. Each fuel represents one lap of the gravity-matter circuit.
Gravity always wins in the end, because it never gives up, whereas a star's nuclear fuel
supply is finite. The star spirals in towards its end, which is largely dependent on the mass
of the gas cloud from which it formed. Thus “stars live their lives on the brink of disaster”,
always on the inwards gravity-matter spiral.
Temperature and pressure are the dominant factors in a star’s gravity-driven life. They
decide when a star’s life begins, and we will see a star’s life end when the matter it is
made of can no longer sustain the temperature and pressure imposed on it.
the creation of the elements
Oliver Sacks has written how “I was pleased when I was told that we ourselves were
made of the very same elements as composed the sun and stars, that some of my atoms
might once have been in a different star. But it frightened me too, made me feel that my
atoms were only on loan and might fly apart at any time, fly away like the fine talcum
powder I saw in the bathroom”. For the stars have created the elements of our material
world, through a set of 3 basic nuclear reaction processes. These can be summarised as
follows…
D:\116105936.doc
Page 101 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei

enlarging nuclei
o fusion reactions - in which the main nuclear fuels “burnt” are: hydrogen, helium,
carbon, neon, oxygen and silicon, whereby nuclei combine into bigger clusters to
release binding energy, and generate heat
o single nucleon captures - including

neutron capture processes, where a nucleus picks up stray neutrons: the slow
"s"-process and the rapid "r"-process, and


proton capture – the p-process.
fragmenting nuclei - the "spallation" process, in which nuclei in space are broken into
smaller fragments by cosmic rays
In addition, any unstable nuclei will seek stability by a beta-decay process, whereby
protons and neutrons interchange by the weak interaction (described in the last chapter).
3.12.4
Describing nuclear reactions
How can we describe nuclear reactions? One of the important stellar reactions is the
"triple alpha" process, whereby three quartets of nucleons, each one (2p,2n), fuse
together into a cluster of a dozen, (6p,6n)…
3 (2p,2n)  (6p,6n)
This is how the nuclear reactions in stars should, strictly, be described. This is the world of
protons and neutrons, that operates on its own terms. However, the nuclides created by
the stars are the foundations for our atomic world, made of elements we are familiar with.
Thus the nuclide (2p,2n) will become helium-4, with a total of 4 nucleons in its nucleus.
The nuclide (6p,6n) will become carbon when it is ejected from the star into space. When
this carbon atom aggregates with others it becomes the material carbon we know - the
"lead" in a pencil, the diamond in a ring, or your burnt toast under the grill. For now, it is
just a cluster of 6 protons and 6 neutrons - a total of 12 nucleons, jostling with other
nuclear clusters in the heat and pressure of a star’s core.
So the nuclear reactions will be described like this…
3 helium-4 (2p,2n)  carbon-12 (6p,6n)
D:\116105936.doc
Page 102 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
including the chemical element name as well as the nucleons. While this is cumbersome,
we can see the elements created, and also follow the arithmetic of the protons and
neutrons.
This reaction is also called helium burning. We normally think of burning as a chemical
reaction with oxygen like a bonfire or gas flame. But this is a nuclear reaction, in which
helium nuclei are consumed and transformed to something else. We can also think of the
reactions as "nuclear Lego”, where nuclei can be built up or broken down, just like
assemblies of standard Lego bricks. As with the first nuclear reactions in the cosmic
fireball, I'll show the nuclear reactions as moves in nucleon-space, and figure 3.50 shows
the major nuclear reactions in this way.
protons, p
gain a helium-4
nucleus (2p,2n)
- an alpha particle
+2
gain a
proton
beta-minus
decay: n  p
+1
lose a neutron
gain a neutron
-1
lose a
proton
beta-plus
decay: p  n
-2
-2
-1
+1
+2
neutrons, n
Figure 3.50: Some of the possible nuclear reactions in a star, shown as moves in
nucleon-space.
Humanity has grown up with the stars in the heavens: “In the great cities of the world, we
have detached ourselves from night. If you are a city dweller who doesn't believe this,
travel at least a hundred miles into the countryside, mount the highest hill, and stare at the
sky. It is not the same sky at all. … On a clear night in the mountains, you become part of
the sky. The stars reach out and touch you, and suddenly, you feel the embrace of a
galaxy”.
Our recent understanding of stars as nuclear forges is the latest and truest conception of
their nature. Perhaps as a heritage of the other ways we've seen them in the past, we
tend to anthropomorphise them; we speak of their life-cycles, of their birth and inevitable
D:\116105936.doc
Page 103 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
death. We may find ourselves thinking of their purpose as being to create the heavy
elements, so we might talk of a star burning its nuclear fuel to generate the heat
necessary to halt the compression due to gravity. We will see that the sequence actually
is: the crush of gravity heats up the star's interior…which ignites a nuclear fuel…which
generates heat…which temporarily halts further compression…until that fuel runs
out…and the cycle repeats itself with the next nuclear fuel that “burns” at a higher
temperature…until all the available fuel is consumed.
Through gravity, the stars recreate the conditions in the fireball soon after the big bang.
But whereas the fireball cooled within minutes, the stars can maintain enormously high
temperatures and pressures for billions of years, and thereby accomplish what the entire
universe could not - the creation of all the viable clusters of nucleons, all the elements of
the periodic table - the basis for our atomic existence.
We're ready now for our newly-formed star's first nuclear reaction.
3.12.5
Hydrogen-burning - the proton-proton chain
no free neutrons
The cosmic fireball reactions easily started with protons combining with free neutrons, and
then numerous reaction steps to produce helium-4 (see section 3.11.6). But our new star
contains no free neutrons, so the nuclear reactions must start with only protons. We've
seen how neutrons are vital to stable clusters, so the star must somehow generate its
neutrons from the protons. Once a proton has changed to a neutron, it can join with
another proton to make hydrogen-2 (deuterium), and things can go on from there.
We'll focus on one proton, in the midst of the other protons dashing about, and repelling
them, due to their mutual electrostatic repulsion (figure 3.51).
D:\116105936.doc
Page 104 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
2. a proton needs ~1 MeV to reach
the top of the "potential hill"
3. a proton with an
average kinetic energy of
~1 keV can only get a little
way up the "potential hill"
before it is repelled
p+
4. a proton with ~10 KeV
kinetic energy can get
close enough for its
matter-wave to penetrate
through the hill, so it can
tunnel through
p+
p+
1. a proton surrounded by
a "potential hill", which
repels other protons
Figure 3.51: A single proton surrounded by a potential “hill”, repelling other
protons; the vertical axis is not to scale
This is the reverse of alpha-decay, where alpha particles escape from a large nucleus by
quantum tunnelling through the wall of the “energy well” that confines them. In the figure
above, our single proton is not so much in a well, as surrounded by a potential energy
“hill”, arising from the repulsion between protons. Other protons “climb” part-way up this
hill, come to a stop, and then “roll” back down again, as they are repelled. Here the
protons are shown as simple particle matter-waves.
a proton tunnels in
Even at 10 million degrees the average proton has around 1 keV of kinetic energy,
nowhere near enough to get over the energy hill; so it “climbs” part way up the slope, and
is then repelled. However, at any moment the protons have different speeds. Think of the
dodgem cars at the fair; at any one moment, a few cars will have stopped because of
head-on collisions, and a few will be moving extra fast due to collisions from cars behind.
In the same way, there is a distribution of speeds among the protons. About 1 in 10 million
protons has 10 times the average kinetic energy, and can climb further up the hill and get
closer. Even this rare proton has only about 1/100th the energy needed to get to the top of
the electric potential hill. But if it is on an exact collision course with the other proton, then
it can briefly get close enough that its matter-wave extends through the potential hill into
the inside; that is, there is a small but finite chance that the proton can tunnel through the
potential hill, and join the other proton inside. So, out of a multitude of proton collisions
one will result in a proton-proton pair.
the weak interaction transforms a proton to a neutron
D:\116105936.doc
Page 105 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
But we have seen that the nuclear force is not quite strong enough to bind a pair of
protons – there’s no such thing as a diproton. However, if one of the protons were to
transform to a neutron the pn combination would be stable. While this is energetically
favourable, it can only happen by the weak interaction, which is very slow, or to put it
another way, is very unlikely to happen. So we mostly see our two protons almost
immediately fly apart. However, there is just a tiny chance that, in the very short time they
are together, one of the protons transforms to a neutron, in which case we now have a
stable nucleus of hydrogen-2, or deuterium. We write this simple nuclear reaction as…
p + p  hydrogen-2(p,n) + e+ (positron) +  (neutrino)
neutrinos from the sun
All this is taking place now, deep in the heart of a star like our sun. How do we know it's
happening? The proof is in the neutrinos produced, which were first detected in 1988.
“The multitude of neutrinos generated by nuclear reactions in the solar core pass through
a half million miles of sun as if it were a thin sheet of glass, emerging in a matter of
seconds. They reach Earth in eight minutes, and pass through it with no resistance,
either. About 500 billion neutrinos from the sun fall on each square inch of ground in a
second. Our bodies are pierced by them unceasingly, day and night. They leave not a
trace.”
a very slow nuclear reaction
The star has taken the first step on the long road of building up bigger nuclei. It looks so
simple - there's no hint of its incredible difficulty. This reaction is so rare that it has never
been observed in laboratory experiments. Even in our own sun it takes an average proton
around 10 billion trillion collisions, lasting about 14 billion years, to to bind with another
proton and transform to a neutron at the same time. However, this unbelievable slowness
ensures that stars like our sun consume their fuel very gradually over billions of years,
thus ensuring that life can evolve slowly on Earth under steady conditions. But for now,
the universe and organic life just wait.
flipping a coin
There is an old story about a student who flips a coin to decide what to do that evening: if
heads, then go to the cinema; if tails, then go to the pub, and if the coin should stand on
its edge - then study. We've seen the three enormous obstacles that must be overcome:
(1) only very few protons have enough energy to get near enough to another proton; (2) of
D:\116105936.doc
Page 106 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
these, very few tunnel through the potential wall; and (3) it is extremely unlikely that either
of the pair of protons will transform into a neutron in the very short time before they repel
and separate. Thus, the p-p reaction is “just about the most inefficient nuclear reaction
imaginable”. Flipping a coin and getting three "edges" in a row is easy by comparison!
But, it's not impossible, only unlikely. If pairs of protons collide enough times, then the
nuclear reaction will happen. If you flip a coin enough times, then you will get not only an
edge, but at some point, three edges in a row will occur. And a star contains so many
hydrogen protons which are undergoing huge numbers of collisions each second, that the
proton-proton reaction happens in abundance. Our own sun "burns" some 600 billion kg
of protons each second - that's over 3 x 1035 protons - or nearly 400 trillion trillion trillion,
and it contains enough protons to do this for about 10 billion years.
We saw how a few of the lightest nuclei were assembled in the cosmic fireball, in the first
few minutes after the big bang, but this was very limited because of the rapid expansion
and cooling. “Stars, on the other hand, stay dense and hot for millions or even billions of
years”, long enough for the rarest, most unlikely of nuclear reactions to occur.
the proton-proton chain
So our star has fused two protons; this is the first stage in a series of reactions called the
proton-proton chain – figure 3.52.
D:\116105936.doc
Page 107 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
2. …another proton
collides…is gripped by the
strong nuclear force…and a
nucleus of He-3 (2p,n) is
made…
1. Two protons collide, and
one of them transforms to a
neutron - to make a nucleus
of deuterium (p,n)…
positron
e+
neutrino

3. …another He-3 joins
with this to make stable
He-4 (2p,2n)…releasing 2
protons in the process.
gamma
ray photon

p
p n
p
p n
p
p
p p
n n
p
p n
p
p
p
p n
p

p

e+
recycled
This can be summarised as…
4
p
p p
n n
He-4
(2p,2n)
+
2e+
+
positrons (antielectrons) meet
electrons and
mutually
annihilate to
produce gamma
radiation
2
neutrinos
leave the star
and fly through
space at
nearly the
speed of light
+
2
+
gamma ray
photons slowly
make their way
though the star's
outer layers and
leave as sunlight
16.7 MeV
the fomation
of one
nucleus of
helium-4
releases
16.7 MeV of
heat energy
Figure 3.52: The nuclear reactions in the proton-proton chain
The three steps in the proton-proton chain successively create deuterium (p,n), then
helium-3(2p,n), and finally helium-4(2p,2n). The two protons left over at the end are
recycled back into the mix. The reaction series creates two photons which will leave the
star as light radiation. The positrons will meet electrons and will annihilate to create more
radiation. The neutrinos scarcely interact with matter and escape from the star into space,
taking some of the reaction energy with them. Once two protons have been fused to make
hydrogen-2, then subsequent progress is much faster. Steps 2 and 3 only require the
strong nuclear force to bind the larger nuclei together, and can occur within minutes.
D:\116105936.doc
Page 108 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
where does the energy come from?
This is the summary of the hydrogen burning reaction…
4 protons + 2 electrons
 1 helium-4(2p,2n) + 2 neutrinos + energy
(4x938.3) + (2x0.5) = 3754.2 MeV  3727.4 MeV
mass loss = 26.8 MeV
The reaction sequence produces two positrons, which annihilate with two electrons back
to radiation energy, so these two electrons are included in the energy accounting. Four
unbound nucleons become bound into a nucleus, and the energy of their binding, which is
released as heat, is “paid for” by the particles’ loss of mass. Thus the mass loss of the
system of particles gives us the energy released.
Hydrogen burning, the rearrangement of four nucleons into one nucleus, releases 26.8
MeV of energy, about 6.7 MeV per nucleon. This reaction, when free nucleons are first
bound into a nucleus, releases an enormous amount of energy. We’ll see that subsequent
reactions, rearranging already bound nucleons into progressively bigger nuclei, are not
nearly so productive.
hydrogen burning sustains the star’s weight
This, then, is the proton-proton chain, otherwise known as “hydrogen burning". “The
energy liberated in these [nuclear] reactions yields a pressure … which opposes
compression due to gravitation. Thus an equilibrium is reached for the energy produced,
the energy liberated by radiation, temperature and pressure”. The reaction releases
energy, so the nuclei in the core get hotter, that is, move faster, thereby exerting a higher
pressure, which can support the weight of the star and halt the star’s contraction.
In the same way, the moving air particles inside a car tyre support the weight of the
vehicle. We deliberately increase the pressure in a tyre by pumping more air particles into
it, but the core pressure in a star depends on its temperature, which is governed by the
nuclear reactions taking place there.
A star of one solar mass burns its hydrogen steadily in its core at a temperature of ~15
million degrees and a density of ~150 g/cm3. It’s worthwhile pausing to think about
density, because this will be an important factor in the star’s life. Gold is one of the
densest metals, with a density of close to 20 g/cm3 – that is a gold lump the size of the tip
of your little finger would have a mass of ~20 grams. So, the hydrogen burning in the core
of a star is compressed to about 8 times the density of gold.
D:\116105936.doc
Page 109 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
nuclear reactions are self-regulating
Stellar nuclear reactions are inherently self-regulating – as long as there is nuclear fuel
available to release energy. “If the core gets hotter, the pressure increases, causing the
gas to expand against gravity, thus cooling the core.” Conversely, if the reaction slows
down, the core cools, the pressure drops, and the star contracts. This then heats up the
core, and the nuclear reaction speeds up again. This is sometimes called a “natural
thermostat”, and this suggests that the star somehow “sets” the temperature, but of
course, it’s not like that. A star compacts, and its core heats up until a nuclear reaction
starts that can generate the heat energy to halt the compaction. We’ll see soon what
happens when a heat generating nuclear reaction is not available. Stars in the hydrogen
burning phase are very stable; a star the mass of our sun will burn hydrogen steadily for
about 10 billion years, though we will see later that this time depends on the star’s mass.
the star’s radiation takes a long time to escape
What is surprising is that energy is generated so slowly in the sun's core that a human
sized volume of our sun burns its nuclear fuel slower than a human converts food into
energy. Then this energy takes a very long time to escape. The proton-proton chain
reaction produces high energy gamma ray photons. Each photon's way is blocked by the
nuclei in the star's core, like a crowd of shoppers in a busy market. The photons bounce
back and forth in a zig-zag path, repeatedly scattered by the charged nuclei, and finally
emerge from the sun's surface with a range of smaller energies, spanning the
electromagnetic spectrum. After working their way out of the sun (a distance of about 700
million metres) in about 30,000 years - light would travel that distance in space in just 2
seconds - they then reach the Earth (200 times further) in about 8 minutes.
the proton-proton chain in nucleon space
Figure 3.53 shows the proton-proton chain as a set of moves in nucleon space.
D:\116105936.doc
Page 110 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
protons, p
lithium-Li,3
He-3 (2p,n)
He-4 (2p,2n)
helium-He 2
H-1 (p)
(2) + p
(3) 2He-3He-4+2p
H-2
(p,n)
Hydrogen-H 1
H-3 (p,2n)
(1) p + pH-2(p,n)
free
protons
(4) the 2 protons
are recycled
back into the mix
0
NO free
neutrons!
0
1
2
3
neutrons, n
Figure 3.53: The proton-proton chain as a set of three moves in nucleon space
With no help from free neutrons, protons have worked their way diagonally in nucleon
space to helium-4. We will see the further stellar nuclear reactions work their way steadily
up and to the right in nucleon space, creating ever larger nuclides.
entering the nuclear valley
Thus the hydrogen-burning star has entered the nuclear valley. The entire universe got
little further than this, though that was with the temperature and pressure rapidly falling.
We can look through the helium “pass”, and glimpse the nuclear terrain which the stars
will explore (figure 3.54).
D:\116105936.doc
Page 111 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the end of the arc of stable nuclides
a proton – a
hydrogen nucleus
Figure 3.54: A view of the nuclear valley through the helium “pass”. The central
black arc of stable nuclides separates the blue neutron-rich nuclides from the
orange proton-rich nuclides. The largest nuclides form the peak in the far distance.
3.12.6
The carbon (CNO) cycle
a richer gas mix
The first generation stars, starting with only hydrogen and helium, have only the
enormously difficult proton-proton cycle as their first nuclear reaction. However, stars of
later generations form from a richer mix that includes heavier nuclei from previous
generations of stars. One of these, carbon, present even in small amounts, can greatly
speed up the production of helium - figure 3.55.
D:\116105936.doc
Page 112 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
protons, p
5. second beta decay, pn
oxygen-O, 8
O-15
(8p,7n)
2. first beta
decay, pn
N-13
(7p,6n)
nitrogen-N, 7
O-16
(8p,8n)
7. …and then the O16 nucleus splits
…into He-4 and C12…
…and the cycle starts
again…
N-14
(7p,7n)
6. capture
a fourth
proton…
N-15
(7p,8n)
1. capture a
proton
carbon-C, 6
C-12
(6p,6n)
3 and 4.
capture two
more protons
C-13
(6p,7n)
…back
to He-4
boron-B, 5
5
6
7
8
9
neutrons, n
Figure 3.55: The CNO cycle produces helium: the C-12 nucleus follows a zig-zag
path through nucleon space, capturing 4 protons and decaying twice, to end up as
O-16, which emits a He-4 nucleus, and so returns to C-12. Proton capture is
accompanied by gamma ray photon emission, and beta decay is accompanied by
the emission of positrons and neutrinos – these are not shown here.
A carbon-12 nucleus captures a series of 4 protons, 2 of which decay to neutrons. The net
result is the gain of 2 protons and 2 neutrons, and moving 2 squares diagonally, landing
on oxygen-16. This nucleus fragments to helium-4, and carbon-12, which undergoes the
cycle again. This cycle is incredibly fast compared to the proton-proton cycle, and can
produce a helium nucleus in about a day. We can see why this is called the CNO cycle,
since it creates in sequence the nuclei of carbon, nitrogen and finally oxygen. It's
described here because of its rôle in building hydrogen rapidly into helium. It's also the
major source of nitrogen in the universe - the same nitrogen we breathe on Earth. The
CNO cycle certainly powers many of the stars shining now, because they contain some
carbon and other nuclear species that had been produced by earlier generations of stars.
The first generation of stars would have been powered solely by the proton-proton chain.
D:\116105936.doc
Page 113 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
once around the gravity-matter circuit
We have now described the major energy generating process of the majority of stars, in
which hydrogen burns, building up helium in the star's core, like ashes in a fireplace. Our
star has consumed the first of its nuclear fuels, and has gone once around the gravitymatter circuit (see back to figure 3.49). Let’s follow our star round the circuit again, and
see how the ashes from the first reaction become the fuel for the second.
3.12.7
Helium burning - the " triple alpha" process
when the hydrogen runs out
The accumulation of helium ash and the depletion of hydrogen slow down the hydrogenburning reaction. The weight of the outer layers can no longer be supported, and the star
contracts, further heating up the core. This is the profound difference between nuclear
“burning” in a star and our familiar chemical combustion. In a chemical combustion
reaction on Earth, once the fuel is consumed, the fire simply goes out. In a star,
everything is driven by gravity, like the inexorable demands of a blackmailer. Gravity
produces the enormous temperatures and pressures needed to start and sustain nuclear
reactions. The energy thus produced halts the star's contraction - as long as the reaction
has fuel to continue. When the fuel runs out, that reaction ceases, and gravity, now
unopposed, compresses the star further, driving up the temperature and starting the next
reaction in the series.
When the hydrogen burning ends, the temperature of the star’s core rises to ~120 million
degrees or more, its density increases to ~20 kg/cm3, the mass of a dozen house bricks in
your little fingertip, and the helium that has accumulated now ignites. The outer layers
also get hotter, so hydrogen-burning spreads to the layer outside the core.
improbability and coincidence
If we think of the star's reactions as "nuclear Lego", then we'll be comfortable with heliumburning. If we can put together singles, pairs and trios of nucleons to make quartets, then
we can put together quartets to make bigger nuclei - all we need is higher temperatures
and energies - right? We've seen how a star accomplishes the fantastically improbable
first step of the proton-proton chain with ease. Like flipping a coin and getting an "edge"
this is merely improbable - flip enough coins enough times and you're certain to get
"edges". Helium burning, however, depends on a set of conditions that appear quite
coincidental. There is a universe of difference between improbability and coincidence.
D:\116105936.doc
Page 114 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
no stable clusters of 5 or 8 nucleons
The problem can be stated simply: there are no stable nuclei of 5 or 8 nucleons. A star
can't add a proton to helium-4, because a cluster of 5 is unstable. Similarly, two helium-4
nuclei can't fuse to make a cluster of 8. There appears to be no route to the heavy
elements, that comprise our physical world.
beryllium-8 is unstable but with a long lifetime
In the now very hot star's core two helium-4 nuclei have enough energy to collide and
stick. This gives a beryllium-8 nucleus, which is unstable, but with a mass only very
slightly greater than the original pair of helium nuclei, it has a unexpectedly long lifetime about 10-15 s. This seems vanishingly short to us, but it's quite long-lived in the nuclear
world, and so there is a small but sustained population of beryllium-8 nuclei in the star's
core - about one for every billion helium-4 nuclei. With trillions of collisions occurring each
second, this lifetime is long enough for about 10,000 encounters with a helium-4 nucleus,
in which the two might possibly fuse to make a carbon-12 nucleus. This is shown in figure
3.56.
protons, p
oxygen-O, 8
nitrogen-N, 7
4. …some of which joins with
another He-4 to make O-16.
There are no
stable clusters of
5 or 8 nucleons
carbon-C, 6
3. …to make stable C-12…
boron-B, 5
beryllium-Be, 4
2. …unstable Be-8, which lasts just long
enough for a third He-4 nucleus to join…
lithium-Li, 3
helium-He, 2
1. Two He-4 nuclei join to make…
hydrogen-H, 1
1
2
3
4
5
6
7
8
neutrons, n
Figure 3.56: The helium burning process: helium-4 nuclei successively fuse,
creating an alpha-particle “staircase” in nucleon-space. (The cell outlined in red
was the centre of the display)
D:\116105936.doc
Page 115 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
This shows the two main reactions in the helium-burning phase: the ‘triple-alpha’ process
in which three He-4 nuclei fuse to make a C-12 nucleus, and this C-12 then fusing with
another alpha particle to make O-16. Much of the oxygen in the universe comes from this
second nuclear reaction.
Beryllium-8 is like a stepping stone that starts crumbling under your weight, but holds up
just long enough for you to jump to the next stable position. The degree of instability of
beryllium-8 is critical. If it were more stable then on reaching the helium ignition
temperature the star's nuclear reactions would quickly transform all helium to carbon. The
sudden release of energy would blow the star apart, and this would prevent the
appearance of all elements heavier than carbon in our universe.
the excited carbon-12 nucleus
But with so few beryllium-8 nuclei available, it's vital that the collisions between beryllium8 and helium-4 have a high chance of leading to fusion. In the normal run of things even
10,000 collisions aren't enough. This is where coincidence first steps in. It turns out that
the carbon-12 nucleus can exist in an excited energy state which is just right for the fusion
of beryllium-8 and helium-4.
The combined mass-energy of the beryllium-8 and helium-4 nuclei is 7.37 MeV. They will
fuse most easily, if the product nucleus has a favoured energy of vibration that is close to,
but less than this value. It turns out that the carbon-12 nucleus has a favoured excited
state at an energy of 7.65 MeV - close but just too high. However, the temperature of 100
million degrees inside the star core is enough to raise the combined energy of beryllium-8
and helium-4 to just above the excited carbon-12 energy, so the fusion reaction can occur
with its greatest efficiency.
resonance in swings and radios
Think of pushing a child on a swing. The pendulum of the swing has a natural frequency,
say 1 complete swing every 2 seconds. We know that if we match the frequency of our
pushing to the swing’s natural frequency then we will transfer our energy to the swing
most quickly and build up a large amplitude. Our push and the swing have the same
frequency – we say they are in resonance. When we tune a non-digital radio, we alter the
natural frequency of the oscillating tuning circuit to match the frequency of the radio
signal. This “resonance allows a radio to be tuned so it is millions of times more sensitive
to radio waves of a certain frequency than signals of all other frequencies”.
D:\116105936.doc
Page 116 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
resonance and the formation of carbon-12
In a similar way, the carbon-12 nucleus has an excited resonant state at just the right
energy for it to be formed rapidly by the fusion of beryllium-8 and helium-4. We see nuclei
grow in steps of helium-4 nuclei, ascending an alpha-particle “staircase” in nucleon space.
Lawrence Krauss explains the cosmic significance of this: “Once carbon has been formed,
the gateway to creation of all the heavy elements that dominate our own existence on
Earth is opened”.
the energy yield of helium burning
The net result of the triple-alpha reaction is to combine three helium nuclei into one
carbon nucleus…
3 He-4(2p,2n)

(3x3727.4) = 11,182.2

C-12(6p,6n)
11,174.9
mass loss = 7.3 MeV
So the rearrangement of 12 nucleons from three nuclei into one yields 7.3 MeV energy, or
about 0.6 MeV per nucleon. This is a lot less than the 6.7 MeV per nucleon produced by
hydrogen burning.
a trio of cosmic coincidences
Carbon-12's precise excited energy state is a remarkable coincidence, but there's more.
The newly made carbon-12 nuclei can combine with helium-4 to make oxygen-16. If this
reaction is too easy then all the carbon-12 will be transformed to oxygen-16, with
consequences for subsequent carbon-based life forms. If this reaction is too slow, then
there will be insufficient oxygen produced for these life forms to breathe. So, “A delicate
balancing trick is required if neither oxygen nor carbon is to be over-abundant in the
universe, at the expense of the other. ... Life is possible because nature has fine-tuned
the properties of three atomic nuclei. Beryllium-8 is unusually long-lived for an unstable
nucleus. Carbon-12 possesses the exact energy state needed to promote its production.
And oxygen-16 lacks an energy state that would promote its production at the expense of
carbon”.
We've thus seen two nuclear fusion processes: first, hydrogen burning, which is highly
improbable, and second, helium burning, which is contingent on a set of coincidences.
And now you and I, oxygen-breathing carbon-based life forms, are here to consider them.
after helium burning
D:\116105936.doc
Page 117 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
The star has now burned its first two nuclear fuels, and is developing a concentric layered
structure, with a core of carbon-12 and oxygen-16, surrounded by a layer of burning
helium, which itself is surrouded by burning hydrogen. The sequence of burning reactions
is spreading outwards through the star, so each layer comprises the ashes of the previous
reaction, that is now proceeding in the next layer out.
We’ve seen two of the nuclear fusion reactions; it’s now time to look at the other
processes that build up nuclei.
3.12.8
Filling in the gaps - capturing protons and neutrons
We have seen earlier the major role played by helium-4 nuclei in the alpha-decay of large
unstable nuclei, and we now see its rôle in the fusion reactions that assemble larger
nuclei in stars. The small size and great stability of the helium-4 nucleus, makes it a
favoured unit of currency in nuclear transactions, both fission and fusion. But how do the
stars create the intermediate nuclei, ones that are not multiples of (2p,2n)? And also, how
are the large nuclei, way beyond the stable maximum of lead, created? These are the
work of the nucleon capture processes.
The p-process – making proton-rich nuclides
We know that stars are rich in protons, so they will be constantly colliding with any nuclei
that are formed. If the energy of collision is enough to overcome their mutual repulsion,
then the two can fuse – the nucleus “captures” the proton. The first step in the CNO cycle
is an example of proton capture, which moves the nucleus one step upwards in nucleon
space (see back to figure xXx).
carbon-12 (6p,6n) + p  nitrogen-13 (7p,6n)
However, this nucleus is unstable, and decays by the weak interaction…
nitrogen-13 (7p,6n)  carbon-13 (6p,7n) + e+ + 
and another form of carbon has been made.
In principle, proton capture can produce the proton-rich nuclei, that lie above the arc of
stable nuclei in nucleon space. Unstable nuclides will undergo beta-plus decay and move
back towards the stable arc, as we saw in the CNO cycle. Because nuclei and protons
repel each other, direct proton capture plays a minor rôle in building up nuclei, and is
associated with light nuclei (small charges) and high temperatures (high collision
energies).
D:\116105936.doc
Page 118 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Proton-rich nuclides can be produced by the removal of one or more neutrons from a
nucleus, following a collision with a high energy photon or neutrino, and this is the way
that large proton-rich nuclides are created.
Neutron capture processes
it all depends on the rate of neutron arrival
In contrast with protons, a stable nucleus can easily capture an uncharged neutron, and
this shifts its proton:neutron balance, moving it one step up the neutron-rich slope of the
nuclear valley. In time, this nucleus will undergo beta-minus decay (np) and move
diagonally back down the slope towards the arc of stability (see back to section zZz).
However, if the nucleus is hit by a second neutron before it has time to decay, then it is
driven another step away from the arc of stability, and further up the nuclear valley slope.
If the nucleus is subjected to a series of neutron collisions, then there is a competition
between the rate of neutron capture and the rate of nucleus decay. If the rate of neutron
arrival is slow, with a long time between captures, then the nucleus will be able to decay
back to the arc of stability before the next neutron arrives. If the rate of neutron arrival is
fast, then the nucleus will be driven far up the slope of the nuclear valley, and be unable
to return to the stable arc at the valley bottom. We see both neutron capture processes at
work in stars, the slow “s-process” and the rapid “r-process”, and they build up quite
different sequences of nuclei - figure 3.57.
D:\116105936.doc
Page 119 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the unstable r-process
nuclei decay back to
the stable arc
the rapid r-process:
very short delays
between neutron
captures, so the nuclei
cannot undergo betaminus (np) decay,
and are driven up the
nuclear valley slope
arc of stable
nuclides
beta-plus (pn)
decays are part of
the s process
the slow s-process: the long delays
between neutron captures allow the
nucleus time to undergo beta-minus
(np) decay back to the line of stability
slow s-process
beta-plus (pn) decay
the s- and rprocesses are
shown here
starting from the
same iron-56
(26p,30n)
nuclide
rapid r-process
Figure 3.57: The s- and r-processes, starting at iron-56, in the nuclear valley. The
path of the s-process meanders along the nuclear valley bottom, but the path of the
r-process runs up the neutron-rich slope, and even goes outside the valley.
The slow s-process
A star is a rich mix of nuclear reactions, some of which produce free neutrons. A major
source of neutrons is the carbon-13 mentioned above, which can step out of the endless
CNO cycle (figure 3.55) and undergo this reaction…
carbon-13(6p,7n) + helium-4(2p,2n)  oxygen-16(8p,8n) + n
Neutrons from reactions like this hit and are captured by other nuclei. It is a very slow
process – there may be hundreds or even thousands of years between successive
captures. This means that any unstable nucleus formed will very likely undergo betaminus decay (n  p) before the next neutron comes along. Notice that neutron capture
may take a nuclide into the beta-minus (np) decay zone, but it decays back to stability
before another neutron comes along. “The effect is that the nucleus works its way along
the floor of the energy valley towards heavier and heavier nuclei”. The s-process creates
D:\116105936.doc
Page 120 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
nuclei up to a size of about 90 nucleons, somewhat bigger than the most tightly bound
iron-group of nuclides, and well short than the largest effectively stable nucleus – bismuth
209.
the rapid r-process
If we imagine the s-process nuclides as produced by a dripping “neutron tap”, then in
contrast, the r-process nuclides are produced by a dam-burst of neutrons. Instead of a
wait of thousands of years between neutron captures, there may now be a thousand
captures per second. Each nucleus is hit by neutrons so frequently that it is driven high up
the slope of the nuclear valley, to become so unstable that its half life is comparable with
the time between successive neutron captures. The neutron deluge starts with an iron
‘seed’ nucleus, and produces a string of nuclei high up on the neutron-rich slope, moving
at great speed up the nuclear valley towards the largest nuclei at the end. The nuclides
high up the valley slopes have extremely short half lives, mere fractions of a second, and
so they can “move” through nucleon space very quickly.
Once the neutron deluge stops, each nuclide descends to the arc of stability at the valley
bottom, following a diagonal path set by beta-minus decay.
the three nucleon capture processes in nucleon-space
We can summarise the work of the three processes of nucleon capture on the nuclide
chart - figure 3.58.
D:\116105936.doc
Page 121 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the slow s-process creates nuclei
along the arc of stability, but only as
far as the biggest stable nucleus
the rapid proton
capture rp-process
creates light
proton-rich nuclei
50
the rapid r-process creates
neutron-rich nuclei up to
the heaviest known nuclei
82
126
magic neutron numbers
iron ‘seed’ nucleus, Fe-26p,30n), the start of the r-process
Figure 3.58: The three nucleon capture processes, shown in nucleon-space
The p-process creates many nuclides on the proton-rich side of the arc of stability. The sprocess creates many of the stable nuclei, as it meanders along the nuclear valley
bottom, taking perhaps thousands of years on each step, and going no further than the
largest stable nucleus. About half of the abundances of the elements between iron and
bismuth are produced by the s-process. The r-process creates, in a only few seconds, a
long arc of highly unstable, neutron-rich nuclides, all the way up to the largest known
nuclei. The arc is kinked where the nuclei have magic numbers of neutrons, conferring
additional stability. We will see later how the violent death of a large star provides the
explosive conditions that create the r-process nuclei.
a bigger context
We have looked at the how the nuclei at the two extremes of the nuclear valley are
created - the light nuclei by the burning of the first two fuels, and the heavy nuclei by the
explosive r-process. These processes mark the beginning and end of a star’s lifetime. We
are ready now to see these reactions in a bigger context, and look at the life cycles of
stars.
3.12.9
The life cycles of stars
stars of different masses
D:\116105936.doc
Page 122 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
The Hertzsprung-Russell (HR) diagram plots a star's total energy output (magnitude or
luminosity - with our sun set at 1.0) against its surface temperature (calculated from its
spectrum - hence the "spectral class" plotted on the x-axis) – figure 3.59.
mass: 30 Ms
Lifetime: 1 My
T ~25 MK
mass: 15 Ms
Lifetime: 10 MY
mass: 1 Ms
Lifetime: 10 BY
T ~12 MK
mass: 0.3 Ms
Lifetime: 800 BY
T ~10 MK
Key
"30Ms" = 30 solar masses
"800By" = 800 billion years
"60My" = 10 million years;
"25MK" = 25 million degrees Kelvin
Figure 3.59: The Hertzsprung-Russell (HR) diagram plots luminosity on the right
hand y-axis against surface temperature on the x-axis. Stars burning hydrogen lie
on the diagonal main sequence; massive stars are bright, hot and short lived, and
small stars are dim and cool and have long lives.
All the stars burning hydrogen lie on the main sequence line, running diagonally across
the diagram. At the lower left are the white dwarf stars - very hot, but small, so they don't
emit much radiation energy. In contrast, the red giant stars in the upper right are cooler,
but so big that they emit large amounts of energy.
Larger stars run hotter and have shorter lives
Small objects, of less than ~0.1 solar masses, do not ignite their hydrogen, and only glow
in the infra-red as they slowly relase their gravitational energy, thus being somewhere
between stars and giant planets. A star needs to start out with at least ~0.1 solar masses
to achieve a core temperature of 10 miilion degrees and start hydrogen burning, and such
a star would then slowly burn hydrogen for an estimated 800 billion years, longer than the
current age of the universe.
The greater a star’s mass, the faster it must burn its nuclear fuel in the core, to support
the greater weight of the surrounding layers, and so we see that more massive stars on
the main hydrogen-burning sequence have shorter lifetimes. Thus, a star with 30 solar
D:\116105936.doc
Page 123 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
masses consumes its hydrogen (protons) in only about 60 million years, and has 10,000
times the energy output of our own sun. So, “the more fuel a star starts off with, the
sooner it runs out. This is because the more massive the star is, the hotter it needs to be
to balance its gravitational attraction”. Meanwhile, “the lower-mass stars consume their
hydrogen fuel in a very frugal manner … [and] ... scrape along from year to year while
spending virtually nothing. In contrast, the most massive stars bear an uncomfortable
resemblance to rich and profligate heirs, who run shamelessly through a multimillion dollar
estate over the course of a single weekend”.
Our sun - the life cycle of a star of 1 solar mass
Figure 3.60 shows the life cycle of our sun - a modest star of 1 solar mass.
7: …so there is a
helium/carbon core
inside big planetary
nebula…
6: …swells to a red
giant…helium is now burning
in the core…the outer layers
are being blown away…
The end of the sun's life
5: The hydrogen in the core is
consumed, and now starts burning
in the outer layers…the sun leaves
the main sequence…
3: …H-burning starts,
and the new sun joins
the Main Sequence…
4: …where it steadily
burns hydrogen in its
core for about 10
billion years.
8: …which has now blown
away, exposing the core as a
white dwarf, with about half the
sun's original mass…
9: …which now slowly
dims and cools.
2: …collapses
under gravity
and heats up,
so that….
1: A large cool
gas cloud…
Figure 3.60: The stages in our sun's life cycle
Our sun is about half way through its 10 billion year phase of hydrogen burning, with a
core temperature of about 15 million degrees. We've seen that when the hydrogenburning reaction slows, the core heats up further, so hydrogen burning spreads outwards,
and heats up the outer layers of the star, which then swell. The surface area increases
faster than the energy output, so the star surface cools, and becomes redder, even
D:\116105936.doc
Page 124 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
though it is emitting more luminous energy. The star is moving through its red giant
phase. Our sun is expected to increase its luminosity 10,000 times and expand 200 times,
thus engulfing the Earth, though its surface will cool slightly to 4,000K. The core
temperature, however, continues to rise under gravitational contraction, until at ~120
million degrees helium ignites, burns for about 100 million years, and builds up carbon in
the sun’s core. With hydrogen continuing to burn steadily outwards through the star, the
outer layers are inflated and the star separates into a highly dense core of helium/carbon
surrounded by a swollen nebulous cloud of gas, rich in carbon and nitrogen, most of
which is blown away.
a white dwarf star
The core is not massive enough to raise its temperature high enough to ignite carbon, the
next fuel after helium. So, with no further nuclear reactions accessible, it shrinks and
becomes a white dwarf star, that just cools slowly for billions of years. A white dwarf star
is very hot, but, being small is not very luminous, so it is at the bottom left in the HR
diagram.
matter under pressure
Even a modest sized star, like our sun, has a mass of about 2 x 10 30 kg – two thousand
billion billion billion kilograms. The matter at the star’s centre must support this enormous
weight. How does matter behave under such extreme pressures?
We have seen earlier that in the recombination event, when the universe cooled to a
temperature of ~3,000 degrees, each nucleus gathered enough electrons in “orbit” around
itself, to make a neutral atom. But for atoms in the hot core of a star, the electrons have
too much energy to remain in settled orbits, so the “nuclei are immersed in a sea of free
electrons that tend to cluster near the nucleus”.
degenerate electrons and the Chandrasekhar mass
The matter in the core of a modest sized star is supported by this sea of free electrons,
which exert an ’electron degeneracy pressure’, which is itself a consequence of the
uncertainty principle. The uncertainty in the positions of the electrons is equivalent to tiny
motions, and these exert a pressure. In a white dwarf star, “this pressure is exerted by
electrons that are crowded together as the collapse of the star squeezes atoms together
until they overlap”. We can think of the electrons' matter-waves starting to overlap each
other, and because the electrons are fermions, they can't occupy the same volume, they
D:\116105936.doc
Page 125 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
must always be separate. So “the material of the star stiffens and, even without the
benefit of additional energy, supports its own weight“.
This degenerate electron pressure brings the gravitational contraction to an end in a
moderate sized star, such as a white dwarf. However, the greatest mass that degenerate
electrons can support is about 1.4 solar masses, known as the Chandrasekhar mass, in
which matter is compressed to a density of about 1000 kg/cm3, that is, 1 tonne/cm3, or
about the mass of a small car in your fingertip.
degenerate neutrons and neutron stars
In stars up to ~1.4 solar masses the electrons are viable independent particles, and can
sustain the pressure at the core. However, in a more massive star, the pressure forces
the particles so close together that the protons in the nuclei are able to capture the free
electrons, and form neutrons by the reaction…
p+ + e -  n0 + 
with the excess energy being carried away by the newly created neutrinos.
Thus, the star’s structure of separate nucleons and electrons is destroyed, and “all nuclei
decay and only a ‘puree’ of neutrons is left”. The star’s core becomes, in effect, a single
gigantic nucleus, composed only of degenerate neutrons - a neutron star. A typical
neutron star has the mass of two suns crammed into a sphere 20 km in diameter. The
density is now hard to imagine – approaching 1 billion tonnes of matter in each cubic
centimetre. We can think perhaps of a moderate sized mountain compressed into your
fingertip, seemingly an impossible condition, but this is the density of the nucleus of every
atom our world is made of. Nucleons are independent entities with a viable physical
presence at these extreme pressures. Yet there is a limit to what even nuclear matter can
withstand, and the greatest mass that degenerate neutrons can support is ~2-3 solar
masses.
black holes
Beyond this mass, physical matter undergoes total collapse into a black hole, from which
no light can escape. The theory of General relativity predicts that space-time is distorted
by gravity. Thus in a gravity field clocks tick and crystals vibrate more slowly. This is a tiny
effect in the Earth’s gravity field, but big enough that the highly accurate clocks on the
satellites used in the GPS navigation system need regular correction. The enormously
strong gravity field near a black hole slows down time so much that radiation from it is so
D:\116105936.doc
Page 126 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
redshifted that it disappears from view.Thus a stellar core that collapses to a blck hole has
reached the "end of time", and is "cut off forever from the rest of the universe", for not
even light can escape.
nuclear fuel thresholds
we can thus see a series of thresholds for nuclear fuel burning and stellar fates – figure
3.61.
core pressure ~1 ton/cm3
the maximum for
degenerate electrons
core temperature (millions of degrees K)
ends up <1.4 Ms
white dwarf
core pressure ~1billion tons/cm3
the maximum for degenerate
neutrons
ends up 1.4 to-~3 Ms
neutron star
ends up >~3 Ms
black hole
10000
silicon burns: 3,300 MK
oxygen burns: 2,000 MK
neon burns: 1,500 MK
1000
carbon burns: 700MK
helium burns: 120MK
100
oxygen burns: 2,000 MK
hydrogen burns: 15MK
10
hydrogen burns: 15MK
carbon burns: 700MK
helium burns: 120MK
1
0
5
10
15
20
25
initial star mass (solar masses)
Figure 3.61: A summary of stellar nuclear fuels and fates (Ms stands for solar mass)
This graph outlines the story so far, and also shows the way ahead. The more mass a star
starts out with the more nuclear fuels it can burn. A star with more than ~0.1 solar masses
can ignite hydrogen, and with more than ~0.25 solar masses it can burn helium to carbon.
The graph shows that stars with more than ~4 solar masses can then ignite carbon, with
more than ~8 solar masses they can ignite oxygen, and with more than ~15 solar masses
they can make it all the way to the last fuel, silicon.
A star’s outer layers are blown away in the process of burning its nuclear fuel, especially
helium, and so it ends up as a much smaller remnant. A star starting out with ~8 solar
D:\116105936.doc
Page 127 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
masses will lose so much material that a core of ~1.4 solar masses is all that remains.
This is the most that can be supported by degenerate electrons, so stars up to ~8 solar
masses end their lives as white dwarf stars, composed mainly of carbon and oxygen.
Stars with initially between ~8 and 20 solar masses end up as neutron stars, and beyond
this they end up as black holes.
going all the way
It's time now to look at a massive star, in which the nuclear reactions will consume all the
available fuels, and end up with a neutron star core. We will see how the enormous
temperatures and pressures developed by a massive star bring about its inevitable
destruction, but also enrich the universe.
3.12.10
The alpha-process - burning all the way to iron
We will follow the life of a large star, starting out with about 15 solar masses, which has
burned helium-4 largely to carbon-12, with some oxygen-16, by the triple-alpha process.
Once helium-4 is depleted in the star's core, gravitational compaction pushes up the
pressure and temperature, and now heavier nuclei are built up by the alpha process. This
is a mix of nuclear reactions, broadly based on transactions involving helium-4 nuclei, at
steadily increasing temperatures - up to more than 3 billion degrees. These enormous
temperatures introduce two new features to nuclear fuel burning. The first is the entrance
of neutrinos into the stellar economy. We have seen that a star in the hydrogen burning
phase loses a small portion of its energy by the emission of neutrinos, in addition to light
radiation. With rising temperatures this becomes more significant, and with a core
temperature more than ~500 million degrees, neutrino losses come to dominate the star’s
‘energy budget’. Second, at the very high temperatures of the alpha-process the thermal
photons have enough energy to break up nuclei, and this ‘photo-dissociation’ becomes an
important feature of nuclear burning reactions. So in addition to the normal fusion
process, whereby two smaller nuclei merge into a larger one, we also see a progressive
rearrangement of nucleons, involving the capture of protons, neutrons and alpha particles.
The alpha process has four distinct stages, distinguished by their principal fuels - carbon,
neon, oxygen, and finally silicon. The first and third burn by simple fusion reactions; the
second and last involve the break up of nuclei by thermal photons.
carbon burning
D:\116105936.doc
Page 128 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
At a temperature of ~700 million degrees, the carbon-12 nuclei have enough energy to
overcome the mutual repulsion of their 6 protons, and they collide and undergo a set of
competing reactions, producing a mix of different nuclides. We’ll look at this stage in some
detail, because it illustrates the important features of the alpha process – figure 3.62.
Ne-20
+
(10p,10n)
He-4
(2p,2n)
O-16 + n
(8p,8n)
C-12
(6p,6n)
Na-23 +
(11p,12n)
energy-bearing
neutrinos escape
from the stellar core
p
C-13 + e+ + 
(6p,7n)
C-12
(6p,6n)
beta-plus (pn) decay
C-12
(6p,6n)
N-13
(7p,6n)
Mg-24
(12p,12n)
e+ + eO-16 + He-4 + He-4
(8p,8n) (2p,2n) (2p,2n)
carbon-burning core
700 million degrees
electronpositron pair
 + 
neutrino
antineutrino
pair
helium-burning shell
180 million degrees
hydrogen-burning shell
35 million degrees
Figure 3.62: The mix of nuclear reactions that constitute carbon burning. The main
reactions produce neon (Ne-20) and sodium (Na-23), but magnesium (Mg-24) and
oxygen (O-16) are also directly produced. Further reactions produce oxygen and
also neutrinos. The most energetic photons have enough energy to create electronpositron pairs, a few of which then create neutrino-antineutrino pairs.
The diagram shows a tumble of nuclear reactions, producing a variety of nuclei in the size
range 16-24. Some of the reactions involve single nucleons. Thus, a loose proton reacts
with a carbon-12 nucleus to give nitrogen-13, which is unstable and undergoes beta-plus
(pn) decay to carbon-13, which then reacts with a helium-4 nucleus to give oxygen-16.
The main nuclear products of carbon burning are: oxygen-16, neon-20, sodium-23, and
magnesium-24.
neutrinos enter the scene
D:\116105936.doc
Page 129 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
The star’s core has now regained the temperature of the universe when it was about 3
minutes old (see back to section 3.11.6). At 700 million degrees, the average photon
energy is about 60 keV, that is, about one eighth of the mass-energy of an electron or
positron. The fastest thermal photons have enough energy to sustain a substantial
population of electron-positron pairs, either directly, or by colliding with a nucleus in the
core. Furthermore, “when the electrons meet and annihilate with positrons, a neutrinoantineutrino pair is occasionally produced. These neutrinos escape the star with ease and
force the burning to go faster to replenish the loss”. Figure 3.62 shows that some of the
nuclear reactions themselves produce additional neutrinos. From the carbon-burning
stage onwards “the dominant energy loss from the star is due to neutrinos streaming out
directly from the stellar nuclear furnace, rather than by [light] photons from the surface”.
Even a main sequence star, burning hydrogen, loses a small fraction of its energy – about
6% - as neutrinos (see back to section 3.12.5). The carbon-burning star core loses ~84%
of its energy as neutrinos, leaving very little to oppose gravitational collapse, and forcing
the nuclear burning reactions to go even faster. Moreover, the nuclear reactions are
yielding less energy. The rearrangement of 24 nucleons from two carbon nuclei into one
neon and one helium nucleus…
C-12 (6p,6n) + C-12 (6p,6n)  Ne-20 (10p,20n) + He-4 (2p,2n)
mass loss 4.7 MeV
releases only 4.7 MeV, about 0.2 MeV per nucleon. So the star is losing energy at a faster
rate, and the nuclear fuel is yielding less energy. Temperatures and densities are starting
to stretch the imagination. The temperature is approaching one million degrees, and the
core density is ~240 kg/cm3 , around a quarter ton of mass in your little fingertip).
summary of carbon burning
Helium burning produced nuclei ~12-16 nucleons in size, and carbon burning has raised
this to a size range of ~16-24 nucleons. We are seeing a mix of fusion reactions, and also
the nuclei reshuffling themselves, mainly by transferring alpha particles, but also single
nucleons. The star is building up a multi-layer structure, with the latest reaction in the
core, surrounded by the previous reactions in successive outer layers.
neon burning and photo-dissociation
When carbon is depleted, the star compacts further, driving the core temperature to about
1,500 million (1.5 billion) degrees, and now the most energetic photons start breaking the
up the neon-20 nuclei, that were assembled in the carbon burning phase.
D:\116105936.doc
Page 130 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
neon-20 (10p,10n) +  photon  oxygen-16 (8p,8n) + helium-4 (2p,2n)
The loose helium then reacts with surviving neon…
helium-4 (2p,2n) + neon-20 (10p,10n)  magnesium-24 (12p,12n)
This photo-dissociation process - breaking up by light - has reshuffled the two neon
nuclei. We will see this process become increasingly significant as the star continues to
heat up. At the end of the neon-burning stage the star’s core contains mainly oxygen-16,
magnesium-24 and silicon-28.
oxygen burning
The core temperature now rises to ~2,000 million (2 billion) degrees, and compresses to a
density of ~7 tonnes/cm3, which is sufficient for oxygen nuclei to overcome their mutual
repulsion and react directly, with two main outcomes:
1)
oxygen-16 (8p,8n) + oxygen-16 (8p,8n)  silicon-28 (14p,14n) + helium-4 (2p,2n)
2)
or…
 sulphur-32 (16p,16n)
Oxygen burning also creates a large range of nuclides up to ~40 nucleons in size, with the
alpha-nuclei being favoured, due to their higher binding energies. Thus a number of
familiar elements appear: phosphorus, sulphur, chlorine, argon, potassium and calcium.
silicon burning
The core temperature now exceeds 3 billion degrees, and the density is about 40
tonnes/cm3, the mass of around 40 cars in your little fingertip. Now the thermal photons
have so much energy that the processes of photo-dissociation and fusion are finely
balanced. A simple but helpful view is of highly energetic photons breaking up Si-28
nuclei, and the fragments - protons, neutrons and alpha particles - then being
incorporated into larger nuclei. Nuclei in the size range 28-65 nucleons thus compete for
survival under the barrage of photons, and the larger nuclei are favoured because they
are more tightly bound. So, “nuclei with smaller binding energies are destroyed by photodissociation in favor of their more tightly bound neighbors, and many nuclear reactions
involving alpha-particles, protons, and neutrons interacting with all the nuclei … take
place”. This has been called "nuclear melting" to distinguish it from nuclear burning.
antarctic penguins
D:\116105936.doc
Page 131 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
At the South Pole, the male Emperor penguins huddle together for warmth, as they guard
their eggs through the antarctic winter. The bigger the huddle, the more penguins there
are in the warm interior. In a similar way, the nucleons in a stellar core gather in
progressively bigger and more tightly bound huddles, against the rising fury of the
photons.
growth v. fragmentation
The nuclear melting process thus broadly favours the tightly bound iron-group of nuclides,
at the peak of the binding energy curve, with 56-60 or so nucleons. Silicon burning
produces a clutch of nuclides in this range: isotopes of chromium, manganese, iron,
cobalt and nickel. However, in the enormously hot stellar core the processes of nuclear
fragmentation and growth are finely balanced, and the viability of a nuclear huddle
depends as much on its resistance to photo-dissociation as on its binding energy. Under
these circumstances, the two main products of silicon burning are Ni-56, the most tightly
bound of the series of alpha-nuclei, and “the natural end of the alpha process”, and Fe-56,
with the third highest binding energy of all the nuclides. The balance of products is
sensitive to the neutron population in the stellar core, so that a small neutron excess
favours the formation of the more Ni-56(28p,28n), and a large excess favours the neutronrich Fe-56(26p,30n), which has an excess of 4 neutrons.
a poor energy yield
The transformation of two silicon-28(14p,14n) nuclei into one nickel-56(28p,28n) nucleus
yields little energy…
Si-28 (14p,14n) + Si-28 (14p,14n)  Ni-56 (28p,28n)
mass loss = 10.9 MeV
This rearrangement of 56 nucleons yields 10.9 MeV, that is, only 0.2 MeV per nucleon.
Ni-56 is unstable, and undergoes beta-plus decay twice to Fe-56...
nickel-56 (28p,28n)  n + cobalt-56 (27p,29n)

n + iron-56 (26p,30n)
Thus Fe-56 is the favoured nuclide in the star’s core, either produced by silicon burning or
by the decay of Ni-56. Iron-56 has an excess of 4 neutrons, and we'll see these neutrons
play an important rôle later.
silicon is the last nuclear fuel
The iron-group of nuclides have the highest binding energies of all the nuclides, and are
at the peak of the binding energy curve. We’ve seen how the nuclear burning reactions
D:\116105936.doc
Page 132 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
fuse smaller nuclei into bigger reduce mass and hence release binding energy to support
the star’s weight. However, this only works as far as the iron-group nuclides. Beyond
these, further enlargement increases mass, and takes energy from the star, rather than
donate it. So, with the “burning” of silicon, the star has used its last nuclear fuel.
The main steps in the alpha process are shown in nucleon space, in figure 3.63.
Ni-56 has the largest binding
energy of the alpha-nuclei,
and is favoured by a small
neutron excess
protons, p
zinc-Zn,30
copper-Cu,29
nickel-Ni,28
cobalt-Co,27
iron-Fe,26
manganese-Mn,25
chromium-Cr,24
vanadium-V,23
titanium-Ti,22
scandium-Sc,21
calcium-Ca,20
potassium-K,19
argon-A,18
chlorine-Cl,17
sulphur-S,16
phosphorus-P,15
silicon-Si,14
aluminium-Al,13
magnesium-Mg,12
sodium-Na,11
neon-Ne,10
fluorine-F,9
oxygen-O,8
nitrogen-N,7
carbon-C,6
boron-B,5
beryllium-Be,4
lithium-Li,3
helium-2,He
hydrogen-H,1
clusters containing…
56
60
nucleons
nucleons
3. …finally, silicon burning creates
nuclides in the tightly bound iron
group, mainly Ni-56 and Fe-56…
2. …oxygen burning
creates Si-28 and
other nuclides, up to
~40 nucleons…
1. C-12 burning
creates O-16
and nuclides up
to ~24
nucleons…
iron-56 has the third
largest binding energy
of all the nuclides, and
is favoured by a large
neutron excess
the line of alphanuclei, where
proton and neutron
numbers are equal
C-12 (6p,6n)
5
10
15
20
25
neutrons, n
Figure 3.63: The alpha process starts with carbon-12 and builds ever larger nuclei,
up to the iron group of nuclides
the alpha-nucleus staircase
The dotted line in the diagram marks the series of alpha-nuclei, that have equal numbers
of protons and neutrons. We can see the alpha process as a kind of stair-case, working in
regular steps diagonally upwards in nucleon space. The tightly bound helium-4 nucleus is
the preferred currency of "trade", so the alpha-nuclei are favoured, and we'll see later that
they are noticeably more abundant in the universe than the ones in between. The major
D:\116105936.doc
Page 133 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
fuels in the alpha process - helium-4 (2p,2n), carbon-12 (6p,6n), oxygen-16 (8p,8n), neon20 (10p,10n) and silicon-28 (14p,14n) – are all even-even nuclei, and two of them are also
doubly magic. This means they all have larger binding energies than nuclei in between,
and so are the major fuels in the alpha-process.
a roll call of familiar elements
The proton axis in figure 3.63 is a roll call of elements, many of them familiar - the fluoride
in toothpaste, the sodium in salt, the aluminium in a saucepan, the silicon in beach sand
and computer "chips", the chlorine in bleach and swimming pools, the calcium in our teeth
and bones, the chromium in shiny electroplate, the iron in a car body and in our red blood
cells, the cobalt in the magnets in our earphones and in vitamin B12. The foundations of
the substances of our material world and of ourselves are emerging from the stellar
inferno.
an onion-like structure
The star has now developed a structure of concentric layers, like an onion, with a different
nuclear fuel burning in each layer (figure 3.64).
15
300
million
stellar surface
hydrogen burning to helium-4, 35 million K
(also nitrogen)
12
9
6
5,000
3
1,000
0
Included
mass
(solar
masses)
0
distance
from
centre
(km)
helium-4 burning to carbon-12, 200 million K
(also oxygen, neon)
carbon-12 burning to oxygen-16, 800 million K
(also neon, sodium, magnesium)
oxygen-16 burning through to silicon-28, 2
billion K
(also phosphorus, sulphur, chlorine, argon,
potassium, calcium)
silicon-28 burning through to nickel-56, 3.3
billion K
(also titanium, chromium, manganese, iron,
cobalt)
iron-56 and nickel-56 core, 7 billion K
centre of the star
Figure 3.64: The interior of a star of 15 solar masses at the end of silicon burning.
The blue scale shows how much of the star’s mass is inside that point – for
D:\116105936.doc
Page 134 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
example, there are about 3 solar masses of matter inside the helium burning shell;
the yellow scale gives the approximate distance from the star’s centre.
This cross-section gives us the star’s nuclear burning history. The main fuels are laid out
in the sequence of their ignition, with subsidiary reactions building up the nuclei that are
not in the alpha sequence. At the base of each layer a burning front is moving outwards,
igniting the nuclear ash produced by the layer above. The silicon burning reaction has
accumulated around 1-2 solar masses of nickel and iron at the star’s core, and continues
to add to this. The nuclear fuels are burning in a dense small core about the size of the
earth, at the centre of a huge, bloated star, that may be as big as the orbit of Jupiter.
3.12.11
running out of nuclear fuel
climbing the binding energy curve
The star has in its lifetime gone six times around the gravity-matter circuit, and burned the
major nuclear fuels: hydrogen, helium, carbon, neon, oxygen, and finally silicon. Each
burning rearranges the nucleons into bigger nuclear configurations, which are more tightly
bound. The binding energy released in each stage of burning has sustained the star, and
temporarily halted its gravitational collapse. We can follow the sequence on the nuclear
Binding Energy/nucleon (MeV)
binding energy curve – figure 3.65.
9
8
burning oxgen  silicon
30 months at 2,000 MK
Energy: ~0.3 MeV/nucleon
7
6
burning carbon  neon
2,000 years at 700 MK
Energy: ~0.2 MeV/nucleon
5
4
burning silicon  nickel/iron
18 days at 3,300 MK
Energy: ~0.2 MeV/nucleon
burning helium  carbon
2 million years at 180 MK
Energy: ~0.6 MeV/nucleon
3
2
burning hydrogen (4p)  helium (He-4) + 27 MeV
11 million years at 35 million degrees (MK)
Energy: ~7 MeV/nucleon
1
0
0
10
20
30
40
50
nucleons
60
Figure 3.65: The sequence of fusion reactions generating heat in a star of 15 solar
masses.
D:\116105936.doc
Page 135 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
As long as the fusion reaction moves the arrangement of nucleons further up the binding
energy curve, mass will be lost, energy will be released, and the star’s collapse is
deferred for a little longer. Hydrogen burning, the first reaction, that binds the free protons,
yields a huge amount of energy, ~27 MeV for binding 4 nucleons, about 7 MeV per
nucleon. After helium, the curve is much less steep, and the burning reactions have to
rearrange more and more nucleons to get even a small energy yield.
Moreover, the star is losing energy at an ever faster rate, through neutrino emission, and
so each successive nuclear fuel lasts a fraction of the time of the previous one. The
figures in the graph show that from hydrogen to oxygen, each fuel lasts only about one
tenth as long as the preceding fuel.
In the star's onion-like structure we can see the retreat of matter under the inexorable
pressure of the star’s gravity. Successive reactions throw new fuels on the nuclear fire,
but these yield less and less energy. Finally, the nuclear rearrangements of the alpha
process produce nuclei in the iron group, with around 56-60 nucleons, and having the
largest binding energies of all the nuclides. Enlarging these will take energy in rather than
give it out. There is no viable way that a nuclide like iron-56 can be made to yield further
binding energy; the star’s core has finally exhausted its nuclear fuel.
the “death” of a star
The gas cloud that started maybe 10 million years before as the merest breath of slightly
cooler hydrogen and helium in almost empty space, has become a turmoil of heavy nuclei
and high energy photons of radiation. The nuclear reactions that for so long have powered
this star and kept its enormous gravity at bay, have now run their course. Millions of years
have homed in on a single day, and on the last seconds at the end of the last day.
Lawrence Krauss captures this moment well for a star originally of 30 solar masses: “For
10 million years all of the nuclear reactions holding the star up against gravitational
collapse have been leading to this single last gasp. Almost 10 million years of hydrogen
burning, followed by 1 million years of helium burning, 100,000 years of carbon, 10,000
years of oxygen, and then a single day for the rest of the trip. Once it is over there is no
hope. In fact, the dense inner core of the star, now surrounded like an onion by shells of
oxygen, carbon, helium and hydrogen, is about to undergo one of the most traumatic
events in all the visible universe”.
a technical note - why not nickel-62?
D:\116105936.doc
Page 136 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
I don’t feel I‘ve understood why the main products of silicon burning are Ni-56 (~60th in the
binding energy rankings) and Fe-56 (3rd highest binding energy). If binding energy is so
important in the competitive melée of the super-hot core, then why not Ni-62, Fe-58, and
then Fe-56, which are ranked one, two and three?
Fewell suggests that the Ni-62 nuclide is preferentially photo-dissociated. And certainly,
resistance to break-up by photons would be as important as binding energy, in deciding
which nuclides are produced by silicon burning.
Wallerstein and WHW suggest that the proton/neutron ratio plays a rôle in deciding which
nuclides are made. WHW describe how the main Si-burning products are those nuclides
that are most tightly bound at the value of neutron excess pertaining in the core. Both
sources say that a small neutron excess in the core leads to the preferential formation of
Ni-56, with its proton/neutron ratio (Z/N) of 1.0. If the neutron excess increases, then the
preferred products will be the neutron-rich isotopes of iron, Fe-54, Fe-56 or Fe-58. WHW
state that for still greater neutron/proton ratios “the equilibrium shifts to heavier isotopes”,
and note that the most tightly bound nucleus is Ni-62.
Wallerstein (fig. 24) relates binding energies to the proton/neutron ratio: “Fe-56 could be
made as itself in equilibrium if the ratio of neutrons to protons in the nucleosynthetic
environment were around 0.87. … In fact, nature seems to have chosen to assemble
most of the solar system’s iron-group nuclei in matter that had equal numbers of neutrons
and protons. In this case Ni-56 was made and later decayed to Fe-56.” (fig. 24 caption)
The figure below plots binding energies against cluster size for three sets of nuclides, with
different proton/neutron (Z/N) ratios.
The bottom graph shows that Ni-56 is the most strongly bound of the alpha-nuclei, that
have equal number of protons and neutrons (Z=N, so Z/N=1.0). The second (blue) plot
shows how a slight neutron enrichment has increased binding energies, and that Fe-56 is
the most tightly bound nuclide of this group. Further neutron enrichment (the green plot)
has not really increased binding energies, and now it’s Fe-58 and Ni-62 that are the
tightest bound.
D:\116105936.doc
Page 137 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Z/N = 1.0
Z/N = 0.85 to 0.90
Fe-56(26p,30n)
Z/N = 0.87
8800
Z/N = 0.80 to 0.85
Fe-58(26p,32n)
Z/N = 0.81
Ni-62(28p,34n)
Z/N = 0.82
BE per nucleon (keV)
8750
8700
Ni-56(28p,28n)
Z/N = 1.0
8650
8600
8550
8500
52
54
56
58
60
62
64
66
nucleons
Perhaps the more neutron-rich nuclides, even with their large binding energies, simply
cannot be made unless the stellar environment is sufficiently neutron-rich to start with. So
silicon burning in a core with very few excess neutrons produces mainly Ni-56, with equal
numbers of protons and neutrons. Increasing the neutron excess enables the production
of Fe-56 and the even more neutron-rich Fe-58. Ni-62 is then competing with these two
nuclides that have almost the same binding energies, but are smaller, and hence easier to
form by aggregation.
So, maybe this is why Ni-62, the most tightly bound nuclide of all, is not produced in
quantity in the star’s core: it is more susceptible to photo-dissociation, there needs to be a
bigger neutron excess in the core, and if there are the excess neutrons available to make
it, then it is in competition with two other nuclides, Fe-56 and Fe-58, that are equally
tightly bound but significantly smaller.
3.12.12
Supernova
A star’s continued existence as a self-supporting entity depends on a continuing nuclear
burning reaction, and the ability of the matter in its core to withstand the conditions there.
A star that is massive enough to make it to the silicon burning stage imposes
temperatures and pressures that atomic matter, comprising nucleons and electrons, can
D:\116105936.doc
Page 138 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
only just withstand. If the core gives way, then the entire stellar edifice collapses. The core
and the outer layers of the star now go very different ways. We'll first tell the core's tale.
The Core's Tale
The life cycle of the core of a massive star “can be thought of as just one long contraction,
beginning with the star’s birth, burning hydrogen on the main sequence, and ending with
the formation of a neutron star or black hole. Along the way, the contraction ‘pauses’,
sometimes for millions of years, as nuclear fusion provides the energy necessary to
replenish what the star is losing to radiation and neutrinos.” Eventually a nickel-iron core
of about 1.5 solar masses builds up, in effect, a white dwarf inside the massive star. The
temperature is so high that even the tightly bound iron-group nuclei are on the edge of
being fragmented by the high energy photons, and the pressure is so high that the
degenerate electrons can only just support it.
The iron-group nuclides are at the top of the binding energy curve, and so no further
energy can be released by nuclear fusion reactions. The star has exhausted its nuclear
fuel at the core, yet continues to lose energy at an enormous rate through the emission of
neutrinos. Consequently, the core contracts, and its temperature and pressure increase
even further, and the matter in the core finally gives way. Two processes then occur that
remove energy from the core, speeding its collapse.
the unravelling of iron-56
The core temperature rises as high as 7 billion degrees, and the thermal photons now
break up the iron-56 nuclei back to alpha particles from which they were formed…
Fe-56 (26p,30n)  13 He-4 (2p,2n) + 4 n
mass gain 124 MeV
This rearranges the nucleons into less tightly bound nuclei, with a mass gain, which is
‘paid for’ by taking energy from the star’s core, so it collapses faster.
electron capture and ‘neutronisation’
As the increasing core pressure pushes the density past 10,000 tonnes/cm 3, the electrons
are “squeezed into iron-group nuclei”, initiating the ‘neutronisation’ reaction, in which an
electron and a proton combine to make a neutron, and emit a neutrino.…
e- + p +  n0 + 
D:\116105936.doc
- ~0.8 MeV
Page 139 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Electrons and protons effectively disappear, leaving only neutrons. This removes all the
electrons that have so far supported the star's weight, and also consumes energy to
create a burst of neutrinos, which leave the core.
doomed either way
Both these processes are inevitable, for failure in one respect triggers failure in the other.
These two events, the unravelling of iron-56 nuclei and the neutronisation reaction,
remove a huge amount of energy from the star, maybe as much energy as it has radiated
in its entire lifetime. The star’s core collapses so fast that it falls inward at about one
quarter of the speed of light, shrinking from about the size of the Earth (~12,000 km
diameter) to a sphere about 60 km in diameter, in around 1 second.
a neutron star
What happens to the core now depends on its mass. If it is more than ~3 solar masses it
collapses completely to a black hole, the "ultimate triumph of gravity over matter”. But for
cores less than ~3 solar masses, the outcome is a sphere of neutrons, a neutron star;
with a density of around 200 million tonnes/cm 3 – “a mass in excess of a million Earths
confined to a region the size of a small city, and with a mass of Manhattan contained in
each cubic centimetre of material”.
Neutron stars are “at the limit of density that matter can have, the subsequent step being
a black hole”. This state of matter would seem beyond comprehension, yet a neutron star
resembles a giant atomic nucleus, and it has about the same density as the tiny nuclear
nuggets that are at the heart of every atom, from hydrogen to lead. But, whereas an
atomic nucleus, with its specific balance of protons and neutrons, has an identity, a
neutron star, has none. During the star’s lifetime, the protons and neutrons in its core
were continually being rearranged in ever-bigger nuclei as star burned its series of fuels.
Now that rich diversity of nuclei has been unravelled by the temperature, and crushed to
neutrons by the pressure. “Matter this compact loses its atomic identity. Electrons blend
with protons; intervening space effectively disappears.” A neutron star represents matter
in its most lumpen condition, faceless and nameless, imprisoned by its own gravity.
As far as the stellar core is concerned the star's patient "journey-work", building large
nuclei over millions of years, has been undone in an instant. It is “as if the real business of
the star, the conjuring of nuclei, was now monumentalised as a giant nuclear tombstone”.
But the core is only a fraction of the entire star. Its collapse has produced a huge amount
D:\116105936.doc
Page 140 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
of energy, creating enormous temperatures and numbers of neutrinos and neutrons, and
these now erupt into the star’s outer layers, and the creation of nuclei continues there.
The outer layers’ tale
It is not easy to convey the scale of a supernova explosion, for it is not just the sheer
amount of energy, but also the speed of its release.
a gravity-powered neutrino explosion
A nickel/iron core of ~1.4 solar masses, collapsing to a neutron star, converts about 10%
of its mass to energy, releasing about 1046 J in a few seconds. The supernova’s light
output can increase by a factor of 100 billion, so that for a few days it can outshine the
100 billion or so stars in its galaxy.
But this visible light is only 1% or so of the supernova’s total energy output. The other
99% of the energy is in the form of neutrinos, one for each proton in the star’s core, about
1058 in total, streaming out from the core at almost the speed of light.
We have already encountered neutrinos, and seen that they “are the closest thing to
nothing that one can imagine. …they have … virtually no effect whatsoever on ordinary
matter… under ordinary conditions, a neutrino could penetrate millions of miles of lead as
if it were window glass.” About 500 billion neutrinos from the sun hit every square inch of
ground every second, and pass through the Earth with no resistance. "Our bodies are
pierced by them unceasingly, day and night. They leave not a trace."
But now deluge of neutrinos produced by the supernova is enormous, about 10 58 in a few
seconds, and the inner layers of the star are so compressed that the densely packed
nuclei deflect the streams of neutrinos. As far as the neutrinos are concerned, “the star
offers ony a slight resistance – just a bit more than if nothing was there at all”. They
expend only about 1% of their total energy in penetrating the star’s outer layers, but even
this amounts to about 1044 J, more energy than the Sun emits in its entire lifetime. This
rips protons and neutrons from the star’s neutronising core, and creates a shock wave
that blows away the star’s outer layers in the supernova explosion.
The numbers on their own perhaps convey little, but pause to consider that only 1% of the
total energy released in the supernova explosion is enough to blow away the star’s outer
layers, at least several solar masses, at speeds of thousands of kilometres per second. A
supernova has been summarised as a “gravity-powered neutrino explosion”. In fact,
without neutrinos, the core mass would just keep increasing from layers of star falling on it
D:\116105936.doc
Page 141 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
until it collapsed as a black hole, and no supernovae would occur. It’s a nice irony that
gravity is by far the weakest of all the physical interactions, and neutrinos are by far the
least substantial of all matter particles.
two sets of reactions
The nickel/iron core with a radius of ~1,000 km, has collapsed to a neutron star with a
radius of ~10 km or so, and a temperature of ~100 billion degrees. The photo-dissociation
of iron has produced free neutrons, and the neutronisation reaction has released a flood
of neutrinos, which rip protons and neutrons free from the neutron star’s surface. These
neutrinos drive a hot neutron-rich ‘wind’ lasting about 10 seconds, through the inner layers
of the star, inducing two main types of reactions there. The reactions in the innermost
layers rapidly assemble nuclei from free protons and neutrons, while a little further out, the
nuclear fuels that have remain unconsumed undergo explosive nuclear burning. Figure
3.66 shows the main nuclear reactions in these layers.
D:\116105936.doc
Page 142 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Temperature, in billions of degrees
1. A ‘wind’ of neutrinos blows for ~10 s, carrying
free protons and neutrons out into the star.
Distance from centre, km
100,000
the turbulent neutrino wind
stirs up the material in the star
2
5
13,000
10,000
5,000
3,700
1,000
5. …Fe-group nuclei capture n  rprocess nuclides (T~0.1bK, t~2 s).
4. …alphas combine  Fe-group
nuclei + free n (T~6bK, t~0.55 s)…
10
100
3. …p + n combine  alpha particles +
free neutrons (T~10bK, t~0.5 s)…
Explosive burning, in a few
seconds, of super-heated
Si, O and C.
For T>~5 billion degrees,
complete burning to irongroup nuclides.
For T between ~2 and 5
billion degrees, incomplete
burning to a diverse mix of
nuclides.
2. Neutrinos rip free protons and
neutrons from the neutron star (t=0)…
10
100
neutron
star
1
~1.4
~3
included mass (solar masses)
Figure 3.66: The nuclear reactions in the outer layers of a star in a supernova
explosion; bK stands for billion degrees.
rapid nucleus assembly – the r-process
The neutrino wind leaves the neutron star with a temperature of ~50 billion degrees, then
it cools, and at ~10 billion degrees, and about half a second after the start of the neutrino
wind, the free protons and neutrons it carries have started assembling into alpha particles.
Very shortly after this, as the temperature drops further, these alpha particles themselves
combine into larger nuclei, especially those in the iron-group, that have the greatest
binding energies.
The wind now carries iron-group nuclei, alpha particles and neutrons. In the next couple of
seconds, as the temperature falls below 1 billion degrees, the iron-group nuclei rapidly
capture neutrons, to build up the r-process nuclei. There can be as many as 100 neutrons
for each iron-group nucleus, and the neutron flux is so great that “the time to capture a
D:\116105936.doc
Page 143 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
neutron is generally less than one-thousandth of a second”. This drives the nuclei high up
the slope on the neutron-rich side of the nuclear valley, until they can hold no more
neutrons. When the neutron flux ends, these unstable nuclei undergo a series of betaminus decays, maybe as many as twenty, taking them down to the nuclear valley floor
(see back to the section on the r-process). It is this r-process that produces, in a couple of
seconds, an array of nuclides heavier than zinc, containing from about 70-250 nucleons.
The amount produced is small, only about one millionth of a solar mass, but this is about
one third of the mass of the Earth, and includes some rare nuclides.
explosive burning
All nuclear fuels will burn very fast if they are heated well above their normal temperature
of burning in a stellar core. A fuel that will burn steadily for millions of years, will undergo
explosive burning in seconds if super-heated. This explosive burning can occur at
temperatures above 2 billion degrees – about 13,000 km out from the neutron star - and
will go to completion if the temperature exceeds 5 billion degrees – about 4,000 km out
(see figure xXx). Much of the shells that have been steadily burning carbon, oxygen and
silicon lie within 13,000 km, and so these fuels will undergo explosive burning - partial in
some regions, complete in others. Any nuclei within ~4,000 km of the neutron star will in a
few seconds be processed all the way to iron-group nuclei.
Explosive burning yields the same nuclides as stable core burning, but the presence of
free alpha particles and protons and neutrons results in a rich diversity of nuclides, from
~20 up to ~90 nucleons, well beyond the iron-group, and rich in protons and neutrons.
a few nuclides are formed by fragmentation
A tiny fraction of the torrent of neutrinos knock single protons or neutrons out of nuclei on
their way through the star’s outer layers. In this way boron-11 is created from carbon-12…
 + C-12 (6p,6n)  B-11 (5p,6n) + p
and similarly, neon-20 (10p,10n) loses a proton to make fluorine-19 (9p,10n).
High energy photons can eject neutrons, to produce proton-rich isotopes of many
elements, such as mercury, tungsten, barium and xenon.
spallation in deep space - the last nucleosynthesis mechanism
We now look at the last stellar mechanism for creating nuclides, and this occurs, not
inside stars, but outside them, and operates, not by fusion, but by fragmentation. The
D:\116105936.doc
Page 144 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
supernova explosion blows the nuclei in the star’s outer layers into space at enormous
speeds, up to near the speed of light. These high speed nuclei, known as cosmic rays,
are predominantly single hydrogen protons and helium nuclei, and travel through ‘empty
space’ for very long times, typically 10 million years.
However, the space between the stars, the interstellar medium, is by no means empty,
but contains a scattering of particles of different kinds, typically around a few tens to a few
hundreds in a cubic centimetre. This is better than any vacuum that can be made on
Earth; compare it with ‘thin air’, which has nearly 30 billion billion gas particles in a cubic
centimetre. The particles are primarily atoms of hydrogen and helium, from the brief burst
of nucleus formation after the big bang. There is also a small proportion of larger nuclei,
produced by earlier stars and supernovae. Finally, there are a large number of molecules
of compounds, such as water, formaldehyde and ethanol, in great clouds between the
stars.
The high speed cosmic ray particles, on their long journey through space, collide with
larger nuclei, and break them into smaller nuclear fragments; this is the spallation
process. The fragmentation of carbon and oxygen nuclei is the source of small nuclides,
such as Li-6 (3p,3n), Be-9 (4p,5n) and B-10 (5p,5n). None of these nuclides can be
formed in stars; they are all destroyed before hydrogen burning starts.
Thus the massive star’s last act, as it explodes as a supernova, is the creation of rare light
nuclides by spallation collisions, that will occur far away and long after the star has
ceased to exist.
Supernova SN1987A
The first outward sign of a supernova is the burst of neutrinos, followed by the blast wave,
and the rapidly expanding shell of gas - within a day this is an incandescent ball a billion
miles across.
The neutrinos from the explosion of supernova SN1987A, a star of ~15 solar masses in
the large Magellanic cloud about 160,000 light years away, were detected on Earth at
07.35 GMT on 23 February 1987, followed some hours later by the visible light. This was
the first supernova observed in that year, hence its designation, and also the first to be
visible to the naked eye for nearly 400 years - the previous one being in 1667.
Two neutrino detectors picked up a total of 19 neutrinos passing through Earth in about
12 seconds. The probabilities of detecting these extremely elusive particles is minute, so
D:\116105936.doc
Page 145 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
these few captures tell us that a total of about one hundred thousand trillion neutrinos
(~1017) passsed through the two detectors. From the captured numbers and energies we
can calculate that the supernova emitted about 1058 neutrinos, at a temperature of over 10
billion degrees, carrying a total energy of around 10 46 Joules, “more energy than an entire
spiral galaxy gives off in a year’s time. … The neutrinos give us a glimpse into the interior
of a hellish fireball. It verges on the miraculous - 19 flashes of light in subterranean
darkness reflect the blazing heart of a supernova.”
The neutrinos were emitted when the star's core collapsed to a neutron star; the light was
emitted when the blast wave broke through the star's surface, maybe a day later. For the
next 160,000 years the neutrinos raced through space towards their meeting point with
Earth, with the light from the blast wave, only a few hours behind. By the time neutrinos
and Earth converged at the same location, we had evolved from the stone age, and had
learned how to make telescopes and neutrino detectors, and derive mathematical models
of supernovae.
The blast wave from SN1987A blew threw outer layers of the star at speeds up to about
40,000 km/second, more than one tenth the speed of light. These layers contained a huge
amount of radioactive nickel-56, and the decay processes, first to cobalt-56 and then to
iron-56, “produced gamma rays (energetic photons) that heated the surrounding gas to
incandescence”. Six months later, when the debris had thinned, a rich mix of heavy
elements were detected, including, iron, calcium, strontium, nickel, cobalt, argon, carbon,
oxygen, neon, sodium, magnesium, silicon, sulphur, chlorine, potassium and calcium. It's
estimated that the the amount of nickel alone was about 7% of the mass of our sun, or
about 23,000 Earth masses. Thus we see supernovae as not only creators, but also
distributors of the elements.
Figure 3.67 shows SN1987A about 20 years after the explosion. The shock wave, about 1
light year across, is clearly visible, but the distribution of the elements is not yet apparent.
D:\116105936.doc
Page 146 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
A pair of bright stars in
SN19878A's galaxy, the
Large Magellanic Cloud
Debris from the supernova
blast. Hidden inside this is a
neutron star or black hole
The supernova shock wave is
colliding with a sphere of
material ejected from the star
maybe 20,000 years before it
exploded.
The ring is here about 1 light
year across, so debris has been
hurled into space at speeds of up
to 20 million miles/hour - about
9,000 km/s, "seeding" space with
heavy elements
Figure 3.67: SN1987A 20 years after the explosion.
Anoher supernova, Cassiopeia A, around 300 years after its explosion, clearly shows the
elements in the expanding star remnants, now about 10 light years across – figure 3.68.
a
b
c
d
D:\116105936.doc
Page 147 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Figure 3.68: X-ray images of the supernova remnant Cassiopeia A, viewed about
300 years after the light from its explosion reached Earth. The expanding cloud of
debris is about 10 light years across.
(a) broadband image, using all X-ray wavelengths, showing the gas at about 50
million degrees
(b) Image using selected X-rays to show the silicon nuclei in the hot gas - this
shows a bright jet of silicon-rich gas breaking out of the left side
(c) the distribution of calcium nuclei
(d) the distribution of iron nuclei
The colours show the intensity of the X-rays: from yellow as the most intense, then
through red and purple to green as the least intense.
And finally, “where did the energy come from to produce the sound and fury which is a
supernova explosion? Energy is conserved: who paid the debts at the end? Answer:
Gravity! … The ultimate energy source in the stars which produce the greatest amount of
energy is gravity power.”
3.13
Review
3.13.1
Seeding inter-stellar space - the "cosmic stock-pot"
enriching the universe
About 100 million supernovae have erupted in the Milky Way galaxy since it was formed
about 10 billion years ago. They and other stars have steadily enriched our galaxy in all
the elements beyond helium. These have been added and stirred into the mix like the
ingredients of a family stock-pot. Just as the meals taken from the stock-pot reflect the
history of what has been put in, so the compositions of new stars and solar systems
reflect the growing enrichment of the galaxy “Dust to dust, atom to atom, the cycle of
stellar life and death progresses - with a larger fraction of heavy elements in each
successive generation.”
The earliest stars, starting with only hydrogen and helium-4 could make only a limited
range of primary nuclei: for example, carbon-12, oxygen-16, silicon-28, calcium-40 and
iron-56. It's only later generations of stars, that start with these primary nuclei already in
the mix, that can go on to make secondary nuclei, neutron-rich isotopes of the primary
nuclei, for example, carbon-13(6p,7n) and oxygen-18(8p,10n).
"Supernovas are the engines of creation. Not only do they give birth to new elements, but
they scatter those elements to the currents of space." The amounts of heavy elements
that one supernova can produce are awesome; for example, about 1,500 Earth masses of
sodium, 5,000 of aluminium and 30,000 of magnesium. Supernovae also disturb the local
D:\116105936.doc
Page 148 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
gas clouds which can trigger the formation of new stars and solar systems from the now
enriched galactic mix.
our solar system
Our solar system, formed around a third generation star, contains about 74% hydrogen
and 24% helium, nearly the same mix as in the very earliest universe. But now the heavy
elements add a small but crucial 2% to this mix. The large planet Jupiter, with about 25
times Earth's gravity, is close to the composition of the overall solar system. The Earth,
with its weak gravity, was not strong enough to hold on to these light gases, so they blew
away to leave our rocky planet, containing just about every nuclide created in the big bang
and the stars.
3.13.2
Reviewing nucleosynthesis processes
a stellar time-line
Figure 3.69 shows the life of a large star as it accelerates towards its destruction as a
-hours
-minutes
-seconds
-days
Supernova explosion
explosive burning
r-process - rapid n capture
Temperature, K
10 billion
alpha process
burning C, Ne, O, Si - up to Fe
1 billion
100 million
-years
-thousands
of years
-millions of
years
-billions of
years
Timescale:
supernova.
He-burning
s-process
Slow neutron capture
H-burning
10 million
1 billion
1 million
1 thousand
0
Time, years
Figure 3.69: Scales of temperature and time for the stellar nuclear processes up to
supernova
D:\116105936.doc
Page 149 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
We can see the temperature climb a thousand-fold, and the time scale shrink from billions
of years to mere seconds.
3.13.3
Flooding the nuclear valley
We have viewed the creation of the nuclides as a series of moves across the nuclear
chess board, starting from just a single proton in the bottom left corner – recall the view
into the nuclear valley through the helium “pass” (figure 3.54). Figure 3.70 gives an
overview of the individual processes.
p-process - makes unstable
proton-rich nuclei, which
decay back towards stability
Fusion processes – making bigger
nuclides
hydrogen burning to helium…
…helium burning to carbon…
…carbon burning through neon,
oxygen, silicon, to iron.
the r-process
makes nuclei
beyond the last
stable nucleus
(r-process) - rapid neutron capture
makes unstable neutron-rich nuclei,
which decay back towards stability
(s-process) – slow neutron capture makes
stable nuclides from iron to lead-208
iron-56
Start here, with single protons!
Figure 3.70: The activity of nuclide building surges up the nuclear valley
We can see the sequence of fusion reactions building up single protons to nuclides in the
iron-group. The slow s-process works its way along the arc of stability as far as the last
stable nuclide. The p- and r-processes make the proton- and neutron-rich nuclides, on
either side of the arc of stability. Only the spallation process is not shown here.
We can imagine the energy level rising with the temperature in the star, and like a tide,
flooding the "nuclear valley", with waves of nucleosynthesis surging ever higher. Neutron
and proton capture processes wash up the valley sides to produce unstable nuclei, which
then decay back towards stability. When the nuclei are ejected into the coolness of space
D:\116105936.doc
Page 150 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
the stable clusters of protons and neutrons remain. In a similar way, when the tide
recedes, it leaves a series of rock pools, small stable bodies of water in isolated hollows.
the array of nuclide species
Every single nuclide is an individual, descended from a series of ancestors via a
sequence of nuclear reactions. Some are ancient, only a few minutes less old than the
universe itself; some are only a few thousand years old, like the nuclei made in supernova
1987A; some have taken millions of years to assemble by the infinite patience of stars;
some were made in a fraction of a second in the convulsions of a supernova; and a few
are here as the fragments of something larger, that was broken up by collisions in deep
space. All have their different stories to tell; each is as unique as a biological species “the personalities of [tantalum-181] and of [iron-56] differ as dramatically as those of the
seagull and the tiger”.
3.13.4
Nuclides and the stellar eco-system
We think of the the populations of living species in terms of a dynamic ecological balance.
Each species feeds on other species lower down the food chain, and is itself devoured by
"predator" species higher up. Species populations then reflect the numbers that can coexist within this ecological balance.
Nuclei within stars also have populations (usually called abundances). They are created
by the fusion of smaller nuclei, and are "consumed" by being fragmented, or capturing
protons or neutrons, or by being incorporated into yet bigger nuclei. The population of a
nuclide species is decided by the balance of such nuclear reactions. Nuclear populations
vary hugely; for example, carbon is nearly 100 million times more abundant than gold in
the solar system. The different conditions that can exist within stars create environments –
that we can perhaps think of as habitats - that favour one group of nuclei rather than
another, with consequences for their populations. We have seen how a big star develops
an onion-like structure, shortly before it explodes as a supernova. We can see this
perhaps as a range of habitats, within which certain nuclei are favoured, with nuclei in one
habitat "feeding" on the product of the next layer up, thus the carbon-burning layer
consumes the ashes of the helium burning in the layer above.
3.14
The emergent atom
3.14.1
From nuclides to atoms
D:\116105936.doc
Page 151 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
With the life cycle of a massive star, and its end as a supernova, we have seen all the
reactions that create the nuclides. We now imagine them blown out into cool interstellar
space, and gathered into a newly forming solar system, no longer subject to the enormous
temperatures and pressures in a star. The unstable nuclides will undergo decay in their
own time, but the stable nuclides, ejected from the stellar ecosystem, are isolated by their
mutual repulsion, and will remain fixed and unchanging.
Since every proton carries one unit of positive charge, each nuclide carries as many units
of positive charge as it has protons. As the nuclides ejected from the star cool in space
each attracts a number of negative electrons, equal to the number of protons, to achieve
overall neutrality. Each nuclide acquires and tethers a number of electrons, that would
otherwise be roaming freely, thus creating a new physical entity, an atom - figure 3.71.
the strong force binds a
colour-neutral trio of uud
quarks into a proton
the nuclear force binds a
group of protons and
neutrons into a nuclear
cluster, overcoming the
protons’ mutual repulsion
the electrons are
electrically attracted to
the nucleus, overcoming
their mutual repulsion
-
u
~5 MeV
removing a
neutron requires
an energy input
of ~5 MeV...
p+
n0
0
0 n
p+ n p+
n0
u
d
the mutual attraction of
all nucleons means a
nuclide has no centre...
...whereas an
electron can be
removed with
only ~5 eV
~5 eV
... whereas an
atom has a
central nucleus
the numbers of protons and
electrons are equal, so the atom is
electrically neutral
Figure 3.71: Quarks are bound by the strong colour force into nucleons, which are
bound by the nuclear force into nuclei, which bind electrons to themselves.
parallels and contrasts
D:\116105936.doc
Page 152 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
This diagram shows the parallels and contrasts between the worlds of quarks, nuclides
and atoms. First, the colour charges of quarks are balanced in a single proton or neutron,
so it has overall colour neutrality. Similarly, the electrical charges are balanced in an atom
so it has overall electrical neutrality. Second, in a stable nuclide, the nuclear force binds
protons and neutrons together against the mutual repulsion of the protons. In a similar
way, the electrons are bound to the central nuclide against their mutual repulsion.
However, because the nuclide holds all the positive charge, the atom has a centre,
whereas a nuclide cluster has none.
disparities in mass and energy
The atom displays big disparities in mass and in energy. The nuclide shown here is a
quite small inhabitant of the nuclear valley, only 7 nucleons, about 6,500 MeV, but this
hugely outweighs the combined mass of the electrons, which is a mere 1.5 MeV. The
electrons are like party balloons tied to a bag of bricks. There are also big differences in
energy; it takes about a million times less energy to remove an electron from a neutral
atom than to remove a neutron from the nuclide. This is an indication of how much weaker
the electrical force is compared to the nuclear force. If we think of nuclides as deaf to
anything less than a shout or a bellow, then an atom will respond to the merest whisper.
At the moderate temperatures on Earth, where the interaction energies are small, the
electrons will readily respond where the nuclide will not. The low energy interactions
between atoms are now mediated wholly by their attendant electrons. We can perhaps
think of the electrons tethered to the nuclide in figure 3.71 as resembling a bunch of party
balloons tied to a bag of bricks. A breeze blows, the balloons sway and gently bump into
each other, but the bag of bricks is indifferent to these gentle interactions.
the nuclide is now an atomic nucleus
The nuclide is no longer in its normal habitat, interacting with other nuclides, and with an
identity defined by its mix of protons and neutrons. It is now the inert centre of a cluster of
tethered electrons, whose number is fixed by the proton number, with the neutron number
now irrelevant - it is now an atomic nucleus. We have seen how very soon after the big
bang the quarks were bound up into protons and neutrons, and have never been seen
free since that time. The same has now happened to the nuclides, for under Earth’s gentle
conditions they are inaccessible, each buried deep inside its atom.
the electrons are now confined
D:\116105936.doc
Page 153 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
The electrons were created by the time the universe was a second old. In themselves,
they appear negligible - a swarm of light particles, mutually repelling each other, and
apparently incapable of any corporate activity. However, each nuclide, by its positive
charge, gathers a specific number of electrons around itself. In this unnatural arrangement
the electrons are forced to co-exist in a confined space, contrary to their urge to be
separate. This is the atom, a conflict of inclinations, a totally novel construct in the
emergent universe.
We have seen how a cluster of nucleons in a nuclide form pairs and closed shells of
magic numbers, making a structured community. We will see in the next chapter how, with
nuclei acting as stable foundations, electrons also form structured, hierarchical
communities, and create wonderful patterns of intricacy and elegance, opening up a
whole new level of complex possibilities.
the chemical elements
So, instead of the 2-dimensional array of nuclides as we have seen them in the nuclear
valley, we now have a series of about 100 neutral atoms, that is, tethered clusters of 1 to
about 100 electrons. These are the chemical elements. Thus we shift our focus from the
nuclides, as they have been made in the big bang and the stars, to the elements, as they
interact chemically on Earth, mediated by their electrons. It is sometimes said that the
elements were created in the stars, and this is approximately true. It is only the nuclides
that are made in the stars, but each then captures the electrons to create an atom that
behaves as a chemical element.
from the 2-D nuclear valley to a 1-D series of elements
We’ll use some specific elements to illustrate the shift from a world of nuclides to a world
of atoms and chemical elements - figure 3.72.
D:\116105936.doc
Page 154 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
8
Nuclides acquire
electrons to match the
proton number. So...
C-14 (6p,8n) decays to N-14 (7p,7n)
protons
Cluster
size...
9
...8 electronsoxygen
N-13 (7p,6n) decays
to C-13 (6p,7n)
7
...7 electronsnitrogen
8
6
...6 electronscarbon
7
5
the isotopes of lithium,
Li, all have 3 protons
6
4
...5 electonsboron
...4 electronsberyllium
5
3
...3 electronslithium
2
...2 electronshelium
1
...1 electronhydrogen
Li-5 (3p,2n)
ejects a proton
to become He4 (2p,2n)
1
2
Li-8 (3p,5n) undergoes
beta-decay to (4p,4n),
which then alphadecays to He-4 (2p,2n)
Li-6(3p,3n) and Li-7(3p,4n)
are stable isotopes
3
4
5
6
7
8
9
neutrons
Figure 3.72: How the chemical elements are related to the nuclides, and how
nuclear decay transforms one element to another.
The diagram shows the 2-dimensional array of the smallest nuclides in the nuclear valley,
where diagonal lines connect members of a nuclear cluster “family”. However, in Earth’s
low energy environment, we only discriminate the number of electrons that are tethered
around each nuclide. Thus all nuclides containing 3 protons, shown in the horizontal box,
will each gather three electrons, to make atoms that are chemically identical, and distinct
from atoms with different numbers of electrons. In this way we distinguish the atomic
elements, depending on their number of tethered electrons, and the first few are listed up
the side of the diagram. So we move from a 2-dimensional array of nuclides in the nuclear
valley, to a 1-dimensional series of atomic elements.
isotopes - same proton number, different neutron number
Thus atoms with the same number of electrons, whilst chemically identical, can have
different nuclides at their centres. An element can have a number of isotopes - atoms with
the same number of protons, but a different number of neutrons. These different nuclides
may have very different ancestries.
D:\116105936.doc
Page 155 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
lithium old and new
The element lithium, defined by its 3 electrons in each atom, has a number of isotopes,
shown in the horizontal box in figure 3.72, of which only 2 are stable - Li-6 (3p,3n) and Li7 (3p,4n). The lithium-6 isotope is only made by cosmic rays from dying supernovae
undergoing spallation reactions in deep space. In contrast, much of lithium-7 was made in
the last nuclear reaction in the cooling fireball after the big bang. Thus, in a lithium battery
in a calculator or a mobile phone, some of the nuclei will be quite recent, maybe just predating the solar system, while others will will be only a few minutes younger than the
universe itself.
lithium can transform to helium
The nuclides in the other lithium isotopes are highly unstable, for example, (3p,2n) and
(3p,5n) each decay to the nuclide (2p,2n), which creates atoms with only 2 electrons,
making the element helium. Speaking in chemical terms, we say that these unstable
isotopes of lithium decay to helium. This would appear to be magic, or nonsense - how
can a soft grey metal become a light gas? But we can see that the protons and neutrons
in the atom’s central nuclide rearrange themselves, and electrons are then acquired or
discarded to match the new proton number. With their enormously bigger energies, the
nuclides take precedence, and the electrons arrange themselves accordingly.
carbon and nitrogen
Figure 3.72 shows some other element transmutations. The elements carbon and
nitrogen are defined by the number of electrons in their atoms, 6 and 7, respectively. The
loss or gain of just one electron will transform one element into the other. Thus the
neutron-rich nuclide in the isotope C-14 (6p,8n) undergoes beta-minus decay to become
(7p,7n), one of the two stable isotopes of nitrogen. Conversely, the proton-rich isotope N13 (7p,6n) decays to become the stable isotope C-13 (6p,7n). The atomic transformations
involve gaining or losing one electron, and because electrons can be removed from atoms
so easily, there are always temporarily free electrons available, if needed.
the transmutation of the elements
These specific examples show how the decay of an unstable nuclide decays and adjusts
its electronic “clothing”, removing or adding “garments” to maintain neutrality. The atomic
elements are not immutable, but can be transformed from one to another, driven by the
enormous energies contained within the atomic nucleus.
D:\116105936.doc
Page 156 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Oliver Sacks writes: “The feeling of the elements’ stability and invariance was crucial to
me psychologically, for I felt them as fixed points, as anchors, in an unstable world. But
now, with radioactivity, came transformations of the most incredible sort. What chemist
would have conceived that out of uranium, a hard tungsteny metal, there could come an
alkaline earth metal like radium; an inert gas like radon; a tellurium-like element,
polonium; radioactive forms of bismuth and thallium; and, finally, lead - exemplars of
almost every group in the periodic table? … Radioactivity did not alter the realities of
chemistry, or the notion of elements; it did not shake the idea of their stability and identity.
What it did do was hint at two realms in the atom - a realtively superficial and accessible
realm governing chemical reactivity and combination, and a deeper realm, inaccessible to
all the usual chemical and physical agents and their relatively small energies, where any
change produced a fundamental alteration of the element’s identity.”
the volcano
We can imagine standing on the edge of a volcano, looking down to the rich plant life
growing on the volcanic slopes, and also down into the crater full of red hot magma. The
plant life is suited to the cool lower slopes, and would be destroyed if exposed to the heat
of the volcanic core. The magma churns deep in the crater, and takes no account of the
plant life above. Standing on the lower slopes, you would have no hint of the huge
energies in the red hot rock underneath. If the volcano erupts, people die and the plant life
is destroyed, but after a while the people return and plant life re-establishes itself, and
things go on as before.
An atom is similar to a volcano, in that there is a huge turbulent nuclear core, deep
beneath the superficial electrons. If the nucleus is stable, there is no hint of this enormous
energy. But when an unstable nucleus erupts, this huge energy reveals itself, and the
electrons are disrupted. But then, like the volcano, electrons gather round the new
nucleus and neutrality is achieved once more. With the volcano, things go on as before,
but with another layer of fertile soil. But with nuclear decay, the balance of the nucleons,
and specifically the number of protons has changed. Consequently, the number of
electrons in the atom changes, and a new atom, with its own chemical identity appears.
The protons and neutrons in their nuclear world, take no account of what the electrons are
doing "outside". A cluster of nucleons is only concerned with its own stability, so it will
decay by any process that is energetically suitable - interchanging protons and neutrons,
or down-sizing by ejecting an alpha particle (2p,2n). The electrons regroup around the
D:\116105936.doc
Page 157 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
new nucleus, and a “new” atom, with its own physical and chemical identity appears. The
nucleons pay no more attention to the electrons, than a volcano heeds the humans living
on its slopes.
Element abundances and ancestries
Figure 3.73 shows how the different nucleosynthesis processes have created the
abundances of the elements in our solar system.
Big bang - the first 3 elements (hydrogen to lithium) in the first three minutes
Nuclear fusion
Burning H to He to C to O to Si to Fe/Ni
- elements 1 - 26 (hydrogen to iron)
Iron peak: the most
tightly bound nuclei
Spallation: cosmic rays in deep space
- elements 3 - 5 (lithium to boron)
Neutron capture: by the slow sprocess, and the rapid r-process in
supernovae - elements beyond iron
Figure 3.73: The abundances of the elements in the solar system.
The atomic number, on the x-axis, gives the number of protons in the nuclei, which
determines the element.
The vertical scale gives relative abundances, referenced to 1million silicon nuclei.
This is a logarithmic scale, so the silicon abundance is 6 since log 1,000,000 = 6.
Thus oxygen (O) at about 7 on the abundance scale is about 10 times more
abundant than silicon (Si). Fluorine at 3 has about For every million silicon there
are ~1,000 atoms of fluorine (log 1,000 = 3), and only ~1 atom of beryllium (log 1 =
0). It may help to view the positive numbers on the y-scale as the number of zeros
after 1.
Many of the elements are named with their standard chemical symbol.
Hydrogen and helium, the big bang nuclei, vastly outnumber all others. After them, there
is a steady downward trend to larger and rarer nuclei, that is punctuated by the dip for the
light elements (containing 3-5 protons, that is, Z=3-5) and the peak for the highly stable
iron-group nuclei (Z=~26). Abundances decrease steadily after the iron-group peak, since
“element formation beyond this point costs energy”. The graph covers all the elements up
to uranium (Z=92), with most being identified with their chemical symbol. Only elements
43, technetium, and 61, promethium are missing, since they have no stable isotopes.
D:\116105936.doc
Page 158 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
“Only the first five lightest elements owe their abundances to origins outside stars - the
first three in the big bang and the fourth and fifth (beryllium and boron) by cosmic ray
interactions with interstellar atoms. From the stars came all the rest. From atomic number
6 (carbon) to atomic number 94 (plutonium) we look to the stars This range of atomic
numbers includes all the common elements of human experience on Earth, save for the
hydrogen within water that blesses the earth’s surface.”
The abundances are clues to the natures of the nuclides in the elements, and the
processes that made them. The prominent “iron-peak” is due to the extreme stability of
these nuclei – “it’s very easy to make iron, for example; but it turns out to be very hard to
make fluorine or boron”. The elements lithium to boron (Z=3-5) do not feature in stellar
fusion processes, and were made by spallation processes, as fragments of carbon and
oxygen nuclei in deep space. Fluorine (Z=9) is comparably rare, and if abundances were
represented as dwellings, then “fluorine would be a shack between two mansions” –
between oxygen and neon. Like the elements lithium to boron, fluorine is largely made by
a spallation process, the ejection of a nucleon from neon (Z=10) in a supernova
explosion.
Elements with more stable nuclei tend, understandably, to be more abundant. We have
seen that nucleons pair up, so nuclides with even numbers of protons or neutrons tend to
be more stable. Thus the abundances follow a zig-zag path across the graph, with the
elements with even numbers of protons (such as C, O, Ne, Si, S…) being consistently
more abundant than their odd-number neighbours. We have seen that nuclides are more
stable if they contain “magic numbers” of nucleons - 2, 8, 20, 28, 50, 82 or 126. The graph
shows a slight abundance peak for elements 50-55, for these have around 50 protons and
82 neutrons. Similarly, lead, with 82 protons and 126 neutrons, has a large abundance.
Every different isotope of every element has its own genealogy - its nuclear ancestry.
Donald Clayton writes, “If I could write an epic poem, I would lyricize over the history of
the universe writ small by their natural abundances. I would rhapsodize over the puzzling
arrangements at different times and places of the thousand or so different isotopes of
some ninety chemical elements. These different arrangements speak of distant past
events”.
Lawrence Krauss describes a very plausible history of a nucleus of oxygen: “Each atom of
oxygen on Earth, by its very existence, suggests a veritable treasure trove of detailed
history: the life and death of millions of stars, the slow dynamic evolution of our galaxy,
D:\116105936.doc
Page 159 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
and indeed the history of matter from well before the galaxies existed. Our oxygen atom
began life as 16 particles”, which fused to create 4 He-4 nuclei, 3 of which then made a C12 nucleus. “Finally, 2 particles, the nucleus of carbon and the nucleus of helium, are
brought together from originally disparate parts of the cosmos, with completely different
individual histories, to make a single nucleus, the nucleus of oxygen.”
3.14
The emergent atom
We are now in a position to review the series of emergent physical structures that
culminate in an atom – figure 3.74.
D:\116105936.doc
Page 160 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
… an atom - a stable cluster
of electrons, around a
massive nucleus, containing
up to ~100 electrons.
e-
e-
A stable atom has emerged, a
tethered cluster of a number of
electrons equal to the number
of protons in the nucleus, and
sensitive to very small
energies
Z+
e…create and sustain…
Continuous interactions between
nucleus and electrons, mediated by
photons…
(or…“photons bind electrons to the
nucleus”)

Z+
e
c
… a nuclide – a cluster of protons
and neutrons, with the mutual
repulsion of the protons overcome by
the stronger nuclear attraction
between all nucleons. About 300
varieties of nuclide cluster are stable.
free electrons are stable,
but are isolated by their
mutual repulsion.
We don’t (usually)
consider the protons and
neutrons inside the
nucleus.
A stable nuclide has emerged,
with a large mass, and carrying
as many units of positive
charge as it contains tethered
protons
p+
n0
0
0 n
+ n
p
p+
n0
…create and sustain…
Continuous interactions between
protons and neutrons, mediated by
pions…
(or…“pions bind protons and
neutrons”)

+
p
n0

b
… a colour-neutral quark trio
(baryon); only 1 baryon is stable –
the proton, the neutron is nearly
stable.
A stable cluster of quarks
has emerged, as a proton,
with a large mass, and
carrying a unit of positive
electric charge
u
u
free protons are stable, but
are isolated by their mutual
repulsions, and unstable
neutrons cannot exist in
isolation.
We don’t consider the
quarks inside protons and
neutrons.
d
…create and sustain…
Continual interactions
between quarks, mediated
by gluons carrying colour
charge…
(or…“gluons bind quarks”)
u
quarks and gluons cannot
exist in isolation
u
d
a
D:\116105936.doc
Page 161 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Figure 3.74: Successive stages in the emergence of the atom. From bottom to
top…(a) the emergent nucleon, (b) the emergent nuclide, and (c) the emergent
atom.
emergent nucleons
We have seen how the colour-neutral quark trios we know as protons and neutrons,
emerge from the continual interactions between quarks, mediated by the gluons that carry
colour-charge.
emergent nuclides
Protons and neutrons are each surrounded by a “cloud” of virtual pions, that “flutter”
between them if they are close enough, creating “an invisible, evanescent web … binding
them together”. It is perhaps ironic that neutrons, in themselves unstable, are essential to
the stability of all nuclides. Thus a nuclide emerges from the continual interactions
between all the nucleons, mediated by pions. The nuclear force, an extension of the
colour force operating inside each nucleon, binds protons against their mutual repulsion,
into compact massive centres of electric charge, with up to ~100 units of positive charge.
emergent atoms
The atom emerges from the electrical attraction between a positively charged nuclide and
electrons, mediated by the exchange of photons of electromagnetic radiation. The nuclide
thus tethers a cluster of electrons, against their mutual repulsion, and creates the range of
chemical elements, based on atoms with up to ~100 electrons.
a hierarchy of emergent structures
Thus we have a hierarchy of emergent structures, each one building on the one below,
and introducing a totally novel feature into the universe. So the strong colour force creates
a stable cluster of quarks, a nucleon, that can have 1 unit of positive electric charge; the
residual nuclear force binds these into clusters, carrying up to ~100 units of positive
charge; and these provide the foundation for the electromagnetic force to gather clusters
of up to ~100 electrons.
Each level in the hierarchy is created and sustained by continual activity in the level
below. At each level in the hierarchy, the activity “underneath” is subsumed in a new
emergent behaviour: quarks “disappear” inside nucleons, and nuclides “disappear” inside
atoms.
atoms can interact and yet keep their identity
D:\116105936.doc
Page 162 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
A nuclide has no centre, so small nuclides can fuse to make one larger cluster, and lose
their identities in the process. But the electrons in an atom respond to very small energies,
about one millionth of those involved in nuclide interactions. So atoms can interact and
join together through their electrons, and yet retain their unique identities within a larger
structure. A nucleus binds as many electrons as it has protons, to make a neutral atom.
Thus the nucleus of an atom decides its identity as an element, yet the nucleus plays no
part in the chemical interactions of the electrons. Electrons can be removed with very
small energies, and the nucleus remains unchanged; the elemental identity is fixed
regardless of any electronic rearrangements. In a similar way, someone remains the
same individual regardless of what clothes they wear.
3.15
The next chapter
In the next chapter we will leave behind quarks, gluons and nucleons, and look at the
behaviour of communities of tethered electrons – we will enter the realm of the chemical
elements. The 3,000 or so known nuclides can create only about 300 stable
combinations, which then become the nuclei of only ~100 different elements. Yet we will
see these few elements create millions of different stable combinations – the different
compound substances of our material world. We will see the amazing structures that
communities of electrons can create. “We could get rid of every electron in our bodies,
and we would never notice the difference if we stood on a scale. Yet, in spite of their puny
heft, electrons may be the most important particle in nature, at least to us, because they
determine almost every observable aspect of our existence.”
D:\116105936.doc
Page 163 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
References
Fred Adams and Greg Laughlin, "The Five Ages of the Universe", Touchstone, 2000
Jim Al-Khalili, “Everything and Nothing”, broadcast on BBC4; programme 1, “Everything”,
21 March2011, and programme 2, “Nothing”, 28 March 2011. http://www.bbc.co.uk/bbcfour/
Jonathan Allday, "Quarks, Leptons and the Big Bang", Institute of Physics publishing,
2003
AME 2003a, “The AME2003 atomic mass evaluation (I): Evaluation of input data,
adjustment procedures”, A.H. Wapstra, G. Audi and C. Thibault, Nuclear Physics A,
volume 729 (2003) p.129–336
The paper is available in the file Ame2003a.pdf from…
http://www.nndc.bnl.gov/amdc/masstables/Ame2003/filel.html
AME 2003b, “The AME2003 atomic mass evaluation (II): Tables, graphs and references”,
G. Audi, A.H. Wapstra and C. Thibault, Nuclear Physics A, volume 729 (2003) p.337–676
The paper is available in file Ame2003b.pdf from…
http://www.nndc.bnl.gov/amdc/masstables/Ame2003/filel.html
This data is the basis for the “mass.mas03” table, available as an ASCII file from…
http://www.nndc.bnl.gov/amdc/masstables/Ame2003/filel.html or…
http://www.nndc.bnl.gov/amdc/web/masseval.html
The mass.mas03 table gives values for mass excess, binding energy/nucleon and atomic
mass (amu), and can be opened in Notepad, and converted to EXCEL format.
Arnold B. Arons, "Development of Concepts of Physics", Addison-Wesley, 1965.
Philip Ball, "The Elements - a very short introduction", Oxford, 2004
John Barrow and Frank Tipler, “The Anthropic Cosmological Principle”, Oxford University
Press, 1986.
Carlos Bertulani, "Nuclear Physics in a Nutshell", Princeton University Press, 2007.
John Tyler Bonner, "Why size matters", Princeton University Press, 2006
Bill Bryson, "A short history of nearly everything", Doubleday, 2003
Margaret Burbidge, Geoffrey Burbidge, William Fowler and Fred Hoyle, “Synthesis of the
elements in the stars”, Reviews of Modern Physics, vol. 29, p.547-650, 1957.
D:\116105936.doc
Page 164 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Wikipedia article at…http://en.wikipedia.org/wiki/B%C2%B2FH (accessed 24 March 2011),
and the full paper available from…http://prola.aps.org/pdf/RMP/v29/i4/p547_1 (Accessed 5
March 2011).
Deborah Cadbury, “The Dinosaur Hunters”, Fourth Estate, 2001
Marcus Chown, "Afterglow of Creation", Arrow, 1993
Marcus Chown, "The Magic Furnace", Jonathan Cape, 1999
Donald Clayton, “Handbook of Isotopes of the Cosmos”, Cambridge, 2003.
Frank Close, "Particle Physics - a very short introduction", Oxford, 2004
Frank Close, "The New Cosmic Onion", Taylor and Francis, 2007
Frank Close, Michael Marten and Christine Sutton, "The Particle Odyssey", Oxford, 2004
Ken Croswell, "The Alchemy of the Heavens", Oxford University Press, 1996
Paul Davies, "The Goldilocks Enigma", Allen Lane, 2006
Armand Delsemme, "Our Cosmic Origins", Cambridge, 1998
Ken Dobson, David Grace and David Lovett, "Physics", Colllins Educational, 1997
John Emsley, "Nature's Building Blocks", Oxford, 2001
Timothy Ferris, "Coming of Age in the Milky Way", The Bodley Head, 1988
Timothy Ferris, "The Whole Shebang", Weidenfeld and Nicolson, 1997
M. P. Fewell, “The atomic nuclide with the highest mean binding energy”, American
Journal of Physics, volume 63, p.653 ,1995, accessed through Google Scholar, August
2010, also referenced in http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin2.html#c1
and http://en.wikipedia.org/wiki/Type_II_supernova
Richard Feynman, "QED - the strange theory of light and matter", Penguin, 1990
Richard Feynman, "The pleasure of finding things out", Penguin, 2000
Richard Feynman, Robert Leyton and Matthew Sands, “Lectures on Physics”, Volumes I
to III, Addison-Wesley, 1963.
W. A. Fowler, Nobel Prize address, at…
http://nobelprize.org/nobel_prizes/physics/laureates/1983/fowler-lecture.pdf
Harald Fritzsch, “Quarks - the stuff of matter”, Penguin, 1992
D:\116105936.doc
Page 165 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Harald Fritzsch, “Elementary Particles - building blocks of matter”, World Scientific, 2005
Robert Gilmore, "The Wizard of Quarks", Copernicus books, 2001
George Greenstein, “Frozen Star”, Futura, 1986
John Gribbin, "Q is for Quantum - an Encyclopedia of Particle Physics", The Free Press,
1998
John Gribbin, "Stardust", Allen Lane, 2000
John Gribbin, "The Universe - a biography", Penguin, 2006
John Gribbin, "Famous for 14 minutes 49 seconds", New Scientist, vol. 137, 13 March
1993, p. 14
Moo-Young Han, "Quarks and Gluons: a century of particle charges", World Scientific,
1999.
Roger Harrison, ed., "Book of Data", published for the Nuffield Foundation by Penguin
Books, 1973
Stephen Hawking, "A brief History of Time", Bantam Press, 1988
Stephen Hawking, "The Universe in a Nutshell", Bantam Press, 2001
Tony Hey and Patrick Walters, "The New Quantum Universe", Cambridge, 2003
Kris Heyde, "From Nucleons to the atomic nucleus - perspectives in Nuclear Physics",
Springer, 1998
Ben Jonson, “The Alchemist”, 1610.
http://www.gutenberg.org/dirs/etext03/lchms10.txt
http://en.wikipedia.org/wiki/The_Alchemist_(play) (both accessed 20 April 2011)
William J. Kaufman III and Roger A. Freedman, "Universe", W.H.Freeman, 5th edition,
1999
Michael Kent, "Advanced Biology", Oxford, 2000
Lawrence Krauss, "Atom", Abacus, 2002
Don Lincoln, “The Quantum Frontier: the large hadron collider”, The Johns Hopkins
University Press, 2009.
Lucretius, “On the Nature of Things”, Translated by Ronald Latham, Penguin, 1951
D:\116105936.doc
Page 166 of 168
12/02/2016
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Katharina Lodders, "Solar System Abundances and Condensation Temperatures of the
Elements", The Astrophysical Journal, volume 591, p. 1220–1247, 2003.
The full paper is available at…http://solarsystem.wustl.edu/publications/2000-2009/p2003/
and the graph is available at…
http://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements (both accessed 6 March
2011)
Laurence Marschall, "The Supernova story", Princeton University Press, 1988
Stephen Mason, "Chemical Evolution", Clarendon Press, 1991
J.MacDonald and D.J.Mullan, “Big Bang Nucleosynthesis: The Strong Nuclear Force
meets the Weak Anthropic Principle”, Physical Review D, volume 80, p. 043507, 2009;
available at http://arxiv.org/PS_cache/arxiv/pdf/0904/0904.1807v2.pdf (accessed 6 April 2011).
Reference 4 in http://en.wikipedia.org/wiki/Dineutron (accessed 6 April 2011)
Ray Mackintosh, Jim Al-Khalili, Björn Jonson, and Teresa Peña, “Nucleus: A trip into the
heart of matter”, The Johns Hopkins University Press, 2001
George Marx, “Life in the Nuclear Valley”, Physics Education, 2001, vol. 36, p.375
Michael Nelkon and Philip Parker, "Advanced Level Physics", Heinemann, 4th edition,
1977
NUBASE 2003, “The NUBASE evaluation of nuclear and decay properties”, G. Audi, O.
Bersillon, J. Blachot and A.H. Wapstra, Nuclear Physics A volume 729 (2003) p.3–128
The full paper is available in Nubase2003.pdf, or just the data in ASCII format, in
nubtab03.asc, from…
http://www.nndc.bnl.gov/amdc/nubase/filel.html or from….
http://www.nndc.bnl.gov/amdc/web/nubase_en.html
The table nubtab03 gives values of mass excess, and can be opened in Notepad, and
converted to EXCEL format. The table nubtab03 and mass.mas03 give the same mass
excess values of nuclides.
Oliver Sacks, “Uncle Tungsten: Memories of a Chemical Boyhood”, Picador, 2002.
Carl Sagan, "Cosmos", Macdonald Futura, 1980
Joseph Silk, "On the shores of the unknown", Cambridge, 2005
D:\116105936.doc
Page 167 of 168
12/02/2016
dBG
“The Emergent Universe - how did we get from there to here?”
Chapter 3: Protons and Neutrons – Nuclei
Simon Singh, "Big Bang", Fourth Estate, 2004
Timothy Paul Smith, “Hidden Worlds”, Princeton University Press, 2003
George Smoot, "Wrinkles in Time - the imprint of creation", Little, Brown and company,
1993
George Smoot, Nobel Prize address, “Cosmic Microwave Background Radiation
Anisotropies: Their discovery and utilisation”, published in Reviews of Modern Physics,
vol. 79, 1349, 2007, and also available at: http://rmp.aps.org/pdf/RMP/v79/i4/p1349_1
(accessed 7 Feb 2011)
Edwin F. Taylor and John Archibald Wheeler, “Spacetime Physics”, W.H.Freeman, 1966
Stuart Ross Taylor, "Destiny or Chance", Cambridge, 1998
Robert Turnbull, "The Structure of Matter", Blackie, 1979
George Wallerstein (editor), "Synthesis of the elements in stars: forty years of progress"
http://cococubed.asu.edu/papers/wallerstein97.pdf (accessed 21 Jan. 2011)
Steven Weinberg, "The First Three Minutes", Basic Books, 2nd edition, 1993
Walt Whitman, Song of Myself, Leaves of Grass (1881-82)
http://www.whitmanarchive.org/published/LG/1881/poems/27 (accessed 20 April 2011)
Chandra Wickramasinghe, “Cosmic Dragons: life and death on our planet”, Souvenir
Press, 2001
William S. C. Williams, "Nuclear and Particle Physics", Clarendon Press, 1991,
updated1997
WHW, or Stan Woosley, A. Heger and Thomas Weaver, “The evolution and explosion of
massive stars”, Reviews of Modern Physics, vol.74, October 2002, p 1015-1071
(shortened to ‘WHW’ in the text)
Stan Woosley and Thomas Janka, "The physics of core-collapse supernovae", Nature
Physics,
volume
1,
p.
ph/papers/0601/0601261.pdf,
147-154,
and
2005;
also
available
as
an
at
http://arxiv.org/ftp/astro-
external
link
in
http://en.wikipedia.org/wiki/Silicon_burning (both accessed 21 Jan. 2011)
Stan Woosley and Thomas Weaver, "The great supernova of 1987", in "Stars and
Galaxies: Citizens of the Universe", readings from Scientific American, Edited by Donald
E. Osterbrock, W. H. Freeman, 1990.
D:\116105936.doc
Page 168 of 168
12/02/2016