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Introduction:
In this introduction to nuclear physics we will firstly summarise some basic
information about the atom, and its nucleus, and present some standard
notation.
We will then discuss some fundamental and necessary concepts of
mass and energy, and develop the idea of binding energy.
We will then look at the statistics, and mechanisms, for radioactivity.
This will lead us to considerations of nuclear reactions, and in particular,
nuclear fusion and fission.
Finally, we will see the application of these ideas in nuclear reactors.
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The Nucleus:
The Atom:
Consists of a small, compact, positively-charged nucleus, surrounded
by a distribution of negatively-charged electrons. The negative electrons are kept
in the vicinity of the nucleus by their Coulomb attraction to the positive nucleus.
The Nucleus:
is composed of two types
of particle – positively-charged protons
and un-charged neutrons.
-electrons
+nucleus
+
Nomenclature:
Atomic Number = Z = Number of protons in nucleus
Neutron Number = N = Number of neutrons in nucleus
Mass Number = A = Number of neutrons + protons
in nucleus
A =Z+N
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+
proton
neutron
3
Significance of Z & A:
Significance of Z:
Protons and electrons have equal, but opposite, charges. Thus, in a neutral atom,
there must be one electron for each proton. The Z of an atom, therefore determines
its total number of electrons.
Electrons are able to exist only in certain discrete energy levels about the nucleus.
As the number of electrons increases, these levels fill up sequentially starting
from the lowest. As a level fills, the next electron must go into the level of next
highest energy.
The chemistry of an atom is determined by how its energy levels are filled.
Thus Z determines the chemistry of an atom. Each chemical element is
characterised by a unique value of Z.
Significance of A:
Protons and neutrons have similar masses, which is about 2000 times
greater than that of an electron. Thus A is a measure of the total mass of an atom.
It does not determine chemical properties.
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Notation:
The type of nucleus is represented by the number of protons, Z, and the
total number of particles, A, from which it is composed, and is written:
A
Z
X
Examples:
p = proton
n = neutron
where
and
Z = atomic number
A = mass number
X = the chemical symbol for the element
1
1
H (Hydrogen nucleus = 1 p)
4
2
He (Helium nucleus = 2 p + 2 n)
6
3
Li (Lithium nucleus = 3 p + 3 n)
56
26
Fe (Iron nucleus = 26 p + 30 n)
235
92
U (Uranium nucleus = 92 p + 143 n)
(Note: sometimes the Z is omitted – example –4He (Helium-4), 56Fe (Iron-56))
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Nomenclature:
Nucleus:
This is the central part of a particular atom, consisting of protons and neutrons.
Nuclide:
This is a type of nucleus.
A
Z
X can represent a particular nucleus, or a " nuclide"
Nucleon:
A nucleon is a component particle of a nucleus.
Protons and neutrons are nucleons.
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Isotopes:
Nucleii with a given number of protons occur with a range of numbers of neutrons.
Nuclides with – the same Z
but
different A
are called different isotopes of an element
Examples:
1
1
H (Hydrogen), 21H (Deuterium), 31H (Tritium) : isotopes of Hydrogen.
11
6
C, 126 C, 136 C, 146 C : isotopes of Carbon.
232
92
U,
233
92
(where
U,
235
92
U,
236
92
U,
238
92
U,
239
92
U : isotopes of Uranium.
= unstable & radioactive)
Many isotopes occur naturally, and naturally-occurring elements are generally a mixture
of different isotopes. The more radioactive isotopes may have occurred naturally in the past,
but have long since decayed to other species. These isotopes can be produced artificially.
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Units:
In nuclear physics special non-SI units are used, which are more suited to the
microscopic scale of the nucleus.
Mass: unified mass unit, u:
Mass of a single Carbon-12 nucleus is defined to have a mass of exactly 12 u.
1 u = 1.660559 x 10-27 kg
Charge: electronic charge, e:
1 electronic charge = e = charge on one electron
1 e = 1.602 x 10-19 coulomb
Energy: electron volt, eV:
1 eV
= the work done to move an electron through a potential of 1 Volt.
= charge x potential
= 1.602 x 10-19 joule
= (1.602 x 10-19 ) x 1
1 keV = 103 eV
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1 MeV = 106 eV
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Mass & Charge for Proton, Neutron, Electron:
Particle
mass
(kg)
(u)
Proton (p)
1.007 276 470
1.672 621 71 x 10-27
charge
(e)
+1
1.674 92 x 10-27
0
Electron (e) 5.485 799 03 x 10-4 9.109 56 x 10-31
-1
Neutron (n) 1.008 664 904
The masses above are for an isolated particle.
Note: masses of p and n are both about 1 u.
Mass of p ~ 2000 times mass of e.
Charges of p and e are equal but opposite.
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Mass-Energy Equivalence:
As was discussed in the notes on Quantum Physics, the total energy, E, the
rest mass, m0 , and the momentum, p, of a particle are related by the following
equation from the theory of relativity -
(
E = (pc ) + m 0 c
2
2
)
2 2
............(N1)
8
8
where c = speed of light = (2.9979 x 10 ) m/s ~ (3.0 x 10 ) m/s.
If the particle is at rest, p = 0, and equation (N1) becomes -
E = m 0 c ............(N2 )
2
Mass, m0 , is equivalent to a proportional amount of energy, E . The constant
of proportionality is the square of the speed of light.
Because c is a large number, a small amount of mass, is equivalent
to a very large amount of energy.
Mass-energy equivalence is fundamental to the understanding nuclear processes.
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Direct Evidence:
Mass can be converted into energy, and energy can be converted into mass.
A direct example of such conversions involves the electron.
Energy into Mass :
If a photon of sufficiently high energy interacts with a nucleus, the photon
can disappear, and be replaced by two particles, an electron and an antielectron
(an antielectron is also called a positron). This is called pair production. Energy
of the photon has been converted into the mass of the two particles.
(For any particle, there is an antiparticle, whose quantum properties are all
exactly opposite to those of the particle.)
Mass into Energy:
If an electron and a positron collide, they can mutually
annihilate and be replaced by two photons (pure electromagnetic energy).
This is called “pair annihilation”. The particles disappear, and their masses are
converted into electromagnetic energy.
There are many other situations where such mass-energy conversions occur,
as we shall see.
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Example: Carbon-12:
Find the energy equivalent of 1 atom of Carbon-12 ( mass = 12.00 u).
In SI Units:
1 u = 1.660559 x 10-27 kg
c = 2.9979x108 m/s
E = m0 c2 = [12.00x(1.66x10-27)] . (3.00x108)2 = 1.79x10-9 J = 1.79 nJ
In Nuclear Units (MeV & u):
Since:
E [Joule] = m [kg] (c [m/s]) 2
1 eV = 1.6022 x 10-19 joule
E [in MeV] (1.6022x10-19)x106 ={ m [in u] (1.6605x10-27 )} x (2.9979x108)2
Giving: E [MeV] = 931.5 m [u] when the constants are combined
In practice we use this form:
E [in MeV] = 931.5 m [in u]
Here: E [MeV] = 931.5 m [u] = 931.5 x 12.00
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= 1.12x104 MeV
= 11.2 GeV
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Strong Nuclear Force:
The equivalence between mass and energy is essential to the
understanding of the physics of the nucleus.
The protons in a nucleus are positive, and repel each other strongly by
Coulomb repulsion. They are held together by another much stronger
attractive nuclear force, called the “strong nuclear force”. This force attracts
any proton or neutron, to any other proton or neutron. It is a short range
force, and therefore its action is essentially confined to the nucleus.
Consider a nucleus of Deuterium 12H, which is composed of
1 proton plus 1 neutron, held together by the “strong nuclear force”.
F
Let’s pull this nucleus apart. We will need to apply
equal, but opposite forces, F , to the two nucleons.
In this process we do work on the nucleus.
F
p n
But the nucleons are stationary after being separated, and thus
have no kinetic energy. There is no potential energy, since the strong force
field is negligible outside the nucleus. No energy has escaped from the
system. So where did the work go that we did on the system?
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Answer:
The work we did has turned into mass!
The total mass of the separated proton (p) and neutron (n) is greater than
The mass of the deuterium nucleus.
If the p & n were recombined to form a deuterium nucleus again, this
mass increase would convert back into energy.
The mass difference, δm, and its energy equivalent, E, are related by Numerically:
E = δm. c2
For deuterium nucleus: measured mass of
2
1 H
= 2.014102 u
(p mass) + (n mass) = 1.007825 u +1.008665 u = 2.016490 u
Difference = δm = 0.002388 u
Energy equivalent of δm = W = (0.002388 u) x 931.5 = 2.22 MeV
This is the work we did to strip the nucleus.
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Binding Energy:
This work that we must do to split up the nucleus,
is called the “binding energy ”(BE) of the nucleus.
It is energy we must input to the nucleus to separate it.
By convention, we take the separated state to be the zero of
nuclear potential energy for the nucleus.
If we must add energy to the nucleus to bring it to a zero energy separated
state, then the binding energy of the nucleus must be negative.
(binding energy) + (input energy)= 0
binding energy = -(input energy)
negative energy
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BE/nucleon
As the total number of nucleons in a nucleus increases, the total BE of
the nucleus also increases. However, due to quantum effects within the
nucleus, the total binding energy does not increase in proportion to the
number of nucleons (= the mass number A).
We usually consider the average BE per nucleon = BE/A
The BE per nucleon is a measure of the average energy required to
extract a nucleon from a given nucleus. It is therefore a measure of
the stability of the nucleus.
The greater the BE/nucleon, the more stable the nuclide.
By measuring isotopic masses, we can plot a curve that represents the
nuclear stability of a range of nuclides.
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BE/nucleon Curve:
region of max stability
The most stable
nuclei occur for
maximum
BE/nucleon
Iron-56 is one of
the most stable.
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How Did We Calculate the BE/nucleon Curve?:
(SI units)
Mass
difference ≡ δm ≡
for ZAX
mass of
mass of
mass of
−
+
protons
neutrons
nucleus
= ( Z .m p + ( A − Z )mn − m X )
where m p is mass of p; mnis mass of n; m X is mass of X;
BE of
nucleus ZAX
BE per nucleon
A
Z
for nucleus X
≡
= δm.c 2
(Total BE )
(Number of nucleons)
=
δm.c 2
A
We calculate the BE/nucleon curve from experimentally measured masses of the
nuclei of the various isotopes. These can be measured to high precision, using
mass spectrometry.
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Comments on the BE/nucleon Curve:
From about 2.2 MeV/A (=MeV per
nucleon) for deuterium (H-2), the
curve rises steeply to a maximum of
about 8.5 MeV/A near iron (Fe-56).
On the rising edge are peaks at He-4,
C-12, and O-16, which are particularly
stable nuclides.
Beyond iron, the curve drops slowly to
uranium (U-235). As uranium is
approached, the stability of nuclides
decreases. Uranium itself is unstable
and radioactive. Elements with higher
atomic number, because of their
increasing instability, do not occur
naturally.
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Nuclear Processes:
We are now in a position to consider some important processes involving nuclei.
Nuclear processes generally involve changes in the structure of a
nucleus. Changes in mass and energy are involved.
We will consider these under three headings.
Radioactivity This where unstable nuclei spontaneously decay, with the
emission of small particles, at high energy.
Nuclear Reactions Here changes in nuclear structure are initiated when particles
are fired at nuclei. New nuclei are produced.
Fission and Fusion reactions These are special nuclear reactions, where either smaller
nuclei fuse into a larger nucleus, or a large nucleus splits
into two smaller fragments.
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Radioactivity:
Some nuclei are unstable, and spontaneously decay to other nuclei by
emitting some of their component particles.
Such nuclides are called radioisotopes, and are said to be radioactive.
The emitted particles exit the nucleus with significantly large kinetic
energies.
There are 4 mechanisms for natural radioactivity.
α-emission: an α-particle is emitted from the nucleus.
β-emission: a β -particles is emitted.
γ-emission: a γ-particle is emitted.
electron capture: an inner orbital electron is captured, and a
neutrino (symbol ν) emitted.
Let’s look firstly, at the nature of each of these emitted particles.
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α & β Particles:
An α-particle consists of a group of 2 p + 2 n, combined as a helium nucleus
4
( 2 He). We saw earlier, that the binding energy/nucleon for this grouping of
protons and neutrons is high, which is why they tend to hold together as a particle.
It is a relatively heavy particle, which interacts strongly with any matter in its path.
A β -particle is an electron. There are two types of electron:
β− is a negative electron (sometimes called a negatron). These are identical to
the orbital electrons of an atom.
β+ is a positive electron (called a positron). It is the antiparticle
of the negative electron. Each of its quantum properties is the opposite
of those of the negative electron. It has an equal, but opposite charge.
It has the same mass but is made of antimatter. If a negative electron and a
positron collide, they annihilate each other and their two masses convert
completely into energy.
(In fact, each type of elementary particle has its own corresponding antiparticle.)
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γ’s & ν’s:
A γ -particle is a lump of pure high-frequency electromagnetic energy.
It has zero charge and zero rest mass.
A neutrino (ν
ν) is lump of non-electromagnetic energy, with zero charge
and very small rest mass.
Note: The α-particle is a relatively large particle, and α-decay
is a common decay mechanism for larger nuclides ( say A
>144 ), but not for smaller nuclides. β-decay occurs over a
wide range of A values.
γ-decay is generally associated with either α- or β-decay.
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Statistics of Radioactive Decay:
We deal firstly with the statistics of the decay process, for any of the
radioactivity mechanisms (α, β, γ, or EC).
It is not possible to predict precisely when a particular nucleus will decay,
as the process is probabilistic. However we can specify the probability of a decay.
This probability will depend on the particular nuclide & decay mechanism.
All nucleii of a given nuclide have the same decay probability, for a given
mechanism.
Thus, for a large population of a given nuclide, the number of decays that occur,
per unit time, is proportional to the total number, N, of nucleii present.
Let dN be the number of decays that occur in a time interval dt.
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Decay-Rate Equation:
(rate of decay) ≡ (nr. of decays per second)
∝ (nr. of active nucleii remaining)
dN
∝N
dt
dN
therefore
=− N
dt
where ≡ constant of proportionality called the decay constant.
≡ probability of a decay per nucleus per unit time
(where the negative sign is required, since dN is a
decrease for a decay, and is thus negative.)
(This is a simple differential equation, which can be solved for N.)
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Solution of Equation:
dN
=− N
dt
1 dN
transposing
=−
N dt
N
t
1 dN
and
.dt = − .dt (Integrating both sides with t)
N dt
N0
0
where N 0 ≡ N at t = 0
N
thus ln(
)=- t
N0
and N = N 0 e -
t
Radioactive samples decay exponentially
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Decay Curve - Half Life:
We can plot the solution:
Half-life = T1/2
N = N 0e −
= time taken for half
the original nuclei to
decay.
t
N = N 0e −
Puting N =
t
1
N 0 at t = T1/ 2 gives
2
1
N 0 = N 0 e − T1/2
2
− ln(2) = − T1/2
T1/2 =
T1/2
2T1/2
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3T1/2
ln(2)
=
0.693
(Half-life in terms
of Decay const.)
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After n Half Lives:
During any time interval of T1/2 ,the number of active nuclei halves.
It doesn’t matter when you start the time interval.
If N = N 0 at t = 0
N0
then N =
at t = T1/2
2
N0
N = 2 at t = 2T1/2
2
N0
N = 3 at t = 3T1/2
2
Summary:
The Number of
N0
active nucleii remaining = n
2
after n half - lives
N0
N = n at t = nT1/2
2
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Some Half-lives:
Half-lives vary greatly.
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Activity:
Activity is a measure of the level of radioactivity of a sample. This is the
average number of decays occurring per second.
Activity ≡ R ≡ Decay Rate ≡ Decays/second ≡
dN
dt
dN
= N (where || are used since dN is negative)
therefore R ≡
dt
= N0e − t since N = N 0e −
t
thus R = R 0e − t sin ce R ≡ N
where R 0 ≡ (R at t = 0) ≡ Initial Activity
Activity decreases exponentially
in proportion to the number, N, of active nucleii remaining.
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Units for Activity:
Bequerel (Bq)
This is the SI unit for the activity, R, of a radioactive sample.
1 Bq = 1 decay per second
1 kBq = 103 Bq
1 MBq = 106 Bq
Curie (Ci):
This is an older non-SI unit for activity, that is still sometimes used.
1 Ci is the activity of 1 gram of pure Radium ( 88226Ra )
1 Ci = 3.7 x
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1010 Bq
1 mCi = 10-3 Ci
1 µCi = 10-6 Ci
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Example:
has a half-life of 10 minutes. If the initial number of 713N nuclei
6
in a sample is 1024x10 nuclei, how many will be left after 50 minutes?
7
13N
By Table:
Using the fact that the number of active nuclei halves after every
half-life:
t (min)
0
10
20
30
40 50
N (M nucleii) 1024 512 256 128 64 32
Answer:
32x106 nuclei remain.
By Formula:
N = N 0e-
where =
t
ln 2
T1
2
Here T1 = 10 min, N 0 = 1024 ×106 , t = 50 min
2
−
∴ N = N0 e
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ln 2
t
T1
2
(
)
= 1024 ×10 e
6
50 )
-ln 2.( 10
= 32 ×106 nuclei Same answer!
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Example:
has a half-life of (1.6 x 103 ) years.
What is the activity of a sample containing (3 x 1012) nuclei ?
88
236Ra
Answer:
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41 Bq
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Decay Mechanisms:
We deal next with the detailed mechanisms for α-, β-, γ-decay
and for electron capture (EC).
These processes involve changes in the nuclear structure of the
unstable parent nucleus.
This parent nucleus is a system to which conservation laws apply.
The total charge, the total mass-energy, and the total momentum
of this system (including any emitted particles) must be the same,
before and after the decay.
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α-Decay:
α-Decay: In this decay, 2 protons + 2 neutrons are ejected from the
nucleus as an α-particle, which is a Helium-4 nucleus.
General
A
Z
X →(( ZA−−42))Y + 24He
α-particle
Examples:
α-decay of Radium226:
Rn (radon) is itself unstable:
α-decay of Uranium-238:
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226
88
Ra →
222
86
Rn →
222
86
218
84
Rn + He
4
2
Po + He
4
2
U → Th + He
238
92
234
90
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4
2
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Conservation of Z & A:
The following two rules apply to nuclear changes:
Conservation of charge:
The total number of
The total number of
=
protons on the LHS
protons on the RHS
Z=
LHS
Z
RHS
atomic number,
Z, is conserved
Conservation of nucleons:
Total number of
Total number of
=
nucleons on LHS
nucleons on RHS
A=
LHS
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A
RHS
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mass number,
A, is conserved
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Q for a Nuclear Process:
Definition of Q:
This is the amount of energy released by the process.
This energy can only come from the conversion of mass into
energy during the process.
If δm is the mass lost during the process,
then Q = δm.c 2
where δm ≡ (mass of the reactants ) - (mass of products )
=
m−
LHS
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m
RHS
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Q for α-Decay:
Q for an α-decay:
For X → Y + α
Q = δm.c 2
= (m X − mY − mα ).c 2
where mX = mass of X
mY = mass of Y
mα = mass of α
In α-decay, the Q goes mainly into the KE of the α. (Α small
amount will go into the recoil energy of nucleus Y.)
Ιf δm = (mx – my – mα) is not greater than zero, there will be no
energy for the emission of the α, and consequently the decay
will not happen.
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Example:
Polonium-210 α-decays to Lead with a half-life of 138.38 days.
Find the Q for this process.
210
84
4
Po→ 206
Pb
+
82
2 He
The atomic masses of nuclides are given in tables (see Serway Table A-3)
δm = mPo − mPb − mHe
= 209.98286 − 205.97446 − 4.002603
= 0.005797u
From Q = δm.c 2 we get Q( MeV ) = 931.50 × δm(u )
= 931.50 × 0.005797
= 5.400 MeV
Most of this energy (5.400MeV) goes into the KE of the ejected α.
Some goes into the recoil KE of the Pb-206 nucleus.
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β-Decay:
Instabilities within the nucleus can result in the ejection of either an
+
electron (β ), or its antiparticle the positron (β ), from the nucleus. Another
particle, called a neutrino (ν), or its antiparticle, the antineutrino ( ), is
always emitted with the β. A neutrino is emitted with the β+, and an
antineutrino, with the β . Both these particles carry energy and
momentum.
General:
antineutrino
with electron
neutrino with
positron
Examples:
Note:
conservation of
A and Z apply.
The properties of the positron are precisely opposite to those of the electron
– equal but opposite charge, equal antimass, spin ½ .
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Properties of the Neutrino (
0
0 ν
):
The ν was first postulated in 1936 to explain anomalies in the conservation
of energy and momentum, in β decay. However, because of its weak
interaction with matter, was not directly experimentally confirmed until 1956.
A neutrino (ν
ν) is an elementary particle which has zero charge.
has a small rest mass, m0 ( 0.05 eV < m0 < 0.3 eV )
travels near the speed of light c
has momentum p
2
2
2 2
has energy E given by E = (cp) + ( m0c ) , where m0 is small.
has angular momentum (spin = 1/2), directed in the opposite
direction to its momentum, p.
interacts very weakly with matter and is consequently difficult to
detect (a ν can pass transparently right through planet Earth).
Neutrinos interact only by the weak or gravitational forces – not by
strong or electromagnetic forces.
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More Neutrino Properties:
Recent experiments show that there are three types (or flavours) of
neutrino – electron neutrino, muon neutrino, and tau neutrino. As
a neutrino travels through space, it can oscillate between these three
types. This is called neutrino flavour oscillation.
The neutrinos in β decay, are electron neutrinos.
The anti-neutrino, because it is the anti-particle of the neutrino, must
have properties, that are precisely opposite to those of the neutrino. This
means that the charge will also be zero, the rest mass will be equal, but
antimatter, and the direction of its spin, must be in the same direction as
its momentum, p, since the spin vector for the neutrino is oppositely directed to
its momentum.
β-decay of the neutron:
Although the neutron is stable inside the nucleus, outside the nucleus it
is unstable, and β-decays to a proton, with a half life of 10.6 minutes -
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Atomic Mass Values & Q:
The masses listed in tables, are traditionally atomic mass values, which
include the masses of the orbital electrons of the neutral atom. For
example, the atomic mass for (24He) = 4.002603 u is the mass of the
nucleus plus the mass of the two orbital electrons.
(atomic mass) = (nuclear mass) + (mass of orbital electrons)
The masses we need for Q calculations are the masses of the nuclear species
taking part in the nuclear change – the nuclear masses.
In the case of α-decay, the number of orbital electrons on the LHS of the
decay reaction, is equal to the total number on the RHS.
For example:
226
88
4
Ra → 222
Rn
+
86
2 He
has 88 on the LHS,
and (86+2)=86 on the RHS.
Thus the orbital electron masses cancel out in the calculation of Q, and the
atomic mass values will give the same result as using the nuclear mass
values. Thus the atomic mass values are generally used.
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In β-decay:
In β-decay and EC, electrons are directly involved in the nuclear change,
and we must either use nuclear masses, or use atomic masses, but take
careful account of the electrons.
Using nuclear masses:
nuclear mass = mass of nucleus
= (mass of neutral atom) – (total mass of orbital electrons)
For a neutral atom:
The number of orbital electrons = the number of protons = Z
Let
mX = neutral atomic mass of X
mY = neutral atomic mass of Y
me = mass of electron
Note: we assume the rest mass of neutrino to be negligible.
+
-
Let’s see what happens in the calculation of Q for β and β decay.
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Q For β-decay
orbital
electrons & β
cancel out!
electrons don’t
cancel out
To get the Q:
or
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Q = δm.c2
Q(MeV) = 931.50 δm(u)
Nuclear (© F.Robilliard)
For β-decay to occur,
Q must be > 0
47
γ-decay
If a nucleus is left in an excited state, it can emit its excitation energy in the
form of a quantum of electromagnetic energy - the γ-photon.
X → X+ γ
A
Z
General :
∗
A
Z
0
0
(The superscript * represents the fact that the nucleus X is in an excited state.)
γ-decay is often associated with other decay mechanisms, which leave
the nucleus in an excited state.
Example
followed by
12
5
B→126C ∗ + β − + υ
[ β - decay]
C ∗ →126C + γ
[γ - decay ]
12
6
Wave-particle duality allows us to treat the photon as a particle, that travels at the
speed of light, c, and has zero charge, zero rest mass, energy = E = hf (Planck’s
equation), momentum = E/c, spin = 1. [see notes on Quantum Physics]
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Electron Capture (EC):
In this process, an (inner) orbital electron is captured by the
nucleus, and a neutrino is emitted.
EC is closely connected with β+-decay
General EC :
Example of EC :
X + e→ Y + ν
A
Z
0
−1
A
Z −1
0
0
Be+ −10e→37 Li + 00ν
7
4
δm = (nuclear mass X) + (electron mass) - (nuclear mass Y)
= [m X − Zme ] + [me ] − [mY − ( Z − 1) me ]
= m X − mY
And :
Q = δm.c 2
Or
Q(Mev) = 931.50 δm(u)
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Nuclear Reactions:
As we have seen, unstable nuclei can decay spontaneously, producing
a change in nuclear structure – radioactivity!.
It is also possible to trigger changes in nuclear structure by firing a
high energy particle at a nucleus.
This is a nuclear reaction, in which a new nucleus is formed, with the
emission of second particle.
X + a → b +Y
Target nucleus X is bombarded by high energy particle
a. Structural changes result in the emission of a
second particle b, leaving a product nucleus Y.
This is often written:
X (a, b)Y
Let’s look at some examples.
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Examples:
Lithium-7 bombarded by a proton:
7
3
Li +11H → 24He+ 24He
7
3
Li ( p, α ) 24He
This was the first nuclear reaction produced by artificially accelerated
particles.
Berylium-9 bombarded by an alpha particle:
9
4
Be+ He→ n + C
4
2
9
4
1
0
12
6
Be(α , n)126C
This reaction was used to experimentally confirm the existence of the
neutron.
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More Examples:
Neutron bombardment of Boron-10:
10
5
B + 01n→ 24He+ 37Li
10
5
B(n, α )37Li
Absorption of a gamma photon, leading to the
photodisintegration of the absorbing nucleus:
25
12
24
Mg + γ →11H +11
Na
25
12
24
Mg (γ , p)11
Na
where the 1124Na is unstable and undergoes β-decay 24
11
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24
Na →12
Mg + β − + ν
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53
Q For a Nuclear Reaction:
For:
X (a, b)Y
Q = δm.c , where
m = (mX + ma-mY -mb )
2
Note: as we have noted earlier, nuclear masses should be used to calculate δm.
In practice, as we have seen, neutral atomic masses ( those usually found in
tables, and which include the masses of orbital electrons) are used, since
the orbital electron masses are the same on both sides of the reaction,
and consequently cancel out.
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Example:
7
3
Li ( p, α ) 24He
δm = mLi + m p − mα − mHe
= mLi + mH − 2mHe
= 7.016004 + 1.007825 − 2(4.002603)
= 0.018623 u
Note: that we need to use the atomic masses of H, and He, rather than the
masses of isolated protons and alpha particles, so that the orbital electron
masses will cancel out.
Q( MeV ) = 931.50 × δm(u )
= 931.50 × 0.018623
= 17.35 MeV
This energy will take the form of the kinetic energy of the two α-particles.
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Exothermic & Endothermic Reactions:
There are two types of nuclear reaction: those where energy is
produced, and those where is is absorbed.
Exothermic: Q >0 :energy is produced.
Endothermic: Q<0 : energy is absorbed.
In exothermic reactions, the evolved energy usually takes the
form of the KE of the product particles.
For endothermic reactions to proceed, energy greater than the
Q must be supplied to the reaction, generally in the form of
the KE of the reacting particles. This energy is a threshold
energy for the reaction.
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Nuclear Fusion and Fission :
These are two important classes of nuclear reaction.
Fusion is where two smaller nuclei join
together to form a single larger nucleus
Fission is the inverse operation, where a
larger nucleus splits into two smaller nuclei.
To understand the important energy implications of these two processes, we
need to revisit the Binding Energy/Nucleon curve.
S
Q
R
Fusion: could involve fusing two nuclei
near P to form a single larger nucleus
near Q, or fusing two nuclei near S to form
a single larger nucleus near R.
Fission: could involve splitting a single
P
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nucleus near R into two smaller nuclei near
S, or splitting a single nucleus near Q into
two smaller nuclei near P.
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57
S
Q
P
Increase in BE :
R
Of the possibilities shown by the arrows in the
figure, two are exothermic, and consequently of
interest in power generation –
Fusion P
Q
Fission R
S
The other two are endothermic - & not of interest.
Both these reactions involve going up the
BE/nucleon curve – that is, both involve an
increase in BE/nucleon.
In fusion and fission, the number of nucleons composing the reactants, and the
number composing the products, is the same. If the BE/nucleon has increased
during the reaction, the total overall BE of the products is greater than that of the
reactants. The total BE has increased.
But BE is the work we need to put into a nucleus to strip it into into its constituent
nucleons.
BE can only increase, if the original nuclei have lost the extra energy
during the reaction, that we now need to add, to strip them. That is,
the reaction must be exothermic (Q > 1)
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Fusion and Fission:
Example of fusion P
Q:
A ( 21 H) nucleus collides with a (11 H) nucleus to form a ( 23 He) nucleus :
2
1
H +11H → 23He + (energy )
Example of fission R
S:
236
92
U can spontaneously fission into fragments
90
37
Rb and 143
55 Cs with the emission of 3 neutrons:
1
U →3790Rb+143
Cs
+
3
(
55
0 n) + (energy)
236
92
( Note: most fissions
are not spontaneous
like this example, but
are triggered by the
absorption of
neutrons.)
Mass:
In both the above examples, the total mass of the reactants is greater
than the total mass of the products. This difference in mass has been
converted into the energy that is released.
We will conclude by looking, in more detail, at two important cases nuclear fusion in stars, and fission in nuclear reactors.
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Nuclear Fusion In Stars:
The energy of stars, including the sun, is produced by
exothermic nuclear fusion processes in stellar cores.
One mechanism for this is called the p-p cycle, and consists
of three steps:
1
1
H +11H →12H +10β + + 00ν
.................(1)
1
1
H +12H → 23He+ 00γ
.................(2)
3
2
He+ 23He→ 24He + 2(11H ) .................(3)
Overall reaction : 2(1) + 2(2) + (3) :
4(11 H)→ 42 He + 2(10β + ) + 2( 00ν ) + 2( 00γ )
The overall effect is to fuse hydrogen into helium with the
evolution of energy. Positrons, neutrinos and gammas are also
produced. The overall Q is about 25 MeV.
Research is ongoing to achieve nuclear fusion efficiently on Earth.
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Nuclear Fission in Reactors:
Nuclear fission can be exothermic, and therefore offers possibilities
for energy generation. However, there are a number of problems
that need to be overcome, before this is possible on a large scale.
If a neutron is fired at a large nucleus, the neutron can be absorbed,
producing an unstable new nucleus. This unstable nucleus can fission
into two fragments, with the release of excess neutrons, plus energy.
This is a n-induced fission.
Because the excess neutrons produced by the fission, can go on to
produce further fissions, this process has the potential to be self
sustaining – that is, we don’t have to keep providing the neutrons.
Uranium-235 is the only naturally-occurring nuclide that will undergo
n-induced fission. There are artificially produced nuclides such as
Plutonium-239, and Uranium-233 that will also fission in this way.
Fission of heavy nuclei can be triggered by bombardment with
particles other than neutrons, such as protons, deuterons, alphas,
or gamma photons, but such fissions are not self sustaining.
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Fission of Uranium-235:
1
0
n + U → U → X + Y + neutrons + Q
235
92
236
92
*
X and Y are the “fission fragments”. There are many different X, Y
pairs possible, for U-235 fission, for example:
1
0
236
*
141
92
1
n+ 235
U
→
U
→
Ba
+
Kr
+
3
(
92
92
56
36
0 n) + Q
1
0
n+ U → U → Xe+ Sr + 2( n) + Q
235
92
236
92
*
140
54
94
38
1
0
The fission fragments are generally radioactive, and will
subsequently decay.
The number of neutrons will vary, depending on the X and Y. The
average number of neutrons over all possible X, Y pairs is ~2.5.
Q is the energy released
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Q for:
1
0
δm
n + U → Ba + Kr + 3( n) + Q
235
92
141
56
92
36
1
0
= mn + mU – mBa – mKr – 3mn
= 235.0439231 – 140.9144069 – 91.92615313 -2(1.008664915)
= 0.1860332 u
Q(MeV) = 931.50.δm(u)
= 931.50 x 0.1860332
= 173.290 MeV
(Q is the energy generated by 1 fission.)
Compare:
(energy released by 1 molecule of octane in petrol combustion)
= 200 eV
Fission is a highly energetic process!
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Sustained Nuclear Fission:
Because of its high Q in fission, and because it occurs naturally,
uranium-235 can be used for large scale energy generation, provided
that the reaction can be made self-sustaining.
Since each fission of uranium-235 produces, on the average, 2.5
neutrons, and it takes only one neutron to induce another fission, a
sustained nuclear reaction is possible, assuming that we lose no more
that an average of 1.5 neutrons per fission.
Management of the neutrons is important. We need to minimise losses,
that can occur through absorption by non-fissionable nuclei, and
maximise absorption by fissionable nuclei. Apart from being absorbed
and causing fission, the following are ways in which neutrons can interact
with nuclei.
1. elastic collisions (no absorption).
2. absorption by neutron radiative capture (n, γ)
Both of these are important to achieve a sustainable nuclear fission reaction.
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1. Elastic Collisions:
Rather than be absorbed, neutrons can collide elastically with some nuclei.
A fast neutron will lose some of its KE to the nucleus, in such a collision.
After many collisions, with many nuclei, the neutron approaches the same
average thermal KE as the nuclei themselves. We say that the neutrons
have been thermalised, or moderated.
The fraction of its KE lost by a neutron in a collision, will be greatest for
nuclei whose masses are comparable to that of the neutron.
Unfortunately some of these nuclei, such as H-1, have a high probability
for neutron absorption.
Materials that efficiently transfer neutron KE to their nuclei by elastic
collisions, are called moderators.
Moderators slow down neutrons.
It turns out, that the probability of absorption by fissionable nuclei increases
as the energy of the neutron decreases, making moderation useful.
Common moderator materials are heavy water ( D2O ,where
D=Deuterium), beryllium (Be), and graphite (12C).
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2. Neutron Radiative Capture:
Some nuclei can absorb neutrons, in a collision, and not fission.
Such an absorption leaves the resultant compound nucleus in an excited state.
The nucleus can get rid of this excess energy by emitting it as a
gamma photon.
General: 1
0
Example:
n+ X →
A
Z
A+1
Z
X →
*
X +γ
A+1
Z
or:
A
Z
X (n, γ )
A+1
Z
X
Cd (n, γ ) Cd
113
48
114
48
This process removes neutrons from a nuclear reaction.
To achieve sustained fission, such losses need to be minimised. Once
the reaction has been started, however, controlling the neutron flux is
critically important, and can be achieved through such absorption.
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Reproduction Constant, K:
For a given mass of fissionable material, neutrons can be lost to fission by
1. (n, γ ) absorption
2. escape from the surface of the material.
Taking these factors into account, we can define the reproduction
constant, K -
the average number of n' s from each
K≡
fission that cause another fission
For a self-sustained nuclear reaction, K=1
If K<1 the reaction will die out.
If K>1 the reaction will increase in neutron flux, and energy
output, and could result in a nuclear meltdown, or, in the
extreme, a nuclear explosion.
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Critical Mass
fissionable
material
For neutrons produced by fissions in a mass
of fissionable material of dimension d:
d
- the net production rate (production rate – absorption loss rate) is
~proportional to the volume of the material, which is ~proportional to d3
- the loss rate is ~proportional to the surface area of
the material, which is ~ proportional to d2
3
2
Thus as d increases, d increases faster than d . More n’s are produced by
the increasing volume, but a smaller fraction of them is lost by leakage
through the surface.
The critical mass is the mass for which K = 1, and a self-sustained
nuclear reaction is achieved in the mass of fissionable material.
A nuclear explosion can be caused when the mass is suddenly
increased beyond critical – super-critical mass!
In this case, we get a runaway chain reaction.
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Chain Reaction For Super-critical Mass (K=2):
The n’s released in a n-induced fission can collide with other fissionable
nuclei to produce further fissions, which produce further n’s, which
cause yet more fissions.
We have assumed
that an average of 2
n
n’s from each fission
n
induce another fission
n
(K=2).
n
n
In this case, the
n
n
number of fissions
n
increases as 2n
where n is the
n
n
n
number of the
fission = 1,2,3,4…
n
n
This is a positive
n
feedback, avalanche
n
process called a
chain reaction.
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Controlled Nuclear Reaction:
If the chain reaction can be controlled, the energy released by fission,
can be utilised for energy generation.
This involves control of the neutron flux producing new fissions.
We need to design the reactor for an excess of neutron flux (K>1),
which can be diminished to K = 1 by the controllable absorption of
excess neutrons.
These conditions have been achieved in the nuclear reactor.
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Nuclear Reactors
The following diagram shows a simplified nuclear reactor cross-section:
Moderator
Fuel rods
Control rods
Shielding
Base
There are several design and operational issues that need to be taken into
account for the reactor to go critical, and then be controllable in operations.
These factors relate to neutron flux.
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Fuel Rods:
The essential component in the fuel rods is enriched uranium.
Naturally occurring uranium consists of a mixture of two isotopes 238U (99.3%) and 235U (0.7%). However only the 235U is fissionable by
thermal neutrons.
Furthermore, the 238U absorbs neutrons, thereby removing neutrons
from the fuel rods, and stifling the chain reaction.
For the reactor to go critical, the fraction of 235U must be increased
(enriched) up to a few %, which requires special technology.
Plutonium (94239Pu) is also fissionable, and can be used in fuel rods.
Pu-239 does not occur naturally, but can be produced from U-238 by
n-absorption, followed by two steps of β-decay.:
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238
92
239
U + 01n →92
U∗
239
92
239
U ∗ →93
Np +
239
93
Np →
239
94
Pu +
0 -1
0
-1
-
+
+
0
0
0
0
Nuclear (© F.Robilliard)
(T = 23min )
(T = 2.3days)
1
2
1
2
72
Moderator:
Neutrons produced by fissions in one fuel rod, can produce fissions
in other fuel rods.
The fuel rod array is immersed in a moderator, which slows these
neutrons to thermal energies in this transit between rods.
The probability of n-absorption by 235U increases as the neutrons
slow, whereas, for the 238U, the opposite is the case - absorption
decreases for slower n’s.
Thus, the probability of fission in uranium-235 is significantly
enhanced by the action of the moderator.
Graphite and heavy water (deuterium dioxide) are used as
moderator materials.
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Control Rods:
If we make the fuel rods sufficiently enriched, and immerse the fuel
rod array in a moderator, the reactor core will go super critical. That
is, we will have a meltdown situation!!
To prevent this, we must reduce the neutron flux in the reactor core.
This is achieved by inserting rods composed of a material that will
readily absorb neutrons.
These control rods are inserted into the reactor core from the top.
The level of n-absorption depends on how far into the core the rods
are inserted,
Thus we are able to control the reproduction factor, K, of the reactor.
K = 1 is the operational condition.
K < 1 will shut down the reactor.
Cadmium is an efficient neutron absorber used in control rods.
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Power Generation:
The heat of the reactor core is used for electrical power generation.
This is commonly achieved by using heavy water as the
moderator. This is cycled through the core and then through a heat
exchanger, where its heat energy is used to generate steam, before
returning to the core.
The steam so produced, is then used to drive a steam turbine, which
is connected to a generator for electric power generation.
The steam is then condensed and recycled to the heat exchanger.
So that the moderator material does not vaporise, it is operated
at high pressure.
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