alternative sources of energy: nuclear fission and fusion

Alternative Sources of Energy:

Nuclear Fission and Fusion

Thesis by Gari Jose Ciodaro Guerra

Degree: Physicist

Supervisor:Neelima Govind Kelkar

Faculty of Science

Department of Physics

June 25, 2013

Acknowledgments

I would like to thank my assessor during this process, Neelima Kelkar, who guided

me, and helped me with all my questions. To my parents, Garibaldi Ciodaro and

Lida Rosa Guerra whose hope and effort, during all states of my life lead me to this

moment. Finally, I would like to thank my grandfather and grandmother for loving

me the way they did.

3

Contents

Acknowledgments 3

1 Introduction 9

2 Essential concepts in nuclear physics 15

2.1 Nuclear Radii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 General Nuclear Reactions . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Cross Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.4 Binding energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5 The Bethe Weizsacker Semi Empirical Mass Formula . . . . . . . . . 24

2.5.1 Construction of the formula . . . . . . . . . . . . . . . . . . . 25

2.5.2 Some implications of the mass formula . . . . . . . . . . . . . 30

2.6 Nuclear Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.7 Nuclear Fission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.7.1 Spontaneous Fission . . . . . . . . . . . . . . . . . . . . . . . 32

2.7.2 Neutron Induced Fission . . . . . . . . . . . . . . . . . . . . . 34

2.7.3 Symmetry of Fission Fragments . . . . . . . . . . . . . . . . . 36

2.7.4 The Emitted Neutron . . . . . . . . . . . . . . . . . . . . . . . 38

3 Nuclear Fission Reactors 43

3.1 Thermal Reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.1.1 Heterogenic Arrangement . . . . . . . . . . . . . . . . . . . . 45

3.1.2 Breeder Reactor . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.1.3 Boiling Water Reactor BWR . . . . . . . . . . . . . . . . . . . 48

3.1.4 Fast Breeder Reactors . . . . . . . . . . . . . . . . . . . . . . 49

5

6 Contents

3.1.5 Gas Cooled Reactor . . . . . . . . . . . . . . . . . . . . . . . 50

3.1.6 High Temperature Gas Reactor . . . . . . . . . . . . . . . . . 50

3.2 Nuclear Reactors World Wide . . . . . . . . . . . . . . . . . . . . . . 52

4 Nuclear Fusion 55

4.1 Key points of fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.1.1 Energy distribution of fragments during fusion . . . . . . . . . 56

4.1.2 Reaction Rate and important concepts . . . . . . . . . . . . . 57

4.2 Fusion Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.3 Important Fusion Reactions . . . . . . . . . . . . . . . . . . . . . . . 60

4.3.1 Advanced Fusion Fuels . . . . . . . . . . . . . . . . . . . . . . 60

4.3.2 Main Controlled Fusion Fuels . . . . . . . . . . . . . . . . . . 61

4.3.3 P-P cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4.3.4 Carbon Nitrogen Oxygen cycle . . . . . . . . . . . . . . . . . 63

4.3.5 Carbon Carbon Reactions . . . . . . . . . . . . . . . . . . . . 63

5 International Thermonuclear Experimental Reactor 65

5.1 Principal in Nuclear Fusion Reactor . . . . . . . . . . . . . . . . . . . 65

5.1.1 The Lawson Criteria . . . . . . . . . . . . . . . . . . . . . . . 66

5.2 The Tokamak Design . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

5.2.1 Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

A General Nuclear Reactions Calculations 75

List of Figures

1.1 One possibility for U-236 to fission . . . . . . . . . . . . . . . . . . . 11

1.2 Fusion as main form of energy . . . . . . . . . . . . . . . . . . . . . . 11

2.1 Experimental Charge Density (e fm−3) as a Function of r(fm)[2] . . . 16

2.2 Flux of Particles[3] . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Cross Section, a)Elastic Scatteing b)Exothermic 1 c)Exothermic 2

d)Inelastic scatteing . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.4 Binding Energy per Nucleon[4] . . . . . . . . . . . . . . . . . . . . . . 24

2.5 Well potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.6 Illustrations of the different terms in the Bethe-Weizscker formula.[5] 28

2.7 Contributions of the terms[6] . . . . . . . . . . . . . . . . . . . . . . . 29

2.8 Nuclear Chart[7] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.9 Potencial Energy vs Average Distant of separation[2] . . . . . . . . . 33

2.10 Fertile Material[9] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.11 Percent yield of fragments in the spontaneous fission of Californium-

252 [11] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.12 Cross-section for the production of a mass number A fragment in

Uranium-238 [11] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.13 Fragments Distribution for Bismuth[12] . . . . . . . . . . . . . . . . . 38

2.14 Possible reactions for Uranium-235 + thermal neutron . . . . . . . . 39

2.15 CNO Cycle[13] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.1 Cross-Section for some nuclei [14] . . . . . . . . . . . . . . . . . . . . 44

3.2 schematic view of the heterogenic arrangement . . . . . . . . . . . . . 45

3.3 Outgoing neutrons per interaction[15] . . . . . . . . . . . . . . . . . . 46

7

8 List of Figures

3.4 Breeding Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.5 Schematic view of the neutron reflector . . . . . . . . . . . . . . . . . 47

3.6 Boiling Water Reactor [16] . . . . . . . . . . . . . . . . . . . . . . . . 48

3.7 Magnox [16] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.8 Two versions of the HTGR[17]. . . . . . . . . . . . . . . . . . . . . . 52

3.9 Word distribution of nuclear reactors[18] . . . . . . . . . . . . . . . . 53

4.1 Deuterium-Tritium[13] . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.2 Coulomb Barrier[19] . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.3 Proton+Boron fusion reaction [26] . . . . . . . . . . . . . . . . . . . . 61

4.4 p-p Cycle[25] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4.5 CNO Cycle[13] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.6 Important Fusion Reactions[24] . . . . . . . . . . . . . . . . . . . . . 64

5.1 Magnets [28] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.2 Vacuum Vessel [28] . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.3 Blanket[28] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

5.4 Divertor[28] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5.5 External heating systems[28] . . . . . . . . . . . . . . . . . . . . . . . 71

5.6 Diagnostics[28] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

5.7 cryostat[28] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

Chapter 1

Introduction

The quest for alternative sources of energy has been one of the biggest challenges

of all time. From the beginning of the industrialization era levels of green house

gases have been growing at a dramatic rate. According to the United States envi-

ronmental protection agency [1] the amount of carbon dioxide (fossil fuel related)

has grown 16 times from 1900 to 2008. They also report that energy supply, industry

and transportation contribute 26%, 19% and 13% of the total green house emissions

respectively.

The need for alternative (renewable) forms of energy is indisputable. The present

document will deal with one such option, namely, nuclear power. This type of energy

in particular has generated much controversy, specially because of the accidents that

nuclear plants have had in the past. For example, due to Fukushima disaster, Ger-

many closed seven nuclear power plants. It is estimated that for every one put out

of service 11 million tons of carbon dioxide emission added to the current balance,

so any member of the scientific community would be wondering if the ”solution”

was better than the problem, especially if the models of climate change are right,

they predict that a rise of two Celsius can be manageable but a three Celsius change

would be disastrous. Countries like Holland and Belgium would be under water as a

result of the permafrost melting, the ecosystem at the poles would collapse as well as

natural cycles of mountain water could be disturbed. Without international legisla-

tion and commitment any effort to control climate change is futile. For example, tar

sands of Canada would produce twice as much green house emission than conven-

9

10 Chapter 1. Introduction

tional Saudi oil 1 extraction, it is recipe for disaster and nobody seems to care about

it. Generally speaking most sources of renewable energy, like solar, wind, biofuel

are not sufficient to cover the world’s hunger for energy, at least for the moment, so

nuclear power still seems to be a good option.

The first nuclear power mechanism that comes to mind, fission, was discovered in

1938 after five decades of work in radioactivity and nuclear theory, it can produce

vast amounts of energy in spite of the fact that out of 1kg of uranium only 0.87g

would transform into energy.

E = mc2 = 0.87× 10−3 ×(3× 108

)2= 7.81013J/kg (1.1)

While through chemical reaction, 1 kg of gasoline would produce:

E = 5× 107J/kg (1.2)

As we can see, nuclear fission releases energy six orders of magnitude bigger than

gasoline and produces virtually zero gas emission. The problem with fission is that

it is very unstable, to have an idea of why that is, we need to explore the general

process. Basically a neutron is place near a uranium or plutonium radioactive nu-

cleus, just with enough energy to make it unstable, this kind of reaction is called

neutron capture, this would cause the nucleus to break apart, releasing an amount

of energy proportional to the mass defect2. The trouble is that, it also liberates on

average other three neutrons, inducing an uncontrolled chain reaction (this is the

principle of the atomic bomb). Those unwanted neutrons must be removed in order

to control the system, requiring sophisticated engineering techniques vulnerable to

human error and the unpredictability of the real world.

An other difficulty with fission is the waste products, lets see for instance the

reaction of uranium 236:

1the reason for it, principally oil from tar sand, is thicker and required more heating to make ittransportable.

2This concept would be developed in later chapters.

11

Figure 1.1: One possibility for U-236 to fission

The long half-life of cesium-137, makes it an example of radioactive waste from

a fission reactor. Long exposure to radioactive materials can cause the DNA to mu-

tate, producing cancer, another yet not less important issue with fission.

To summarize the above discussion; fission plays a major role in keeping the

energy supply in the world, controlling in an indirect way the green house emission.

On the other hand, the resulting products of it, can be dangerous to human health in

case of leaks and accidents. So unfortunately, without any innovation this technology

seems doomed, at least in the producing energy field. However we cannot just shut

down all nuclear power plants at once, it has to be done at the same rate as we find

other sources of clean energy to replace them.

Figure 1.2: Fusion as main form of energy

Nuclear fusion as we can see, has a long range influence in many of the earth

12 Chapter 1. Introduction

processes, even fossil fuel like coal and petroleum[2] were created by photosynthesis,

so it can be thought to be a kind of solar stored energy. Making it a sustainable

source of energy has proved to be a formidable challenge, so difficult to accomplish

that it could bring together countries which are otherwise enemies. In general nuclear

fusion is the union of two light nuclei, just as fission is the division of two heavy ones,

the main difference between those two nuclear phenomena is that fission is prevented

by surface tension of the nucleus 3 and fusion is prevented by the coulomb repulsion

force from the positively charged protons. In order to overcome this force the best

chance is introducing heat to the system, that last one detail is the real problem of

fusion, in many experiment the amount of energy used to heat it is bigger than the

output energy, but scientists have not yet renounced to the idea of a clean abundant

source of energy as fusion is. The principal fusion fuels are the isotopes of hydrogen,

Deuterium, one proton one neutrons, Tritium, one proton two neutrons and 3He two

protons one neutron, the most common fusions reactions are the following.

D + T → 4He(3.52MeV) + n(14.06MeV) (1.3)

D +D → T (1.01MeV) + p(3.02MeV) (1.4)

D +D → 3He(0.82MeV) + n(2.45MeV) (1.5)

D + 3He→ 4He(3.67MeV) + p(14.67MeV) (1.6)

Deuterium is abundant in nature, since it is possible to find 1g of it in 300 liters

of sea water, on the other hand Tritium is unstable and is commonly found in the

atmosphere formed via cosmic ray interaction, it has half-life of 12.3 years, and

it decays thought beta interaction with average energy of 0.0057MeV and without

producing a gamma ray, so it does not happen in useful quantities on the surface

of the earth. However Tritium can be obtain relatively easily in heavy water. Fi-

nally 3He, this isotope can be found on the surface of the moon, but mining the

moon is not a clear option for the near future, so early generations of fusion reactors

would aim to produce the D+T reaction, since it has the biggest exothermic energy

for the neutron. It would have to obtain the tritium fuel via neutron reaction in

3If we apply the liquid drop model

13

Lithium. If the D-T reaction were made possible about “10 to the 12”GJ of energy

could be produced, that is about 10 to the power 11 times the entire world elec-

tricity production per year. A fusion reactor would require a metric ton of fuel per

GW, and would produce helium as waste, while if wanted to produce the same GW

of energy with coal, we would need 2 million times more fuel, what is more since

the weight of the fusion fuels are nothing compared with the weight of coal, the

transportations infrastructure would suffered a lot less. Compared to fission, fusion

power plants could not be cloaked to produce materials for nuclear bombs, since in

fusion the gamma radiation is absent, any trace of it would indicate an undeniable

plan of making a bomb, problems such as Irans uranium enrichment could easily

be classified as a thread or not. The use of the land is another point in favor of

fusion, while solar or biomass need vast amount of land fusion can do it in relatively

smaller areas. Another problem that fusion has is the highly energetic neutron.

While in fission neutron energy distribution is of a few electron-volts, we can see

from the D-T reaction that almost all neutrons are created with 14.1 MeV, which

implies that any material used as first wall in the reaction would have to survive the

atomic dislocations and nuclear reactions induced by these highly energetic neutrons.

ITER (International Thermonuclear Experimental Reactor) is doing the most

ambitious project in this field. They claim that for 2022, the first commercial ver-

sion of a nuclear fusion reactor would be ready. In order to reach the temperatures

required to overcome coulomb repulsion (20-40 KeV) they are going to use a tech-

nique called magnetic confinement, in which the plasma (charged gas) follows the

path of the magnetic field generated by a giant solenoid. The design is based on a

Russian device called the tokamak.

Chapter 2

Essential concepts in nuclear

physics

2.1 Nuclear Radii

The radii of proton and neutron are known to be of the order of 1fm. If we think of

nucleons as solid spherical particles, locked inside a solid sphere then we can write:

A ∼=43πR3

43πr3→ R ∼= r0A

1/3 (2.1)

Where r was changed to r0, to acknowledge the fact that even if the nucleon

spheres are packed, there should be empty space between them, so that the vol-

ume of the nucleus should be bigger than the sums of each sphere, as consequence,

r0 > r = 1fm.

It is possible to check the above approximation, using electron scattering. An

electron is thrown against a target, interacting electromagnetically with it, accept-

ing that proton and neutron densities have the same shape1, the charge distribution

would be identically the same as the mass distribution, allowing as to determine the

radius of the nucleus.

1This is necessary, because neutron do not have charge to deflect the electron path

15

16 Chapter 2. Essential concepts in nuclear physics

R.Hofstadter and his collaborator, developed the above technique, finding the

charge or mass density per unit volume described by:

ρ(r) =ρ0

1 + e(r−R1/2)/a(2.2)

Where ρ0, R1/2 and a are fitting parameters. Equation 2.2 has the form of a

Fermi-distribution.

Figure 2.1: Experimental Charge Density (e fm−3) as a Function of r(fm)[2]

It is clear that figure2.1 show us that 2.2 is not totally true, if the charge density

had been constant for 0 < r < R and then zero for r > R then, that would have been

the case. We see that, for A > 40, with ρ0 = 0.15 nucleon fm−3, the charge density

is nearly independent of A, this phenomena is known as saturation of nuclear force,

a concept needed to developed nuclear models.

2.2. General Nuclear Reactions 17

2.2 General Nuclear Reactions

In general we have a nuclear reaction as follows:

a+X → Y + c (2.3)

An incident particle interacts with a target nucleus, producing a new nucleus and

another particle. If particle “a” does not exist then we have a decay. We can use

energy conservation to develop some general properties about the above reaction.

Ei = Ea + c2 (m0a +m0X) (2.4)

Ef = Ec + c2 (m0Y +m0c) + EY (2.5)

Equating and rearranging:

c2 (m0a +m0X − (m0Y +m0c)) = ∆Ekinetic = Ec + Ey − Ea = Q (2.6)

We see that if we are talking about masses this is equal to the mass defect ; if we are

talking about energy we have the binding energy. If Q is bigger than zero we have an

exothermic reaction, the new nucleus would be lighter and the kinetic energy would

be bigger, on the other hand if the Q is negative, then we have an endothermic reac-

tion. In order to produce an endothermic reaction the projectile would have to have

a minimum energy corresponding to the mass of the new prodructs. To calculate

this threshold, we make use of the invariance of the dot product in the Minkowski

space, using frames that ease the calculation, getting:

Eth =(m0Y c

2 +m0cc2) 2 − (m0a

2c4 +m0X2c4)

2m0Xc2(2.7)

Equation A.6 is the total threshold energy, so:

Eth = EthKintic +m0ac2 (2.8)

EthKinetic =(m0Y c

2 +m0cc2) 2 − (m0ac

2 +m0Xc2) 2

2m0Xc2(2.9)


Using A.5 and taking into account that the binding energy is a lot less than rest en-

ergy of the target nucleus, we arrive at an approximated expression for the threshold

energy2.

EthKinetic = −Q(m0a +m0X

m0X

)(2.10)

2.3 Cross Sections

Let us consider a beam of particles passing through an area A, with a width dx as

figure 2.2 shows

Figure 2.2: Flux of Particles[3]

Here we can define:

I =(Number of Particles)

m2sArea = φA (2.11)

If we assume a reaction occur between the incident particles and the number N

of nuclei in the volume3 A.dx, then the reaction rate will be proportional to those

two parameters4:

2To see all the algebraic developed please refer to appendix A3N can be expressed as the number of atoms per unit volume times dxA4provided the nuclei act independently

2.3. Cross Sections 19

Rate ∝ Nφ (2.12)

The proportionality constant is known as the cross section:

σ =event rate per nucleus

incident flux(2.13)

So it can be inferred that σ is the equivalent to the probability of the process to

occur. Calculation of cross sections is a difficult task, since it requires the knowledge

of the interaction involved, however we can still try to see some of its general features.

Let us analyze the cross section of a general reaction like A.1. The initial quantum

mechanical state is defined by the initial particle’s momentum, angular momentum

and polarization. The final states, however form a continuum, just as in atomic

physics, when the leaving photon of an exited atom does not have a prefered direction

of escaping, so every possibility must be taken into account. Using Fermi Golden

rule for scattering5 we can write the transition probability per unit time as:

w =2π

~<| Hif |2>

dn

dE(2.14)

Where <| Hif |2> is the average square value of the matrix element connecting

the initial and final states. This term is sometime called the amplitude for the

process. It contains all the dynamical information. The other part is called the

density of the final states or phase space available if recall statistical physics. At

first sight, equation 2.14 seems to give a divergent result, since as stated before, the

density of final states form a continuum, it means that dndE

goes to infinity. To avoid

this, let us enclose the system in box of volume Ω, it would restrict the levels of the

final states as follows:

dn

dE=

2Ω

(2π)2~3p2cdpcdE

(2.15)

5Pag 208 Introduction To Elementary Particles. David Griffiths


It is almost impossible to calculate Hif explicitly since it depends on the nuclear

structure, however we shall infer some of its general properties. In scattering theory

from quantum mechanics, we have in general, a initial state represented by a plane

waves φi and spherical waves φf of the form eikr/(Ω)1/2 representing the final state,

if we call U , the interaction energy, we would have by definition:

Hif =

∫φf∗Uφidτ

∼=< U >V

Ω(2.16)

where V is the volume of the nucleus. Equation 2.16 is incomplete, due to

quantum mechanical nature of the phenomena, a tunnel probability of crossing the

coulomb barrier for charged particles, must be taken into account. This term is

called the Gamow factor:

Ga =1

~

∫(2ma (Va − Ea)) 1/2 dr ∼= πZXZa

e2

~va(2.17)

Therefore we get:

Hif∼=< U >

V

Ωe−Ga−Gc (2.18)

using the definition of probability, that is, favorable events over the total, we can

combine equations 2.12 with 2.15 as:

rate

N= w (2.19)

Replacing all the values in 2.19, and using dE = vcdpc we obtain

σX→Y =1

π~4Ω2 <| Hif |2>

p2c

vavc(2IY + 1) (2Ic + 1) (2.20)

2.3. Cross Sections 21

where va is the relative velocity of a with respect to A, vc velocity of c with

respect to Y, and pb, momentum of b. The final part of the equation was another

correction due to spin degeneracy. In spite of its complexity it is possible to deduce

very important properties about any nuclear interaction, since equation A.1 was, in

every way, general.

• Elastic scattering.The two interacting particles are neutral, conservation of

energy leads to va = vb. At low energy, for uncharged particle we have σ

constant. This reaction can be seen as to ”back where we started”, for every

practical purpose.

• Exothermic reaction 1. If the incoming particle is uncharged and energet-

ically small, we have that p2cvavc∝ 1/va since vc ∼= constant, accordingly to the

difference of million electrons volts, between Q and the energy of a slow neutron

for example. The matrix of transitional states is proportional to exponential

of the Gamow factor, for uncharged particles Ga = 0 so e−Gc ∼= constant. The

above considerations tell us that, the cross section for this kind of interaction

has a 1/va behavior.

σX→Y ≈ 1/va (2.21)

Slow neutron-fission enter in this category, We see that due to this behavior,

the smaller the kinetic energy of the neutron the bigger the probability of hap-

pening, but as we will see in the fission section, the neutron energy cannot be

smaller than activation energy of a giving nuclei, therefore limiting the cross

section.

• Exothermic reaction 2. If the incoming particle is charged, we will have the

same 1/vb behavior as before, but this time the Gamow factor would play a

major role in the cross section, since now, none of them vanished. Therefore:

σX→Y ∝ (1/va)e−2Ga (2.22)


Fusion enters in this category, we can see that the heavier the interacting nuclei

or particles the probability lower its value exponentially, while increasing the

kinetic energy improves the chance of occurrence. By replacing some values

for D + D reaction, we see that in order to have reasonable chance of hap-

pening we would need a kinetic energy of about 10KeV, using the Boltzmann

constant we see that we would need temperature of about 116099275.4 K, to

have a reasonable chance of occurrence for this reaction, that is why fusion is

so energetically challenging, this would be span in the fusion section.

• Inelastic scattering(n,n’).

The incoming particle leaves the nucleus in a exited state, the process is en-

dothermic. The incoming particle has a energy just above the thresholds energy

equation 2.10, that is vn ≈ constant, but v′n would be proportional to square

root of the excess energy above the threshold, which lead us:

σX→Y ∝ (energy excess)1/2 (2.23)

in figure 2.3, the energy starts in the threshold, this is the cross section for a

fast neutron induced fission.

2.4. Binding energy 23

Here we can observe the qualitative behavior of the cross section vs energy, for

the explained types of scattering.

Figure 2.3: Cross Section, a)Elastic Scatteing b)Exothermic 1 c)Exothermic 2d)Inelastic scatteing

2.4 Binding energy

The binding energy is one of the most important concepts in nuclear energy, since

it tell us whether fusion or fission are plausible as a source of energy. The binding

energy of a nucleus is the difference between the rest energy of its constituents and

the rest energy of the actual nucleus.

Q = B(A,Z) = Nmnc2 + Zmpc

2 −m(A,Z)c2 (2.24)

We have saturation of binding energy near iron, moving in the graphics we see that

fusionable nuclei are theoretically from lighter to the iron nucleus, and fissionable

are from the heavier to the iron nucleus.


Figure 2.4: Binding Energy per Nucleon[4]

2.5 The Bethe Weizsacker Semi Empirical Mass

Formula

One of the biggest challenges of theoretical nuclear physics was to get a formula

that accounts for the binding energies of the nuclei. In 1935 Bethe and Weizsacker

formulated a semi empirical equation that relies on the liquid drop analogy, which

basic assumption is that the nucleus is a charged, nonpolar liquid drop held together

by the nuclear force. 6, where we can observe the following properties:

• the charge is to a good approximation equally distributed in the nucleus, this

implies that the nucleus cannot be compressed when it is at its ground state

(Constant average binding energy).

• the nuclear force is a short range type.

• every nucleon is equally bound except the ones on the surface, since they have

less neighbouring nucleons.

6http://courses.chem.indiana.edu/c460/documents/SEC8LDModel000.pdf

2.5. The Bethe Weizsacker Semi Empirical Mass Formula 25

2.5.1 Construction of the formula

• Volume Term: there is a term due to the fact that for nuclear force, neutrons

and protons are basically the same, it is written as follows:

Evol = avA (2.25)

The reason for calling it the volume term is because A, the atomic mass, is

proportional to the volume, just as the liquid droplet has the same energy prop-

erties despite the number of molecules that form it (Average binding energy

independent of A)

Volume = 4πR3

3= 4πr3

(A1/3

)3= 4πr3A (2.26)

It is clear that in the absent of any other consideration, equation 2.26 would

give us a 16 eV biding energy per nucleon (r=1.2 fm).

• Surface Term: There is a surface term that would prevent the nucleus from

spontaneous fission (explained later), this term makes the shape of the nucleus

a spherical one, since it minimizes the energy. It would be negative since it

is essentially a subtraction from the volume term regarding the limits of the

drop, keeping in mind the analogy, we have:

Es = −σArea = −4πσR2 = −4πr2A2/3 (2.27)

where sigma is the tension constant.

Es = −acA2/3 (2.28)

The surface term plays a major role in heavy nuclei, since the number of

nucleons in their boundaries, compare to those in the bulk, is big.

• Coulomb Term Due to the present of positively charged particles as proton,

there is Coulomb repulsion energy. Ignoring the other types of interactions, we

have that the amount of work needed for arranging the protons in a nucleus is:


dw =4

3πR3ρ ∗ 4πr2dr ∗ ρ

4πε0R(2.29)

where ρ, the charge density for r < R is:

ρ =Ze

43πR3 (2.30)

Itegrating 2.29 from 0 to R

W =3

5

Z2e2

4πε0R(2.31)

if we subtract the energy needed to arrange the first proton7, that is putting

z=1 in 2.30, we obtain:

W =3

5

e2

4πε0R

(Z2 − Z

)=

3

5

e2

4πε0RZ(Z − 1) (2.32)

Expressing 2.32 in terms of A, and setting constant:

EC = −acZ(Z − 1)

A1/3(2.33)

Due to the Z2 dependence of 2.33, large nuclei lose a considerably large amount

of biding energy due to this therm. It is somehow clear, that for large stable

nuclei to exist, it is necessary to overcome this term, this can be achieved by

adding neutron to heavy nuclei, increasing the nuclear interaction.

• Symmetric term. Light stable nuclei are accepted to have z< 20 and a z/N

radius which tends to one. For heavy stable nuclei we have 20 < z < 83 and

a z/N radius tending to 1.5. Acknowledging the mentioned fact, it is clear

the semi empirical formula must have a term which account for it, that is the

symmetric term. To help us understand the phenomena, let’s pretend we have

a two potential well with the same energy levels, one occupied by neutrons and

7There is no work in bringing the first proton


one occupied by protons. Both proton and neutrons are fermions, so they have

to follow the Pauli Exclusion Principle:

Figure 2.5: Well potential

From figure 2.5 we have:

∆E = E(N 6= Z)− E(N = Z) (2.34)

Where we deduced:

∆E =(N − Z)

4

2

ε (2.35)

Putting 2.35 with a constant coefficient, and dividing it by A square

Esym = −aa(N − Z)

A2

2

(2.36)

• Parity term. There was observed a small but systematically different between

the values calculated with the other terms of semi-empirical formula and the


ones actually observed. They found that for even Z - even N nuclei gain binding

energy, odd-A nuclei are not affected and odd Z - odd N nuclei lose binding

energy, giving rise to:

EP = apδ

A1/2(2.37)

where δ = 1 for even-even, δ = 0 for odd-even,even-odd combinations, and

δ = −1 odd odd.

• Final form.Adding all the terms we finally obtain the Bethe-Weizscker for-

mula.

B(A,Z) = avA− asA2/3 − acZ(Z − 1)

A1/3− aa

(N − Z)

A

2

+ apδ

A1/2(2.38)

Figure 2.6: Illustrations of the different terms in the Bethe-Weizscker formula.[5]


The coefficients can be obtained by fitting the formula for known nuclei.

av = 15.753MeV (2.39)

as = 17.804MeV (2.40)

ac = 0.7103MeV (2.41)

aa = 23.69MeV (2.42)apA1/2

= 33.5 (2.43)

If we plot all terms, the combined contribution, and the experimental data, the

following is obtained.

Figure 2.7: Contributions of the terms[6]


2.5.2 Some implications of the mass formula

General principles that can be deduced from mass formula

• If we derived the mass formula with respect to Z and A, setting the result

to zero in order to find the minimum, It would be realized that all nuclei are

thermodynamically driven to Fe, that is, the losses in binding energy due to

Coulomb interaction and surface term are balanced near Fe-56.

• For nuclei with high A, the primary losses in binding energy are due to the

large coulomb interaction, therefore splitting heavy nuclei by fission leads to

more stable products, releasing energy in the process.

• For nuclei with low A, the losses are mostly caused by the surface term, that

is, the smaller the nucleus the more nucleons would be in the surface, there-

fore adding two light nuclei would give a daughter nucleus whose fractions of

nucleons on the surface would be smaller, giving rise to more stable nucleus.

This is the principle of fusion, a process that runs the physics of stars.

• The symmetry term favor N/Z=1. This behavior is observed for light nuclei

up to Ca-40. For heavier nuclei the ratio is bigger. This tendency is due to

fact that larger A for a given Z lowers the coulomb energy loss i.e the square

dependence of coulomb term.

• The parity term indicates that even numbers of neutrons and protons gain

additional binding relative to those with an odd nucleon.

2.6. Nuclear Stability 31

2.6 Nuclear Stability

We see that N/Z ratio makes a straight line for light nuclei, but as we increase the

mass, the line is not straight anymore. The ratio changes to 1.5, the dark squares

indicate more stable nuclei and the zig-zag lines above and below them are the proton

and neutron drip lines respectively.

Figure 2.8: Nuclear Chart[7]

2.7 Nuclear Fission

As was stated in the introduction, nuclear fission is one of the most important con-

cepts with practical implications in physics. There are many ways fission can occur,

such as photo-fission, neutron induced-fission and spontaneous fission, every one of

them is governed by the liquid drop model of the previous section. Disregarding the

type, some features of the reaction can be inferred.

Let us explore the theoretical concept of symmetric fission 8. Using the liquid

drop model we have:

B(A,Z)− 2B(A/2, Z/2) = 17.2A2/3(1− 21/3

)+ 0.70

Z(Z + 1)

A1/3

(1− 2−2/3

)(2.44)

8Asymmetric fission is more likely to happen. Stability of magic number


Equation 2.44 is positive for heavy nuclei, telling us that the difference, would be

the energy release in the reaction, however it does not tell us how it happened.

2.7.1 Spontaneous Fission

First identified by Soviet physicists in 1940 [8], it is a rare type of decay, that is, the

probability is very small compared to alpha decays in nuclei having Z ≈ 92, such

as Uranium, but is the main channel for heavier nuclei such as californium-254. To

understand this phenomenon, let us study B(A,Z) from equation 2.44, under small

deformations of its surface, while keeping its volume constant. The shape of the

nucleus is initially spherical, if we deform it in small amount, its shape would change

to an elipse. That is, major axis a = R(1 + ε), minor b = R(1 + ε)−0.5 with constant

volume V = 43πab2, then the new surface would be:

S ≈ 4πR2(1 + (2/5)ε2

)(2.45)

where we identify the factor ε as the one which accounts for the deformation.

This leaves us with ES = 17.2A2/3(1 + 25ε2). The Coulomb term also changes. By

calculating the electrostatic energy we get:

Ec ≈3

5

e2Z(Z − 1)

R

(1− ε2

5

)(2.46)

Writing the above results in a more familiar form

ES = E0S

(1 +

2

5ε2)

(2.47)

EC = E0C

(1− 1

5ε2)

(2.48)

Equation 2.47 and 2.48 represent the energy due to the deformation, where the

0 superscript indicates the energy term of the spherical nuclei. If we now calculate

the energy change during this process we get:

∆EC = E0C − EC (2.49)

2.7. Nuclear Fission 33

∆ES = ES − E0S (2.50)

We can infer then, that if 2.49 and 2.50 are equal, we have a nucleus that has a

balance of the nuclear surface energy term and the coulomb energy term, so equating

them we obtain:

E0C

2E0S

= 1 = acZ(Z − 1)

2A1/3A2/3(2.51)

Which leads to:Z2

A≈ 50 (2.52)

Eex = 6.8MeV > 6.6MeV (2.53)

Equation 2.52 tell us that any nuclide with a ratio Z2/A less than 50 is stable.

this establishes an upper limit for nuclei, since above this number every nucleus

should undergo spontaneous fission, However due to quantum tunnel effect near this

number spontaneous fission could occur, but as we move further from this point the

cross section decreases rapidly.

Figure 2.9: Potencial Energy vs Average Distant of separation[2]


The difference between the energy of the spherical ground state and lowest excited

state energy is called EA, activation energy, a parameter used to determine if the

nucleus would fission.

2.7.2 Neutron Induced Fission

We saw in the previous section that for fission to occur with reasonable chance, an

energy above of EA must be given to the system. Let us see the example of Uranium-

235 when it captures a slow(thermal) neutron forming the compound excited state

Uranium-236 :

235U + n→(

236U)∗

(2.54)

The excitation energy of the Uranium-236 can be calculated directly:

Eex =(m((

236U)∗)−m ( 236U

))c2 (2.55)

Taking into account that the entering neutron had essentially zero kinetic energy,

we can infer the mass of the compound U*-236 nucleus:

m((

236U)∗)

= m(

235U)

+mn = 235.04392u+ 1.008665u = 236.05258u (2.56)

Replacing in equation 2.55 we find that Eex = 6.5MeV . The activation energy

for Uranium-236 is 6.2MeV , so we can conclude that Uranium-235 can be fissioned

through a thermal neutron. If we repeat the above calculation for Uranium-238, we

find Eex = 4.8MeV . The activation energy for Uranium-239 is 6.6MeV , so if we

wanted U-238 to fission, the missing 2MeV energy should be supplied by the neutron

through its kinetic energy.

Fissile material

Fissile materials are a group of nuclei that can fission through any slow neutron

reaction. Thermal neutrons are in equilibrium with the surroundings and have nearly

zero kinetic energy. In order for a material to be classified as such, their fission should


occur only with neutron absorption, that is, the energy imparted by the presence

of the new neutron should be enough to surpass the activation energy therefore not

adding kinetic energy to the system. Examples of fissile nuclei are Uranium-235,

Uranium-233 and Plutonium-239.

Fissionable material

Fissionable materials are all nuclei that can fision through either low-energy thermal

(slow) neuntros or high-energy (fast) neutrons. Therefore all fissile material fall into

this category, but now the nuclei that react only through high energetic neutron, are

also included. For example, Uranium-238 is a fissionable but not a fissile material,

while Uranium-235 is both.

Fertile material

All materials that can undergo transmutation during a reaction, to become fissile.

We can see some examples:

Figure 2.10: Fertile Material[9]


2.7.3 Symmetry of Fission Fragments

As we saw earlier the cross section indirectly tells us whether a particular reaction has

more chance to happene than another. This implies that fission reaction can yield

different results every time it takes place with different nuclei. Such behavior gives

rise to figure 2.11, we observe two peaks for the spontaneous fission of Californium,

one for A ≈ 110 and another for A ≈ 140. This isotope has A = 252, therefore

those peaks correspond to asymmetric fragments. The valley(A ≈ 125) is for the

symmetric result , so one can conclude that for heavy nuclei(Of the order A=238),

asymmetric fission is more likely to happen than symmetric.

Figure 2.11: Percent yield of fragments in the spontaneous fission of Californium-252[11]

The reason for such behavior is the shell structure of the nuclei9,that is, quali-

tative speaking, the probability of symmetric fission is proportional to the excited

energy of the fragments. Low energy interaction reduces the probability of symmet-

ric fission while high energy interaction should increases it. This can be verified with

the experimental data obtained for Uranium-238 bombarded with protons in range

of energy from 10MeV to 200MeV :

9The shell structure, is nuclear model that uses the Pauli exclusion principle, to organized thenuclear energy level in a way similar to the electrons for the atom.


Figure 2.12: Cross-section for the production of a mass number A fragment inUranium-238 [11]

It can be seen From figure 2.12 that for energies above 150 MeV the minimum

probability for symmetric distributions disappear, leaving a flat of probability for

fission fragments between A = 80 and A = 140. Such behavior is a hallmark of highly

fissionable nuclide. For lighter nuclei such as Radium, a three peak distribution

is found near the threshold energy for fission, making symmetric fission equally

probable to very asymmetric fission. For even lighter nuclei such as Bismuth, the


dominance of symmetric fission begin to arise as figure 2.13 shows.

Figure 2.13: Fragments Distribution for Bismuth[12]

It has been observed that for spontaneous fission and low energy neutron in-

duced fission only two fragments arise, however if the alpha particle is considered a

third fragment, then it is commonly observed that its energy(15MeV peak) is big-

ger than the common alpha emission. Its existence is explained by the reduction

in the Coulomb barrier due to the neck formation during the splitting (see figure

2.9). Some highly energetic particles can leave the nucleus during fission, if those

leaving particles are of the order of 5MeV average biding energy per nucleon, then a

phenomena called nuclear fragmentation occurs, due to the unbalance created by the

escaping particle. A theoretical description of that high energetic reaction cannot be

treated with the Bethe-Weizsacker formula, so we would not consider such scenario.

2.7.4 The Emitted Neutron

Emission of neutrons is our main concern in the energy production field; they carry

the energy that would be used to power the reactor. However, it is not necessary


that the neutron colliding with the nucleus will produce a fission reaction. Let us

see again the slow neutron interaction for Uranium-235:

Figure 2.14: Possible reactions for Uranium-235 + thermal neutron

Each possibility of figure 2.14, carries a cross section associated with it. They can

be divided into two groups absorption (fission and capture) and scattering (elastic

and inelastic). The total cross section would be:

σtotal = σfission + σcapture + σelastic + σinelastic (2.57)

In a reactor, we would need the fission cross section to be dominant. Fission

and capture have a behavior enclosed in the exothermic reaction 10 while elastic and

inelastic scattering were treated separately. We are concerned with the absorption

group since it will tell us the actual availability of neutron that can be used to have

a chain reaction, the following relation was stated experimentally:

ηA =σfission

σfission + σcapture

η (2.58)

For Uranium-235, reacting with thermal neutron, η, the average number of neu-

trons released is 2.4, so if we have natural Uranium 0.72% of 235, we would have the

number of available neutrons to induce fission ηA:

ηA =0.72σfission ( 235U)

0.72σabsorption ( 235U) + 99.28σcapture ( 238U)η = 1.328 (2.59)

10See 1.3 Cross section


For pure Uranium-235 ηA would be 2.06. As we can see, the availability of neutron

for chain reaction can vary through the enrichment.

Fusion

Another nuclear interaction of great interest in the energy production field is fusion,

capable of producing large amount of energy without the safety and environmental

concerns giving by fission. One of the most important advantages over fission is the

availability of the material to produce it, basically the hydrogen isotopes.

Reactions from 1.6 are some examples of fusion, other are:

n+ 6Li→ T + 4He + 4.8MeV (2.60)

D + 6Li→ 4He + 4He + 22.4MeV (2.61)

p+ 11B → 3 4He + 8.8MeV (2.62)

p+ p→ D + e+ + νe + 0.42MeV (2.63)

The first three are called terrestrial reactions, the fourth is the basic reaction

running in the sun. Most of the theoretical approach has been done already during

the topics above discussed, to summarize:

• From the curve of binding energy we see that, if we combine two light nuclei,

the binding energy per nucleon would increase, the difference of those values,

is the energy that can theoretically be used for building a reactor.

• The liquid drop model explains that if we have a light nucleus its suface energy

would be somehow big, that is, the primary losses in binding energy for light

nuclei are due to surface energy, since the fraction of nucleons in the surface

compared to those in the core is big, therefore adding two light nuclei would

give a daughter nucleus whose fractions of nucleons on the surface would be

smaller, giving rise to more stable nucleus.

• Fusion is an exothermic reaction, explained in the cross-section part, it was said

there, that for fusion reaction to have an acceptable cross-section highly tem-


peratures and densities must be achieved, since the Coulomb barrier between

the interacting initial particles must be overcome.

In the following chapters we will discuss the missing rather technical concept

of nuclear fusion, as are, reaction rates depending in the moderator, magnetic con-

finement, and inertial confinement by laser etc. For now let us see just the stellar

part.

Fusion Processes in the Sun

Stars are essentially giant fusion reactors, within them, there a is battle between

gravitational attraction due to its large mass, and the expanding force generated by

the fusion of light nuclei. There are two notable solar cycles, the first producing

about 25MeV is:

Figure 2.15: CNO Cycle[13]

In 2.15 the Carbon nucleus is not destroyed during the process, it only plays the

role of catalyst. The overall process is to combine four protons into an alpha particle,

expelling neutron and gamma rays. The other important cycle is:


1H + 1H → 2H + e+ + ν (2.64)

2H + 1H → 3He + γ (2.65)

3He + 3He→ 4He + 2 1H + γ (2.66)

In the stars the temperatures are about 15 million degree centigrade, giving

nuclear energies of the order of 1KeV. The particles in this giant confinement follow

a Boltzmann distribution of velocities. All atoms are ionized and no free neutrons

exist. We saw that the cross section for this scenario has a great dependence on the

Gamow factor, as result , the particles in the Boltzmann distribution with the highest

speed would react preferably, however this velocity distribution must be take into

account in the cross section, that is, introducing a factor e−E/KT . It can immediately

be inferred that even though the Gamow factor would favor high energy interaction,

the new factor would reduce the cross section for such increasing energy. That is

why is so difficult to maintain the rate of reaction, in the sun is achievable because

the number of events are as huge, so in principle, no matter how small the cross

section is, there always will be an acceptable occurrence of events. These cycles of

fusion are at the moment impossible on earth, since we do not have the mass of the

sun to maximize the probability, nor its temperature to help the Gamow factor. The

most promising for fusion on earth is the Tritium+ Deuterium reaction; it would be

treated in detail in the following chapters.

Chapter 3

Nuclear Fission Reactors

There are many types of fission reactors, each of them having different advantages,

risk, and costs. In this chapter we would go through some of them, making special

emphasis in the ones that are being used in the industry. There are some critical

parameters to take into account when nuclear reactors are approached, defined by

Dr.Michael P from MIT, those are:

• Temperature inlet and Temperature outlet

• Efficiency

• Thermo Cycle

• Moderator

• Neutron spectrum

• Materials

• Power density

• Special features

• Safety

• Commercial readiness

• Coolant

43

44 Chapter 3. Nuclear Fission Reactors

• Fuel

According to the type of fission they used, we can classify them into two groups,

Thermal and Fast reactors.

3.1 Thermal Reactors

As we saw in the fission section, the lower the neutron energy the biggest the cross

section for fission, depending on the target nuclei this increment would change, at

the same time that this is happening, the absorption cross section also increases as

well, but at a lower rate (see figure 3.1)3.1

(a) Neutron Induced Fission (b) Neutron Absorption

Figure 3.1: Cross-Section for some nuclei [14]

After the fission process takes place, high energy neutrons leave the reaction, in

order to lower the energy of this neutron to thermal levels, as required by the cross

section, a moderator is used. To lower the energy of neutron two ways might be

taken, elastic or inelastic scattering, however the cross section for elastic scattering

is preferred, since it does not decrease with the energy as the neutron gets near to

the thermal level(see figure 2.3), the mean logarithmic energy lost per interaction is

giving by:

ξ = 1 + 2

(A− 1

A+ 1

)2(

1−(A− 1

A+ 1

)2)−1

ln

(A− 1

A+ 1

)(3.1)

3.1. Thermal Reactors 45

As can be inferred from equation 3.1 a moderator with low atomic mass A is

better. Moreover it should have a small cross section for neutron capture. Some

examples of moderator are, Hydrogen ξ=1, deuterium with ξ=0.725 and Helium

ξ=0.425. It is clear that the energy lost per collision is less as the atomic mass

number increases; this is due to the kinetic energy conservation, with the energy

being divided between the moderator nuclei and the fast neutron.

3.1.1 Heterogenic Arrangement

In most cases the moderator is separated from the fuel. In the fuel region the predom-

inant dynamic is the diffusion of fast neutron, this mass transfer to the moderator

begins the elastic interaction that would send back those neutrons to the thermal

state. However as we saw earlier in figure 3.1, there is an energetic band before

the fission cross section, where the absorption of neutron is more likely to happen.

To avoid this, the moderator is placed around the fuel so that, only the surface of

the fuel interact with those neutrons leaving the core, where the fission rate must

be maintained at almost constant rate, nearly intact. This is called self-shielding.

Keeping the volume of the fuel constant while increasing the surface would result in

a better shielding1.

Figure 3.2: schematic view of the heterogenic arrangement

1Light-Water Reactors uses this arrangement.


3.1.2 Breeder Reactor

Depending on the production of neutrons, a reactor can be, theoretically self sus-

tained. Let’s recall equation 2.58.

ηA =σfission

σfission + σcapture

η (3.2)

Figure 3.3: Outgoing neutrons per interaction[15]

Depending on the energy as figure 3.3 tells us, the number of outgoing neutrons

can be bigger. The number of newly generated fissile nuclides per consumed fissile

atom is called Conservation ratio. It means we can produce fissile materials at the

same time as we consume fissile material. One example is the Uranium-238. If it is

present in the core, it can be transformed into Plutonium-239, another fissile nucleus.

For a thermal reactor to be considered as breeder, equation 3.2 should be bigger than

2.


Under appropriate operating conditions, the neutrons given off by fission reactions

can “breed” more fuel from otherwise non-fissionable isotopes. The most common

such reaction is that of Plutonium-239 from non-fissionable uranium-238.

Figure 3.4: Breeding Reaction

Figure 3.5: Schematic view of the neutron reflector

The reflector is usually made from the same material as the moderator; its func-

tion is to send back to the moderator and core, neutrons that try to leave the reactor.

If they pass the reflector, then a shielding material prevents this neutron, from leav-

ing the reactor and isolates it from outside radiation. To extract the energy a neutron

absorber is used. It also controls the power output. This is called neutron absorber.

There are several types of configurations or combinations of those components, and

we will study them when we investigate about specific types of reactors. Generally

speaking, in every reactor the fuel decreases and fission products accumulate, which

would eventually decreases the effective neutron multiplication factor. To control it,


the reactor is initialized with k > 1 and then adjusted to k = 1 using the neutron

absorber so that the reactor maintain in its critical state over a long term.

Another important part is the coolant. The coolant in a nuclear reactor is used

to remove the heat from the nuclear reactor core. The most commonly used coolant

is water.

3.1.3 Boiling Water Reactor BWR

As was stated before, this reactor uses thermal neutron to produce energy, through

fission reaction in the core. In general, this reactor uses as coolant, very pure water.

It travels from the bottom to the reactor vessel, increasing energy through heat

transfer during the process; this creates a mixture of steam a droplets, which are

separated, before the steam reaches the Steam line. Once purified, the steam goes

to a turbine which is connected to a electric generator, responsible of transforming

kinetic energy into electricity. The unused steam is sent into a condenser, where

the operators can decide if it sent back to the river for example, or pump into the

reactor for re circulation. The Jet pump and Recirculation pump are used to control

the amount of coolant entering the reactor, therefore giving a control to the power

output of the plant.

Figure 3.6: Boiling Water Reactor [16]


3.1.4 Fast Breeder Reactors

This type of reactor uses high energy neutrons to produce the fission reaction, if it

takes no place, then the scatter neutron will transform a not fissile nucleus into one

that is, making the reactor very sustainable. Lets mention some of its advantages in

respect to the thermal reactor.

Advantages

• Nuclear wastes have less radiotoxicity and half live.

• The fast neutron reaction produce more neutron per fission than a thermal

reactor, therefore giving the possibility of creating more fissile materials than

consumed.

• The fuel problem would be solved since a lot of materials can react through

fast fission reaction.

Advantages

• Designing this kind of reactor is more difficult since there are no moderator

to stabilized the reaction. At these energies the absorption cross section is

too small to even consider the possibility control rod implantation. Through

thermal expansion of the fuel2, or neutral poison3, stabilization can be achieved,

but it is technologically more challenging than thermal reactors.

• The cost of the fuel increases due to the high enrichment fissionable material.

In a thermal reactor the required enrichment is about 5%, in fast reactors

would be about 20%. This is a direct consequence of the large critical mass

necessary to start the chain reaction for fast neutron.

• Most experiment has used Sodium as coolant giving its high heat capacity.

However is very corrosive, and ignitable, giving rise to safety concerns.

2Depended respond of the reactor to the flight time of the neutrons across the core.3Neutron Poison is a substance with large neutron absorption cross-section.


3.1.5 Gas Cooled Reactor

As its name indicates, this reactor is cooled with gas, mostly CO2 and Helium.

It also uses graphite as moderator. In 1955 the first GGR was built in England,

it was called MAGNOX. The reason for its name is due to the special Mag-

nesium alloy (Magnox), of which the fuel cladding was made. It uses natural

Uranium as fuel and it is no longer built.

Figure 3.7: Magnox [16]

3.1.6 High Temperature Gas Reactor

The GCR was built in the 1950s in France and England, then the AGR was

devolved in the 1970s mostly in England. The idea of obtaining better effi-

ciency if higher temperature was achieved comes from those early days of gas

reactors. One motivation for it, was that for higher temperatures in respect

to the AGR, fossil-fired steam conditions can be attained, therefore improving

electricity production. Another motivations was the possibility of broader ap-

plications for nuclear power, such as providing industrial process heat.


All that was mentioned, motivated the development of the HTGR, a reac-

tor that Has Graphite as moderator, a ceramic fuel, and inert Helium gas as

coolant. The most remarkable characteristic of this type of reactor is form of

the fuel. Spherical kernels of Uranium, coated with multiple layers of refrac-

tory materials, after the fission occurs, these spheres acts like tiny think-walled

containment vessel. In terms of security the refractory materials gives an extra

way of limiting the probability of catastrophic t temperature- rising.

The U.S.A version of the HTGR, possesses an annular geometry with both

inner and outer graphite reflector. The spherical fuel, is contained inside rods,

in a somehow combination with previous reactor. the rods are inside a steel

vessel, connected through a cross duct to a steam generator, which is in another

vessel in a lower level of the reactor core, as the figure 3.8 shows4.

One the most important feature of this reactor is its safety concept. There are

three ways of preventing overheating incidents. The first one is the cross duct

that connect the heated gas to a condenser that in case of failure can be used

to send heat to the ambient, however, is the plant is not working properly the

mechanism might not be available, In that case a second mechanism, would ac-

tivate. Using heat exchanger full of water, a great amount of heat can be taken

to the ambient. If neither of the above control the situation, the RCCS (the

reactor cavity cooling system) which basically is an exhaust that is concentric

to the core, the RCCS brigs fresh to air to the core, cooling the whole plant.

This feature is only possibly because of the spherical enclosure that acts like

barrier for dangerous radiation. Despite its high safety operation, its output

power is lower, for the time being, than older reactors.

4The main difference of these two reactor is way the manage its fuel: The U.S.A version isrefuelled off-line, while the German has a continuous way of working


(a) U.S.A HTGR (b) Germany HTGR

Figure 3.8: Two versions of the HTGR[17].

3.2 Nuclear Reactors World Wide

According to the World Nuclear Association, there are 68 reactors currently under

construction in 15 countries. However studies from previous years had predicted

that there should have been around 30 countries, which is a difference of 50%. The

reason for the discrepancy is believed to be, the Fukushima nuclear disaster, causing

that important countries, such as Germany and Switzerland stand out.

It was predicted that for years from 2007 to 2009, 17 civilian reactors would get into

service, but only 5 actually did. As was stated in the introduction, shutting down

the reactor without any plan for supplying the energetic deficit created, would lead

3.2. Nuclear Reactors World Wide 53

to a remarkable increase in fossil fuel usage.

Figure 3.9: Word distribution of nuclear reactors[18]

As can be seen in 3.9, the lack of nuclear reactors coincides with those countries

of lower development. Giving the tendency of leaving the technology, countries

like Colombia, might never benefit from the knowledge enrichment, nuclear power

produce.

Chapter 4

Nuclear Fusion

In the previous chapters we have mentioned some key points about fusion. The

following section treats those key points in brief as preamble to the more detail

explanation of the fusion mechanism used in I.T.E.R and others projects. The final

section would be dedicated to analyze the viability of fusion energy in the years to

come.

4.1 Key points of fusion

As we saw in the chapter 2, the fusion process is governed energetically by the bind-

ing energy, and its probability of happening by the cross section. In fusion we are

concerned with the formation of light elements from even lighter components below

Iron, where the biding energy reaches its maximum.

Figure 4.1: Deuterium-Tritium[13]

55

56 Chapter 4. Nuclear Fusion

One of the most promising fusion reactions for energy production is the Deu-

terium Tritium; let us calculate the energy associated with this reaction.

21D +3

1 T →42 He +1

0 n (4.1)

Calculating the the mass defect of equations of 4.1, we get:

∆m = 0.0187mH (4.2)

which leads to:

E = mc2 = 0.0187mHc2 = 2.8184× 10−12 J (4.3)

which is about 17.56MeV . An estimated of 1.210∗1026 reaction can occur per kg

of Deuterium/Tritium mixture, multiplying those numbers we can get about 4GW

for 24 hours, it would be enough to fill the need of about 100000 U.S.A citizen. This

might not look so impressive, but keep in mind that there was just 1kg of fuel.

4.1.1 Energy distribution of fragments during fusion

As stated for fission, the energy distribution for fusion is not symmetric, using energy

and momentum conservation we get:

1

2mAv

2A +

1

2mBv

2B = Efus (4.4)

Using the center of mass frame, the momentum conservation would yield to:

mAvA +mBvB = 0 (4.5)

Combining the two equations we obtain:

EA =mB

mA +mB

Efus (4.6)

EB =mA

mB +mA

Efus (4.7)

4.1. Key points of fusion 57

So the energy released by kinetic gain, would depend on the masses of the frag-

ments. For reaction 4.1, we know that Helium is about 4 times more heavy than

neutron, therefore just 20% would be acquired by Helium while 80% would go to the

neutron.

4.1.2 Reaction Rate and important concepts

In the second chapter we analyze some principles properties of the cross-section, re-

action rate in general, and some of its characteristic for fission and fusion reaction.

Now It shall be done for fusion specifically.

Let us state again the following:

x1 + x2 → x3 + x4 (4.8)

Using the center of mass frame we can write the center mass energy:

Ecm =1

2meffv

2 (4.9)

where v = v1− v2. During experiments measurements are done in the laboratory

frame, as a beam-target cross section, that is using the energy E1 as:

E1 = Ecm(m1 +m2)

m2

(4.10)

The cross-section is related as:

σ(E) = σbeam (E1) (4.11)

Defining n1 as the density of target nuclei, and having that all move at the same

velocity, and the same relative velocity between x1 and x2 per pair, we have:

n2σ(v) (4.12)

which is the probability of the reaction of nucleus 1, per unit path. Making a

simple units analysis we conclude:


Probability of reaction

unit path× unit path

unit time= n2σ(v)v (4.13)

The reactivity, is the probability of reaction per unit time per unit density of

target nuclei, that is σv. In reality, the assumption of equals velocities per pair of

interacting particles is not true, therefore we have:

< σv>=

∫σ(v)vf(v)dv (4.14)

where f(v) is the distribution of velocities.

Fusion fuel is usually a mixtures of elements, so it is possible to define:

R12 =f1f2

1 + δ12

n2 < σv > (4.15)

which is the volumetric reaction rate, f1 f2 are the atomic fraction species, and

Kronecker delta, would take away the double counting when we have the same atomic

species.

4.2 Fusion Cross Section

An approximated model for the behavior for two interacting nucleons can be seen in

the following figure:

Figure 4.2: Coulomb Barrier[19]

4.2. Fusion Cross Section 59

Where r < rn the interacting nuclei would feel the strong nuclear force. Beyond

this point, we can approximate the physical phenomena as a potential well of finite

height, which has been observed to be U0 = 30 − 45MeV . Combining the above

equations we obtain:

vb =Z2Z1

(A1/3 + A1/3)MeV (4.16)

If we follow classical mechanics, no nuclei having less than this energy could fuse,

this is called the Coulomb barrier, as was stated in the previous chapters. Therefore

for energies less e < vb would yield a turning point of:

r = z1z2e2

ε2(4.17)

However using quantum mechanics, it is possible to penetrate this barrier, making

fusion possible, for smaller energies, let us defined:

σ = σgeometry × τ ×R (4.18)

where we have that τ is the transparency barrier and R contains all the nuclear

interaction information.

σgeometry ≈1

ε(4.19)

τ is known as the Gamow factor for non resonant reactions:

τ ≈ τG = e

(−√

EG/E)

(4.20)

With Gamow energy as:

E6 = (παfz1z2) 22meffc2 (4.21)

With as α = 1/137.04 as the constant of fine structure. Using 4.20 and 4.19,we

can conclude that the tunneling probability decreases rapidly with the atomic num-

ber, giving a fare explanation of why fusion for energy production is focus on the

very lightest atoms.


The reaction characteristics R as was said before, contains all the information

of the specific nuclear interaction. This term is relatively big for strong interaction,

smaller for electromagnetic interaction and even smaller, as much as 20 times, for

weak interaction. For most reactions, the variation of R(E) is small, when compare

with the variation of the Gamow factor. In conclusion, we can write the cross-section

as:

σ(E) =S(E)

Ee

(−√

Eg/E)

(4.22)

Where S(E) is the astrophysical s factor, which for most reaction is a slowly

variation function of E. Therefore giving the astrophysical S factor, for E = 0, it is

possible to account an approximated yet useful value for the cross section.

4.3 Important Fusion Reactions

Fusion reactions can be classified in advance fusion fuels, main controlled fusion fuels,

p-p cycle, Carbon Nitrogen Oxygen cycle, Carbon Carbon reactions. The differences

between each other are basically the mass of its initials constituents, cross-sections,

and the field of interest (research or energy production).

4.3.1 Advanced Fusion Fuels

This group is for reactions between hydrogen isotopes and light nuclei, like Helium,

Lithium and Boron. The cross-section is small compared to the main controlled

fusion fuels groups, which tell us that the Gamow energy it is relatively big, that for

low energies. At high energies the cross sections are intermediate between that of

DD and that of DT. One reaction with special research interest is t Proton-Boron

reaction:

p+ 11B → 3α + 8.6MeV (4.23)

Reaction 4.23 is interesting because no radioactive material is involved, being

particularly appealing for energy production, and special fuel applications, the prob-

lem is that it exhibits a very narrow resonance at E=148 KeV, making difficult to

4.3. Important Fusion Reactions 61

maintain the reaction.

Figure 4.3: Proton+Boron fusion reaction [26]

4.3.2 Main Controlled Fusion Fuels

This group consists of the Hydrogen isotopes, Deuterium and Tritium. Giving it low

Z, the Gamow energy is relatively small, therefore a relatively large tunnel penetra-

bility, with large astrophysical factor.

The DT reaction:

D + T → α(3.5MeV) + n(14.1MeV) (4.24)

At nearly 64 KeV(small energy)[19] It reaches its maximum cross-section, which

is the largest of the here studied reactions. Due to its broad resonance having the

astrophysical factor value for E=0, tell us nothing, since in the interval of interest

its value varies greatly, and cannot be approximated as other reactions.

DD reaction

D +D → T (1.01Mev) + p(3.03MeV) (4.25)


D +D → 3He(0.82MeV) + n(2.45MeV) (4.26)

For energies from 10 to 100 KeV, the cross section is about 2 orders of magnitude

smaller than that for DT, and is nearly the same for both reactions.

TT reaction

T + T → α + 2n+ 11.3MeV (4.27)

It is important to notice that the energy associated to the three products, is

not uniquely determined by conservation laws. Regarding to the cross sections, it is

about the same as that of a DD reaction.

4.3.3 P-P cycle

This the main source of energy of the sun. Among all known fusion reactions, the

first two reactions of the cycle, pp and pD have the lowest Gamow energy, having

cross sections much smaller than those already explained reactions. Due to the fact

that the p-p involves a low probability beta decay, the astrophysical facto is 25 orders

of magnitude smaller than the present in the DT reaction[19].

Figure 4.4: p-p Cycle[25]

4.3. Important Fusion Reactions 63

4.3.4 Carbon Nitrogen Oxygen cycle

The S factors are not very small, but the Gamow energy is near to 40 MeV, therefore

the cross section is even smaller than that of the p-p cycle at low temperature(in

the sun). In conclusion the CNO cycle prevails over the p-p at temperature bigger

than 1.5 KeV[19], before that limit, the p-p cycle would happen more often that the

CNO.

Figure 4.5: CNO Cycle[13]

4.3.5 Carbon Carbon Reactions

This is the main source of energy for some white dwarfs. The S factor is very large,

and even at high energies as 100 KeV the cross section is very small, nearly zero[19],

so theoretically, this reaction is only possible for stars densities above 109g/cm3.

Therefore is not yet considered as a possibility for energy production on earth.

Using the explanations provided in the two last section, about the astrophysical

factor and the Gamow energy, table 4.6 shows some important values, regarding

equation 4.22.


Figure 4.6: Important Fusion Reactions[24]

Chapter 5

International Thermonuclear

Experimental Reactor

The first attempt to construct a commercial version of a nuclear fusion reactor, is

being developed at ITER. It would bring together science and industry as it has

never before, through the quest of a potentially limitless, sustainable, safe and se-

cure source of energy: fusion [28]. The cost of is around 13 billion Euro.

The project benefits of the knowledge gained in past experiences such as the Eu-

ropean Fusion Programme.

It is currently being built in the south of France.It uses the Tokamak technology,

a torodial device that uses complex magnetic fields to confine and compress the ex-

tremely hot fusion plasma, the biggest ever build. [28]

5.1 Principal in Nuclear Fusion Reactor

Building a fusion nuclear reactor has two principal branches of research, one called in-

ertial confinement fusion and the most promising one, magnetic confinement fusion.

In any case, the more appealing fuel is the T-D reactions, and the D-D reactions,

since their cross sections are relatively big at achievable temperatures. The problem

65

66 Chapter 5. International Thermonuclear Experimental Reactor

with the T-D reactions, is the that Tritium is not naturally available, but it can

be bred from the neutron emitted by T-D reaction, which would then reacts with

Litium-6, that then decays into Tritium and Helium-4. D-T, is the fuel chosen for

ITER.

At large temperatures (e.g. 106Degree Celsius for Hydrogen1 ) the Bremsstrahlung

radiation becomes a important source of energy loss, in addition to the conventional

ones, heat conduction and convection. For the radiation problem, J.D. Lawson de-

veloped a criterion that took after his name. This concept tells us if a fusion reactor

is at its ignition point, meaning that the heating of the plasma by the products of

fusion is enough to maintain the temperature in the plasma itself. The conduction

problem is handled better by the magnetic confinement, which ITER is using. Finally

the convection problem, is simply solved by taking away the medium of propagation

(air), by creating a vacuum vessel.

5.1.1 The Lawson Criteria

Expressing again equation 4.15, and multiplying it by the energy release per reaction,

we obtain the power per unit volume.

PR = n1n2 < σv > (T )ER (5.1)

In this model only light radiation is important, so Lawson estimated the power

loss as:

PB = 1.4 ∗ 10−34n1n2T1/2 W

cm3(5.2)

Equating 5.1 and 5.2 we get the critical temperature, where the plasma system

is self sustain. However in practice the “L” coefficient is preferred:

PR = n1n2 < σv > (T )ER > PB = 1.4 ∗ 10−34n1n2T1/2 W

cm3(5.3)

Rearranging we get and introducing ne(electron density) and τE(confinementtime)

1At this temperature, H is completely ionized [28]

5.2. The Tokamak Design 67

which is the energy content per unit volume, get arrives at:

neτE > L =12

ER

kBT

< σv >(5.4)

The right part of the equation 5.4 is a function of the temperature, finding its

minimum, we arrive at the Larson criterion. For the D-T reaction this minimum is

near 1.5× 1020s/m3, which occur at T=25 KeV[27].

5.2 The Tokamak Design

Is a device which is at the core of the nuclear fusion reactor being built in ITER.

It uses torodial and polodian magnetic field lines to keep the heated plasma in

the shape of a torus, this is done because stable plasma equilibrium is required

by magnetohydrodynamics, MDH. The last model is the simplest way to describe a

plasma, is uses the analogy of a somehow liquid behavior, having a special parameter

called “beta”, which tell us the relation between plasma pressure and the intensity

of the magnetic field. Having a stable plasma equilibrium at high β is critical for

commercial viability of a fusion reactor. β limits the possible magnetic configurations

and output power of it. There are several forms of instabilities, but it will not be

treated here. Instead it is possible to say that the plasma is in equilibrium when

there are no net forces capable of accelerating any part of it. If we have the plasma

in equilibrium, the next requirement is the stability, that is, if a small perturbation

happens, will the plasma oscillate with a steady amplitude? or will it damp out the

oscillation?. Clearly the second condition is the desired one.


5.2.1 Components

Now we shall explore the specific components of the nuclear fusion reactor in ITER.

• Magnets: Its fuction is to give shape and control the plasma. It is located

between the cryostat and the vacuum vessel. Shield from the scaping neutrons

originated in the fusion process. Its consist in three main components, the

toroidal field system, the poloidan field system and the central solenoid2. Ad-

ditional coils might be installed in the future to control edge localized modes.

The ELMs are highly energetic fluctuation near the plasma edges. If they are

not controlled, it will affect the efficiency of the reactor.

Figure 5.1: Magnets [28]

Table 5.1: Specifications

Magnets Quantity Material

Torodial fiel coils 18 Nb3SnPolodial fields coils 6 NbTi

Central solenoid 1 Nb3Sn

2It is used to induce the main plasma current through the oscillation of its own current.



Torodial field coils

Magnetic energy (GJ) Magnetic field (T)41 11,8

Weigh (Tons) Total Lenght(km)6540 80000

• Vacuum Vessel: It made of hermetically-sealed steel container, that serves

as hosting of the plasma(without direct contact). It is one of the key parts

for steaming out power, since it dictate the volume capability of the fusion

reactor. It is used in double layer steel configuration, and possesses, in the

inner surfaces, blankets, which would provide shielding for the vessel and for

magnets, as was stated before. These blankets can also be used for breeding

the Litium-6, to produce Tritium. It would also have cavities where cooling

water can passes to control the overall temperature.

Figure 5.2: Vacuum Vessel [28]


Vacuum vessel

Internal diameter(m) Weigh (Tons)6 5000

External diameter(m) High(m)19 11


• Blanket: It acts as the cooling structure, shielding the magnets-vacuum vessel

and a way to breed Tritium as mentioned before. When the neutron collides

with the blanket, their kinetic energy is transformed into heat that is then

transport by the coolants (water). When ready to commercial use, this heat

could be used as energy for the public.

Figure 5.3: Blanket[28]


Blanket

Segments S. Dimension (mxm)440 1x1.5

Weigh (Tons)4.6

• Divertor: Is the main energy extractor of the reactor, it consist of 54 remotely-

removable cassettes, each of them act like a target for the plasma particles to

strike, transforming the kinetic energy into heat. The resulting temperatures

would be around 3000 Celsius. Very few materials can last 3and maintain its

structural stability at such temperatures, Therefore the ITER project will be

a pioneer in the testing of materials under extreme conditions. Initially ITER

would use Carbon fiber-reinforced carbon composite, because of high thermal

3It should last as much as the ITER project, 20 years


conductivity, and then Tungsten, which would last longer with a lower rate of

erosion.

Figure 5.4: Divertor[28]

• External Heating Systems: As explained in above sections, the fusion reactions

will only take place with a sufficient probability at very elevated temperatures,

and then accordingly to the Lawson Criteria, to achieve the self sustained-

stable plasma, even more energy has to be put into the system. In the case of

ITER, temperatures of about 150 million Celsius should be achieved. In order

to accomplish that, three mechanism of external heating must be used, neutral

beam injection, ion cyclotron heating,electron cyclotron heating. The first one

uses a beam of highly energetic neutral particles that will collide (without

Coulomb repulsion) with plasma particles, transferring its energy. The second

one uses radio waves in a similar manner the home microwaves transfer heat

to the food. The last one uses electromagnetic radiation of 170 GHz, to make

resonance of electrons, which would produce heat.

Figure 5.5: External heating systems[28]

• Diagnostics: Gathering data for research, here every possible instrument is

going to be placed, to take data.


Figure 5.6: Diagnostics[28]

• Cryostat: Super cooling the magnets for better performance.

Figure 5.7: cryostat[28]


Cryostat

Height(m) Width(m)29.3 28.6

The ITER project is currently in development, according to time table, during

the present year, they started the construction of the complex that would hold the

Tokamak, and is expected to begin operations with Deuterium and Tritium in 2027.

In years to come, the progress made in the ITER project promises to change the

way we see the sources of energies, and if it achieved, it would change forever our


relationship with environment, as well as the global economic order.

ITER has been designed to produce 500MW of output power for 50MW input

power, that is ten times the energy put in. There are seven participating memebers:

European Union, India, Japan, China, Russia, South Korea and U.S.A. The success

of ITER will lead to the building of DEMO- another fusion nuclear plant. The

success of DEMO will be followed by the commercialization phase.

Appendix A

General Nuclear Reactions

Calculations

In general we have a nuclear reaction as follows:

a+X → Y + c (A.1)

An incident particle interacts with a target nucleus, pruducing a new nucleus

and another particle, if the the particle ”a” does not exist then we have a dacay. we

can use energy conservation to develop some general porperties of the above reaction.

Ei = Ea + c2 (m0a +m0X) (A.2)

Ef = Ec + c2 (m0Y +m0c) + EY (A.3)

Ea + c2 (m0a +m0X) = Ec + c2 (m0Y +m0c) + EY (A.4)

Rearranging we have:

75

76 Appendix A. General Nuclear Reactions Calculations

c2 (m0a +m0X − (m0Y +m0c)) = ∆Ekinetic = Ec + Ey − Ea = Q (A.5)

We see that if we are talking about masses this is equal to the mass defect, if

we are talking about energy we have the biding energy. If Q is bigger than cero we

have an exotermic reaction, the new nucleus would be lighter and the kinetic energy

woulg be bigger, on the other hand if the Q is negative, then we have an endotermic

reaction, in other to pruduce an endotermic reation the proyectil would have to have

minimun energy as the following calculation shows:

The relativistic cuandrimomentum is:

P ν = m0γ(c,u)

=

(E

c,p

)(A.6)

According to the dot product defined in minkowski space, we have:

ηνuPνP µ =

(E

c

)2

−

P 2 = Invariant (A.7)

Now we calculate the energy before and after the reaction, the first one in the

targeted nucleus inertical frame, the second one in the center of mass frame.

Ei = Eth +moXc2 =

√m0a

2c4 + Pia2c2 +moXc

2 (A.8)

Ef = m0Y c2 +m0Cc

2 (A.9)

Using property of dot product invariance we are able to write the conservation

law regartless the frame:

77

1

c2

(√m0a

2c4 + Pia2c2 +m0Xc

2)

2 − Pia2 =

1

c2

(m0Y c

2 +m0cc2)

2 (A.10)

Eth2 + 2Ethm0Xc

2 +m0X2c4 =

(m0Y c

2 +m0cc2)

2 + Pia2c2 (A.11)

m0a2c4 + Pia

2c2 + 2Ethm0Xc2 +m0X

2c4 =(m0Y c

2 +m0cc2)

2 + Pia2c2 (A.12)

Eth =(m0Y c

2 +m0cc2) 2 − (m0a

2c4 +m0X2c4)

2m0Xc2(A.13)

equation 22 is the total threshold energy, so:

Eth = EthKintic +m0ac2 (A.14)

EthKinetic =(m0Y c

2 +m0cc2) 2 − (m0ac

2 +m0Xc2) 2

2m0Xc2(A.15)

On de the other hand we have equation 14, operating it we get:

(m0Y c

2 +m0cc2)

2 =(m0ac

2 +m0Xc2)

2 − 2Q(m0ac

2 +m0Xc2)

+Q2 (A.16)

Replacing 25 into 24 we obtein:

EthKinetic = −Q(m0ac

2 +m0Xc2

m0Xc2− Q

2m0Xc2

)(A.17)

We know that the binding energy is a lot less than rest energy of the nucleus, so

we can aproximate eqation 27, giving finally:

EthKinetic = −Q(m0a +m0X

m0X

)(A.18)

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alternative sources of energy: nuclear fission and fusion

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