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We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic number (the number of protons in the nucleus) and A is the mass number (the number of neutrons and protons in the nucleus). If we denote the number of neutrons in the nucleus by N, then we have ; A= Z + N For example 235 U 92 is the isotope Uranium-235 with a nucleus consisting of 92 protons and 235 – 92 = 143 neutrons. 1 H 1 Is a proton, the nucleus of the isotope Hydrogen-1, whereas Hydrogen-2 is the nucleus of the isotope Deuterium (heavy Atomic and Nuclear Structure

We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

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Page 1: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

We will denote any isotope by using the notation A

XZ

where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic number (the number of protons in the nucleus) and A is the mass number (the number of neutrons and protons in the nucleus). If we denote the number of neutrons in the nucleus by N, then we have ;

A= Z + N

For example 235 U

92is the isotope Uranium-235 with a nucleus consisting of 92 protons and 235 – 92 = 143 neutrons.

1 H

1

Is a proton, the nucleus of the isotope Hydrogen-1, whereas Hydrogen-2 is the nucleus of the isotope Deuterium (heavy hydrogen) which consists of a proton and a neutron bound together.

Atomic and Nuclear Structure

Page 2: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

The sum of the masses of the individual masses of the protons, neutrons and electrons, gives the total constituent mass of an atom. The rest masses of these three fundamental particles are:

Rest mass of electron me = 0.0005486 uRest mass of proton mp = 1.007276 uRest mass of neutron mn = 1.008665 u

where u is the atomic mass unit = 1.660539x10-27 kg

If one compares this with the total mass of the actual atom (with the bound nucleus) one discovers that the latter is smaller, and the difference is the so-called nuclear mass defect mdefect. Einstein (1905) made the celebrated proposal (since substantiated by experiment) that this mass defect is equivalent to energy, namely the binding energy of the nucleus, so that

EB = mdefectc2

where c is the speed of light (3x108 ms-1). Because energy conversions are required in this sort of topic, we can express c2 in energy/mass units namely 931 MeV/u, where 1 eV = 1.602 x10-19 J.

The Mass Defect and Nuclear Binding Energy

Page 3: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

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Page 5: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

What is fusion ?

D2 + T3 He4 + n1 + 17.6 MeV

First fusion reactions discovered…

energy gain ~ 450:1

by an Australian, Sir Mark Oliphant

(and Lord Rutherford), 1932

Why D-T? It has the lowest energy barrier of all fusion reactions

Basis of Fusion Energy Science : Can we exploit this reaction (and other low activation barrier reactions), to produce base-load energy ?

Page 6: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

1 4 0 0 04 H He + 2 e +3 +2 Q = 24.8 MeV

1 2 1 0 0

which is actually made up of the following steps; 1 1 2 0 0

Step 1: H + H H + e + Q = 0.4 MeV 1 1 1 1 0

2 1 3 0Step 2: H + H He + Q = 5.5 MeV

1 1 2 0Note that step 2 must occur twice before step 3 can take place

3 3 4 1 0Step 3: He + He He + 2H + Q = 13.0 Me V

2 2 2 1 0 For the overall balance, steps 1 and 2 occur twice giving the energy release of 2x(0.4 + 5.5) + 13 MeV = 24.8 MeV. Actually there is an alternative step 3 but the information above gives an indication

Nuclear Processes in the Sun

Page 7: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

Each reaction is characterised by its energy release and its rate.Total energy release

(MeV)

2D + 2D 3He(0.82) + 1n(2.45) 50% 3.27

2D + 2D 3T(1.00) + 1p(3.03) 50% 4.03

2D + 3T 4He(3.52) + 1n(14.08) 17.6*

2D + 3He 1p(14.7) + 4He(3.7) 18.4

3T + 3T 4He(1.24) + 1n(5.03) + 1n(5.03) 11.3* This was the first fusion reaction performed in the laboratory. Work carried out by Oliphant et al. (1934)

One gram of deuterium undergoing the first 2 (D-D) reactions releases about 1011 J i.e. roughly 25,000 kWh per gramme mass of the reacting nuclei (a similar yield to that for the fission of Uranium). Side reactions (3 and 4) increase this energy yield by about a factor 5.

Fusion Reactions 1/2

Page 8: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

Total energy release (MeV)

3T + 3He 1p(5.38) + 4He(4.76) + 1n(5.38) 51% 12.1

3T + 3He 2D(9.54) + 4He(4.76) 43% 14.3

3T + 3He 5He(11.9) + 1p(2.4) 6% 14.3

3He + 3He 1p(5.73) + 1p(5.73) + 4He(1.44) 12.9

1p + 6Li 3He(2.3) + 4He(1.72) 4.02

3He + 6Li 1p(12.4) + 4He(2.89) + 4He(2.89) 16.09

1p + 11B 4He(2.89) + 4He(2.89) + 4He(2.89) 8.67

Fusion Reactions 2/2

Page 9: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

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Page 10: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

The plasma state : the fourth state of matter

• plasma is an ionized gas • 99.9% of the visible universe is in a plasma state

Inner region of the M100 Galaxy in the

Virgo Cluster, imaged with the Hubble

Space Telescope Planetary Camera at

full resolution. A Galaxy of Fusion Reactors.

• Fusion is the process that powers the sun and the stars

Page 11: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic
Page 12: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic
Page 13: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

+-

Particle acceleration

(eg. Linear accelerator, pyroelectric crystal,

beam-target [laser-block], beam-beam )

Inertial compression (confinement)

(eg. laser target fusion)

Catalytic process

(eg. Muon catalysis)

Confinement (plasma near thermal equil)

(gravitational, electric, magnetic)

Aim : Overcome electrostatic repulsion between like charges

How of fusion reactions

nuclearaccelerator

Page 14: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

For a steady-state fusion reactor the input power equals the output power. For net energy production, the electrical output (after conversion of the reactor thermal output (Eo) to electricity with efficiency e) must be more than sufficient to feedback energy (which is converted at efficiency i) to provide the input energy (Ein) to sustain the reactor.

This condition is ei Eo > Ein (5.4-1)Now, let Ec = fusion energy from charged-particle productsEn = fusion energy from neutronsEr = fusion fuel radiation lossEth = fusion thermal energy (3nkBTV, where V is the volume of plasma)El = fusion fuel energy loss (Eth + Er)For a steady state of the fusion fuel, the heating input must balance the losses i.e.Ein + Ec = El (5.4-2)Assuming that all (the losses and fusion neutron) energy can be recovered, thenEl + En = Eo (5.4-3)

The Conditions for Net Fusion Energy Production 1/2

Page 15: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

Combining these three equations (5.4-1) to (5.4-3) to eliminate El, we have

Ec + En > Eing() (5.4-4)

where g() = 1/(ei) – 1. Now Ec + En is the total energy released by fusion reactions Ef. Using this result (Ef/g() > Ein) and the definition of El we have

Ef/g() + Ec - Er > Eth (5.4-4)

Now a fusion reactor is characterized by an energy confinement time (i.e. representative of the time it takes for the energy in the plasma to be lost by processes other than radiation, if there is no power input). Let this time be represented by . The plasma power loss (independent of radiation losses) are then Eth/. In terms of the corresponding powers (E/(V)= P), then (5.4-4) can be expressed as

n > (3kBT)/Pf/(g()n2) + Pc/(n2) – Pr/(n2) (5.4-5)

Conditions for Net Fusion Energy Production 2/2

Page 16: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

• Achieve sufficiently high

ion temperature Ti

exceed Coulomb barrier

density nD energy yield

energy confinement time E

nD ETi>3 1021 m-3 keV s

• “Lawson” ignition criteria :

Fusion power > heat loss

• At these extreme conditions matter exists in the plasma state

100 million °C

Conditions for fusion power in confined systems

+ + +

D T

Fusion power & heat loss = f(Ti,nD,E)

+

--

Page 17: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

Progress in magnetically confined fusion

Q = Pout /Pheat ~1

• “Breakeven” regime :

Eg. Joint European Tokamak : 1983 -

1997 : Q=0.7, 16.1MW fusion 1997- : steady-state, adv.

confinement geometries

Q=1: Breakeven

D2 + T3 He4 (3.5 MeV) + n1 (14.1 MeV)

• “Burning” regime : ITER

≥ PheatQ>5 ITERPout

Q=5: Burning

• “Ignition” regime, Q∞ : Power Plant.

Q=: Ignition

Page 18: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

D-T fusion fuels are very abundant

Relative Abundance

1D : 6000H 1T : 1017 H.

Manufactured: Li+n→ He + T

1Li:106 H (earth)

1Li:1000H (solar system)

Deuterium Tritium Lithium

NB: 99.9885% of all matter is H

• According to the DOE (2001), world energy usage = 13.5 TW

• Estimated Earth reserves are : 6 x 108 TW years of D-T, 2 x 1011 TW years of D-D

T. J. Dolan, Fus. Res., 2000

Page 19: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

Present ferritic technology allows a reduction of >3,000 over 100 years

100% recycling is possible after 100 years

Using future Vanadium alloy structures, fusion is 1,000,000x less radioactive after 30 years than fission.

http://fi.neep.wisc.edu http://www.ofes.fusion.doe.gov

Comparison of fission and fusion radioactivity after decommissioning

Low level waste, compared to fission

Page 20: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

Why exploit fusion-power on Earth?

Fossil Fission

Fusion Renew.

Base-load energy generation ?

High energy density Power grid stability Low greenhouse emitter Universal accessibility of fuel Large scale availability ~50

yrs

Terrorist Potential Low High Low

Nuclear non proliferation - -Radioactive waste - 104 yrs 100 yrs -

Page 21: We will denote any isotope by using the notation A X Z where X is the element’s chemical symbol (e.g. H for Hydrogen, He for Helium etc.), Z is the atomic

What are Australia’s energy needs?

• Our cities and industry are powered by large-scale base load supply Erraring power station : 2.64GW – Australia’s largest power plant

Australia is a hot, dry, sparsely-populated, resource-rich nation

• Our population is 20 million- 1.2% growth rate (2006),>10 million live in Sydney, Melbourne, and Brisbane

Nearby Tomago Aluminium smelter (Hunter Valley)