1. 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 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 U 92
is the isotope Uranium-235 with a nucleus consisting of 92 protons and 235 – 92 = 143 neutrons. 1 H
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.

2. The Mass Defect and Nuclear Binding Energy
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 = u
Rest mass of proton mp = u
Rest mass of neutron mn = u
where u is the atomic mass unit = x10-27 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 = x10-19 J.

5. What is fusion ? First fusion reactions discovered…
D2 + T3  He4 + n MeV
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 ?

6. Nuclear Processes in the Sun
4 H  He + 2 e +3  +2  Q = 24.8 MeV
which is actually made up of the following steps;
Step 1: H + H  H + e +  Q = 0.4 MeV
Step 2: H + H  He +  Q = 5.5 MeV
Note that step 2 must occur twice before step 3 can take place
Step 3: He + He  He + 2H +  Q = 13.0 Me V
For the overall balance, steps 1 and 2 occur twice giving the energy release of 2x( ) + 13 MeV = 24.8 MeV. Actually there is an alternative step 3 but the information above gives an indication

7. Each reaction is characterised by its energy release and its rate.
Fusion Reactions 1/2
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 He(3.52) + 1n(14.08) *
2D + 3He 1p(14.7) + 4He(3.7)
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.

8. Fusion Reactions 2/2
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) p(5.73) + 4He(1.44) 12.9
1p + 6Li 3He(2.3) He(1.72)
3He + 6Li 1p(12.4) He(2.89) + 4He(2.89)
1p + 11B 4He(2.89) + 4He(2.89) + 4He(2.89) 8.67

10. 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

13. How of fusion reactions
Aim : Overcome electrostatic repulsion between like charges
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)
nuclear
accelerator  + -

14. The Conditions for Net Fusion Energy Production 1/2
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
ei Eo > Ein (5.4-1)
Now, let
Ec = fusion energy from charged-particle products
En = fusion energy from neutrons
Er = fusion fuel radiation loss
Eth = 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, then
El + En = Eo (5.4-3)

15. Conditions for Net Fusion Energy Production 2/2
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/(ei) – 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)

16. Conditions for fusion power in confined systems- - + + + + D T
Achieve sufficiently high
ion temperature Ti
 exceed Coulomb barrier
density nD  energy yield
energy confinement time E
100 million °C
“Lawson” ignition criteria :
Fusion power & heat loss = f(Ti,nD,E)
Fusion power > heat loss
nD ETi>3  1021 m-3 keV s
At these extreme conditions matter exists in the plasma state

17. Progress in magnetically confined fusion
“Ignition” regime, Q∞ : Power Plant.
Q=: Ignition
Q = Pout /Pheat ~1
“Breakeven” regime :
Eg. Joint European Tokamak :
1997 : Q=0.7, 16.1MW fusion : steady-state, adv confinement geometries
Q=1: Breakeven
D2 + T3 He4 (3.5 MeV) + n1 (14.1 MeV)
“Burning” regime : ITER
≥ Pheat
Q>5 ITER
Pout
Q=5: Burning

18. D-T fusion fuels are very abundant
NB: % of all matter is H
Deuterium
Tritium
Lithium
Relative Abundance
1D : 6000H
1T : 1017 H.
Manufactured: Li+n→ He + T
1Li:106 H (earth)
1Li:1000H (solar system)
According to the DOE (2001), world energy usage = 13.5 TW
Estimated Earth reserves are : x 108 TW years of D-T, 2 x TW years of D-D
T. J. Dolan, Fus. Res., 2000

19. Low level waste, compared to fission
Comparison of fission and fusion radioactivity after decommissioning
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.

20. 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
Nuclear non proliferation -
Radioactive waste
104 yrs
100 yrs

21. What are Australia’s energy needs?
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
Our cities and industry are powered by large-scale base load supply Erraring power station : 2.64GW – Australia’s largest power plant
Nearby Tomago Aluminium smelter (Hunter Valley)