PHY401: Nuclear and Particle Physics
Lecture 19, Monday, October 5, 2020 Dr. Anosh Joseph IISER Mohali Nuclear Fusion
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Nuclear Fusion
Let us look at the plot of average binding energy per nucleon against mass number.
It has a maximum at A ≈ 56.
And after that it slowly decreases for heavier nuclei.
The decrease is much quicker for lighter nuclei.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Nuclear Fusion
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Nuclear Fusion
Lighter nuclei are less tightly bound than medium-size nuclei (exception: magic nuclei).
Thus, in principle, energy could be produced by two light nuclei fusing to produce a heavier, and more tightly bound nucleus.
This is process of inverse to fission.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Nuclear Fusion
Just as for fission, the energy released comes from difference in the binding energies of initial and final states.
This process is called nuclear fusion.
Very attractive as a potential source of power...
... because of the far greater abundance of stable light nuclei in nature than very heavy nuclei.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Nuclear Fusion
Thus fusion would offer enormous potential for power generation, if ...
...the huge practical problems could be overcome.
Fusion processes also explain how stars are formed.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
Practical problem to obtaining fusion has its origin in Coulomb repulsion.
It inhibits two nuclei getting close enough together to fuse.
We have the Coulomb potential
1 ZZ 0e2 VC = 0 . (1) 4π0 (R + R )
Z and Z 0: atomic numbers of the two nuclei
R and R 0: their effective radii.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
(R + R 0): can be treated as the distance of closest approach.
For a medium and heavy nuclei
1 R = 1.2A 3 fm. (2)
Inserting this in VC
e2 h cZZ 0 V = C π h h 1 1 i 4 0 c 1.2 A 3 + (A 0) 3 fm ZZ 0 = 1.198 MeV. (3) h 1 1 i A 3 + (A 0) 3
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
Set A ≈ A 0 ≈ 2Z ≈ 2Z 0. Then
5 VC ≈ 0.15A 3 MeV. (4)
For example, with A ≈ 8 we have VC ≈ 4.8 MeV.
This much energy has to be supplied to overcome the Coulomb barrier.
This is a relatively small amount of energy.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
Could be achieved by simply colliding two accelerated beams of light nuclei?
But in practice nearly all particles would be elastically scattered.
Only practical way is to heat a confined mixture of the nuclei to supply enough thermal energy to overcome Coulomb barrier.
Can estimate the temperature needed from E = kBT.
For an energy of 4.8 MeV, this implies a temperature of 5.6 × 1010 K.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
This is well above even the typical temperature of 107 K found in stellar interiors.
Fusion actually occurs at a lower temperature than this estimate.
Due to a combination of two reasons.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
The first and most important is the phenomenon of quantum tunneling.
Thus the full height of Coulomb barrier does not have to be overcome.
Encountered a similar problem in the context of α decay.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
Can draw on that analysis here to get some results.
Probability of barrier penetration depends on a number of factors.
Most important is the Gamow factor, which is a function of relative velocities and charges of reaction products.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
Probability for tunneling is proportional to
e−G(E), (5)
The function G(E) may be written as r E G = G , (6) E where 2 2 EG = 2mc (παZ1Z2) . (7)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
m: reduced mass of two fusing nuclei that have electric charges Z1e and Z2e.
Thus probability of barrier penetration increases as E increases.
Nevertheless, the probability of fusion is still extremely small.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
As an example, if we consider fusion of two protons at a typical stellar temperature of 107 K, we find EG ≈ 490 keV (8) and E ≈ 1 keV. (9)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
Hence the probability of fusion is proportional to
1 e−(EG /E) 2 ≈ e−22 ≈ 10−9.6. (10) This is a very large suppression factor.
Thus actual fusion rate is still extremely slow.
The other reason that fusion occurs at a lower temperature than expected is the following.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
A collection of nuclei at a given mean temperature, whether in stars or elsewhere, will have a Maxwellian distribution of energies about the mean.
There will be some with energies substantially higher than mean energy.
Nevertheless, even a stellar temperature of 108 K corresponds to an energy of only about 10 keV.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Coulomb Barrier
So, fraction of nuclei with energies of order 1 MeV in such a star would only be of order
−43 exp(−E/kBT) ∼ exp(−100) ∼ 10 . (11)
This is a minute amount.
We now examine the interplay of these two factors.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
Product of the increasing barrier penetration factor with energy, and the Maxwellian decreasing exponential implies that...
... in practice fusion takes place over a rather narrow range of energies.
To see this let us consider the fusion between two types of nuclei, a and b, having number densities na and nb and at a temperature T.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
Assume that temperature is high enough so that the nuclei form a plasma.
Assume uniform values of number densities and temperature.
Also assume that velocities of the two nuclei are given by Maxwell-Boltzmann distribution.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
Thus probability of having two nuclei with a relative speed v in the range v to v + dv is
r 3 2 m 2 −mv2 P(v)dv = exp v2dv, (12) π kBT 2kBT m: reduced mass of the pair.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
Denoting σab as fusion cross-section and taking the average
hσabvi ≡ σabvP(v)dv, (13) Z0∞ we have the fusion rate per unit volume
Rab = nanbhσabvi. (14)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
The fusion cross-section may be written as " r # S(E) E σ (E) = exp − G . (15) ab E E
Function S(E) contains the details of the nuclear physics.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
Plugging in the quantities we get
1 3 8 2 1 2 Rab = nanb πm kBT " 1 # E E 2 × S(E) exp − − G dE.(16) k T E Z0∞ B
S(E): a slowly varying function of E.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
Thus behavior of integrand is dominated by behavior of exponential term.
Falling exponential of the Maxwellian energy distribution combines with rising exponential of quantum tunneling effect...
... to produce a maximum in the integrand at E = E0, where
1 1 3 E = E (k T)2 . (17) 0 4 G B
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
Fusion takes place over a relatively narrow range of 1 energies E0 ± 2 ∆E0, where
4 1 5 6 6 ∆E0 = 1 1 EG (kBT) . (18) 3 2 2 3
A schematic illustration of the interplay between these two effects is shown in figure below.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion Reaction Rates
1 RHS of figure: function exp[−E/kBT − (EG/E) 2 ] for the real example of the fusion of two protons at a temperature of 2 × 107 K.
We have
EG = 496 keV, kBT = 1.7 keV, (19) so that fusion is most likely at
E0 = 7.2 keV. (20)
Rate depends very strongly on both temperature and nuclear species.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Nucleosynthesis and Stellar Evolution
Nuclear fusion plays a vital role in evolution of astrophysical objects of many types.
It determines how heavier elements are formed from lighter ones.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
Current theory of the origin of the universe - the Big Bang theory - starts with the creation of the fundamental particles, quarks, leptons, and others, each with their own antiparticle.
How this particle-antiparticle symmetry evolved into what we see today, where matter is overwhelmingly made of particles, is not yet fully understood.
But within 10−2 s after the Big Bang, quarks in the rapidly cooling primordial plasma coalesced to form protons and neutrons.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
Then the reactions
− e + p n + νe (21) were possible.
Initially these reactions could proceed in either direction with equal rates, despite the neutron-proton mass difference.
The reason is the particles had high kinetic energies derived from the extreme temperature.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
As the universe continued to expand and cool, within one second, particles had insufficient energy to maintain equilibrium and only the decay
− n → p + e + ν¯e (22) was possible, with a half-life of about 10 minutes.
The neutrinos escaped to become relics of the Big Bang.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
However, before neutrons could decay, within a few minutes, most of them were captured by protons to form the bound state deuterium, d = np, via the reactions n + p d + γ (23) where again initially the two reactions were in equilibrium.
However, as the universe cooled further the fraction of photons having sufficient energy to photo-dissociate deuterium was reduced.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
So deuterium production started to dominate.
Production of deuterium in sufficient quantities marked the start of nucleosynthesis.
1 Let us use the notation for nuclei, with p = 1H, 1 2 n = 0H, and d = 1H then helium-3 can be produced via deuterium reactions with hydrogen:
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
2 1 3 1H + 1H → 2H + γ, (24) or in deuteron-deuteron collisions
2 2 3 1 1H + 1H → 2He + 0H. (25)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
3 The latter interaction can also produce tritium 1H, with a lifetime of 12 years. That is,
3 2 4 1 1H + 1H → 2H + 0H. (26)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
Next, helium-4 can be produced via a number of reactions involving protons, neutrons, deuterium, tritium, and helium-3, including
3 2 4 1 1H + 1H → 2He + 0H,
2 3 4 1 1H + 2He → 2He + 1H. (27)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
4 1 4 2He + 0H → 2He + γ,
3 1 4 1H + 1H → 2He + γ,
2 2 4 1H + 1H → 2He + γ.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
In principle, successive sequences analogous to those above could synthesize the next highest nuclei.
At the temperature prevalent at this stage (T ≈ 109 K) the nuclei energies are less than 0.1 MeV.
So these reactions have very low probabilities.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
Fusion can produce mass-7 nuclei via the reactions
4 3 7 2He + 1H → 3Li + γ,
4 3 7 2He + 2He → 4Be + γ. (28)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
But the Coulomb barriers to be overcome are about 1 MeV.
Three minutes after the Big Bang, the hadron constituents of the universe were: protons 75%, helium 24%, and in addition there were small quantities of deuterium, with traces of lithium and beryllium nuclei.
The present measurement of helium-4 indicates good agreement and better agreement for helium-3.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Big-bang Nucleosynthesis (BBN)
But for lithium-7, there is a significant discrepancy between BBN and WMAP/Planck, and the abundance derived from Population II stars.
The discrepancy is a factor of 2.4-4.3 below the theoretically predicted value.
Indicates that the current BBN model has to be modified.
Carbon nucleus can be formed in stars at lower temperatures, and so we next turn to considering stellar nucleosynthesis.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
Stellar evolution involves the release of gravitational potential energy through contraction of the nuclear material of the star.
This increases its temperature to a point where a particular fusion reaction can occur.
Energy produced flows to the surface of the star and temporarily halts the gravitational contraction.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
When all the nuclear ‘fuel’ for the reaction is ‘burnt’, contraction restarts and, provided the mass of the star is great enough, ...
...the temperature of the star again rises to a point where the products of the first stage of burning are used as the fuel for the next set of fusion reactions, and so on.
In the case of the Sun, for example, hydrogen is initially burned by the so-called proton-proton cycle.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
This has more than one branch, but one of these, the PP-I cycle, is dominant.
It starts with the fusion of hydrogen nuclei to produce nuclei of deuterium
1 1 2 + 1H + 1H → 1H + e + νe + 0.42 MeV, (29) which is the primary source of the solar neutrinos.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
The deuterium then fuses with more hydrogen to 3 produce 2He
1 2 3 1H + 1H → 2He + γ + 5.49 MeV. (30)
3 4 And finally, two 2He nuclei fuse to form 2He
3 3 4 1 2He + 2He → 2H + 2( 1H) + 12.86 MeV. (31)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
The relatively large energy release in the last reaction 3 is again because 2He is a doubly magic nucleus and so is very tightly bound.
The first of these reactions (involving proton proton fusion), being a weak interaction, proceeds at an extremely slow rate, and sets the scale for the long lifetime of the Sun.
Combining these equations, we have overall
1 4 + 4( 1H) → 2He + 2e + 2νe + 2γ + 24.68 MeV. (32)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
The nuclear end product is helium-4.
Because the temperature of the Sun is ∼ 107 K, all its material is in the form of a plasma.
The positrons produced in reaction (p-p fusion) will annihilate with electrons in the plasma to release a further 1.02 MeV of energy per positron and so the total energy released is 26.72 MeV.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
However, of this each neutrino will carry off 0.26 MeV on average, which is lost into space.
Thus on average, 6.55 MeV of electromagnetic energy is radiated from the Sun for every proton consumed in the PP-I chain.
The PP-I chain is not the only fusion cycle contributing to the energy output of the Sun, but it is the most important.
(About 84% of the total luminosity of the Sun is from PP-I chain.)
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
When hydrogen has been depleted as a fuel, helium burning starts.
At the lower temperatures in stars the following rare reaction occurs
4 8 12 ∗ 2He + 4Be → 6 C . (33)
12 ∗ Here 6 C is an excited state of carbon.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
A very small fraction of the latter will decay to the ground state, so that overall we have
4 12 3( 2He) → 6 C + 7.27 MeV. (34)
Once carbon is present, another cycle, called the CNO chain, can start.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
Although this contributes only about 3% of the Sun’s energy output, it plays an important role in the evolution of stellar objects of greater mass.
12 13 14 In the presence of any of the nuclei 6 C, 6 C, 7 N, or, 15 7 N, hydrogen will catalyze burning via the reactions
12 1 13 6 C + 1H → 7 N + γ + 1.95 MeV, 13 13 + 7 N → 6 C + e + νe + 1.20 MeV,
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
13 1 13 6 C + 1H → 7 N + γ + 7.55 MeV,
14 1 15 7 N + 1H → 8 O + γ + 7.34 MeV, 15 15 + 8 O → 7 N + e + νe + 1.68 MeV,
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
15 1 12 4 7 N + 1H → 6 C + 2He + 4.96 MeV.
Thus overall in the CNO cycle we have
1 4 + 4( 1H) → 2He + 2e + 2νe + 3γ + 24.68 MeV.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
These and other fusion chains all produce electron neutrinos as final state products, and using detailed models of the Sun, the flux of such neutrinos at the surface of the Earth can be predicted.
However, the actual count rate is far lower than the theoretical expectation.
This is the solar neutrino problem.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
The solution lies in the phenomenon of neutrino oscillations, where some electron neutrinos are converted to neutrinos of other flavors in their passage from the Sun to Earth.
Fusion processes continue to synthesize heavier elements until the core of the stellar object is composed mainly of nuclei with A ≈ 56.
That is, the peak of the binding energy per nucleon curve.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
At which point fusion ceases to be a way of creating heavier elements.
Heavier nuclei are produced mainly by the accretion of neutrons in collisions.
These reactions are intrinsically implausible and the process is very slow - hence the name s-process.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
The majority of stars die quietly, accumulating more neutrons and expanding until they eventually collapse slowly under gravity.
However, if part of a binary system, they may attract hydrogen and helium from their companion star and the accreted matter may ignite reactions in which protons are created to create unstable proton-rich nuclei.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
This is the so-called rp-process.
Another scenario, which is very rare, is where a very massive star collapses very rapidly in just a few seconds, resulting in a spectacular explosion called a supernova.
During this process ultra-rapid reactions (called the r-process) involving neutron-rich nuclei create elements as heavy as uranium, which are then distributed throughout the universe.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Stellar Nucleosynthesis
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Figure: The Crab Nebula, a six-light-year-wide expanding remnant of a star’s supernova explosion. Japanese and Chinese astronomers recorded this violent event in 1054 CE. Stellar Nucleosynthesis
We are stardust!
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
The exothermic reactions
2 2 3 1H + 1H → 2He + n + 3.27 MeV. (35)
2 2 3 1H + 1H → 1He + p + 4.03 MeV. (36) suggest that deuterium might be a suitable fuel for a fusion reactor.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
Deuterium is present in huge quantities in sea water and is easy to separate at low cost.
The effective energy produced by the fusion process will be reduced by the heat radiated by the hot plasma.
The mechanism for this is predominantly electron bremsstrahlung.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
Only when the energy deposited in the plasma by the alpha particles exceeds the radiation losses would the reaction be self-sustaining.
This is referred to as the ignition point.
A very high particle density or a long confinement time, or both, is required to sustain the plasma.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
At the temperatures required for fusion, any material container will vaporize.
The central problem is how to contain the plasma for a sufficiently long time for the reaction to take place.
The two main methods are magnetic confinement and inertial confinement.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
Both techniques present enormous technical challenges.
In practice, most work has been done on magnetic confinement.
In this approach, the plasma is confined by magnetic fields and heated by electromagnetic fields.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
The simplest configuration is a toroidal field produced by passing a current through a doughnut-shaped solenoid.
Most practical realizations of these ideas are devices called tokamaks.
The largest and most powerful tokamak in existence is the Joint European Torus (JET) in the UK
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
Nuclear fusion holds the promise of unlimited power without the problem of radioactive waste.
But the road to realization of this goal is long and we are far from the end.
Considerable progress has been made towards the goal of reaching the ignition point.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
For example, in 1997 JET produced 16 MW of fusion power from a total input of 24 MW, a ‘gain’ of about 0.7, although only for a few seconds.
In 2013 a small real gain, i.e. greater than 1, was fleetingly achieved, the first for any fusion facility.
A major new tokamak machine, the International Thermonuclear Experimental Reactor (ITER), is being built in France by a global collaboration.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
ITER, although considerably larger than JET, has the same basic torus-like configuration.
The project has experienced a number of problems, not always scientific, and the first plasma is now planned for 2025.
Operation with tritium-deuterium fuel is not anticipated until 2035.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
The aim is to produce 500 MW of power sustained for up to 1000 seconds, compared to JET’s peak of 16 MW for less than a second.
There are other groups in institutes around the world using small tokamaks to explore different approaches to producing a sustainable plasma.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali Fusion reactors
Figure: Schematic diagrams showing: (a) the main magnetic field components of the JET tokamak; (b) how these elements are incorporated into the JET device.
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali References
I B. R. Martin and G. Shaw, Nuclear and Particle Physics: An Introduction, 3rd edition, Wiley (2019).
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali End
PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali