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PHY401: Nuclear and Particle Physics

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PHY401: Nuclear and Particle Physics 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) Z01 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 Z01 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.
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