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Home , Fission barrier, Nuclear fusion

28 3 8. FISSION AND FUSION where rnucl = 48.x 10 nuclei /m = nuclei density of . For example, the average of a prompt from fission is 2 MeV and s tot ª 7barns (Figs.8.1 and Induced fission – fissile materials 8.2) so that l ª 3cm. A 2 MeV neutron will travel this distance in about 1.5x10-9 s. For A ª 240, ª 5–6 MeV. If a neutron with zero enters nucleus to form compound nucleus, latter will have an excitation energy above its ground 235 Consider an explosive release of energy in a nuclear , using U (c = 1). From Fig.8.1, state = neutron’s in that ground state. Example: zero-energy neutron entering 235 236 neutron with this energy has probability of about 18% to induce fission in interaction with U U 235 forms with excitation energy 6.46 MeV >> fi compound nucleus U . Otherwise it will scatter and lose energy, so that the probability for a further interaction undergoes fission. To induce fission in 238U requires neutron with kinetic energy > 1.4 MeV. 239 239 will be somewhat increased (cross-section increases with decreasing energy). If neutron does Binding energy of last neutron in U is 4.78 MeV < fission threshold of U . Differences not escape outside the target, the most probable number of collisions it will make before in binding energy of the last neutron in even-A (even-N/even-Z) and odd-A nuclei given by inducing fission is about 6-7 fi moving a distance of 7cm in a time t ª10-8 s before pairing term in SEMF. Explains why for odd-A nuclei 233UUPuPu,,235 239 ,241 fission may p 92 92 94 94 inducing fission. Not all will induce fission. (Some will escape from the surface and be induced by zero-energy neutrons, whereas 232Th,,238 U240 Pu ,242 Pu require an 90 92 94 94 some will under radiative capture.) If probability that newly created neutron induces fission is energetic neutron. q, each neutron on average leads to creation of ()nq -1 additional neutrons in time tp . If there are Nt() neutrons present at time t, then there will be Commonly used in reactors is uranium. ( = 99.3% 238U + 0.7% 235U ) 235 s tot and s f , for neutrons incident on U shown in Fig.8.1. Most important features are: Nt()()()()+=dd t Nt[]11 + nq - ttp 1. E < 0.1 eV, s for 235U >> s for 238U ; fission fraction large ( ~%84 ) (rest mainly tot tot at tt+ d . In the limit as d t Æ 0, radiative capture with formation of excited state of 236U ) dN ()nq -1 2. 0.1 eV < E < 1 keV, cross-sections for both isotopes show prominent peaks due to = Nt() dt t resonant capture of neutron. p 235 with solution 3. E > 1 keV, ssf tot for U still significant. In both s tot mainly due to contributions from and inelastic excitation of nucleus. Nt()=- N (01 )exp([] nq ) ttp

Widths of the low-energy resonances in 235U cross-section are ~ 1eV , and compound Thus number increases or decreases exponentially. For 235U , number increases exponentially nucleus formed at these resonances decay predominantly by fission fi fission takes place in if qn>ª104. . If dimensions of the metal are << 7 cm, q will be small and chain reaction h -14 a time of order t ff=ªG 10 s after neutron absorption. Fission fragments are usually will die out. However, a sufficiently large brought together at t = 0 will have q > 04. -8 highly excited and quickly boil off neutrons. The average number of these so-called prompt (there will be neutrons present at t = 0 arising from ) and since tp ª10 s, neutrons for 235U is n ª 25. . Decay products will decay by chains of b -decays and some of explosion will occur in microsecond. For simple sphere of 235U , critical radius at which resulting nuclei give off further neutrons – delayed component – subject to mean delay of nq = 1 is about 8.7 cm; critical mass = 52 kg. Problem: energy released as assembly becomes about 13s and may occur many years later fi biological hazard of . critical will blow apart unless special steps are made to prevent this. One solution: fissile material (subcritical sphere of 239Pu) surrounded by chemical explosives Fission chain reactions shaped so that when they explode the resulting shock wave implodes the to Define become supercritical. number of neutrons produced in the (n + 1)th stage of fission k ∫ number of neutrons produced in then th stage of fission from : nuclear reactors Several distinct types of reactor. Main elements of thermal reactor shown in Fig.8.3 with k = 1 fi process is critical (sustained reaction can occur – ideal for operation of power core in Fig.8.4. Latter consists of fissile material (fuel elements), control rods and plant); k < 1 fi process is subcritical (reaction will die out); k > 1 fi process is supercritical moderator. Natural uranium often used as fuel. A 2 MeV neutron has very little chance of (energy will grow very rapidly, leading to an uncontrollable explosion) inducing fission in 238U . Much more likely to scatter inelastically, leaving nucleus in excited state. After couple of such collisions energy of neutron will be below threshold of 1.4 MeV 235 238 Assume uranium is mixture of the two isotopes U and U in the ratio cc:(1- ), then for inducing fission. Neutron with its energy < 1.4 MeV has to find a nucleus of 235U . 235 238 average neutron total cross-section is sstot=+-cc tot s()1 tot and mean free path, i.e. (mean Chances very small unless its energy has been reduced to < 0.1 eV, where cross-section is distance neutron travels between interactions) is large. Before that happens it is likely to have been captured into one of the 238U resonances. Thus, either fuel must be enriched with a greater fraction of 235U (2%–3% is common in l = 1(rsnucl tot ) commercial reactors), or some method must be devised to overcome this problem. Role of moderator: (surrounds fuel elements – >> fuel elements) main purpose to slow down fast neutrons produced by elastic collisions in moderator. Absorption into resonances

8.1 8.2 238 of U is thereby avoided. Must have negligible cross-section for absorption and be 1 ZZ¢ e2 V = inexpensive. ( DO2 , or graphite used.) C 4pe0 RR+ ¢ In nuclear , delayed neutrons of no consequence because the explosion will have RA12. 13/ fm taken place long before they would have been emitted; in power reactor they are of crucial ZZand ¢ = atomic numbers of the two nuclei, RRand ¢ = radii ( = ): importance, because the fuel rods can remain in reactors for several years. With delayed 2 neutrons, each fission leads to ()nnq+-d 1 additional neutrons (dn = number of delayed Ê e ˆ hcZZ¢ 1 197MeVfm ZZ¢ [] V = = C Á h ˜ 13// 13 13// 13 neutrons per fission). In steady state, ()nnq+-=d 10. By retracting or inserting control Ë 4pe0 c¯ 12.()[]AA+ ¢ fm 137 1.2fm AA+ ()¢ rods (made of cadmium – high absorption cross-section for neutrons) number of neutrons available to induce fission can be regulated. Key to maintaining constant k value of unity setting (e.g) AAª ¢ ªª22 Z Z¢, gives VAª 53/ 8MeV. With AVªª84,MeV has to be and therefore constant power output. Presence of delayed neutrons vital, because although C C supplied to overcome Coulomb barrier. Only practical way is to mixture of nuclei to lifetime of a prompt neutron may be as much as 10-3 s in reactor, rather than 10-8 s for pure 235 supply enough . Necessary temperature estimated from EkT= B , ( kB = U , this would still be a very short time to change q and hence avoid a nuclear catastrophe! 10 7 Reactor design ensures that nq -<10, so can only become critical in presence of delayed Boltzmann’s constant). For energy of 2 MeV, temperature is ~ 3x10 K >> 10 K found in neutrons. Thus time scale to manipulate control rods becomes that of delayed neutrons, stellar interiors. Fusion actually occurs at lower temperature by combination of fact that (adequate). Important for that dq dT < 0 fi increase in temperature energy distribution is Maxwellian (with high-energy tail), and tunnelling. (See Fig.8.5) leads to fall in reaction rate. Rest is conventional engineering. From SEMF, fission of single 235 -11 Stellar fusion U nucleus releases ~200 MeV ( 32.x 10 ). (180 MeV is ‘prompt’ energy.) One Energy of due to reactions, foremost is PPI cycle: gram of 235U has about 6x1023 235ª 3x10 21 ; if fission were complete would yield 11 6 11 2 + total energy of about 10 joules, (1 MW-day). 3x10 times > yield obtained by burning 1 HH+Æ He + +ne +042.MeV gram of . In practice only about 1% of the energy content of natural uranium can be extracted. 12HH+Æ 3 He ++g 549.MeV

Fast – no large volume of moderator; no large density of thermal neutrons 33He+Æ He 4 He +21286() 1 H + .MeV established. Proportion of fissile material ~ 20%; fast neutrons used to induce fission. Fuel ( 239Pu) is obtained by chemical separation from spent fuel rods of a thermal reactor: Large energy release in the last reaction is because 4He is doubly magic nucleus. Overall, 239 239 --239 92UNpeÆ++Æ93 nnee(.36days )94 Pu ++ e 14 + 4222()HHeeÆ++++nge 24 . 68MeV n = 2.96 for 239Pu. Core: 20% 239Pu and 80% 238U ( – obtained from spent Because temperature of Sun is ~107 K, all material is fully ionised () fi fuel rods) surrounded by blanket of 238U where more plutonium is made. Can produce more produced annihilate with to release further 1.02 MeV of energy per . fissile 239Pu than consumed (‘breeder’). Hence total energy released is 26.72 MeV. Of this, each will carry off 0.26 MeV on average (lost into space). Thus on average, 6.55 MeV of electromagnetic energy is radiated Major problem: generation of radioactive waste – no totally satisfactory solution available. In from Sun for every consumed in PPI chain. Another cycle that plays an important principle could “neutralise” long-lived fission fragments by using resonance capture of role in the evolution of some stellar objects, is CNO chain (Contributes ~ 3% of energy neutrons to convert to short-lived, or stable, nuclei. Example: 99Tc () has large output of Sun.) Needs presence of any of the nuclei 12CCN,,13 14 or15 N: resonant cross-section for to 100Ru (stable ). However, amount of 6 6 7 7 radioactive waste very large and neutron energy has to be matched to particular waste 12 1 13 material, which has to be separated and prepared fi increase cost of energy production. CH+Æ N ++g 195. MeV 13 13 + NCeÆ+++e 120. MeV Nuclear fusion: Coulomb barrier n BA maximum at A ª 60; slowly decreases for heavier nuclei. For lighter nuclei, decrease 13CH+Æ 1 14 N ++g 755.MeV is much quicker, so in general lighter nuclei less tightly bound than medium size nuclei fi energy produced if two nuclei fuse to produce heavier nucleus. Energy released comes 14 1 15 from difference in binding of initial and final states (nuclear fusion). Energy NH+Æ O ++g 734. MeV released per fusion smaller than in fission. More than balanced by far greater abundance of 15ONe 15 + 168. MeV Æ+++ne stable light nuclei. Problem: Coulomb repulsion inhibits two nuclei getting close enough and together to fuse. This is 15NH+Æ 1 12 CHe + 4 +496. MeV

8.3 8.4 ( 12C, for example, could be present because the 4He produced in PPI cycle could fuse via 3727()412HeÆ+ C .MeV.) Thus, overall:

14 + 4223()HHeeÆ++++nge 24 . 68 MeV

These, and other, fusion chains all produce and from detailed models of Sun the flux of such neutrinos at surface of Earth can be predicted. Actual count rate << theoretical expectation. (solar neutrino problem). Possible solution: neutrino oscillations (see Sec 6) – only possible if neutrinos have mass.

Fusion reactors 2 PP reactions far too slow. Coulomb barriers for deuteron 1H are same as for proton and the exothermic reactions 2 2 3 1 HH+Æ1 2 Hn ++327. MeV

2 2 3 1 HH+Æ1 1 H ++ p403. MeV suggest might be suitable fuel for fusion reactor. Deuterium also present in huge quantities in sea water and is easy to separate at low cost. Even better reaction in terms of energy output is: 2 3 4 1HH+Æ1 2 He ++n17. 62 MeV

Advantages: heat of reaction is greater; cross-section is much larger. Principal disadvantage: does not occur naturally (it has a mean life of only 17.7 years) – increases cost. Working energy where deuterium-tritium reaction has reasonable cross-section is about 20 keV, i.e. 3x108 K. At these temperatures any material container will vapourise and so central problem is how to contain plasma for sufficiently long times for reaction to take place. Two main methods: magnetic confinement (ingredients are confined by electromagnetic fields); inertial confinement (small pellets of ‘fuel’ are imploded by bursts of pulsed beams). Although reactions have been observed, there is a very long way to go before a power plant could be built.

To achieve temperature T in deuterium-tritium plasma, there has to be an input of energy 43rdB()kT 2 per unit volume. Here rd is the number density of deuterium and the factor of 4 comes about because rd is equal to the number density of tritium ions and the electron density is twice this, giving 4 rd particles per unit volume. Reaction rate in plasma 2 31- is rd W, where W (measured in ms ) is related to the average cross-section for fusion reaction. If plasma is confined for time tc, then, per unit volume of plasma,

energy output r 2 W tMeV(.17 6 ) L ∫=dc ª()10--19 ms31 t energy input 6 kT dc rdB where experimental value Wª10-22 has been used. For a useful device,r L > 1, which implies 19 -3 rdct > 10 ms

This is the fi either very high particle density or long confinement time, or both, is required. 8.5