Nuclear Fusion and Radiation

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1 Nuclear Fusion and Radiation Lecture 3 (Meetings 5 & 6) Eugenio Schuster schuster@lehigh.edu Mechanical Engineering and Mechanics Lehigh University Nuclear Fusion and Radiation p. 1/39

2 Fusion Reactions It was previously noted (lecture 1) that for the D T fusion reaction to take place a relative kinetic energy of 0.36MeV would be required to overcome the Coulomb barrier. However, experimental observations indicate that substantial fusion of D T takes place at energies well below the barrier height. Classical mechanics cannot explain these observations and we must turn to quantum mechanics to understand what is happening. Consider particles approaching a rectangular potential barrier: Figure 1: Potential barrier Nuclear Fusion and Radiation p. 2/39

3 Fusion Reactions According to classical mechanics the incident particles would be totally reflected. However, according to quantum mechanics there can be partial reflection and partial transmission of the incident beam. Figure 2: Potential barrier Using a graphical wave representation: Figure 3: Potential barrier Nuclear Fusion and Radiation p. 3/39

4 Fusion Reactions Calculation of the transmission probability is complicated and will not be considered here. It can be shown that the probability for transmission, P T, has the dependencies, P T = f(e,u,a) The theory predicts significant transmission at energies well below the barrier height. The probability for a given fusion reaction to take place depends on the probability for transmission (with formation of a compound nucleus) and the probability for the given exit channel to occur. Nuclear Fusion and Radiation p. 4/39

5 Fusion Reactions Figure 4: Head-on nuclear fusion reaction (HSMK) Nuclear Fusion and Radiation p. 5/39

6 Fusion Reactions For example, 2 1D+ 3 1 T ( 5 2He) exhibits two exit channels: Nuclear Elastic Scattering: Fusion: ( 5 2He) 2 1 D T ( 5 2He) 4 2 He+ 1 0 n The combined probabilities for compound nucleus formation and de-excitation by a given exit channel represents the reaction probability. Nuclear Fusion and Radiation p. 6/39

7 Cross Sections Historically reaction probabilities have been designated as reaction cross sections. We first give a functional definition for the cross section σ. Consider a test particle of species 1 (e.g., D) moving in a medium of target particles of species 2 (e.g., T). For this case we make the following definition: Reaction Probability per Test Particle per Unit Length Traveled in the Medium σn 2 where σ is the cross section for reaction, and n 2 is the number density of the medium. Note that σ has dimensions of l 2 (e.g., cm 2, m 2 ) and n 2 has dimensions of #/l 3 (e.g., #/cm 3, #/m 3 ), and σn 2 has units of (#)/l or l 1, that is, probability/unit length traveled by the test particle in the medium. Nuclear Fusion and Radiation p. 7/39

8 Cross Sections We now give a geometrical interpretation for the cross section. Figure 5: Cross section geometrical interpretation Nuclear Fusion and Radiation p. 8/39

9 Cross Sections If each target particle has cross sectional area, σ, then P = σ n 2l 3 l 2 = σn 2 l and the probability per unit length traveled by the test particle in the medium is P = σn 2 l just as we had before. If we have n 1 test particles per unit volume of the medium (in addition to n 2 target particles per unit volume) and if the velocity of the test particles (with respect to the target particles) is v, then the total path length traveled by the n 1 particles per sec per unit volume is n 1 v. Nuclear Fusion and Radiation p. 9/39

10 Cross Sections Thus, the reaction rate, the number of reactions between particles 1 and 2 per unit volume of medium per sec is Reaction Rate = (σn 2 )(n 1 v) σn 2 = the reaction probability per unit length traveled by one particle. n 1 v = the total path length traveled by all particles of type 1 per unit volume per sec. In the CGS system the reaction rate, R, has the following units: ( )( ) 1 1 (cm ) R = (σn 2 )(n 1 v) = (cm 2 ) cm 3 cm 3 = # sec cm 3 sec Note that the quantity n 1 v is designated the flux φ in neutron transport calculations. Nuclear Fusion and Radiation p. 10/39

11 Cross Sections The determination of σ as a function of E is the work of nuclear physicists. Experimental values of cross sections are needed in general and theory only provides a guide. Sometimes we can get an order of magnitude estimate for a cross section by use of the geometric argument σ πr 2. For example, consider the D T reaction for which R DT cm and, thus, σ πr cm 2 This, of course, is NOT an accurate cross section value but indicates the order of magnitude of nuclear cross sections, cm 2. For this reason this dimension has been given a name as follows cm 2 1barn Nuclear Fusion and Radiation p. 11/39

12 Cross Sections Consider first the fusion reaction 1 1H H 2 1 D +β + +ν (Q = 1.44MeV) The cross section for this reaction is too small to be observed in the laboratory. However, the cross section can be inferred from stellar processes and can be estimated based on theory ( 10 9 b). The two reactions 1 1H D 3 2He+γ 1 1H T 4 2He+γ (Q = 5.5MeV) (Q = 19.8MeV) have small, but observed cross sections ( 10 6 b). These cross sections are too small to be of interest for fusion power. Nuclear Fusion and Radiation p. 12/39

13 Cross Sections Next consider the reactions 2 1D D 3 2He+ 1 0 n (Q = 3.3MeV) 2 1D D 3 1T H (Q = 4.03MeV) 2 1D T 4 2He+ 1 0 n (Q = 17.6MeV) 2 1D He 4 2He+ 1 1 H (Q = 18.3MeV) The cross section versus deuteron energy for these reactions is given in Fig. 6. The following points are noted: σ DT is by far the largest cross section. σ DT is significant well below 0.36MeV (360keV ), the approximate Coulomb barrier height. σ DT at 110keV is 5b, which is approximately an order of magnitude larger than πr 2. Nuclear Fusion and Radiation p. 13/39

14 Cross Sections Figure 6: D D, D T and D He 3 cross sections. Nuclear Fusion and Radiation p. 14/39

15 Fueling Strategy As we shall see later in the course, under the proper conditions some of the fusion reactions listed above have large enough cross sections to yield reaction rates which are interesting for fusion power production. However, we must first address a more fundamental question: Are there sufficient reserves of fusion fuels for a fusion energy economy? Let us examine this question in terms of a D T based fusion power economy. What are the fuel requirements? Deuterium is a stable isotope and can be recovered from water ( atom %) relatively easily. Tritium, on the other hand, is radioactive decaying by beta decay to helium-3, 3 1 T β 3 2 He, T 1/ years. Nuclear Fusion and Radiation p. 15/39

16 Fueling Strategy Because of the short half-life of T, there are not sufficient amounts of T in nature to sustain a D T fusion power economy. Therefore, T for D T fusion power must be manmade. Fortunately this appears possible using the following strategy: 2 1 D+3 1 T 4 2 He+1 0 n 1 0n+ 6 3 Li(7.5%) 4 2He+ 3 1 T 1 0n+ 7 3 Li(92.5%) 1 0n+ 4 2 He+ 3 1 T As we shall show later, we can design a fusion reactor so that it will breed the tritium it consumes by using the D T generated neutrons and lithium. Thus, the fuel requirement for T becomes a resource requirement for lithium. Lithium and deuterium resources are sufficient for 1000 s of years of energy production via D T fusion reactors. Nuclear Fusion and Radiation p. 16/39

17 Other Energy Units In examining energy resources it is useful to introduce a quantity of energy called the terawatt yr (TW YR) defined as 1TW YR (J/sec)(365)(3600)(24)(sec/yr) J Consider the following energy consumption points: 1975 : US consumed 2.6TW YR WORLD consumed 8.5TW YR 2025 : US consumption 4.5TW YR WORLD consumption 27TW YR Nuclear Fusion and Radiation p. 17/39

18 Other Energy Units List of countries by electricity consumption List of countries by total primary energy consumption Energy in the United States Nuclear Fusion and Radiation p. 18/39

19 Fusion Gain in Ideal Reactor So far everything looks promising for a D T fusion reactor so let us try to construct an idealized fusion reactor. Consider the following system: Figure 7: Idealized fusion reactor. Nuclear Fusion and Radiation p. 19/39

20 Fusion Gain in Ideal Reactor The probability for a D T fusion reaction per cm of D + path length in the target, P/l, is given by P/l = σn 2 = ( cm 2 ) }{{} σ 0.2g/cm 3 3g/mol ( #/mol) }{{} n 2 =n T On this basis we can easily make the target thick enough to achieve a reaction probability approaching unity. Let us define an energy gain, G o, for our system: = 0.2 # cm G o = Fusion Energy/Incident D Energy of Incident D = 17.6MeV/D-T Fusion 0.1MeV/Incident D = 176 THIS IS A VERY HIGH GAIN. THE REACTOR LOOKS VERY ATTRACTIVE. IS THERE ANYTHING WRONG WITH THIS ANALYSIS? Nuclear Fusion and Radiation p. 20/39

21 Competing Processes We neglected the effect of Coulomb scattering in our calculation of G o. Coulomb or Rutherford scattering (elastic scattering between charged particles) is the primary source of competition for fusion. In Coulomb scattering the incident D+ looses energy and because the fusion cross section decreases as the incident particle energy decreases, the probability for fusion decreases. If the Coulomb scattering cross section is large with respect to the fusion cross section, the D+ will be down-scattered in energy before it can engage in significant fusion reaction. Nuclear Fusion and Radiation p. 21/39

22 Coulomb Scattering Cross Sections The detailed derivation of the Coulomb scattering cross section can be found in the textbook (Principles of Fusion Energy, Chapter 3). We shall use some simple arguments here to estimate σ CS. We would expect significant Coulomb scattering when two nuclei feel the presence of the Coulomb potential relative to their mutual kinetic energy. Consider the following schematic representation. Nuclear Fusion and Radiation p. 22/39

23 Coulomb Scattering Cross Sections Figure 8: Coulomb potential. Nuclear Fusion and Radiation p. 23/39

24 Coulomb Scattering Cross Sections If the kinetic energy of particle 2 corresponds to point a, particle 2 will not feel the Coulomb potential, that is, since T a >> E c (R) the Coulomb potential is not strong enough to deflect (scatter) particle 2 significantly. On the other hand, if the kinetic energy of particle 2 corresponds to point b, T b E C (R) and particle 2 will feel the Coulomb potential. In this case, particle 2 will be deflected significantly. Based on these arguments we can define the effective range, R eff, of the Coulomb potential of particle 1 with respect to particle 2 by the following condition: E 1,2 (R eff ) = T 1,2 Z 1Z 2 4πǫ o R eff = T 1,2 Nuclear Fusion and Radiation p. 24/39

25 Coulomb Scattering Cross Sections Taking Z 1 = Z 2 = C and ǫ o = F/m, R eff = Z 1Z 2 4πǫ o T = m = m (T = T 1,2 ) T(J) T(keV) If we employ a geometric interpretation for σ CS : σ CS π ( Reff 2 ) 2 = π 4 ( ) 2 T 2 (kev) m 2 = T 2 (kev) barns The above expression gives a very good estimate for large angle (significant deflection) Coulomb scattering events when T is the center of mass energy of the two particles: T 1 2 M ov 2 = 1 2 M 1 M 2 M 1 +M 2 V 2 (V relative velocity) Nuclear Fusion and Radiation p. 25/39

26 Coulomb Scattering Cross Sections It must be emphasized that this expression is only an estimate. lt does, however, give reasonable values for large angle scattering. Moreover, it gives the correct energy dependence, σ CS 1/T 2, for all Coulomb scattering events (large and small angle). Nuclear Fusion and Radiation p. 26/39

27 Energetic Deuterons on Cold Targets Consider the following situation: Figure 9: Energetic vs. cold deuteron/triton collision Nuclear Fusion and Radiation p. 27/39

28 Energetic Deuterons on Cold Targets For the deuteron on deuteron collision case M o = M DM D = M2 D = M D M D +M D 2M D 2 Thus, T 1 2 M ov 2 = 1 2 M D 2 V 2 D = 1 2 T D σ D D(cold) CS = ( T b = D 2 ) 2 TD 2 b For our idealized reactor, T D = 100keV, σ D D(cold) CS = b = 6.4b Note that σ D D f 0.03b at 100keV. Therefore, σ CS /σ f 200. Nuclear Fusion and Radiation p. 28/39

29 Energetic Deuterons on Cold Targets For the deuteron on triton collision case M o = M DM T = M D(3M D /2) M D +M T M D +(3M D /2) = 3M2 D /2 5M D /2 = 3M D 5 Thus, T 1 2 M ov 2 = 1 2 3M D 5 V 2 D = 3 5 T D σ D T(cold) CS = ( 3T D 5 ) b = TD 2 b For our idealized reactor, T D = 100keV, σ D T(cold) CS = b = 4.4b Note that σ D T f 5b at 100keV. Therefore, σ CS /σ f 1. Nuclear Fusion and Radiation p. 29/39

30 Energetic Deuterons on Cold Targets We will now estimate the energy gain of our idealized fusion reactor when D T Coulomb scattering is taken into account. Consider the following: Figure 10: Incident D beam on T target Nuclear Fusion and Radiation p. 30/39

31 Energetic Deuterons on Cold Targets As the beam penetrates into the target, beam particles can engage in two types of interactions with target particles: (1) D T fusion (2) D T Coulomb scattering Note that if fusion takes place, a deuteron is removed from the incident beam. Figure 11: D T fusion reaction Nuclear Fusion and Radiation p. 31/39

32 Energetic Deuterons on Cold Targets If scattering takes place, the scattered deuteron (with less energy!!!) is still available for fusion or additional scattering. Figure 12: D T scattering Nuclear Fusion and Radiation p. 32/39

33 Energetic Deuterons on Cold Targets After Coulomb scattering, a beam particle has been degraded in energy and: (1) the probability for fusion has decreased (2) the probability for Coulomb scattering has increased Figure 13: Fusion vs. scattering probabilities Nuclear Fusion and Radiation p. 33/39

34 Energetic Deuterons on Cold Targets The following table gives specific values E(keV) σ f (b) σ CS (b) σ f /σ CS Thus, as the beam penetrates into the target, the beam energy is degraded and the probability for fusion decreases dramatically. Let us try to estimate the effect of beam degradation on the fusion energy gain, G. Nuclear Fusion and Radiation p. 34/39

35 Energetic Deuterons on Cold Targets Figure 14: Beam degradation Remember that in slide 10 we defined the reaction rate as R = (σn 2 )(n 1 v) }{{} φ where φ denotes the particle flux. Therefore, the number of fusion reactions at E o in dx about x is given by (we neglect fusion reactions at E D < E O ): R(x)dx = I(x) }{{} φ(x) σ f }{{} σ n }{{} n 2 dx [I] = [nv] = # cm 2 s Nuclear Fusion and Radiation p. 35/39

36 Energetic Deuterons on Cold Targets The total number of fusion reactions in the target per cm 2 s is t 0 R(x)dx The fusion energy gain, G, is thus, t 0 G = QR(x)dx I o E o Now, t 0 R(x)dx = t 0 I(x)σ f ndx = σ f n t 0 I(x)dx To calculate I(x) consider the reduction in beam intensity, di, in dx about x, di = I(x)σ f ndx I(x)σ CS ndx Nuclear Fusion and Radiation p. 36/39

37 Energetic Deuterons on Cold Targets Figure 15: Beam degradation The Coulomb scattering term in this equation assumes that each scattering event degrades the incident deuteron energy below the point at which it could engage in a fusion reaction. A better assumption would be: di = I(x)σ f ndx I(x) σ CS s ndx where s is the number of Coulomb scattering collisions required for substantial degradation of E D. Nuclear Fusion and Radiation p. 37/39

38 Energetic Deuterons on Cold Targets Thus, di ( dx = I(x)n σ f + σ CS s ) I(x) = di dx 1 n ( σ f + σ CS s ) Now, t 0 R(x)dx = σ f n t 0 I(x)dx = σ f n n ( σ f + σ CS s ) t 0 ( di ) dx dx lf we take the thickness t large enough to totally attenuate I o, t 0 R(x)dx = σ f I ( o σf + σ ) CS s Nuclear Fusion and Radiation p. 38/39

39 Energetic Deuterons on Cold Targets Thus, G = σ f I ( o σf + σ ) Q = CS I s o E o σ f Q E o ( σf + σ CS s Now we return to the D on T (cold) case and make the most pessimistic assumption about s, i.e., s = 1. Then, ) G = 5 ( ) = 0.53G o 93 Thus, D T scattering degrades G by only about a factor of 2 and the resulting gain is still substantial. WHAT IS STILL MISSING? WHAT MUST BE CONSIDERED IN ADDITION TO D T (COLD) COULOMB SCATTERING? Nuclear Fusion and Radiation p. 39/39

Nuclear Binding Energy

Nuclear Binding Energy Nuclear Energy Nuclei contain Z number of protons and (A - Z) number of neutrons, with A the number of nucleons (mass number) Isotopes have a common Z and different A The masses of the nucleons and the