Fusion Energy Course RRY115 An Essay about Fusion Reactions

Transcription

1 Fusion Energy Course RRY115 An Essay about Fusion Reactions

2 Abstract The understanding of why a fusion reaction actually is exothermic when it comes to light nucleus and endothermic when it comes to heavier nucleus can be explained by a simple subatomic model of the strong interactions together with the coulomb forces and statistical observations of dierent nuclide. When calculating energies required for a fusion to take place for a deuterium and tritium nucleus (to overcome the repelling coulomb forces) one can be doubtful about the possible theories for fusion being able to take place, when not introduced to the quantum tunneling eects, considering the sun's low temperature. 1

3 Contents 1 Nuclear binding energy The semi empirical mass formula Calculation of the energy released in a D-T reaction Quantum mechanics 5.1 Tunneling eect Nuclear reaction cross section Fusion cross section Conclusions 8

4 Introduction Fusion energy is a hot research topic today as the society demands for more sustainable, high producing and environmentally safe energy production. Although it seems that a working commercial fusion reactor is a couple of decades away to be realized, it has also a military interest. A brief explanation of how energy is released in a fusion reaction, together with cross sections and general fusion theory will be presented. An explanation of the quantum tunneling eects and how it makes fusion take place in our daily surroundings is also covered. 1 Nuclear binding energy As already known, nuclear binding energy is holding the nucleus together forming a nuclide. This is nothing less than a potential energy which is of main interest when it comes to nuclear energy production. 1.1 The semi empirical mass formula The semi empirical mass formula is dened as M(Z, A) = Zm( 1 H) + Nm n B(Z, A)/c, where it states that the combined mass of the protons and neutrons in a nucleus is the same as the free nucleons minus the binding energy, B. The total binding energies of a nucleus is expressed with several terms each regarding a specic nuclear force contribution and correction terms. [1][] The strong interaction between nucleons has a rather short range and therefore only the closest neighbor nucleons contributes to a binding energy. Because the nuclear density is nearly constant, the contribution of each nucleon is summed. The binding energy contribution from the strong interaction between nucleons is therefore a v A, where A is the atomic mass (observe that protons and neutrons are considered having the same contribution to the strong interaction forces) and a v is a constant. Because of geometrical reasons, a volume is associated with a closed surface. The nucleons at the surface of the nucleus does not have the same amount of surrounding nucleons as inside the core. As the volume of the nucleus is proportional to the amount of nucleons A, the surface is proportional to A /3. A correction term of a s A /3 is therefore added. The coulomb repulsion of the protons in the nucleus is also considered which is reducing the binding energy due to the repelling forces. Assuming that the nucleus is a uniformly charged sphere one can calculate that 3 5 (1/4πɛ 0R 0 )Q /R is the potential energy in the nucleus. Considering a nite number of distinct charged protons (there are Z-1 protons interacting with a single proton in the nucleus, there are Z protons all having the same forces applied to them) we can substitute Q = e Z(Z Z) and by knowing that the radius of the nucleus is proportional to A 1/3 Z(Z 1), the coulomb repulsion energy becomes a C where A 1/3 a C is a constant. An observed relationship of the stable isotopes is that Z = A/, in other words the same amount of neutrons and protons is present in the nucleus. The asymmetry term takes this into account. Pauli's exclusion principle states that two fermions can not have the same quantum state in a nucleus. Protons and 3

5 neutrons are dierent types of particles and will therefore occupy dierent quantum states. If a nucleus have more neutrons than protons (rough approximation, there are underlying quantum mechanical calculations for this which are omitted here) there will be neutrons with higher energy than the most energetic proton. A neutron can turn into a proton through β-decay. This is because the nucleus falls into a lower total energy state which means it has a higher binding energy per nucleon when the relation between neutrons and protons are symmetrical. (N Z) The asymmetry term is a (A Z) sym A = a sym A where N is the number of neutrons. The last term that contributes to the binding energy is the so called pairing eect term. More than 97 % of the stable nuclide have an even number of protons and neutrons. The binding energy seems to be higher per nucleon in nucleus containing even number of neutrons and protons. This is because of the Pauli exclusion principle stating that a nucleus would have a lower energy (higher binding energy) if the number of protons with spin up is equal to the number of protons with spin down. The same applies for neutrons. The pairing term is stated to be δ = +a p A 3/4 if Z, N is even 0 if A is odd a p A 3/4 if Z, N is odd where a p is a constant. The calculations is omitted here. The total binding energy can now be expressed as B = a v A a s A /3 Z(Z 1) (A Z) a C a A 1/3 sym A + δ. The constants a v = 15.5 MeV, a s = 16.8 MeV, a C = 0.7 MeV, a sym = 3 MeV and a p = 34 MeV are chosen to t the experimental curve in Figure 1. As can be seen in the gure, lighter nuclide have lower binding energy per nucleon up to about 56 Fe. Therefore, a fusion of two light nuclide leads to a compound where each of the nucleons have higher binding energy than before the fusion. This binding energy increase is usually released as kinetic energy. 1. Calculation of the energy released in a D-T reaction The reaction for a D-T fusion is D + T 4 He + n. The total mass before the reaction to occur is = u and the mass of 4 He and a neutron, n, is = u. The mass dierence before and after the reaction is therefore = u which is corresponding to the increase in the binding energy per nucleon. This energy is given to the 4 He and n as kinetic energy that equals = 17, MeV. [3] With the aid of the fundamental laws of momentum and energy conservation, we can show that E4 He 3.5 MeV. Assume the deuterium and tritium are at rest when merged together, in other words the total momentum before the reaction is zero, therefore the total momentum after must be zero. The alpha particle and neutron produced by the fusion will therefore have the relation m α v α = m n v n E n = m αv α m n = m α m n E α by the conservation of momentum. As noted before, the kinetic energy released by the fusion reaction is 17.6 MeV, which the alpha particle and neutron is 4

6 Figure 1: An experimental data on the binding energies per nucleon of dierent isotopes. carrying; E n + E α = mnv n + mαv α = mα m n E α + E α = 17.6 MeV E α = 17.6m n m n+m α 3.5 MeV. The rest of the energy is carried by the neutron, 14.1 MeV. In Table 1 dierent fusion reaction releasing energies are compared. [4] Reaction Energy (M ev ) p + p D 0.93 H + D 3 He He + 3 He 4 He + p 1.86 D + T 4 He + n 17.6 D + 3 He 4 He + p 18.3 D + D 3 He + n (50%) 3.7 D + D T + p (50%) 4.03 Table 1: Fusion reactions and the released energy. The two dierent D + D reactions have an equal probability of occurrence. Quantum mechanics In quantum mechanical theory all particles have a wave motion (particle wave dualism). The time independent Schrödinger equation h d ψ m dr + V (r)ψ(r) = 5

7 Eψ(r) describes the non relativistic particle motion, where E is the particle energy, m the mass of the particle and V (r) the potential along the particle motion. [5].1 Tunneling eect In classical physics, if E < V (r) the particle will reect at the barrier. It can not overcome the barrier as long as it has an energy lower than the potential energy. In quantum theory, it exists a solution to the Schrödinger equation beyond the potential barrier, meaning that the barrier can be penetrated by the particle. The square of the solution to the Schrödinger equation of the outgoing particle wave divided by the square of the incoming particle wave, P = ψ o (r)/ψ i (r) 1 =, (1) V0 4 E(V o E) sinh ( m(v 0 E)/ h a) describes the probability of the particle to be penetrating a rectangular potential barrier. See Figure. [] Figure : The particle waves are shown. Clearly demonstrates the tunneling eect. The nuclei in the sun does not have enough energy (temperature) to classically overcome the coulomb barrier of two positively charged nuclei. Fortunately, due to the quantum mechanical tunneling eect, this is still possible and likely to occur due to the particle density attained by the gravity forces are high enough so that the Lawson criteria can be fullled. [6][7] 3 Nuclear reaction cross section Cross section of a nuclear reaction describes the probability of a nuclear reaction being able to occur. Usually, cross section is denoted as σ and is measured in barns (1 barn=10 4 cm ). 6

8 3.1 Fusion cross section The coulomb barrier Figure 3 is the potential barrier between two charged nuclei. It follows the coulomb potential V = e Z iz j 4πɛ 0(r) up until the radius where the strong interaction is more dominant than the repelling coulomb force. Figure 3: The coulomb barrier potential. This kind of potential is approximated with thin rectangular potentials with the width dr and height e Z iz j 4πɛ 0(r). The probability (Equation 1) is approximated and leads to dp = exp( dr (m/ h )(V (r) Q)). Summation over r yields P = e G where G, the Gamow factor, is further approximated to G = e πz iz j 4πɛ. 0 hv The energy dependence of the cross section makes it proportional to v and e G, σ 1 v e G. When taking into account that the velocity of particles in gas are distributed according to the Maxwell-Boltzmann velocity distribution, one can express the proportion of particles found with the velocity v in a gas at a certain temperature, n(v) e mv /kt. This leads to being able calculating the expected reaction rate <σv> 1 0 v e G e mv /kt v dv. Dierent reaction rates at dierent temperatures for several fusion reactants can be found in Figure 4. [] 7

9 Figure 4: The reaction rate of fusions at dierent temperatures. 4 Conclusions As can be seen in Figure 4, the D-T reaction is overall the best fusion reactor fuel due to the high reaction rate at lower temperatures. The raw material of deuterium is abundantly available. Although, it also has problematic issues that has to be handled. The neutron bombardment can not be magnetic conned and therefore could possibly destroy the inside antennas and other diagnostic tools in the reactor vessel. The choosing of materials have to be done wisely. Another issue is the 4 He products being an impurity after they have released all their energy to the plasma. Bremsstrahlung losses are considered low when it comes to D-T reactions (due to the low operating temperatures). [4] Fusion energy releases a lot more energy per weight than any other known technical in use energy source today. [8] There are lots of deuterium in seawater and fusion requires less (in volume) of raw material to achieve a certain amount of kinetic energy than gasoline and other fuel types. Compared to ssion, it does not generate any extremely long lived harmful radioactive isotopes. Unfortunately, it is several decades left before a commercial fusion reactor is ready to be built. 8

10 References [1] G. Choppin, J-O. Liljenzin, J. Rydberg, Radiochemistry and Nuclear Chemistry - 3rd ed., Butterworth-Heinemann, Woburn, MA, 001, pp [] S.K. Krane, Introductory nuclear physics, John Wiley & Sons inc., Hoboken, 1988, pp. 15-3, 65-70, [3] J. Magill, G. Pfennig, J. Galy, Karlsruher Nuklidkarte - 7th ed., 006. [4] T. Fülöp, G. Papp, I. Pusztai, Fusion energy - Lecture notes, Chalmers university of technology, Nuclear engineering, Department of Applied Physics, 011. [5] P. Apell, B. Lundqvist, G. Niklasson, Kvantfysik del 1, Institute of applied and fundamental physics, CTH, 008. [6] J. D. Lawson, Some Criteria for a Power Producing Thermonuclear Reactor, Proceedings of the Physical Society B, Volume 70 (1957), pp. 6. [7] R. Nave, Hyperphysics, < (visited 8, February, 01). [8] R. Nave, Hyperphysics, < (visited 8, February, 01). 9