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14.a. Nuclear Reactions
14.b. Fission
14.c. Fusion
A nuclear reaction is induced by the bombardment of nuclei with a particle beam. The schematics of such a reaction are
a + X -> b + Y,
where X and Y are nuclei, a is the incoming particle and b is an outgoing particle. These reactions are also written sometimes as X(a,b)Y. An example is 13C(p,n) 13N.
Nuclear reaction are governed by conservation laws: Energy-momentum, nucleon number (A), and charge (Q).
The incoming particle carries kinetic energy Ka. The outgoing nucleus and particle also carry kinetic energies KY and Kb. The target nucleus, X, is supposed to be at rest. So the energy balance of the reaction is
Ka + c2( MX + Ma ) = Kb + KY + c2( MY + Mb )
The energy produced in the reaction is
Q = Kb + KY - Ka.
If Q > 0 then the reaction converts mass into energy and is exothermic. If Q < 0, then energy is converted into mass and the reaction is endothermic. Endothermic reactions have a threshold energy (minimal Ka at which the reaction occurs), which is larger than the mass energy difference c2( MY + Mb - MX - Ma ), because of momentum conservation.
Nuclear Cross Sections.
Suppose a number of particles reacting with target nuclei is DR out of R0 incoming particles over unit area. Then DR/R0 is the probability of nuclear reactions. This should be equal to the fraction of total area covered by the target nuclei, N s. where N is the number of target nuclei over unit area. I.e. DR = R0 N s. Now N, the number of target nuclei is over unit area is equal to the density of target nuclei, n, times the thickness of the target, N = n x. Thus DR/R0 = n x s. This gives a possibility to experimentalists to measure nuclear cross sections.
Take now an infinitesimal layer of target dx, the decrease in the intensity of the beam is
-dR = R n x s.
This is a differential equation very similar to the one we solved for radioactive decays. The solution is similar too:
R = R0 e- n x s
The decrease of the initial number of particles is exponential, and it is controlled by the cross section and the density of nuclei. Cross sections are usually very strongly dependent on the primary energy (Ka). Obviously the cross section is of dimension of area, m2. The unit of cross section is 1 barn = 10-28 m2. This is a very large cross section. Often used is mb (millibarn). The largest known cross section is the low energy neutron capture cross section of 113Cd:
n + 113Cd -> 114Cd + g. s = 104 b = 10-24 m2.
This is obviously orders of magnitude larger than the geometric cross section: sgeom = p R2 = (4.83 x 1.2 f)2 x 3.14 = 1.05 x 10-28 m2 ~ 1 b. Therefore 113Cd is used to stop thermal neutrons (e.g. in a fast switch off of a nuclear reactor). Thermal neutrons have average energies of O(k Troom). In nuclear reactors it is important to slow neutrons down to thermal velocities, because the capture cross section initiating fission has a large cross section at thermal velocities only. Therefore one needs moderators (not Cadmium!) to slow them down. Usually boron, graphite, heavy water, or light water are used. These slow down neutrons by multiple elastic scattering.
Nuclei in the range 50<A<80 have the highest binding energies per nucleon, approximately 8.5 MeV/nucleon. Lighter nuclei have smaller binding energies because of surface effects. Their surface to volume ratio is much larger, consequently a larger fraction of the nucleons are near the surface and have smaller binding energies. Binding energy/ nucleon goes as low as 1 MeV for the deuteron. Nuclei heavier than those in the above range also have smaller binding energies because the positive Coulomb repulsion energy. Thus, when forming larger nuclei from smaller ones (such as p or d) or when splitting heavy nuclei (such as U, or Pt) mass energy is transformed into kinetic energy.
There is another difference between heavy nuclei and medium range nuclei, besides the differing binding energy per nucleon. The ratio of neutrons/protons is lower in the medium range therefore at splitting some of the neutrons become superfluous and emitted as free neutrons. The easiest way to split a heavy nucleus is by neutrons. Neutrons are not repulsed by the positively charged nucleus and as such they penetrate the nucleus with ease. Typical neutron induce fission reaction of 23592U are
10n + 23592U -> 14054Xe + 9438Sr + 2 10n,
10n + 23592U -> 13250Sn + 10142Mo + 3 10n,
10n + 23592U -> 14156Ba + 9236Kr + 3 10n.
It is easy to find out the number of neutrons produced if we require that both the total A and total Z balance on the side of the equation. It is natural to produce several neutrons in fission reactions because the fission products have much lower Z (thought the total Z is obviously conserved) and at lower Z the ratio of N/Z is smaller for the most stable isotopes.
The energy production can be roughly estimated by taking 7.5 as the average binding energy for the U nucleus and 8.5 for the fission products. Then in the last reaction we have
8.5x(141+92)-7.5x235 = 208 MeV
In other words every fission produces 208 MeV heat. How much is produced by a kg of 23592U? The number of 23592U nuclei in one kg is N = NA 1000g / (235 g) = 2.56x1024 nuclei, each producing 208 MeV provides
E = 208MeVx 2.56x1024= 2.37x107 kWh. This would supply 10MW for 2370 hrs. This is the same energy as that of 20,000 tons of TNT.
Notice that in the above fission reactions more neutrons are produced than used. The factor between the two is denoted by K. Avaraging over all fission reactions K=2.5 for 23592U. The fact that K>1 makes a chain reaction possible. neutrons produced by one nucleus can be used to split more nuclei. One problemis that the neutrons produced carry energies up to 100 MeV or more. Such energetic neutrons are not likely to cause fission. Neutrons need to be slowed down before they can split other nuclei. In a sufficiently large chunk of 23592U they can be slowed down by multiple collisions so that they produce new fissions. The minimal size of such a chunk is called the critical mass.
In nature the abundance of 23592U is only 0.7%. Most of the nuclei are of the isotope 23892U. If a nucleus is split and neutrons are produced then chances are they are absorbed by 23892U nuclei, which do not likely undergo fission and do not produce new neutrons. They rather produce the elemnents Pt or Np. In a nuclear reactor one wishes to keep K slightly above 1. This can be easily done if the concentration of 23592U is slightly increased from natural levels.
Due to surface effects light nuclei, with the notable exception of 4He, usually have much smaller binding energies per nucleon then heavier ones. This implies that if they fuse together they go into a lower energy state and kinetic energy is produced. That is the basis of fusion reastions. Fusion reactions produce the energy radiation of stars, among others of the sun. Starting mostly from H heavier atoms are fused together to create more stable nuclei, mostly He. The ceration of He nuclei happens in a series of nuclear reaction which, obviously, should contain beta-decays to transform some of the protons into neutrons that are required in He and other heavier atoms. Examples for such chains are the following:
p + p -> 21d + e+ + n
p + 21d -> 32He + g
p + 32He -> 42He + e+ + n or
32He + 32He -> 42He + p + p
This is called the proton cycle that produces 42He = a-particles from protons. Such a cycle is equivalent to the following single reaction
p + p + p + p -> 42He + e+ + n + e+ + n + 24MeV
In other words, the production of every a-particle releases 24 MeV kinetic energy. This is the energy that produces the heat of the sun, the energy of H-bombs and of future fusion reactors.
In future fusion reactors a more promising series of nuclear reactions would start from deuterium, heavy hydrdogen that can be found in sea water in abundance. The reaction one would use are
d + d -> 32He + n + 3.3 MeV
d + d -> 31H + p + 4 MeV ( 31H is tritium a heavy, radioactive isotope of hydrogen)
d + 32He -> 42He + n + 17.6 MeV
The major obstacle of producing fusion power is the coulomb repulsion of all of the particles on the left hand side of all of the listed reactions. Before nuclear forces start to act between the two fusing nuclei they are repelled by each other. This repulsion must be overcome by a sufficient amount of kinetic energy.
Suppose that the two deuterons must approach each other to a distance of 10f so that the attractive nuclear forces would start to act between them. What should be the temperature of the plasma, formed from electrons and deuterons (no deuterium can exist at that temperature, so atoms get ionized), that the fusion reaction would start? The repulsion energy at a distance of 10f is EC = k e2 / R = 1.44 MeV f / 10 f = 0.144 MeV. To have a substantial fraction of deuterons half of that energy the temperature of the plasma must be 3kT /2~ 0.07 MeV, T = 40 x 300 x 70,000 K = 560 Million K, that is a pretty high temperature, much higher than the one inside the sun.
In fact, the true ignition temperature is a little lower, it is around 4 x 108 K. The problem is that even if we succeed raising the temperature of the plasma to this temperature a lot of energy is lost to the environment due to various mechanisms so that the temperature drops again. Also a certain density must be maintained so that there would be a reasonable chance that two deuterons meet. Again, the natural tendency of the plasma (think of entropy) is to diffuse and thus stop the fusion reaction. So the upshot is that it is not enough to reach and keep the ignition temperature but one has to satisfy the Lawson criterion to achieve a positive energy output,
n t ³ 1014 s cm-3 (for d-t plasma)
n t ³ 1016 s cm-3 (for d-d plasma)
where n is the density and t is the confinement time. Both the constant and the ignition temperature are substantially lower for d-t plasma, but t (tritium) is radioactive and has a fairly short half-lifetime (12.3 years) so its abundance is extremely low. It is expensive to use t. The Lawson criterion is essentially the condition of the output energy be larger than the energy we need to put into the fusion reactor to heat the plasma to ignition temperature. That energy per unit volume is obviously Einput = const. n, where n is the density. The output energy, however increases with n2, because the probability of collisions is proportional to n2. The output energy is also proportional to t, the time that we are able to maintain the plasma at the ignition temperature. Thus Eoutput = const' t n2. Then the Lawson criterion follows from Eoutput > Einput .
There are various ways that physicists try to satisfy the lawson criteria and achieve ignition temperature. The first successful design is the "tokomak" invented by Russians. This is a torus, which, using complicated magnetic fields confines the plasma. There is a toroidal current outside the torus that has a dual role of heating the plasma and of creating the magnetic field confining the plasma. The biggest problems with this design is the existence of various instabilities in the plasma that must be overcome. Current designs are on the verge of achieving breakeven condition and ignition temperature.
The other very promising design is the inertial confinement machine. This plans to maintain the plasma for short periods of time only, but at very high densities. The high densities are achieved by concenrated strike of a large number of high powered laser beams on a pellet made of solid d-t mixture. Again, the current designs are at the verge of satisfying the Lawson criterion and reaching ignition temperature. It is expected that by the second decade of the new millenium we will be close to build commercial fusion reactors.