Nuclear reactions: simple and clear. Physics problem solving: nuclear reactions
For a long time, a person did not leave the dream of the mutual transformation of elements - more precisely, of the transformation various metals into one. After realizing the futility of these attempts, the point of view about the inviolability of chemical elements was established. And only the discovery of the structure of the nucleus at the beginning of the 20th century showed that the transformation of elements into one another is possible - but not by chemical methods, that is, by affecting the outer electron shells of atoms, but by interfering with the structure of the atomic nucleus. Phenomena of this kind (and some others) are related to nuclear reactions, examples of which will be discussed below. But first, it is necessary to recall some basic concepts that will be required in the course of this discussion.
General concept of nuclear reactions
There are phenomena in which the nucleus of an atom of one or another element interacts with another nucleus or some elementary particle, that is, it exchanges energy and momentum with them. Such processes are called nuclear reactions. Their result may be a change in the composition of the nucleus or the formation of new nuclei with the emission of certain particles. In this case, options such as:
- the transformation of one chemical element into another;
- fusion, that is, the fusion of nuclei, in which the nucleus of a heavier element is formed.
The initial phase of the reaction, determined by the type and state of the particles entering it, is called the inlet channel. Exit channels are the possible pathways that a reaction will take.
Rules for recording nuclear reactions
The examples below demonstrate the ways in which it is customary to describe reactions involving nuclei and elementary particles.
The first method is the same as that used in chemistry: the initial particles are placed on the left side, and the reaction products are placed on the right. For example, the interaction of a beryllium-9 nucleus with an incident alpha particle (the so-called neutron discovery reaction) is written as follows:
9 4 Be + 4 2 He → 12 6 C + 1 0 n.
The upper indices indicate the number of nucleons, that is, the mass numbers of nuclei, the lower indices, the number of protons, that is, atomic numbers. The sums of both on the left and right sides must match.
An abbreviated way of writing the equations of nuclear reactions, often used in physics, looks like this:
9 4 Be (α, n) 12 6 C.
The general form of such an entry is: A (a, b 1 b 2 ...) B. Here A is the target nucleus; a - incident particle or nucleus; b 1 , b 2 and so on - light reaction products; B is the final core.
Energy of nuclear reactions
In nuclear transformations, the energy conservation law is fulfilled (along with other conservation laws). Wherein kinetic energy particles in the input and output channels of the reaction can differ due to changes in the rest energy. Since the latter is equivalent to the mass of the particles, the masses before and after the reaction will also be different. But the total energy of the system is always conserved.
The difference in the rest energy of the particles entering into the reaction and leaving it is called the energy yield and is expressed in a change in their kinetic energy.
In processes involving nuclei, three types of fundamental interactions are involved - electromagnetic, weak and strong. Thanks to the latter, the nucleus has such an important feature as a high binding energy between its constituent particles. It is significantly higher than, for example, between the nucleus and atomic electrons or between atoms in molecules. This is evidenced by a noticeable mass defect - the difference between the sum of the masses of nucleons and the mass of the nucleus, which is always less by a value proportional to the binding energy: Δm = E St / c 2 . The mass defect is calculated using a simple formula Δm = Zm p + Am n - M i, where Z is the nuclear charge, A is the mass number, m p is the mass of the proton (1.00728 a.m.u.), m n is the mass of the neutron ( 1.00866 amu), M i is the mass of the nucleus.
When describing nuclear reactions, the concept of specific binding energy is used (that is, per nucleon: Δmc 2 /A).
Binding energy and nuclear stability
The most stable, that is, the highest specific binding energy, are nuclei with a mass number from 50 to 90, for example, iron. This “peak of stability” is due to the non-central nature of nuclear forces. Since each nucleon interacts only with its neighbors, it is weaker bound on the surface of the nucleus than inside. The fewer interacting nucleons in the nucleus, the lower the binding energy, so light nuclei are less stable. In turn, as the number of particles in the nucleus increases, the Coulomb repulsive forces between protons increase, so that the binding energy of heavy nuclei also decreases.
Thus, for light nuclei, the most probable, that is, energetically favorable, are fusion reactions with the formation of a stable medium-mass nucleus, while for heavy nuclei, on the contrary, decay and fission processes (often multistage), as a result of which more stable products are also formed. These reactions are characterized by a positive and often very high energy yield, which accompanies an increase in the binding energy.
Below we consider some examples of nuclear reactions.
Decay reactions
Nuclei can undergo spontaneous changes in composition and structure, during which some elementary particles or fragments of the nucleus, such as alpha particles or heavier clusters, are emitted.
So, during alpha decay, which is possible due to quantum tunneling, an alpha particle overcomes the potential barrier of nuclear forces and leaves the parent nucleus, which, accordingly, reduces the atomic number by 2, and the mass number by 4. For example, the nucleus of radium-226, emitting alpha particle, turns into radon-222:
226 88 Ra → 222 86 Rn + α (4 2 He).
The decay energy of the radium-226 nucleus is about 4.87 MeV.
Beta decay, conditioned, occurs without changing the number of nucleons (mass number), but with an increase or decrease in the nuclear charge by 1, when emitting an antineutrino or neutrino, as well as an electron or positron. An example of a nuclear reaction of this type is the beta-plus decay of fluorine-18. Here, one of the protons of the nucleus turns into a neutron, a positron and a neutrino are emitted, and fluorine turns into oxygen-18:
18 9 K → 18 8 Ar + e + + v e .
The energy of the beta decay of fluorine-18 is about 0.63 MeV.
Nuclear fission
Fission reactions have a much greater energy yield. This is the name of the process in which the nucleus spontaneously or forcedly breaks up into fragments close in mass (usually two, rarely three) and some lighter products. The nucleus is divided if its potential energy exceeds the initial value by a certain amount, called the fission barrier. However, the probability of a spontaneous process, even for heavy nuclei, is low.
It increases significantly when the nucleus receives the corresponding energy from the outside (when a particle enters it). The neutron penetrates most easily into the nucleus, since it is not subject to electrostatic repulsion forces. The hit of a neutron leads to an increase in the internal energy of the nucleus, it is deformed with the formation of a neck and fissions. The fragments fly apart under the action of Coulomb forces. An example of a nuclear fission reaction shows uranium-235 absorbing a neutron:
235 92 U + 1 0 n → 144 56 Ba + 89 36 Kr + 3 1 0 n.
The splitting into barium-144 and krypton-89 is just one of options fission of uranium-235. This reaction can be written as 235 92 U + 1 0 n → 236 92 U* → 144 56 Ba + 89 36 Kr + 3 1 0 n, where 236 92 U* is a highly excited compound nucleus with high potential energy. Its excess, along with the difference in the binding energies of the parent and daughter nuclei, is released mainly (about 80%) in the form of the kinetic energy of the reaction products, and also partially in the form of the potential energy of fission fragments. The total fission energy of a massive nucleus is about 200 MeV. In terms of 1 gram of uranium-235 (assuming that all the nuclei have reacted), this is 8.2 ∙ 10 4 megajoules.
chain reactions
The fission of uranium-235, as well as nuclei such as uranium-233 and plutonium-239, is characterized by one important feature- the presence of free neutrons among the reaction products. These particles, penetrating into other nuclei, in turn, are able to initiate their fission, again with the emission of new neutrons, and so on. This process is called a nuclear chain reaction.
Flow chain reaction depends on how the number of emitted neutrons of the next generation correlates with their number in previous generation. This ratio k = N i /N i -1 (here N is the number of particles, i is the serial number of the generation) is called the neutron multiplication factor. For k< 1 цепная реакция не идет. При k >1 the number of neutrons, and hence the number of fissile nuclei, increases like an avalanche. An example of a nuclear chain reaction of this type is the explosion of an atomic bomb. For k = 1, the process is stationary, as exemplified by a reaction controlled by neutron-absorbing rods, in nuclear reactors.
Nuclear fusion
The greatest energy release (per one nucleon) occurs during the fusion of light nuclei - the so-called fusion reactions. To enter into a reaction, positively charged nuclei must overcome the Coulomb barrier and approach at a distance of strong interaction, not exceeding the size of the nucleus itself. Therefore, they must have extremely high kinetic energy, which means high temperatures (tens of millions of degrees and above). For this reason, fusion reactions are also called fusion reactions.
An example of a nuclear fusion reaction is the formation of helium-4 with the emission of a neutron during the fusion of deuterium and tritium nuclei:
2 1 H + 3 1 H → 4 2 He + 1 0 n.
Here, an energy of 17.6 MeV is released, which, per nucleon, is more than 3 times the energy of uranium fission. Of these, 14.1 MeV falls on the kinetic energy of the neutron and 3.5 MeV - the nucleus of helium-4. Such a significant value is created due to the huge difference in the binding energies of the nuclei of deuterium (2.2246 MeV) and tritium (8.4819 MeV) on the one hand, and helium-4 (28.2956 MeV) on the other.
In nuclear fission reactions, the energy of electrical repulsion is released, while in fusion, energy is released due to the strong interaction - the most powerful in nature. This determines such a significant energy yield of this type of nuclear reactions.
Examples of problem solving
Consider the fission reaction 235 92 U + 1 0 n → 140 54 Xe + 94 38 Sr + 2 1 0 n. What is its energy output? AT general view the formula for its calculation, which reflects the difference between the rest energies of particles before and after the reaction, is as follows:
Q \u003d Δmc 2 \u003d (m A + m B - m X - m Y + ...) ∙ c 2.
Instead of multiplying by the square of the speed of light, you can multiply the mass difference by a factor of 931.5 to get the energy value in megaelectronvolts. Substituting the appropriate values into the formula atomic masses, we get:
Q = (235.04393 + 1.00866 - 139.92164 - 93.91536 - 2∙1.00866) ∙ 931.5 ≈ 184.7 MeV.
Another example is the fusion reaction. This is one of the stages of the proton-proton cycle - the main source of solar energy.
3 2 He + 3 2 He → 4 2 He + 2 1 1 H + γ.
Let's use the same formula:
Q = (2 ∙ 3.01603 - 4.00260 - 2 ∙ 1.00728) ∙ 931.5 ≈ 13.9 MeV.
The main share of this energy - 12.8 MeV - falls in this case on the gamma photon.
We have considered only the simplest examples of nuclear reactions. The physics of these processes is extremely complex, they are very diverse. The study and application of nuclear reactions is of great importance both in the practical field (power engineering) and in fundamental science.
>> Nuclear reactions
§ 106 NUCLEAR REACTIONS
Atomic nuclei undergo transformations during interactions. These transformations are accompanied by an increase or decrease in the kinetic energy of the particles involved in them.
Nuclear reactions called changes in atomic nuclei when they interact with elementary particles or with each other. You have already familiarized yourself with examples of nuclear reactions in § 103. Nuclear reactions occur when particles come close to the nucleus and fall into the sphere of action of nuclear forces. Like-charged particles repel each other. Therefore, the approach of positively charged particles to nuclei (or nuclei to each other) is possible if a sufficiently large kinetic energy is imparted to these particles (or nuclei). This energy is imparted to protons, deuterium nuclei - deuterons, -particles and other heavier nuclei with the help of accelerators.
For the implementation of nuclear reactions, this method is much more efficient than using helium nuclei emitted by radioactive elements. Firstly , with the help of accelerators, particles can be imparted an energy of the order of 10 5 MeV, i.e., much greater than that which a-particles have (maximum 9 MeV). Secondly , you can use protons that do not appear in the process of radioactive decay (this is advisable because the charge of protons is half the charge of -particles, and therefore the repulsive force acting on them from the nuclei is also 2 times less). Thirdly , it is possible to accelerate nuclei heavier than helium nuclei.
The first nuclear reaction on fast protons was carried out in 1932. It was possible to split lithium into two particles:
The theory of relativity says that mass is special shape energy. It follows that it is possible to convert mass into energy and energy into mass. At the intraatomic level, such reactions take place. In particular, a certain amount of mass itself may well turn into energy. This happens in several ways. First, the nucleus can decay into a number of smaller nuclei, this reaction is called "decay". Secondly, smaller nuclei can easily combine to make a larger one - this is a fusion reaction. In the universe, such reactions are very common. Suffice it to say that the fusion reaction is the source of energy for stars. But the decay reaction is used by humanity because people have learned to control these complex processes. But what is a nuclear chain reaction? How to manage it?
What happens in the nucleus of an atom
A nuclear chain reaction is a process that occurs when elementary particles or nuclei collide with other nuclei. Why "chain"? This is a set of successive single nuclear reactions. As a result of this process, there is a change in the quantum state and nucleon composition of the original nucleus, even new particles appear - reaction products. The nuclear chain reaction, whose physics allows one to study the mechanisms of interaction of nuclei with nuclei and with particles, is the main method for obtaining new elements and isotopes. In order to understand the flow of a chain reaction, one must first deal with single ones.
What is needed for the reaction
In order to carry out such a process as a nuclear chain reaction, it is necessary to bring the particles (a nucleus and a nucleon, two nuclei) closer together at a distance of the strong interaction radius (about one fermi). If the distances are large, then the interaction of charged particles will be purely Coulomb. In a nuclear reaction, all laws are observed: conservation of energy, momentum, momentum, baryon charge. A nuclear chain reaction is denoted by the symbol set a, b, c, d. The symbol a denotes the original nucleus, b the incoming particle, c the new outgoing particle, and d the resulting nucleus.
Reaction energy
A nuclear chain reaction can take place both with the absorption and release of energy, which is equal to the difference in the masses of the particles after the reaction and before it. The absorbed energy determines the minimum kinetic energy of the collision, the so-called threshold of a nuclear reaction, at which it can freely proceed. This threshold depends on the particles involved in the interaction and on their characteristics. On the initial stage all particles are in a predetermined quantum state.
Implementation of the reaction
The main source of charged particles that bombard the nucleus is the one that gives beams of protons, heavy ions and light nuclei. Slow neutrons are obtained through the use of nuclear reactors. To fix incident charged particles, different types of nuclear reactions, both fusion and decay, can be used. Their probability depends on the parameters of the particles that collide. This probability is associated with such a characteristic as the reaction cross section - the value of the effective area, which characterizes the nucleus as a target for incident particles and which is a measure of the probability that the particle and the nucleus will enter into interaction. If particles with a nonzero spin take part in the reaction, then the cross section directly depends on their orientation. Since the spins of the incident particles are not completely randomly oriented, but more or less ordered, all corpuscles will be polarized. The quantitative characteristic of oriented beam spins is described by the polarization vector.
reaction mechanism
What is a nuclear chain reaction? As already mentioned, this is a sequence of more simple reactions. The characteristics of the incident particle and its interaction with the nucleus depend on the mass, charge, and kinetic energy. The interaction is determined by the degree of freedom of the nuclei, which are excited during the collision. Gaining control over all these mechanisms allows you to carry out such a process as a controlled nuclear chain reaction.
Direct reactions
If a charged particle that hits the target nucleus only touches it, then the duration of the collision will be equal to the distance necessary to overcome the radius of the nucleus. Such a nuclear reaction is called a direct reaction. General characteristic for all reactions of this type is the excitation of a small number of degrees of freedom. In such a process, after the first collision, the particle still has enough energy to overcome the nuclear attraction. For example, such interactions as inelastic neutron scattering, charge exchange, and refer to direct. The contribution of such processes to the characteristic called "total cross section" is quite negligible. However, the distribution of the products of the passage of a direct nuclear reaction makes it possible to determine the probability of emission from the beam direction angle, the selectivity of the populated states, and to determine their structure.
Pre-equilibrium emission
If the particle does not leave the region of nuclear interaction after the first collision, then it will be involved in a whole cascade of successive collisions. This is actually just what is called a nuclear chain reaction. As a result of this situation, the kinetic energy of the particle is distributed among the constituent parts of the nucleus. The state of the nucleus itself will gradually become much more complicated. During this process, a certain nucleon or a whole cluster (a group of nucleons) can concentrate energy sufficient for the emission of this nucleon from the nucleus. Further relaxation will lead to the formation of statistical equilibrium and the formation of a compound nucleus.
chain reactions
What is a nuclear chain reaction? This is her sequence constituent parts. That is, multiple consecutive single nuclear reactions, caused by charged particles, appear as reaction products in the previous steps. What is a nuclear chain reaction? For example, the fission of heavy nuclei, when multiple fission events are initiated by neutrons obtained during previous decays.
Features of a nuclear chain reaction
Among all chemical reactions chains are widely used. Particles with unused bonds play the role of free atoms or radicals. In a process such as a nuclear chain reaction, the mechanism of its occurrence is provided by neutrons, which do not have a Coulomb barrier and excite the nucleus upon absorption. If the necessary particle appears in the medium, then it causes a chain of subsequent transformations that will continue until the chain breaks due to the loss of the carrier particle.
Why the carrier is lost
There are only two reasons for the loss of the carrier particle of a continuous chain of reactions. The first consists in the absorption of a particle without the process of emission of a secondary one. The second is the departure of the particle beyond the limit of the volume of the substance that supports the chain process.
Two types of process
If in each period of the chain reaction only a single carrier particle is born, then this process can be called unbranched. It cannot lead to the release of energy on a large scale. If there are many carrier particles, then this is called a branched reaction. What is a nuclear chain reaction with branching? One of the secondary particles obtained in the previous act will continue the chain started earlier, while the others will create new reactions that will also branch. This process will compete with the processes leading to the break. The resulting situation will give rise to specific critical and limiting phenomena. For example, if there are more breaks than purely new chains, then self-sustaining of the reaction will be impossible. Even if it is excited artificially by introducing the required number of particles into a given medium, the process will still decay with time (usually rather quickly). If the number of new chains exceeds the number of breaks, then a nuclear chain reaction will begin to spread throughout the substance.
Critical situation
The critical state separates the region of the state of matter with a developed self-sustaining chain reaction, and the region where this reaction is impossible at all. This parameter is characterized by equality between the number of new circuits and the number of possible breaks. Like the presence of a free carrier particle, the critical state is the main item in such a list as "conditions for the implementation of a nuclear chain reaction." The achievement of this state can be determined by a number of possible factors. A heavy element is excited by just one neutron. As a result of a process such as a nuclear fission chain reaction, more neutrons are produced. Therefore, this process can produce a branched reaction, where neutrons will act as carriers. In the case when the rate of neutron captures without fission or escapes (loss rate) is compensated by the rate of multiplication of carrier particles, then the chain reaction will proceed in a stationary mode. This equality characterizes the multiplication factor. In the above case, it is equal to one. Due to the introduction between the rate of energy release and the multiplication factor, it is possible to control the course of a nuclear reaction. If this coefficient is greater than one, then the reaction will develop exponentially. Uncontrolled chain reactions are used in nuclear weapons.
Nuclear chain reaction in energy
The reactivity of a reactor is determined by a large number of processes that occur in its core. All these influences are determined by the so-called reactivity coefficient. The influence of a change in the temperature of graphite rods, coolants or uranium on the reactivity of the reactor and the intensity of such a process as a nuclear chain reaction are characterized by temperature coefficient(for coolant, for uranium, for graphite). There are also dependent characteristics for power, for barometric indicators, for steam indicators. To maintain a nuclear reaction in a reactor, it is necessary to convert some elements into others. To do this, you need to take into account the conditions for the flow of a nuclear chain reaction - the presence of a substance that is able to divide and release from itself during decay a certain number of elementary particles, which, as a result, will cause the fission of the remaining nuclei. Uranium-238, uranium-235, plutonium-239 are often used as such substances. During the passage of a nuclear chain reaction, the isotopes of these elements will decay and form two or more others. chemical substances. In this process, the so-called "gamma" rays are emitted, an intense release of energy occurs, two or three neutrons are formed, capable of continuing the reactions. There are slow neutrons and fast ones, because in order for the nucleus of an atom to disintegrate, these particles must fly at a certain speed.
Discovery of the neutron and its properties
Nuclear reactions under the action of neutrons take special place in nuclear physics. Due to the fact that the neutron has no electric charge, it freely penetrates into any atomic nucleus and causes nuclear reactions. Consider first the properties of the neutron.
The neutron was discovered following Rutherford's prediction in 1920.
In the experiments of Bethe and Becker (1930), beryllium nuclei were irradiated with α-particles and neutral radiation was registered, the nature of which was not determined.
α + Be → neutral radiation (what?, γ?).
In the experiments of Joliot-Curie (1932), α-particles were directed to a beryllium target, and then to a paraffin one, in order to determine the nature of neutral radiation. The release of protons was observed after the paraffin target. The scheme of experience is shown below.
α + Be → paraffin → p
Recoil protons with Е р = 4.3 MeV were registered. The question arose: under the action of what particles were they formed?
If they were caused by γ-quanta, then the energy of γ-quanta E γ should have been ~ 50 MeV. γ-quanta with such an energy could not appear from this reaction.
Chadwick analyzed these experiments and suggested that as a result of the reaction, neutral particles with a mass comparable to that of a proton are emitted. Then he set up an experiment in a cloud chamber and observed the nitrogen recoil nuclei. He compared these results with the results of the Joliot-Curie experiments, in which recoil protons from paraffin were registered, and determined the mass of this neutral particle from the laws of conservation of energy
and momentum
m 1 v = m 1 v 1 + m p v p ; 
where N is the nitrogen nucleus; v 1 is the speed of the neutral particle after the collision; m 1 is the mass of the neutral particle. It turned out to be close to the mass of a proton
Thus, it became clear that in the Joliot-Curie experiments, a reaction took place in which neutral particles, neutrons, were emitted:
α + 9 Be → 12 C+ n.
When they hit the paraffin, they knocked out recoil protons with an energy of Ер = 4.3 MeV.
The properties of the neutron, obtained from numerous experiments, are presented below:
mass − m n c 2 = 939.5 MeV, m n = 1.008665 a.u. eat.,
magnetic moment - μ n = -1.91μ i,
spin − J = ћ/2,
lifetime − τ n = (10.61 ±0.16) min,
r.m.s. radius −
Nuclear reactions not only provide new information about the nature and properties of nuclear forces, but are also used in practice in national economy and in military affairs. This primarily applies to nuclear reactions under the action of neutrons at low energies.
11.4 Neutron sources
Neutron sources are various nuclear reactions.
 Rice. 88: Neutron spectrum. |
1. A mixture of radium with beryllium (sometimes polonium with beryllium) is used, where the reaction takes place
α + 9 Be → 12 C+ n + 5.5 MeV.
The kinetic energy of the neutron T is distributed over the spectrum
(Fig. 88).
The decay of Ra produces α-particles with energies of 4.8 MeV and 7.7 MeV. They react with 9 Be and generate a neutron flux. The neutron energy spread is due to the fact that α-particles of different energies create neutrons of different energies. The carbon nucleus 12 C is formed in the ground and excited states.
The neutron yield is ~ 10 7 neutrons per 1 g Ra per second. At the same time, γ-rays are emitted.
2. Other sources of neutrons are photonuclear reactions (γ,n), which produce slow and monochromatic neutrons.
γ + 2 H → p + n, Q = -2.23 MeV.
ThC" (208 Tl) is used. It emits γ-quanta with Е γ ~ 2.62 MeV and Е n ~ E p; T n ~20 keV.
3. Photodisintegration of Be by photons with energy E γ = 1.78 MeV
γ + 9 Be → 8 Be + n, Q = -1.65 MeV; T n ~ 100 keV.
4. Emission of neutrons under the action of accelerated deuterons with E d = 16 MeV in the reaction
2 H + 9 Be → 10 B + n + 4.3 MeV.
E n = 4 MeV, output 10 6 neutrons per second.
5. Reaction 2 H + 2 H → 3 He + n + 3.2 MeV,
D + D (ice from heavy water), i?n = 2.5 MeV.
6. Irradiation with tritium deuterons
2 H + 3 H → 4 He + n + 17.6 MeV.
Since this reaction is exothermic, the deuterons are accelerated to the energy E d = 0.3 MeV in gas discharge tubes. Monochromatic neutrons with Е n ~ 14 MeV are formed.
This neutron source is used in geology.
7. In stripping reactions under the action of deuterons with E d ~ 200 MeV, n c
E n ~ 100 MeV.
11.5 Nuclear reactors, nuclear chain reaction
The most powerful source of neutrons is nuclear reactors, devices in which a controlled fission chain reaction is maintained.
Fission of U nuclei occurs in the reactor and neutrons with Е n from 0 to 13 MeV are formed, the intensity of the source is 10 19 neutrons/s cm2. The fission process proceeds under the influence of neutrons, which freely penetrate the nuclei due to the absence of the Coulomb potential barrier.
During nuclear fission, radioactive fragments are formed and 2-3 n are emitted, which again react with U nuclei; there is a chain process (Fig. 89).
n + 235 U → 236 U → 139 La + 95 Mo + 2n
Rice. 89: Illustration of 235U nuclear fission.
To describe the fission process of 235 U, a liquid drop model is used, in which the Weizsäcker formula works. After the neutron hits the uranium nucleus, there is competition between the surface energy of the new nucleus and the Coulomb repulsion energy. As a result, under the action of Coulomb forces, the nucleus is divided into two lighter nuclei.
Energy Q released during nuclear fission (A,Z)
(A,Z) → 2(A/2,Z/2) + Q,
calculated using the Weizsäcker formula
Q = 2ε(A/2,Z/2) − ε(A,Z) = (1 − 2 1/3) a sim A 2/3 + (1 − 2 2/3) a cool Z 2 A -1/3 ;
Q (MeV) \u003d -4.5A 2/3 + 0.26 Z 2 A -1/3, ε - specific binding energy: E St / A. For the nucleus 235 U Q = 180 MeV.
In order for the nucleus to split, energy E > E a must be introduced into it, where E a 90: Potential energy nucleus depending on the distance to the center of the nucleus (solid curve), E 0 is the ground state, E 0 + E a is the excited state, E a is the activation energy
(Fig. 90).
The measure of the ability of nuclei to fission is the ratio of the energy of the Coulomb repulsion of protons to the energy of surface tension:
where Z 2 /A is the fission parameter, the larger it is, the easier the nucleus is divided; Z 2 /A = 49 critical value of the division parameter.
An illustration of the nuclear fission process is shown in fig. 91.
In a nuclear reactor, the process of nuclear fission is repeated many times as a result of the formation of many generations of fission. In the first fission event of 235 U, 2.4 neutrons are produced on average. The lifetime of one generation is ~ 10 s. If there is a birth of K generations, then ~ 2 K neutrons are formed after a time of ~ 2·10 -6 s. If K = 80, the number of neutrons will be 2 80 ~ 10 24 - this will result in the fission of 10 24 atoms (140 g of uranium). The energy released in this case, 3·10 13 W, is equal to the energy generated by burning 1000 tons of oil.
Rice. 91: The process of nuclear fission occurring in a nuclear reactor.
In fission reactions, energy is released in the form of heat. Removal of heat from the reactor is carried out by a coolant, to which special requirements are imposed. It should have a high heat capacity, weakly absorb neutrons, and have low chemical activity. We won't discuss design features elements of a nuclear reactor. We only note that when thermal neutrons hit the 235 U nucleus, fast neutrons are formed, and the reaction proceeds only on slow neutrons. Therefore, it is necessary to slow down fast neutrons. This happens in the moderator. Carbon or heavy water is used as a moderator. Stopping the fission process is implemented with the help of cadmium nuclei, which capture the resulting neutrons. Thus, the design of a nuclear reactor necessarily includes a neutron moderator (carbon) and cadmium rods that absorb the generated neutrons.
The reactors use natural uranium 238U (99.3%) and enriched 235U (0.7%). 235 U is divided under the action of thermal neutrons. 238 U is used in fast neutron reactors.
The processes occurring in the reactor are characterized by the following probabilities:
ν is the number of generated fast neutrons;
ε is the multiplication factor of fast neutrons;
P is the probability for a neutron to reach thermal energy;
ƒ is the probability of neutron capture during deceleration;
σ t /σ tot is the probability of triggering a fission reaction.
The product of these probabilities gives an estimate of the multiplication factor k of thermal neutrons in a nuclear reactor:
A chain reaction goes on if k > 1; the quantities included in the multiplication factor have the following values: ν = 2.47; ε = 1.02; P = 0.89; ƒ = 0.88; σt /σtot = 0.54.
Thus, k ∞ = 1.07 for a reactor of infinite dimensions. In real conditions to eff< k ∞ , т.к. часть нейтронов уходит из реактора.
In fast neutron reactors (239 Pu and 238 U), the following process occurs:
As a result of this reaction, 239 Pu is reproduced. The resulting plutonium reacts with a neutron: n + 239 Pu, ν = 2.41 neutrons are formed.
The number of 239 Ri nuclei doubles every 7-10 years.
The fission reaction of atomic nuclei is used to produce atomic energy. Nuclear reactors operate in many nuclear power plants.
11.6 Fusion reactions, synthesis of light nuclei
Another source of atomic energy can be the synthesis of light atomic nuclei. Light nuclei are less tightly bound, and when they merge into a heavy nucleus, more energy is released. In addition, thermonuclear reactions are cleaner due to the lack of accompanying radioactive emissions than fission chain reactions.
The following fusion reactions can be used to obtain thermonuclear energy:
d + d = 3 He + n + 4 MeV,
d + d = t + p + 3.25 MeV,
d + t = 4 He + n + 17.b MeV,
3 He + d = 4 He + p + 18.3 MeV,
6 Li + 2di = 2 4 He + 22.4 MeV. J
The energy of the nuclei entering into the reaction must be sufficient to overcome the Coulomb potential barrier. On fig. 92 shows the energy dependence of the cross sections of some reactions. As can be seen from the figure, the synthesis of deuterium d and tritium t nuclei is the most preferable. In this fusion reaction, the Coulomb potential barrier is low and the interaction cross section is large at low energies of the merging nuclei. For the reaction to proceed, it is necessary to have a sufficient concentration of these nuclei per unit volume and a sufficient temperature of the heated plasma.
The number of merging events R ab per unit time per unit volume is determined by the relation
R ab = n a n b w ab (T).
w ab (T) = σ ab v ab ,
where n a , n b is the number of nuclei a, b; σ ab is the effective cross section of the reaction, v ab is the relative velocity of particles in the plasma, Т is the temperature. As a result of the reaction, energy is released
W = Rab Qab τ,
where R ab is the number of merging events, Q ab is the energy released in 1 event, τ is the time.
Let n a \u003d n b \u003d 10 15 nuclei / cm 3, T \u003d 100 keV. Then W ~ 10 3 W/cm 3 s.
In a self-sustaining thermonuclear reaction, more energy must be released than is spent on heating and confining the plasma. The cost of heating n a = n b = 2n particles to a temperature T: 3n·kT: k - Boltzmann's constant. Thus, the following condition must be satisfied:
n 2 w ab Q ab τ > 3nkТ
(released energy > heating energy).
Lawson formulated the following condition for the d + t fusion reaction:
nτ > 10 14 s cm -3 ,
where nτ is the retention parameter. On fig. 93 shows the dependence of this parameter on temperature. The reaction proceeds if nτ > ƒ(T). The temperature T ~ 2·10 8 K corresponds to an energy of 10 keV. The minimum value of the retention parameter nτ = 10 14 s/cm 3 for the reaction d + t is reached at a temperature of 2 10 8 K.
Rice. 93: Dependence of holding parameters on temperature. The shaded area ƒ(Т) is the zone of controlled thermonuclear fusion for the reaction d + t. − Parameter values achieved at various facilities by 1980.
For other reactions:
The confinement of a plasma having the necessary conditions for the reaction to proceed, is implemented in Tokamak-type installations using a magnetic field. Such installations operate in Russia and in a number of other countries. As can be seen from fig. 93, the regime of controlled thermonuclear fusion has not yet been achieved.
Attempts are being made to obtain the conditions necessary for thermonuclear fusion using laser facilities. In this case, a small volume containing deuterium and tritium nuclei is compressed from all sides by laser radiation. In this case, the nuclei of deuterium and tritium are heated to the desired temperature. Laser fusion requires the introduction of a factor of 100, because there is a lot of useless energy going to pump the laser.
Attempts to carry out controlled thermonuclear fusion in laboratory conditions encounter a number of difficulties.
- 1. Until now, it has not been possible to obtain a stable regime of high-temperature plasma.
- 2. Energy losses in plasma are high even due to low concentrations of impurities of atoms with large Z.
- 3. The "problem of the first wall" in the Tokamak, which limits the reactor plasma, has not been solved (the neutron flux destroys it).
- 4. In nature, there is no radioactive tritium t with a half-life of T 1/2 = 12.5 years, so there is a problem of tritium reproduction in the reaction
n + 7 Li = α + t + n.
Until now, it has not been possible to overcome these difficulties and obtain a controlled thermonuclear fusion reaction.
AT vivo thermonuclear fusion reactions take place on the sun and in stars.
Literature
- 1. Shirokov Yu.M., Yudin N.P. Nuclear physics. -M.: Nauka, 1972.
- 2. Kapitonov I.M. Introduction to nuclear and particle physics. -M.: UPPS, 2002.
Topics of the USE codifier: nuclear reactions, fission and fusion of nuclei.
In the previous sheet, we repeatedly talked about the splitting of the atomic nucleus into its component parts. But how can this be achieved in reality? As a result of which physical processes can you break the core?
Observations of radioactive decay under changing environmental conditions - namely, at various pressures and temperatures, in electrical and magnetic fields- showed that the rate of radioactive decay does not depend on these conditions. All these factors are not capable of causing any transformations of chemical elements into each other. Obviously, the energy changes here are too small to affect the atomic nucleus - so the wind blowing over a brick house is not able to destroy it.
But you can destroy the house artillery shell. And Rutherford in 1919 decided to use the most powerful "shells" that were then available. These were -particles emitted with an energy of about 5 MeV at radioactive decay uranium. (As you remember, these are the same shells with which he bombarded a sheet of gold foil eight years ago in his famous experiments that gave rise to the planetary model of the atom.)
True, the transformation of gold into others chemical elements were not observed in those experiments. The core of gold itself is very strong, and besides, it contains quite a lot of protons; they create a strong Coulomb field that repels the -particle and does not let it get too close to the nucleus. But in order to break the nucleus, the projectile must get close to the nucleus so that nuclear forces turn on! Well, time a large number of prevents protons - maybe take a lighter nucleus, where there are few protons?
Rutherford bombarded nitrogen nuclei and as a result carried out the first in the history of physics nuclear reaction:
(1)
On the right side of (1) we see reaction products is an oxygen isotope and a proton.
It became clear that to study nuclear reactions it is necessary to have projectile particles high energy. This opportunity is given accelerators elementary particles. Accelerators have two major advantages over natural "radioactive guns".
1. Accelerators can accelerate any charged particles. This is especially true for protons, which do not appear during the natural decay of nuclei. Protons are good because they carry a minimum charge, which means they experience the smallest Coulomb repulsion from the target nuclei.
2. Accelerators make it possible to achieve energies that are several orders of magnitude higher than the energy of α-particles during radioactive decay. For example, at the Large Hadron Collider, protons are accelerated to energies of several TeV; this is a million times more than 5 MeV for -particles in reaction (1) carried out by Rutherford.
So, with the help of protons that passed through the accelerator, in 1932 it was possible to break the lithium nucleus (while obtaining two -particles):
(2)
Nuclear reactions made it possible to artificially transform chemical elements.
In addition, new, previously unknown particles began to be detected in the reaction products. For example, when beryllium was irradiated with -particles in the same 1932, the neutron was discovered:
(3)
Neutrons are remarkably suitable for splitting nuclei: having no electric charge, they freely penetrate into the nucleus. (At the same time, it is not necessary to accelerate neutrons - slow neutrons penetrate nuclei more easily. It turns out that neutrons even need to be slowed down, and this is done by passing neutrons through ordinary water.) So, when nitrogen is irradiated with neutrons, the following reaction occurs:
(4)
Energy yield of a nuclear reaction
Discussing the binding energy, we saw that as a result of nuclear processes, the mass of a system of particles does not remain constant. This, in turn, leads to the fact that the kinetic energy of the products of a nuclear reaction differs from the kinetic energy of the initial particles.
First of all, we recall that the total energy of a mass particle is the sum of its rest energy and kinetic energy:
Let a nuclear reaction occur as a result of the collision of particles, the products of which are particles and:
(5)
The total energy of the particle system is conserved:
(6)
The kinetic energy of the initial particles is . The kinetic energy of the reaction products is . energy output nuclear reaction is the difference between the kinetic energies of the reaction products and the initial particles:
From (6) we easily obtain:
(7)
If class="tex" alt="(!LANG:Q > 0"> , то говорят, что реакция идёт !} with energy release more kinetic energy of initial particles. From (7) we see that in this case the total mass of the reaction products less
If , then the reaction is with energy absorption: kinetic energy of reaction products less kinetic energy of initial particles. The total mass of the reaction products in this case more the total mass of the initial particles.
Thus, the terms "release" and "absorption" of energy should not cause confusion: they refer only to kinetic particle energy. The total energy of a system of particles, of course, remains unchanged in any reaction.
1. Using the table of masses of neutral atoms, we find and expressed in a. e.m. (to find the mass of the nucleus, do not forget to subtract the mass of electrons from the mass of a neutral atom).
2. We calculate the mass of the initial particles, the mass of the reaction products and find the mass difference .
3. Multiply by and get the value expressed in MeV.
We will now consider in detail the calculation of the energy yield using two examples of the bombardment of lithium nuclei: first - by protons, then - by particles.
In the first case, we have the reaction (2) already mentioned above:
The mass of a lithium atom is a. e. m. The mass of an electron is a. e. m. Subtracting from the mass of an atom the mass of its three electrons, we obtain lithium nucleus mass :
A. e. m.
The mass of a proton is a. e. m., so that the mass of the initial particles:
A. e. m.
Let's move on to the reaction products. The mass of a helium atom is a. e. m. Subtract the mass of electrons and find helium nucleus mass :
A. e. m.
Multiplying by , we get the mass of the reaction products:
A. e. m.
The mass, as we see, has decreased; this means that our reaction proceeds with the release of energy. Mass difference:
A. e. m.
Released energy:
MeV.
Now let's look at the second example. When lithium nuclei are bombarded with -particles, the following reaction occurs:
(8)
The masses of the original nuclei are already known to us; it remains to calculate their total mass:
A. e. m.
From the table we take the mass of the boron atom (it is equal to a. e. m.); subtract the mass of five electrons and get the mass of the nucleus of the boron atom:
A. e. m.
The mass of a neutron is a. e. m. Find the mass of the reaction products:
A. e. m.
This time the weight has increased. class="tex" alt="(!LANG:(m_2 > m_1)"> !}, that is, the reaction proceeds with the absorption of energy.
The mass difference is:
A. e. m.
Energy yield of the reaction:
MeV.
Thus, the MeV energy is absorbed in reaction (8). This means that the total kinetic energy of the reaction products (boron and neutron nuclei) is less by MeV than the total kinetic energy of the initial particles (lithium nuclei and -particles). Therefore, for this reaction to take place in principle, the energy of the initial particles must be at least MeV.
Nuclear fission
By bombarding uranium nuclei with slow neutrons, the German physicists Hahn and Strassmann discovered the appearance of elements in the middle part periodic system: barium, krypton, strontium, rubidium, cesium, etc. So it was discovered nuclear fission uranium.
On fig. 1 we see the process of nuclear fission (image from oup.co.uk.). Capturing a neutron, the uranium nucleus splits into two fragment, and two or three neutrons are released.
Rice. 1. Fission of the uranium nucleus
Fragments are the nuclei of radioactive isotopes of elements in the middle of the periodic table. Usually one of the fragments is larger than the other. For example, during the bombardment of uranium, such combinations of fragments can occur (as they say, the reaction proceeds according to the following channels).
Barium and krypton:
Cesium and rubidium:
Xenon and strontium:
In each of these reactions, a very large energy is released - of the order of MeV. Compare this value with the energy yield of reaction (2) found above, equal to MeV! Where does this amount of energy come from?
Let's start with the fact that due to the large number of protons (pieces) packed in the uranium nucleus, the Coulomb repulsive forces that expand the nucleus are very large. Nuclear forces, of course, are still able to keep the nucleus from disintegration, but the powerful Coulomb factor is ready to have its say at any moment. And such a moment comes when a neutron gets stuck in the nucleus (Fig. 2 - image from the site investingreenenergy.com).
Rice. 2. Deformation, vibrations and rupture of the core
The trapped neutron causes the nucleus to deform. Fluctuations in the shape of the nucleus will begin, which can become so intense that the nucleus will stretch into a “dumbbell”. The short-range nuclear forces that hold together a small number of neighboring nucleons of the isthmus cannot cope with the electrical repulsion forces of the halves of the dumbbell, and as a result, the nucleus will burst.
The fragments will scatter with great speed - about the speed of light. They take away most released energy (about MeV from ).
The fission of heavy nuclei can be interpreted from the point of view of the plot of dependence of the specific binding energy of the nucleus on its mass number already known to us (Fig. 3).
Rice. 3. Fission of heavy nuclei is energetically favorable
The color highlights the region in which the specific binding energy reaches the greatest value MeV/nucleon. This is the region of the most stable nuclei. To the right of this region, the specific binding energy gradually decreases to MeV/nucleon at the uranium nucleus.
The process of transformation of less stable nuclei into more stable ones is energetically favorable and is accompanied by the release of energy. During the fission of the uranium nucleus, as we see, the specific binding energy increases by approximately MeV/nucleon; this energy is just released in the process of fission. Multiplying this by the number of nucleons in the uranium nucleus, we obtain approximately the same MeV of the energy output, which was mentioned above.
Nuclear chain reaction
The appearance of two or three neutrons in the process of fission of the uranium nucleus - the most important fact. These "first generation" neutrons can hit new nuclei and cause them to fission; as a result of the fission of new nuclei, neutrons of the "second generation" will appear, which will fall into the following nuclei and cause their fission; neutrons of the "third generation" will arise, which will lead to the fission of the next nuclei, etc. So it goes nuclear chain reaction, during which a huge amount of energy is released.
For a nuclear chain reaction to proceed, it is necessary that the number of released neutrons in the next generation be not less than number neutrons in the previous generation. Value
called neutron multiplication factor. Thus, the chain reaction goes under the condition class="tex" alt="(!LANG:k > 1"> . Если , то цепная реакция не возникает.!}
In case of class="tex" alt="(!LANG:k > 1"> происходит лавинообразное нарастание числа освобождающихся нейтронов, и цепная реакция становится !} unmanageable. This is how an atomic bomb explodes.
What happens in nuclear reactors controlled fission chain reaction with multiplication factor . The steady flow of a controlled chain reaction is ensured by introducing into the reactor core (that is, into the area where the reaction takes place) special control rods that absorb neutrons. When the rods are fully inserted, their absorption of neutrons is so great that the reaction does not occur. During the start-up of the reactor, the rods are gradually withdrawn from the core until the released power reaches the required level. This level is carefully controlled, and when it is exceeded, devices are turned on that introduce control rods back into the core.
thermonuclear reaction
Along with the fission reaction of heavy nuclei, the reverse process in a certain sense turns out to be energetically possible - synthesis of light nuclei, that is, the fusion of nuclei of light elements (located at the beginning periodic table) to form a heavier nucleus.
In order for the fusion of nuclei to begin, they need to be brought close together - so that nuclear forces come into play. For such an approach, it is necessary to overcome the Coulomb repulsion of the nuclei, which sharply increases with decreasing distance between them. This is possible only when the kinetic energy of the nuclei is very high, which means that at a very high temperature (tens and hundreds of millions of degrees). Therefore, the nuclear fusion reaction is called thermonuclear reaction.
As an example of a thermonuclear reaction, we give the fusion reaction of deuterium and tritium nuclei (heavy and superheavy isotopes of hydrogen), as a result of which a helium nucleus and a neutron are formed:
(9)
This reaction proceeds with the release of energy equal to MeV (try to do the calculations yourself and get this value). This is a lot, considering that only nucleons are involved in the reaction! Indeed, per nucleon, reaction (9) releases an energy of approximately MeV, while the fission of a uranium nucleus releases “only” MeV per nucleon.
Thus, thermonuclear reactions serve as a source of even more energy than nuclear fission reactions. From a physical point of view, this is understandable: the energy of a nuclear fission reaction is basically the kinetic energy of the fragments accelerated electric repulsive forces, and in nuclear fusion, energy is released as a result of the acceleration of nucleons towards each other under the action of much more powerful nuclear forces of attraction.
Simply put, during the fission of nuclei, the energy of the electrical interaction is released, and during the synthesis of nuclei, the energy of the strong (nuclear) interaction.
In the interiors of stars, temperatures are reached that are suitable for the fusion of nuclei. The light of the Sun and distant stars carries the energy released in thermonuclear reactions - during the fusion of hydrogen nuclei into helium nuclei and the subsequent fusion of helium nuclei into the nuclei of heavier elements located in the middle part of the periodic system. The direction of thermonuclear fusion is shown in fig. four ; the synthesis of light nuclei is energetically favorable, since it is directed towards increasing the specific binding energy of the nucleus.
Rice. 4. Synthesis of light nuclei is energetically favorable
An uncontrolled thermonuclear reaction is carried out during an explosion hydrogen bomb. First the built-in explodes atomic bomb- needed to create high temperature in the first stage of a thermonuclear explosion. When the required temperature is reached in the thermonuclear fuel of the bomb, fusion reactions begin, and the explosion of the hydrogen bomb itself occurs.
The implementation of a controlled thermonuclear reaction remains an unsolved problem, on which physicists have been working for more than half a century. If it is possible to achieve a controlled flow of thermonuclear fusion, then humanity will have at its disposal a virtually unlimited source of energy. It's extremely important task facing current and future generations - in the light of the threatening prospect of depletion of the oil and gas resources of our planet.
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