And the decay is represented by the reaction equation GDS. Radioactive transformations

You already know that in the middle of the 20th century. the problem arose of finding new sources of energy. In this regard, thermonuclear reactions attracted the attention of scientists.

• Thermonuclear reaction is the fusion reaction of light nuclei (such as hydrogen, helium, etc.), occurring at temperatures from tens to hundreds of millions of degrees.

Creation high temperature necessary to impart sufficiently large kinetic energy to the nuclei - only under this condition will the nuclei be able to overcome the forces of electrical repulsion and get close enough to fall into the zone of action of nuclear forces. At such small distances, the forces of nuclear attraction significantly exceed the forces of electrical repulsion, due to which synthesis (i.e., fusion, association) of nuclei is possible.

In § 58, using the example of uranium, it was shown that energy can be released during the fission of heavy nuclei. In the case of light nuclei, energy can be released during the reverse process - during their fusion. Moreover, the reaction of fusion of light nuclei is energetically more favorable than the reaction of fission of heavy nuclei (if we compare the released energy per nucleon).

An example of a thermonuclear reaction is the fusion of hydrogen isotopes (deuterium and tritium), resulting in the formation of helium and the emission of a neutron:

This is the first thermonuclear reaction that scientists have managed to carry out. It was implemented in thermonuclear bomb and was of an uncontrollable (explosive) nature.

As already noted, thermonuclear reactions can occur with the release of large quantity energy. But in order for this energy to be used for peaceful purposes, it is necessary to learn how to conduct controlled thermonuclear reactions. One of the main difficulties in carrying out such reactions is to contain high-temperature plasma (almost completely ionized gas) inside the installation, in which nuclear fusion occurs. The plasma should not come into contact with the walls of the installation in which it is located, otherwise the walls will turn into steam. Currently, very strong magnetic fields are used to confine plasma in a confined space at an appropriate distance from the walls.

Thermonuclear reactions play an important role in the evolution of the Universe, in particular in the transformations chemicals in it.

Thanks to thermonuclear reactions occurring in the depths of the Sun, energy is released that gives life to the inhabitants of the Earth.

Our Sun has been radiating light and heat into space for almost 4.6 billion years. Naturally, at all times, scientists have been interested in the question of what is the “fuel” due to which the Sun produces a huge amount of energy for such a long time.

There were different hypotheses on this matter. One of them was that energy in the Sun is released as a result chemical reaction combustion. But in this case, as calculations show, the Sun could exist for only a few thousand years, which contradicts reality.

The original hypothesis was put forward in the middle of the 19th century. It was that the increase internal energy and the corresponding increase in the temperature of the Sun occurs due to a decrease in its potential energy under gravitational compression. It also turned out to be untenable, since in this case the lifespan of the Sun increases to millions of years, but not to billions.

The assumption that the release of energy in the Sun occurs as a result of thermonuclear reactions occurring on it was made in 1939 by the American physicist Hans Bethe.

They also proposed the so-called hydrogen cycle, i.e. a chain of three thermonuclear reactions leading to the formation of helium from hydrogen:

where is a particle called a “neutrino”, which means “little neutron” in Italian.

To produce the two nuclei needed for the third reaction, the first two must occur twice.

You already know that, in accordance with the formula E = mс 2, as the internal energy of a body decreases, its mass also decreases.

To imagine the colossal amount of energy the Sun loses as a result of the conversion of hydrogen into helium, it is enough to know that the mass of the Sun decreases by several million tons every second. But, despite the losses, the hydrogen reserves on the Sun should last for another 5-6 billion years.

The same reactions occur in the interiors of other stars, the mass and age of which are comparable to the mass and age of the Sun.

Questions

1. What reaction is called thermonuclear? Give an example of a reaction.
2. Why are thermonuclear reactions only possible at very high temperatures?
3. Which reaction is energetically more favorable (per nucleon): the fusion of light nuclei or the fission of heavy ones?
4. What is one of the main difficulties in carrying out thermonuclear reactions?
5. What is the role of thermonuclear reactions in the existence of life on Earth?
6. What is the source of solar energy according to modern ideas?
7. How long should the supply of hydrogen on the Sun last, according to scientists’ calculations?

This is interesting...

Elementary particles. Antiparticles

Particles that make up atoms various substances- electron, proton and neutron - called elementary. The word "elementary" implied that these particles are primary, simplest, further indivisible and unchangeable. But it soon turned out that these particles are not immutable at all. They all have the ability to transform into each other when interacting.

Therefore, in modern physics the term “elementary particles” is usually used in a different way. exact value, and for the name large group tiny particles matter that is not atoms or atomic nuclei (the exception is the proton, which is the nucleus of a hydrogen atom and at the same time belongs to elementary particles).

Currently, more than 350 different elementary particles. These particles are very diverse in their properties. They may differ from each other in mass, sign and size electric charge, lifetime (that is, the time from the moment a particle is formed until the moment it transforms into some other particle), penetrating ability (that is, the ability to pass through matter) and other characteristics. For example, most particles are “short-lived” - they live no more than two millionths of a second, while the average lifetime of a neutron outside the atomic nucleus is 15 minutes.

The most important discovery in the field of elementary particle research was made in 1932, when the American physicist Carl David Anderson discovered a trace of an unknown particle in a cloud chamber placed in a magnetic field. Based on the nature of this trace (radius of curvature, direction of bending, etc.), scientists determined that it was left by a particle, which is like an electron with a positive electric charge. This particle was called a positron.

It is interesting that a year before the experimental discovery of the positron, its existence was theoretically predicted by the English physicist Paul Dirac (the existence of just such a particle followed from the equation he derived). Moreover, Dirac predicted the so-called processes of annihilation (disappearance) and the birth of an electron-positron pair. Annihilation is that an electron and a positron disappear upon meeting, turning into γ-quanta (photons). And when a γ-quantum collides with any massive nucleus, an electron-positron pair is born.

Both of these processes were first observed experimentally in 1933. Figure 166 shows the tracks of an electron and a positron formed as a result of the collision of a γ-quantum with a lead atom during the passage of γ-rays through a lead plate. The experiment was carried out in a cloud chamber placed in a magnetic field. The same curvature of the tracks indicates the same mass of particles, and the curvature in different sides- O opposite signs electric charge.

Rice. 166. Tracks of an electron-positron pair in a magnetic field

In 1955, another antiparticle was discovered - the antiproton (the existence of which also followed from Dirac's theory), and a little later - the antineutron. An antineutron, like a neutron, has no electrical charge, but it undoubtedly belongs to antiparticles, since it participates in the process of annihilation and the birth of a neutron-antineutron pair.

The possibility of obtaining antiparticles led scientists to the idea of ​​​​creating antimatter. Antimatter atoms should be built in this way: in the center of the atom there is a negatively charged nucleus, consisting of antiprotons and antineutrons, and positrons revolve around the nucleus. In general, the atom is neutral. This idea also received brilliant experimental confirmation. In 1969, at the proton accelerator in Serpukhov, Soviet physicists obtained nuclei of antihelium atoms.

At present, antiparticles of almost all known elementary particles have been experimentally discovered.

Chapter summary. The most important

Below are given physical concepts and phenomena. The sequence of presentation of definitions and formulations does not correspond to the sequence of concepts, etc.

Copy the names of the concepts into your notebook and write them in square brackets. serial number definition (formulation) corresponding to this concept.

• Radioactivity;
• nuclear (planetary) model of the structure of the atom;
• atomic nucleus;
• radioactive transformations of atomic nuclei;
• experimental methods for studying particles in atomic and nuclear physics ;
• nuclear forces;
• nuclear binding energy;
• mass defect of the atomic nucleus;
• chain reaction;
• nuclear reactor ;
• environmental and social problems problems arising from the use of nuclear power plants;
• absorbed dose of radiation.

1. Registration of particles using a Geiger counter, studying and photographing particle tracks (including those participating in nuclear reactions) in a cloud chamber and a bubble chamber.
2. The forces of attraction acting between nucleons in the nuclei of atoms and significantly exceeding the forces of electrostatic repulsion between protons.
3. The minimum energy required to split a nucleus into individual nucleons.
4. Spontaneous emission of radioactive rays by atoms of certain elements.
5. A device designed to carry out a controlled nuclear reaction.
6. Consists of nucleons (i.e. protons and neutrons).
7. Radioactive waste, the possibility of accidents, promotion of the proliferation of nuclear weapons.
8. An atom consists of a positively charged nucleus located at its center, around which electrons orbit at a distance significantly greater than the size of the nucleus.
9. Transformation of one chemical element in the other, during α- or β-decay, as a result of which the nucleus of the original atom undergoes changes.
10. The difference between the sum of the masses of the nucleons forming a nucleus and the mass of this nucleus.
11. A self-sustaining fission reaction of heavy nuclei, in which neutrons are continuously produced, dividing more and more new nuclei.
12. Energy ionizing radiation, absorbed by the emitted substance (in particular, body tissues) and calculated per unit mass.

Test yourself

7.1. Phenomenological consideration. Alpha decay is a spontaneous process of transformation of a nucleus ( A, Z) to the core ( A– 4, Z– 2) with the emission of a helium-4 nucleus ( α -particles):

According to condition (5.1), such a process is possible if the α-decay energy

Expressing the rest energy of the nucleus through the sum of the rest energies of nucleons and the binding energy of the nucleus, we rewrite inequality (7.1) in the following form:

Result (7.2), which includes only the binding energies of nuclei, is due to the fact that during α decay not only total number nucleons, but also the number of protons and neutrons separately.

Let us consider how the α-decay energy changes E α when changing mass number A. Using the Weizsäcker formula for nuclei lying on the theoretical line of stability, one can obtain the dependence presented in Fig. 7.1. It can be seen that, within the framework of the droplet model, α decay should be observed for nuclei with A> 155, and the decay energy will monotonically increase with increasing A.

The same figure shows the real relationship E α from A, constructed using experimental data on binding energies. Comparing the two curves, you can see that the drip model conveys only the general trend of change E α. In fact, the lightest radionuclide that emits alpha particles is 144 Nd, i.e. the actual region of α-radioactivity is somewhat wider than predicted by the semi-empirical formula. In addition, the dependence of the decay energy on A is not monotonic, but has maxima and minima. The most pronounced maxima occur in the areas A= 140-150 (rare earth elements) and A= 210-220. The appearance of maxima is associated with the filling of the neutron and proton shells of the daughter nucleus to the magic number: N= 82 and Z= 82. As is known, filled shells correspond to anomalously high binding energies. Then, according to the model of nucleon shells, the energy of α-decay of nuclei with N or Z, equal to 84 = 82 + 2, will also be abnormally high. Due to the shell effect, the region of α-radioactivity begins with Nd ( N= 84), and for the vast majority of α-active nuclei Z 84.

An increase in the number of protons in the nucleus (at constant A) promotes α-decay, because increases the relative role of Coulomb repulsion, which destabilizes the nucleus. Therefore, the energy of α decay in a series of isobars will increase with increasing number of protons. An increase in the number of neutrons has the opposite effect.

For nuclei overloaded with protons, β + -decay or electron capture may become competing processes, i.e. processes leading to a decrease Z. For nuclei with an excess of neutrons, the competing process is β – -decay. Starting from mass number A= 232, spontaneous fission is added to the listed decay types. Competing processes can occur so quickly that it is not always possible to observe α-decay against their background.

Let us now consider how the decay energy is distributed between fragments, i.e. α-particle and daughter nucleus, or recoil core. It's obvious that

, (7.3)

Where T α– kinetic energy of the α-particle, T i.o.– kinetic energy of the daughter nucleus (recoil energy). According to the law of conservation of momentum (which is zero in the state before decay), the resulting particles receive momentum equal to absolute value and opposite in sign:

Let's use Fig. 7.1, from which it follows that the α-decay energy (and therefore the kinetic energy of each particle) does not exceed 10 MeV. The rest energy of an α particle is about 4 GeV, i.e. hundreds of times more. The rest energy of the daughter nucleus is even greater. In this case, to establish the relationship between kinetic energy and momentum, one can use the relation of classical mechanics

Substituting (7.5) into (7.3) we obtain

. (7.6)

From (7.6) it follows that the bulk of the decay energy is carried away by the lightest fragment - the α-particle. Yes, when A= 200 the daughter core returns only 2% of E α.

The unambiguous distribution of decay energy between two fragments leads to the fact that each radionuclide emits alpha particles of strictly defined energies, or, in other words, alpha spectra are discrete. Thanks to this, a radionuclide can be identified by the energy of α-particles: the spectral lines serve as a kind of “fingerprint”. Moreover, as experiment shows, α-spectra very often contain not one, but several lines of different intensities with similar energies. In such cases we talk about fine structureα spectrum (Fig. 7.2).

To understand the origin of the fine structure effect, remember that the α-decay energy is nothing more than the difference between the energy levels of the mother and daughter nuclei. If the transition occurred only from the ground state of the mother nucleus to the ground state of the daughter nucleus, the α-spectra of all radionuclides would contain only one line. Meanwhile, it turns out that transitions from the ground state of the mother nucleus can also occur in excited states.

The half-lives of α-emitters vary widely: from 10 – 7 seconds to 10 17 years. On the contrary, the energy of emitted α-particles lies in a narrow range: 1-10 MeV. Relationship between decay constant λ and energy of α-particles Tα is given Geiger's law–Nettola, one of the recording forms of which is:

, (7.7)

Where WITH 1 and WITH 2 – constants that change little when moving from nucleus to nucleus. In this case, an increase in the energy of α-particles by 1 MeV corresponds to a decrease in the half-life by several orders of magnitude.

7.2. Passage of α-particles through a potential barrier. Before the advent of quantum mechanics, no theoretical explanation was given for such a sharp dependence λ from Tα. Moreover, the very possibility of alpha particles escaping from the nucleus with energies significantly lower than the height of the potential barriers that were proven to surround the nuclei seemed mysterious. For example, experiments on the scattering of α-particles of 212 Po with an energy of 8.78 MeV on uranium showed that near the uranium nucleus there are no deviations from Coulomb’s law; however, uranium emits alpha particles with an energy of only 4.2 MeV. How do these α-particles penetrate through a barrier whose height is at least 8.78 MeV, and in reality even higher?..

In Fig. 7.3 shows the dependence of potential energy U positively charged particle from the distance to the nucleus. In the area r > R between the particle and the nucleus there are only electrostatic repulsion forces, in the region r < R More intense nuclear attractive forces prevail, preventing the particle from escaping from the nucleus. Resulting curve U(r) has a sharp maximum in the region r ~ R, called Coulomb potential barrier. Barrier height

, (7.8)

Where Z 1 and Z 2 – charges of the emitted particle and daughter nucleus, R– radius of the nucleus, which in the case of α-decay is taken equal to 1.57 A 1/3 fm. It is easy to calculate that for 238 U the height of the Coulomb barrier will be ~ 27 MeV.

The emission of α-particles (and other positively charged nucleon formations) from the nucleus is explained by quantum mechanical tunneling effect, i.e. the ability of a particle to move in a classically forbidden region between turning points, where T < U.

In order to find the probability of a positively charged particle passing through a Coulomb potential barrier, we first consider a rectangular barrier of width a and heights V, onto which a particle with energy falls E(Fig. 7.4). Outside the barrier in regions 1 and 3, the Schrödinger equation looks like

,

and in internal area 2 how

.

The solution is plane waves

.

Amplitude A 1 corresponds to a wave incident on the barrier, IN 1 – wave reflected from the barrier, A 3 – a wave that has passed through the barrier (since the transmitted wave is no longer reflected, the amplitude IN 3 = 0). Since E < V,

magnitude q– purely imaginary, and the wave function under the barrier

.

The second term in formula (7.9) corresponds to an exponentially growing wave function, and therefore growing with increasing X probability of detecting a particle under the barrier. In this regard, the value IN 2 can't be big compared to A 2. Then, putting IN 2 is simply equal to zero, we have

. (7.10)

Transparency coefficient D barrier, i.e. the probability of finding a particle that was originally in region 1 in region 3 is simply the ratio of the probabilities of finding the particle at points X = A And X= 0. For this, knowledge of the wave function under the barrier is sufficient. As a result

. (7.11)

Let us further imagine a potential barrier of arbitrary shape as a set N rectangular potential barriers with height V(x) and width Δ x(Fig. 7.5). The probability of a particle passing through such a barrier is the product of the probabilities of passing all barriers one after another, i.e.

Then, considering barriers of infinitesimal width and passing from summation to integration, we obtain

(7.12)

Limits of integration x 1 and x 2 in formula (7.12) correspond to classical turning points at which V(x) = E, while the movement of the particle in the regions x < x 1 and x > x 2 is considered free.

For the Coulomb potential barrier, the calculation D according to (7.12) can be carried out exactly. This was first done by G.A. Gamow in 1928, i.e. even before the discovery of the neutron (Gamow believed that the nucleus consists of alpha particles).

For an α particle with kinetic energy T in the potential of the species u/r the expression for the barrier transparency coefficient takes next view:

, (7.13)

and the meaning ρ is determined by equality T = u/ρ . Integral in the exponent after substitution ξ = r 1/2 takes a form convenient for integration:

.

The latter gives

If the height of the Coulomb barrier is significantly greater than the energy of the α particle, then ρ >> R. In this case

. (7.14)

Substituting (7.14) into (7.13) and taking into account that ρ = BR/T, we get

. (7.15)

In the general case, when the height of the Coulomb barrier is comparable to the energy of the emitted particle, the transparency coefficient D is given by the following formula:

, (7.16)

where is the reduced mass of two flying particles (for an α-particle it is very close to its own mass). Formula (7.16) gives for 238 U the value D= 10 –39, i.e. the probability of α-particle tunneling is extremely low.

Result (7.16) was obtained for the case central spread particles, i.e. such when an α-particle is emitted by the nucleus strictly in the radial direction. If the latter does not take place, then the angular momentum carried away by the α-particle not equal to zero. Then when calculating D an adjustment related to the presence of additional centrifugal barrier:

, (7.17)

Where l= 1, 2, 3, etc.

Meaning U c(R) is called the height of the centrifugal barrier. The existence of a centrifugal barrier leads to an increase in the integral in (7.12) and a decrease in the transparency coefficient. However, the centrifugal barrier effect is not too great. Firstly, since the rotational energy of the system at the moment of expansion U c(R) cannot exceed the α-decay energy T, then most often, and the height of the centrifugal barrier does not exceed 25% of the Coulomb barrier. Secondly, it should be taken into account that the centrifugal potential (~1/ r 2) decreases much faster with distance than the Coulomb one (~1/ r). As a result, the probability of emission of an α-particle with l≠ 0 has practically the same order of magnitude as for l = 0.

Possible values l are determined by the selection rules for angular momentum and parity, which follow from the corresponding conservation laws. Since the spin of the α particle is zero and its parity is positive, then

(indices 1 and 2 refer to the mother and daughter nuclei, respectively). Using rules (7.18), it is not difficult to establish, for example, that α-particles of 239 Pu (Fig. 7.2) with an energy of 5.157 MeV are emitted only during central expansion, while for α-particles with energies of 5.144 and 5.016 MeV l = 2.

7.3. α-decay rate. The probability of α-decay as a complex event is the product of two quantities: the probability of the formation of an α-particle inside the nucleus and the probability of leaving the nucleus. The process of α-particle formation is purely nuclear; it is quite difficult to calculate accurately, since it has all the difficulties of a nuclear problem. Nevertheless, for the simplest assessment, we can assume that α-particles in the nucleus exist, as they say, “in ready-made form.” Let v– the speed of the α-particle inside the nucleus. Then it will appear on its surface n once per unit time, where n = v/2R. Let us assume that, in order of magnitude, the radius of the nucleus R equal to the de Broglie wavelength of the α particle (see Appendix B), i.e. , Where . Thus, considering the decay probability as the product of the barrier transparency coefficient and the frequency of collisions of an α-particle with the barrier, we have

. (7.19)

If the barrier transparency coefficient satisfies relation (7.15), then after substitution and logarithm (7.19) we obtain the Geiger-Nettall law (7.7). Taking the energy of α particles T << IN, we can approximately determine how the coefficients of formula (7.7) depend on A And Z radioactive nucleus. Substituting the height of the Coulomb barrier (7.8) into (7.15) and taking into account that during α-decay Z 1 = Z α= 2 and μ ≈ M α, we have

,

Where Z 2 – charge of the daughter nucleus. Then, taking logarithm (7.19), we find that

,

.

Thus, WITH 1 depends very weakly (logarithmically) on the mass of the nucleus, and WITH 2 depends linearly on its charge.

According to (7.19), the frequency of collisions of an α particle with a potential barrier is about 5·10 20 s –1 for most α-radioactive particles. Consequently, the value that determines the α-decay constant is the barrier transparency coefficient, which strongly depends on energy, since the latter is included in the exponent. This is due to the narrow range in which the energies of α-particles of radioactive nuclei can change: particles with energies above 9 MeV fly out almost instantly, while at energies below 4 MeV they live in the nucleus for so long that α-decay is very difficult to detect.

As already noted, α-radiation spectra often have a fine structure, i.e. the energy of emitted particles takes not one, but a whole series of discrete values. The appearance of particles with lower energy in the spectrum ( short-run) corresponds to the formation of daughter nuclei in excited states. By virtue of law (7.7), the yield of short-range α particles is always significantly less than the yield of particles of the main group. Therefore, the fine structure of α-spectra is associated, as a rule, with transitions to rotationally excited levels of non-spherical nuclei with low excitation energy.

If the decay of the mother nucleus occurs not only from the ground state, but also from excited states, one observes long-distanceα particles. An example is the long-range α-particles emitted by the nuclei of the polonium isotopes 212 Po and 214 Po. Thus, the fine structure of α-spectra in some cases carries information about the levels of not only daughter, but also mother nuclei.

Taking into account the fact that the α-particle does not exist in the nucleus, but is formed from its constituent nucleons (two protons and two neutrons), as well as a more rigorous description of the movement of the α-particle inside the nucleus, require a more detailed consideration of the physical processes occurring in the nucleus. In this regard, it is not surprising that α-decays of nuclei are divided into lightweight And detainees. A decay is called facilitated if formula (7.19) is satisfied quite well. If the actual half-life exceeds the calculated half-life by more than an order of magnitude, such decay is called delayed.

Facilitated α decay is observed, as a rule, in even-even nuclei, and delayed decay is observed in all others. Thus, the transitions of the odd nucleus 235 U into the ground and first excited states 231 Th slow down almost a thousand times. If not for this circumstance, this important radionuclide (235 U) would have been so short-lived that it would not have survived in nature to this day.

Qualitatively delayed α-decay is explained by the fact that the transition to the ground state during the decay of a nucleus containing an unpaired nucleon (with the lowest binding energy) can only occur when this nucleon becomes part of an α-particle, i.e. when another pair of nucleons breaks. This way of forming an alpha particle is much more difficult than its construction from already existing pairs of nucleons in even-even nuclei. For this reason, there may be a delay in the transition to the ground state. If, on the other hand, an α particle is nevertheless formed from pairs of nucleons already existing in such a nucleus, the daughter nucleus should end up in an excited state after decay. The last reasoning explains the rather high probability of transition to excited states for odd nuclei (Fig. 7.2).

Most atomic nuclei are unstable. Sooner or later they spontaneously (or, as physicists say, spontaneously) decay into smaller nuclei and elementary particles, which are commonly called decomposition products or child elements. Decaying particles are usually called starting materials or parents. All of the chemical substances we are familiar with (iron, oxygen, calcium, etc.) have at least one stable isotope. ( Isotopes are called varieties of a chemical element with the same number of protons in the nucleus - this number of protons corresponds to the atomic number of the element - but a different number of neutrons.) The fact that these substances are well known to us indicates their stability - which means they live long enough , in order to accumulate in significant quantities in natural conditions without breaking down into components. But each of the natural elements also has unstable isotopes - their nuclei can be obtained in the process of nuclear reactions, but they do not live long because they quickly decay.

Nuclei of radioactive elements or isotopes can decay in three main ways, and the corresponding nuclear decay reactions are named by the first three letters of the Greek alphabet. At alpha decay A helium atom consisting of two protons and two neutrons is released - it is usually called an alpha particle. Since alpha decay entails a decrease in the number of positively charged protons in an atom by two, the nucleus that emitted the alpha particle turns into the nucleus of an element two positions lower from it in the periodic table. At beta decay the nucleus emits an electron and the element moves one position forward according to the periodic table (in this case, essentially, a neutron turns into a proton with the radiation of this very electron). Finally, gamma decay - This decay of nuclei with the emission of high-energy photons, which are commonly called gamma rays. In this case, the nucleus loses energy, but the chemical element does not change.

However, the mere fact of instability of one or another isotope of a chemical element does not mean that by collecting together a certain number of nuclei of this isotope, you will get a picture of their instantaneous decay. In reality, the decay of the nucleus of a radioactive element is somewhat reminiscent of the process of frying corn when making popcorn: the grains (nucleons) fall off the “cob” (kernel) one at a time, in a completely unpredictable order, until all of them fall off. The law describing the reaction of radioactive decay, in fact, only states this fact: over a fixed period of time, a radioactive nucleus emits a number of nucleons proportional to the number of nucleons remaining in its composition. That is, the more grains-nucleons still remain in the “undercooked” cob-kernel, the more of them will be released during a fixed “frying” time interval. When we translate this metaphor into the language of mathematical formulas, we get an equation describing radioactive decay:

d N = λN d t

where d N— number of nucleons emitted by a nucleus with total number of nucleons N in time d t, A λ - experimentally determined radioactivity constant test substance. The above empirical formula is a linear differential equation, the solution of which is the following function, which describes the number of nucleons remaining in the nucleus at a time t:

N = N 0 e - λt

Where N 0 is the number of nucleons in the nucleus at the initial moment of observation.

The radioactivity constant thus determines how quickly the nucleus decays. However, experimental physicists usually measure not it, but the so-called half-life nucleus (that is, the period during which the nucleus under study emits half of the nucleons it contains). For different isotopes of different radioactive substances, half-lives vary (in full accordance with theoretical predictions) from billionths of a second to billions of years. That is, some nuclei live almost forever, and some decay literally instantly (here it is important to remember that after the half-life, half of the total mass of the original substance remains, after two half-lives - a quarter of its mass, after three half-lives - one eighth, etc. .d.).

As for the emergence of radioactive elements, they are born in different ways. In particular, the ionosphere (the thin upper layer of the atmosphere) of the Earth is constantly bombarded by cosmic rays consisting of high-energy particles ( cm. Elementary particles). Under their influence, long-lived atoms are split into unstable isotopes: in particular, from stable nitrogen-14 in the earth’s atmosphere, the unstable isotope carbon-14 with 6 protons and 8 neutrons in the nucleus is constantly formed ( cm. Radiometric dating).

But the above case is rather exotic. Much more often, radioactive elements are formed in reaction chains nuclear fission . This is the name given to a series of events during which the original (“mother”) nucleus decays into two “daughter” (also radioactive), which, in turn, decay into four “granddaughter” nuclei, etc. The process continues until until stable isotopes are obtained. As an example, let's take the isotope uranium-238 (92 protons + 146 neutrons) with a half-life of about 4.5 billion years. This period, by the way, is approximately equal to the age of our planet, which means that approximately half of the uranium-238 from the composition of the primary matter of the formation of the Earth is still found in the totality of elements of earthly nature. Uranium-238 turns into thorium-234 (90 protons + 144 neutrons), which has a half-life of 24 days. Thorium-234 turns into palladium-234 (91 protons + 143 neutrons) with a half-life of 6 hours - etc. After more than ten decay stages, the stable isotope of lead-206 is finally obtained.

There is a lot that can be said about radioactive decay, but a few points deserve special mention. Firstly, even if we take a pure sample of any one radioactive isotope as a starting material, it will decay into different components, and soon we will inevitably get a whole “bouquet” of different radioactive substances with different nuclear masses. Secondly, the natural chains of reactions of atomic decay reassure us in the sense that radioactivity is a natural phenomenon, it existed long before man, and there is no need to take it upon ourselves and blame human civilization alone for the fact that there is background radiation on Earth. Uranium-238 has existed on Earth since its very inception, has decayed, is decaying - and will continue to decay, and nuclear power plants accelerate this process, in fact, by a fraction of a percent; so they do not have any particularly harmful effects on you and me in addition to what is provided by nature.

Finally, the inevitability of radioactive atomic decay poses both potential problems and potential opportunities for humanity. In particular, in the chain of reactions of the decay of uranium-238 nuclei, radon-222 is formed - a noble gas without color, odor and taste, which does not enter into any chemical reactions, since it is not capable of forming chemical bonds. This inert gas, and it literally oozes from the depths of our planet. Usually it has no effect on us - it simply dissolves in the air and remains there in a slight concentration until it breaks down into even lighter elements. However, if this harmless radon remains in an unventilated room for a long time, then over time its decay products will begin to accumulate there - and they are harmful to human health (if inhaled). This is how we get the so-called “radon problem”.

On the other hand, the radioactive properties of chemical elements bring significant benefits to people if we approach them wisely. Radioactive phosphorus, in particular, is now injected to produce a radiographic picture of bone fractures. The degree of its radioactivity is minimal and does not cause harm to the patient’s health. Entering the bone tissue of the body along with ordinary phosphorus, it emits enough rays to record them on light-sensitive equipment and take pictures of a broken bone literally from the inside. Surgeons, accordingly, have the opportunity to operate on a complex fracture not blindly and at random, but by studying the structure of the fracture in advance using such images. In general, applications radiography there are countless numbers in science, technology and medicine. And they all work on the same principle: the chemical properties of an atom (essentially, the properties of the outer electron shell) make it possible to assign a substance to a certain chemical group; then, using the chemical properties of this substance, the atom is delivered “to the right place”, after which, using the property of the nuclei of this element to decay in strict accordance with the “schedule” established by the laws of physics, the decay products are recorded.

E. Resenford, together with the English radiochemist F. Soddy, proved that radioactivity is accompanied by the spontaneous transformation of one chemical element into another.
Moreover, as a result of radioactive radiation, the nuclei of atoms of chemical elements undergo changes.

DESIGNATION OF THE ATOMIC NUCLEUS

ISOTOPES

Among the radioactive elements, elements were discovered that were chemically indistinguishable, but different in mass. These groups of elements were called "isotopes" ("occupying one place in the periodic table"). The nuclei of atoms of isotopes of the same chemical element differ in the number of neutrons.

It has now been established that all chemical elements have isotopes.
In nature, all chemical elements, without exception, consist of a mixture of several isotopes, therefore, in the periodic table, atomic masses are expressed in fractional numbers.
Isotopes of even non-radioactive elements can be radioactive.

ALPHA - DECAY

Alpha particle (nucleus of a helium atom)
- characteristic of radioactive elements with a serial number greater than 83
.- the law of conservation of mass and charge number is necessarily satisfied.
- often accompanied by gamma radiation.

Alpha decay reaction:

During the alpha decay of one chemical element, another chemical element is formed, which in the periodic table is located 2 cells closer to its beginning than the original one

Physical meaning of the reaction:

As a result of the emission of an alpha particle, the charge of the nucleus decreases by 2 elementary charges and a new chemical element is formed.

Offset rule:

During the beta decay of one chemical element, another element is formed, which is located in the periodic table in the next cell after the original one (one cell closer to the end of the table).

BETA - DECAY

Beta particle (electron).
- often accompanied by gamma radiation.
- may be accompanied by the formation of antineutrinos (light electrically neutral particles with high penetrating power).
- the law of conservation of mass and charge number must be fulfilled.

Beta decay reaction:

Physical meaning of the reaction:

A neutron in the nucleus of an atom can turn into a proton, electron and antineutrino, as a result the nucleus emits an electron.

Offset rule:

FOR THOSE WHO ARE NOT TIRED YET

I suggest writing the decay reactions and handing in the work.
(make a chain of transformations)

1. The nucleus of which chemical element is the product of one alpha decay
and two beta decays of the nucleus of a given element?

The structure and properties of particles and atomic nuclei have been studied for about a hundred years in decays and reactions.
Decays represent the spontaneous transformation of any object of microworld physics (nucleus or particle) into several decay products:

Both decays and reactions are subject to a number of conservation laws, among which the following laws should be mentioned, firstly:

In the future, other conservation laws operating in decays and reactions will be discussed. The laws listed above are the most important and, what is especially significant, are performed in all types of interactions.(It is possible that the law of conservation of baryon charge does not have such universality as conservation laws 1-4, but its violation has not yet been discovered).
The processes of interactions between objects of the microworld, which are reflected in decays and reactions, have probabilistic characteristics.

Decays

Spontaneous decay of any object of microworld physics (nucleus or particle) is possible if the rest mass of the decay products is less than the mass of the primary particle.

Decays are characterized decay probabilities , or the inverse probability of average life time τ = (1/λ). The quantity associated with these characteristics is also often used half-life T 1/2.
Examples of spontaneous decays

; π 0 → γ + γ; π + → μ + + ν μ ;

(2.4)

n → p + e − + e ; μ + → e + + μ + ν e ;

(2.5)

In decays (2.4) there are two particles in the final state. In decays (2.5) there are three.
We obtain the decay equation for particles (or nuclei). The decrease in the number of particles (or nuclei) over a time interval is proportional to this interval, the number of particles (nuclei) at a given time and the probability of decay:

Integration (2.6) taking into account the initial conditions gives the relationship between the number of particles at time t and the number of the same particles at the initial time t = 0:

The half-life is the time during which the number of particles (or nuclei) decreases by half:

Spontaneous decay of any object of microworld physics (nucleus or particle) is possible if the mass of the decay products is less than the mass of the primary particle. Decays into two products and into three or more are characterized by different energy spectra of the decay products. In the case of decay into two particles, the spectra of the decay products are discrete. If there are more than two particles in the final state, the spectra of the products are continuous.

The difference in the masses of the primary particle and the decay products is distributed among the decay products in the form of their kinetic energies.
The laws of conservation of energy and momentum for decay should be written in the coordinate system associated with the decaying particle (or nucleus). To simplify the formulas, it is convenient to use the system of units = c = 1, in which energy, mass and momentum have the same dimension (MeV). Conservation laws for this decay:

From here we obtain for the kinetic energies of the decay products

Thus, in the case of two particles in the final state kinetic energies of products are determined definitely. This result does not depend on whether the decay products have relativistic or non-relativistic velocities. For the relativistic case, the formulas for kinetic energies look somewhat more complicated than (2.10), but the solution to the equations for the energy and momentum of two particles is again unique. This means that in the case of decay into two particles, the spectra of the decay products are discrete.
If three (or more) products appear in the final state, solving the equations for the laws of conservation of energy and momentum does not lead to an unambiguous result. In case if there are more than two particles in the final state, the spectra of the products are continuous.(In what follows, using the example of -decays, this situation will be considered in detail.)
In calculating the kinetic energies of nuclear decay products, it is convenient to use the fact that the number of nucleons A is conserved. (This is a manifestation baryon charge conservation law , since the baryon charges of all nucleons are equal 1).
Let us apply the obtained formulas (2.11) to the -decay of 226 Ra (the first decay in (2.4)).

Mass difference between radium and its decay products
ΔM = M(226 Ra) - M(222 Rn) - M(4 He) = Δ(226 Ra) - Δ(222 Rn) - Δ(4 He) = (23.662 - 16.367 - 2.424) MeV = 4.87 MeV. (Here we used tables of excess masses of neutral atoms and the relation M = A + for masses, etc. surplus masses Δ)
The kinetic energies of helium and radon nuclei resulting from alpha decay are equal to:

,
.

The total kinetic energy released as a result of alpha decay is less than 5 MeV and is about 0.5% of the rest mass of the nucleon. The ratio of the kinetic energy released as a result of the decay and the rest energies of particles or nuclei - criterion for the admissibility of using the nonrelativistic approximation. In the case of alpha decays of nuclei, the smallness of kinetic energies compared to rest energies allows us to limit ourselves to the nonrelativistic approximation in formulas (2.9-2.11).

Problem 2.3. Calculate the energies of particles produced in meson decay

The decay of the π + meson occurs into two particles: π + μ + + ν μ. The mass of the π + meson is 139.6 MeV, the mass of the μ muon is 105.7 MeV. The exact value of the muon neutrino mass ν μ is not yet known, but it has been established that it does not exceed 0.15 MeV. In an approximate calculation, we can set it equal to 0, since it is several orders of magnitude lower than the difference between the pion and muon masses. Since the difference between the masses of the π + meson and its decay products is 33.8 MeV, for neutrinos it is necessary to use relativistic formulas for the relationship between energy and momentum. In further calculations, the small mass of the neutrino can be neglected and the neutrino can be considered an ultrarelativistic particle. Laws of conservation of energy and momentum in the decay of the π + meson:

m π = m μ + T μ + E ν
|p ν | = | p μ |

E ν = p ν

An example of two-particle decay is also the emission of a -quantum during the transition of an excited nucleus to a lower energy level.
In all two-particle decays analyzed above, the decay products have an “exact” energy value, i.e. discrete spectrum. However, a deeper consideration of this problem shows that the spectrum of even the products of two-particle decays is not a function of energy.

.

The spectrum of decay products has a finite width Γ, which is larger the shorter the lifetime of the decaying nucleus or particle.

(This relation is one of the formulations of the uncertainty relation for energy and time).
Examples of three-body decays are -decays.
The neutron undergoes -decay, turning into a proton and two leptons - an electron and an antineutrino: np + e - + e.
Leptons themselves, for example, the muon, also experience beta decays (the average lifetime of a muon
τ = 2.2 ·10 –6 sec):

.

Conservation laws for muon decay at maximum electron momentum:
For the maximum kinetic energy of the muon decay electron, we obtain the equation

The kinetic energy of the electron in this case is two orders of magnitude higher than its rest mass (0.511 MeV). The momentum of a relativistic electron practically coincides with its kinetic energy, indeed

p = (T 2 + 2mT) 1/2 = )