As already noted (see § 138), nucleons are firmly bound in the nucleus of an atom by nuclear forces. To break this connection, i.e., to completely separate the nucleons, it is necessary to expend a certain amount of energy (to do some work).
The energy required to separate the nucleons that make up the nucleus is called the binding energy of the nucleus. The magnitude of the binding energy can be determined on the basis of the law of conservation of energy (see § 18) and the law of proportionality of mass and energy (see § 20).
According to the law of conservation of energy, the energy of nucleons bound in a nucleus must be less than the energy of separated nucleons by the value of the binding energy of the nucleus 8. On the other hand, according to the law of proportionality of mass and energy, a change in the energy of a system is accompanied by a proportional change in the mass of the system
where c is the speed of light in vacuum. Since in the case under consideration is the binding energy of the nucleus, the mass of the atomic nucleus must be less than the sum of the masses of the nucleons that make up the nucleus, by a value called the mass defect of the nucleus. Using formula (10), one can calculate the binding energy of a nucleus if the mass defect of this nucleus is known
Masses are currently atomic nuclei defined since a high degree accuracy by means of a mass spectrograph (see § 102); the masses of the nucleons are also known (see § 138). This makes it possible to determine the mass defect of any nucleus and calculate the binding energy of the nucleus using formula (10).
As an example, let us calculate the binding energy of the nucleus of a helium atom. It consists of two protons and two neutrons. The mass of the proton is the mass of the neutron Therefore, the mass of the nucleons that form the nucleus is The mass of the nucleus of the helium atom Thus, the defect of the helium atomic nucleus is
Then the binding energy of the helium nucleus is
The general formula for calculating the binding energy of any nucleus in joules from its mass defect will obviously have the form
where is the atomic number, A is the mass number. Expressing the mass of nucleons and the nucleus in atomic mass units and taking into account that
one can write the formula for the binding energy of the nucleus in megaelectronvolts:
The binding energy of the nucleus per nucleon is called the specific binding energy. Therefore,
At the helium core
The specific binding energy characterizes the stability (strength) of atomic nuclei: the more v, the more stable the nucleus. According to formulas (11) and (12),
We emphasize once again that in formulas and (13) the masses of nucleons and nuclei are expressed in atomic mass units (see § 138).
Formula (13) can be used to calculate the specific binding energy of any nuclei. The results of these calculations are presented graphically in Figs. 386; the ordinate shows the specific binding energies in the abscissa is the mass numbers A. It follows from the graph that the specific binding energy is maximum (8.65 MeV) for nuclei with mass numbers of the order of 100; for heavy and light nuclei, it is somewhat less (for example, uranium, helium). The specific binding energy of the hydrogen atomic nucleus is zero, which is quite understandable, since there is nothing to separate in this nucleus: it consists of only one nucleon (proton).
Every nuclear reaction is accompanied by the release or absorption of energy. The dependence graph here A allows you to determine at what transformations of the nucleus energy is released and at what - its absorption. During the fission of a heavy nucleus into nuclei with mass numbers A of the order of 100 (or more), energy (nuclear energy) is released. Let us explain this with the following discussion. Let, for example, the division of the uranium nucleus into two
atomic nuclei ("fragment") with mass numbers Specific binding energy of the uranium nucleus specific binding energy of each of the new nuclei To separate all the nucleons that make up the atomic nucleus of uranium, it is necessary to expend energy equal to the binding energy of the uranium nucleus:
When these nucleons combine into two new atomic nuclei with mass numbers 119), an energy equal to the sum of the binding energies of the new nuclei will be released:
Consequently, as a result of the fission reaction of the uranium nucleus, nuclear energy will be released in an amount equal to the difference between the binding energy of new nuclei and the binding energy of the uranium nucleus:
The release of nuclear energy also occurs during nuclear reactions of a different type - when several light nuclei combine (synthesis) into one nucleus. Indeed, let, for example, the fusion of two sodium nuclei into a nucleus with a mass number takes place.
When these nucleons combine into a new nucleus (with a mass number of 46), energy will be released, equal to energy links of the new core:
Consequently, the reaction of the synthesis of sodium nuclei is accompanied by the release of nuclear energy in an amount equal to the difference between the binding energy of the synthesized nucleus and the binding energy of sodium nuclei:
Thus, we come to the conclusion that
the release of nuclear energy occurs both in the fission reactions of heavy nuclei and in the reactions of fusion of light nuclei. The amount of nuclear energy released by each reacted nucleus is equal to the difference between the binding energy 8 2 of the reaction product and the binding energy 81 of the original nuclear material:
This provision is extremely important because it is based on industrial methods obtaining nuclear energy.
Note that the most favorable, in terms of energy yield, is the reaction of fusion of hydrogen or deuterium nuclei
Since, as follows from the graph (see Fig. 386), in this case the difference in the binding energies of the synthesized nucleus and the initial nuclei will be the largest.
The nucleons inside the nucleus are held together by nuclear forces. They are held by a certain energy. It is quite difficult to measure this energy directly, but it can be done indirectly. It is logical to assume that the energy required to break the bond of nucleons in the nucleus will be equal to either more than that the energy that holds the nucleons together.
Binding Energy and Nuclear Energy
This applied energy is already easier to measure. It is clear that this value will very accurately reflect the value of the energy that keeps the nucleons inside the nucleus. Therefore, the minimum energy required to split the nucleus into individual nucleons is called nuclear binding energy.
Relationship between mass and energy
We know that any energy is directly proportional to the mass of the body. Therefore, it is natural that the binding energy of the nucleus will also depend on the mass of the particles that make up this nucleus. This relationship was established by Albert Einstein in 1905. It is called the law of the relationship between mass and energy. In accordance with this law, the internal energy of a system of particles or the rest energy is directly proportional to the mass of the particles that make up this system:
where E is energy, m is mass,
c is the speed of light in vacuum.
Mass defect effect
Now suppose that we have broken the nucleus of an atom into its constituent nucleons, or that we have taken a certain number of nucleons from the nucleus. We expended some energy on overcoming nuclear forces, as we were doing work. In the case of the reverse process - the fusion of the nucleus, or the addition of nucleons to an already existing nucleus, the energy, according to the law of conservation, on the contrary, will be released. When the rest energy of a system of particles changes due to any processes, their mass changes accordingly. Formulas in this case will be as follows:
∆m=(∆E_0)/c^2 or ∆E_0=∆mc^2,
where ∆E_0 is the change in the rest energy of the system of particles,
∆m is the change in the particle mass.
For example, in the case of the fusion of nucleons and the formation of a nucleus, we release energy and reduce the total mass of nucleons. Mass and energy are carried away by the emitted photons. This is the mass defect effect.. The mass of a nucleus is always less than the sum of the masses of the nucleons that make up this nucleus. Numerically, the mass defect is expressed as follows:
∆m=(Zm_p+Nm_n)-M_i,
where M_m is the mass of the nucleus,
Z is the number of protons in the nucleus,
N is the number of neutrons in the nucleus,
m_p is the free proton mass,
m_n is the mass of a free neutron.
The value ∆m in the above two formulas is the value by which the total mass of the particles of the nucleus changes when its energy changes due to rupture or fusion. In the case of synthesis, this quantity will be the mass defect.
| Parameter name | Meaning |
| Article subject: | Mass defect and nuclear binding energy |
| Rubric (thematic category) | Radio |
Studies show that atomic nuclei are stable formations. This means that there is a certain connection between nucleons in the nucleus.
The mass of nuclei can be determined very accurately using mass spectrometers - measuring instruments that separate beams of charged particles (usually ions) with different specific charges using electric and magnetic fields Q/t. Mass spectrometric measurements showed that the mass of the nucleus is less than the sum of the masses of its constituent nucleons. But since any change in mass (see § 40) must correspond to a change in energy, then, therefore, a certain energy must be released during the formation of the nucleus. The opposite also follows from the law of conservation of energy: in order to divide the nucleus into its component parts, it is extremely important to spend the same amount of energy, ĸᴏᴛᴏᴩᴏᴇ is released during its formation. Energy that is extremely important to expend. to split the nucleus into individual nucleons, it is customary to call nuclear binding energy(see § 40).
According to expression (40.9), the binding energy of nucleons and nuclei
E St = [Zmp +(A–Z)m n–m i] c 2 , (252.1)
where m p, m n, m i are the masses of the proton, neutron and nucleus, respectively. The tables usually do not give masses. m i nuclei and masses T atoms. For this reason, the formula for the binding energy of the nucleus is
E St = [Zm H +(A–Z)m n–m] c 2 , (252.2)
where m N is the mass of a hydrogen atom. Because m N more m p , by the amount me, then the first term in square brackets includes a lot Z electrons. But since the mass of an atom T different from the mass of the nucleus m i just on the mass of electrons, then calculations using formulas (252 1) and (252.2) lead to the same results. Value
Δ T = [Zmp +(A–Z)m n] –m i (252.3)
called mass defect kernels. The mass of all nucleons decreases by this amount when an atomic nucleus is formed from them. Often, instead of the binding energy, one considers specific bond energyδE St is the binding energy per nucleon. It characterizes the stability (strength) of atomic nuclei, ᴛ.ᴇ. the more δE St, the more stable the core. The specific binding energy depends on the mass number A element (Fig. 45). For light nuclei ( A≥ 12) the specific binding energy increases steeply up to 6 ÷ 7 MeV, undergoing whole line jumps (for example, for H δE St= 1.1 MeV, for He - 7.1 MeV, for Li - 5.3 MeV), then more slowly increases to maximum value 8.7 MeV for elements with A= 50 ÷ 60, and then gradually decreases for heavy elements (for example, for U it is 7.6 MeV). Note for comparison that the binding energy of valence electrons in atoms is about 10 eV (10 -6 times less).
The decrease in the specific binding energy during the transition to heavy elements is explained by the fact that with an increase in the number of protons in the nucleus, their energy also increases. Coulomb repulsion. For this reason, the bond between nucleons becomes less strong, and the nuclei themselves become less strong.
The most stable are the so-called magic cores, in which the number of protons or the number of neutrons is equal to one of the magic numbers: 2, 8, 20, 28, 50, 82, 126. Particularly stable doubly magic cores, in which both the number of protons and the number of neutrons are magical (there are only five of these nuclei: He, O, Ca, Pb).
From fig. 45 it follows that the nuclei of the middle part of the periodic table are the most stable from an energy point of view. Heavy and light nuclei are less stable. This means that the following processes are energetically favorable:
1) fission of heavy nuclei into lighter ones;
2) the fusion of light nuclei with each other into heavier ones.
Both processes release enormous amounts of energy; these processes are currently carried out practically (fission reaction and thermonuclear reactions).
The mass defect and the binding energy of the nucleus - the concept and types. Classification and features of the category "Mass defect and binding energy of the nucleus" 2017, 2018.
isotopes
isotopes- varieties of atoms (and nuclei) of one chemical element with different amount neutrons in the nucleus. Chemical properties atoms depend practically only on the structure of the electron shell, which, in turn, is determined mainly by the charge of the nucleus Z(that is, the number of protons in it) and almost does not depend on its mass number A(that is, the total number of protons Z and neutrons N). All isotopes of the same element have the same nuclear charge, differing only in the number of neutrons.
An example of isotopes: 16 8 O, 17 8 O, 18 8 O - three stable isotopes of oxygen.
88. The structure of the atomic nucleus. subatomic particles. Elements. isotopes.
An atom consists of a nucleus and an electron cloud surrounding it. located in the electronic cloud electrons bear negative electric charge. Protons, which are part of the nucleus, carry positive charge.
In any atom, the number of protons in the nucleus is exactly equal to the number of electrons in the electron cloud, so the atom as a whole is a neutral particle that does not carry a charge.
An atom can lose one or more electrons, or vice versa - capture other people's electrons. In this case, the atom acquires a positive or negative charge and is called ion.
The outer dimensions of an atom are those of the much less dense electron cloud, which is about 100,000 times the diameter of the nucleus.
In addition to protons, the nucleus of most atoms contains neutrons that carry no charge. The mass of a neutron practically does not differ from the mass of a proton. Together, protons and neutrons are called nucleons.
Binding Energy and Nucleus Mass Defect
Nucleons in the nucleus are firmly held by nuclear forces. In order to remove a nucleon from the nucleus, a lot of work must be done, i.e., significant energy must be imparted to the nucleus.
The binding energy of the atomic nucleus E st characterizes the intensity of the interaction of nucleons in the nucleus and is equal to the maximum energy that must be expended to divide the nucleus into separate non-interacting nucleons without imparting kinetic energy to them. Each nucleus has its own binding energy. The greater this energy, the more stable the atomic nucleus. Accurate measurements of the masses of the nucleus show that the rest mass of the nucleus m i is always less than the sum of the rest masses of its constituent protons and neutrons. This mass difference is called the mass defect:
It is this part of the mass Dm that is lost when the binding energy is released. Applying the law of the relationship between mass and energy, we obtain:
* c2 (short here, multiply by C squared)
where is the speed of light in vacuum.
Other important parameter The nucleus is the binding energy per nucleon of the nucleus, which can be calculated by dividing the binding energy of the nucleus by the number of nucleons it contains:
This value represents the average energy that needs to be expended to remove one nucleon from the nucleus, or the average change in the binding energy of the nucleus when a free proton or neutron is absorbed into it.
On fig. a graph of the experimentally established dependence of Eb on A is shown.
As can be seen from the explanatory figure, at small mass numbers, the specific binding energy of nuclei increases sharply and reaches a maximum at (about 8.8 MeV). Nuclides with such mass numbers are the most stable. With further growth, the average binding energy decreases, however, in a wide range of mass numbers, the energy value is almost constant (MeV), from which it follows that we can write .
This behavior of the average binding energy indicates the property of nuclear forces to reach saturation, that is, the possibility of interaction of the nucleon with only a small number of "partners". If the nuclear forces did not have the saturation property, then within the radius of action of nuclear forces, each nucleon would interact with each of the others and the interaction energy would be proportional to , and the average binding energy of one nucleon would not be constant for different nuclei, but would increase with increasing .
90.Theories of the structure of the atomic nucleus
In the process of development of physics, various hypotheses of the structure of the atomic nucleus were put forward. The most famous are the following:
· Drop model of the nucleus - proposed in 1936 by Niels Bohr.
Drop model of the nucleus- one of the earliest models of the structure of the atomic nucleus, proposed by Niels Bohr in 1936 within the framework of the theory of the compound nucleus, developed by Yakov Frenkel and, later, by John Wheeler, on the basis of which Karl Weizsacker was the first to obtain a semi-empirical formula for the binding energy of the nucleus of an atom, named in his honor Weizsäcker formula.
According to this theory, the atomic nucleus can be represented as a spherical uniformly charged drop of special nuclear matter, which has some properties, such as incompressibility, saturation of nuclear forces, "evaporation" of nucleons (neutrons and protons), resembles a liquid. In this connection, some other properties of a liquid drop can be extended to such a core-drop, for example, surface tension, drop fragmentation into smaller ones (nucleus fission), merging of small drops into one large one (nucleus synthesis).
· Shell model of the nucleus - proposed in the 30s of the XX century.
In the shell model of an atom, electrons fill the electron shells, and once the shell is filled, the binding energy for the next electron is significantly reduced.
· Generalized Bohr-Mottelson model.
O. m. i. it was proposed based on the assumption of independent motion of nucleons in a field with a slowly varying potential. Nucleons int. filled shells form a "core" that has collective degrees of freedom and is described using the liquid drop model (see Fig. drop model of the nucleus). The nucleons of the outer, unfilled shells, interacting with the surface of this drop, form a common, as a rule, non-spherical, self-consistency. potential. The adiabaticity of the change in this potential makes it possible to separate the single-particle motion of nucleons occurring in a fixed way. potential, from collective movement, leading to a change in shape and orientation cf. kernel fields. This approach is similar to the separation of the motion of electrons and nuclei in molecules.
Cluster kernel model
Model of nucleon associations
Optical model of the nucleus
Superfluid core model
Statistical kernel model
nuclear forces
Nuclear forces are forces that hold nucleons in the nucleus, which are large attractive forces that act only at small distances. They have saturation properties, in connection with which the exchange character is attributed to the nuclear forces. Nuclear forces depend on spin, do not depend on electric charge and are not central forces.
radioactive decay
radioactive decay(from lat. radius"beam" and activus"effective") - a spontaneous change in the composition of unstable atomic nuclei (charge Z, mass number A) by emitting elementary particles or nuclear fragments. The process of radioactive decay is also called radioactivity, and the corresponding elements are radioactive. Substances containing radioactive nuclei are also called radioactive.
All have been found to be radioactive. chemical elements with an atomic number greater than 82 (that is, starting with bismuth), and many lighter elements (promethium and technetium do not have stable isotopes, and some elements, such as indium, potassium or calcium, have some natural isotopes that are stable, while others are radioactive) .
natural radioactivity- spontaneous decay of the nuclei of elements found in nature.
Relative atomic mass A of a chemical element (it is given along with the symbol of the element and its ordinal number in each cell periodic system D. I. Mendeleev) is the average value of the relative isotope masses, taking into account the isotope content. The relative atomic mass actually shows how many times the mass of a given atom is greater than the mass of 1/12 of the carbon isotope. Like any relative value, Ar is a dimensionless quantity.
Per unit of atomic mass ( atomic mass unit - a.m.u.) is currently accepted as 1/12 of the mass of the 12 C nuclide. A mass of 12.0000 amu is attributed to this nuclide. true value atomic mass unit is 1.661 10-27 kg.
The masses of the three fundamental particles, expressed in amu, have the following values:
the mass of a proton is 1.007277 a.m.u., the mass of a neutron is 1.008665 a.m.u., the mass of an electron is 0.000548 a.m.u.
1.9.4. mass defect
If you calculate the mass of any isotope (isotope mass) by summing the masses of the corresponding number of protons, neutrons and electrons, the result will not give an exact match with the experiment. The discrepancy between the
the measured and experimentally found values of isotope masses are called
mass defect.
So, for example, the isotopic mass of one of the isotopes of chlorine 35 Cl, obtained by adding the masses of seventeen protons, eighteen neutrons and seventeen electrons, is:
17 1.007277 + 18 1.008665 + 17 0.000548 = 35.289005 amu
However, exact experimental determinations of this quantity give the result 34.96885 a.m.u. The mass defect is 0.32016 amu.
Explanations for the mass defect phenomenon can be given using the ideas formulated by Albert Einstein in the theory of relativity. The mass defect corresponds to the energy required to overcome the repulsive forces between protons.
In other words, the mass defect is a measure of the binding energy of nuclear particles. If it were possible to divide the nucleus into its constituent nucleons, then the mass of the system would increase by the amount of the mass defect. The binding energy shows the difference between the energy of nucleons in the nucleus and their energy in the free state, i.e. The binding energy is the energy required to separate the nucleus into its constituent nucleons.
The binding energy can be calculated by A. Einstein's formula:
E = mc2 ,
where: m is the mass in kg, s is the speed of light - 2.9979 108 m/s, E is the energy in J. For example, the binding energy for one mole (4 g) of the nuclide4 He (molar
mass defect is 3.0378 10-5 kg) is equal to:
∆ Е = (3.0378 10-5 kg/mol) (2.9979 108 m/s)2 = 2.730 1012 J/mol covalent bond more than
10 million times. To obtain such energy through chemical reaction tens of tons of the substance would have to be used.
Since the binding energy is extremely high, it is customary to express it in megaelectronvolts (1 MeV = 9.6 1010 J/mol) per nucleon. Thus, the binding energy per nucleon in the 4 He nucleus is approximately 7 MeV, and in the 35 Cl nucleus it is 8.5 MeV.
1.9.5. nuclear forces
The nucleus of an atom is a special object for study. Even with a superficial examination of it, there are many perplexities. Why don't the protons that make up the nucleus repel each other? elementary laws electrostatics? The simplest calculation using Coulomb's law shows that at nuclear distances, two protons should repel each other with a force of about 6000 N, and they are attracted to each other with a force 40 times greater than this value. Moreover, this force equally acts both between two protons and between two neutrons, as well as between a proton and a neutron, i.e. completely independent of the charge of the particles.
Obviously, nuclear forces represent a completely different class of forces; they cannot be reduced to electrostatic interactions. The energy that accompanies nuclear reactions is millions of times greater than the energy that characterizes chemical transformations.
Application of principles quantum mechanics to the description of the motion of electrons gives very satisfactory results at the present time. Can this theory be used to model the processes occurring in the nucleus of an atom? The most important feature nuclear forces is extremely small radius of their action. Indeed, the motion of an electron occurs in a region of space estimated by values of the order of 10-8 cm, and all intranuclear phenomena occur at distances of the order of 10-12 cm and less. These values are slightly larger than the intrinsic sizes of nucleons. The ratio of scales characterizing the motion of an electron on the one hand and intranuclear phenomena on the other can be compared in order of magnitude with the same ratio
for the macrocosm, which obeys the laws of classical mechanics, and the microcosm, which lives according to the laws of quantum mechanics.
With such a small size of the nucleus, almost the entire mass of the atom is concentrated in it. Knowing the approximate volume of the nucleus and the mass of the atom, it is possible to estimate the density of nuclear matter: it exceeds the average density of ordinary matter by a factor of 2 1017 and amounts to a value of the order of 1013 - 1014 g/cm3. An attempt to really understand such quantities leads to the following illustration: with a similar density of matter, the volume of a match head (approximately 5 mm3) should contain a mass equal to the mass of 1 million tons of water. If such a match head fell to the surface of the Earth, it would pierce everything rocks and penetrated to the center of the planet.
1.9.6. Nuclear transformations
The transformations of atomic nuclei due to their interactions with elementary particles or with each other are called nuclear reactions.
Spontaneous nuclear fission natural radioactivity- accompanied by three types of radiation.
Alpha radiation is a stream of nuclei of helium atoms with a charge of +2 and a mass number of 4 (4 He). positive charge of these particles explains the fact that alpha rays are deflected in electric field towards the negatively charged plate, and relatively big size helium atoms justifies a significantly lower penetrating power compared to the other two types of radiation.
Obviously, when such a particle is emitted, the nucleus loses two protons and two neutrons. The loss of two protons reduces the atomic number by two units, hence the result is the formation of a new chemical element.
For example, the nuclide of radium-226, when an alpha particle is lost, turns into a nuclide of radon-222, which can be represented as equations nuclear reaction :
88 Ra → 86 Rn +2 He.
When compiling such equations, one should observe the equality of the sums of atomic numbers and the sums of mass numbers on the left and right sides (charge and mass conservation must be ensured).
In some cases, an abbreviated form of writing the nuclear reaction equation is also used: the initial nuclide is written on the left, the final nuclide on the right, in parentheses between them, first, the particle that causes this transformation is indicated, and then emitted as a result of it. In this case, letter designations are used for such particles: α (alpha particle), p (proton), n (neutron), d (deuterium nucleus - deuteron), etc. For example, for the alpha decay discussed above:
Ra (-, α ) Rn.
The sign "-" indicates the absence of a bombarding particle (the decay of the nucleus occurs spontaneously).
Beta radiation, in turn, is divided into β - (it is usually called
are simply β-radiation) and β + -radiation. β - radiation is a stream of electrons moving at a speed close to the speed of light. These electrons come from the decay of a neutron:
90 Th → 91 Pa + -1 e.
The nuclides of thorium-234 and protactinium-234 have the same mass numbers. Such nuclides are called isobars.
The occurrence of β + radiation is due to the transformation of a proton into a neutron, accompanied by the emission of a positron - elementary particle, which is analogous to an electron, but has a positive charge:
19 K → 18 Ar ++1 e.
Gamma radiation is hard electromagnetic radiation with shorter wavelengths than X-ray. It does not deviate in electrical and magnetic fields and has a high penetrating power.
The emission of γ-rays accompanies α- and β-decay, as well as the process of electron capture by the nucleus. In the latter case, the nucleus captures an electron from a low energy level (K- or L-electron), and one of the protons turns into a neutron:
1 p + -1 e | → 0n. |
|
The mass number of the nuclide does not change, but the atomic number decreases by one, for example:
23 V + -1 e → 22 Ti.
Unstable, spontaneously decaying nuclides are called ra-
dionuclides or radioactive isotopes . Their decay continues until stable isotopes are formed. Stable isotopes are no longer subject to radioactive decay, so they persist in nature. Examples are 16O and 12C.
half-life An unstable isotope is called the time during which its radioactivity is halved compared to the original. Half-lives can range from millionths of a second to millions of years (Table 1.2).
Table 1.2
Half-lives of some isotopes
Half life |
||||
3 10-7 s |
||||
5.7 103 | ||||
4.5 109 | ||||
1.39 1010 years |
||||
Many radioactive decay reactions are constituent parts more complex sequential nuclear reactions - the so-called series of radioactive transformations or radioactive series.
Each transformation in such a series leads to the formation of an unstable isotope, which in turn undergoes radioactive decay. The parent nuclide is called parent isotope, and the resulting daughter isotope. At the next stage, the daughter isotope becomes the parent isotope, turning into the next child, and so on. This chain of successive transformations continues until a stable isotope is the result of a nuclear reaction.
Thus, the radioactive series of uranium begins with the isotope 238 U and, as a result of fourteen successive reactions of nuclear decay, ends with the stable isotope 206 Pb. In this case, the total mass loss is 32 units.
Both stable and unstable nuclides can be produced by nuclear reactions by bombarding nuclei with high-energy particles. Per-
voe artificial nuclear transformation carried out by E. Rutherford: in 1915
du, passing alpha rays through nitrogen, he received a stable isotope of oxygen 17 O. In 1935, Irene and Frederic Joliot-Curie proved that as a result of the bombardment of aluminum with alpha particles, a radioactive isotope of phosphorus is formed, emitting positrons. For the discovery artificial radioactivity scientists were awarded the Nobel Prize.
During nuclear reactions, a nuclear target is bombarded with protons, neutrons, electrons, which leads to a change in the nuclear composition and the formation of a new chemical element. The bombarding particles must have high kinetic energy to overcome the electrostatic repulsive forces from the target. Therefore, particles are accelerated to high speeds in special installations called accelerators (there are two main types: a linear accelerator and a cyclotron).
Table 1.3 |
||||||||
Nuclear reactions | ||||||||
Complete Equation | Short form |
|||||||
(α ,p) | ||||||||
7 N +2 He | → 8 O | 14 N (α ,p)17 O |
||||||
(α ,n) | ||||||||
13 Al +2 He → 15 P +0 n | 27 Al (α ,n)30 P |
|||||||
11 Na +1 H → 12 Mg +0 n | 23 Na(p,n)23 Mg |
|||||||
(p, a ) | ||||||||
4 Be +1 H → 3 Li +2 He | 9 Be (p,α )6 Li |
|||||||
7 N +1 H→ 8 O +γ | 14 N (p,γ )15 O |
|||||||
15P+1H → 15P+1H | 31P (d,p)32P |
|||||||
13 Al +1 H → 14 Si +0 n | 27Al(d,n)28Si |
|||||||
7 N +0 n → 6 C +1 H | 14N(n,p)14C |
|||||||
27 Co +0 n→ 27 Co +γ | 59 Co (n,γ )60 Co |
|||||||
(n, a) | ||||||||
13 Al +0 n → 11 Na +2 He | 27Al(n,α)24Na |
|||||||
Artificial nuclear transformations can be classified according to the type of particles bombarding and emitted as a result of the reaction (Table 1.3.).
With the help of nuclear reactions, new chemical elements with atomic numbers 99 and more were synthesized. For this purpose, a nuclear target is bombarded with heavy particles, for example, 7 N or 12 C. Thus, the element einsteinium was obtained by bombarding uranium-238 with nitrogen-14 nuclei:
REPEAT MATERIALS
Atom dimensions: ≈ 10 -8 cm Nucleus dimensions: ≈ 10 -12 – 10 -13 cm
Density of nuclear matter: ≈ 10 14 g / cm 3
subatomic particles
opening (date) |
|||||
ELECTRON | 9.110 10-28 | Thompson (1897) |
|||
1.673 10-24 | Rutherford (1914) |
||||
1.675 10-24 | Chadwick (1932) |
||||
quantum numbers
Name | Designation | Accepted | What characterizes |
||
values | |||||
energy |
|||||
Orbital | 0, 1, 2, ...n–1 | orbital shape, |
|||
energy |
|||||
sublevel |
|||||
Magnetic | –ℓ,..,–1,0,+1,..,+ ℓ | spatial |
|||
orientation |
|||||
orbitals |
|||||
Spin | +½ , -½ | own |
|||
electron |
|||||
Electronic formulas of atoms
To write the electronic formula of an atom, you need to know the following:
1. notation: nℓх (n is the number of the energy level: 1,2,3,..., ℓ is the letter designation of the sublevel: s, p, d, f; x is the number of electrons). Examples: 5s2 - two electrons per s - sublevel of the fifth energy level (n = 5, ℓ = 0), 4d8 - eight electrons on the d-sublevel of the fourth energy level (n = 4, ℓ = 2).
2. Sequence of filling energy sublevels : 1s< 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f...
(each sublevel is filled only after the previous one in this row is completely built up).
3. Maximum capacity of sublevels:
Example: the electronic formula of the chlorine atom is the distribution of seventeen electrons of a given atom over energy sublevels and has the form:
17Cl 1s2 2s2 2p6 3s2 3p5
Short form of writing an electronic formula : electrons finding-
in fully built-up energy levels are represented by the symbol of the corresponding noble gas, followed by the distribution of the remaining electrons.
Example: short electronic formula of the chlorine atom:
17Cl 3s2 3p5
Distribution of electrons in quantum cells
quantum cells
s-sublevel
p-sublevel
d-sublevel
f-sublevel
In accordance with Hund's rule: initially, each electron is given a separate quantum cell (unpaired electrons with parallel spins), the following electrons enter already occupied cells, for them the values of ms are opposite sign are paired electrons).
Notation: ms = +½ ,↓ ms = -½
Examples: 6 electrons occupy quantum cells of the f-sublevel:
f-sublevel
for nine electrons, the scheme takes the form:
f-sublevel
Electronic graphic formulas of atoms
17Cl | |||||||||||||
2p6 |
Valence electrons- electrons of the outer energy level, as well as the penultimate d-sublevel, if it is not completely built up.
Nuclide designations:
the superscript is the mass number of the nuclide, the subscript is the atomic number of the corresponding element.
Example: isotope of chlorine:
17Cl
Abbreviation: 36 Cl
Composition of the nucleus Number of protons - atomic number, serial number element in the period
D. I. Mendeleev’s dical system; The number of neutrons is the difference between the mass number and the number of
Example: number of protons and neutrons for the isotope of chlorine
17 Cl is: number of protons = 17, number of neutrons = 36-17= 19.
Isotopes - one atomic number, different atomic masses (the nucleus contains the same number protons, different number of neutrons)
Nuclear reactions
On the left and right sides of the nuclear reaction equation, a balance must be maintained between:
– sums of mass numbers (superscripts),
– sums of atomic numbers (subscripts).
Example:
Abbreviated form of the nuclear reaction equation:
on the left - the original nuclide,
on the right is the final nuclide,
in parentheses between them: the particle causing the given transformation, then the particle emitted as a result of it.
Letter designations:α (alpha particle), p (proton), n (neutron), d (deuterium nucleus - deuteron), etc.
Example: 23 Na (p,n) 23 Mg for the reaction
11 Na +1 H → 12 Mg +0 n