The nuclei of atoms are strongly bound systems of a large number of nucleons.
For the complete splitting of the nucleus into its constituent parts and their removal at large distances from each other, it is necessary to expend a certain amount of work A.

The binding energy is the energy equal to the work that must be done to split the nucleus into free nucleons.

E bonds = - A

According to the law of conservation, the binding energy is simultaneously equal to the energy that is released during the formation of a nucleus from individual free nucleons.

Specific binding energy

This is the binding energy per nucleon.

Except for the lightest nuclei, the specific binding energy is approximately constant and equal to 8 MeV/nucleon. Elements with mass numbers from 50 to 60 have the maximum specific binding energy (8.6 MeV/nucleon). The nuclei of these elements are the most stable.

As nuclei are overloaded with neutrons, the specific binding energy decreases.
For the elements at the end of the periodic table, it is equal to 7.6 MeV/nucleon (for example, for uranium).

Release of energy as a result of nuclear fission or fusion

In order to split the nucleus, it is necessary to expend a certain amount of energy to overcome the nuclear forces.
In order to synthesize a nucleus from individual particles, it is necessary to overcome the Coulomb repulsive forces (for this, energy must be expended to accelerate these particles to high speeds).
That is, in order to carry out the splitting of the nucleus or the fusion of the nucleus, some energy must be expended.

During nuclear fusion at short distances, nuclear forces begin to act on nucleons, which induce them to move with acceleration.
Accelerated nucleons emit gamma quanta, which have an energy equal to the binding energy.

At the output of the nuclear fission reaction or fusion, energy is released.

It makes sense to carry out nuclear fission or nuclear synthesis, if the resulting, i.e. the energy released as a result of splitting or fusion will be greater than the energy expended
According to the graph, the gain in energy can be obtained either by fission (splitting) of heavy nuclei, or by fusion of light nuclei, which is done in practice.

mass defect

Measurements of the masses of nuclei show that the mass of the nucleus (Mn) is always less than the sum of the rest masses of the free neutrons and protons composing it.

During nuclear fission: the mass of the nucleus is always less than the sum of the rest masses of the formed free particles.

In the synthesis of the nucleus: the mass of the formed nucleus is always less than the sum of the rest masses of the free particles that formed it.

The mass defect is a measure of the binding energy of an atomic nucleus.

The mass defect is equal to the difference between the total mass of all nucleons of the nucleus in the free state and the mass of the nucleus:

where Mm is the mass of the nucleus (from the reference book)
Z is the number of protons in the nucleus
mp is the rest mass of a free proton (from the handbook)
N is the number of neutrons in the nucleus
mn is the rest mass of a free neutron (from the handbook)

The decrease in mass during the formation of a nucleus means that the energy of the system of nucleons decreases.

Nucleus Binding Energy Calculation

The nuclear binding energy is numerically equal to the work that must be expended to split the nucleus into individual nucleons, or the energy released during the synthesis of nuclei from nucleons.
The measure of the nuclear binding energy is the mass defect.

The formula for calculating the binding energy of a nucleus is Einstein's formula:
if there is some system of particles that has mass, then a change in the energy of this system leads to a change in its mass.

Here, the binding energy of the nucleus is expressed as the product of the mass defect and the square of the speed of light.

In nuclear physics, the mass of particles is expressed in atomic mass units (a.m.u.)

in nuclear physics, it is customary to express energy in electronvolts (eV):

Let's calculate the correspondence of 1 a.m.u. electronvolts:

Now the calculation formula for the binding energy (in electronvolts) will look like this:

EXAMPLE OF CALCULATION OF THE BINDING ENERGY OF THE NUCLEI OF A HELIUM ATOM (He)

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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:

where m n is the mass of a hydrogen atom.

Such a replacement is convenient for calculations, and the calculation error arising in this case is insignificant. If we substitute Dt in a.m.u. into the formula for the binding energy then for E St can be written:

Important information about the properties of nuclei is contained in the dependence of the specific binding energy on the mass number A.

Specific binding energy E beats - the binding energy of the nucleus per 1 nucleon:

On fig. 116 shows a smoothed graph of the experimentally established dependence of E beats on A.

The curve in the figure has a weakly expressed maximum. Elements with mass numbers from 50 to 60 (iron and elements close to it) have the highest specific binding energy. The nuclei of these elements are the most stable.

It can be seen from the graph that the reaction of fission of heavy nuclei into the nuclei of elements in the middle part of the D. Mendeleev table, as well as the reactions of fusion of light nuclei (hydrogen, helium) into heavier ones are energetically favorable reactions, since they are accompanied by the formation of more stable nuclei (with large E sp) and, therefore, proceed with the release of energy (E > 0).

Since most nuclei are stable, there is a special nuclear (strong) interaction between nucleons - attraction, which ensures the stability of nuclei, despite the repulsion of like-charged protons.

The binding energy of the nucleus is a physical quantity equal to the work that must be done to split the nucleus into its constituent nucleons without imparting kinetic energy to them.

It follows from the law of conservation of energy that the same energy should be released during the formation of a nucleus, which must be expended in the splitting of the nucleus into its constituent nucleons. The binding energy of the nucleus is the difference between the energy of all nucleons in the nucleus and their energy in the free state.

Binding energy of nucleons in an atomic nucleus:

where, are the masses of the proton, neutron and nucleus, respectively; is the mass of a hydrogen atom; is the atomic mass of the substance.

Mass corresponding to binding energy:

is called the nuclear mass defect. The mass of all nucleons decreases by this amount when a nucleus is formed from them.

The specific binding energy is the binding energy per nucleon: . It characterizes the stability (strength) of atomic nuclei, i.e. the more, the stronger the core.

The dependence of the specific binding energy on the mass number is shown in the figure. The most stable nuclei of the middle part of the periodic table (28<A<138). В этих ядрах составляет приблизительно 8,7 МэВ/нуклон (для сравнения, энергия связи валентных электронов в атоме порядка 10эВ, что в миллион раз меньше).

With the transition to heavier nuclei, the specific binding energy decreases, since with an increase in the number of protons in the nucleus, the energy of their Coulomb repulsion increases (for example, for uranium it is 7.6 MeV). Therefore, the bond between nucleons becomes less strong, the nuclei themselves become less strong.

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 implemented practically; nuclear fission reactions and nuclear fusion reactions.

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 there 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 an amount 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

At present, the masses of atomic nuclei have been determined with a high degree of 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), an energy equal to the binding energy of the new nucleus will be released:

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, since industrial methods for obtaining nuclear energy are based on it.

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.

Studies show that atomic nuclei are stable formations. This means that there is a certain connection between nucleons in the nucleus. The study of this connection can be carried out without drawing on information about the nature and properties of nuclear forces, but based on the law of conservation of energy. Let's introduce some definitions.

The binding energy of the nucleon in the nucleus called a physical quantity equal to the work that must be done to remove a given nucleon from the nucleus without imparting kinetic energy to it.

Complete core binding energy is determined by the work that must be done to split the nucleus into its constituent nucleons without imparting kinetic energy to them.

It follows from the law of conservation of energy that during the formation of a nucleus, an energy equal to the binding energy of the nucleus must be released from its constituent nucleons. Obviously, the binding energy of the nucleus is equal to the difference between the total energy of the free nucleons that make up the given nucleus and their energy in the nucleus. From the theory of relativity it is known that there is a relationship between energy and mass:

E \u003d mc 2. (250)

If through ΔE sv denote the energy released during the formation of the nucleus, then this release of energy, according to formula (250), should be associated with a decrease in the total mass of the nucleus during its formation from composite particles:

Δm = ΔE sv / since 2 (251)

If denoted by m p , m n , m I the masses of the proton, neutron and nucleus, respectively, then ∆m can be determined by the formula:

Dm = [Zm p + (A-Z)m n]- m I . (252)

The mass of nuclei can be very accurately determined using mass spectrometers - measuring instruments that separate beams of charged particles (usually ions) with different specific charges using electric and magnetic fields q/m. Mass spectrometric measurements showed that, indeed, the mass of the nucleus is less than the sum of the masses of its constituent nucleons.

The difference between the sum of the masses of the nucleons that make up the nucleus and the mass of the nucleus is called nuclear mass defect(formula (252)).

According to formula (251), the binding energy of nucleons in a nucleus is determined by the expression:

ΔЕ CB = [Zm p+ (A-Z)m n - m I ]with 2 . (253)

The tables usually do not give the masses of the nuclei m I, and the masses of atoms m a. Therefore, for the binding energy, the formula is used

ΔE SW =[Zm H+ (A-Z)m n - m a ]with 2 (254)

where m H- mass of a hydrogen atom 1 H 1 . As m H more m p, by the value of the electron mass me, then the first term in square brackets includes the mass Z of electrons. But since the mass of an atom m a different from the mass of the nucleus m I just on the mass Z of electrons, then calculations using formulas (253) and (254) lead to the same results.

Often, instead of the binding energy of the nucleus, one considers specific bond energydЕ CB is the binding energy per nucleon of the nucleus. It characterizes the stability (strength) of atomic nuclei, i.e., the more dЕ CB, the more stable the core . The specific binding energy depends on the mass number BUT element. For light nuclei (A £ 12), the specific binding energy rises steeply to 6 ¸ 7 MeV, undergoing a series of jumps (see Figure 93). For example, for dЕ CB=1.1 MeV, for -7.1 MeV, for -5.3 MeV. With a further increase in the mass number dE, SW increases more slowly to a maximum value of 8.7 MeV for elements with BUT=50¸60, and then gradually decreases for heavy elements. For example, for 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). On the curve of dependence of the specific binding energy on the mass number for stable nuclei (Figure 93), the following patterns can be noted:

A) If we discard the lightest nuclei, then in a rough, so to speak zero approximation, the specific binding energy is constant and equal to approximately 8 MeV per

nucleon. The approximate independence of the specific binding energy from the number of nucleons indicates the saturation property of nuclear forces. This property is that each nucleon can only interact with a few neighboring nucleons.

b) The specific binding energy is not strictly constant, but has a maximum (~8.7 MeV/nucleon) at BUT= 56, i.e. in the area of ​​iron nuclei, and falls to both edges. The maximum of the curve corresponds to the most stable nuclei. It is energetically advantageous for the lightest nuclei to merge with each other with the release of thermonuclear energy. For the heaviest nuclei, on the contrary, the process of fission into fragments is beneficial, which proceeds with the release of energy, called atomic energy.