The nucleons in an atomic nucleus are bound together by nuclear forces; therefore, in order to divide the nucleus into its individual protons and neutrons, it is necessary to expend a lot of energy. This energy is called the binding energy of the nucleus.

The same amount of energy is released when free protons and neutrons combine to form a nucleus. Therefore, according to Einstein's special theory of relativity, the mass atomic nucleus must be less than the sum of the masses of free protons and neutrons from which it was formed. This mass difference Δm, corresponding to the energy core communicationsEsv, is determined by the Einstein relation:

Eb = с 2 Δm. (37.1)

The binding energy of atomic nuclei is so high that this mass difference is quite accessible to direct measurement. With the help of mass spectrographs, such a mass difference has indeed been found for all atomic nuclei.

The difference between the sum of the rest masses of free protons and neutrons, of which the nucleus is formed, and the mass of the nucleus is called the mass defect of the nucleus. The binding energy is usually expressed in megaelectronvolts (MeV) (1 MeV=10 6 eV). Since the atomic mass unit (a.m.u.) is 1.66 * 10 -27 kg, you can determine the energy corresponding to it:

E \u003d mc 2, E amu \u003d 1.66 * 10 -27 * 9 * 10 16 J,

E amu = (1.66 * 10 -27 * 9 * 10 16 J) / (1.6 * 10 -13 J / MeV) = 931.4 MeV.

The binding energy can be measured directly from the energy balance in the nuclear fission reaction. Thus, the binding energy of the deuteron was determined for the first time during its splitting by γ-quanta. However, from formula (37.1), the binding energy can be determine much more precisely, since with the help of a mass spectrograph masses of isotopes can be measured with an accuracy of 10 -4%.

Let us calculate, for example, the binding energy of the helium nucleus 4 2 He (α-particles). Its mass in atomic units is M (4 2 He) = 4.001523; proton mass mр=1.007276, neutron mass mn=1.008665. Hence the mass defect of the helium nucleus

Δm \u003d 2 / mp + 2mn - M (4 2 He),

Δm \u003d 2 * 1.007276 + 2 * 1.008665-4.001523 \u003d 0.030359.

Multiplying byE a.u.m = 931.4 MeV, we get

Eb = 0.030359 * 931.4 MeV ≈ 28.3 MeV.

Using a mass spectrograph, the masses of all isotopes were measured and the mass defect and binding energy of nuclei were determined. The binding energies of the nuclei of some isotopes are given in Table. 37.1. With the help of such tables, energy calculations of nuclear reactions are performed.

If the total mass of nuclei and particles formed in any nuclear reaction, less than the total mass of the initial nuclei and particles, then in such a reaction the energy corresponding to this decrease in mass is released. When the total number of protons and the total number of neutrons are conserved, the decrease in the total mass means that the total mass defect increases as a result of the reaction and the nucleons in the new nuclei are even more strongly bound to each other than in the original nuclei. The released energy is equal to the difference between the total binding energy of the formed nuclei and the total binding energy of the original nuclei, and it can be found using the table without calculating the change in the total mass. This energy can be released into environment in the form of kinetic energy of nuclei and particles or in the form of γ-quanta. An example of a reaction accompanied by the release of energy is any spontaneous reaction.

Let's carry out an energy calculation of the nuclear reaction of the transformation of radium into radon:

226 88 Ra → 222 86 Rn + 4 2 He.

The binding energy of the original nucleus is 1731.6 MeV (Table 37.1), and the total binding energy of the formed nuclei is 1708.2 + 28.3 = 1736.5 MeV and is 4.9 MeV more than the binding energy of the original nucleus.

Consequently, this reaction releases an energy of 4.9 MeV, which is mainly the kinetic energy of the α-particle.

If as a result of the reaction nuclei and particles are formed, the total mass of which is greater than that of the initial nuclei and particles, then such a reaction can proceed only with the absorption of energy corresponding to this increase in mass, and will never occur spontaneously. The amount of absorbed energy is equal to the difference between the total binding energy of the initial nuclei and the total binding energy of the nuclei formed in the reaction. In this way, it is possible to calculate what kinetic energy a particle or another nucleus must have in a collision with a target nucleus in order to carry out this kind of reaction, or to calculate the required value of a γ-quantum for the splitting of a nucleus.

Thus, the minimum value of the γ-quantum required for the splitting of the deuteron is equal to the binding energy of the deuteron 2.2 MeV, since in this reaction:

2 1 H + γ → 1 1 H + 0 n 1

a free proton and a neutron are formed (Eb = 0).

A good agreement of this kind of theoretical calculations with the results of experiments shows the correctness of the above explanation of the defect in the mass of atomic nuclei and confirms the principle established by the theory of relativity, the proportionality of mass and energy.

It should be noted that the reactions the transformation of elementary particles occurs (for example, β-decay), are also accompanied by the release or absorption of energy corresponding to a change in the total mass of particles.

An important characteristic of the nucleus is the average binding energy of the nucleus per nucleon, Eb/A (Table 37.1). The larger it is, the stronger the nucleons are interconnected, the stronger the nucleus. From Table. 37.1 shows that for most nuclei the value of Eb/A is approximately 8 MeV per nucleon and decreases for very light and heavy nuclei. Among the light nuclei, the helium nucleus stands out.

The dependence of the value of Eb/A on the mass number of the nucleus A is shown in fig. 37.12. In light nuclei, a large fraction of nucleons is located on the surface of the nucleus, where they do not fully use their bonds, and the value of Eb/A is small. As the mass of the nucleus increases, the ratio of surface to volume decreases and the fraction of nucleons located on the surface decreases.. Therefore, Eb/A grows. However, as the number of nucleons in the nucleus increases, the Coulomb repulsive forces between protons increase, weakening the bonds in the nucleus, and the value of Eb/A for heavy nuclei decreases. Thus, the value of Eb/A is maximum for cores of medium mass (at A = 50-60), therefore, they are distinguished by the greatest strength.

this implies important conclusion. In the reactions of fission of heavy nuclei into two medium nuclei, as well as in the synthesis of a medium or light nucleus from two lighter nuclei, nuclei are obtained that are stronger than the initial ones (with a larger value of Eb/A). This means that energy is released during such reactions. This is the basis for obtaining atomic energy in the fission of heavy nuclei and thermonuclear energy - in the fusion of nuclei.

The nucleons in an atomic nucleus are bound together by nuclear forces; therefore, in order to divide the nucleus into its individual protons and neutrons, it is necessary to expend a lot of energy. This energy is called the binding energy of the nucleus.

The same amount of energy is released when free protons and neutrons combine to form a nucleus. Therefore, according to Einstein's special theory of relativity, the mass of an atomic nucleus must be less than the sum of the masses of free protons and neutrons from which it was formed. This mass difference corresponding to the binding energy of the nucleus is determined by the Einstein relation (§ 36.7):

The binding energy of atomic nuclei is so high that this mass difference is quite accessible to direct measurement. With the help of mass spectrographs, such a mass difference has indeed been found for all atomic nuclei.

The difference between the sum of the rest masses of free protons and neutrons, of which the nucleus is formed, and the mass of the nucleus is called the mass defect of the nucleus.

The binding energy is usually expressed in mega-electronvolts (MeV). Since the atomic mass unit (a.m.u.) is equal to kg, we can determine the energy corresponding to it:

The binding energy can be measured directly from the energy balance in the nuclear fission reaction. Thus, the binding energy of the deuteron was determined for the first time during its splitting by y-quanta. However, from formula (37.1), the binding energy can be determined much more accurately, since with the help of a mass spectrograph it is possible to measure the masses of isotopes with an accuracy of .

Let us calculate, for example, the binding energy of the helium nucleus. Its mass in atomic units is equal to the mass of the proton and the mass of the neutron. Hence the mass defect of the helium nucleus

Multiplying by MeV, we get

Using a mass spectrograph, the masses of all isotopes were measured and the mass defect and binding energy of nuclei were determined. The binding energies of the nuclei of some isotopes are given in Table. 37.1. With the help of such tables, energy calculations of nuclear reactions are performed.

Table 37.1. (see scan) Binding energy of atomic nuclei

If the total mass of the nuclei and particles formed in any nuclear reaction is less than the total mass of the initial nuclei and particles, then the energy corresponding to this decrease in mass is released in such a reaction. When the total number of protons and the total number of neutrons are conserved, the decrease in the total mass means that the total mass defect increases as a result of the reaction and the nucleons in the new nuclei are even more strongly bound to each other than in the original nuclei. The released energy is equal to the difference between the total binding energy of the formed nuclei and the total binding energy of the original nuclei, and it can be found using the table without calculating the change in the total mass. This energy can be released into the environment in the form of the kinetic energy of nuclei and particles or in the form of y-quanta. An example of a reaction accompanied by the release of energy is any spontaneous reaction.

Let's carry out an energy calculation of the nuclear reaction of the transformation of radium into radon:

The binding energy of the original nucleus is 1731.6 MeV (Table 37.1), and the total binding energy of the formed nuclei is equal to MeV and is 4.9 MeV greater than the binding energy of the original nucleus.

Consequently, in this reaction, an energy of 4.9 MeV is released, which mainly constitutes the kinetic energy of the a-particle.

If as a result of the reaction nuclei and particles are formed, the total mass of which is greater than that of the initial nuclei and particles, then such a reaction can proceed only with the absorption of energy corresponding to this increase in mass, and will never occur spontaneously. The amount of absorbed energy is equal to the difference between the total binding energy of the initial nuclei and the total binding energy of the nuclei formed in the reaction. In this way, one can calculate what kinetic energy a particle or other nucleus must have in a collision with a target nucleus in order to carry out this kind of reaction, or calculate the required value of the -quantum for the splitting of any nucleus.

So, the minimum value of the -quantum necessary for the splitting of the deuteron is equal to the binding energy of the deuteron 2.2 MeV, since

in this reaction:

free proton and neutron are formed

A good agreement of this kind of theoretical calculations with the results of experiments shows the correctness of the above explanation of the defect in the mass of atomic nuclei and confirms the principle of proportionality of mass and energy established by the theory of relativity.

It should be noted that reactions in which the transformation of elementary particles occurs (for example, -decay) are also accompanied by the release or absorption of energy corresponding to a change in the total mass of the particles.

An important characteristic of the nucleus is the average binding energy of the nucleus per nucleon (Table 37.1). The larger it is, the stronger the nucleons are interconnected, the stronger the nucleus. From Table. 37.1 shows that for most nuclei the value is about 8 MeV per. nucleon and decreases for very light and heavy nuclei. Among the light nuclei, the helium nucleus stands out.

The dependence of the value on the mass number of the nucleus A is shown in fig. 37.12. In light nuclei, a large fraction of nucleons is located on the surface of the nucleus, where they do not fully use their bonds, and the value is small. As the mass of the nucleus increases, the ratio of surface to volume decreases and the fraction of nucleons located on the surface decreases. Therefore, it is growing. However, as the number of nucleons in the nucleus increases, the Coulomb repulsive forces between protons increase, weakening the bonds in the nucleus, and the size of heavy nuclei decreases. Thus, the value is maximum for nuclei of medium mass (hence, they are distinguished by the greatest strength.

An important conclusion follows from this. In the reactions of fission of heavy nuclei into two medium nuclei, as well as in the synthesis of a medium or light nucleus from two lighter nuclei, nuclei are obtained that are stronger than the original ones (with a larger value. Hence, energy is released during such reactions. This is based on the production of atomic energy during the fission of heavy nuclei ( § 39.2) and thermonuclear energy - in the fusion of nuclei (§ 39.6).

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 or greater than 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, consequently, 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. As m N more m p , by the amount me, then the first term in square brackets includes the mass 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 BUT element (Fig. 45). For light nuclei ( BUT≥ 12) the specific binding energy increases steeply up to 6 ÷ 7 MeV, undergoing a number of 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 BUT= 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).

Decrease specific energy connection 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 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.

Atomic nucleus. mass defect. The binding energy of the atomic nucleus

The atomic nucleus is the central part of the atom, in which all the positive charge and almost all the mass is concentrated.

The nuclei of all atoms are made up of particles called nucleons. Nucleons can be in two states - in an electrically charged state and in a neutral state. A nucleon in a charged state is called a proton. The proton (p) is the nucleus of the lightest chemical element- hydrogen. The proton charge is equal to the elementary positive charge, which is equal in magnitude to the elementary negative charge q e = 1.6 ∙ 10 -19 C., i.e. charge of an electron. A nucleon in a neutral (uncharged) state is called a neutron (n). The masses of nucleons in both states differ little from each other, i.e. m n ≈ m p .

Nucleons are not elementary particles. They have a complex internal structure and consist of even smaller particles of matter - quarks.

The main characteristics of an atomic nucleus are charge, mass, spin and magnetic moment.

Core charge is determined by the number of protons (z) that make up the nucleus. The nuclear charge (zq) is different for different chemical elements. The number z is called the atomic number or charge number. The atomic number is the atomic number of a chemical element in periodic system elements of D. Mendeleev. The charge of the nucleus also determines the number of electrons in the atom. The number of electrons in an atom determines their distribution over energy shells and subshells and, consequently, all physicochemical characteristics atom. The nuclear charge determines the specifics of a given chemical element.

Core mass The mass of a nucleus is determined by the number (A) of nucleons that make up the nucleus. The number of nucleons in the nucleus (A) is called the mass number. The number of neutrons (N) in the nucleus can be found if from total number nucleons (A) subtract the number of protons (z), i.e. N=F-z. In the periodic table, up to its middle, the number of protons and neutrons in the nuclei of atoms is approximately the same, i.e. (А-z)/z= 1, by the end of the table (А-z)/z= 1.6.

The nuclei of atoms are usually denoted as follows:

X - symbol of a chemical element;

Z is the atomic number;

A is the mass number.

When measuring the masses of nuclei simple substances it was found that most of the chemical elements are composed of groups of atoms. Having the same charge, the nuclei of different groups differ in masses. The varieties of atoms of a given chemical element, which differ in the masses of nuclei, are called isotopes. Isotope nuclei have the same number protons, but different number neutrons ( and ; , , , ; , , ).

In addition to the nuclei of isotopes (z - the same, A - different), there are nuclei isobars(z - different, A - the same). ( and ).

Masses of nucleons, nuclei of atoms, atoms, electrons and other particles in nuclear physics it is customary to measure not in "KG", in atomic mass units (amu - otherwise called the carbon mass unit and denoted by "e"). For the atomic mass unit (1e), 1/12 of the mass of the carbon atom is taken 1e = 1.6603 ∙ 10 -27 kg.

Nucleon masses: m p -1.00728 e, m n =1.00867 e.

We see that the mass of the nucleus expressed in "e" will be written as a number close to A.

Spin of the nucleus. The mechanical angular momentum (spin) of the nucleus is equal to the vector sum of the spins of the nucleons that make up the nucleus. The proton and neutron have a spin equal to L = ± 1/2ћ. Accordingly, the spin of nuclei with an even number of nucleons (A even) is an integer or zero. The spin of a nucleus with an odd number of nucleons (A odd) is half-integer.

The magnetic moment of the nucleus. The magnetic moment of the nucleus (P m i) of the nucleus compared with the magnetic moment of the electrons filling electron shells atom is very small. On the magnetic properties atom, the magnetic moment of the nucleus does not affect. The unit of measurement of the magnetic moment of the nuclei is the nuclear magneton μ i = 5.05.38 ∙ 10 -27 J/T. It is 1836 times less than the magnetic moment of the electron - the Bohr magneton μ B = 0.927 ∙ 10 -23 J / T.

The magnetic moment of the proton is equal to 2.793 μ i and is parallel to the spin of the proton. The magnetic moment of the neutron is equal to 1.914 μ i and is antiparallel to the spin of the neutron. The magnetic moments of the nuclei are of the order of the nuclear magneton.

To split a nucleus into its constituent nucleons, a certain amount of work must be done. The value of this work is a measure of the binding energy of the nucleus.

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

In the reverse process of the formation of a nucleus, the same energy should be released from the constituent nucleons. This follows from the law of conservation of energy. Therefore, the binding energy of the nucleus is equal to the difference between the energy of the nucleons that make up the nucleus and the energy of the nucleus:

ΔE \u003d E nuk - E i. (one)

Taking into account the relationship between mass and energy (E = m ∙ c 2) and the composition of the nucleus, we rewrite equation (1) as follows:

ΔЕ = ∙ s 2 (2)

Value

Δm \u003d zm p + (A-z) m n - M i, (3)

Equal to the difference between the masses of the nucleons that make up the nucleus and the mass of the nucleus itself, is called the mass defect.

Expression (2) can be rewritten as:

ΔЕ = Δm ∙ s 2 (4)

Those. mass defect is a measure of the binding energy of a nucleus.

In nuclear physics, the mass of nucleons and nuclei is measured in amu. (1 amu = 1.6603 ∙ 10 27 kg), and energy is usually measured in MeV.

Considering that 1 MeV = 10 6 eV = 1.6021 ∙ 10 -13 J, we find the energy value corresponding to the atomic mass unit

1.a.u.m. ∙ s 2 = 1.6603 ∙10 -27 ∙9 ∙10 16 = 14.9427 ∙ 10 -11 J = 931.48 MeV

Thus, the binding energy of the nucleus in MeV is

ΔE sv = Δm ∙931.48 MeV (5)

Taking into account that tables usually give not the mass of nuclei, but the mass of atoms, for the practical calculation of the mass defect, instead of formula (3)

enjoy another

Δm \u003d zm H + (A-z)m n - M a, (6)

That is, the mass of the proton was replaced by the mass of the light hydrogen atom, thereby adding z electron masses, and the mass of the nucleus was replaced by the mass of the atom M a, thereby subtracting these z electron masses.

The binding energy per nucleon in a nucleus is called the specific binding energy.

(7)

The dependence of the specific binding energy on the number of nucleons in the nucleus (on the mass number A) is given in Fig.1.