Why is the nucleus of an atom stable? What holds neutrons, which have no charge, and positively charged protons inside it?

This phenomenon cannot be explained in terms of electromagnetic influence between charged particles. Neutrons do not carry a charge, so electromagnetic forces do not act on them. Well, protons, positively charged particles, should repel each other. But this doesn't happen. Particles do not fly apart and the nucleus does not disintegrate. What forces force nucleons to stick together?

Nuclear forces

The forces that hold protons and neutrons inside the nucleus are called nuclear forces. Obviously, they must significantly exceed the electrostatic forces of repulsion and the forces of gravitational attraction of particles. Nuclear forces are the most powerful of all forces existing in nature. It has been experimentally established that their magnitude is 100 times greater than the forces of electrostatic repulsion. But they act only at a short distance, inside the nucleus. And if this distance is even by a very small amount greater than the diameter of the core, the action nuclear forces stops, and the atom begins to disintegrate under the influence of electrostatic repulsion forces. Therefore these forces short-acting.

Nuclear forces are forces of attraction. They do not depend on whether the particle has a charge or not, since inside the nucleus they hold both charged protons and non-charged neutrons. The magnitude of these forces is the same for a pair of protons, a pair of neutrons, or a neutron-proton pair. The interaction of nuclear forces is called strong interaction.

Nuclear binding energy. Mass defect

Thanks to nuclear forces, the nucleons in the nucleus are bound very tightly. In order to break this connection, you need to do work, that is, spend a certain amount of energy. The minimum energy required to separate a nucleus into individual particles is called nuclear binding energy atom. When individual nucleons combine into the nucleus of an atom, energy is released equal in magnitude to the binding energy. This energy is enormous. For example, if you burn 2 cars coal, then the energy will be released, which can be obtained by synthesizing only 4 g of the chemical element helium.

How to determine the binding energy?

It is obvious to us that the total mass of an orange is equal to the sum of the masses of all its slices. If each slice weighs 15 g, and there are 10 slices in an orange, then the weight of the orange is 150 g. By analogy, it would seem that the mass of the nucleus should be equal to the sum of the masses of the nucleons of which it consists. In reality, everything turns out to be wrong. Experiments show that the mass of the nucleus is less than the sum of the masses of the particles included in it. How is this possible? Where does some of the mass disappear to?

Let us recall the law of equivalence of mass and energy, which is also called the law of the relationship between mass and energy and is expressed by Einstein’s formula:

E= mc 2 ;

Where E – energy, m - weight, With – speed of light.

m = E/c 2 .

According to this law, mass does not disappear, but turns into energy released when nucleons combine to form a nucleus.

The difference between the masses of a nucleus and the total mass of the individual nucleons included in it is called mass defect and denote Δ m .

A mass at rest contains a huge store of energy. And when nucleons combine into a nucleus, energy is released ΔE = Δm c 2 , and the mass of the nucleus decreases by the amount Δ m. That is, the mass defect is a value equivalent to the energy that is released during the formation of a nucleus.

Δ m = ΔE/c 2 .

Mass defect can be defined in another way:

Δ m = Z m p + N m n - M i

Where Δ m – mass defect,

M i – core mass,

m p – proton mass,

m n – neutron mass,

Z – number of protons in the nucleus,

N – number of neutrons in the nucleus.

M i< Z m p + N m n .

It turns out that everyone has a mass defect chemical elements with the exception of protium, a hydrogen atom, which has only one proton and no neutrons in its nucleus. And the more nucleons in the nucleus of an element, the greater the mass defect for it.

Knowing the masses of particles that interact in nuclear reaction, as well as the particles that are formed as a result, it is possible to determine the amount of nuclear energy released and absorbed.

In order for atomic nuclei to be stable, protons and neutrons must be contained within the nuclei. huge forces, many times greater than the forces of Coulomb repulsion of protons. The forces that hold nucleons in the nucleus are called nuclear . They represent a manifestation of the most intense type of interaction known in physics - the so-called strong interaction. Nuclear forces are approximately 100 times greater than electrostatic forces and tens of orders of magnitude greater than the forces of gravitational interaction between nucleons. An important feature of nuclear forces is their short-range nature. Nuclear forces manifest themselves noticeably, as Rutherford's experiments on the scattering of α-particles showed, only at distances on the order of the size of the nucleus (10 -12 -10 -13 cm). At large distances, the action of relatively slowly decreasing Coulomb forces manifests itself.

Based on experimental data, we can conclude that protons and neutrons in the nucleus behave identically with respect to strong interaction, i.e., nuclear forces do not depend on the presence or absence of an electric charge on the particles.

The most important role in nuclear physics is played by the concept nuclear binding energy .

The binding energy of a nucleus is equal to the minimum energy that must be expended to completely split the nucleus into individual particles. From the law of conservation of energy it follows that the binding energy is equal to the energy that is released during the formation of a nucleus from individual particles.

The binding energy of any nucleus can be determined using precise measurement its mass. Currently, physicists have learned to measure the masses of particles - electrons, protons, neutrons, nuclei, etc. - with very high accuracy. These measurements show that mass of any nucleus M Ialways less than the sum of the masses of its constituent protons and neutrons:

This energy is released during the formation of a nucleus in the form of γ-quanta radiation.

As an example, let's calculate the binding energy of a helium nucleus; for example, the ionization energy is 13.6 eV.

In tables it is customary to indicate specific binding energy , i.e. binding energy per nucleon. For a helium nucleus, the specific binding energy is approximately 7.1 MeV/nucleon. In Fig. 6.6.1 shows a graph of the specific binding energy versus mass number A. As can be seen from the graph, the specific binding energy of nucleons is not the same for different atomic nuclei. For light nuclei, the specific binding energy first increases steeply from 1.1 MeV/nucleon for deuterium to 7.1 MeV/nucleon for helium. Then, having undergone a series of jumps, the specific energy slowly increases to a maximum value of 8.7 MeV/nucleon for elements with mass number A= 50-60, and then decreases relatively slowly for heavy elements. For example, for uranium it is 7.6 MeV/nucleon.

The decrease in specific binding energy upon transition to heavy elements is explained by an increase in the energy of Coulomb repulsion of protons. In heavy nuclei, the bond between nucleons weakens, and the nuclei themselves become less strong.

When stable lungs nuclei, where the role of the Coulomb interaction is small, the number of protons and neutrons Z And N turn out to be the same (, , ). Under the influence of nuclear forces, proton-neutron pairs are formed. But for heavy nuclei containing a large number of protons, due to the increase in the Coulomb repulsion energy, additional neutrons are required to ensure stability. In Fig. Figure 6.6.2 is a diagram showing the number of protons and neutrons in stable nuclei. For nuclei following bismuth ( Z> 83), due to the large number of protons, complete stability is generally impossible.

From Fig. 6.6.1 it is clear that the most stable from an energy point of view are the nuclei of the elements in the middle part of the periodic system. This means that there are two possibilities for obtaining a positive energy yield from nuclear transformations:

1. division of heavy nuclei into lighter ones;

2. fusion of light nuclei into heavier ones.

Both of these processes release enormous amounts of energy. Currently, both processes have been carried out practically: fission reactions and thermonuclear reactions.

Let's do some estimations. Let, for example, a uranium nucleus be divided into two identical nuclei with mass numbers of 119. For these nuclei, as can be seen from Fig. 6.6.1, specific binding energy is about 8.5 MeV/nucleon. The specific binding energy of the uranium nucleus is 7.6 MeV/nucleon. Consequently, the fission of a uranium nucleus releases energy equal to 0.9 MeV/nucleon or more than 200 MeV per uranium atom.

Let's now consider another process. Let, under certain conditions, two deuterium nuclei merge into one helium nucleus. The specific binding energy of deuterium nuclei is 1.1 MeV/nucleon, and the specific binding energy of helium nuclei is 7.1 MeV/nucleon. Consequently, the synthesis of one helium nucleus from two deuterium nuclei will release an energy equal to 6 MeV/nucleon or 24 MeV per helium atom.

It should be noted that the synthesis of light nuclei, compared to the fission of heavy nuclei, is accompanied by approximately 6 times greater energy release per nucleon.

Nuclear binding energy
Binding energy

Nuclear binding energy – the minimum energy required to split a nucleus into its constituent nucleons (protons and neutrons). The nucleus is a system of bound nucleons, consisting of Z protons (mass of a proton in a free state m p) and N neutrons (mass of a neutron in a free state m n). In order to split a nucleus into its component nucleons, a certain minimum energy W, called binding energy, must be expended. In this case, a nucleus at rest with mass M transforms into a set of free resting protons and neutrons with a total mass Zm p + Nm n. The energy of a nucleus at rest is Mc 2. Energy of released nucleons at rest (Zm p + Nm n)с 2. In accordance with the law of conservation of energy Mc 2 + W = (Zm p + Nm n)c 2. Or W = (Zm p + Nm n)c 2 - Ms 2. Since W > 0, then M< (Zm p + Nm n), т.е. масса, начального ядра, в котором нуклоны связаны, меньше суммы масс свободных нуклонов, входящих в его состав.
W increases with increasing number A of nucleons in the nucleus (A = Z + N). It is convenient to deal with the specific binding energy ε = W/A, i.e. average binding energy per nucleon. For most nuclei ε ≈ 8 MeV (1 MeV = 1.6·10 -13 J). To break a chemical bond, energy is needed 10 6 times less.

Communication energy

Bond energy serves as a measure of the strength of any chemical bond. To break a chemical bond, it is necessary to expend energy equal in magnitude to the energy that was released during the formation of the chemical bond.

The amount of energy released when a molecule is formed from atoms, called bond formation energy or just the energy of connection.

Bond energy is expressed in kJ/mol, for example:

H + H ® H 2 + 435 kJ.

Naturally, the same amount of energy must be spent to break chemical bonds in 1 mole of hydrogen. Therefore, the higher the binding energy, the stronger the bond. For example, E SV (H 2) = 435 kJ/mol, and E SV (N 2) = 942 kJ/mol. And, indeed, the bond in the nitrogen molecule (as shown earlier, triple) is much stronger than the bond in the hydrogen molecule.

Bond cleavage can be carried out homolytically (with the formation of neutral atoms) and heterolytically (with the formation of ions), and the energy of cleavage may vary.

NaCl (g) = Na (g) + Cl g – 414 kJ

For molecules of the same type, the length of a chemical bond can also serve as a characteristic of the bond strength: after all, what shorter length connections, topics more degree overlapping electron clouds.

Thus, the bond lengths ℓ (HF) = 0.092 nm and ℓ (HJ) = 0.162 nm indicate greater bond strength in the hydrogen fluoride molecule, which is confirmed in practice.

It should be noted that experimentally determined bond lengths characterize only the average distance between atoms, since atoms in molecules and crystals vibrate around the equilibrium position.

The overlap of electron clouds, leading to the formation of a chemical bond, is possible only if they have a certain mutual orientation. The overlap region is also located in a certain direction towards the interacting atoms. Therefore they say that A covalent chemical bond has directionality. In this case, three types of bonds can arise, which are called s- (sigma), p- (pi) and d- (delta) bonds.

In the cases of formation of H 2 and Cl 2 molecules discussed above, the overlap of electron clouds occurs along the straight line connecting the centers of the atoms. Covalent bond, formed as a result of overlapping electron clouds along a line connecting the centers of atoms, is called an s-bond. An s-bond is formed (Fig. 3) when s – s – clouds (for example, H2), рх – рх – clouds (Cl 2), s – px (HF) overlap.

Rice. 3. s-bonds in molecules H 2 (a), Cl 2 (b), HF (c)

When p-electron clouds interact, oriented perpendicular to the axis connecting the centers of atoms (p y - and p z - clouds), two overlapping regions are formed, located on both sides of the axis. This position corresponds to the formation of a p-bond.

p-bondis a bond for which the connecting electron cloud has a plane of symmetry passing through the atomic nuclei.

p-bonds do not exist by themselves: they are formed in molecules that already have s-bonds, and leads to the appearance of double and triple bonds.

Thus, in the N2 molecule, each nitrogen atom has three unpaired

2р – electrons. One cloud from each nitrogen atom participates in the formation of an s-bond (p x – p x - overlap).

Clouds p y - and p z - directed perpendicular to the s-connection line can overlap with each other only with the lateral sides of the “dumbbells”. This overlap leads to the formation of two p-bonds, i.e. the bond in the N2 molecule is triple. However, these connections are energetically unequal: the degree of overlap of p x – p x – clouds is much higher than p y – p y and p z – p z. And, indeed, the energy of a triple bond is lower than triple the energy of a single s-bond, and when chemical reactions First of all, p-bonds are broken.

Rice. 4. Various cases of p-bond formation

Absolutely anyone chemical substance consists of a certain set of protons and neutrons. They are held together due to the fact that the binding energy of the atomic nucleus is present inside the particle.

A characteristic feature of nuclear attractive forces is their very high power at relatively small distances (from about 10 -13 cm). As the distance between particles increases, the attractive forces inside the atom weaken.

Reasoning about binding energy inside the nucleus

If we imagine that there is a way to separate protons and neutrons from the nucleus of an atom in turn and place them at such a distance that the binding energy of the atomic nucleus ceases to act, then this must be very hard work. In order to extract its components from the nucleus of an atom, one must try to overcome intra-atomic forces. These efforts will go towards splitting the atom into the nucleons it contains. Therefore, we can judge that the energy of the atomic nucleus is less than the energy of the particles of which it consists.

Is the mass of intra-atomic particles equal to the mass of an atom?

Already in 1919, researchers learned to measure the mass of the atomic nucleus. Most often, it is “weighed” using special technical instruments called mass spectrometers. The operating principle of such devices is that the characteristics of particle motion are compared with different masses. Moreover, such particles have the same electrical charges. Calculations show that those particles that have different masses move along different trajectories.

Modern scientists have determined with great accuracy the masses of all nuclei, as well as their constituent protons and neutrons. If we compare the mass of a particular nucleus with the sum of the masses of the particles it contains, it turns out that in each case the mass of the nucleus will be greater than the mass of individual protons and neutrons. This difference will be approximately 1% for any given chemical. Therefore, we can conclude that the binding energy of an atomic nucleus is 1% of its rest energy.

Properties of intranuclear forces

Neutrons that are inside the nucleus are repelled from each other by Coulomb forces. But the atom does not fall apart. This is facilitated by the presence of an attractive force between particles in an atom. Such forces, which are of a nature other than electrical, are called nuclear. And the interaction of neutrons and protons is called strong interaction.

Briefly, the properties of nuclear forces are as follows:

• this is charge independence;
• action only over short distances;
• as well as saturation, which refers to the retention of only a certain number of nucleons near each other.

According to the law of conservation of energy, the moment nuclear particles combine, energy is released in the form of radiation.

Binding energy of atomic nuclei: formula

For the above calculations, the generally accepted formula is used:

E St=(Z·m p +(A-Z)·m n -MI)·c²

Here under E St refers to the binding energy of the nucleus; With- speed of light; Z-number of protons; (A-Z) - number of neutrons; m p denotes the mass of a proton; A m n- neutron mass. M i denotes the mass of the nucleus of an atom.

Internal energy of nuclei of various substances

To determine the binding energy of a nucleus, the same formula is used. The binding energy calculated by the formula, as previously stated, is no more than 1% of the total energy of the atom or rest energy. However, upon closer examination, it turns out that this number fluctuates quite strongly when moving from substance to substance. If you try to determine its exact values, they will differ especially for the so-called light nuclei.

For example, the binding energy inside a hydrogen atom is zero because it contains only one proton. The binding energy of a helium nucleus will be 0.74%. For nuclei of a substance called tritium, this number will be 0.27%. Oxygen has 0.85%. In nuclei with about sixty nucleons, the intraatomic bond energy will be about 0.92%. For atomic nuclei with greater mass, this number will gradually decrease to 0.78%.

To determine the binding energy of the nucleus of helium, tritium, oxygen, or any other substance, the same formula is used.

Types of Protons and Neutrons

The main reasons for such differences can be explained. Scientists have found that all nucleons contained inside the nucleus are divided into two categories: surface and internal. Inner nucleons are those that find themselves surrounded by other protons and neutrons on all sides. The superficial ones are surrounded by them only from the inside.

The binding energy of an atomic nucleus is a force that is more pronounced in the inner nucleons. Something similar, by the way, happens with the surface tension of various liquids.

How many nucleons fit in a nucleus

It was found that the number of internal nucleons is especially small in the so-called light nuclei. And for those that belong to the lightest category, almost all nucleons are regarded as surface ones. It is believed that the binding energy of an atomic nucleus is a quantity that should increase with the number of protons and neutrons. But even this growth cannot continue indefinitely. With a certain number of nucleons - and this is from 50 to 60 - another force comes into play - their electrical repulsion. It occurs even regardless of the presence of binding energy inside the nucleus.

The binding energy of an atomic nucleus in various substances used by scientists to release nuclear energy.

Many scientists have always been interested in the question: where does energy come from when lighter nuclei merge into heavier ones? In fact, this situation is similar to atomic fission. In the process of fusion of light nuclei, just as it happens during the fission of heavy ones, nuclei of a more durable type are always formed. To “get” all the nucleons contained in them from light nuclei, it is necessary to expend less energy than what is released when they combine. The converse is also true. In fact, the energy of fusion, which falls on a certain unit of mass, may be greater than the specific energy of fission.

Scientists who studied nuclear fission processes

The process was discovered by scientists Hahn and Strassman in 1938. At the Berlin University of Chemistry, researchers discovered that in the process of bombarding uranium with other neutrons, it turns into lighter elements that are in the middle of the periodic table.

Lise Meitner also made a significant contribution to the development of this field of knowledge, to whom Hahn at one time invited her to study radioactivity together. Hahn allowed Meitner to work only on the condition that she would conduct her research in the basement and never go to the upper floors, which was a fact of discrimination. However, this did not prevent her from achieving significant success in research of the atomic nucleus.