Negative and positive energy of a particle. Abstract: Can energy be negative? Potential energy of a body in a gravitational field

5. States with negative energy. Positive electron

The equations of Dirac theory exhibit special properties, allowing solutions corresponding to the states of a particle whose energy can be negative. An electron in one of these states must have some rather strange properties. To increase his speed, energy must be taken away from him. And, on the contrary, to stop him, you need to give him some energy. In an experiment, an electron has never behaved so strangely. Therefore, it was quite legitimate to believe that states with negative energy, the existence of which is allowed by Dirac’s theory, are not actually realized in nature. One might say that in this sense the theory gives too much, at least at first glance.

The fact that Dirac's equations allow for the possibility of the existence of states with negative energy is undoubtedly a result of their relativistic nature. Indeed, even in the relativistic dynamics of the electron, developed by Einstein within the framework of the special theory of relativity, the possibility of motion with negative energy is revealed. However, at that time the difficulty in Einstein's dynamics was not very serious, because it, like all previous theories, assumed that all physical processes were continuous. And since the electron’s own mass is finite, it always has finite internal energy in accordance with the relativistic principle of equivalence of mass and energy. Since this internal energy cannot disappear, we cannot continuously move from a state with positive to a state with negative energy. Thus, the assumption of continuity of physical processes completely excludes this kind of transition.

Therefore, it is enough to assume that at the initial moment of time all electrons are in states with positive energy to see that the state always remains the same. The difficulty becomes much more serious in Dirac mechanics, for it is quantum mechanics, allowing for the existence of discrete transitions in physical phenomena. It can be easily seen that transitions between states with positive and negative energy are not only possible, but should also occur quite often. Klein gave an interesting example of how an electron with positive energy, entering a region where a rapidly changing field operates, can leave this region in a state with negative energy. Consequently, the fact that an electron with negative energy had never been discovered experimentally turned out to be very dangerous for Dirac’s theory.

To get around this difficulty, Dirac came up with a very ingenious idea. Noting that, according to the Pauli principle, which we will talk about in the next chapter, there cannot be more than one electron in one state, he assumed that in the normal state of the surrounding world, all states with negative energy are occupied by electrons. It follows that the density of electrons with negative energy is the same everywhere. Dirac hypothesized that this uniform density could not be observed. At the same time, there are more electrons than are needed to fill all states with negative energy.

This excess is represented by electrons with positive energy, which is what we can observe in our experiments. In exceptional cases, an electron with negative energy can, under the influence of an external force, transform into a state with positive energy. In this case, an observed electron instantly appears and at the same time a hole, an empty space, is formed in the distribution of electrons with negative energy. Dirac showed that such a hole can be observed experimentally and should behave like a particle with a mass equal to the mass of the electron and a charge equal to it, but of opposite sign. We will think of it as an anti-electron, a positive electron. This unexpectedly formed hole cannot exist for long. It will be filled with an electron with positive energy, which will undergo a spontaneous transition to an empty state with negative energy, accompanied by radiation. So, Dirac explained the unobservability of states with negative energy and at the same time predicted the possibility, albeit rare and ephemeral existence, of positive electrons.

Undoubtedly, Dirac's hypothesis was very simple, but at first glance it seemed somewhat artificial. It is possible that a large number of physicists would remain somewhat skeptical in this regard if experiment did not immediately prove the existence of positive electrons, the characteristic properties of which Dirac had just predicted.

Indeed, in 1932, first the subtle experiments of Anderson, and then Blackett and Occhialini, discovered that the decay of atoms under the influence of cosmic rays produces particles that behave exactly like positive electrons. Although it was still impossible to state absolutely strictly that the mass of the new particles was equal to the mass of the electron, and their electric charge was equal and opposite in sign to the charge of the electron, subsequent experiments made this coincidence more and more likely. Further, it turned out that positive electrons tend to quickly disappear (annihilate) when they come into contact with matter, and annihilation is accompanied by radiation. The experiments of Thibault and Joliot-Curie seemed to leave no doubt on this issue.

The exceptional circumstances under which positive electrons appear and their ability to annihilate, shortening their lifetime, are precisely the properties that Dirac foresaw. Thus, the situation turned out to be the opposite: the existence of solutions to the Dirac equations with negative energy not only does not call them into question, but, on the contrary, shows that these equations predicted the existence and described the properties of positive electrons.

Nevertheless, we must admit that Dirac's ideas about holes lead to serious difficulties regarding the electromagnetic properties of the vacuum. It is likely that Dirac's theory will be reformed and establish greater symmetry between both types of electrons, with the result that the idea of holes, together with the difficulties associated with it, will be abandoned. At the same time, there is no doubt that the experimental discovery of positive electrons (now called positrons) represents a new and remarkable confirmation of the ideas underlying Dirac mechanics. The symmetry between both types of electrons, which is established as a result of a more thorough study of some analytical features of the Dirac equations, is of great interest and undoubtedly will play an important role in the further development of physical theories.

From the book Physical Chemistry: Lecture Notes author Berezovchuk A V

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This issue has never been specifically discussed in detail in the so-called stable textbooks. It was considered too difficult for high school students. At the same time, “by default” students (and often teachers) believe that energy can only be a positive quantity. This leads to misunderstandings when analyzing energy conversion in various processes. For example, how can we explain that when water is boiled, all the energy imparted to the substance goes to evaporation, while the average kinetic energy of particle motion does not change, and the interaction energy of particles becomes equal to zero? Where does the energy coming from the heater disappear? Many such examples can be given. But it is more expedient not to remain silent that the energy of interaction between bodies can be both positive and negative. The difficulties in understanding this provision are far-fetched. After all, even primary school students understand that the ambient temperature can be both positive and negative! Moreover, schoolchildren quite easily perceive the existence, along with the Kelvin scale, of other temperature scales (Celsius, Fahrenheit, Reaumur). Thus, the idea that the numerical value of some physical quantity depends on a conventionally chosen origin of its reference is not incomprehensible to a high school student.

Selecting the potential energy reference point

We will show how to explain to students that when studying mechanical phenomena, in many cases it is convenient to choose the level of reference for potential energy so that it will have a negative value.

Analysis of energy transformation implies a more detailed familiarization of students with its forms. Any textbook reports that a body of mass m, moving relative to a chosen reference frame with some speed v, has kinetic energy Ekin = mv2/2 in this frame. If in some frame of reference the body is motionless, then its kinetic energy is equal to zero. Therefore, the kinetic energy of a body is called the energy of motion. Unlike other characteristics of motion, such as speed v or momentum p = mv, kinetic energy is not related to the direction of motion. It is a scalar quantity. It is advisable to invite students to independently show that the kinetic energy of a body and a system of bodies cannot be a negative quantity.

The nature of potential energy can be completely different. In the case of a mathematical pendulum (a material point of mass m suspended on a weightless inextensible thread of length l), it is associated with the attraction of the pendulum’s load by the Earth. It is this gravitational interaction that reduces the speed of the load as it moves upward. In the case of a tennis ball hitting a wall, potential energy is associated with the deformation of the ball. What the energy of interaction of the load with the Earth and the energy of deformation have in common is that such energy can be converted into kinetic energy and vice versa.

However, not all processes are reversible. For example, when a hammer hits a piece of lead, the kinetic energy of the hammer seems to disappear without a trace - the hammer almost does not bounce back after the impact. In this case, the kinetic energy of the hammer is converted into heat and its subsequent irreversible dissipation.

Let's take a closer look at the concept of potential energy. The nature of potential energy is different, so there is no single formula for calculating it. Of all types of interaction, we most often encounter the gravitational interaction of the Earth and bodies located near its surface, so first of all we should dwell on the discussion of the features of gravitational interaction.

What is the formula for calculating the potential energy of interaction of the Earth with bodies located near its surface? The answer is suggested by the oscillations of the pendulum. Please note (Fig. 1): points B, at which kinetic energy is completely converted into latent (potential) form, and point A,

where the kinetic energy of the pendulum is completely restored, lie at different heights above the Earth's surface. Huygens also found out that the height h of the pendulum's rise to point B is proportional to the square of its speed v2max at the bottom point A. Leibniz estimated the amount of latent (potential) energy at points B by the mass m of the pendulum's load and the height h of its rise during oscillations. Accurate measurements of the maximum speed vmax and height h show that the equality is always satisfied:

where g  10 N/kg = 10 m/s2. If, in accordance with the law of conservation of energy, we assume that all the kinetic energy of the pendulum is converted at points B into the energy of gravitational interaction of its load with the Earth, then the energy of this interaction must be calculated using the formula:

This formula hides a conditional agreement: the position of the interacting bodies, at which the energy of their interaction En is conventionally considered equal to zero (zero level), is chosen so that in this position the height h = 0. But when choosing the zero level, physicists are guided only by the desire to simplify the solution to the limit tasks. If for some reason it is convenient to assume that the potential energy is equal to zero at a point at a height h0  0, then the formula for potential energy takes the form:

Ep = mg(h – h0).

Consider a stone falling from a cliff (Fig. 2). It is necessary to determine how the kinetic energy Ek of the stone and the potential energy En of its interaction with the Earth changes as it falls. Suppose that at the edge of the cliff (point A) the speed of the stone is zero.

When a stone falls, its friction with the air is small, so we can assume that there is no dissipation of energy and its transformation into heat. Consequently, according to the law of conservation of energy, when a stone falls, the sum of the kinetic and potential energy of the system of bodies Earth + stone does not change, i.e.

(Ek + Ep)|B = (Ek+E0)|A.

Let us note the following.

1. According to the conditions of the problem at point A, the speed of the stone is zero, therefore Ek| A = 0.

2. It is convenient to choose the zero level of potential energy of interaction between the stone and the Earth in such a way as to simplify the solution of the problem to the utmost. Since only one fixed point is indicated - the edge of rock A - it is reasonable to take it as the origin and put Ep| A = 0. Then the total energy (Ek + Ep)|A = 0. Consequently, by virtue of the law of conservation of energy, the sum of the kinetic and potential energies of the stone and the Earth remains equal to zero at all points of the trajectory:

(Ek + Ep)|B = 0.

The sum of two non-zero numbers is equal to zero only if one of them is negative and the other is positive. We have already noted that kinetic energy cannot be negative. Therefore, from the equality (Ek + Ep)|B = 0 it follows that the potential energy of interaction of a falling stone with the Earth is a negative quantity. This is due to the choice of the zero potential energy level. We took the edge of the rock as the zero reference point for the coordinate h of the stone. All points through which the stone flies lie below the edge of the cliff, and the values of the h coordinates of these points lie below zero, i.e. they are negative. Consequently, according to the formula En = mgh, the energy En of the interaction of a falling stone with the Earth must also be negative.

From the equation of the law of conservation of energy Ek + En = 0 it follows that at any height h down from the edge of the rock, the kinetic energy of the stone is equal to its potential energy taken with the opposite sign:

Ek = –En = –mgh

(It should be remembered that h is a negative value). Graphs of the dependence of potential energy Ep and kinetic energy Ek on coordinate h are shown in Fig. 3.

It is also useful to immediately examine the case when a stone is thrown upward at point A with a certain vertical speed v0. At the initial moment, the kinetic energy of the stone is Ek = mv02/2, and the potential energy, by convention, is zero. At an arbitrary point in the trajectory, the total energy is equal to the sum of the kinetic and potential energies mv2/2 + mgh. The law of conservation of energy is written as:

mv02/2 = mv2/2 + mgh.

Here h can have both positive and negative values, which corresponds to the stone moving upward from the throwing point or falling below point A. Thus, for certain values of h the potential energy is positive, and for others it is negative. This example should show the student the convention of assigning potential energy a certain sign.

After familiarizing students with the above material, it is advisable to discuss the following questions with them:

1. Under what condition is the kinetic energy of a body equal to zero? potential energy of the body?

2. Explain whether the graph in Fig. 1 corresponds to the law of conservation of energy of the system of bodies Earth + stone. 3.

3. How does the kinetic energy of a thrown ball change? When does it decrease? is it increasing?

4. Why, when a stone falls, its potential energy turns out to be negative, but when a boy rolls down a hill, it is considered positive?

Potential energy of a body in a gravitational field

The next step involves introducing students to the potential energy of a body in a gravitational field. The energy of interaction of a body with the gravitational field of the Earth is described by the formula En = mgh only if the gravitational field of the Earth can be considered uniform, independent of coordinates. The gravitational field is determined by the law of universal gravitation:

where R is the radius vector drawn from the center of mass of the Earth (taken as the origin) to a given point (recall that in the law of gravity, bodies are considered point-like and motionless). By analogy with electrostatics, this formula can be written as:

and call it the vector of the gravitational field intensity at a given point. It is clear that this field changes with distance from the body creating the field. When can a gravitational field be considered homogeneous with sufficient accuracy? Obviously, this is possible in a region of space whose dimensions h are much smaller than the distance to the center of the field R. In other words, if you are considering a stone falling from the top floor of a house, you can safely ignore the difference in the value of the gravitational field on the top and bottom floors. However, when studying the movement of planets around the Sun, one cannot assume that the planet is moving in a uniform field, and one should use the general law of gravitation.

You can derive a general formula for the potential energy of gravitational interaction between bodies (but do not ask students to reproduce this conclusion, although they, of course, should know the final formula). For example, let's consider two stationary point bodies of masses m1 and m2, located at a distance R0 from each other (Fig. 4). Let us denote the energy of gravitational interaction of these bodies by En0. Let us further assume that the bodies have moved slightly closer to the distance R1. The energy of interaction of these bodies became En1. According to the law of conservation of energy:

Ep = Ep1 – Ep0 = Fthrust. avg s,

where Fthrust cр – the value of the average gravitational force in the section s = R1 – R0 of the body moving in the direction of the force. According to the law of universal gravitation, the magnitude of the force is:

If the distances R1 and R0 differ little from each other, then the distance Rav2 can be replaced by the product R1R0. Then:

In this equality En1 corresponds to corresponds . Thus:

We have obtained a formula that indicates two features of the potential energy of gravitational interaction (it is also called gravitational energy):

1. The formula itself already contains the choice of the zero level of potential gravitational energy, namely: the energy of gravitational interaction of bodies becomes zero when the distance between the bodies in question is infinitely large. Please note that this choice of zero value of the energy of gravitational interaction of bodies has a clear physical interpretation: when the bodies move infinitely far from each other, they practically cease to interact gravitationally.

2. Since any real distance, for example between the Earth and a rocket, of course, the energy of gravitational interaction with such a choice of reference point is always negative.

In Fig. Figure 5 shows a graph of the dependence of the energy of gravitational interaction of the rocket with the Earth on the distance between the center of the Earth and the rocket. It reflects both features of gravitational energy that we talked about: it shows that this energy is negative and increases towards zero as the distance between the Earth and the rocket increases.

Communication energy

The knowledge acquired by students that energy can be both positive and negative quantities should find its application in the study of the binding energy of particles of a substance in its various states of aggregation. For example, students can be offered the following qualitative reasoning.

We have already seen that particles of matter always move chaotically. It was by endowing particles with the ability to move in this way that we were able to explain a number of natural phenomena. But then why don’t tables and pencils, walls of houses and ourselves scatter into separate particles?

We have to assume that particles of matter interact and are attracted to each other. Only a sufficiently strong mutual attraction of particles can hold them near each other in liquids and solids and prevent them from quickly scattering in different directions. But why then do the particles in gases not stay close to each other, why do they fly apart? Apparently, in gases the interconnection of particles is not sufficient to retain them.

In mechanics, to evaluate the interaction (connection) of bodies, we used such a physical quantity as the potential energy of interaction. In the kinetic theory of matter, the connection between particles of matter is characterized by the energy of their interaction Ec (this energy is not always potential). The fact that particles in liquids and solids hold each other, but not in gases, suggests that the binding energy of particles with each other in these media is different.

Gas. In a gas, the distance between particles is large and their connection is weak. The particles occasionally collide with each other and with the walls of the vessel. The collisions are elastic in nature, i.e. total energy and total momentum are conserved. In the intervals between collisions, particles move freely, i.e. do not interact. It is reasonable to assume that the interaction (bond) energy of particles in a gas is approximately zero.

Liquid. In a liquid, particles are brought closer together and partially touch. Their mutual attraction is strong and is characterized by the binding energy Ecw (water). To tear off one molecule from the bulk of the liquid, it is necessary to perform work A > 0. As a result, the molecule will become free, as in a gas, i.e. its binding energy can be considered equal to zero. According to the law of conservation of energy, Ecw (water) + A = 0, from which Ecw (water) = –A< 0.

To determine the numerical value of the energy Eb(water) of particles in water, let us turn to experiment. Already everyday observations suggest: in order to evaporate water boiling in a kettle, you need to burn a certain amount of wood or gas. In other words, work needs to be done. Using a thermometer, you can make sure that the temperature of boiling water and the temperature of the steam above it are the same. Consequently, the average energy of particle motion in boiling water and in steam is the same. The thermal energy transferred to boiling water from the fuel is converted into the interaction energy of particles of evaporating water. This means that the energy Eb of particles in boiling water is less than in water vapor. But in a pair Ec(pair) = 0, therefore, the energy of interaction of particles in a liquid is less than zero, i.e. negative.

Measurements using calorimeters show that to evaporate 1 kg of boiling water at normal atmospheric pressure, about 2.3  106 J of energy must be transferred to it. Part of this energy (approximately 0.2  106 J) is spent so that the resulting water vapor can displace air particles from a thin layer above the surface of the liquid. The rest of the energy (2.1  106 J) goes to increase the binding energy of water particles during their transition from liquid to vapor (Fig. 6). Calculations show that 1 kg of water contains 3.2  1025 particles. Dividing the energy 2.1  106 J by 3.2  1025, we obtain: the binding energy Eb of each water particle with other particles during its transition from liquid to vapor increases by 6.6  10–20 J.

Solid. To melt and turn ice into water, you need to do work or transfer a certain amount of heat to the ice. Binding energy of water molecules in the solid phase Eb< 0, причем эта энергия по модулю больше, чем энергия связи молекул воды в жидкой phase. When ice melts, its temperature remains 0 °C; The water formed during melting has the same temperature. Therefore, in order to transfer a substance from a solid to a liquid state, it is necessary to increase the interaction energy of its particles. To melt 1 kg of ice that has already begun to melt, you need to expend 3.3  105 J of energy (Fig. 7). Almost all of this energy is used to increase the binding energy of particles during their transition from ice to water. Sharing energy

3.3  105 J per number of 3.2  1025 particles contained in 1 kg of ice, we find that the interaction energy Eb of ice particles is 10–20 J less than in water.

So, the interaction energy of vapor particles is zero. In water, the binding energy of each of its particles with other particles is approximately 6.6  10–20 J less than in steam, i.e. Eb(water) = –6.6  10–20 J. In ice, the binding energy of each particle with all other ice particles is 1.0  10–20 J less than in water (and accordingly 6.6  10– 20 J + 1.0  10–20 J = 7.6  10–20 J less than in water vapor). This means that in ice Ec(ice) = –7.6  10–20 J.

Consideration of the features of the interaction energy of particles of a substance in various states of aggregation is important for understanding the transformation of energy during transitions of a substance from one state of aggregation to another.

Let us give, in particular, examples of questions that students can now answer without much difficulty.

1. Water boils at a constant temperature, absorbing energy from the flame of a gas burner. What happens when this happens?

A) The energy of movement of water molecules increases;

B) the interaction energy of water molecules increases;

C) the energy of movement of water molecules decreases;

D) the interaction energy of water molecules decreases.

(Answer: B.)

2.When melting ice:

A) the kinetic energy of a piece of ice increases;

B) the internal energy of ice increases;

C) the potential energy of a piece of ice decreases;

D) the internal energy of ice decreases.

(Answer: B.)

Until now, we have considered the energy of interaction between bodies attracting each other. When studying electrostatics, it is useful to discuss with students the question of whether the interaction energy of particles is positive or negative when they repel each other. When particles repel each other, there is no need to impart energy to them in order to move far away from each other. The interaction energy is converted into the energy of motion of flying particles and decreases to zero as the distance between particles increases. In this case, the interaction energy is a positive value. The identified features of interaction energy can be consolidated when discussing the following issues:

1. Is the energy of interaction between two oppositely charged balls positive or negative? Justify your answer.

2. Is the energy of interaction between two similarly charged balls positive or negative? Justify your answer.

3. Two magnets approach each other with like poles. Does the energy of their interaction increase or decrease?

Communication energy in the microcosm

According to the concepts of quantum mechanics, an atom consists of a nucleus surrounded by electrons. In the reference frame associated with the nucleus, the total energy of the atom is the sum of the energy of electron motion around the nucleus, the energy of the Coulomb interaction of electrons with a positively charged nucleus, and the energy of the Coulomb interaction of electrons with each other. Let's consider the simplest of atoms - the hydrogen atom.

It is believed that the total energy of an electron is equal to the sum of the kinetic energy and the potential energy of the Coulomb interaction with the nucleus. According to Bohr's model, the total energy of an electron in a hydrogen atom can only take on a certain set of values:

where E0 is expressed in terms of world constants and the mass of the electron. It is more convenient to measure the numerical values of E(n) not in joules, but in electron volts. The first allowed values are:

E(1) = –13.6 eV (energy of the ground, most stable state of the electron);

E(2) = –3.4 eV;

E(3) = –1.52 eV.

It is convenient to mark the entire series of allowed values of the total energy of a hydrogen atom with dashes on the vertical energy axis (Fig. 8). Formulas for calculating possible values of electron energy for atoms of other chemical elements are complex, because Atoms have many electrons that interact not only with the nucleus, but also with each other.

Atoms combine to form molecules. In molecules, the picture of the movement and interaction of electrons and atomic nuclei is much more complex than in atoms. Accordingly, the set of possible values of internal energy changes and becomes more complex. The possible values of the internal energy of any atom and molecule have some features.

We have already clarified the first feature: the energy of an atom is quantized, i.e. can only take a discrete set of values. The atoms of each substance have their own set of energy values.

The second feature is that all possible values E(n) of the total energy of electrons in atoms and molecules are negative. This feature is associated with the choice of the zero level of interaction energy between the electrons of an atom and its nucleus. It is generally accepted that the interaction energy of an electron with a nucleus is zero when the electron is removed at a large distance and the Coulomb attraction of the electron to the nucleus is negligible. But in order to completely tear an electron away from the nucleus, you need to expend some work and transfer it to the nucleus + electron system. In other words, in order for the energy of interaction between an electron and a nucleus to become zero, it must be increased. And this means that the initial energy of interaction between the electron and the nucleus is less than zero, i.e. negative.

The third feature is that those made in Fig. 8, the marks of possible values of the internal energy of an atom are terminated at E = 0. This does not mean that the energy of the electron + nucleus system cannot, in principle, be positive. But when it reaches zero, the system ceases to be an atom. Indeed, at the value E = 0, the electron is removed from the nucleus, and instead of a hydrogen atom, there is an electron and a nucleus that are not connected to each other.

If the detached electron continues to move with kinetic energy Ek, then the total energy of the system of no longer interacting particles ion + electron can take on any positive values E = 0 + Ek.

Issues for discussion

1. What components make up the internal energy of an atom?

2. Why did we consider the energy of an atom only using the example of the hydrogen atom?

3. What conclusions about the features of the internal energy of an atom follow from its quantum mechanical model?

4. Why do we consider the internal energy of an atom or molecule to be negative?

5. Can the energy of an ion + electron group be positive?

Familiarity with the internal energy of an atom will not only consolidate knowledge about the possibility of negative values of potential energy, but also explain a number of phenomena, for example, the phenomenon of the photoelectric effect or the emission of light by atoms. Finally, the knowledge gained will allow students to discuss a very interesting question about the interaction of nucleons in the nucleus.

It has been established that the atomic nucleus consists of nucleons (protons and neutrons). A proton is a particle with a mass 2000 times greater than the mass of an electron, carrying a positive electrical charge (+1). As is known from electrodynamics, charges of the same sign repel each other. Therefore, electromagnetic interaction pushes protons apart. Why doesn't the core fall apart into its component parts? Back in 1919, by bombarding nuclei with α-particles, E. Rutherford found out that in order to knock a proton out of the nucleus, the α-particle must have an energy of about 7 MeV. This is several hundred thousand times more energy than is required to remove an electron from an atom!

As a result of numerous experiments, it was established that particles inside the nucleus are connected by a fundamentally new type of interaction. Its intensity is hundreds of times greater than the intensity of the electromagnetic interaction, which is why it was called the strong interaction. This interaction has an important feature: it has a short range and “turns on” only when the distance between nucleons does not exceed 10–15 m. This explains the small size of all atomic nuclei (no more than 10–14 m).

The proton-neutron model of the nucleus allows one to calculate the binding energy of nucleons in the nucleus. Let us recall that according to measurements it is approximately equal to –7 MeV. Let's imagine that 4 protons and 4 neutrons combined to form a beryllium nucleus. The mass of each neutron is mn = 939.57 MeV, and the mass of each proton is mp = 938.28 MeV (here we use the system of units accepted in nuclear physics, in which mass is measured not in kilograms, but in equivalent energy units, recalculated using Einstein’s relation E0 = mc2). Consequently, the total rest energy of 4 protons and 4 neutrons before they combine into a nucleus is 7511.4 MeV. The rest energy of the Be nucleus is 7454.7 MeV. It can be represented as the sum of the rest energy of the nucleons themselves (7511.4 MeV) and the binding energy of nucleons with each other Eb. That's why:

7454.7 MeV = 7511.4 MeV + Ev.

From here we get:

Ep = 7454.7 MeV –7511.4 MeV = –56.7 MeV.

This energy is distributed over all 8 nucleons of the beryllium nucleus. Consequently, each of them accounts for approximately –7 MeV, as follows from the experiments. We again found that the binding energy of mutually attracted particles is a negative quantity.

It is often believed that there are two antagonistic life energies that can mutually destroy each other. It is believed that a person is usually charged with positive vital energy, and when he receives a charge of negative vital energy, he becomes ill, he may get sick or even pass away into another world altogether.

Is it so?

This approach, from a physical point of view, contains contradictions. For example, a person who carries negative vital energy must somehow isolate it from positive energy, otherwise these two energies will interact with each other and the person who carries negative vital energy must suffer first.

In general, if negative and positive life energy are distributed in the space surrounding us, then they should mutually destroy each other, forming lifeless spaces.

If negative vital energy is generated from something, then it is of the same nature as positive vital energy, which is generated from the same thing, but acts in such a way that it causes the body to lose vital energy.

We generally need to look at this question more broadly.

Any loss of vital energy by the body negatively affects well-being and health in general. Losses can occur as a result of various reasons.

1. Physical overload.
2. Stress overload.
3. Mental overload.
4. Diseases.
5. Energy vampirism.
6. Open mental programming.
7. Hidden mental programming.

In the case of physical, stressful and mental overload, everything is clear - this is the direct use of vital energy for its intended purpose, and consumption leads to a decrease in reserves. Diseases also lead to loss of vital energy.

Diseases can be either a consequence of a combination of circumstances (injuries, infections and their consequences, genetic predisposition), or a manifestation of a deficiency of vital energy, that is, a consequence of any of the remaining six points or their combinations.

In the case of energy vampirism, part of the vital energy is withdrawn in favor of the energy vampire. As a result, there is less vital energy in the human body. Accordingly, your health worsens and the risk of disease increases.

Methods of open and hidden mental programming are very dangerous.

If we consider the methods of open mental programming, they are usually used in the process of human communication. These are ordinary psychological methods of influencing the mental sphere of a person.

Any communication between two people is mutual mental programming. This mental programming can have both positive and negative effects depending on the attitudes that people have when communicating. If you are praised and sincerely expressed sympathy and support, then it is natural that the effect on your mental sphere will be positive.

If you are scolded, criticized, humiliated, proven to be incompetent, then this introduces negative elements of programming into your mental sphere, which has a destructive effect on it and leads to a loss of vital energy.

Open mental programming, which is carried out through direct contact between people, is never pure, based only on verbal formulas. Words are the key to the resonant interaction of the subconscious.

The spoken word, both in the subconscious of the speaker and in the subconscious of the listener, evokes similar images that interact on the subconscious level, establishing mental subconscious contact, leading to the exchange of vital energy. The more and brighter such images are generated, the stronger the contact on the subconscious level, the more intense the energy exchange occurs.

If a psychological attack is carried out with strong emotional and verbal expression, then this leads to the introduction of destructive programs into the consciousness and subconscious of the victim of the attack, which, with regular exposure, can seriously damage the psyche and lead to catastrophic energy losses. The simplest example of such a destructive program is the proverb - “if you tell a person a hundred times that he is a pig, then the hundred and first time he will grunt.”

A similar attack can be carried out without direct psychological contact. The creation of a destructive mental program and its introduction into the mental sphere of the victim is carried out using ritual, hypnotic and other techniques. As a result of this implementation, both general losses of vital energy and those of its blocks that are responsible for certain areas of consciousness or internal organs of the body occur.

Usually such destructive program blocks are called negative energy. Naturally, such a name is logically incorrect. These destructive programs can just as easily be called negative programming.

Such programming is extremely dangerous for the person who composes such programs, since if he makes mistakes, he himself can become a victim of such programming. These programs can project their action onto the programmer according to the principle: “don’t dig a hole for someone else, you will fall into it yourself.”

N.K. Gladysheva, IOSO RAO, school No. 548, Moscow

Selecting the potential energy reference point

Ep = mg(h – h0).

(Ek + Ep)|B = (Ek+E0)|A.

Let us note the following.

1. According to the conditions of the problem at point A, the speed of the stone is zero, therefore Ek| A = 0.

(Ek + Ep)|B = 0.

Ek = –En = –mgh

(It should be remembered that h is a negative value). Graphs of the dependence of potential energy Ep and kinetic energy Ek on coordinate h are shown in Fig. 3.

mv02/2 = mv2/2 + mgh.

After familiarizing students with the above material, it is advisable to discuss the following questions with them:

1. Under what condition is the kinetic energy of a body equal to zero? potential energy of the body?

2. Explain whether the graph in Fig. 1 corresponds to the law of conservation of energy of the system of bodies Earth + stone. 3.

3. How does the kinetic energy of a thrown ball change? When does it decrease? is it increasing?

4. Why, when a stone falls, its potential energy turns out to be negative, but when a boy rolls down a hill, it is considered positive?

Potential energy of a body in a gravitational field

V.Yu. Mishin

Tuberculin diagnostics- a diagnostic test to determine the presence of specific sensitization of the human body to MBT, caused either by infection or artificially - vaccination with the BCG vaccine strain.

Old Koch tuberculin(Alt Tuberculin Koch - ATK) is a water-glycerol extract of a tuberculosis culture of human and bovine MBT, grown in meat-peptone broth with the addition of a 4% glycerol solution.

However, tuberculin obtained in this way contains protein derivatives of meat and peptone that are part of the medium, which leads to nonspecific reactions that complicate diagnosis. Therefore, ATK has found limited use in recent years. Available in 1 ml ampoules containing 100,000 TE.

More specific and free of ballast substances is purified protein derivative(Purified Protein Derivative - PPD), obtained by American scientists F. Seibert and S. Glenn (F. Seibert, S. Glenn) in 1934. This preparation represents the filtrate of a heat-killed substance purified by ultrafiltration, precipitated with trichloroacetic acid, washed with alcohol and ether, and dried in a vacuum from a frozen state. cultures of Mycobacterium tuberculosis of human and bovine types.

In our country, domestic dry purified tuberculin was produced in 1939 under the leadership of MA Linnikova at the Leningrad Research Institute of Vaccines and Serums, which is why this tuberculin is called PPD-L.

PPD-L is available in two forms:

purified tuberculin in standard dilution- ready-to-use colorless transparent liquid in ampoules of 3 ml with an activity of 2 TE in 0.1 ml. It is a solution of tuberculin in a 0.85% sodium chloride solution with the addition of Tween-80, which is a detergent and ensures the stability of the biological activity of the drug, and 0.01% quinosol as a preservative. Standard solutions of tuberculin are also prepared, containing 5 TE, YUTE, 100 TE in 0.1 ml of solution;
dry purified tuberculin in the form of a white powder in ampoules of 50,000 TE in one package with a solvent - carbolized saline solution.

Activity any tuberculin expressed in tuberculin units (THOSE). The national standard for tuberculin PPD-L was approved in 1963; 1 TU of domestic tuberculin contains 0.00006 mg of dry preparation. It is the tuberculin unit that is the basis for regulating the strength of the tuberculin test.

In terms of its biochemical composition, tuberculin is a complex compound, including proteins (tuberculoproteins), polysaccharides, lipid fractions and nucleic acid. The active principle of tuberculin is tuberculoproteins.

From an immunological point of view, tuberculin is a hapten (incomplete antigen), i.e. it does not cause the production of specific antibodies, but in an infected organism it initiates an antigen-antibody response, similar to the reaction to a live or killed MBT culture.

It has now been established that the body's reactions to tuberculin are a classic manifestation of the immunological phenomenon of HRT, which develops as a result of the interaction of an antigen
(tuberculin) with effector lymphocytes having specific receptors on their surface.

In this case, some lymphocytes die, releasing proteolytic enzymes that cause a damaging effect on tissue. An inflammatory reaction occurs not only at the injection site, but also around tuberculosis foci. When sensitized cells are destroyed, active substances with pyrogenic properties are released.

In response to the introduction of tuberculin into the body, those infected and patients with tuberculosis develop injection, general and focal reactions. The body's response to tuberculin depends on the dose and site of administration. Thus, a local (prick) reaction occurs with cutaneous (Pirquet test), intradermal (Mantoux test) administration of the drug, and the appearance of a local, general and focal reaction occurs with subcutaneous administration (Koch test).

Puncture reaction characterized by the appearance of papules (infiltrate) and hyperemia at the site of tuberculin injection. With hyperergic reactions, the formation of vesicles, bullae, lymphangitis, and necrosis is possible. Measuring the diameter of the infiltrate allows you to accurately assess the reaction and reflect the degree of sensitivity of the body to the amount of tuberculin used.

Pathomorphology of tuberculin reaction in the initial stage (first 24 hours) it is manifested by edema and exudation, in later periods (72 hours) - a mononuclear reaction. In hyperergic reactions with pronounced necrosis, specific elements with epithelioid and giant cells are found at the injection site.

General reaction of the infected organism to the administration of tuberculin is manifested by a deterioration in general condition, headache, arthralgia, increased body temperature, changes in hemogram, biochemical, immunological parameters.

Focal reaction characterized by increased perifocal inflammation around the tuberculosis focus. In the pulmonary process, the focal reaction is manifested by increased cough, chest pain, increased amount of sputum, hemoptysis, and radiographically - an increase in inflammatory changes in the area of the specific lesion; with kidney tuberculosis - the appearance of leukocytes and MBT in the urine; in fistulous forms of peripheral lymphadenitis - increased suppuration, etc.

Sensitivity of the human body to tuberculin can be different: negative ( anergy), when the body does not respond to the introduction of tuberculin; weak ( hypoergy), moderate ( normergy) and pronounced ( hyperergy).

The intensity of reactions to tuberculin depends on the severity and virulence of the infection (existence of contact with a patient with tuberculosis, infection with highly virulent MBT strains from a dying patient, etc.), the body's resistance, dose, method and frequency of administration.

If tuberculin is used in large doses and at short intervals, the body’s sensitivity to it increases (Booster effect).

The absence of the body's response to tuberculin (anergy) is divided into primary - in individuals not infected with tuberculosis, and secondary - a condition accompanied by loss of tuberculin sensitivity in individuals infected and sick with tuberculosis.

Secondary anergy develops with lymphogranulomatosis, sarcoidosis, many acute infectious diseases (measles, rubella, scarlet fever, whooping cough, etc.), vitamin deficiency, cachexia, progressive tuberculosis, febrile conditions, treatment with hormones, cytostatics, and pregnancy.

On the contrary, in conditions of exogenous superinfection, in the presence of helminthic infestation, chronic foci of infection, multiple caries, calcifications in the lungs and intrathoracic lymph nodes, and hyperthyroidism, tuberculin tests are enhanced.

Tuberculin diagnostics is divided into mass and individual. Under mass tuberculin diagnostics involve examining healthy groups of children and adolescents using an intradermal Mantoux test with 2 TE PPD-L. Under individual- carrying out differential diagnosis of tuberculosis and nonspecific diseases, determining the nature of tuberculin sensitivity, determining the activity of specific changes.

The goals of mass tuberculin diagnostics are:

identification of persons newly infected with MTB (“turn” of tuberculin tests);
identification of individuals with hyperergic and increasing reactions to tuberculin;
selection of contingents for anti-tuberculosis vaccination with the BCG vaccine of children aged 2 months and older who did not receive vaccination in the maternity hospital, and for revaccination with BCG;
early diagnosis of tuberculosis in children and adolescents;
determination of epidemiological indicators for tuberculosis (infection of the population with MTB, annual risk of infection with MTB).

For mass tuberculin diagnostics, only a single intradermal Mantoux tuberculin test with 2 TE PPD-L is used.

Mantoux test technique. To perform the Mantoux test, disposable one-gram tuberculin syringes are used. 0.2 ml of tuberculin is drawn into the syringe from the ampoule, then the solution is released to the 0.1 ml mark.

The inner surface of the middle third of the forearm is treated with 70° alcohol and dried with sterile cotton wool. The needle is inserted with the cut upward into the upper layers of the stretched skin (intradermal) parallel to its surface. After inserting the needle hole into the skin, 0.1 ml of solution (2 TE PPD-L) is injected from a syringe, i.e. 1 dose. With the correct technique, a papule in the form of a “lemon peel” is formed in the skin, measuring at least 7-9 mm in diameter and whitish in color.

Mantoux test recording technique. The Mantoux test is assessed after 72 hours by measuring (mm) the diameter of the infiltrate transverse to the axis of the forearm.

When performing the Mantoux test, the reaction is considered:

negative - complete absence of infiltrate and hyperemia or the presence of only an injection mark (infiltrate with a diameter of 0-1 mm);
doubtful - the presence of an infiltrate of 2-4 mm or only hyperemia of any size;
positive - the presence of infiltrate with a diameter of 5 mm or more;
hyperergic - the presence of an infiltrate with a diameter of 17 mm or more in children and adolescents, in adults - 21 mm or more. In the presence of vesicles, necrosis, lymphangitis, regardless of the size of the infiltrate, the reaction is considered hyperergic.

The Mantoux test with 2 TE PPD-L is given to children and adolescents annually, starting at 12 months, regardless of the previous result. The sample is administered by a specially trained nurse. All test results are recorded in the medical record.

With systematic tuberculin diagnostics, the doctor can analyze the dynamics of tuberculin tests and identify the moment of MBT infection - the transition of a previously negative test to a positive one (not associated with BCG vaccination), the so-called “turn” of tuberculin tests; an increase in tuberculin sensitivity and the development of hyperergy to tuberculin.

All children and adolescents from the risk groups listed above, which are identified by the results of mass tuberculin diagnostics, are registered with a phthisiatrician for 1-2 years. They undergo an examination, including an X-ray of the respiratory system (longitudinal tomograms if indicated), general clinical tests of blood and urine, and their surroundings are examined in order to early diagnose the disease and find the source of their infection. In order to prevent the development of the disease, infected children and adolescents are given prophylactic (preventive) treatment.

At the ages of 7 and 14 years, children who have a negative Mantoux test result with 2 TU PPD-L and no contraindications to the vaccine are necessarily revaccinated with the BCG vaccine in order to create artificial active anti-tuberculosis immunity in them.

Goals of mass tuberculin diagnostics:

differential diagnosis of post-vaccination and infectious allergies to tuberculin;
differential diagnosis of tuberculosis and other diseases;
determination of the threshold of individual sensitivity to tuberculin;
determination of the activity of the tuberculosis process;
assessment of the effectiveness of anti-tuberculosis treatment.

For individual tuberculin diagnostics, in addition to the Mantoux test with 2 TU PPD-L, the Mantoux test with various doses of tuberculin, the Koch test, etc. are used.

Post-vaccination immunity (post-vaccination allergy). In the context of mandatory mass vaccine prevention of tuberculosis, many children and adolescents have anti-tuberculosis immunity due to the introduction of the vaccine, and also respond positively to
tuberculin (post-vaccination allergy).

When deciding what exactly is associated with positive tuberculin sensitivity, one should take into account the nature of the test itself, the time period that has passed since the administration of the BCG vaccine, the number and size of BCG scars, and the presence of contact with a patient with tuberculosis.

For post-vaccination tuberculin sensitivity characterized by a gradual decrease in the size of the infiltrate every year and the transition 2-3-4 years after vaccination to dubious and negative results. The papule is often flat, ill-defined, on average 7-10 mm in diameter, and does not leave behind long-term pigmentation.

When infected with MBT persistent preservation or even increase in sensitivity to tuberculin is observed. The papule is tall, bright, clearly defined, the pigment spot persists for a long time. The average diameter of the infiltrate is 12 mm; the presence of a hyperergic reaction indicates MBT infection.

Koch test used when conducting individual tuberculin diagnostics, most often for the purpose of differential diagnosis of tuberculosis with other diseases and determining its activity. Tuberculin during the Koch test is administered subcutaneously, most often starting with 20 TU. If the result is negative, increase the dose to 50 TE, and then to 100 TE. If there is no reaction to subcutaneous injection of 100 TE, then the diagnosis of tuberculosis is removed.

When performing the Koch test, the local (in the area of tuberculin injection), focal (in the area of the specific lesion) and general reaction of the body, as well as blood changes (hemotuberculin and proteinotuberculin tests) are taken into account. Preliminary blood and plasma parameters are determined before tuberculin administration and 48 hours after it.

The general reaction is characterized by an increase in body temperature by 0.5 ° C, symptoms of intoxication;
focal - exacerbation of tuberculous changes;
local - the formation of an infiltrate at the site of tuberculin injection with a diameter of 10-20 mm.

Hemotuberculin test is considered positive if there is an increase in ESR by 6 mm per hour or more, an increase in the number of leukocytes by 1000 or more, a shift in the leukocyte formula to the left, a decrease in lymphocytes by 10% or more.

Protein tuberculin test assessed as positive if there is a decrease in albumin and an increase in a- and y-globulins by 10% of the initial data. The Koch test is also combined with immunological tests of blast transformation, macrophage migration, etc.

The Koch test is considered positive if any three or more indicators change. It should be remembered that the focal reaction is of greatest importance in the evaluation of this test.

Negative and positive energy of a particle. Abstract: Can energy be negative? Potential energy of a body in a gravitational field

Benefits and features of using kefir face mask

Benefits and features of using kefir face mask