Planetary model of the atom. Summary: Planetary model of the atom Isoelectronic sequence of hydrogen
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Planetary model of the atom
19. In the planetary model of the atom, it is assumed that the number
1) electrons in orbits is equal to the number of protons in the nucleus
2) protons is equal to the number of neutrons in the nucleus
3) electrons in orbits is equal to the sum of the numbers of protons and neutrons in the nucleus
4) neutrons in the nucleus is equal to the sum of the numbers of electrons in orbits and protons in the nucleus
21. The planetary model of the atom is substantiated by experiments on
1) dissolution and melting of solids 2) gas ionization
3) chemical production of new substances 4) scattering of α-particles
24. The planetary model of the atom is substantiated
1) calculations of the motion of celestial bodies 2) experiments on electrification
3) experiments on the scattering of α -particles 4) photographs of atoms in a microscope
44. In Rutherford's experiment α -particles are scattered
1) the electrostatic field of the atomic nucleus 2) the electron shell of the target atoms
3) the gravitational field of the atomic nucleus 4) the target surface
48. In Rutherford's experiment, most of the α-particles freely pass through the foil, practically without deviating from rectilinear trajectories, because
1) the nucleus of the atom has a positive charge
2) electrons are negatively charged
3) the nucleus of an atom has small (compared to an atom) size
4) α-particles have a large (compared to atomic nuclei) mass
154. What statements correspond to the planetary model of the atom?
1) The nucleus is in the center of the atom, the charge of the nucleus is positive, the electrons are in orbits around the nucleus.
2) The nucleus is in the center of the atom, the charge of the nucleus is negative, the electrons are in orbits around the nucleus.
3) Electrons - in the center of the atom, the nucleus revolves around the electrons, the charge of the nucleus is positive.
4) Electrons - in the center of the atom, the nucleus revolves around the electrons, the charge of the nucleus is negative.
225. The experiments of E. Rutherford on the scattering of α-particles showed that
A. almost all the mass of an atom is concentrated in the nucleus. B. the nucleus has a positive charge.
Which of the statements is correct?
1) only A 2) only B 3) both A and B 4) neither A nor B
259. What idea of \u200b\u200bthe structure of the atom corresponds to the model of the atom of Rutherford?
1) The nucleus is at the center of the atom, the electrons are in orbits around the nucleus, the charge of the electrons is positive.
2) The nucleus is at the center of the atom, electrons are in orbits around the nucleus, the charge of electrons is negative.
3) The positive charge is evenly distributed over the atom, the electrons in the atom vibrate.
4) The positive charge is evenly distributed over the atom, and the electrons move in the atom in different orbits.
266. What idea about the structure of the atom is correct? Most of the mass of an atom is concentrated
1) in the nucleus, the electron charge is positive 2) in the nucleus, the nuclear charge is negative
3) in electrons, the charge of electrons is negative 4) in the nucleus, the charge of electrons is negative
254. What idea of \u200b\u200bthe structure of the atom corresponds to the model of the atom of Rutherford?
1) The nucleus is in the center of the atom, the charge of the nucleus is positive, most of the mass of the atom is concentrated in electrons.
2) The nucleus is in the center of the atom, the charge of the nucleus is negative, most of the mass of the atom is concentrated in the electron shell.
3) The nucleus is in the center of the atom, the charge of the nucleus is positive, most of the mass of the atom is concentrated in the nucleus.
4) The nucleus is in the center of the atom, the charge of the nucleus is negative, most of the mass of the atom is concentrated in the nucleus.
Bohr's postulates
267. The diagram of the lowest energy levels of atoms of a rarefied atomic gas has the form shown in the figure. At the initial moment of time, the atoms are in a state with energy E (2) According to Bohr's postulates, this gas can emit photons with energy
1) 0.3 eV, 0.5 eV and 1.5 eV 2) only 0.3 eV 3) only 1.5 eV 4) any in the range from 0 to 0.5 eV
273. The figure shows a diagram of the lowest energy levels of the atom. At the initial moment of time, the atom is in a state with energy E (2). According to Bohr's postulates, a given atom can emit photons with energy
1) 1 ∙ 10 -19 J 2) 3 ∙ 10 -19 J 3) 5 ∙ 10 -19 J 4) 6 ∙ 10 -19 J
279. What determines the frequency of a photon emitted by an atom according to the model of the Bohr atom?
1) the difference between the energies of stationary states 2) the frequency of revolution of the electron around the nucleus
3) the de Broglie wavelength for an electron 4) Bohr's model does not allow it to be determined
15. The atom is in a state with energy E 1< 0. Минимальная энергия, необходимая для отрыва электрона от атома, равна
1) 0 2) E 1 3) - E 1 4) - E 1/2
16. How many photons of different frequencies can be emitted by hydrogen atoms in the second excited state?
1) 1 2) 2 3) 3 4) 4
25. Suppose that the energy of gas atoms can take only those values \u200b\u200bthat are indicated in the diagram. Atoms are in a state with energy e (3). What energy can a given gas absorb?
1) any within the range from 2 ∙ 10 -18 J to 8 ∙ 10 -18 J 2) any, but less than 2 ∙ 10 -18 J
3) only 2 ∙ 10 -18 J 4) any greater than or equal to 2 ∙ 10 -18 J
29. When a photon with an energy of 6 eV is emitted, the charge of an atom
1) does not change 2) increases by 9.6 ∙ 10 -19 C
3) increases by 1.6 ∙ 10 -19 C 4) decreases by 9.6 ∙ 10 -19 C
30. Light with a frequency of 4 ∙ 10 15 Hz consists of photons with an electric charge equal to
1) 1.6 ∙ 10 -19 C 2) 6.4 ∙ 10 -19 C 3) 0 C 4) 6.4 ∙ 10 -4 C
78. An electron of the outer shell of an atom first passes from a stationary state with energy E 1 to a stationary state with energy E 2, absorbing a photon with a frequency v 1 . Then it goes from state E 2 to a stationary state with energy E s, absorbing a photon with frequency v 2 > v 1 . What happens when an electron passes from the E 2 state to the E 1 state.
1) emission of light with frequency v 2 – v 1 2) absorption of light by frequency v 2 – v 1
3) emission of light with frequency v 2 + v 1 4) absorption of light by frequency v 2 – v 1
90. The energy of a photon absorbed by an atom during the transition from the ground state with energy E 0 to an excited state with energy E 1 is equal to (h is Planck's constant)
95. The figure shows the energy levels of an atom and indicates the wavelengths of photons emitted and absorbed during transitions from one level to another. What is the wavelength for photons emitted during the transition from the E 4 level to the E 1 level, if λ 13 \u003d 400 nm, λ 24 \u003d 500 nm, λ 32 \u003d 600 nm? Express your answer in nm and round to the nearest whole number.
96. The figure shows several energy levels of the electron shell of the atom and indicates the frequencies of photons emitted and absorbed during transitions between these levels. What is the minimum wavelength of photons emitted by an atom at any
possible transitions between levels E 1, E 2, ez and E 4, if v 13 \u003d 7 ∙ 10 14 Hz, v 24 \u003d 5 ∙ 10 14 Hz, v 32 \u003d 3 ∙ 10 14 Hz? Express your answer in nm and round to the nearest whole number.
120. The figure shows a diagram of the energy levels of the atom. Which of the transitions between energy levels marked with arrows is accompanied by absorption of a quantum of the minimum frequency?
1) from level 1 to level 5 2) from level 1 to level 2
124. The figure shows the energy levels of the atom and indicates the wavelengths of photons emitted and absorbed during transitions from one level to another. It was experimentally established that the minimum wavelength for photons emitted during transitions between these levels is λ 0 \u003d 250 nm. What is the value of λ 13 if λ 32 \u003d 545 nm, λ 24 \u003d 400 nm?
145. The figure shows a diagram of possible values \u200b\u200bof the energy of atoms of a rarefied gas. At the initial moment of time, the atoms are in a state with energy E (3). Possible emission of photons with energy
1) only 2 ∙ 10 -18 J 2) only 3 ∙ 10 -18 and 6 ∙ 10 -18 J
3) only 2 ∙ 10 -18, 5 ∙ 10 -18 and 8 ∙ 10 -18 J 4) any from 2 ∙ 10 -18 to 8 ∙ 10 -18 J
162. The energy levels of an electron in a hydrogen atom are given by the formula Е n \u003d - 13.6 / n 2 eV, where n \u003d 1, 2, 3, .... When an atom passes from the E 2 state to the E 1 state, the atom emits a photon. Once on the surface of the photocathode, the photon knocks out the photoelectron. The wavelength of light corresponding to the red border of the photoelectric effect for the material of the photocathode surface, λ cr \u003d 300 nm. What is the maximum possible speed of a photoelectron?
180. The figure shows several of the lowest energy levels of the hydrogen atom. Can an atom in the E 1 state absorb a 3.4 eV photon?
1) yes, in this case, the atom passes into the state Е 2
2) yes, in this case the atom passes into the state E 3
3) yes, while the atom is ionized, decaying into a proton and an electron
4) no, the photon energy is not enough for the transition of an atom to an excited state
218. The figure shows a simplified diagram of the energy levels of the atom. Numbered arrows indicate some of the possible transitions of the atom between these levels. Establish a correspondence between the processes of absorption of light of the longest wavelength and emission of light of the longest wavelength and the arrows indicating the energy transitions of the atom. For each position of the first column, select the corresponding position of the second and write down the selected numbers in the table under the corresponding letters.
226. The figure shows a fragment of the atomic energy level diagram. Which of the transitions between energy levels marked with arrows is accompanied by the emission of a photon with maximum energy?
1) from level 1 to level 5 2) from level 5 to level 2
3) from level 5 to level 1 4) from level 2 to level 1
228. The figure shows the four lower energy levels of the hydrogen atom. What transition corresponds to the absorption of a photon with an energy of 12.1 eV by an atom?
1) E 3 → E 1 2) E 1 → E 3 3) E 3 → E 2 4) E 1 → E 4
238. An electron with a momentum p \u003d 2 ∙ 10 -24 kg ∙ m / s collides with a resting proton, forming a hydrogen atom in a state with energy E n (n \u003d 2). In the process of atom formation, a photon is emitted. Find the frequency v of this photon, neglecting the kinetic energy of the atom. The energy levels of an electron in a hydrogen atom are given by the formula, where n \u003d 1,2, 3, ....
260. The diagram of the lowest energy levels of the atom has the form shown in the figure. At the initial moment of time, the atom is in a state with energy E (2). According to Bohr's postulates, an atom can emit photons with energy
1) only 0.5 eV 2) only 1.5 eV 3) any less than 0.5 eV 4) any within the range from 0.5 to 2 eV
269. The figure shows the diagram of the energy levels of the atom. What number indicates the transition that corresponds radiationphoton with the lowest energy?
1) 1 2) 2 3) 3 4) 4
282. Radiation of a photon by an atom occurs at
1) the motion of an electron in a stationary orbit
2) the transition of an electron from the ground state to an excited
3) the transition of an electron from an excited state to the ground state
4) all listed processes
13. Emission of photons occurs during the transition from excited states with energies E 1\u003e E 2\u003e E 3 to the ground state. For the frequencies of the corresponding photons v 1, v 2, v 3, the following relation is valid
1) v 1 < v 2 < v 3 2) v 2 < v 1 < v 3 3) v 2 < v 3 < v 1 4) v 1 > v 2 > v 3
1) greater than zero 2) equal to zero 3) less than zero
4) more or less than zero depending on the state
98. An atom at rest has absorbed a photon with an energy of 1.2 ∙ 10 -17 J. In this case, the momentum of the atom
1) did not change 2) became equal to 1.2 ∙ 10 -17 kg ∙ m / s
3) became equal to 4 ∙ 10 -26 kg ∙ m / s 4) became equal to 3.6 ∙ 10 -9 kg ∙ m / s
110. Suppose that the diagram of the energy levels of atoms of some substance has the form,
shown in the figure, and the atoms are in a state with energy E (1). An electron moving with a kinetic energy of 1.5 eV collided with one of these atoms and bounced off, gaining some additional energy. Determine the momentum of the electron after the collision, assuming that the atom was at rest before the collision. The possibility of emission of light by an atom upon collision with an electron should be neglected.
111. Suppose that the diagram of the energy levels of the atoms of a certain substance has the form shown in the figure, and the atoms are in a state with energy E (1). An electron, having collided with one of these atoms, bounced off, acquiring some additional energy. The momentum of an electron after a collision with an atom at rest turned out to be equal to 1.2 ∙ 10 -24 kg ∙ m / s. Determine the kinetic energy of the electron before the collision. Disregard the possibility of emission of light by an atom in a collision with an electron.
136. The π ° -meson with a mass of 2.4 ∙ 10 -28 kg decays into two γ-quanta. Find the modulus of the momentum of one of the formed γ-quanta in the reference frame where the primary π ° -meson is at rest.
144. The vessel contains rarefied atomic hydrogen. The hydrogen atom in the ground state (E 1 \u003d - 13.6 eV) absorbs a photon and is ionized. An electron ejected from an atom as a result of ionization moves away from the nucleus at a speed of v \u003d 1000 km / s. What is the frequency of the absorbed photon? Disregard the energy of thermal motion of hydrogen atoms.
197. A resting hydrogen atom in the ground state (E 1 \u003d - 13.6 eV) absorbs a photon with a wavelength λ \u003d 80 nm in vacuum. At what speed does the electron that has escaped from the atom as a result of ionization move away from the nucleus? Disregard the kinetic energy of the formed ion.
214. A free pion (π ° -meson) with a rest energy of 135 MeV moves with a speed v, which is much less than the speed of light. As a result of its decay, two γ quanta were formed, one of them propagating in the direction of the pion's motion, and the other in the opposite direction. The energy of one quantum is 10% more than the other. What is the speed of a pion before decay?
232. The table shows the energy values \u200b\u200bfor the second and fourth energy levels of the hydrogen atom.
| Level number | Energy, 10 -19 J |
| -5,45 | |
| -1,36 |
What is the energy of a photon emitted by an atom in the transition from the fourth level to the second?
1) 5.45 ∙ 10 -19 J 2) 1.36 ∙ 10 -19 J 3) 6.81 ∙ 10 -19 J 4) 4.09 ∙ 10 -19 J
248. An atom at rest emits a photon with an energy of 16.32 ∙ 10 -19 J as a result of the transition of an electron from an excited state to a ground state. As a result of recoil, the atom begins to move forward in the opposite direction with a kinetic energy of 8.81 ∙ 10 -27 J. Find the mass of the atom. The speed of an atom is considered small compared to the speed of light.
252. The vessel contains rarefied atomic hydrogen. The hydrogen atom in the ground state (E 1 \u003d -13.6 eV) absorbs a photon and is ionized. An electron ejected from an atom as a result of ionization moves away from the nucleus at a speed of 1000 km / s. What is the wavelength of the absorbed photon? Disregard the energy of thermal motion of hydrogen atoms.
1) 46 nm 2) 64 nm 3) 75 nm 4) 91 nm
257. The vessel contains rarefied atomic hydrogen. The hydrogen atom in the ground state (E 1 \u003d -13.6 eV) absorbs a photon and is ionized. The electron ejected from the atom as a result of ionization moves away from the nucleus at a speed of v \u003d 1000 km / s. What is the energy of the absorbed photon? Disregard the energy of thermal motion of hydrogen atoms.
1) 13.6 eV 2) 16.4 eV 3) 19.3 eV 4) 27.2 eV
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What is it? This is Rutherford's model of the atom. It is named after the New Zealand-born British physicist Ernest Rutherford, who heralded the discovery of the nucleus in 1911. In his experiments with scattering alpha particles on thin metal foil, he found that most alpha particles passed directly through the foil, but some bounced off. Rutherford suggested that in the area of \u200b\u200bthe small area from which they bounced off, there is a positively charged nucleus. This observation led him to describe the structure of the atom, which, with corrections for the quantum theory, is accepted today. Just as the Earth revolves around the Sun, the electric charge of an atom is concentrated in the nucleus, around which electrons of opposite charge revolve, and the electromagnetic field keeps the electrons in orbit of the nucleus. Therefore, the model is called planetary.
Before Rutherford, there was another model of the atom - Thompson's model of matter. There was no nucleus in it, it was a positively charged "cake" filled with "zest" - electrons that freely rotated in it. By the way, it was Thompson who discovered electrons. In a modern school, when they begin to get acquainted with, they always start with this model.
Atom models of Rutherford (left) and Thompson (right)
// wikimedia.org
The quantum model that describes the structure of the atom today is of course different from the one that Rutherford came up with. There is no quantum mechanics in the motion of planets around the Sun, but in the motion of an electron around the nucleus, it is. However, the concept of the orbit still remains in the theory of atomic structure. But after it became known that the orbits are quantized, that is, there is no continuous transition between them, as Rutherford thought, it became incorrect to call such a planetary model. Rutherford took the first step in the right direction, and the development of the theory of the structure of the atom followed the path that he outlined.
Why is this interesting for science? Rutherford's experiment discovered nuclei. But everything that we know about them, we learned after. His theory has evolved over many decades, and it contains answers to fundamental questions about the structure of matter.
Paradoxes were quickly discovered in Rutherford's model, namely: if a charged electron revolves around a nucleus, then it must emit energy. We know that a body that is moving in a circle at a constant speed is still accelerating, because the speed vector is turning all the time. And if a charged particle moves with acceleration, it must radiate energy. This means that it should almost instantly lose it all and fall to the core. Therefore, the classical model of the atom does not fully agree with itself.
Then physical theories began to appear that tried to overcome this contradiction. An important addition to the atomic structure model was made by Niels Bohr. He discovered that there are several quantum orbits around the atom, along which the electron moves. He suggested that the electron does not emit energy all the time, but only when it moves from one orbit to another.
Bohr's Atom Model
// wikimedia.org
And after Bohr's model of the atom, Heisenberg's uncertainty principle appeared, which finally explained why the fall of an electron onto a nucleus is impossible. Heisenberg discovered that in an excited atom, an electron is in distant orbits, and the moment it emits a photon, it falls into the main orbit, losing its energy. The atom, on the other hand, goes into a stable state, in which the electron will rotate around the nucleus as long as nothing excites it from the outside. This is a stable state, beyond which the electron will not fall.
Due to the fact that the basic state of the atom is a stable state, matter exists, we all exist. Without quantum mechanics, we would have no stable matter at all. In this sense, the main question a layman might ask quantum mechanics is why doesn't everything fall at all? Why is all the substance not going to a point? And quantum mechanics is capable of answering this question.
Why know this? In a sense, Rutherford's experiment was repeated again with the discovery of quarks. Rutherford discovered that positive charges - protons - are concentrated in nuclei. What's inside the protons? Now we know that there are quarks inside the protons. We learned this by conducting a similar experiment on deep inelastic electron-proton scattering in 1967 at SLAC (National Accelerator Laboratory, USA).
This experiment was carried out on the same principle as Rutherford's experiment. Then alpha particles fell, and here electrons fell on protons. As a result of the collision, protons can remain protons, or they can be excited due to high energy, and then other particles, such as pi-mesons, can be produced during the scattering of protons. It turned out that this cross section behaves as if there are point components inside the protons. We now know that these point components are quarks. In a sense, this was Rutherford's experience, but on the next level. Since 1967, we already have a quark model. But what will happen next, we do not know. Now you need to scatter something on the quarks and look at what they fall apart. But this is the next step, so far it has not been possible.
In addition, the most important story from the history of Russian science is associated with the name of Rutherford. Petr Leonidovich Kapitsa worked in his laboratory. In the early 1930s, he was banned from leaving the country and forced to remain in the Soviet Union. Upon learning of this, Rutherford sent Kapitsa all the instruments he had in England, and thus helped to create the Institute for Physical Problems in Moscow. That is, thanks to Rutherford, a significant part of Soviet physics took place.
The stability of any system on an atomic scale follows from the Heisenberg uncertainty principle (the fourth section of the seventh chapter). Therefore, a consistent study of the properties of the atom is possible only within the framework of quantum theory. Nevertheless, some results of practical importance can be obtained in the framework of classical mechanics by adopting additional rules for quantizing orbits.
In this chapter, we will calculate the position of the energy levels of the hydrogen atom and hydrogen-like ions. The calculations are based on the planetary model, according to which electrons rotate around the nucleus under the action of the forces of Coulomb attraction. We assume that electrons move in circular orbits.
13.1. Compliance principle
Quantization of angular momentum is applied in the model of the hydrogen atom proposed by Bohr in 1913. Bohr proceeded from the fact that in the limit of small quanta of energy, the results of quantum theory should correspond to the conclusions of classical mechanics. He formulated three postulates.
An atom can be for a long time only in certain states with discrete energy levels E i . Electrons, rotating in corresponding discrete orbits, move with acceleration, but, nevertheless, they do not emit. (In classical electrodynamics, any accelerated moving particle radiates if it has a nonzero charge).
Radiation is emitted or absorbed by quanta during the transition between energy levels:
From these postulates follows the rule for quantizing the angular momentum of an electron
,
where ncan be equal to any natural number:
Parameter n called principal quantum number... To derive formulas (1.1), let us express the level energy in terms of the moment of rotation. Astronomical measurements require knowledge of wavelengths with a sufficiently high accuracy: six correct numbers for optical lines and up to eight in the radio range. Therefore, when studying the hydrogen atom, the assumption of an infinitely large mass of the nucleus turns out to be too rough, since it leads to an error in the fourth significant figure. It is necessary to take into account the motion of the nucleus. To take it into account, the concept is introduced reduced mass.
13.2. Reduced mass
An electron moves around the nucleus under the action of an electrostatic force
,
where r- a vector whose beginning coincides with the position of the nucleus, and the end points to an electron. Recall that Z is the atomic number of the nucleus, and the charges of the nucleus and the electron are equal, respectively Ze and 
... According to Newton's third law, a force equal to - f(it is equal in magnitude and directed opposite to the force acting on the electron). We write down the equations of motion of the electron
.
We introduce new variables: the speed of the electron relative to the nucleus
and the speed of the center of mass
.
Adding (2.2a) and (2.2b), we obtain
.
Thus, the center of mass of a closed system moves uniformly and rectilinearly. Now we divide (2.2b) by m Z and subtract it from (2.2a) divided by m e ... The result is an equation for the relative velocity of the electron:
.
The value included in it
called reduced mass... Thus, the problem of the joint motion of two particles - an electron and a nucleus - is simplified. It is enough to consider the motion of one particle around the nucleus, the position of which coincides with the position of the electron, and its mass is equal to the reduced mass of the system.
13.3. The relationship between energy and torque
The force of the Coulomb interaction is directed along the straight line connecting the charges, and its modulus depends only on the distance r between them. Consequently, equation (2.5) describes the motion of a particle in a centrally symmetric field. An important property of motion in a field with central symmetry is the conservation of energy and angular momentum.
Let us write down the condition that the motion of an electron in a circular orbit is determined by the Coulomb attraction to the nucleus:
.
It follows from it that the kinetic energy
equal to half the potential energy
,
taken with the opposite sign:
.
Total energy E,respectively, is equal to:
.
It turned out to be negative, as it should be for steady states. The states of atoms and ions with negative energy are called bound... Multiplying equation (3.4) by 2 r and replacing the work on the left side mVr at the moment of rotation M, express the speed V after a moment:
.
Substituting the obtained value of the velocity into (3.5), we obtain the required formula for the total energy:
.
Let's pay attention to the fact that the energy is proportional to the even degree of the torque. In Bohr's theory, this fact has important consequences.
13.4. Torque quantization
Second equation for variables V and r we will obtain from the orbit quantization rule, the derivation of which will be performed based on Bohr's postulates. Differentiating formula (3.5), we obtain a relationship between small changes in torque and energy:
.
According to the third postulate, the frequency of an emitted (or absorbed) photon is equal to the frequency of revolution of an electron in its orbit:
.
From formulas (3.4), (4.2) and the connection
between the velocity, the angular momentum and the radius, a simple expression follows for the change in angular momentum during the transition of an electron between neighboring orbits:
.
Integrating (4.3), we obtain
Constant C we will search in the half-open interval
.
The double inequality (4.5) does not introduce any additional restrictions: if FROMgoes beyond the limits (4.5), then it can be returned to this interval, simply by renumbering the values \u200b\u200bof the moment in formula (4.4).
Physical laws are the same in all frames of reference. Let's pass from the right-handed coordinate system to the left-handed one. Energy, like any scalar quantity, will remain the same,
.
The axial vector of the angular momentum behaves differently. As you know, any axial vector changes sign when performing this operation:
There is no contradiction between (4.6) and (4.7), since the energy, according to (3.7), is inversely proportional to the square of the moment and remains the same when the sign changes M.
Thus, the set of negative torque values \u200b\u200bmust repeat the set of its positive values. In other words, for each positive value M n there must be a negative value equal to it in absolute value M – m :
Combining (4.4) - (4.8), we obtain a linear equation for FROM:
,
with decision
.
It is easy to verify that formula (4.9) gives two values \u200b\u200bof the constant FROMsatisfying inequality (4.5):
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The obtained result is illustrated by the table, which shows the series of the moment for three values \u200b\u200bof C: 0, 1/2 and 1/4. You can clearly see that in the last line ( n\u003d 1/4) torque value for positive and negative values ndiffers in absolute value.
Bohr was able to agree with the experimental data by setting the constant C equal to zero. Then the orbital angular momentum quantization rule is described by formulas (1). But also makes sense and meaning Cequal to half. It describes inner momentelectron, or its spin - a concept that will be discussed in detail in other chapters. The planetary model of the atom is often presented starting with formula (1), but historically it was derived from the principle of correspondence.
13.5. Electron orbital parameters
Formulas (1.1) and (3.7) lead to a discrete set of orbital radii and electron velocities, which can be renumbered using the quantum number n:
A discrete energy spectrum corresponds to them. Total energy of an electron E n can be calculated by formulas (3.5) and (5.1):
.
We have obtained a discrete set of energy states of a hydrogen atom or a hydrogen-like ion. State corresponding to value nequal to one is called main,other - excited and if n
very large, then - very agitated. Figure 13.5.1 illustrates formula (5.2) for a hydrogen atom. Dotted line 
the ionization limit is indicated. It is clearly seen that the first excited level is much closer to the ionization limit than to the ground level.
condition. Approaching the ionization limit, the levels in Fig. 13.5.2 are gradually condensed. 
Only a solitary atom has infinitely many levels. In a real environment, various interactions with neighboring particles lead to the fact that the atom has only a finite number of lower levels. For example, in stellar atmospheres, an atom usually has 20-30 states, but in a rarefied interstellar gas hundreds of levels can be observed, but not more than a thousand.
In the first chapter, we introduced Rydberg for dimensional considerations. Formula (5.2) reveals the physical meaning of this constant as a convenient unit for measuring the energy of an atom. She also reveals that Ry is dependent on attitude 
:
.
Due to the large difference between the masses of the nucleus and the electron, this dependence is very weak, but in some cases it cannot be neglected. The numerator of the last formula contains a constant
erg 
eV,
to which the value of Ry tends with an unlimited increase in the mass of the nucleus. Thus, we have specified the Ry unit given in the first chapter.
The moment quantization rule (1.1), of course, is less precise than expression (12.6.1) for the eigenvalue of the operator 
... Accordingly, formulas (3.6) - (3.7) have a very limited meaning. Nevertheless, as we will see below, the final result (5.2) for energy levels coincides with the solution of the Schrödinger equation. It can be used in all cases if the relativistic corrections are negligible.
So, according to the planetary model of the atom, in bound states, the speed of rotation, the radius of the orbit and the energy of the electron take on a discrete series of values \u200b\u200band are completely determined by the magnitude of the main quantum number. States with positive energy are called free; they are not quantized, and all the parameters of the electron in them, except for the angular momentum, can take on any values \u200b\u200bthat do not contradict the conservation laws. The moment of rotation is always quantized.
The planetary model formulas allow calculating the ionization potential of a hydrogen atom or hydrogen-like ion, as well as the wavelength of the transition between states with different values n. You can also estimate the size of the atom, the linear and angular velocities of an electron in its orbit.
The derived formulas have two limitations. First, they do not take into account relativistic effects, which gives an order error ( V/c) 2. The relativistic correction increases with increasing nuclear charge as Z 4 and for the FeXXVI ion is already fractions of a percent. We'll look at this effect at the end of this chapter while staying within the planetary model. Second, in addition to the quantum number n the energy of the levels is determined by other parameters - the orbital and internal moments of the electron. Therefore, the levels are split into several sublevels. The splitting value is also proportional to Z 4 and becomes noticeable for heavy ions.
All features of discrete levels are taken into account in a consistent quantum theory. Nevertheless, Bohr's simple theory turns out to be a simple, convenient, and fairly accurate method for studying the structure of ions and atoms.
13.6 Rydberg's constant
In the optical range of the spectrum, it is usually not the quantum energy that is measured E, and the wavelength is the transition between levels. Therefore, the wavenumber is often used to measure the level energy E / hc, measured in inverse centimeters. Wave number corresponding 
, denoted 
:
cm 
.
Index reminds that the mass of the nucleus in this definition is considered to be infinitely large. Taking into account the finite mass of the nucleus, the Rydberg constant is
.
Heavy nuclei have more energy than light ones. The ratio of the masses of a proton and an electron is
Substituting this value in (2.2), we obtain a numerical expression for the Rydberg constant for the hydrogen atom:
The nucleus of the heavy hydrogen isotope, deuterium, consists of a proton and a neutron, and is approximately twice as heavy as the nucleus of a hydrogen atom, a proton. Therefore, according to (6.2), the Rydberg constant for deuterium R D is greater than hydrogen R H:
It is even higher in the unstable isotope of hydrogen - tritium, the nucleus of which consists of a proton and two neutrons.
In the elements in the middle of the periodic table, the effect of the isotope shift competes with the effect associated with the finite size of the nucleus. These effects have the opposite sign and cancel each other out for elements close to calcium.
13.7. Isoelectronic sequence of hydrogen
According to the definition given in the fourth section of the seventh chapter, ions consisting of a nucleus and one electron are called hydrogen-like. In other words, they refer to the isoelectronic sequence of hydrogen. Their structure qualitatively resembles a hydrogen atom, and the position of the energy levels of ions, the nuclear charge of which is not too large ( Z Z\u003e 20) there are quantitative differences associated with relativistic effects: the dependence of the electron mass on the velocity and spin – orbit interaction.
We will consider the most interesting in astrophysics ions of helium, oxygen and iron. In spectroscopy, the charge of an ion is specified using spectroscopic symbol, which is written in Roman numerals to the right of the chemical element symbol. The number represented by the Roman numeral is one more than the number of electrons removed from the atom. For example, the hydrogen atom is denoted as HI, and the hydrogen-like ions of helium, oxygen and iron, respectively, HeII, OVIII and FeXXVI. For multielectron ions, the spectroscopic symbol coincides with the effective charge that the valence electron "feels".
Let's calculate the motion of an electron in a circular orbit, taking into account the relativistic dependence of its mass on speed. Equations (3.1) and (1.1) in the relativistic case are as follows:
Reduced mass m defined by formula (2.6). We also recall that
.
Multiply the first equation by 
and divide it by the second. As a result, we get
The fine structure constant was introduced in formula (2.2.1) of the first chapter. Knowing the speed, we calculate the radius of the orbit:
.
In the special theory of relativity, kinetic energy is equal to the difference between the total energy of the body and its rest energy in the absence of an external force field:
.
Potential energy U as a function r is defined by formula (3.3). Substituting into expressions for T and U the obtained values \u200b\u200b and r, we get the total energy of the electron:
For an electron rotating in the first orbit of a hydrogen-like iron ion, the value of 2 is 0.04. For lighter elements, it is, accordingly, even less. When 
the decomposition is valid
.
The first term, as it is easy to verify, is equal to the value of energy (5.2) in Bohr's nonrelativistic theory up to notation, and the second is the required relativistic correction. Let us denote the first term as E B then
Let us write out explicitly the expression for the relativistic correction:
So, the relative value of the relativistic correction is proportional to the product 2 Z 4 . Taking into account the dependence of the electron mass on the velocity leads to an increase in the depth of the levels. This can be understood as follows: the absolute value of the energy grows with the mass of the particle, and the moving electron is heavier than the stationary one. Weakening of the effect with increasing quantum number n is a consequence of the slower motion of an electron in an excited state. Strong dependence on Z is a consequence of the high speed of an electron in the field of a nucleus with a large charge. In the future, we will calculate this value according to the rules of quantum mechanics and obtain a new result - the removal of the degeneracy in the orbital angular momentum.
13.8. Highly excited states
The states of an atom or ion of any chemical element in which one of the electrons is at a high energy level are called highly excited, or rydberg.They have an important property: the position of the levels of an excited electron can be described with a sufficiently high accuracy within the framework of the Bohr model. The point is that an electron with a large quantum number n, according to (5.1), is very far from the nucleus and other electrons. Such an electron in spectroscopy is usually called "optical" or "valence", and the rest of the electrons, together with the nucleus - "atomic residue". The structure of an atom with one highly excited electron is shown schematically in Fig. 13.8.1. At the bottom left is an atomic
the remainder: the nucleus and electrons in the ground state. A dotted arrow indicates a valence electron. The distances between all the electrons inside the atomic remnant are much less than the distance from any of them to the optical electron. Therefore, their total charge can be considered almost completely concentrated in the center. Therefore, it can be assumed that the optical electron moves under the action of the Coulomb force directed towards the nucleus, and, thus, its energy levels are calculated by the Bohr formula (5.2). The electrons of the atomic remnant screen the nucleus, but not completely. To take into account partial shielding, the concept was introduced effective chargeatomic residue Z eff. In the considered case of a strongly distant electron, the quantity Z eff is equal to the difference of the atomic number of a chemical element Z and the number of electrons of the atomic residue. Here we restrict ourselves to the case of neutral atoms, for which Z eff \u003d 1.
The position of strongly excited levels is obtained in Bohr's theory for any atom. It suffices to replace in (2.6) 
on the mass of the atomic residue 
which is less than the mass of an atom 
by the value of the electron mass. Using the identity obtained from this
we can express the Rydberg constant as a function of the atomic weight Aof the considered chemical element:
planetary modelatom ... + --- a - \u003d 0; (2.12) h² h ∂t 4πm ∂а а Δβ + 2 (grad аgradβ) - ----- \u003d 0. (2. 13 ) h ∂t For βh φ \u003d - (2.14) 2πm Madelung obtained the equation ...
Chapter 1 nucleons and atomic nuclei
DocumentWill be shown in chapter 8, magnetic ... by Rutherford in 1911 planetarymodelatom, Dutch scientist A. Wang ... have a really increased levelenergy... Nuclei with neutron ... cellulose contains 13 atoms oxygen, 34 atom hydrogen and 3 atom carbon, ...
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Basic educational programEnhancement level qualifications, competencies and level payment ... GIA: 46 46 13 20 13 - 39 7 ... Poem "Vasily Terkin" ( chapters). M.A. Sholokhov Story ... Planetarymodelatom... Optical spectra. Light absorption and emission atoms... The composition of the atomic nucleus. Energy ...
Chapter 4 Differentiation and Self-Organization of Primary Cosmic Baryonic Matter
DocumentQuantity atoms at 106 atoms silicon, ... measure ( level) energy; ... Galimov dynamic model explains well ... 4.2.12-4.2. 13 the relationship is presented ... interconnected planetary system ... the analysis algorithm is presented in chapters 2 and 4. How ...
Moscow State University of Economics, Statistics, Informatics
Discipline abstract: "KSE"
on the topic of :
"Planetary model of the atom"
Completed:
3rd year student
DNF-301 groups
Ruziev Temur
Teacher:
Mosolov D.N.
Moscow 2008
In Dalton's first atomic theory, it was assumed that the world consists of a certain number of atoms - elementary bricks - with characteristic properties that are eternal and unchanging.
These ideas changed drastically after the discovery of the electron. All atoms must contain electrons. But how are the electrons located in them? Physicists could only philosophize based on their knowledge in the field of classical physics, and gradually all points of view agreed on one model proposed by J.J. Thomson. According to this model, an atom consists of a positively charged substance, inside which electrons are interspersed (perhaps they are in intense motion), so that the atom resembles a raisin pudding. Thomson's model of the atom could not be directly verified, but all sorts of analogies testified in its favor.
In 1903, the German physicist Philip Lenard proposed a model of an "empty" atom, inside of which some unknown neutral particles, composed of mutually balanced positive and negative charges, "fly". Lenard even gave a name to his nonexistent particles - dynamids, but the only one whose existence was proved by rigorous, simple and beautiful experiments was Rutherford's model.
The enormous scope of Rutherford's scientific work in Montreal - he published 66 articles personally and jointly with other scientists, not counting the book "Radioactivity" - brought Rutherford the fame of a first-class researcher. He receives an invitation to take a chair in Manchester. On May 24, 1907, Rutherford returned to Europe. A new period of his life began.
The first attempt to create a model of the atom based on the accumulated experimental data belongs to J. Thomson (1903). He believed that the atom is an electrically neutral spherical system with a radius of approximately 10-10 m. The positive charge of an atom is evenly distributed throughout the entire volume of the sphere, and negatively charged electrons are inside it. To explain the line emission spectra of atoms, Thomson tried to determine the arrangement of electrons in an atom and calculate the frequencies of their vibrations around equilibrium positions. However, these attempts were unsuccessful. A few years later, in the experiments of the great English physicist E. Rutherford, it was proved that Thomson's model is incorrect.
English physicist E. Rutherford investigated the nature of this radiation. It turned out that the beam of radioactive radiation in a strong magnetic field was divided into three parts: a-, b- and y-radiation. b-rays are a stream of electrons, a-rays are the nucleus of a helium atom, y-rays are shortwave electromagnetic radiation. The phenomenon of natural radioactivity indicates the complex structure of the atom.
In Rutherford's experiments to study the internal structure of an atom, a gold foil was irradiated with a-particles passing through slits in lead screens at a speed of 107 m / s. a-Particles emitted by a radioactive source are the nuclei of a helium atom. After interacting with the foil atoms, the alpha particles fell on screens covered with a layer of zinc sulfide. When the alpha particles hit the screens, they produced faint flashes of light. The number of flashes was used to determine the number of particles scattered by the foil at certain angles. The counting showed that most of the a-particles pass through the foil without hindrance. However, some a-particles (one in 20,000) deviated sharply from the original direction. Collision of an a-particle with an electron cannot change its trajectory so significantly, since the electron mass is 7350 times less than the a-particle mass.
Rutherford suggested that the reflection of a-particles is due to their repulsion by positively charged particles with masses commensurate with the mass of the a-particle. Based on the results of such experiments, Rutherford proposed a model of the atom: a positively charged atomic nucleus is located in the center of the atom, around which (like planets revolving around the Sun) negatively charged electrons rotate under the action of electric forces of attraction. The atom is electrically neutral: the charge of the nucleus is equal to the total charge of the electrons. The linear size of a nucleus is at least 10,000 times smaller than the size of an atom. This is Rutherford's planetary model of the atom. What keeps an electron from falling on a massive nucleus? Of course, a quick spin around it. But in the process of rotation with acceleration in the field of the nucleus, the electron must radiate part of its energy in all directions and, gradually slowing down, still fall onto the nucleus. This thought did not give rest to the authors of the planetary model of the atom. The next obstacle on the way of a new physical model, it seemed, was to destroy the whole picture of atomic structure, built with such difficulty and proven by clear experiments ...
Rutherford was confident that a solution would be found, but he could not imagine that it would happen so soon. The defect in the planetary atomic model will be corrected by the Danish physicist Niels Bohr. Bohr painfully pondered Rutherford's model and looked for convincing explanations for what obviously happens in nature, despite all doubts: electrons, without falling on the nucleus and not flying away from it, constantly revolve around their nucleus
In 1913, Niels Bohr published the results of lengthy reflections and calculations, the most important of which have since become known as Bohr's postulates: in an atom there is always a large number of stable and strictly defined orbits along which an electron can rush for an infinitely long time, for all the forces acting on it , turn out to be balanced; an electron can pass in an atom only from one stable orbit to another, equally stable. If during such a transition the electron moves away from the nucleus, then it is necessary to communicate to it from the outside a certain amount of energy equal to the difference in the energy supply of the electron in the upper and lower orbits. If an electron approaches the nucleus, then it "dumps" excess energy in the form of radiation ...
Probably, Bohr's postulates would have taken a modest place among a number of interesting explanations of new physical facts obtained by Rutherford, if not for one important circumstance. Bohr, using the relationships he found, was able to calculate the radii of the "allowed" orbits for an electron in a hydrogen atom. Bohr suggested that the quantities characterizing the microworld should quantize
, i.e. they can only take on certain discrete values.
The laws of the microworld are quantum laws!
These laws have not yet been established by science at the beginning of the 20th century. Bohr formulated them in the form of three postulates. complementing (and "saving") Rutherford's atom.
First postulate:
Atoms have a number of stationary states corresponding to certain energies: E 1, E 2 ... E n. Being in a stationary state, an atom does not emit energy, despite the movement of electrons.
Second postulate:
In a stationary state of an atom, electrons move along stationary orbits, for which the quantum relation is fulfilled:
m V r \u003d n h / 2 p (1)
where m V r \u003d L is the angular momentum, n \u003d 1,2,3 ..., h is Planck's constant.
Third postulate:
Radiation or absorption of energy by an atom occurs during its transition from one stationary state to another. In this case, a portion of energy is emitted or absorbed ( quantum
), equal to the difference between the energies of stationary states between which the transition occurs: e \u003d h u \u003d E m -E n (2)
1.from the ground stationary state to the excited state,
2.from an excited stationary state to the ground state.
Bohr's postulates contradict the laws of classical physics. They express a characteristic feature of the microworld - the quantum nature of the phenomena taking place there. The conclusions based on Bohr's postulates are in good agreement with experiment. For example, they explain the regularities in the spectrum of the hydrogen atom, the origin of the characteristic spectra of X-rays, etc. In fig. 3 shows a part of the energy diagram of stationary states of the hydrogen atom. 
The arrows indicate the transitions of the atom leading to the emission of energy. It can be seen that the spectral lines are combined in series, differing in the level to which the atom transitions from other (higher) ones.
Knowing the difference between the energies of the electron in these orbits, it was possible to construct a curve describing the emission spectrum of hydrogen in various excited states and to determine which wavelength the hydrogen atom should emit especially willingly if excess energy was supplied to it from the outside, for example, with the help of bright light of mercury lamps. This theoretical curve completely coincided with the emission spectrum of excited hydrogen atoms measured by the Swiss scientist J. Balmer back in 1885!
Used Books:
- A. K. Shevelev “Structure of nuclei, particles, vacuum (2003)
- A. V. Blagov "Atoms and Nuclei" (2004)
- http://e-science.ru/- portal of natural sciences
