In the planetary model of the atom, it is assumed that. Planetary model of the atom
Moscow State University Economics Statistics Informatics
Abstract on discipline: "KSE"
on the topic of :
"Planetary Model of the Atom"
Completed:
3rd year student
Groups DNF-301
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 building blocks - with characteristic properties, eternal and unchanging.
These ideas changed drastically after the discovery of the electron. All atoms must contain electrons. But how are the electrons arranged in them? Physicists could only philosophize on the basis of their knowledge of classical physics, and gradually all points of view converged on one model proposed by J.J. Thomson. According to this model, an atom is made up of a positively charged substance with electrons interspersed (perhaps in a lot of motion) so that the atom resembles a pudding with raisins. Thomson's model of the atom could not be directly tested, but all sorts of analogies testified in its favor.
In 1903, the German physicist Philipp Lenard proposed a model of an “empty” atom, inside which some neutral particles that were not discovered by anyone “fly”, composed of mutually balanced positive and negative charges. Lenard even gave a name for his non-existent particles - dynamides. However, the only one whose right to exist was proved by strict, simple and beautiful experiments was Rutherford's model.
Huge scope scientific work Rutherford in Montreal - he published 66 articles, both personally and jointly with other scientists, not counting the book "Radioactivity", brought Rutherford fame as a first-class researcher. He receives an invitation to take the 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 the atom is evenly distributed throughout the volume of the ball, and the negatively charged electrons are inside it. To explain the line emission spectra of atoms, Thomson tried to determine the location of electrons in an atom and calculate the frequencies of their oscillations around equilibrium positions. However, these attempts were not successful. A few years later, in the experiments of the great English physicist E. Rutherford, it was proved that the Thomson model was incorrect.
The English physicist E. Rutherford investigated the nature of this radiation. It turned out that a 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 short-wave electromagnetic radiation. The phenomenon of natural radioactivity indicates the complex structure of the atom.
In Rutherford's experiments to study the internal structure of the atom, gold foil was irradiated with alpha particles passing through slots in lead screens at a speed of 107 m/s. a-Particles emitted by a radioactive source are the nuclei of the helium atom. After interacting with foil atoms, the a-particles fell on screens coated with a layer of zinc sulfide. Hitting the screens, a-particles caused weak flashes of light. The number of flashes was used to determine the number of particles scattered by the foil at certain angles. The calculation showed that most of the o-particles pass through the foil without hindrance. However, some α-particles (one of 20,000) deviated sharply from their original direction. The collision of an α-particle with an electron cannot change its trajectory so significantly, since the mass of an electron is 7350 times less than the mass of an α-particle.
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 this kind of experiments, Rutherford proposed a model of the atom: in the center of the atom there is a positively charged atomic nucleus, around which (like planets revolving around the Sun) negatively charged electrons revolve under the action of electric forces of attraction. An atom is electrically neutral: the charge of the nucleus is equal to the total charge of the electrons. The linear size of the 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 onto a massive nucleus? Of course, the rapid rotation 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 decelerating, nevertheless fall onto the nucleus. This thought haunted the authors of the planetary model of the atom. The next obstacle on the way of the new physical model, it seemed, was to destroy the entire picture of the atomic structure, constructed with such difficulty and proven by clear experiments...
Rutherford was sure that a solution would be found, but he could not imagine that it would happen so soon. The defect in the planetary model of the atom will be corrected by the Danish physicist Niels Bohr. Bohr agonized over Rutherford's model and looked for convincing explanations of what obviously happens in nature despite all doubts: electrons, without falling on the nucleus and without 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 the atom there always exists big number stable and strictly defined orbits, along which an electron can rush indefinitely, because all the forces acting on it turn out to be balanced; An electron can move in an atom only from one stable orbit to another equally stable one. If, during such a transition, the electron moves away from the nucleus, then it is necessary to impart to it from the outside a certain amount of energy equal to the difference in the energy reserve of the electron in the upper and lower orbits. If an electron approaches the nucleus, then it “discards” excess energy in the form of radiation ...
Probably, Bohr's postulates would take a modest place among a number of interesting explanations of new physical facts mined by Rutherford House, if not for one important circumstance. Bohr, using the relationships he found, was able to calculate the radii of "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 certain discrete values.
The laws of the microworld are quantum laws!
These laws at the beginning of the 20th century had not yet been established by science. 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 energy values: E 1 , E 2 ...E n . Being in a stationary state, an atom does not radiate energy, despite the movement of electrons.
Second postulate:
In the stationary state of an atom, electrons move along stationary orbits, for which the quantum relation is satisfied:
m V r=n h/2 p (1)
where m V r =L - angular momentum, n=1,2,3..., h- Planck's constant.
Third postulate:
The emission or absorption of energy by an atom occurs when it passes from one stationary state to another. In this case, a portion of energy is emitted or absorbed ( quantum
) equal to the energy difference of the stationary states between which the transition occurs: e = h u = E m -E n (2)
1. from the main stationary state to an excited one,
2. from the excited stationary state to the ground state.
Bohr's postulates contradict the laws of classical physics. They express salient feature microworld - the quantum nature of the phenomena occurring 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. On fig. 3 shows part of the energy diagram of the stationary states of the hydrogen atom. 
The arrows show the transitions of the atom, leading to the emission of energy. It can be seen that the spectral lines are combined into series, which differ in the level to which the transition of the atom occurs from other (higher) ones.
Knowing the difference between the energies of an electron in these orbits, it was possible to construct a curve describing the radiation spectrum of hydrogen in various excited states and to determine what wavelength the hydrogen atom should especially readily emit if excess energy is supplied to it from the outside, for example, with the help of bright mercury light. 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
Read also:
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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 solids 2) gas ionization
3) chemical production new substances 4) scattering of α-particles
24. The planetary model of the atom is justified
1) movement calculations 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 scatter
1) electrostatic field of the atomic nucleus 2) electron shell of target atoms
3) gravitational field of the nucleus of an atom 4) target surface
48. In Rutherford's experiment most ofα-particles freely pass through the foil, practically without deviating from rectilinear trajectories, because
1) the nucleus of an atom has a positive charge
2) electrons have a negative charge
3) the nucleus of an atom has small (compared to the atom) dimensions
4) α-particles have a large (compared to the nuclei of atoms) 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. E. Rutherford's experiments on the scattering of α-particles showed that
A. almost the entire mass of the atom is concentrated in the nucleus. B. the nucleus has a positive charge.
Which statement(s) is(are) correct?
1) only A 2) only B 3) both A and B 4) neither A nor B
259. What idea of the structure of the atom corresponds to Rutherford's model of the atom?
1) The nucleus is in the center of the atom, the electrons are in orbits around the nucleus, the charge of the electrons is positive.
2) The nucleus is in the center of the atom, the electrons are in orbits around the nucleus, the charge of the electrons is negative.
3) The positive charge is evenly distributed over the atom, the electrons in the atom oscillate.
4) The positive charge is evenly distributed throughout the atom, and the electrons move in the atom in different orbits.
266. Which idea of the structure of the atom is correct? Most of the mass of an atom is concentrated
1) in the nucleus, the charge of electrons is positive 2) in the nucleus, the charge of the nucleus 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 the structure of the atom corresponds to Rutherford's model of the atom?
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 scheme 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, 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) 0.3 eV only 3) 1.5 eV only 4) any between 0 and 0.5 eV
273. The figure shows a diagram of the lowest energy levels of an 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 an 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 Bohr model of the atom?
1) energy difference of stationary states 2) frequency of electron revolution around the nucleus
3) the de Broglie wavelength for an electron 4) the Bohr model does not allow to determine it
15. An 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 emit hydrogen atoms in the second excited state?
1) 1 2) 2 3) 3 4) 4
25. Suppose that the energy of gas atoms can only take on the values indicated in the diagram. Atoms are in a state with energy e (3). Photons of what energy can this 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 emitting a photon with an energy of 6 eV, 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 in 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 one . Then it passes from the state E 2 to a stationary state with energy E s, absorbing a photon with a frequency v 2 > v one . What happens during the transition of an electron from the state E 2 to the state E 1.
1) light emission frequency v 2 – v 1 2) light absorption frequency v 2 – v 1
3) light emission frequency v 2 + v 1 4) light absorption 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 (h - 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 level E 4 to level E 1 if λ 13 = 400 nm, λ 24 = 500 nm, λ 32 = 600 nm? Express your answer in nm, and round to the nearest whole number.
96. The figure shows several energy levels electron shell atom and the frequencies of photons emitted and absorbed during transitions between these levels are indicated. What is the minimum wavelength of photons emitted by an atom when any
possible transitions between levels E 1, E 2, e s and E 4, if v 13 \u003d 7 ∙ 10 14 Hz, v 24 = 5 ∙ 10 14 Hz, v 32 = 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 an atom. Which of the transitions between energy levels marked with arrows is accompanied by the absorption of a minimum frequency quantum?
1) from level 1 to level 5 2) from level 1 to level 2
124. 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. It has been experimentally established that the minimum wavelength for photons emitted during transitions between these levels is λ 0 = 250 nm. What is the value of λ 13 if λ 32 = 545 nm, λ 24 = 400 nm?
145. The figure shows a diagram of the possible values of the energy of rarefied gas atoms. At the initial moment of time, the atoms are in a state with energy E (3) . It is possible for the gas to emit 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 = - 13.6/n 2 eV, where n = 1, 2, 3, ... . During the transition of an atom from the state E 2 to the state E 1, the atom emits a photon. Once on the surface of the photocathode, a photon knocks out a photoelectron. The wavelength of light corresponding to the red border of the photoelectric effect for the surface material of the photocathode, λcr = 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 photon with an energy of 3.4 eV?
1) yes, while the atom goes into the state E 2
2) yes, while the atom goes 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 the atom to an excited state
218. The figure shows a simplified diagram of the energy levels of an atom. Numbered arrows mark some possible transitions of the atom between these levels. Establish a correspondence between the processes of absorption of light of the largest wavelength and emission of light of the largest 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 diagram of the energy levels of the atom. 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 momentum p = 2 ∙ 10 -24 kg ∙ m/s collides with a proton at rest, forming a hydrogen atom in a state with energy E n (n = 2). During the formation of an atom, a photon is emitted. Find the frequency v 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 =1,2, 3, ....
260. The scheme of the lowest energy levels of an 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 0.5 to 2 eV
269. The figure shows a diagram of the energy levels of an atom. What number indicates the transition that corresponds to radiation photon with the lowest energy?
1) 1 2) 2 3) 3 4) 4
282. Emission of a photon by an atom occurs when
1) the movement of an electron in a stationary orbit
2) the transition of an electron from the ground state to an excited one
3) the transition of an electron from an excited state to the ground
4) all listed processes
13. Photon emission occurs during the transition from excited states with energies E 1 > E 2 > E 3 to the ground state. For the frequencies of the corresponding photons v 1 , v 2 , v 3 , the 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) greater or less than zero depending on the state
98. An atom at rest 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 scheme of energy levels of 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 moving with a kinetic energy of 1.5 eV collided with one of these atoms and bounced off, acquiring 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 in a collision with an electron is neglected.
111. Suppose that the scheme of energy levels of 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 colliding with one of these atoms bounced off, acquiring some additional energy. The momentum of an electron after a collision with a resting atom turned out to be equal to 1.2 ∙ 10 -24 kg ∙ m/s. Determine the kinetic energy of the electron before the collision. The possibility of emission of light by an atom in a collision with an electron is neglected.
136. A π°-meson with a mass of 2.4 ∙ 10 -28 kg decays into two γ-quanta. Find the momentum modulus of one of the resulting γ -quanta in the reference frame where the primary π ° meson is at rest.
144. There is rarefied atomic hydrogen in a vessel. The hydrogen atom in the ground state (E 1 = - 13.6 eV) absorbs a photon and is ionized. An electron escaping from an atom as a result of ionization moves away from the nucleus at a speed v = 1000 km/s. What is the frequency of the absorbed photon? Neglect 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 in vacuum with a wavelength λ \u003d 80 nm. With what speed does an electron flown out of an atom as a result of ionization move away from the nucleus? Neglect 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 pion motion, and the other in the opposite direction. The energy of one quantum is 10% more than the other. What is the speed of the pion before decay?
232. The table shows the energy values for 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 during 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 the 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. Consider the speed of an atom to be small compared to the speed of light.
252. A vessel contains rarefied atomic hydrogen. The hydrogen atom in the ground state (E 1 = -13.6 eV) absorbs a photon and is ionized. An electron escaping 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 an absorbed photon? Neglect the energy of thermal motion of hydrogen atoms.
1) 46 nm 2) 64 nm 3) 75 nm 4) 91 nm
257. A vessel contains rarefied atomic hydrogen. The hydrogen atom in the ground state (E 1 = -13.6 eV) absorbs a photon and is ionized. An electron escaping from an atom as a result of ionization moves away from the nucleus at a speed v = 1000 km/s. What is the energy of the absorbed photon? Neglect the energy of thermal motion of hydrogen atoms.
1) 13.6 eV 2) 16.4 eV 3) 19.3 eV 4) 27.2 eV
1 | | | |
The stability of any system on an atomic scale follows from the Heisenberg uncertainty principle (fourth section of the seventh chapter). Therefore, a consistent study of the properties of an atom is possible only within the framework of quantum theory. However, some important results practical value, can also be obtained within classical mechanics, by adopting additional orbit quantization rules.
In this chapter, we will calculate the position of the energy levels of the hydrogen atom and hydrogen-like ions. The calculation is based on the planetary model, according to which electrons revolve around the nucleus under the influence of the Coulomb attraction forces. We assume that electrons move in circular orbits.
13.1. Conformity principle
Angular momentum quantization is used in the model of the hydrogen atom proposed by Bohr in 1913. Bohr proceeded from the fact that, in the limit of small energy quanta, the results of quantum theory should correspond to the conclusions of classical mechanics. He formulated three postulates.
An atom can exist for a long time only in certain states with discrete energy levels. E i . Electrons, rotating in the corresponding discrete orbits, move with acceleration, but, nevertheless, they do not radiate. (In classical electrodynamics, any accelerated particle radiates if it has a non-zero charge).
Radiation comes out or is absorbed by quanta during the transition between energy levels:
From these postulates follows the rule of quantization of the moment of rotation of the electron
,
where n can be equal to any natural number:
Parameter n called principal quantum number. To derive formulas (1.1), we express the level energy in terms of the moment of rotation. Astronomical measurements require knowledge of wavelengths with a sufficiently high accuracy: six correct digits 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 digit. the motion of the nucleus must be taken into account. To take it into account, the concept reduced mass.
13.2. Reduced mass
An electron moves around the nucleus under the influence of an electrostatic force
,
where r- a vector, the beginning of which coincides with the position of the nucleus, and the end points to the electron. Recall that Z is the atomic number of the nucleus, and the charges of the nucleus and electron are equal, respectively Ze and 
. According to Newton's third law, a force acts on the nucleus equal to - f(it is equal in absolute value and directed opposite to the force acting on the electron). Let us write down the equations of electron motion
.
We introduce new variables: the speed of an electron relative to the nucleus
and the speed of the center of mass
.
Adding (2.2a) and (2.2b), we get
.
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 electron velocity:
.
The quantity included in it
called reduced mass. Thus, the task of joint movement two particles - an electron and a nucleus - is simplified. It suffices to consider the motion around the nucleus of one particle, 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. 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 torque.
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 this 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 stable states. The states of atoms and ions with negative energy are called related. Multiplying equation (3.4) by 2 r and replacing the product on the left side mVr at the moment of rotation M, let's express the speed V in a moment:
.
Substituting the obtained velocity value into (3.5), we obtain the desired formula for the total energy:
.
Note that the energy is proportional to an even power 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 carried out on the basis of Bohr's postulates. Differentiating formula (3.5), we obtain a connection between small changes in momentum and energy:
.
According to the third postulate, the frequency of the emitted (or absorbed) photon is equal to the frequency of the electron in orbit:
.
From formulas (3.4), (4.2) and the relation
between the speed, torque and radius follows a simple expression for the change in the angular momentum during the transition of an electron between adjacent orbits:
.
Integrating (4.3), we obtain
Constant C we will search in a semi-open interval
.
The double inequality (4.5) introduces no additional restrictions: if WITH goes beyond (4.5), then it can be returned to this interval by simply renumbering the moment values in formula (4.4).
Physical laws are the same in all frames of reference. Let's move from a right-handed coordinate system to a left-handed one. Energy, like any scalar quantity, will remain the same,
.
The axial torque vector behaves differently. As is known, any axial vector changes sign when performing the specified operation:
There is no contradiction between (4.6) and (4.7), since, according to (3.7), the energy is inversely proportional to the square of the moment and remains the same when changing sign M.
Thus, the set of negative torque values must repeat the set of its positive values. In other words, for every 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 linear equation for WITH:
,
with a solution
.
It is easy to see that formula (4.9) gives two values of the constant WITH satisfying inequality (4.5):
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The result is illustrated by a table that shows the series of the moment for three values of C: 0, 1/2 and 1/4. It is clearly seen that in the last line ( n=1/4) torque value for positive and negative values n differs in absolute value.
Bohr managed to obtain agreement with the experimental data by setting the constant C equal to zero. Then the orbital momentum quantization rule is described by formulas (1). But it also makes sense C equal to half. It describes internal moment electron, or spin- a concept that will be discussed in detail in other chapters. The planetary model of the atom is often stated starting with formula (1), but historically it was derived from the correspondence principle.
13.5. Electron Orbit 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:
They correspond to a discrete energy spectrum. Total electron energy 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 a value n equal to one is called basic, other - excited and if n
very large, then - very excited. Figure 13.5.1 illustrates formula (5.2) for the hydrogen atom. dotted line 
the ionization limit is indicated. It is clearly seen that the first excited level is much closer to the ionization boundary than to the ground level.
condition. Approaching the ionization boundary, the levels in Fig. 13.5.2 gradually thicken. 
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, under the conditions of stellar atmospheres, an atom usually has 20–30 states, but hundreds of levels, but not more than a thousand, can be observed in a rarefied interstellar gas.
In the first chapter, we introduced a rydberg based on dimensional considerations. Formula (5.2) reveals physical meaning this constant as a convenient unit for measuring the energy of an atom. Moreover, it shows that Ry depends on the relation 
:
.
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 is the constant
erg 
eV,
to which the value of Ry tends with an unlimited increase in the mass of the nucleus. Thus, we have refined the unit of measurement Ry given in the first chapter.
The momentum quantization rule (1.1) is, of course, less precise than expression (12.6.1) for eigenvalue operator 
. Accordingly, formulas (3.6) - (3.7) have a very limited meaning. Nevertheless, as we shall see below, the final result (5.2) for the 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 rotation speed, the radius of the orbit and the energy of the electron take on a discrete series of values and are completely determined by the value of the principal quantum number. States with positive energy are called free; they are not quantized, and all electron parameters in them, except for the moment of rotation, can take on any values that do not contradict the laws of conservation. Torque is always quantized.
The formulas of the planetary model make it possible to calculate the ionization potential of a hydrogen atom or a hydrogen-like ion, as well as the wavelength of the transition between states with different values n. One can also estimate the size of an atom, linear and angular velocity movement of an electron in an 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 as the nuclear charge increases as Z 4 and for the FeXXVI ion is already fractions of a percent. At the end of this chapter, we will consider this effect, remaining within the framework of the planetary model. Secondly, 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 amount of splitting is also proportional Z 4 and becomes noticeable in heavy ions.
All features of discrete levels are taken into account in consistent quantum theory. However, simple theory Bohr is a simple, convenient and sufficiently accurate method for studying the structure of ions and atoms.
13.6 Rydberg 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 reciprocal centimeters. Wave number corresponding to 
, denoted 
:
cm 
.
The index reminds us 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 equal to
.
In heavy nuclei, it is greater than in light ones. The mass ratio of the proton and electron is
Substituting this value into (2.2) we obtain the numerical expression for the Rydberg constant for the hydrogen atom:
The nucleus of the heavy isotope of hydrogen - 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 that of hydrogen R H:
It is even higher for the unstable isotope of hydrogen - tritium, the nucleus of which consists of a proton and two neutrons.
For elements in the middle of the periodic table, the isotopic shift effect competes with the effect associated with the finite size of the nucleus. These effects have the opposite sign and compensate each other 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 whose nuclear charge is not too large ( Z Z > 20), there appear quantitative differences associated with relativistic effects: the dependence of the electron mass on the velocity and the spin–orbit interaction.
We will consider the most interesting ions of helium, oxygen and iron in astrophysics. In spectroscopy, the charge of an ion is given by spectroscopic symbol, which is written in Roman numerals to the right of the symbol chemical element. The number represented by a Roman numeral is one more than the number of electrons removed from the atom. For example, a hydrogen atom is designated as HI, and the hydrogen-like ions of helium, oxygen, and iron, respectively, are HeII, OVIII, and FeXXVI. For multielectron ions, the spectroscopic symbol coincides with the effective charge that the valence electron "feels".
Let us calculate the motion of an electron in a circular orbit, taking into account the relativistic dependence of its mass on velocity. Equations (3.1) and (1.1) in the relativistic case look like this:
Reduced mass m is defined by formula (2.6). Recall also that
.
Multiply the first equation by 
and divide it by the second. As a result, we get
Constant fine structure is introduced in the formula (2.2.1) of the first chapter. Knowing the speed, we calculate the radius of the orbit:
.
In the special theory of relativity, the 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 determined by formula (3.3). Substituting into expressions for T and U the obtained values of 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 equal to 0.04. For lighter elements, it is, accordingly, even less. At 
fair decomposition
.
It is easy to see that the first term is, up to notation, equal to the energy value (5.2) in the nonrelativistic Bohr theory, and the second is the desired relativistic correction. We denote the first term as E B , then
Let us write out in explicit form the expression for the relativistic correction:
So, the relative value of the relativistic correction is proportional to the product 2 Z 4 . Accounting for the dependence of the electron mass on velocity leads to an increase in the level depth. This can be understood as follows: the absolute value of the energy grows with the mass of the particle, and a moving electron is heavier than a stationary one. Weakening of the effect with increasing quantum number n is a consequence of the slower motion of the electron in the 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 quantity according to the rules of quantum mechanics and obtain new result- removal of degeneracy in orbital 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 aroused, or Rydberg. They possess important property: the position of the levels of an excited electron can be described with a sufficiently high accuracy in terms of the Bohr model. The fact is that an electron with a large value of the quantum number n, according to (5.1), is very far from the nucleus and other electrons. In spectroscopy, such an electron is usually called "optical", or "valence", and the remaining electrons, together with the nucleus, are called "atomic residue". Schematically, the structure of an atom with one highly excited electron is shown in Fig. 13.8.1. At the bottom left is the atomic
remainder: nucleus and electrons in the ground state. The dotted arrow points to the valence electron. The distances between all electrons within an atomic residue are much less than the distance from any of them to an 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 using the Bohr formula (5.2). The electrons of the atomic residue shield the nucleus, but not completely. To take into account partial screening, the concept is introduced effective charge atomic residue Z eff . In the considered case of a strongly distant electron, the quantity Z eff is equal to the difference in the atomic number of the chemical element Z and the number of electrons in the atomic residue. Here we restrict ourselves to the case of neutral atoms, for which Z ff = 1.
The position of strongly excited levels is obtained in Bohr's theory for any atom. It suffices to replace in (2.6) 
per atomic mass 
, which is less than the mass of an atom 
by the mass of the electron. With the help of the identity obtained from here
we can express the Rydberg constant as a function of atomic weight A considered chemical element:
planetary modelsatom... + --- a -- = 0; (2.12) h² h ∂t 4πm ∂a a Δβ + 2(grad agradβ) – ----- = 0. (2. 13 ) h ∂t For βh φ = -- (2.14) 2πm Madelung obtained the equation...
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Main educational programRaise level qualifications, competencies and level payment... GIA: 46 46 13 20 13 - 39 7 ... The poem "Vasily Terkin" ( chapters). M.A. Sholokhov Story... planetarymodelatom. Optical spectra. Absorption and emission of light atoms. Compound atomic nucleus. Energy ...
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What's this? This is Rutherford's model of the atom. It is named after the New Zealand-born British physicist Ernest Rutherford, who in 1911 announced the discovery of the nucleus. In the course of his experiments on the scattering of alpha particles by thin metal foil, he found that most of the alpha particles passed directly through the foil, but some bounced off. Rutherford suggested that in the region of the small area from which they bounced, there is a positively charged nucleus. This observation led him to a description of the structure of the atom, which, with corrections for 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 around the nucleus. Therefore, the model is called planetary.
Before Rutherford, there was another model of the atom, the Thompson model of matter. It did not have a nucleus, it was a positively charged "cupcake" filled with "raisins" - electrons that freely rotated in it. By the way, it was Thompson who discovered electrons. V modern school when they begin to get acquainted with, they always start with this model.
Models of the atom by 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 there is quantum mechanics in the motion of an electron around the nucleus. However, the concept of an orbit has still remained in the theory of the structure of the atom. 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 went along the path that he outlined.
Why is this interesting for science? Rutherford's experiment discovered nuclei. But everything we know about them, we learned later. His theory has been developed 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 the nucleus, then it must radiate energy. We know that a body that moves in a circle at a constant speed is still accelerating because the velocity 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 all of it and fall onto the core. Therefore, the classical model of the atom is not fully consistent with itself.
Then physical theories began to appear that tried to overcome this contradiction. An important addition to the model of the structure of the atom was made by Niels Bohr. He discovered that around the atom there are several quantum orbits along which the electron moves. He suggested that the electron does not radiate energy all the time, but only when moving from one orbit to another.
Bohr model of the atom
// wikimedia.org
And after the Bohr model of the atom, the Heisenberg uncertainty principle appeared, which finally explained why the fall of an electron on the nucleus is impossible. Heisenberg discovered that in an excited atom, an electron is in distant orbits, and at the moment when it emits a photon, it falls into the main orbit, having lost its energy. The atom, on the other hand, goes into a stable state, in which the electron will rotate around the nucleus until nothing excites it from the outside. This is a stable state, beyond which the electron will not fall.
Due to the fact that the ground state of the atom is a stable state, matter exists, we all exist. Without quantum mechanics, we wouldn't have stable matter at all. In this sense, the main question that a non-specialist can ask quantum mechanics is why does everything not fall at all? Why doesn't all matter come together to a point? AND quantum mechanics able to answer this question.
Why know this? In a sense, Rutherford's experiment was repeated again in the discovery of quarks. Rutherford discovered that positive charges - protons - are concentrated in nuclei. What's inside protons? We now know that inside protons are quarks. We learned this by doing a similar experiment on deep inelastic scattering of electrons by protons 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 a collision, protons can remain protons, or they can be excited due to high energy, and then other particles, such as pi-mesons, can be born during the scattering of protons. It turned out that this cross section behaves as if there are point components inside the protons. Now we know that these point components are quarks. In a sense, it 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 see what they fall apart on. But this is the next step, so far this has not been done.
In addition, the most important plot from the history of Russian science is associated with the name of Rutherford. Pyotr 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 that 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.
