What is it? This is the model of the Rangeford atom. It is named after the British physics of the New Zealand origin of Ernest Rostford, who in 1911 announced the opening of the nucleus. In the course of its experiments on the scattering of alpha particles on a thin metal foil, he found that most alpha particles were directly passed through foil, but some bounced. Rutherford suggested that in the area of \u200b\u200bthe small area from which they bounced, there is a positively charged core. This observation led it to the description of the structure of the atom, which, with amendments to quantum theory, is accepted today. Similarly, the Earth rotates around the Sun, the electric charge of the atom focuses in the kernel, around which electrons rotate the opposite charge, and the electromagnetic field holds electrons in the orbit of the kernel. Therefore, the model is called the planetary.
Before Rutherford, there was another atom model - the model of substance Thompson. It did not have a nucleus, she represented a positively charged "cupcake" filled with "highlights" - electrons, which in it freely rotated. By the way, it is Thompson who opened electrons. In modern school, when they start to get acquainted with, always start with this model.
Models of the atom of Rutherford (left) and Thompson (right)
// wikimedia.org.
A quantum model that today describes the structure of an atom, of course, differs from the one that Rutherford came up with. There are no quantum mechanics in the movement around the Sun, and in the motion of the electron around the kernel it is. However, the concept of the orbit has so far 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, called such a model of the planetary became incorrect. Rutherford made the first step in the right direction, and the development of the theory of the building of the atom went along the way he planned.
What is it interesting for science? Reforda's experiment opened the kernel. But all that we know about them, we learned after. His theory developed for many decades, and it lies answers to fundamental questions about the structure of matter.
Paradoxes quickly discovered in the reforde model, namely: if the charged electron rotates around the kernel, then it should emit energy. We know that the body, which moves in a circle with a constant speed, is still accelerated, because the speed vector turns all the time. And if the charged particle moves with acceleration, it should emit energy. This means that it should almost instantly lose her all and fall on the core. Therefore, the classical atom model is not fully consistent with itself.
Then the physical theories began to appear, which tried to overcome this contradiction. An important addition in the model of the structure of the atom introduced Niels Bohr. He found that there are several quantum orbits around the atom for which the electron moves. He suggested that the electron emits the energy not all the time, but only moving from one orbit to another.
Model atom boron
// wikimedia.org.
And after the Borov model atom, the principle of the uncertainty of Heisenberg appeared, who finally explained why the electron drop on the kernel is impossible. Heisenberg found that an electron electron is in distant orbits in an excited atom, and at the moment when he emits a photon, it falls on the main orbit, losing its energy. The at the same time passes into a stable state, in which the electron will rotate around the kernel until it excites it outside. This is a stable state, further which the electron will not fall.
Due to the fact that the main state of the atom is a steady state, matter exists, we all exist. Without quantum mechanics, we would not have a steady matter at all. In this sense, the main question that a non-specialist can ask a quantum mechanic - why doesn't all fall at all? Why is all the substance not going to the point? And quantum mechanics can answer this question.
Why do you know? In some sense, Rutherford experiment repeated again when the quarks are opened. Rutherford discovered that positive charges - protons - focused in the nuclei. And what inside the protons? Now we know that there are quarks inside the protons. We learned this by conducting a similar experiment on deep inelastic scattering of electrons on protons in 1967 in the SLAC (National Acceleration Laboratory, USA).
This experiment was carried out by the same principle as the experiment of Rutherford. Then the alpha particles were falling, and here the electrons fell on protons. As a result of the collision, the protons may remain protons, and may be excited due to high energy, and then, when scattering protons, other particles can be born, such as Pi-Mesons. It turned out that this section behaves as if inside the protons there are point components. Now we know that these point components are quarks. In some sense it was the experience of Rutherford, but already at 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 quarks and watch what they are broken. But this is the next step, until it fails.
In addition, with the name of Rutherford, the most important plot from the history of domestic science is connected. Peter Leonidovich Kapitsa worked in his laboratory. In early 1930 he was forbidden to leave the country, and he was forced to stay in the Soviet Union. Having learned about it, Rutherford sent his capital all the devices that were in England in England, and thus helped create the Institute of Physical Problems in Moscow. That is, thanks to Rutherford, a significant part of Soviet physics took place.
Moscow State University Economics Statistics Computer Science
Abstract for discipline: "ks"
on the topic :
"Planetary atom model"
Performed:
Student 3 courses
DNF-301 groups
Ruziev Temur
Teacher:
Mosolov D.N.
Moscow 2008
In the first atomic theory of Dalton, it was assumed that the world consists of a certain number of atoms - elementary bricks - with characteristic properties, eternal and unchanged.
These ideas have changed decisively after opening an electron. All atoms must contain electrons. But how are electrons in them are located? Physics could only philosophize, based on their knowledge in the field of classical physics, and gradually, all points of view came together on one model proposed by J.J. Thomson. According to this model, an atom consists of a positively charged substance, the electrons are engaged inside (possibly, they are in intensive movement), so the atom resembles pudding with raisins. The Thomson atom model could not be checked directly, but all sorts of analogies testified in her favor.
The German physicist Philipp Lenard in 1903 proposed a model of the "empty" atom, inside which "fly" some nobody detected neutral particles composed of mutually balanced positive and negative charges. Lenard even gave the name for its non-existent particles - dynamides. However, the only right to the existence of which was proved by strict, simple and beautiful experiments, was the Rutherford model.
A huge scope of the scientific work of Rutherford in Montreal - they were published both personally and together with other scientists 66 articles, not counting the books "Radioactivity," the glory of the first-class researcher brought Ruther. He gets an invitation to take the department in Manchester. On May 24, 1907, Rutherford returned to Europe. A new period of his life began.
The first attempt to create an atom model based on accumulated experimental data belongs to J. Thomson (1903). He believed that the atom is an electronic spray-shaped electronic system with a radius of approximately 10-10 m. A positive atom charge is evenly distributed throughout the bulk volume, and negatively charged electrons are inside it. To explain the timing spectra of the emission of Thomson atoms, tried to determine the location of electrons in the atom and calculate the frequencies of their oscillations near the equilibrium positions. However, these attempts were not crowned with success. A few years later, in the experiments of the Great English Physics, E. Rutinford has been proven that the Thomson model is incorrect.
English physicist E. Rutinford explored the nature of this radiation. It turned out that the bundle of radioactive radiation in a strong magnetic field was divided into three parts: a-, b- and y-radiation. B-rays are the flow of electrons, a-rays - the kernel of the helium atom, U-rays - short-wave electromagnetic radiation. The phenomenon of natural radioactivity indicates the complex structure of the atom.
In the experiments of Rutinford on the study of the internal structure of the Golden foil atom, the A-particles passing through the slots in lead screens at a speed of 107 m / s. A-particles emitted by a radioactive source are the kernel of the helium atom. After interaction with the foil atoms, the A-particles fell on the screens covered with a layer of sulfur zinc. Having hitting the screens, and the particles caused light flashes of light. According to the amount of flashes, the number of particles scattered by the foil to certain angles was determined. Counting showed that most OS particles passes the foil freely. However, some A-particles (one of 20,000) were abruptly deviated from the initial direction. The wage of the OS particle with an electron cannot significantly change its trajectory, since the electron mass is 7350 times smaller than the mass of the A-particle.
Rutherford suggested that the reflection of A-particles is due to their repulsion to positively charged particles with masses, commensurate with a mass of a-particle. Based on the results of this kind of experiments, Rutherford proposed a model of an atom: in the center of the atom there is a positively charged atomic nucleus, around which (like planets treated around the Sun) rotate under the action of the electrical attraction forces of negatively charged electrons. Atom electrophelene: The charge of the nucleus is equal to the total charge of electrons. Linear kernel size at least 10,000 times less than the size of the atom. Such is the planetary model of the atom on Rangeford.What does the electron from falling on the massive core? Of course, rapid rotation around it. But in the process of rotation with acceleration in the core field, the electron should be emitted part of its energy in all directions and, gradually thoring, still fall on the core. This thought did not give rest to the authors of the planetary model of the atom. Another obstacle on the way of a new physical model seemed to have had to destroy all the picture of the nuclear structure built and proven with clear experiments ...
Rutherford was sure that the decision would be found, but he could not assume that this would happen so soon. The defect of the planetary model of the atom will correct the Danish physicist Niels Bohr. Bor painfully thoughtfully over the Rutherford model and was looking for convincing explanations to the fact that it was obviously happening in nature, contrary to all doubts: electrons, not falling on the core and not flying away from him, constantly rotate around their kernel
In 1913, Nils Bor published the results of long reflection and calculations, the most important of which began to be called the postulates of Bor: in the atom there is always a large number of sustainable and strictly defined orbits, according to which the electron can rush in infinitely long, for all the forces acting on it turn out to be balanced; The electron can move in an atom only with one stable orbit to another, as stable. If, with such a transition, the electron is removed from the nucleus, then it is necessary to inform him from the outside of a certain amount of energy equal to the difference in the electron energy reserve on the upper and lower orbit. If the electron approaches the kernel, then he "resets" in the form of radiation ...
Probably, the postulates of Bor would occupy a modest place among a number of interesting explanations of new physical facts extracted by the report house, if it were not for one important circumstance. Bor with the help of relations found by him managed to calculate the radii of "allowed" orbits for an electron in the hydrogen atom. Bor suggested that the magnitles characterizing the micrometer should quantize
. They can take only certain discrete values.
The laws of the micromyr - quantum laws!
These laws at the beginning of the 20th century have not yet been established by science. Bohr formulated them in the form of three postulates. Supplementing (and "saving") atom of Rutherford.
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 energy atom does not radiate, despite the movement of electrons.
Second postulate:
In the stationary state of the electrons atom move in stationary orbits, for which a quantum ratio is performed:
m · v · r \u003d n · h / 2 · p (1)
where m · v · r \u003d l is the moment of the impulse, n \u003d 1,2,3 ..., H-constant plank.
The third postulate:
The radiation or absorption of the energy atom occurs when it moves from one inpatient state to another. In this case, the energy portion is emitted or absorbed ( quantum
) equal to the difference in the energy of stationary states, between which the transition takes place: E \u003d H · U \u003d E M -E n (2)
1. The main stationary state in the initiated,
2. From the excited stationary state in the main one.
Bohr's postulates contradict the laws of classical physics. They express the characteristic feature of the microworld - the quantum nature of the phenomena occurring there. Conclusions based on the postulates of boron are well consistent with the experiment. For example, they explain the patterns in the spectrum of the hydrogen atom, the origin of the characteristic spectra of X-ray rays, etc. In fig. 3 shows a part of the energy diagram of stationary states of the hydrogen atom. 
The arrows show the atom transitions leading to energy radiation. It can be seen that the spectral lines are combined in a series that differ in what level from other (higher) is the transition of an atom.
Knowing the difference between the electron energies in these orbits, it was possible to construct a curve describing the range of hydrogen radiation in various excited states and determine the waves of which length should be particularly willing to emit a hydrogen atom if the excessive energy is inverted to it, for example, using bright light with mercury Lamps. This theoretical curve fully coincided with the spectrum of radiation of excited hydrogen atoms, measured by the Swiss scientist Ya. 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
The stability of any system in atomic scales follows from the principle of Guisenberg uncertainty (the 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 results that have important practical importance can also be obtained within the framework of classical mechanics, adopting additional quantization rules of orbits.
In this chapter, we 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 rotate around the core under the action of the forces of the Coulomb attraction. We believe that the electrons move along the orbits of a circular form.
13.1. Principle of conformity
The quantization of the angular momentum is used in the model of the hydrogen atom proposed by Bohr in 1913. Bor 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.
Atom can be only in certain states with discrete energy levels. E. i. . Electrons, rotating at the appropriate discrete orbits, move accelerated, but, nevertheless, they do not emit. (In classical electrodynamics, every accelerated moving particle radiates if it has a different charge).
The radiation is either absorbed by the quanta during the transition between the energy levels:
Of these postulates, the quantization rule of the electron rotation
,
where n.it can be equal to any natural number:
Parameter n. called the main quantum number. For the output of formulas (1.1), we will express energy energy through the moment of rotation. Astronomical measurements require the knowledge of wavelengths with a sufficiently large accuracy: six faithful digits for optical lines and up to eight - in the radio band. Therefore, when studying the hydrogen atom, the assumption of an infinitely large mass of the kernel is too coarse, since it leads to an error in the fourth significant digit. It is necessary to take into account the movement of the kernel. For his accounting is introduced presented mass.
13.2. Presented mass
The electron is moving around the core under the action of electrostatic power
,
where r.- The vector, the beginning of which coincides with the position of the kernel, and the end indicates an electron. Recall that Z. - this is the nuclear number of the nucleus, and the charges of the nucleus and the electron are equal, respectively Ze. and 
. According to the third law of Newton, the core acts the force equal to - f.(It is equal to the module and is directed opposite to the force acting on the electron). Write an electron equation
.
We introduce new variables: electron speed relative to the kernel
and speed center
.
Folding (2.2a) and (2.2b), we get
.
Thus, the mass center of the closed system is moving evenly and straight. Now we divide (2.2b) on m. Z. and subtract it from (2.2a) divided by m. e. . As a result, an equation is obtained for the relative velocity of the electron:
.
The magnitude of it
called the resulting mass. Thus, the task of the joint movement of two particles - electron and nucleus is simplified. It is enough to consider the movement around the kernel of one particle, the position of which coincides with the position of the electron, and its mass is equal to the sum of the system.
13.3. Communication between the energy and the moment of rotation
The strength of the Coulomb interaction is directed along the straight, connecting charges, and its module depends only on distance r. Between them. Consequently, equation (2.5) describes the movement of the particle in the central symmetric field. An important property of movement in the field with central symmetry is to preserve the energy and moment of rotation.
We write down the condition that the electron movement along a circular orbit is determined by Coulomb attraction to the kernel:
.
It follows that kinetic energy
equal to half of potential energy
,
brought with a back sign:
.
Full energy E,respectively, equal to:
.
It turned out negative, as it should be for sustainable states. The states of atoms and ions with negative energy are called connected. Multiplying equation (3.4) by 2 r. and replacing the work in the left part m.V.r. At the time of rotation M., Express speed V. after moment:
.
Substituting the resulting speed value in (3.5), we obtain the desired formula for full energy:
.
We draw attention to the fact that the energy is proportional to the degree of rotation. In Bor's theory, this fact has important consequences.
13.4. Quantization of the moment of rotation
Second equation for variables V. and r. We will get from the quantization rule of the orbits, the output of which will be executed, based on the postulates of boron. Differentiating formula (3.5), we get a link between small changes in torque and energy:
.
According to the third postulate, the frequency of the radiated (or absorbed) photon is equal to the electron circulation frequency in orbit:
.
From formulas (3.4), (4.2) and communications
between speed, torque and radius implies a simple expression to change the moment of pulse when the electron transitions between adjacent orbits:
.
Integrating (4.3), we get
Constanta C. We will search in half open interval
.
Double inequality (4.5) does not contribute any additional restrictions: if WITHit goes beyond (4.5), then it can be returned to this interval, simply renumbering the moment value in formula (4.4).
Physical laws are the same in all reference systems. Let us turn from the legal coordinate system to Levovintovaya. Energy, as any scalar value, will remain the same,
.
Otherwise, the axial vector of rotation is behaved. As you know, any axial vector when performing the specified operation changes the sign:
Between (4.6) and (4.7) there is no contradiction, since the energy, according to (3.7), is inversely proportional to the square square and remains the same when changing the sign M..
Thus, the set of negative values \u200b\u200bof the moment must repeat the set of its positive values. In other words, for each positive value M. n. Be sure to be equal to it in the module negative value M. – m. :
Combining (4.4) - (4.8), we obtain a linear equation for WITH:
,
with decision
.
It is easy to make sure that formula (4.9) gives two values \u200b\u200bof the constant WITHsatisfying inequality (4.5):
.
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The result obtained illustrates the table in which the number of torque for three values \u200b\u200bC: 0, 1/2 and 1/4 are given. It is clearly seen that in the last line ( n.\u003d 1/4) The magnitude of the moment of rotation for positive and negative values n.differs in absolute value.
Coincidence with experimental data Bor managed to get, putting a constant C. equal to zero. Then the quantization rule of the orbital moment is described by formulas (1). But also makes sense and meaning C.equal half. It describes domestic momentelectron, or his spin - The concept that will be discussed in detail in other chapters. Often the planetary model of the atom is set out from formula (1), but historically it has been derived from the principle of conformity.
13.5. Electron orbit parameters
Formulas (1.1) and (3.7) leads to a discrete set of radii of orbit and electron velocities, which can be rented using a quantum number n.:
It corresponds to the discrete energy spectrum. Full energy electron E. n. It can be calculated by formulas (3.5) and (5.1):
.
We received a discrete set of energy states of the hydrogen atom or hydrogen-like ion. Meaning n.equal to one called mainother - excited and if n.
very large, then - highly excited. Figure 13.5.1 illustrates formula (5.2) for hydrogen atom. Dotted line 
the ionization boundary is indicated. It is clearly seen that the first excited level is much closer to the ionization border than the main
status. Approaching the ionization border, the levels in Fig.13.5.2 are gradually condensed. 
Infinitely many levels have only a solitary atom. In a real medium, various interactions with neighboring particles lead to the fact that the atom has only a finite number of lower levels. For example, under conditions of star atmospheric, an atmosphere is usually 20-30 states, but hundreds of levels can be observed in a rarefied interstellar gas, but not more than a thousand.
In the first chapter, we introduced the Redberg, based on the considerations of dimension. Formula (5.2) reveals the physical meaning of this constant as a convenient unit of measurement of the energy of the atom. In addition, it shows that Ry depends on the relationship 
:
.
By virtue of a large difference in the masses of the nucleus and the electron, this dependence is very weak, but in some cases it is impossible to neglect it. In the numerator of the last formula worth a constant
erg 
eV,
to which the value of Ry is striving for an unlimited increase in the mass of the kernel. Thus, we clarified the Ry measurement unit given in the first chapter.
The quantization rule of the moment (1.1) is, of course, is less accurate than the expression (12.6.1) for its own value 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. They can be used in all cases, if relativistic corrections are negligible.
So, according to the planetary model of the atom, in the bound states, the speed of rotation, the radius of the orbit and the electron energy take the discrete number of values \u200b\u200band are fully determined by the magnitude of the main quantum number. Conditions with positive energy are called free; They are not quantized, and all electron parameters in them, besides the moment of rotation, can take any values \u200b\u200bthat do not contradict the laws of preservation. The moment of rotation is always quantized.
The formulas of the planetary model make it possible to calculate the ionization potential of the 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, linear and angular velocity of the electron motion in orbit.
Released formulas have two limitations. First, they do not take into account the relativistic effects, which gives an error of order ( V./c.) 2. Relativistic amendment increases as the charge of the kernel is increasing as Z. 4 And for Ion Fexxvi is already the interests of the 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 levels of levels are determined by other parameters - orbital and internal points of the electron. Therefore, the levels are split into several sublevels. The cleavage 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, the simple theory of boron turns out to be simple, convenient and fairly accurate method of studying the structure of ions and atoms.
13.6. Received Rydberg
In the optical range of the spectrum is usually not measured by the energy of a quantum E., and the wavelength transition between levels. Therefore, the wave number is often used to measure the level of level E / HC.measured in reverse centimeters. Wave number 
, denotes 
:
cm 
.
The index recalls that the mass of the core in this definition is considered infinitely large. Taking into account the ultimate mass of the kernel of Rydberg equal
.
At hard nuclei, it is more than the lungs. The ratio of the mass of the proton and electron is equal
Substituting this value in (2.2) we obtain the numerical expression of the Ridberg permanent expression for the hydrogen atom:
The core of the heavy isotope of hydrogen - deuterium - consists of a proton and neutron, and approximately half the heavier of the hydrogen atom - proton. Therefore, according to (6.2), Ridberg's permanent deuterium R. D more than hydrogen R. H:
Even above, it has a unstable hydrogen isotope - tritium, the kernel of which consists of a proton and two neutrons.
In the middle elements of the Mendeleev table, the effect of isotopic shift competes with the effect associated with the elder sizes of the kernel. 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 kernel and one electron are called hydrogen-like. In other words, they relate to the isoelectronic sequence of hydrogen. Their structure qualitatively resembles a hydrogen atom, and the position of the energy levels of ions, the noner charge is not too large ( Z. Z\u003e 20) Quantitative differences associated with relativistic effects are emerging: the dependence of the electron mass on the speed and spin-orbital interaction.
We will look at the most interesting helium, oxygen and iron ions in astrophysics. In spectroscopy, the charge of the ion is set using spectroscopic symbol, which is written by the Roman numerals to the right of the symbol of the chemical element. The number depicted by the Roman number per unit exceeds the amount of electron removed from the atom. For example, a hydrogen atom is denoted as Hi, and hydrogen-like helium, oxygen and iron ions, respectively, HEII, OVIII and FEXXVI. For multi-electronic ions, the spectroscopic symbol coincides with an effective charge, which "feels" valence electron.
Calculate the movement of the electron in a circular orbit, taking into account the relativistic dependence of its mass from speed. Equations (3.1) and (1.1) in a relativistic case look like this:
Presented mass m. defined by formula (2.6). Recall also that
.
Multiply the first equation on 
And divide it to the second. As a result, we get
The permanent structure was introduced in formula (2.2.1) of the first chapter. Knowing speed, calculate the radius of orbits:
.
In a special theory of relativity, kinetic energy is equal to the difference in the total energy of the body and its mine energy in the absence of an external power field:
.
Potential energy U. as a function r. Determined by formula (3.3). Substituting in expressions for T. and U. The obtained values \u200b\u200b and r., I get the full electron energy:
For an electron rotating at the first orbit of a hydrogen-like iron ion, the value 2 is 0.04. More easily elements, respectively, even less. For 
Fair decisions
.
The first term, as easy to ensure, is accurate to the designation equal to the value of energy (5.2) in the nonrelativistic theory of boron, and the second is a desired relativistic amendment. Denote the first term as E. B, then
Drink out express expression for relativistic amendment:
So, the relative magnitude of the relativistic amendment is proportional to the work 2 Z. four . Accounting for the dependence of the mass of the electron from the speed leads to an increase in the depth of levels. This can be understood as follows: the absolute value of energy grows together with a mass of the particle, and the moving electron is heavier fixed. Weakening effect with increasing quantum number n. It is a consequence of a slower movement of an electron in an excited state. Strong dependence OT Z. it is a consequence of high electron speed in a core field with a large charge. In the future, we will calculate this magnitude according to the rules of quantum mechanics and we obtain a new result - removal of degeneration by the orbital moment.
13.8. Highly excited states
The states of the atom or the ion of any chemical element in which one of the electrons is at a high energy level, called high frequented, or redbergovsky.They have an important property: the position of the levels of the excited electron with a sufficiently high accuracy can be described within the Bohr model. The fact is that the electron with a large value of the quantum number n.According to (5.1), it is very far from the nucleus and other electrons. Such an electron in spectroscopy is called "optical", or "valent", and the remaining electrons together with the core - the "atomic residue". Schematically, the structure of an atom with one strongly excited electron is shown in Fig.13.8.1. Left at the bottom is atomic
residue: core and electrons are basically condition. The dotted arrow indicates a valence electron. The distances between all electrons inside the atomic residue are much smaller than the distance from any of them to the optical electron. Therefore, their total charge can be considered almost completely focused in the center. Therefore, it can be assumed that the optical electron is moving under the action of the Coulomb force directed towards the kernel, and thus its energy levels are calculated using the Bora formula (5.2). The electrons of the atomic residue shielded the kernel, but not completely. Concept introduced for partial shielding effective chargeatomic residue Z. EFF. In the case under consideration, a strongly remote electron value Z. EFF equal to the difference in the atomic number of the chemical element Z. and the number of electrons of the atomic residue. Here we limit ourselves to the case of neutral atoms for which Z. Eff \u003d 1.
The position of strongly excited levels is obtained in the theory of boron for any atom. Enough in (2.6) replace 
on the mass of the atomic residue 
which is less than the mass of the atom 
By the magnitude of the mass of the electron. Using the identity from here
we can express the permanent Rydberg as a function of atomic weight A.the chemical element under consideration:
planetary modelsatom ... + --- A - \u003d 0; (2.12) H² H ∂T 4πm ∂a Δβ + 2 (GRAD Agradβ) - ----- \u003d 0. (2. 13 ) H ∂T at βh φ \u003d - (2.14) 2πM Madoung received an equation ...
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See also:
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Planetary model atom
19. In the planetary model of the atom, it is assumed that the number
1) electrons in orbits are equal to the number of protons in the kernel
2) protons equal to the number of neutrons in the kernel
3) electrons in orbits are equal to the sum of the numbers of protons and neutrons in the core
4) neutrons in the kernel equal to the amount of electron numbers in orbits and protons in the kernel
21. The planetary atom model is justified by the experiments on
1) dissolution and melting solid bodies 2) gas ionization
3) chemical production of new substances 4) scattering α-particles
24. The planetary model of the atom is justified
1) calculations of the movement of the celestial tel 2) by electrification experiences
3) experiments on the scattering α-particles 4) photos of atoms in the microscope
44. In the experience of Rangeford α-Pacifics dispersed
1) electrostatic field of the nuclei of atom 2) electronic shell of target atoms
3) the gravitational field of the nuclei of the atom 4) the surface of the target
48. In Refordford's experience, most of the α-particles are fluently passing through foil, practically without deviating from straight trajectories, because
1) the atom core has a positive charge
2) electrons have a negative charge
3) the atom core has small (compared to atom)
4) α-particles have a greater (compared to the atomic nuclei) mass
154. What approvals correspond to the planetary model of the atom?
1) The kernel is in the center of the atom, the kernel charge is positive, electrons in orbits around the kernel.
2) The kernel - in the center of the atom, the nucleus charge is negative, electrons in orbits around the kernel.
3) electrons - in the center of the atom, the kernel turns around electrons, the nucleus charge is positive.
4) electrons - in the center of the atom, the kernel turns around electrons, the charge of the nucleus is negative.
225. Experiments E. Rutherford on the scattering of α-particles showed that
A. Almost all the mass of the atom focuses in the kernel. B. The kernel has a positive charge.
What (s) from the statements correctly (s)?
1) only a 2) only b 3) and a, and b 4) nor, nor b
259. What is the idea of \u200b\u200bthe structure of the atom corresponds to the model of the atom of the reforde?
1) The kernel is in the center of the atom, electrons in orbits around the kernel, the electron charge is positive.
2) the kernel in the center of the atom, electrons in orbits around the kernel, the electron charge is negative.
3) A positive charge is evenly distributed across atom, electrons in the atom make oscillations.
4) A positive charge is evenly distributed across atom, and electrons move in atom in different orbits.
266. What is the idea of \u200b\u200bthe structure of the atom right? Most of the mass of the atom focuses
1) in the kernel, electron charge 2) in the kernel, the charge of the nucleus is negative
3) in electrons, electron charge is negative 4) in the kernel, electron charge is negative
254. What is the idea of \u200b\u200bthe structure of the atom corresponds to the model of the Rangeford atom?
1) the kernel - in the center of the atom, the nucleus charge is positive, most of the mass of the atom focuses in electrons.
2) the kernel - in the center of the atom, the nucleus charge is negative, most of the mass of the atom focuses in the electronic shell.
3) the kernel in the center of the atom, the nucleus charge is positive, most of the mass of the atom focuses in the kernel.
4) the kernel in the center of the atom, the nucleus charge is negative, most of the mass of the atom focuses in the kernel.
Choose Bora.
267. The diagram of the lower energy levels of the atoms of the sparse atomic gas has the appearance in the figure. At the initial moment of time, atoms are in a state with energy E (2) according to the postulates of boron, 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 lower energy levels of the atom. At the initial moment of time, the atom is in a state of Energy E (2). According to the postulates of Bor, this 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 the photon emitted by the atom according to the model of the boron atom?
1) the difference between the energy of stationary states 2) the frequency of the electron circulation around the kernel
3) de Brogly wavelength for electron 4) The boron model does not allow it to determine
15. 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 gas atomic energy can only take the values \u200b\u200bthat are listed on the diagram. Atoms are in a state with Energy E (3). What kind of energy can be absorbed by this gas?
1) anywhere from 2 ∙ 10 -18 j to 8 ∙ 10 -18 J 2) Anyone, but less than 2 ∙ 10 -18 J
3) only 2 ∙ 10 -18 J 4) any, greater or equal to 2 ∙ 10 -18 J
29. When emissing a photon with an energy of 6 eV atom charge
1) does not change 2) increases by 9.6 ∙ 10 -19
3) increases by 1.6 ∙ 10 -19 KL 4) decreases by 9.6 ∙ 10 -19
30. Light with a frequency of 4 ∙ 10 15 Hz consists of photons with an electrical charge equal
1) 1.6 ∙ 10 -19 KL 2) 6.4 ∙ 10 -19 KL 3) 0 KL 4) 6.4 ∙ 10 -4 CL
78. The electron of the outer shell of the atom first moves from the stationary state with the E 1 E 1 in a stationary state with E 2, absorbing photon frequency v. one . It then moves from the state E 2 to a stationary state with the energy E Z, absorbing photon frequency v. 2 > v. one . What happens when the electron is transition from the state E 2 to the state of E 1.
1) light radiation frequency v. 2 – v. 1 2) light absorption frequency v. 2 – v. 1
3) radiation light frequency v. 2 + v. 1 4) light absorption frequency v. 2 – v. 1
90. The photon energy absorbed by the atom during the transition from the main state with the energy E 0 into the excited state with the energy E 1 equal to (H is a constant plank)
95. The figure shows the energy levels of the atom and indicate the wavelengths of photons emitted and absorbed when transition from one level to another. What is the wavelength for photons emitted during the transition from the level E 4 to the level E 1, if λ 13 \u003d 400 nm, λ 24 \u003d 500 nm, λ 32 \u003d 600 nm? Answer express in Nm, and round up to whole.
96. The figure shows several energy levels of the electronic shell of the atom and indicate the frequencies of photons emitted and absorbed when transitions between these levels. What is the minimum wavelength of photons emitted by atom any
possible transitions between levels e 1, e 2, e z 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? Answer express in Nm and round up to whole.
120. The figure shows the diagram of the energy levels of the atom. Which of the transition arrows marked between the energy levels is accompanied by the absorption of the minimum frequency quanta?
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 indicate the wavelengths of photons emitted and absorbed when transition from one level to another. It is experimentally established that the minimum wavelength for photons emitted when 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 the scheme of possible values \u200b\u200bof the energy of the atomic gas atoms. At the initial moment of time, atoms are in a state with Energy E (3). It is possible to emit gas 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 of 2 ∙ 10 -18 to 8 ∙ 10 -18 J
162. The levels of electron energy in the hydrogen atom are set by the formula E n \u003d - 13.6 / N 2 eV, where n \u003d 1, 2, 3, .... When the atom transition from state E 2 to the state of E 1, the atom emits a photon. Once on the surface of the photocathode, the photon knocks out a photoelectron. The wavelength of light corresponding to the red border of the photo effect for the material of the photocathode surface, λ kr \u003d 300 nm. What is the maximum possible photoelectron speed?
180. The figure shows the slightly lower levels of the energy of the hydrogen atom. Can an atom located in a state of E 1, absorb a photon with an energy of 3.4 eV?
1) Yes, at the same time the atom goes into the state of E 2
2) Yes, at the same time the atom goes into state E 3
3) Yes, at the same time atom ionizums, decaying the proton and electron
4) no, photon energy is not enough to transition an atom into an excited state
218. The figure shows a simplified diagram of the energy levels of the atom. Numerous arrows marked some possible atom transitions between these levels. Set the correspondence between the light absorption processes of the greatest wavelength and the emission of the light of the greatest wavelength and arrows indicating the energy transitions of the atom. To each position of the first column, select the corresponding position of the second and record the selected numbers in the table under the appropriate letters.
226. The figure shows a fragment of the diagram of the energy levels of the atom. Which of the transition arrows marked between energy levels is accompanied by a photon radiation 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 four lower levels of hydrogen atom. What transition corresponds to the absorption of a photon atom with an energy of 12.1 eV?
1) E 3 → E 1 2) E 1 → E 3 3) E 3 → E 2 4) E 1 → E 4
238. The electron, having a pulse P \u003d 2 ∙ 10 -24 kg ∙ m / s, faces a condensed proton, forming a hydrogen atom in a state with an energy e n (n \u003d 2). In the process of the formation of an atom, a photon is emitted. Find the frequency v. This photon, neglecting the kinetic energy of the atom. The levels of electron energy in the hydrogen atom are set by the formula, where n \u003d 1.2, 3, ....
260. The diagram of the lower energy levels of the atom has the appearance of the picture. At the initial moment of time, the atom is in a state of Energy E (2). According to the postulates of Bora, an atom can emit photons with energy
1) only 0.5 eV 2) only 1.5 eV 3) of any, less than 0.5 eV 4) anywhere from 0.5 to 2 eV
269. The figure shows the diagram of the energy levels of the atom. Which digit is marked by the transition that corresponds to radiationphoton with lowest energy?
1) 1 2) 2 3) 3 4) 4
282. Photon radiation atom occurs when
1) Electron motion in stationary orbit
2) the transition of an electron from the main state into an excited
3) the transition of an electron from an excited state in the main
4) all the following processes
13. Photon radiation occurs when moving 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 ratio 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) more zero 2) is zero 3) less zero
4) more or less zero depending on the state
98. A restless atom absorbed a photon with an energy of 1.2 ∙ 10 -17 J. In this case, the impetus
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 some substance has the form,
the figure shown in the figure, and atoms are in a state of Energy E (1). The electron moving with a kinetic energy of 1.5 eV collided with one of such atoms and bounced off, acquiring some additional energy. Determine the email pulse after the collision, believing that the atom rested before the collision. The possibility of emitting light by atom when a collision with an electron is neglected.
111. Suppose that the scheme of energy levels of atoms of a certain substance has the appearance shown in the figure, and atoms are in a state with an energy E (1). The electron, encountered with one of these atoms, bounced, acquiring some additional energy. The electron pulse after a collision with a resting atom turned out to be equal to 1.2 ∙ 10 -24 kg ∙ m / s. Determine the kinetic electron energy before the collision. The possibility of emitting light by atom when a collision with an electron is neglected.
136. π ° - Meson weighing 2.4 ∙ 10 -28 kg decomposes by two γ -Kvanta. Find the pulse module of one of the resulting γ-quanta in the reference system, where the primary π ° is resting.
144. The vessel is rarefied atomic hydrogen. A hydrogen atom is mainly state (E 1 \u003d - 13.6 eV) absorbs photon and ionizums. The electron, flying out of the atom as a result of ionization, moves away from the nucleus at a speed V \u003d 1000 km / s. What is the frequency of the absorbed photon? The energy of the thermal movement of hydrogen atoms must be neglected.
197. The resting atom of hydrogen is mainly (E 1 \u003d - 13.6 eV) absorbs photon in vacuum with a wavelength λ \u003d 80 nm. What speed moves away from the kernel an electron, flying out of an atom as a result of ionization? The kinetic energy of the ION forming to neglect.
214. Free peony (° ° mezzan) with a mine of 135 MeV moves at a speed V, which is significantly less than the speed of light. As a result of its decay, two γ-quantum were formed, and one of them propagates in the direction of the peony movement, and the other in the opposite direction. The energy of one quantum is 10% more than the other. What is the rate of peony 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 when moving 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. A restless atom radiates a photon with an energy of 16.32 ∙ 10 -19 J as a result of an electron transition from an excited state to the main one. The atom as a result of the return begins to move progressively in the opposite direction with the kinetic energy 8.81 ∙ 10 -27 J. Find a mass of the atom. Atom speed is considered small compared to the speed of light.
252. In the vessel is rarefied atomic hydrogen. The hydrogen atom is mainly state (E 1 \u003d -13,6 eV) absorbs the photon and ionizums. The electron, flying out of the 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? The energy of the thermal movement of hydrogen atoms must be neglected.
1) 46 nm 2) 64 nm 3) 75 nm 4) 91 nm
257. The vessel is rarefied atomic hydrogen. The hydrogen atom is mainly state (E 1 \u003d -13,6 eV) absorbs the photon and ionizums. The electron, flying out of the atom as a result of ionization, moves away from the nucleus at a speed V \u003d 1000 km / s. What is the energy of the absorbed photon? The energy of the thermal movement of hydrogen atoms must be neglected.
1) 13.6 eV 2) 16.4 eV 3) 19.3 eV 4) 27.2 eV
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