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The Nuclear Atom. History 450 BC, Democritus – The idea that matter is composed of tiny particles, or atoms.450 BC, Democritus – The idea that matter.

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Presentation on theme: "The Nuclear Atom. History 450 BC, Democritus – The idea that matter is composed of tiny particles, or atoms.450 BC, Democritus – The idea that matter."— Presentation transcript:

1 The Nuclear Atom

2 History 450 BC, Democritus – The idea that matter is composed of tiny particles, or atoms.450 BC, Democritus – The idea that matter is composed of tiny particles, or atoms. XVII-th century, Pierre Cassendi, Robert Hook – explained states of matter and transactions between them with a model of tiny indestructible solid objects.XVII-th century, Pierre Cassendi, Robert Hook – explained states of matter and transactions between them with a model of tiny indestructible solid objects – Avogadro’s hypothesis that all gases at given temperature contain the same number of molecules per unit volume.1811 – Avogadro’s hypothesis that all gases at given temperature contain the same number of molecules per unit volume – Kinetic theory of gases.1900 – Kinetic theory of gases. Consequence – Great three quantization discoveries of XX century: (1) electric charge: (2) light energy; (3) energy of oscillating mechanical systems.

3 Historical Developments in Modern Physics 1895 – Discovery of x-rays by Wilhelm Röntgen.1895 – Discovery of x-rays by Wilhelm Röntgen – Discovery of radioactivity of uranium by Henri Becquerel1896 – Discovery of radioactivity of uranium by Henri Becquerel 1897 – Discovery of electron by J.J.Thomson1897 – Discovery of electron by J.J.Thomson 1900 – Derivation of black-body radiation formula by Max Plank.1900 – Derivation of black-body radiation formula by Max Plank – Development of special relativity by Albert Einstein, and interpretation of the photoelectric effect.1905 – Development of special relativity by Albert Einstein, and interpretation of the photoelectric effect – Determination of electron charge by Robert Millikan.1911 – Determination of electron charge by Robert Millikan – Proposal of the atomic nucleus by Ernest Rutherford.1911 – Proposal of the atomic nucleus by Ernest Rutherford – Development of atomic theory by Niels Bohr.1913 – Development of atomic theory by Niels Bohr – Development of general relativity by Albert Einstein.1915 – Development of general relativity by Albert Einstein Development of Quantum Mechanics by deBroglie, Pauli, Schrödinger, Born, Heisenberg, Dirac,… Development of Quantum Mechanics by deBroglie, Pauli, Schrödinger, Born, Heisenberg, Dirac,….

4 The Nuclear Atoms There are 112 chemical elements that have been discovered, and there are a couple of additional chemical elements that recently have been reported. Each element is characterized by atom that contains a number of protonsZ, and equal number of electrons, and a number of neutrons N. The number of protons Z is called the atomic number. The lightest atom, hydrogen (H), has Z=1; the next lightest atom, helium (He), has Z=2; the third lightest, lithium (Li), has Z=3 and so forth. Each element is characterized by atom that contains a number of protons Z, and equal number of electrons, and a number of neutrons N. The number of protons Z is called the atomic number. The lightest atom, hydrogen (H), has Z=1; the next lightest atom, helium (He), has Z=2; the third lightest, lithium (Li), has Z=3 and so forth.

5 The Nuclear Atoms Nearly all the mass of the atom is concentrated in a tiny nucleus which contains the protons and neutrons. Typically, the nuclear radius is approximately 1 fm to 10 fm (1fm = m). The distance between the nucleus and the electrons is approximately 0.1 nm=100,000fm. This distance determines the size of the atom.

6 Nuclear Structure An atom consists of an extremely small, positively charged nucleus surrounded by a cloud of negatively charged electrons. Although typically the nucleus is less than one ten-thousandth the size of the atom, the nucleus contains more than 99.9% of the mass of the atom!

7 The number of protons in the nucleus, Z, is called the atomic number. This determines what chemical element the atom is. The number of neutrons in the nucleus is denoted by N. The atomic mass of the nucleus, A, is equal to Z + N. The number of protons in the nucleus, Z, is called the atomic number. This determines what chemical element the atom is. The number of neutrons in the nucleus is denoted by N. The atomic mass of the nucleus, A, is equal to Z + N. A given element can have many different isotopes, which differ from one another by the number of neutrons contained in the nuclei. In a neutral atom, the number of electrons orbiting the nucleus equals the number of protons in the nucleus.

8 Since the electric charges of the proton and the electron are +1 and -1 respectively (in units of the proton charge), the net charge of the atom is zero. Since the electric charges of the proton and the electron are +1 and -1 respectively (in units of the proton charge), the net charge of the atom is zero. At present, there are 112 known elements which range from the lightest, hydrogen, to the recently discovered and not yet to-be-named element 112. All of the elements heavier than uranium are man made. Among the elements are approximately 270 stable isotopes, and more than 2000 unstable isotopes.

9 The Nuclear Atoms Since electrons and protons have equal and opposite charges, and there are an equal number of protons and electrons in an atom, atoms are electrically neutral. Atoms that lose or gain one or more electrons are then electrically charged and are called ions. We will begin our study of atoms by discussing the Bohr’s semiclassical model, developed in 1913 to explain the spectra emitted by hydrogen atoms.

10 Atomic Spectra By the beginning of the 20 th century a large body of data has been collected on the emission of light by atoms of individual elements in a flame or in a gas exited by electrical discharge. Diagram of the spectrometer

11 Atomic Spectra Light from the source passed through a narrow slit before falling on the prism. The purpose of this slit is to ensure that all the incident light strikes the prism face at the same angle so that the dispersion by the prism caused the various frequencies that may be present to strike the screen at different places with minimum overlap.

12 The source emits only two wavelengths, λ 2 >λ 1. The source is located at the focal point of the lens so that parallel light passes through the narrow slit, projecting a narrow line onto the face of the prism. Ordinary dispersion in the prism bends the shorter wavelength through the lager total angel, separating the two wavelength at the screen.

13 In this arrangement each wavelength appears as a narrow line, which is the image of the slit. Such a spectrum was dubbed a line spectrum for that reason. Prisms have been almost entirely replaced in modern spectroscopes by diffraction gratings, which have much higher resolving power.

14 When viewed through the spectroscope, the characteristic radiation, emitted by atoms of individual elements in flame or in gas exited by electrical charge, appears as a set of discrete lines, each of a particular color or wavelength. The positions and intensities of the lines are a characteristic of the element. The wavelength of these lines could be determined with great precision.

15 Atomic Spectra In 1885 a Swiss schoolteacher, Johann Balmer, found that the wavelengths of the lines in the visible spectrum of hydrogen can be represented by formula Balmer suggested that this might be a special case of more general expression that would be applicable to the spectra of other elements.

16 Atomic Spectra Such an expression, found by J.R.Rydberg and W. Ritz and known as the Rydberg-Ritz formula, gives the reciprocal wavelengths as: where m and n are integers with n>m, and R is the Rydberg constant.

17 Atomic Spectra The Rydberg constant is the same for all spectral series. For hydrogen the R H = x 10 7 m -1. For very heavy elements R approaches the value R ∞ = x 10 7 m -1. Such empirical expressions were successful in predicting other spectra, such as other hydrogen lines outside the visible spectrum Such empirical expressions were successful in predicting other spectra, such as other hydrogen lines outside the visible spectrum.

18 Atomic Spectra So, the hydrogen Balmer series wavelength are those given by Rydberg equation with m=2 and n=3,4,5,… Other series of hydrogen spectral lines were found for m=1 (by Lyman) and m=3 (by Paschen).

19 Hydrogen Spectral Series Compute the wavelengths of the first lines of the Lyman, Balmer, and Paschen series.

20 Emission line spectrum of hydrogen in the visible and near ultraviolet. The lines appear dark because the spectrum was photographed. The names of the first five lines are shown, as is the point beyond which no lines appear. H ∞ called the limits of the series.

21 The Limits of Series Find the predicted by Rydberg-Ritz formula for Lyman, Balmer, and Paschen series.

22 A portion of the emission spectrum of sodium. The two very close bright lines at 589 nm are the D 1 and D 2 lines. They are the principal radiation from sodium street lighting.

23 A portion of emission spectrum of mercury.

24 Part of the dark line (absorption) spectrum of sodium. White light shining through sodium vapor is absorbed at certain wavelength, resulting in no exposure of the film at those points. Note that frequency increases toward the right, wavelength toward the left in the spectra shown.

25 Rutherford’s Nuclear Model Many attempts were made to construct a model of the atom that yielded the Balmer and Rydberg-Ritz formulas. It was known that an atom was about m in diameter, that it contained electrons much lighter than the atom, and that it was electrically neutral. The most popular model was that of J.J.Thomson, already quite successful in explaining chemical reactions.

26 Thomson’s model of the atom: A sphere with a positive charge with electrons embedded in it so that the net charge would normally be zero.

27 One difficulty with all such models was that electrostatic forces alone cannot produce stable equilibrium. Thus the charges were required to move and, if they stayed within the atom, to accelerate. However, the acceleration would result in continuous radiation, which is not observed. Thomson was unable to obtain from his model a set of frequencies that corresponded with the frequencies of observed spectra. Thomson was unable to obtain from his model a set of frequencies that corresponded with the frequencies of observed spectra. The Thomson model of the atom was replaced by one based on results of a set of experiments conducted by Ernest Rutherford and his student H.W.Geiger. The Thomson model of the atom was replaced by one based on results of a set of experiments conducted by Ernest Rutherford and his student H.W.Geiger.

28 Rutherford was investigating radioactivity and had shown that the radiations from uranium consist of at least two types, which he labeledα andβ Rutherford was investigating radioactivity and had shown that the radiations from uranium consist of at least two types, which he labeled α and β. He showed, by an experiment similar to that of Thompson, thatq /mfor theα - particleswas half that of the proton. He showed, by an experiment similar to that of Thompson, that q /m for the α - particles was half that of the proton. Suspecting that theαparticles were double ionized helium, Rutherford in his classical experiment let a radioactive substance decay and then, by spectroscopy, detected the spectra line of ordinary helium. Suspecting that the α particles were double ionized helium, Rutherford in his classical experiment let a radioactive substance decay and then, by spectroscopy, detected the spectra line of ordinary helium.

29 Schematic diagram of the Rutherford apparatus. The beam of α - particles is defined by the small hole D in the shield surrounding the radioactive source R of 214 Bi. The α beam strikes an ultra thin gold foil F, and α particles are individually scattered through various angels θ. The experiment consisted of counting the number of scintillations on the screen S as a function of θ.

30 A diagram of the original apparatus as it appear in Geiger’s paper describing the results.

31 An α -particle by such an atom (Thompson model) would have a scattering angle θ much smaller than 1 0.In the Rutherford’s scattering experiment most of the α -particles were either undeflected, or deflected through very small angles of the order 1 0, however, a few α -particles were deflected through angles of 90 0 and more.

32 If the atom consisted of a positively charged sphere of radius m, containing electrons as in the Thomson model, only a very small scattering deflection angle could be observed. Such model could not possibly account for the large angles scattering. The unexpected large angles α-particles scattering was described by Rutherford with these words: It was quite incredible event that ever happened to me in my life. It was as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.

33 Rutherford concluded that the large angle scattering could result only from a single encounter of the α particle with a massive charge with volume much smaller than the whole atom. Assuming this “nucleus” to be a point charge, he calculated the expected angular distribution for the scattered α particles. His predictions on the dependence of scattering probability on angle, nuclear charge and kinetic energy were completely verified in experiments.

34 Rutherford Scattering geometry. The nucleus is assumed to be a point charge Q at the origin O. At any distance r the α particle experiences a repulsive force kq α Q/r 2. The α particle travel along a hyperbolic path that is initially parallel to line OA a distance b from it and finally parallel to OB, which makes an angle θ with OA.

35 Force on a point charge versus distance r from the center of a uniformly charged sphere of radius R. Outside the sphere the force is proportional to Q/r 2, inside the sphere the force is proportional to q I /r 2 = Qr/R 3, where q I = Q(r/R) 3 is the charge within a sphere of radius r. The maximum force occurs at r =R

36 Two α particles with equal kinetic energies approach the positive charge Q = +Ze with impact parameters b 1 and b 2, where b 1 θ 2.

37 The path of α particle can be shown to be a hyperbola, and the scattering angle θ can be relate to the impact parameter b from the laws of classical mechanics. Let the intensity of the incident α particles beam be I 0 particles per second per unit of the area. The number of α particles scattered per second by one nucleus through the angles grater than θ equal the number of α particles per second that have impact parameter less than b(θ). This number is πb 2 /I 0. The quality πb 2, which has the dimension of the area, is called the cross section σ for scattering.

38 The cross section σ is thus defined as the number of particles scattered per nucleus per unit time divided by the incident intensity. The total number of nuclei of foil atoms in the area covered by the beam is nAt, where n is the number of foil atoms per unit volume. A is the area of the beam, and t is the thickness of the foil.

39 The total number of particles scattered per second is obtained by multiplying πb 2 I 0 by the number of nuclei in the scattering foil. Let n be the number of nuclei per unit volume: For a foil of thickness t, the total number of nuclei “seen” by the beam is nAt, where A is the area of the beam. The total number scattered per second through angles grater than θ is thus πb 2 I 0 ntA. If divide this by the number of α particles incident per second I 0 A we get the fraction f scattered through angles grater than θ: f = πb 2 nt

40 On the base of that nuclear model Rutherford derived an expression for the number of α particles ΔN that would be scattered at any angle θ : All this predictions was verified in Geiger experiments, who observe several hundreds thousands α particles.

41 (a)Geiger data for α scattering from thin Au and Ag foils. The graph is in log-log plot to cover over several orders of magnitudes. (b)Geiger also measured the dependence of ΔN on t for different elements, that was also in good agreement with Rutherford formula.

42 (a) If the α particle does not penetrate the nuclear charge, the nuclei can be considered a point charge located at center. (b) If the particle has enough energy to penetrate the nucleus, the Rutherford scattering law does not hold, but would require modification to account for that portion of the nuclear charge “behind” the penetrating α particle.

43 Data from Rutherford’s group showing observed α scattering at a large fixed angel versus values of r d computed from r d = kq α Q/1/2m α v 2 for various kinetic energies.

44 The Size of the Nucleus This equation can be used to estimate the size of the nucleus For the case of 7.7-MeV α particles the distance of closest approach for a head-on collision is

45 According to classical physics, a charge e moving with an acceleration a radiates at a rate (a)Show that an electron in a classical hydrogen atom spirals into the nucleus at a rate (b) Find the time interval over which the electron will reach r = 0, starting from r 0 = 2.00 × 10 –10 m.

46 According to classical physics, a charge e moving with an acceleration a radiates at a rate (a)Show that an electron in a classical hydrogen atom spirals into the nucleus at a rate

47 According to classical physics, a charge e moving with an acceleration a radiates at a rate (b) Find the time interval over which the electron will reach r = 0, starting from r 0 = 2.00 × 10 –10 m. Since atoms last a lot longer than 0.8 ns, the classical laws (fortunately!) do not hold for systems of atomic size.

48 The Bohr’s Postulates Bohr overcome the difficulty of the collapsing atom by postulating that only certain orbits, called stationary states, are allowed, and that in these orbits the electron does not radiate. An atom radiates only when the electron makes a transition from one allowed orbit (stationary state) to another: 1. The electron in the hydrogen atom can move only in certain nonradiating, circular orbits called stationary states.1. The electron in the hydrogen atom can move only in certain nonradiating, circular orbits called stationary states. 2. The photon frequency from energy conservation is given by2. The photon frequency from energy conservation is given by f = (E i – E f ) / h where E i and E f are the energy of initial and final state, h is the Plank’s constant.

49 Such a model is mechanically stable, because the Coulomb potential provides the centripetal force The total energy for a such system can be written as the difference between kinetic and potential energy:

50 The Bohr’s Postulates Combining the second postulate with the equation for the energy we obtain: where r 1 and r 2 are the radii of the initial and final orbits.

51 The Bohr’s Postulates To obtain the frequencies implied by the experimental Rydberg-Ritz formula, it is evident that the radii of the stable orbits must be proportional to the squares of integers.

52 The Bohr’s Postulates Bohr found that he could obtain this condition if he postulates the angular momentum of the electron in a stable orbit equals an integer times ħ=h/2π. Since the angular momentum of a circular orbit is just mvr, the third Bohr’s postulate is: 3. mvr =nh/2π=nħ n=1,2,3………. where ħ=h/2π=1.055 x Js=6.582x10 16 eVs

53 The Bohr’s Postulates The obtained equation mvr = nh/2π=nħ relates the speed of electron v to the radius r. Since we had or We can write or

54 The Bohr’s Postulates Solving for r, we obtain where a 0 is called the first Bohr’s radius And for frequency

55 Bohr’s Postulates Substituting the expression for r in equation for frequency: If we will compare this expression with Z=1 for f=c/λ with the empirical Rydberg-Ritz formula: we will obtain for the Rydberg constant

56 Bohr’s Postulates Using the known values of m, e, and ħ, Bohr calculate R and found his results to agree with the spectroscopy data. The total mechanical energy of the electron in the hydrogen atom is related to the radius of the circular orbit

57 Energy levels If we will substitute the quantized value of r as given by we obtain

58 Energy levels orwhere The energies E n with Z=1 are the quantized allowed energies for the hydrogen atom.

59 Energy levels Transitions between this allowed energies result in the emission or absorption of a photon whose frequency is given by and whose wavelength is

60 Energy levels Therefore is convenient to have the value of hc in electronvolt nanometers: hc = 1240 eV∙nm Since the energies are quantized, the frequencies and the wavelengths of the radiation emitted by the hydrogen atom are quantized in agreement with the observed line spectrum.

61 A hydrogen atom is in its first excited state (n = 2). Using the Bohr theory of the atom, calculate (a) the radius of the orbit, (b) the linear momentum of the electron, (c) the angular momentum of the electron, (d) the kinetic energy of the electron, (e) the potential energy of the system, and (f) the total energy of the system.

62 (a) (b)

63 (c) (d) (f) (e)

64 (a) In the classical orbital model, the electron orbits about the nucleus and spirals into the center because of the energy radiated. (b) In the Bohr model, the electron orbits without radiating until it jumps to another allowed radius of lower energy, at which time radiation is emitted.

65 Energy level diagram for hydrogen showing the seven lowest stationary states and the four lowest energy transitions for the Lyman, Balmer, and Pashen series. The energies of infinite number of levels are given by E n = (-13.6/n 2 )eV, where n is an integer. λ 21 =hc / (E 1 -E 2 )

66 The spectral lines corresponding to the transitions shown for the three series.

67

68 In a hot star, because of the high temperature, an atom can absorb sufficient energy to remove several electrons from the atom. Consider such a multiply ionized atom with a single remaining electron. The ion produces a series of spectral lines as described by the Bohr model. The series corresponds to electronic transitions that terminate in the same final state. The longest and shortest wavelengths of the series are 63.3 nm and 22.8 nm, respectively. (a) What is the ion? (b) Find the wavelengths of the next three spectral lines nearest to the line of longest wavelength.

69 The shortest wavelength,, corresponds to, and the longest wavelength,, to.

70 In a hot star, because of the high temperature, an atom can absorb sufficient energy to remove several electrons from the atom. Consider such a multiply ionized atom with a single remaining electron. The ion produces a series of spectral lines as described by the Bohr model. The series corresponds to electronic transitions that terminate in the same final state. The longest and shortest wavelengths of the series are 63.3 nm and 22.8 nm, respectively. (a) What is the ion? (b) Find the wavelengths of the next three spectral lines nearest to the line of longest wavelength. Divide (1) and (2): (1) (2)

71 In a hot star, because of the high temperature, an atom can absorb sufficient energy to remove several electrons from the atom. Consider such a multiply ionized atom with a single remaining electron. The ion produces a series of spectral lines as described by the Bohr model. The series corresponds to electronic transitions that terminate in the same final state. The longest and shortest wavelengths of the series are 63.3 nm and 22.8 nm, respectively. (a) What is the ion? (b) Find the wavelengths of the next three spectral lines nearest to the line of longest wavelength. From (2): Hence.

72 In a hot star, because of the high temperature, an atom can absorb sufficient energy to remove several electrons from the atom. Consider such a multiply ionized atom with a single remaining electron. The ion produces a series of spectral lines as described by the Bohr model. The series corresponds to electronic transitions that terminate in the same final state. The longest and shortest wavelengths of the series are 63.3 nm and 22.8 nm, respectively. (a) What is the ion? (b) Find the wavelengths of the next three spectral lines nearest to the line of longest wavelength. (b). Setting gives.

73 A stylized picture of Bohr circular orbits for n=1,2,3,4. The radii r n ≈n 2. In high Z-elements (elements with Z ≥12), electrons are distributed over all the orbits shown. If an electron in the n=1 orbit is knocked from the atom (e.g., by being hit by a fast electron accelerated by the voltage across an x-ray tube) the vacancy that produced is filed by an electron of higher energy (i.e., n=2 or higher). The difference in energy between the two orbits is emitted as a photon, whose wavelength will be in the x-ray region of the spectrum, if Z is large enough The difference in energy between the two orbits is emitted as a photon, whose wavelength will be in the x-ray region of the spectrum, if Z is large enough.

74 Moseley’s plot of the square root of frequency versus Z for characteristic x-rays. When an atom is bombarded by high energy electrons, an inner atomic electron is some time knocked out, leaving a vacancy in the inner shell. The K-series x rays are produced by atomic transitions in the n=1 - (K) shell, whereas the L series is produced by transitions to vacancies in the n=2 (L) shell.

75 Characteristic x-ray spectra. (a) Part the spectra of neodymium (Z=60) and samarium (Z=62).The two pairs of bright lines are the Kα and K β lines. (b) Part of the spectrum of the artificially produced element promethium (Z=61).

76 The Franck-Hertz Experiment While investigating the inelastic scattering of electrons, J.Frank and G.Hertz in 1914 performed an important experiment that confirmed by direct measurement Bohr’s hypothesis of energy quantization in atoms. They found that when the electron kinetic energy was 4.9 eV or grater, the vapor emitted ultraviolet light of wavelength 0.25 μm. The experiment involved measuring the plate current as a function of V 0.

77 The Franck-Hertz Experiment Schematic diagram of the Franck-Hertz experiment. Electrons ejected from the heated cathode C at zero potential are drawn to the positive grid G. Those passing through the holes in the grid can reach the plate P and contribute in the current I, if they have sufficient kinetic energy to overcome the small back potential ΔV. The tube contains a low pressure gas of the element being studied.

78 The Franck-Hertz Experiment As V 0 increased from 0, the current increases until a critical value (about 4.9 V for mercury) is reached. At this point the current suddenly decreases. As V 0 is increased further, the current rises again.

79 Current versus acceleration voltage in the Franck-Hertz experiment. The current decreases because many electrons lose energy due to inelastic collisions with mercury atoms in the tube and therefore cannot overcome the small back potential. Current, (mAmp)

80 The regular spacing of the peaks in this curve indicates that only a certain quantity of energy, 4.9 eV, can be lost to the mercury atoms. This interpretation is confirmed by the observation of radiation of photon energy 4.9 eV, emitted by mercury atoms. Current, (mAmp)

81 Suppose mercury atoms have an energy level 4.9 eV above the lowest energy level. An atom can be raised to this level by collision with an electron; it later decays back to the lowest energy level by emitting a photon. The wavelength of the photon should be This is equal to the measured wavelength, confirming the existence of this energy level of the mercury atom. Similar experiments with other atoms yield the same kind of evidence for atomic energy levels.

82 Lets consider an experimental tube filled by hydrogen atoms instead of mercury. Electrons accelerated by V 0 that collide with hydrogen electrons cannot transfer the energy to letter electrons unless they have acquired kinetic energy eV 0 =E 2 – E 1 =10.2eV

83 If the incoming electron does not have sufficient energy to transfer ΔE = E 2 - E 1 to the hydrogen electron in the n=1 orbit (ground state), than the scattering will be elastic.

84 If the incoming electron does have at least ΔE kinetic energy, than an inelastic collision can occur in which ΔE is transferred to the n=1 electron, moving it to the n=2 orbit. The excited electron will typically return to the ground state very quickly, emitting a photon of energy ΔE.

85 Energy loss spectrum measurement. A well-defined electron beam impinges upon the sample. Electrons inelastically scattered at a convenient angle enter the slit of the magnetic spectrometer, whose B field is directed out of the page, and turn through radii R determined by their energy (E inc – E 1 ) via equation

86 An energy-loss spectrum for a thin Al film.

87 Reduced mass correction The assumption by Bohr that the nucleus is fixed is equivalent the assumption that it has an infinity mass. If instead we will assume that proton and electron both revolve in circular orbits about their common center of mass we will receive even better agreement for the values of the Rydberg constant R and ionization energy for the hydrogen. We can take in account the motion of the nucleus (the proton) very simply by using in Bohr’s equation not the electron rest mass m but a quantity called the reduce mass μ of the system. For a system composed from two masses m 1 and m 2 the reduced mass is defined as:

88 Reduced mass correction If the nucleus has the mass M its kinetic energy will be ½Mv 2 = p 2 /2M, where p = Mv is the momentum. If we assume that the total momentum of the atom is zero, from the conservation of momentum we will have that momentum of electron and momentum of nucleus are equal on the magnitude.

89 Reduced mass correction The total kinetic energy is than: The Rydberg constant equation than changed to: The factor μ was called mass correction factor.

90 As the Earth moves around the Sun, its orbits are quantized. (a) Follow the steps of Bohr’s analysis of the hydrogen atom to show that the allowed radii of the Earth’s orbit are given by where M S is the mass of the Sun, M E is the mass of the Earth, and n is an integer quantum number. (b) Calculate the numerical value of n. (c) Find the distance between the orbit for quantum number n and the next orbit out from the Sun corresponding to the quantum number n + 1

91 As the Earth moves around the Sun, its orbits are quantized. (a) Follow the steps of Bohr’s analysis of the hydrogen atom to show that the allowed radii of the Earth’s orbit are given by where M S is the mass of the Sun, M E is the mass of the Earth, and n is an integer quantum number. (b) Calculate the numerical value of n. (c) Find the distance between the orbit for quantum number n and the next orbit out from the Sun corresponding to the quantum number n + 1 Replace the Coulomb force by the gravitational force exerted by the Sun on the Earth

92 Taking,, we find

93 Since n is very large, we can neglect the number 1 This number is much smaller than the radius of an atomic nucleus so the distance between quantized orbits of the Earth is too small to observe.


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