Survey

Document related concepts

Conservation of energy wikipedia, lookup

Quantum electrodynamics wikipedia, lookup

Bohr–Einstein debates wikipedia, lookup

Density of states wikipedia, lookup

Old quantum theory wikipedia, lookup

Nuclear binding energy wikipedia, lookup

Elementary particle wikipedia, lookup

Nuclear structure wikipedia, lookup

Theoretical and experimental justification for the Schrödinger equation wikipedia, lookup

History of subatomic physics wikipedia, lookup

Atomic nucleus wikipedia, lookup

Hydrogen atom wikipedia, lookup

Transcript

Structure of the Atom CHAPTER 4 Structure of the Atom The Atomic Models of Thomson and Rutherford Rutherford Scattering The Classic Atomic Model The Bohr Model of the Hydrogen Atom Successes & Failures of the Bohr Model Characteristic X-Ray Spectra and Atomic Number Atomic Excitation by Electrons Niels Bohr (1885-1962) The opposite of a correct statement is a false statement. But the opposite of a profound truth may well be another profound truth. An expert is a person who has made all the mistakes that can be made in a very narrow field. Never express yourself more clearly than you are able to think. Prediction is very difficult, especially about the future. - Niels Bohr History • 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. • 1811 – Avogadro’s hypothesis that all gases at given temperature contain the same number of molecules per unit volume. • 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. Historical Developments in Modern Physics • 1895 – Discovery of x-rays by Wilhelm Röntgen. • 1896 – Discovery of radioactivity of uranium by Henri Becquerel • 1897 – Discovery of electron by J.J.Thomson • 1900 – Derivation of black-body radiation formula by Max Plank. • 1905 – Development of special relativity by Albert Einstein, and interpretation of the photoelectric effect. • 1911 – Determination of electron charge by Robert Millikan. • 1911 – Proposal of the atomic nucleus by Ernest Rutherford. • 1913 – Development of atomic theory by Niels Bohr. • 1915 – Development of general relativity by Albert Einstein. • 1924+ - Development of Quantum Mechanics by deBroglie, Pauli, Schrödinger, Born, Heisenberg, Dirac,…. The Structure of Atoms There are 112 chemical elements that have been discovered, and there are a couple of additional chemical elements that recently have been reported. Flerovium is the radioactive chemical element with the symbol Fl and atomic number 114. The element is named after Russian physicist Georgy Flerov, the founder of the Joint Institute for Nuclear Research in Dubna, Russia, where the element was discovered. Georgi Flerov (1913-1990) The name was adopted by IUPAC on May 30, 2012. About 80 decays of atoms of flerovium have been observed to date. All decays have been assigned to the five neighbouring isotopes with mass numbers 285–289. The longest-lived isotope currently known is 289Fl with a half-life of ~2.6 s, although there is evidence for a nuclear isomer, 289bFl, with a half-life of ~66 s, that would be one of the longestlived nuclei in the superheavy element region. The Structure of Atoms 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. 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 from 1 fm to 10 fm (1fm = 10-15m). The distance between the nucleus and the electrons is approximately 0.1 nm=100,000fm. This distance determines the size of the atom. 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! 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. Structure of the Atom Evidence in 1900 indicated that the atom was not a fundamental unit: 1. There seemed to be too many kinds of atoms, each belonging to a distinct chemical element (way more than earth, air, water, and fire!). 2. Atoms and electromagnetic phenomena were intimately related (magnetic materials; insulators vs. conductors; different emission spectra). 3. Elements combine with some elements but not with others, a characteristic that hinted at an internal atomic structure (valence). 4. The discoveries of radioactivity, x-rays, and the electron (all seemed to involve atoms breaking apart in some way). The Nuclear Atoms We will begin our study of atoms by discussing some early models, developed in beginning of 20 century to explain the spectra emitted by hydrogen atoms. Atomic Spectra By the beginning of the 20th 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 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. The source emits only two wavelengths, λ2>λ1. 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. 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. 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. 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. 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 n 364.6 2 nm (n 4) 2 n 3,4,5,...... Balmer suggested that this might be a special case of more general expression that would be applicable to the spectra of other elements. 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: 1 1 R 2 2 n m 1 where m and n are integers with n>m, and R is the Rydberg constant. Atomic Spectra The Rydberg constant is the same for all spectral series. For hydrogen the RH = 1.096776 x 107m-1. For very heavy elements R approaches the value R∞ = 1.097373 x 107m-1. Such empirical expressions were successful in predicting other spectra, such as other hydrogen lines outside the visible spectrum. Atomic Spectra So, the hydrogen Balmer series wavelength are those given by Rydberg equation 1 1 R 2 2 n m 1 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). Hydrogen Spectral Series Compute the wavelengths of the first lines of the Lyman, Balmer, and Paschen series. 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. The Limits of Series Find the predicted by Rydberg-Ritz formula for Lyman, Balmer, and Paschen series. A portion of the emission spectrum of sodium. The two very close bright lines at 589 nm are the D1 and D2 lines. They are the principal radiation from sodium street lighting. A portion of emission spectrum of mercury. 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. Nuclear Models 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 10-10m 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. Knowledge of atoms in 1900 Electrons (discovered in 1897) carried the negative charge. Electrons were very light, even compared to the atom. Protons had not yet been discovered, but clearly positive charge had to be present to achieve charge neutrality. Thomson’s Atomic Model Thomson’s “plum-pudding” model of the atom had the positive charges spread uniformly throughout a sphere the size of the atom, with electrons embedded in the uniform background. In Thomson’s view, when the atom was heated, the electrons could vibrate about their equilibrium positions, thus producing electromagnetic radiation. Unfortunately, Thomson couldn’t explain spectra with this model. The 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. 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. Experiments of Geiger and Marsden Rutherford, Geiger, and Marsden conceived a new technique for investigating the structure of matter by scattering a particles from atoms. 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, that q /m for the α - particles was half that of the proton. 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. Beta decay Beta decay occurs when the neutron to proton ratio is too great in the nucleus and causes instability. In basic beta decay, a neutron is turned into a proton and an electron. The electron is then emitted. Here's a diagram of beta decay with hydrogen-3: Beta Decay of Hydrogen-3 to Helium-3. Alpha Decay The reason alpha decay occurs is because the nucleus has too many protons which cause excessive repulsion. In an attempt to reduce the repulsion, a Helium nucleus is emitted. The way it works is that the Helium nuclei are in constant collision with the walls of the nucleus and because of its energy and mass, there exists a nonzero probability of transmission. That is, an alpha particle (Helium nucleus) will tunnel out of the nucleus. Here is an example of alpha emission with americium-241: Alpha Decay of Americium-241 to Neptunium-237 Gamma Decay Gamma decay occurs because the nucleus is at too high an energy. The nucleus falls down to a lower energy state and, in the process, emits a high energy photon known as a gamma particle. Here's a diagram of gamma decay with helium-3: Gamma Decay of Helium-3 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, that q /m for the α - particles was half that of the proton. 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. 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 214Bi . 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. A diagram of the original apparatus as it appear in Geiger’s paper describing the results. An α-particle by such an atom (Thompson model) would have a scattering angle θ much smaller than 10. In the Rutherford’s scattering experiment most of the α-particles were either undeflected, or deflected through very small angles of the order 10, however, a few α-particles were deflected through angles of 900 and more. An α-particle by such an atom (Thompson model) would have a scattering angle θ much smaller than 10. In the Rutherford’s scattering experiment a few α-particles were deflected through angles of 900 and more. Experiments of Geiger and Marsden Geiger showed that many a particles were scattered from thin gold-leaf targets at backward angles greater than 90°. Electrons can’t backscatter a particles. Before After Calculate the maximum scattering angle - corresponding to the maximum momentum change. It can be shown that the maximum momentum transfer to the a particle is: Dpmax 2me va Determine qmax by letting Dpmax be perpendicular to the direction of motion: q max Dpa 2me va pa M a va too small! Try multiple scattering from electrons If an a particle is scattered by N electrons: N = the number of atoms across the thin gold layer, t = 6 × 10−7 m: n= The distance between atoms, d = n-1/3, is: N=t/d still too small! If the atom consisted of a positively charged sphere of radius 10-10 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. Rutherford’s Atomic Model even if the α particle is scattered from all 79 electrons in each atom of gold. Experimental results were not consistent with Thomson’s atomic model. Rutherford proposed that an atom has a positively charged core (nucleus!) surrounded by the negative electrons. Geiger and Marsden confirmed the idea in 1913. Ernest Rutherford (1871-1937) 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. 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 kqαQ/r2. 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. 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/r2, inside the sphere the force is proportional to qI/r2= Qr/R3, where qI = Q(r/R)3 is the charge within a sphere of radius r. The maximum force occurs at r =R Two α particles with equal kinetic energies approach the positive charge Q = +Ze with impact parameters b1 and b2, where b1<b2. According to equation for impact parameter in this case θ1 > θ2. 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. bk qa Q q ma v 2 cot 2 The quantity πb2, which has the dimension of the area , is called the cross section σ for scattering. 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. The total number of particles scattered per second is obtained by multiplying πb2I0 by the number of nuclei in the scattering foil. Let n be the number of nuclei per unit volume: g at 3 N A mol N A at cm n M cm3 g M mol For a foil of thickness t, the total number of nuclei as “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 πb2I0ntA. If divide this by the number of α particles incident per second I0A we get the fraction of α particles f scattered through angles grater than θ: f = πb2nt 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 : DN I 0 Ant r2 kZe 2 Ek 2 2 1 4q sin 2 All this predictions was verified in Geiger experiments, who observe several hundreds thousands α particles. (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. Data from Rutherford’s group showing observed α scattering at a large fixed angle versus values of rd qa Q for various kinetic energies. computed from rd k 1 ma v 2 2 The Size of the Nucleus This equation can be used to estimate the size of the nucleus qa Q rd k 1 2 ma v 2 For the case of 7.7-MeV α particles the distance of closest approach for a head-on collision is (2)(79)(1.44eV nm) 5 14 rd 3 10 nm 3 10 m 6 7.7 10 eV The Classical Atomic Model Consider an atom as a planetary system. The Newton’s 2nd Law force of attraction on the electron by the nucleus is: 2 1 e mv Fe 2 4 0 r r 2 where v is the tangential velocity of the electron: v e e 2 K m v 4 0 mr 4 0 r 1 2 2 1 2 The total energy is then: This is negative, so the system is bound, which is good. The Planetary Model is Doomed From classical E&M theory, an accelerated electric charge radiates energy (electromagnetic radiation), which means the total energy must decrease. So the radius r must decrease!! Electron crashes into the nucleus!? According to classical physics, a charge e moving with an acceleration a radiates at a rate dE 1 e2a 2 dt 6 0 c 3 (a) Show that an electron in a classical hydrogen atom spirals into the nucleus at a rate dr e4 2 2 2 3 2 dt 1 2 0 r m e c (b) Find the time interval over which the electron will reach r = 0, starting from r0 = 2.00 × 10–10 m. 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. 2. The photon frequency from energy conservation is given by f Ei E f h where Ei and Ef are the energy of initial and final state, h is the Plank’s constant. Such a model is mechanically stable , because the Coulomb potential 2 Ze U k r provides the centripetal force 2 k Ze mv F 2 r r 2 1 k Ze 2 mv 2 2r 2 The total energy for a such system can be written as the sum of kinetic and potential energy: 2 2 k Ze k Ze k Ze E 2r r 2r 2 The Bohr’s Postulates Combining the second postulate with the equation for the energy we obtain: E1 E 2 1 k Ze 1 1 f h 2 h r2 r1 2 where r1 and r2 are the radii of the initial and final orbits. The Bohr’s Postulates E1 E 2 1 k Ze 1 1 f h 2 h r2 r1 2 To obtain the frequencies implied by the experimental Rydberg-Ritz formula, 1 c 1 f cR 2 2 n 2 n1 it is evident that the radii of the stable orbits must be proportional to the squares of integers. 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. nh mvr n 2 n=1,2,3………. where ħ=h/2π=1.055 x 10-34J·s=6.582x10 -16eV·s The Bohr’s Postulates The obtained equation mvr = nh/2π=nħ relates the speed of electron v to the radius r. Since we had 2 kZe mv 2 r r 2 or kZe v mr 2 2 We can write 2 v n 2 2 m r 2 2 2 or k Ze n 2 2 m r mr 2 2 The Bohr’s Postulates Solving for r, we obtain 2 2 a0 rn n 2 m k Ze Z 2 where a0 is called the first Bohr’s radius 2 a0 0 . 0529 nm 2 mke Bohr’s Postulates Substituting the expression for r in equation for frequency: 2 4 1 k Ze 1 1 2 mk e Z f 3 2 h r2 r1 4 2 1 1 2 2 n 2 n1 If we will compare this expression with Z=1 for f=c/λ with the empirical Rydberg-Ritz formula: 1 1 R 2 2 n 2 n1 1 2 4 we will obtain for the Rydberg constant R mk e 3 4 c Bohr’s Postulates 2 4 mk e R 4c 3 Using the known values of m, e, and ħ, Bohr calculated 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 1 kZe2 E 2 r Energy levels If we will substitute the quantized value of r as given by 2 2 a0 rn n 2 mkZe Z 2 we obtain 2 2 2 2 2 4 1 kZe 1 kZ e 1 mk Z e En 2 2 2 2 r 2 n a0 2 n Energy levels 1 kZe2 1 kZ 2e 2 1 mk 2 Z 2e 4 En 2 2 r 2 n a0 2 n 2 2 or where E0 En Z 2 n 2 mk 2e 4 1 ke2 E0 13.6eV 2 2 2 a0 The energies En with Z=1 are the quantized allowed energies for the hydrogen atom. Energy level diagram for hydrogen showing the seven lowest stationary states. The energies of infinite number of levels are given by En = (-13.6/n2)eV, where n is an integer. 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. Energy levels Transitions between this allowed energies result in the emission or absorption of a photon whose frequency is given by f Ei E f h and whose wavelength is c hc f Ei E f 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. (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. λ21=hc / (E1-E2) 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 En = (-13.6/n2)eV, where n is an integer. hc 1240eV nm 12 121.5nm E1 E2 (13.6 3.4)eV 1 1 1 R 2 2 R1 2 1.096776 107 m 1 (0.75) 12 m n 2 1 12 1.2156 10 7 m 121.5nm The spectral lines corresponding to the transitions shown for the three series. λ21=hc / (E1-E2) Compute the wavelength of the Hβ spectral line of the Balmer series predicted by Bohr model. A hydrogen atom at rest in the laboratory emits a photon of the Lyman α radiation. (a) Compute the recoil kinetic energy of the atom. (b) What fraction of the excitation energy of the n = 2 state is carried by the recoiling atom? (Hint: Use conservation of momentum.) 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. A stylized picture of Bohr circular orbits for n=1,2,3,4. The radii rn≈n2. 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. (a) (b) (c) 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 Kβ lines. (b) Part of the spectrum of the artificially produced element promethium (Z=61), its Kα and Kβ lines fall between those of Nd and Sm. (c) Part of the spectrum of all three elements. 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. The experiment involved measuring the plate current as a function of V0. 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. The Franck-Hertz Experiment As V0 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 V0 is increased further, the current rises again. They found that when the electron’s kinetic energy was 4.9 eV or greater, the vapor of mercury emitted ultraviolet light of wavelength 0.25 μm. Current, (mAmp) 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) 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. 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 hc 1240 eV nm 253 . 06 nm 0 . 25 m E 4 . 9 eV 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. Lets consider an experimental tube filled by hydrogen atoms instead of mercury. Electrons accelerated by V0 that collide with hydrogen electrons cannot transfer the energy to letter electrons unless they have acquired kinetic energy eV0=E2 – E1=10.2eV If the incoming electron does not have sufficient energy to transfer ΔE = E2 - E1 to the hydrogen electron in the n=1 orbit (ground state), than the scattering will be elastic. If the incoming electron does have at least ΔE kinetic energy, then 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. 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 (Einc – E1) via equation 2me ( Einc E1 ) R eB An energy-loss spectrum for a thin Al film. 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 m1 and m2 the reduced mass is defined as: m1 m 2 m1 m 2 Reduced mass correction If the nucleus has the mass M its kinetic energy will be ½Mv2 = p2/2M, 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. Reduced mass correction The total kinetic energy is then: mM m p p (M m) p p m M m Ek 1 2 M 2m 2Mm 2 M The Rydberg constant equation than changed to: 2 2 2 2 m R R R m 1 M The factor μ was called mass correction factor. 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 r n 2 2 GM S M E 2 where MS is the mass of the Sun, ME 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