Structure of the Atom

Structure of the AtomStructure of the Atom

The Atomic Models of Thomson and RutherfordThe Atomic Models of Thomson and Rutherford Rutherford ScatteringRutherford Scattering The Classic Atomic ModelThe Classic Atomic Model The Bohr Model of the Hydrogen AtomThe Bohr Model of the Hydrogen Atom Successes & Failures of the BohrSuccesses & Failures of the Bohr ModelModel Characteristic X-Ray Spectra and Atomic Characteristic X-Ray Spectra and Atomic

NumberNumber Atomic Excitation by ElectronsAtomic Excitation by Electrons

CHAPTER 4Structure of the AtomStructure of the Atom

The opposite of a correct statement is a false statement. But the opposite of a profound The opposite of a correct statement is a false statement. But the opposite of a profound truth may well be another profound truth. 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 An expert is a person who has made all the mistakes that can be made in a very narrow field. field.

Never express yourself more clearly than you are able to think. Never express yourself more clearly than you are able to think.

Prediction is very difficult, especially about the future. Prediction is very difficult, especially about the future. - Niels Bohr- Niels Bohr

Niels Bohr (1885-1962)Niels Bohr (1885-1962)

HistoryHistory

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

• XVII-th century, Pierre Cassendi, Robert Hook – explained XVII-th century, Pierre Cassendi, Robert Hook – explained states of matter and transactions between them with a model states of matter and transactions between them with a model of tiny indestructible solid objects.of tiny indestructible solid objects.

• 1811 – Avogadro’s hypothesis that all gases at given 1811 – Avogadro’s hypothesis that all gases at given temperature contain the same number of molecules per unit temperature contain the same number of molecules per unit volume.volume.

• 1900 – Kinetic theory of gases.1900 – Kinetic theory of gases.

Consequence – Great three quantization discoveries of XX Consequence – Great three quantization discoveries of XX century: (1) electric charge: (2) light energy; (3) energy of century: (1) electric charge: (2) light energy; (3) energy of oscillating mechanical systems. oscillating mechanical systems.

Historical Developments in Modern PhysicsHistorical Developments in Modern Physics • 1895 – Discovery of x-rays by 1895 – Discovery of x-rays by Wilhelm RWilhelm Röntgenöntgen..• 1896 – Discovery of radioactivity of uranium by 1896 – Discovery of radioactivity of uranium by Henri Henri

BecquerelBecquerel• 1897 – Discovery of electron by 1897 – Discovery of electron by J.J.ThomsonJ.J.Thomson• 1900 – Derivation of black-body radiation formula by 1900 – Derivation of black-body radiation formula by Max Max

Plank.Plank.• 1905 – Development of special relativity by 1905 – Development of special relativity by Albert EinsteinAlbert Einstein, ,

and interpretation of the photoelectric effect.and interpretation of the photoelectric effect.• 1911 – Determination of electron charge by 1911 – Determination of electron charge by Robert MillikanRobert Millikan..• 1911 – Proposal of the atomic nucleus by 1911 – Proposal of the atomic nucleus by Ernest RutherfordErnest Rutherford..• 1913 – Development of atomic theory by 1913 – Development of atomic theory by Niels BohrNiels Bohr..• 1915 – Development of general relativity by 1915 – Development of general relativity by Albert EinsteinAlbert Einstein..• 1924+ - Development of Quantum Mechanics by 1924+ - Development of Quantum Mechanics by deBrogliedeBroglie, ,

Pauli, Schrödinger, Born, Heisenberg, Dirac,….Pauli, Schrödinger, Born, Heisenberg, Dirac,….

There are There are 112 chemical elements112 chemical elements that have that have been discovered, and there are a couple of additional been discovered, and there are a couple of additional chemical elements that recently have been reported. chemical elements that recently have been reported.

FleroviumFlerovium is the radioactive chemical is the radioactive chemical element with the symbol element with the symbol FlFl and atomic number and atomic number 114114. . The element is named after Russian physicist The element is named after Russian physicist Georgy FlerovGeorgy Flerov, the founder of the Joint Institute for , the founder of the Joint Institute for Nuclear Research in Dubna, Russia, where the Nuclear Research in Dubna, Russia, where the element was discovered.element was discovered.

Georgi Flerov (1913-1990)Georgi Flerov (1913-1990)

The Structure of AtomsThe Structure of Atoms

The name was adopted by IUPAC on May 30, 2012.The name was adopted by IUPAC on May 30, 2012. About 80 decays of atoms of About 80 decays of atoms of fleroviumflerovium have been observed to date. All decays have been assigned to the five have been observed to date. All decays have been assigned to the five neighbouring isotopes with mass numbers 285–289. The longest-lived isotope neighbouring isotopes with mass numbers 285–289. The longest-lived isotope currently known is currently known is 289289Fl Fl with a half-life of ~2.6 s, although there is evidence for a with a half-life of ~2.6 s, although there is evidence for a nuclear isomer, nuclear isomer, 289b289bFlFl, with a half-life of ~66 s, that would be one of the longest-, with a half-life of ~66 s, that would be one of the longest-lived nuclei in the superheavy element region.lived nuclei in the superheavy element region.

http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=Fc3oNX8v_PJ_JM&tbnid=iU7D8EArb_NkZM:&ved=0CAgQjRwwAA&url=http%3A%2F%2Ffls2.jinr.ru%2Fflnr%2Fhistory%2Fflerov.html&ei=hyw_UaG-Ioiq8ATQ54DQCw&psig=AFQjCNHm3SN3JBw3-dVwEWc0wNUjhNsKvQ&ust=1363181063615132

The Structure of AtomsThe Structure of Atoms

Each element is characterized by atom that Each element is characterized by atom that contains a number of protonscontains a number of protons ZZ, and equal number , and equal number of electrons, and a number of neutronsof electrons, and a number of neutrons NN. . The The number of protonsnumber of protons ZZ is called theis called the atomic numberatomic number. .

The lightest atom, The lightest atom, hydrogenhydrogen ( (HH), has), has Z=1Z=1; ; the next lightest atom, the next lightest atom, heliumhelium ( (HeHe), has), has Z=2Z=2; the ; the third lightest, third lightest, lithiumlithium ( (LiLi), has), has Z=3Z=3 and so forth.and so forth.

The Nuclear AtomsThe Nuclear Atoms

Nearly all the mass of the atom is Nearly all the mass of the atom is concentrated in a tiny nucleus which contains the concentrated in a tiny nucleus which contains the protons and neutrons. protons and neutrons.

Typically, the nuclear radius is approximately Typically, the nuclear radius is approximately from from 1 fm1 fm to to 10 fm10 fm ((1fm = 101fm = 10-15-15mm). The distance ). The distance between the nucleus and the electrons is between the nucleus and the electrons is approximatelyapproximately 0.1 nm=100,000fm0.1 nm=100,000fm. This distance . This distance determines the determines the sizesize of the atom. of the atom.

Nuclear StructureNuclear Structure

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

The number of protons in the The number of protons in the nucleus,nucleus, ZZ,, is called the is called the atomic atomic numbernumber. This determines what . This determines what chemical element the atom is. The chemical element the atom is. The number of neutrons in the nucleus number of neutrons in the nucleus is denoted by is denoted by NN. The . The atomic massatomic mass of the nucleus, of the nucleus, AA, is, is equal to equal to Z + NZ + N..

A given element can have A given element can have many different isotopes, which many different isotopes, which differ from one another by the differ from one another by the number of neutrons contained in number of neutrons contained in the nuclei. In a neutral atom, the the nuclei. In a neutral atom, the number of electrons orbiting the number of electrons orbiting the nucleus equals the number of nucleus equals the number of protons in the nucleus. protons in the nucleus.

Structure of the AtomStructure of the Atom

Evidence in 1900 indicated that Evidence in 1900 indicated that the atom wasthe atom was notnot a fundamental unit:a fundamental unit:

1.1. There seemed to be too many kinds There seemed to be too many kinds of atoms, each belonging to a distinct chemical of atoms, each belonging to a distinct chemical element (way more than earth, air, water, and fire!).element (way more than earth, air, water, and fire!).

2.2. Atoms and electromagnetic phenomena were Atoms and electromagnetic phenomena were intimately related (magnetic materials; insulators vs. intimately related (magnetic materials; insulators vs. conductors; different emission spectra). conductors; different emission spectra).

3.3. Elements combine with some elements but not with Elements combine with some elements but not with others, a characteristic that hinted at an internal atomic others, a characteristic that hinted at an internal atomic structure (valence).structure (valence).

4.4. The discoveries of radioactivity, The discoveries of radioactivity, x-x-rays, and the rays, and the electron (all seemed to involve atoms breaking apart in electron (all seemed to involve atoms breaking apart in some way).some way).

The Nuclear AtomsThe Nuclear Atoms

We will begin our study of atoms by We will begin our study of atoms by discussing some early models, developed in discussing some early models, developed in beginning of 20 century to explain the spectra beginning of 20 century to explain the spectra emitted by hydrogen atoms.emitted by hydrogen atoms.

Atomic SpectraAtomic SpectraBy the beginning of the 20By the beginning of the 20thth century a large body of data has century a large body of data has been collected on the emission of light by atoms of individual been collected on the emission of light by atoms of individual elements in a flame or in a gas exited by electrical discharge.elements in a flame or in a gas exited by electrical discharge.

Diagram of the spectrometerDiagram of the spectrometer

Atomic SpectraAtomic Spectra

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

The source emits only two wavelengths, The source emits only two wavelengths, λλ22>>λλ11. The source is . The source is

located at the focal point of the lens so that parallel light passes 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 through the narrow slit, projecting a narrow line onto the face of the prism. Ordinary dispersion in the prism bends the shorter the prism. Ordinary dispersion in the prism bends the shorter wavelength through the lager total angel, separating the two wavelength through the lager total angel, separating the two wavelength at the screen. wavelength at the screen.

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

When viewed through the When viewed through the spectroscope, the characteristic radiation, spectroscope, the characteristic radiation, emitted by atoms of individual elements in emitted by atoms of individual elements in flame or in gas exited by electrical charge, flame or in gas exited by electrical charge, appears as a set of discrete lines, each of a appears as a set of discrete lines, each of a particular color or wavelength. particular color or wavelength.

The positions and intensities of the The positions and intensities of the lines are a characteristic of the element. lines are a characteristic of the element. The wavelength of these lines could be The wavelength of these lines could be determined with great precision. determined with great precision.

Emission line spectrum of hydrogen in the visible and near Emission line spectrum of hydrogen in the visible and near ultraviolet. The lines appear dark because the spectrum ultraviolet. The lines appear dark because the spectrum was photographed. The names of the first five lines are was photographed. The names of the first five lines are shown. As is the point beyond which no lines appear,shown. As is the point beyond which no lines appear, HH∞∞

called thecalled the limits of the serieslimits of the series..


In 1885 a Swiss schoolteacher, Johann Balmer, In 1885 a Swiss schoolteacher, Johann Balmer, found that the wavelengths of the lines in the visible found that the wavelengths of the lines in the visible spectrum of hydrogen can be represented by formulaspectrum of hydrogen can be represented by formula

Balmer suggested that this might be a special Balmer suggested that this might be a special case of more general expression that would be case of more general expression that would be applicable to the spectra of other elements.applicable to the spectra of other elements.

,......5,4,3)4(

6.3642

2

nnmn

n


Such an expression, found by J.R.Rydberg Such an expression, found by J.R.Rydberg and W. Ritz and known as the Rydberg-Ritz and W. Ritz and known as the Rydberg-Ritz formula, gives the reciprocal wavelengths as:formula, gives the reciprocal wavelengths as:

where where mm and and nn are integers with are integers with n>mn>m, and , and RR is is the the Rydberg constantRydberg constant. .

22

111

nmR


The The Rydberg constantRydberg constant is the same for all is the same for all spectral series. spectral series.

For hydrogen the For hydrogen the RRH H = 1.096776 x 10 = 1.096776 x 1077mm-1-1..

For very heavy elements For very heavy elements RR approaches the approaches the value value RR∞∞ = 1.097373 x 10 = 1.097373 x 1077mm-1-1..

Such empirical expressions were Such empirical expressions were successful in predicting other spectra, such as successful in predicting other spectra, such as other hydrogen lines outside the visible spectrumother hydrogen lines outside the visible spectrum.


So, the hydrogen So, the hydrogen Balmer seriesBalmer series wavelength wavelength are those given by Rydberg equationare those given by Rydberg equation

with with m=2m=2 and and n=3,4,5,…n=3,4,5,…

Other series of hydrogen spectral lines were found Other series of hydrogen spectral lines were found for for m=1m=1 (by (by LymanLyman) and ) and m=3m=3 (by (by PaschenPaschen). ).

22

111

nmR

Hydrogen Spectral SeriesHydrogen Spectral Series

Compute the wavelengths of the first Compute the wavelengths of the first lines of the lines of the LymanLyman, , BalmerBalmer, and , and PaschenPaschen series. series.

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

called thecalled the limits of the serieslimits of the series..

The Limits of SeriesThe Limits of Series

Find the predicted by Rydberg-Ritz Find the predicted by Rydberg-Ritz formula for Lyman, Balmer, and Paschen formula for Lyman, Balmer, and Paschen series.series.

A portion of the emission spectrum of sodium. The two A portion of the emission spectrum of sodium. The two very close bright lines atvery close bright lines at 589 nm589 nm are theare the DD11 and and DD22 lines. lines.

They are the principal radiation from sodium streetThey are the principal radiation from sodium street lighting. lighting.

A portion of emission spectrum of mercury.A portion of emission spectrum of mercury.

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

Nuclear ModelsNuclear ModelsMany attempts were made to construct a Many attempts were made to construct a

model of the atom that yielded the model of the atom that yielded the Balmer Balmer and and Rydberg-Ritz Rydberg-Ritz formulas. formulas.

It was known that an atom was aboutIt was known that an atom was about 1010--

1010mm in diameter, that it contained electrons much in diameter, that it contained electrons much lighter than the atom, and that it was electrically lighter than the atom, and that it was electrically neutral. neutral.

The most popular model was that of The most popular model was that of J.J.ThomsonJ.J.Thomson, already quite successful in , already quite successful in explaining chemical reactions. explaining chemical reactions.

Knowledge of atoms in 1900

Electrons (discovered in Electrons (discovered in 1897) carried the negative 1897) carried the negative charge.charge.

Electrons were very light, Electrons were very light, even compared to the atom.even compared to the atom.

Protons had not yet been Protons had not yet been discovered, but clearly discovered, but clearly positive charge had to be positive charge had to be present to achieve charge present to achieve charge neutralityneutrality..

In Thomson’s view, when the atom was heated, the In Thomson’s view, when the atom was heated, the electrons could vibrate about their equilibrium positions, electrons could vibrate about their equilibrium positions, thus producing electromagnetic radiation.thus producing electromagnetic radiation.

Unfortunately, Thomson couldn’t explain spectra with this Unfortunately, Thomson couldn’t explain spectra with this model.model.

Thomson’s Atomic ModelThomson’s Atomic Model

Thomson’s “Thomson’s “plum-puddingplum-pudding” ” model of the atom had the model of the atom had the positive charges spread positive charges spread uniformly throughout a sphere uniformly throughout a sphere the size of the atom, with the size of the atom, with electrons embedded in the electrons embedded in the uniform background.uniform background.

The difficulty with all such models was The difficulty with all such models was that electrostatic forces alone cannot produce stable that electrostatic forces alone cannot produce stable equilibrium. Thus the charges were required to move equilibrium. Thus the charges were required to move and, if they stayed within the atom, to accelerate. and, if they stayed within the atom, to accelerate. However, the acceleration would result in continuous However, the acceleration would result in continuous radiation, which is not observed. radiation, which is not observed.

Thomson was unable to obtain from his model a set of Thomson was unable to obtain from his model a set of

frequencies that corresponded with the frequencies of frequencies that corresponded with the frequencies of observed spectra. observed spectra.

The TThe Thomsonhomson model of the atom was model of the atom was

replaced by one based on results of a set of replaced by one based on results of a set of experiments conducted by experiments conducted by Ernest RutherfordErnest Rutherford and his and his student student H.W.GeigerH.W.Geiger. .

Experiments of Geiger and MarsdenExperiments of Geiger and Marsden

Rutherford, Geiger, and Rutherford, Geiger, and Marsden conceived a new Marsden conceived a new technique for investigating technique for investigating the structure of matter by the structure of matter by scattering scattering particles from particles from atoms. atoms.

RutherfordRutherford was investigating radioactivity was investigating radioactivity and had shown that the radiations from uranium and had shown that the radiations from uranium consist of at least two types, which he labeledconsist of at least two types, which he labeled αα andand ββ.

He showed, by an experiment similar to He showed, by an experiment similar to that of that of ThompsonThompson, that, that q /mq /m for thefor the αα - particles - particles was half that of the proton.was half that of the proton.

Suspecting that theSuspecting that the αα particles were particles were double ionized helium, double ionized helium, Rutherford Rutherford in his in his classical experiment let a radioactive substance classical experiment let a radioactive substance decay and then, by spectroscopy, detected the decay and then, by spectroscopy, detected the spectra line of ordinary helium. spectra line of ordinary helium.

Beta decayBeta decay Beta decay occurs when the neutron to proton ratio is too Beta decay occurs when the neutron to proton ratio is too great in the nucleus and causes instability. In basic beta great in the nucleus and causes instability. In basic beta decay, a neutron is turned into a proton and an electron. decay, a neutron is turned into a proton and an electron. The electron is then emitted. Here's a diagram of beta The electron is then emitted. Here's a diagram of beta decay with hydrogen-3:decay with hydrogen-3:

Beta Decay of Hydrogen-3 to Helium-3. Beta Decay of Hydrogen-3 to Helium-3.

Alpha DecayAlpha DecayThe reason alpha decay occurs is because the nucleus has too many The reason alpha decay occurs is because the nucleus has too many protons which cause excessive repulsion. In an attempt to reduce the 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 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 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. 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. That is, an alpha particle (Helium nucleus) will tunnel out of the nucleus. Here is an example of alpha emission with americium-241:Here is an example of alpha emission with americium-241:

Alpha Decay of Americium-241 to Neptunium-237Alpha 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-3Gamma Decay of Helium-3

RutherfordRutherford was investigating radioactivity was investigating radioactivity and had shown that the radiations from uranium and had shown that the radiations from uranium consist of at least two types, which he labeledconsist of at least two types, which he labeled αα andand ββ.

He showed, by an experiment similar to He showed, by an experiment similar to that of that of ThompsonThompson, that, that q /mq /m for thefor the αα - particles - particles was half that of the proton.was half that of the proton.

Suspecting that theSuspecting that the αα particles were particles were double ionized helium, double ionized helium, Rutherford Rutherford in his in his classical experiment let a radioactive substance classical experiment let a radioactive substance decay and then, by spectroscopy, detected the decay and then, by spectroscopy, detected the spectra line of ordinary helium. spectra line of ordinary helium.

Schematic diagram of the Rutherford apparatus. The beam Schematic diagram of the Rutherford apparatus. The beam ofof αα - particles is defined by the small hole- particles is defined by the small hole DD in the shield in the shield surrounding the radioactive sourcesurrounding the radioactive source R R ofof 214214BiBi . The. The αα beam strikes beam strikes an ultra thin gold foil an ultra thin gold foil FF, and , and αα particles are individually scattered particles are individually scattered through various angelsthrough various angels θθ. . The experiment consisted of counting The experiment consisted of counting the number of scintillations on the screenthe number of scintillations on the screen SS as a function ofas a function of θθ..

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

An An αα-particle by such an atom (Thompson model) would have -particle by such an atom (Thompson model) would have

a scattering angle a scattering angle θθ much smaller than much smaller than 1100. . In the Rutherford’s In the Rutherford’s

scattering experiment most of the scattering experiment most of the αα--particles were either particles were either

undeflected, or deflected through very small angles of the undeflected, or deflected through very small angles of the

order order 1100, however, a few , however, a few αα-particles were deflected through -particles were deflected through

angles of angles of 909000 and more.and more.

An An αα-particle by such an atom (Thompson model) would have -particle by such an atom (Thompson model) would have

a scattering angle a scattering angle θθ much smaller than much smaller than 1100. . In the Rutherford’s In the Rutherford’s

scattering experiment scattering experiment a few a few αα-particles were deflected through -particles were deflected through

angles of angles of 909000 and more.and more.

Experiments of Geiger and Marsden Experiments of Geiger and Marsden

Geiger showed that many Geiger showed that many particles were scattered from particles were scattered from thin gold-leaf targets at backward angles greater than 90thin gold-leaf targets at backward angles greater than 90°°..

Electrons Electrons can’t back-can’t back-scatter scatter particles.particles.Calculate the maximum scattering angle - corresponding to the Calculate the maximum scattering angle - corresponding to the maximum momentum change.maximum momentum change.

It can be shown that the maximum It can be shown that the maximum

momentum transfer to the momentum transfer to the particle is: particle is:

Determine Determine maxmax by letting by letting

ppmaxmax be perpendicular to be perpendicular to

the direction of motion:the direction of motion:

Before After

2 vmax ep m

2 v

ve

max

p m

p M

too small!too small!

If an If an particle is scattered by particle is scattered by NN electrons: electrons:

Try multiple scattering from electronsTry multiple scattering from electrons

The distance between atoms, The distance between atoms, d = nd = n-1/3-1/3, is:, is:

NN = the number of atoms across the thin gold layer= the number of atoms across the thin gold layer, t = 6 × 10−7 m:

still too small!still too small!

n =

N = t / d

If the atom consisted of a positively charged If the atom consisted of a positively charged sphere of radiussphere of radius 1010-10 -10 mm, containing electrons as in , containing electrons as in the Thomson model, only a very small scattering the Thomson model, only a very small scattering deflection angle could be observed. deflection angle could be observed.

Such model could not possibly account for Such model could not possibly account for the large angles scattering. The unexpected large the large angles scattering. The unexpected large anglesangles αα-particles -particles scattering was described by scattering was described by Rutherford with these words:Rutherford with these words:

It was quite incredible event that ever It was quite incredible event that ever happened to me in my life. It was as incredible happened to me in my life. It was as incredible as if you fired a 15-inch shell at a piece of tissue as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you. paper and it came back and hit you.

even if the even if the αα particle is scattered from all 79 particle is scattered from all 79 electrons in each atom of gold.electrons in each atom of gold.

Experimental results were not Experimental results were not consistent with Thomson’s atomic consistent with Thomson’s atomic model.model.

Rutherford proposed that an atom Rutherford proposed that an atom has a positively charged core has a positively charged core ((nucleus!nucleus!) surrounded by the ) surrounded by the negative electrons.negative electrons.

Geiger and Marsden confirmed the Geiger and Marsden confirmed the idea in 1913.idea in 1913.

Rutherford’s Atomic ModelRutherford’s Atomic Model

Ernest Rutherford Ernest Rutherford (1871-1937)(1871-1937)

Rutherford concluded that the large angle Rutherford concluded that the large angle scattering could result only from a single encounter scattering could result only from a single encounter of theof the αα particle with a massive charge with volume particle with a massive charge with volume much smaller than the whole atom. much smaller than the whole atom.

Assuming thisAssuming this “nucleus”“nucleus” to be a point to be a point charge, he calculated the expected angular charge, he calculated the expected angular distribution for the scattereddistribution for the scattered αα particles. particles.

His predictions on the dependence of His predictions on the dependence of scattering probability on angle, nuclear charge and scattering probability on angle, nuclear charge and kinetic energy were completely verified in kinetic energy were completely verified in experiments. experiments.

Rutherford Scattering geometry. The nucleus is assumed to Rutherford Scattering geometry. The nucleus is assumed to be a point chargebe a point charge QQ at the originat the origin OO. At any distance. At any distance r r thethe αα particle experiences a repulsive forceparticle experiences a repulsive force kqkqααQ/rQ/r22. . The The αα particle particle

travel along a hyperbolic path that is initially parallel to linetravel along a hyperbolic path that is initially parallel to line OAOA a distancea distance bb from it and finally parallel tofrom it and finally parallel to OBOB, which , which makes an anglemakes an angle θθ with with OAOA..

Force on a point charge versus distanceForce on a point charge versus distance rr from the center from the center of a uniformly charged sphere of radiusof a uniformly charged sphere of radius RR. Outside the . Outside the sphere the force is proportional tosphere the force is proportional to Q/rQ/r22, inside the sphere , inside the sphere the force is proportional tothe force is proportional to qqII/r/r22= Qr/R= Qr/R33, where, where qqII = Q(r/R) = Q(r/R)3 3

is the charge within a sphere of radius is the charge within a sphere of radius rr. The maximum . The maximum force occurs atforce occurs at r =Rr =R

TwoTwo αα particles with equal kinetic energies particles with equal kinetic energies approach the positive chargeapproach the positive charge Q = +ZeQ = +Ze with with impact parametersimpact parameters bb11 andand bb22, where, where bb11<b<b22. . According to equation for impact parameter in According to equation for impact parameter in this casethis case θθ11 > > θθ22..

The path ofThe path of αα particle can be shown to be a particle can be shown to be a hyperbola, and the scattering anglehyperbola, and the scattering angle θθ can be relate to can be relate to the the impact parameterimpact parameter bb from the laws of classical from the laws of classical mechanics.mechanics.

The quantityThe quantity ππbb22, which has the, which has the dimension of dimension of the area , is called the the area , is called the cross sectioncross section σσ for for scattering.scattering.

2cot

2

vm

Qqkb

The The cross sectioncross section σσ is thus defined as the is thus defined as the number of particles number of particles scattered per nucleus scattered per nucleus per unit time divided per unit time divided by the incident by the incident intensity. intensity.

The total number of nuclei of foil atoms in the area covered The total number of nuclei of foil atoms in the area covered by the beam isby the beam is nAtnAt, where, where nn is the number of foil atoms per is the number of foil atoms per unit volume.unit volume. AA is the area of the beam, andis the area of the beam, and t t is the thickness is the thickness of the foil. of the foil.

The total number of particles scattered per second is The total number of particles scattered per second is obtained by multiplyingobtained by multiplying ππbb22II00 by the number of nuclei in the by the number of nuclei in the scattering foil. Letscattering foil. Let n n be the number of nuclei per unit volume:be the number of nuclei per unit volume:

For a foil of thicknessFor a foil of thickness t t, the total number of nuclei as , the total number of nuclei as “seen” by the beam is“seen” by the beam is nAtnAt, where, where AA is the area of the beam. is the area of the beam. The total number scattered per second through angles grater The total number scattered per second through angles grater thanthan θθ is thusis thus ππbb22II00ntAntA. If divide this by the number of. If divide this by the number of αα particles incident per secondparticles incident per second II00AA we get the fraction of we get the fraction of αα particlesparticles f f scattered through angles grater thanscattered through angles grater than θθ: :

f = f = ππbb22ntnt

3

3

cm

at

M

N

mol

gM

mol

atN

cm

g

n AA

On the base of that nuclear model Rutherford On the base of that nuclear model Rutherford derived an expression for the number of derived an expression for the number of αα particles particles ΔΔNN that would be scattered at any angle that would be scattered at any angle θθ ::

All this predictions was verified in Geiger All this predictions was verified in Geiger experiments, who observe several hundreds experiments, who observe several hundreds thousands thousands αα particles. particles.

2sin

1

2 4

22

20

kE

kZe

r

AntIN

(a)Geiger data for (a)Geiger data for αα scattering from thin scattering from thin AuAu and and AgAg foils. The foils. The graph is in graph is in log-loglog-log plot to cover over several orders of plot to cover over several orders of magnitudes.magnitudes.

(b)Geiger also measured the dependence of (b)Geiger also measured the dependence of ΔΔNN on on tt for for different elements, that was also in good agreement with different elements, that was also in good agreement with Rutherford formula Rutherford formula. .

Data from Rutherford’s group showing observedData from Rutherford’s group showing observed αα

scattering at a large fixed angle versus values ofscattering at a large fixed angle versus values of rrdd computed fromcomputed from for various kinetic energies.for various kinetic energies.

2

21

vm

Qqkrd

The Size of the NucleusThe Size of the Nucleus

This equation can be used to estimate the This equation can be used to estimate the size of the nucleussize of the nucleus

For the case of For the case of 7.7-MeV7.7-MeV αα particles the particles the distance of closest approach for a head-on collision isdistance of closest approach for a head-on collision is

mnmeV

nmeVrd

1456

103103107.7

)44.1)(79)(2(

2

21

vm

Qqkrd

2 2

20

1 v

4e

e mF

r r

The Classical Atomic ModelThe Classical Atomic Model

Consider an atom as a planetary system.Consider an atom as a planetary system.

The Newton’s 2The Newton’s 2ndnd Law force of attraction on Law force of attraction on the electron by the nucleus is:the electron by the nucleus is:

where where vv is the tangential velocity of the electron:is the tangential velocity of the electron:

The total energy is then:The total energy is then:

0

v4

e

mr

221 1

2 20

v4

eK m

r

This is negative, so This is negative, so the system is bound, the system is bound, which is good.which is good.

The Planetary Model is DoomedThe Planetary Model is Doomed

From classical E&M theory, an accelerated electric charge From classical E&M theory, an accelerated electric charge radiates energy (electromagnetic radiation), which means the radiates energy (electromagnetic radiation), which means the total energy must decrease. So the total energy must decrease. So the radius radius r r must decrease!!must decrease!!

Electron Electron crashes crashes into the into the

nucleus!?nucleus!?

According to classical physics, a charge According to classical physics, a charge e e moving with an moving with an acceleration acceleration aa radiates at a rate radiates at a rate

(a)(a) Show that an electron in a classical hydrogen atom spirals Show that an electron in a classical hydrogen atom spirals into the nucleus at a rateinto the nucleus at a rate

3

22

06

1

c

ae

d t

d E

32220

2

4

1 2 cmr

e

d t

d r

e

(b) Find the time interval over (b) Find the time interval over which the electron will reach which the electron will reach r r = 0= 0, , starting from starting from rr0 0 = 2.00 × 10= 2.00 × 10–10–10 m m. .

The Bohr’s PostulatesThe Bohr’s PostulatesBohr overcome the difficulty of the collapsing atom by Bohr overcome the difficulty of the collapsing atom by

postulating that only certain orbits, calledpostulating that only certain orbits, called stationary statesstationary states, , are allowed, and that in these orbits the electron does not are allowed, and that in these orbits the electron does not radiate. An atom radiates only when the electron makes a radiate. An atom radiates only when the electron makes a transition from one allowed orbit (stationary state) to another:transition from one allowed orbit (stationary state) to another:

1. 1. The electron in the hydrogen atom can move only in The electron in the hydrogen atom can move only in certain nonradiating, circular orbits called stationary certain nonradiating, circular orbits called stationary states.states.

2. 2. The photon frequency from energy conservation is The photon frequency from energy conservation is given bygiven by

where where EEii andand EEff are the energy of initial and final state,are the energy of initial and final state, hh is the is the

Plank’s constant.Plank’s constant. h

EEf fi

Such a model is mechanically stable , because the Coulomb Such a model is mechanically stable , because the Coulomb potentialpotential

provides the centripetal forceprovides the centripetal force

The total energy for a such system can be written as the The total energy for a such system can be written as the sum of kinetic and potential energy: sum of kinetic and potential energy:

r

ZekU

2

r

k Zem v

r

m v

r

k ZeF

22

1 22

2

2

2

r

k Ze

r

k Ze

r

k ZeE

22

222

The Bohr’s PostulatesThe Bohr’s Postulates

Combining the second postulate with the Combining the second postulate with the equation for the energy we obtain:equation for the energy we obtain:

12

221 11

2

1

rrh

kZe

h

EEf

where where rr11 andand rr22 are the radii of the initial and are the radii of the initial and

final orbits.final orbits.


12

221 11

2

1

rrh

kZe

h

EEf

To obtain the frequencies implied by the To obtain the frequencies implied by the experimental experimental Rydberg-RitzRydberg-Ritz formula, formula,

21

22

11

nncR

cf

it is evident that the radii of the stable orbits must be it is evident that the radii of the stable orbits must be proportional to the squares of integers.proportional to the squares of integers.


Bohr found that he could obtain this condition if he Bohr found that he could obtain this condition if he postulates the angular momentum of the electron in a postulates the angular momentum of the electron in a stable orbit equals an integer timesstable orbit equals an integer times ħ=h/2πħ=h/2π. Since the . Since the angular momentum of a circular orbit is justangular momentum of a circular orbit is just mvrmvr, , the the third Bohr’s postulatethird Bohr’s postulate is: is:

3.3. n=1,2,3……….n=1,2,3……….

wherewhere ħ=h/2π=1.055 x 10ħ=h/2π=1.055 x 10-34-34J·s=6.582x10 J·s=6.582x10 -16-16eV·seV·s

nnh

mvr 2


The obtained equationThe obtained equation mvr = nh/2mvr = nh/2π=nħπ=nħ relates the relates the speed of electronspeed of electron vv to the radiusto the radius rr. . Since we hadSince we had

r

mv

r

kZe 2

2

2

oror

mr

kZev

22

We can writeWe can write

22

222

rmnv

oror

m r

kZe

rmn

2

22

22


Solving forSolving for rr, we obtain, we obtain

Z

an

m k Zenr 02

2

22

wherewhere aa00 is called the is called the first Bohr’s radiusfirst Bohr’s radius

nmm k e

a 0529.02

2

0

Bohr’s PostulatesBohr’s PostulatesSubstituting the expression forSubstituting the expression for r r in equation for in equation for frequency:frequency:

If we will compare this expression withIf we will compare this expression with Z=1Z=1 for for f=c/f=c/λλ with the empirical with the empirical Rydberg-RitzRydberg-Ritz formula: formula:

we will obtain for the we will obtain for the Rydberg constantRydberg constant

21

22

3

422

12

2 11

4

11

2

1

nn

em kZ

rrh

k Zef

21

22

111

nnR

3

42

4 c

emkR

Bohr’s PostulatesBohr’s Postulates

Using the known values ofUsing the known values of mm,, ee,, and and ħħ, , Bohr Bohr calculatedcalculated RR and found his results to agree with the and found his results to agree with the spectroscopy data. spectroscopy data.

The total mechanical energy of the electron in The total mechanical energy of the electron in the hydrogen atom is related to the radius of the the hydrogen atom is related to the radius of the circular orbit circular orbit

3

42

4 c

emkR

r

kZeE

2

2

1

Energy levelsEnergy levels

If we will substitute the quantized value ofIf we will substitute the quantized value of rr as given byas given by

we obtainwe obtain

Z

an

mkZenr 02

2

22

22

422

02

222

2

1

2

1

2

1

n

eZmk

an

ekZ

r

kZeEn


oror

wherewhere

The energiesThe energies EEnn withwith Z=1Z=1 are the quantized allowed are the quantized allowed

energies for the hydrogen atom. energies for the hydrogen atom.

22

422

02

222

2

1

2

1

2

1

n

eZmk

an

ekZ

r

kZeEn

202

n

EZEn

eVa

keemkE 6.13

2

1

2 0

2

2

42

0

Energy level diagram for hydrogen showing the seven lowest Energy level diagram for hydrogen showing the seven lowest stationary states. The energies of infinite number of levels stationary states. The energies of infinite number of levels are given byare given by EEnn = (-13.6/n = (-13.6/n22)eV)eV, where, where nn is an integer. is an integer.

A hydrogen atom is in its first excited state (A hydrogen atom is in its first excited state (n n == 22). Using the ). Using the Bohr theory of the atom, calculate (a) the radius of the orbit, Bohr theory of the atom, calculate (a) the radius of the orbit, (b) the linear momentum of the electron, (c) the angular (b) the linear momentum of the electron, (c) the angular momentum of the electron, (d) the kinetic energy of the momentum of the electron, (d) the kinetic energy of the electron, (e) the potential energy of the system, and (f) the electron, (e) the potential energy of the system, and (f) the total energy of the system. total energy of the system.


Transitions between this allowed Transitions between this allowed energies result in the emission or absorption energies result in the emission or absorption of a photon whose frequency is given byof a photon whose frequency is given by

and whose wavelength isand whose wavelength is

h

EEf fi

fi EE

hc

f

c


Therefore is convenient to have the Therefore is convenient to have the value ofvalue of hchc in in electronvolt nanometers!electronvolt nanometers!

hc = 1240 eVhc = 1240 eV∙nm∙nm

Since the energies are quantized, the Since the energies are quantized, the frequencies and the wavelengths of the frequencies and the wavelengths of the radiation emitted by the hydrogen atom are radiation emitted by the hydrogen atom are quantized in agreement with the observed quantized in agreement with the observed line spectrum. line spectrum.

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

Energy level diagram for hydrogen showing the seven lowest Energy level diagram for hydrogen showing the seven lowest stationary states and the four lowest energy transitions for the stationary states and the four lowest energy transitions for the Lyman, Balmer, and Pashen series. The energies of infinite Lyman, Balmer, and Pashen series. The energies of infinite number of levels are given bynumber of levels are given by EEnn = (-13.6/n = (-13.6/n22)eV)eV, where, where nn is is an integer. an integer.

λ21=hc / (E1-E2)

nmeV

nmeV

EE

hc5.121

)4.36.13(

1240

2112

nmm

mRnm

R

5.121102156.1

)75.0(10096776.12

11

111

712

17222

12

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

λ21=hc / (E1-E2)

Compute the wavelength of the Compute the wavelength of the HHββ spectral line of the spectral line of the

Balmer series predicted by Bohr model. Balmer series predicted by Bohr model.

A hydrogen atom at rest in the laboratory emits a photon of A hydrogen atom at rest in the laboratory emits a photon of the Lyman the Lyman αα radiation. (a) Compute the recoil kinetic energy radiation. (a) Compute the recoil kinetic energy of the atom. (b) What fraction of the excitation energy of the of the atom. (b) What fraction of the excitation energy of the nn = 2 = 2 state is carried by the recoiling atom? (Hint: Use state is carried by the recoiling atom? (Hint: Use conservation of momentum.) conservation of momentum.)

In a hot star, because of the high temperature, an atom can In a hot star, because of the high temperature, an atom can absorb sufficient energy to remove several electrons from the absorb sufficient energy to remove several electrons from the atom. Consider such a multiply ionized atom with a single atom. Consider such a multiply ionized atom with a single remaining electron. The ion produces a series of spectral remaining electron. The ion produces a series of spectral lines as described by the Bohr model. The series lines as described by the Bohr model. The series corresponds to electronic transitions that terminate in the corresponds to electronic transitions that terminate in the same final state. The longest and shortest wavelengths of the same final state. The longest and shortest wavelengths of the series are series are 63.3 nm63.3 nm and and 22.8 nm22.8 nm, respectively. (a) What is the , respectively. (a) What is the ion? (b) Find the wavelengths of the next three spectral lines ion? (b) Find the wavelengths of the next three spectral lines nearest to the line of longest wavelength. nearest to the line of longest wavelength.

A stylized picture of Bohr A stylized picture of Bohr circular orbits forcircular orbits for n=1,2,3,4n=1,2,3,4. . The radiiThe radii rrnn≈n≈n22. .

In highIn high ZZ-elements -elements (elements with(elements with Z ≥12Z ≥12), ), electrons are distributed over electrons are distributed over all the orbits shown. If an all the orbits shown. If an electron in the electron in the n=1n=1 orbit is orbit is knocked from the atom (e.g., knocked from the atom (e.g., by being hit by a fast electron by being hit by a fast electron accelerated by the voltage accelerated by the voltage across an across an x-rayx-ray tube) the tube) the vacancy that produced is filed vacancy that produced is filed by an electron of higher by an electron of higher energy (i.e., energy (i.e., n=2n=2 or higher). or higher).

The difference in energy between the two orbits is emitted as a The difference in energy between the two orbits is emitted as a photon, whose wavelength will be in the photon, whose wavelength will be in the x-rayx-ray region of the region of the spectrum, if spectrum, if ZZ is large enough is large enough.

Characteristic x-ray spectra. (a) Part the spectra ofCharacteristic x-ray spectra. (a) Part the spectra of neodymium neodymium (Z=60)(Z=60) andand samarium (Z=62)samarium (Z=62).The two pairs of bright lines are .The two pairs of bright lines are thethe KKαα and and KKββ lines. (lines. (b) Part of the spectrum of the artificially b) Part of the spectrum of the artificially produced elementproduced element promethiumpromethium (Z=61)(Z=61),, its its KKαα and and KKββ lines fall lines fall between those of between those of NdNd and and SmSm. (c) Part of the spectrum of all . (c) Part of the spectrum of all three elements.three elements.

(a)

(b)

(c)

The Franck-Hertz ExperimentThe Franck-Hertz Experiment

While investigating the inelastic scattering While investigating the inelastic scattering

of electrons, of electrons, J.Frank and G.HertzJ.Frank and G.Hertz in 1914 in 1914

performed an important experiment that confirmed performed an important experiment that confirmed

by direct measurement by direct measurement Bohr’sBohr’s hypothesis of energy hypothesis of energy

quantization in atoms. quantization in atoms.

The experiment involved measuring the plate The experiment involved measuring the plate

current as a function ofcurrent as a function of VV00. .

The Franck-Hertz ExperimentThe Franck-Hertz ExperimentSchematic diagram of the Schematic diagram of the Franck-Hertz experiment. Franck-Hertz experiment. Electrons ejected from the Electrons ejected from the heated cathode heated cathode CC at zero at zero potential are drawn to the potential are drawn to the positive grid positive grid GG. Those . Those passing through the holes passing through the holes in the grid can reach the in the grid can reach the plateplate PP and contribute in and contribute in the current the current II, if they have , if they have sufficient kinetic energy to sufficient kinetic energy to overcome the small back overcome the small back potential potential ΔΔVV. The tube . The tube contains a low pressure contains a low pressure gas of the element being gas of the element being studied.studied.

The Franck-Hertz ExperimentThe Franck-Hertz Experiment

As As VV00 increased from increased from 00, the current , the current

increases until a critical value (about increases until a critical value (about 4.9 V4.9 V for for mercury) is reached. At this point the current mercury) is reached. At this point the current suddenly decreases. As suddenly decreases. As VV00 is increased further, is increased further,

the current rises again.the current rises again.

They found that when the electron’s They found that when the electron’s kinetic energy was kinetic energy was 4.9 eV4.9 eV or greater, the vapor or greater, the vapor of mercury emitted ultraviolet light of wavelength of mercury emitted ultraviolet light of wavelength 0.25 0.25 μμmm..

The current The current decreases because decreases because many electrons lose many electrons lose energy due to energy due to inelastic collisions inelastic collisions with mercury atoms with mercury atoms in the tube and in the tube and therefore cannot therefore cannot overcome the small overcome the small back potential. back potential.

Cur

rent

, (m

Am

p)Current versus acceleration voltage in the Current versus acceleration voltage in the Franck-HertzFranck-Hertz

experiment.experiment.

The regular spacing of the The regular spacing of the peaks in this curve peaks in this curve indicates that only a indicates that only a certain quantity of energy, certain quantity of energy, 4.9 eV4.9 eV, can be lost to the , can be lost to the mercury atoms. This mercury atoms. This interpretation is confirmed interpretation is confirmed by the observation of by the observation of radiation of photon energy radiation of photon energy 4.9 eV4.9 eV, emitted by , emitted by mercury atoms.mercury atoms.

Cur

rent

, (m

Am

p)

Suppose mercury atoms have an energy level Suppose mercury atoms have an energy level 4.9 eV4.9 eV above the lowest energy level. An atom can above the lowest energy level. An atom can be raised to this level by collision with an electron; it be raised to this level by collision with an electron; it later decays back to the lowest energy level by later decays back to the lowest energy level by emitting a photon. The wavelength of the photon emitting a photon. The wavelength of the photon should be should be

mnmeV

nmeV

E

hc 25.006.2539.4

1240

This is equal to the measured wavelength, This is equal to the measured wavelength, confirming the existence of this energy level of the confirming the existence of this energy level of the mercury atom. mercury atom. Similar experiments with other atoms Similar experiments with other atoms yield the same kind of evidence for atomic energy yield the same kind of evidence for atomic energy levels.levels.

Lets consider an experimental tube filled by Lets consider an experimental tube filled by hydrogen atoms instead of mercury. Electrons hydrogen atoms instead of mercury. Electrons accelerated by accelerated by VV00 that collide with hydrogen that collide with hydrogen

electrons cannot transfer the energy to letter electrons cannot transfer the energy to letter electrons unless they have acquired kinetic electrons unless they have acquired kinetic energy energy

eVeV00=E=E22 – E – E11=10.2eV=10.2eV

If the incoming electron does not have If the incoming electron does not have sufficient energy to transfersufficient energy to transfer ΔΔE = EE = E22 - E - E11 to the to the

hydrogen electron in the hydrogen electron in the n=1n=1 orbit (ground state), orbit (ground state), than the scattering will be elastic. than the scattering will be elastic.

If the incoming electron does have at leastIf the incoming electron does have at least ΔΔEE kinetic energy, then an inelastic collision can occur in kinetic energy, then an inelastic collision can occur in whichwhich ΔΔEE is transferred to theis transferred to the n=1n=1 electron, moving it electron, moving it to theto the n=2n=2 orbit. The excited electron will typically orbit. The excited electron will typically return to the ground state very quickly, emitting a return to the ground state very quickly, emitting a photon of energyphoton of energy ΔΔEE. .

Energy loss spectrum measurement. A well-defined Energy loss spectrum measurement. A well-defined electron beam impinges upon the sample. Electrons electron beam impinges upon the sample. Electrons inelastically scattered at a convenient angle enter the slit of the inelastically scattered at a convenient angle enter the slit of the magnetic spectrometer, whosemagnetic spectrometer, whose BB field is directed out of the field is directed out of the page, and turn through radiipage, and turn through radii RR determined by their energydetermined by their energy (E(Eincinc – E– E11)) via equation via equation

eB

EEmR ince )(2 1

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

Reduced mass correctionReduced mass correctionThe assumption by Bohr that the nucleus is fixed is The assumption by Bohr that the nucleus is fixed is

equivalent the assumption that it has an infinity mass.equivalent the assumption that it has an infinity mass.

If instead we will assume that proton and electron both If instead we will assume that proton and electron both revolve in circular orbits about their common center of mass we revolve in circular orbits about their common center of mass we will receive even better agreement for the values of the will receive even better agreement for the values of the Rydberg constant Rydberg constant RR and ionization energy for the hydrogen. and ionization energy for the hydrogen.

We can take in account the motion of the nucleus (the We can take in account the motion of the nucleus (the proton) very simply by using in Bohr’s equation not the electron proton) very simply by using in Bohr’s equation not the electron rest mass rest mass m m but a quantity called the but a quantity called the reduce massreduce mass μμ of the of the system. For a system composed from two masses system. For a system composed from two masses mm11 and and mm22

the reduced mass is defined as:the reduced mass is defined as:

21

21

mm

mm

Reduced mass correctionReduced mass correction

If the nucleus has the massIf the nucleus has the mass MM its kinetic its kinetic energy will beenergy will be ½½MvMv22 = p = p22/2M/2M, where , where p = Mvp = Mv is the is the momentum. momentum.

If we assume that the total momentum of the If we assume that the total momentum of the atom is zero, from the conservation of momentum atom is zero, from the conservation of momentum we will have that momentum of electron and we will have that momentum of electron and momentum of nucleus are equal on the magnitude. momentum of nucleus are equal on the magnitude.

Reduced mass correction

The total kinetic energy is then:The total kinetic energy is then:

The Rydberg constant equation than changed to:The Rydberg constant equation than changed to:

The factor The factor μμ was calledwas called mass correction factor mass correction factor..

22

)(

22

2222 p

Mm

pmM

m

p

M

pEk

Mm

m

mM

mM

1

R

Mm

mRR 1

As the Earth moves around the Sun, its orbits are quantized. (a) Follow the 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 steps of Bohr’s analysis of the hydrogen atom to show that the allowed radii of the Earth’s orbit are given byof the Earth’s orbit are given by

where where MMSS is the mass of the Sun, is the mass of the Sun, MMEE is the mass of the Earth, and is the mass of the Earth, and nn is an is an

integer quantum number. (b) Calculate the numerical value of integer quantum number. (b) Calculate the numerical value of nn. . (c) Find the (c) Find the distance between the orbit for quantum number distance between the orbit for quantum number nn and the next orbit out from and the next orbit out from the Sun corresponding to the quantum number the Sun corresponding to the quantum number n + 1n + 1

2

22

ES MGM

nr

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Structure of the Atom