Structure of the Atom

1
Structure of the Atom
2
CHAPTER 4Structure 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
3
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 Avogadros 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.

4
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,.

5

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 285289. 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 longest-lived nuclei in the
superheavy element region.
6
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 Z1 the
  next lightest atom, helium (He), has Z2 the
  third lightest, lithium (Li), has Z3 and so
  forth.

7
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 nm100,000fm. This distance
  determines the size of the atom.

8
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!
9
     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.
10
Structure of the Atom

• Evidence in 1900 indicated that the atom was not
  a fundamental unit

• There seemed to be too many kinds of atoms, each
  belonging to a distinct chemical element (way
  more than earth, air, water, and fire!).
• Atoms and electromagnetic phenomena were
  intimately related (magnetic materials
  insulators vs. conductors different emission
  spectra).
• Elements combine with some elements but not with
  others, a characteristic that hinted at an
  internal atomic structure (valence).
• The discoveries of radioactivity, x-rays, and the
  electron (all seemed to involve atoms breaking
  apart in some way).

11
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.

12
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
13
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.
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• The source emits only two wavelengths, ?2gt?1.
  The source is located at the focal point of the
  lens so that parallel light passes through the
  narrow slit, projecting a narrow line onto the
  face of the prism. Ordinary dispersion in the
  prism bends the shorter wavelength through the
  lager total angel, separating the two wavelength
  at the screen.

15

• 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.

16

• 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.

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• 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, H8 called
  the limits of the series.

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Atomic Spectra

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

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Atomic Spectra

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

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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
  R8 1.097373 x 107m-1.
• Such empirical expressions were successful in
  predicting other spectra, such as other hydrogen
  lines outside the visible spectrum.

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Atomic Spectra

• So, the hydrogen Balmer series wavelength are
  those given by Rydberg equation
• with m2 and n3,4,5,
• Other series of hydrogen spectral lines were
  found for m1 (by Lyman) and m3 (by Paschen).

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Hydrogen Spectral Series

• Compute the wavelengths of the first lines of
  the Lyman, Balmer, and Paschen series.

23

• 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, H8 called
  the limits of the series.

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The Limits of Series

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

25

• 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.

26

• A portion of emission spectrum of mercury.

27

• 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.
  

28
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.

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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.
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Thomsons Atomic Model

• Thomsons 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 Thomsons view, when the atom was heated, the
  electrons could vibrate about their equilibrium
  positions, thus producing electromagnetic
  radiation.
• Unfortunately, Thomson couldnt explain spectra
  with this model.

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• 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.

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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.

33

• Rutherford was investigating radioactivity
  and had shown that the radiations from uranium
  consist of at least two types, which he labeled a
  and ß.
• He showed, by an experiment similar to that of
  Thompson, that q /m for the a - particles was
  half that of the proton.
• Suspecting that the a 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.

34
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.
35
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
36
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
37

• Rutherford was investigating radioactivity
  and had shown that the radiations from uranium
  consist of at least two types, which he labeled a
  and ß.
• He showed, by an experiment similar to that of
  Thompson, that q /m for the a - particles was
  half that of the proton.
• Suspecting that the a 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.

38

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

39

• A diagram of the original apparatus as it appear
  in Geigers paper describing the results.

40

• An a-particle by such an atom (Thompson model)
  would have a scattering angle ? much smaller than
  10. In the Rutherfords scattering experiment
  most of the a-particles were either undeflected,
  or deflected through very small angles of the
  order 10, however, a few a-particles were
  deflected through angles of 900 and more.

41

• An a-particle by such an atom (Thompson model)
  would have a scattering angle ? much smaller than
  10. In the Rutherfords scattering experiment a
  few a-particles were deflected through angles of
  900 and more.

42
Experiments of Geiger and Marsden

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

43
Electrons cant back-scatter a particles.

• 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
Determine qmax by letting Dpmax be perpendicular
to the direction of motion
too small!
44
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!
45

• 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 a-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.

46
Rutherfords Atomic Model
even if the a particle is
scattered from all 79 electrons in each atom of
gold. Experimental results were not consistent
with Thomsons 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)
47

• Rutherford concluded that the large angle
  scattering could result only from a single
  encounter of the a 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 a particles.
• His predictions on the dependence of scattering
  probability on angle, nuclear charge and kinetic
  energy were completely verified in experiments.

48

• Rutherford Scattering geometry. The nucleus is
  assumed to be a point charge Q at the origin O.
  At any distance r the a particle experiences a
  repulsive force kqaQ/r2. The a 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 ?
  with OA.

49
Rutherford Scattering
Theres a relationship between the impact
parameter b and the scattering angle q.
When b is small, r is small. the Coulomb force
is large. ? can be large and the particle can be
repelled backward.
where
50
Rutherford Scattering

• Any particle inside the circle of area p b02
  will be similarly (or more) scattered.

The cross section s p b2 is related to the
probability for a particle being scattered by a
nucleus (t foil thickness) The fraction of
incident particles scattered is
51

• 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

52

• Two a particles with equal kinetic energies
  approach the positive charge Q Ze with impact
  parameters b1 and b2, where b1ltb2. According to
  equation for impact parameter in this case ?1 gt
  ?2.

53

• The path of a particle can be shown to be a
  hyperbola, and the scattering angle ? can be
  relate to the impact parameter b from the laws of
  classical mechanics.
• The quantity pb2, which has the dimension of
  the area , is called the cross section s for
  scattering.

54

• The cross section s 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.
55

• The total number of particles scattered per
  second is obtained by multiplying pb2I0 by the
  number of nuclei in the scattering foil. Let n be
  the number of nuclei per unit volume
• For a foil of thickness t, the total number of
  nuclei 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 ? is thus
  pb2I0ntA. If divide this by the number of a
  particles incident per second I0A we get the
  fraction of a particles f scattered through
  angles grater than ?
• f pb2nt

56
On the base of that nuclear model
Rutherford derived an expression for the number
of a particles ?N that would be scattered at any
angle ? All this predictions was
verified in Geiger experiments, who observe
several hundreds thousands a particles.
57

• (a)Geiger data for a 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 ?N on
  t for different elements, that was also in good
  agreement with Rutherford
  formula.

58

• Data from Rutherfords group showing observed a
  scattering at a large fixed angle versus values
  of rd
• computed from for various
  kinetic energies.

59
The Size of the Nucleus

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

60
The Classical Atomic Model

• Consider an atom as a planetary system.
• The Newtons 2nd Law force of attraction on the
  electron by the nucleus is

where v is the tangential velocity of the
electron
The total energy is then
This is negative, so the system is bound, which
is good.
61
The Planetary Model is Doomed

• From classical EM 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!?
62

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

(b) Find the time interval over which the
electron will reach r 0, starting from r0
2.00 1010 m.
63
The Bohrs 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
• where Ei and Ef are the energy of initial and
  final state, h is the Planks constant.

64

• Such a model is mechanically stable , because
  the Coulomb potential
• provides the centripetal force
• The total energy for a such system can be
  written as the sum of kinetic and potential
  energy

65
The Bohrs Postulates

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

where r1 and r2 are the radii of the initial
and final orbits.
66
The Bohrs Postulates

To obtain the frequencies implied by the
experimental Rydberg-Ritz formula,
it is evident that the radii of the stable orbits
must be proportional to the squares of integers.
67
The Bohrs 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 hh/2p. Since the angular momentum of a
  circular orbit is just mvr, the third Bohrs
  postulate is
• 3.
  n1,2,3.
• where hh/2p1.055 x 10-34Js6.582x10 -16eVs

68
The Bohrs Postulates

• The obtained equation mvr nh/2pnh relates the
  speed of electron v to the radius r. Since we had

or
We can write
or
69
The Bohrs Postulates

• Solving for r, we obtain

where a0 is called the first Bohrs radius
70
Bohrs Postulates

• Substituting the expression for r in equation for
  frequency
• If we will compare this expression with Z1 for
  fc/? with the empirical Rydberg-Ritz formula
• we will obtain for the Rydberg constant

71
Bohrs Postulates

• Using the known values of m, e, and h, 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

72
Energy levels

• If we will substitute the quantized value of r
  as given by
• we obtain

73
Energy levels

• or
• where
• The energies En with Z1 are the quantized
  allowed energies for the hydrogen atom.

74

• 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.

75

• 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.

76
Energy levels

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

77
Energy levels

• Therefore is convenient to have the value of hc
  in electronvolt nanometers!
• hc 1240 eVnm
• 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.

78

• (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.

79
?21hc / (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.

80
(No Transcript)
81

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

82
?21hc / (E1-E2)
83

• Compute the wavelength of the Hß spectral line
  of the Balmer series predicted by Bohr model.

84

• A hydrogen atom at rest in the laboratory emits
  a photon of the Lyman a 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.)

85

• 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.

86

• A stylized picture of Bohr circular orbits for
  n1,2,3,4. The radii rnn2.
• In high Z-elements (elements with Z 12),
  electrons are distributed over all the orbits
  shown. If an electron in the n1 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., n2 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.
87
(a)
(b)
(c)

• Characteristic x-ray spectra. (a) Part the
  spectra of neodymium (Z60) and samarium
  (Z62).The two pairs of bright lines are the Ka
  and Kß lines. (b) Part of the spectrum of the
  artificially produced element promethium (Z61),
  its Ka and Kß lines fall between those of Nd and
  Sm. (c) Part of the spectrum of all three
  elements.

88
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 Bohrs hypothesis
  of energy quantization in atoms.
• The experiment involved measuring the plate
  current as a function of V0.

89
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 ?V. The tube
contains a low pressure gas of the element being
studied.
90
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 electrons kinetic
  energy was 4.9 eV or greater, the vapor of
  mercury emitted ultraviolet light of wavelength
  0.25 µm.

91
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)
92

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

Current, (mAmp)
93

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

This is equal to the measured wavelength,
confirming the existence of this energy level of
the mercury atom.

Similar experiments with other atoms yield the
same kind of evidence for atomic energy levels.
94

• 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
• eV0E2 E110.2eV

95

• If the incoming electron does not have
  sufficient energy to transfer ?E E2 - E1 to the
  hydrogen electron in the n1 orbit (ground
  state), than the scattering will be elastic.

96

• If the incoming electron does have at least ?E
  kinetic energy, then an inelastic collision can
  occur in which ?E is transferred to the n1
  electron, moving it to the n2 orbit. The excited
  electron will typically return to the ground
  state very quickly, emitting a photon of energy
  ?E.

97

• 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

98

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

99
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
  Bohrs 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

100
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.

101
Reduced mass correction

• The total kinetic energy is then
• The Rydberg constant equation than changed to
• The factor µ was called mass correction factor.

102

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