Quantization of Charge, Light, and Energy

1
Quantization of Charge, Light, and Energy
2
CHAPTER 3Prelude to Quantum Theory

• Discovery of the X-Ray and the Electron
• Determination of Electron Charge
• Line Spectra
• Quantization
• Blackbody Radiation
• Photoelectric Effect
• X-Ray Production
• Compton Effect
• Pair Production and Annihilation

Max Karl Ernst Ludwig Planck (1858-1947)
We have no right to assume that any physical laws
exist, or if they have existed up until now, or
that they will continue to exist in a similar
manner in the future. An important scientific
innovation rarely makes its way by gradually
winning over and converting its opponents. What
does happen is that the opponents gradually die
out. - Max Planck
3

• 1. Quantization of Electric Charge
• 2. Blackbody Radiation
• 3. The Photoelectric Effect
• 4. X-rays and the Compton Effect.

4

• Three great quantization discoveries
• 1. Quantization of Electrical Charge
• 2. Quantization of Light Energy
• 3. Quantization of energy of Oscillating
  Mechanical System

5
Quantization of Electric Charge

• Early measurements of e and e/m.
• The first estimates of the magnitude of electric
  charges found in atoms were obtained from
  Faradays law.
• Faraday passed a direct current through weakly
  conducting solutions and observed the subsequent
  liberation of the components of the solution on
  electrodes.
• Faraday discovered that the same quantity of
  electricity, F, called latter one faraday, and
  equal to about 96,500 C, always decomposed 1
  gram-ionic weight of monovalent ions.

6
Quantization of Electric Charge

• 1F (one faraday) 96,500C
• 1F decomposed 1gram-ionic weight of monovalent
  ions.
• Example If 96,500 C passes through a solution
  of NaCl, 23g of Na appears at the cathode and
  35.5g of Cl appears at the anode.
• 1F NAe - Faradays Law of Electrolysis
• where NA Avogadros number
• e minimum amount of charge, that was
  called an electron

7
Discovery of the the Electron

• In the 1890s scientists and engineers were
  familiar with cathode rays. These rays were
  generated from one of the metal plates in an
  evacuated tube with a large electric potential
  across it.

J. J. Thomson (1856-1940)
Wilhelm Röntgen (1845-1923)
It was surmised that cathode rays had something
to do with atoms. It was known that cathode
rays could penetrate matter and were deflected by
magnetic and electric fields.
8
Discovery of electron Thomsons Experiment.

• Many studies of electrical discharges in gases
  were done in the late 19th century. It was found
  that the ions responsible for gaseous conduction
  carried the same charge as did those in
  electrolysis.
• J.J. Thomson in 1897 used crossed electric and
  magnetic fields in his famous experiment to
  deflect the cathode-rays.
• In this way he verified that cathode-rays must
  consist of charged particles. By measuring the
  deflection of these particles Thomson showed that
  all the particles have the same charge-to-mass
  ratio q/m. He also showed that particles with
  this charge-to-mass ratio can be obtained using
  any material for a source, which means that this
  particles, now called electrons, are a
  fundamental consistent of all matter.

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Thomsons Cathode-Ray Experiment

• Thomson used an evacuated cathode-ray tube to
  show that the cathode rays were negatively
  charged particles (electrons) by deflecting them
  in electric and magnetic fields.

10
Thomsons Experiment e/m

• Thomsons method of measuring the ratio of the
  electrons charge to mass was to send electrons
  through a region containing a magnetic field
  perpendicular to an electric field.

J. J. Thomson
11
Thomsons tube for measuring e/m.
Electrons from the cathode C pass through the
slits at A and B and strike a phosphorescent
screen. The beam can be deflected by an electric
field between plates D and E or by magnetic
field. From measurements of the deflections
measured on a scale on the tubes screen, e/m can
be determined.
12
FEqE
_
-
-
q
FBqvB
FEFB qE qvB

• Crossed electric and magnetic fields. When a
  negative particle moves to the right the particle
  experience a downward magnetic force FBqvB and
  an upward electric force FEqE. If these forces
  are balanced, the speed of the particle is
  related to the field strengths by
• vE/B

13
Thomsons Experiment

• In his experiment Thomson adjusted E - B so that
  the particles were undeflected.
• This allowed him determine the initial speed of
  the particle u E/B. He then turned off the B
  field and measured the deflection of the
  particles on the screen.

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Deflection of the Electron Beam.
Deflection of the beam is shown with the top
plate positive. Thomson used up to 200 V between
the plates. A magnetic field was applied
perpendicular to the plane of the diagram
directed into the page to bend the beam back down
to its undeflected position.
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• With the magnetic field turned off, the beam
  are deflected by an amount yy1y2. y1 occurs
  while the electrons are between the plates, y2
  after electrons leave the region between the
  plates. Lets x1 be the horizontal distance across
  the deflection plates. If the electron moves
  horizontally with speed v0 when it enters the
  region between the plates, the time spent between
  the plates is t1x1/v0 , and the vertical
  component of velocity when it leaves the plates
  is
• where Ey is the upward component of electric
  field. The deflection
• y1 is

16

• The electron then travels an additional
  horizontal distance x2 in the field-free region
  from the deflection plates to the screen. Since
  the velocity of the electron is constant in this
  region, the time to reach the screen is t2x2/v0
  , and the additional vertical deflection is

17

• The total deflection at screen is therefore
• This equation cab be used to determine the
  charge to-mass ratio q/m from measured deflection
  y.

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Determination of Electron Charge

• Millikans oil-drop experiment

Robert Andrews Millikan (1868 1953)

• Millikan was able to show that electrons had a
  particular charge.

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Calculation of the oil drop charge

• Millikan used an electric field to balance
  gravity and suspend a charged oil drop

Turning off the electric field, Millikan noted
that the drop mass, mdrop, could be determined
from Stokes relationship of the terminal
velocity, vt, to the drop density, r, and the air
viscosity, h
Drop radius
and
Thousands of experiments showed that there is a
basic quantized electron charge
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Example

• Electrons pass undeflected through the plates of
  Thomsons apparatus when the electric field is
  3000 V/m and there is a crossed magnetic field of
  1.40 G. If the plates are 4-cm long and the ends
  of the plates are 30 cm from the screen, find the
  deflection on the screen when the magnetic field
  is turned off.
• me9.11 x 10-31kg
• qe-1.6 x 10-19C

21
The Mass Spectrometer

• The mass spectrometer, first designed by
  Francis William Aston in 1919, was developed as
  a means of measuring the masses of isotopes.
• Such measurements are important in determining
  both the presence of isotopes and their abundance
  in nature.
• For example, natural magnesium has been found to
  consist of 78.7 24Mg 10.1 25Mg and 11.2
  26Mg. These isotopes have masses in the
  approximate ratio 242526.

22
Schematic drawing of a mass spectrometer.
Positive ions from an ion source are accelerated
through a potential difference ?V and enter a
uniform magnetic field. The magnetic field is
out of the plane of the page as indicated by the
dots. The ions are bent into a circular arc and
emerge at P2 (a plane of photographic plate or
another ion detector). The radius of the circle
is proportional to the mass of the ion
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The Mass Spectrometer
In the ion source positive ions are formed by
bombarding neutral atoms with X-rays or a beam of
electrons. (Electrons are knock out of the atoms
by the X-rays or bombarding electrons). These
ions are accelerating by an electric field and
enter a uniform magnetic field. If the positive
ions start from rest and move through a potential
difference ?V, the ions kinetic energy when they
enter the magnetic field equals their loss in
potential energy, q?V
24

• The ions move in a semicircle of radius R. The
  velocity of the particle is perpendicular to the
  magnetic field. The magnetic force provides the
  centripetal acceleration v2/R in circular
  motion.
• We will use Newtons second low to relate the
  radius of semicircle to the magnetic field and
  the speed of the particle. If the velocity of the
  particle is v, the magnitude of the net force is
  qvB, since v and B are perpendicular.

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The Mass Spectrometer

• The speed v can be eliminate from equations
• and
• Substituting this for v2
• Simplifying and solving for (m/q)

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Separating Isotopes of Nickel

• A 58Ni ion charge e and mass 9.62 x 10-26kg is
  accelerated through a potential drop of 3 kV and
  deflected in a magnetic field of 0.12T.
• (a) Find the radius of curvature of the orbit of
  the ion.
• (b) Find the difference in the radii of
  curvature of 58Ni ions and 60Ni ions. (assume
  that the mass ratio is 5860).

27
Blackbody Radiation

• One unsolved puzzle in physics in late nineteen
  century was the spectral distribution of so
  called cavity radiation, also referred to as
  blackbody radiation.
• It was shown by Kirchhoff that the most
  efficient radiator of electromagnetic waves was
  also a most efficient absorber. A perfect
  absorber would be one that absorbed all incident
  radiation.
• Since no light would be reflected it is called a
  blackbody.

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• A small hole in the wall of the cavity
  approximating an ideal blackbody. Electromagnetic
  radiation (for example, light) entering the hole
  has little chance of leaving before it is
  completely adsorbed within the cavity.

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Blackbody Radiation

• As the walls of the cavity absorb this incoming
  radiation , their temperature rises and the body
  begin to irradiate.
• A blackbody is a cavity within a material that
  only emits thermal radiation. Incoming radiation
  is absorbed in the cavity.

Blackbody radiation is theoretically interesting
because the radiation properties of the
blackbody are independent of the particular
material. Physicists can study the properties of
irradiated intensity versus wavelength at fixed
temperatures.
30

Radiation emitted by the object at temperature T
that passed through the slit is dispersed
according to its wavelength. The prism shown
would be an appropriate device for that part of
the emitted radiation in the visible region. In
other spectral regions other types of devices or
wavelength-sensitive detectors would be used.
31

• Spectral distribution function R(?) measured
  at different temperatures. The R(?) axis is in
  arbitrary units for comparison only. Notice the
  range in ? of the visible spectrum. The Sun emits
  radiation very close to that of a blackbody at
  5800 K. ?m is indicated for the 5000-K and 6000-K
  curves.

32
Wiens Displacement Law

• The spectral intensity R(?, T) is the total power
  irradiated per unit area per unit wavelength at a
  given temperature.
• Wiens displacement law The maximum of the
  spectrum shifts to smaller wavelengths as the
  temperature is increased.

33
Wiens Displacement Law

• ?mT constant (2.898 x 10-3)m K
• Example The wavelength at the peak of the
  spectral distribution for a blackbody at 4300 K
  is 674 nm (red). What would be the temperature of
  the blackbody that have the peak in intensity at
  420 nm (violet)?
• Solution From the Wiens law, we have
• ?1T1 ?2T2
• (674 x 10-9m)(4300 K) (420 x 10-9m)(T2)
• T26900 K

34

• This law is used to determine the surface
  temperatures of stars by analyzing their
  radiation. It can also be used to map out the
  variation in temperature over different regions
  of the surface of an object. Such a map is called
  thermograph.
• For example thermograph can be used to detect
  cancer because cancerous tissue results in
  increased circulation which produce a slight
  increase in skin temperature.

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• Radiation From the Sun. The radiation emitted by
  the surface of the sun emits maximum power at
  wavelength of about 500 nm. Assuming the sun to
  be a blackbody emitter, (a) what is it surface
  temperature? (b) Calculate ?max for a
  blackbody at room temperature, T300 K.

37
Blackbody Radiation

• In 1879, the Austrian physicist J.Stefan first
  measured the total amount of radiation emitted by
  blackbody at all wavelengths and found it varied
  with absolute temperature.
• It was latter explained through a theoretical
  derivation by Boltzman, so the result became
  known as the Stefan-Boltzman radiation Law
• The total power radiated increases with the
  temperature
• with the constant s experimentally measured to
  be 5.6705 10-8 W / (m2 K4).

38
Blackbody Radiation

• P sT4
• Note that the power per unit area radiated by
  blackbody depends only on the temperature, and
  not of other characteristic of the object, such
  as its color or the material, of which it is
  composed.
• P tells as the rate at which energy is emitted by
  the object. For example, doubling the absolute
  temperature of an object increases the energy
  flows out of the object by factor of 2416.
• An object at room temperature (300 K) will
  double the rate at which it radiates energy as a
  result of temperature increase of only 570.

39

• The calculation of the distribution function
  R(?) involves the calculation of the energy
  density of electromagnetic waves in the cavity.
  The power radiated by the black body is
  proportional to the total energy density U
  (energy per unit volume) of the radiation in the
  cavity. The

proportionality constant can be shown to be c/4,
where c is the speed of the light
40

• The spectral distribution of the power
  proportional to the spectral distribution of the
  energy density in the cavity.
• If u(?)d? is the fraction of the energy per unit
  volume in the cavity in the range d?, then u(?)
  and R(?) are related by

The energy density distribution function u(?) can
be calculated from classical physics.
41

• We can find the number of modes of oscillation
  of the electromagnetic field in the cavity with
  wavelength ? in the interval d? and multiply it
  by average energy per mode.

The result is that the number of modes of
oscillation per unit volume, n(?), is independent
of the shape of cavity and is given by

42
Rayleigh-Jeans Equation

• The number of modes of oscillation per unit
  volume
• According to the classical kinetic theory, the
  average energy per mode of oscillation is kT, the
  same as for a one-dimensional harmonic
  oscillator, where k is the Boltzman constant.
• Classical theory thus predicts for the energy
  density spectral distribution function

43
Rayleigh-Jeans Formula

• This prediction, initially derived by Lord
  Rayleigh using the classical theories of
  electromagnetism and thermodynamics is called
  the Rayleigh-Jeans Law

It approaches the experimental data at longer
wavelengths, but it deviates badly at short
wavelengths.
44

At short wavelength this law predicts that u(?)
becomes large, approaching infinity as ??0,
whereas experiment shows that the distribution
actually approaches zero as ??0.
45
This problem for small wavelengths became known
as the ultraviolet catastrophe and was one of the
outstanding exceptions that classical physics
could not explain.
46
Plancks Radiation Law

• In 1900 the German physicist Max Plank by making
  some unusual assumptions derived a function u(?)
  that agreed with experimental data.
• Planck assumed that the radiation in the cavity
  was emitted (and absorbed) by some sort of
  oscillators. Classically, the electromagnetic
  waves in the cavity are produced by accelerated
  electric charges in the walls vibrating like
  simple harmonic oscillators.

The average energy for simple harmonic oscillator
is calculated classically from Maxwell-Boltzman
distribution function
where A is a constant and f(E) is
the fraction of oscillators with energy E.
47

• Maxwell-Boltzman distribution function
• The average energy is then found, as is any
  weighted average

Planck made two modifications to the classical
theory 1. The oscillators (of electromagnetic
origin) can only have certain discrete energies,
where n is an integer, v is the frequency, and
h is called Plancks constant h 6.6261
10-34 Js. 2. The oscillators can absorb or emit
energy in discrete multiples of the fundamental
quantum of energy given by

48

• The Maxwell-Boltzman distribution than becomes
• where A is determined by normalization condition
  that the sum of all fractions fn must be equal 1.
  

49

• The normalization condition
• The average energy of oscillation is then given
  by discrete-sum equivalent
• To solve this equation let put
• where ye-x.

50

• This sum is a series expansion of (1-y)-1, so
• (1-y)-11yy2y3 ,
• then SfnA(1-y)-11 gives A1-y and
• Note that
• so we have
• since

51
Planks Law

• Multiplying this sum by hv and using A(1-y),
  the average energy is
• Multiplying the numerator and the denominator by
  ex and substituting for x, we obtain

52
Planks Law

• Multiplying this result by the number of
  oscillators per unit volume in the interval d?
  given by n(?)8pc?-4 (the number of modes of
  oscillation per unit volume) we obtain the energy
  distribution function for the radiation in
  cavity
• This function is called Planks Law.

53
Planks Law

• The value of Planks constant h can be
  determined by fitting the function
• to the experimental data, although the direct
  measurement is better, but more difficult. The
  presently accepted value of Plank constant is
• h 6.626 x 10-34 Js 4.136 x 10-15 eVs

54
Comparison of Planks Law and the Rayleigh-Jeans
Law with experimental data at T1600 K. The u(?)
axis is linear.
55
Planks Law

• A dramatic example of an application of
  Plancks law is the test of the predictions of
  the so-called Big Bang theory of the formation
  and expansion of the universe.
• Current cosmological theory suggests that the
  universe originated in an extremely
  high-temperature explosion, one consequence of
  which is to fill out the infant universe with
  radiation that can be approximate with black body
  spectral distribution.
• In 1965, Arno Penzias and Robert Wilson
  discovered radiation of wavelength 7.35 cm
  reaching the Earth with same intensity from all
  directions in space. It was recognized soon as a
  remnant of the Big Bang (relict radiation).

56
The energy density spectral distribution of the
cosmic microwave background radiation. The solid
line is Planks Law with T2.735 K. The
measurements were made by the Cosmic Back Ground
Exploder (COBE) satellite.
57

• Problem 1. Thermal Radiation from the Human
  Body. The temperature of the skin is
  approximately 35C. What is the wavelength at
  which the peak occurs in the radiation emitted
  from the skin?

58

• Problem 2. The Quantized Oscillator. A 2-kg
  mass is attached to a massless spring of force
  constant k25N/m. The spring is stretched 0.4m
  from its equilibrium position and released. (a)
  Find the total energy and frequency of
  oscillation according to classical calculations.
  (b) Assume that the energy is quantized and find
  the quantum number, n, for the system. (c) How
  much energy would be carried away in one-quantum
  change?

59

• Problem 3. The Energy of a Yellow Photon. What
  is the energy carried by a quantum of light whose
  frequency equals 6 x 1014 Hz yellow light? What
  is the wavelength of this light?

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• The surface of the sun has a temperature of
  approximately 5800 K. To good approximation we
  can treat it as a blackbody. (a) What is the
  peak-intensity wavelength ?m? (b) What is the
  total radiated power per unit area? (c) Find the
  power per unit area radiated from the surface of
  the sun in the wavelength range 600.0 to 605.0 nm.

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The Photoelectric Effect

• It is one of the ironies in the history of the
  science that in the same famous experiment of
  Heinrich Hertz in1887 in which he produced and
  detected electromagnetic waves, thus confirmed
  Maxwells wave theory of light, he also
  discovered the photoelectric effect led directly
  to particle description of light.
• It was found that negative charged particles were
  emitted from a clean surface when exposed to
  light.
• P.Lenard in 1900 detected them in a magnetic
  field and found that they had a charge-to-mass
  ratio of the same magnitude as that measured by
  Thompson for cathode rays the particles being
  emitted were electrons.

62
Schematic diagram of the apparatus used by
P.Lenard to demonstrate the photoelectric effect
and to show that the particles emitted in the
process were electrons. Light from the source L
strikes the cathode C.
Photoelectrons going through the hole in anode A
are recorded by the electrometer connected to a.
A magnetic field, indicated by the circular pole
piece, could deflect the particles to an
electrometer connected to ß, enabling the
establishment of the sign of their charge and
their q/m ratio.
63
If some of emitted electrons that reaches an
anode A pass through the small hole, a current
results in the external electrometer circuit
connected to a. The number of electrons,
reaching the
anode, can be increased by making the anode
positive with respect to cathode. Letting V be
the potential difference between A and C the next
picture shows the current versus V for two values
of the intensity of light incident on the
cathode
64

• Photocurrent i versus anode voltage V for light
  of frequency f with two intensities I1 and I2 ,
  where I2gtI1 . At sufficiently large V all emitted
  electrons reach the anode and the current reaches
  its maximum value.
• From experiment it was observed that the maximum
  current is proportional to the light intensity.
  

65

• An expected result if the intensity of
  incident light doubled the number of emitted
  electrons should also double. If intensity of
  incident light is too low to provide electrons
  with energy necessary to escape from the metal,
  no emission of electron should be observed.
• However, for given frequency of the incident
  light, there was no minimum intensity below which
  the current was absent.

66
When V is negative, the electrons are repelled
from the anode and only electrons with initial
kinetic energy mv2/2 greater than eV can reach
the anode. From the graph we can see if the
voltage is less, than V0 no electrons reach
anode. The potential V0 is called the stopping
potential. It related to the maximum kinetic
energy as (mv2/2)max eV0
67
Photo-electric Effect Classical Theory
The kinetic energy of the photoelectrons should
increase with the light intensity and not depend
on the light frequency. Classical theory also
predicted that the electrons absorb energy from
the beam at a fixed rate. So, for extremely low
light intensities, a long time would elapse
before any one electron could obtain sufficient
energy to escape.
Initial observations by Heinrich Hertz 1887
68
Photo-electric effect observations

• The kinetic energy of the photoelectrons is
  independent of the light intensity.
• The kinetic energy of the photoelectrons, for a
  given emitting material, depends only on the
  frequency of the light.

69
Photo-electric effect observations

• There was a threshold frequency of the light,
  below which no photoelectrons were ejected.

The existence of a threshold frequency is
completely inexplicable in classical theory.
70
Photo-electric effect observations

• When photoelectrons are produced, their number
  (not their kinetic energy) is proportional to the
  intensity of light.

(number of electrons)

• Also, the photoelectrons are emitted almost
  instantly following illumination of the
  photocathode, independent of the intensity of the
  light.

71

• The experimental result, that the kinetic energy
  of the ejected electrons is independent of the
  incident light intensity was surprising
  increasing the rate of energy falling on the
  cathode does not increase the maximum kinetic
  energy of the emitted electrons.
• In 1905, Einstein offered an explanation of this
  result he assumed, that energy quantization used
  by Plank in the blackbody problem was a universal
  characteristic of light - Light energy consist of
  discrete quanta of energy h?.

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Einsteins Theory Photons

• Einstein suggested that the electro-magnetic
  radiation field is quantized into particles
  called photons. Each photon has the energy
  quantum
• where ? is the frequency of the light and h is
  Plancks constant.
• Alternatively,

where
73
Einsteins Theory

• Conservation of energy yields

where f is the work function of the metal
(potential energy to be overcome before an
electron could escape).
In reality, the data were a bit more complex.
Because the electrons energy can be reduced by
the emitter material, consider vmax (not v)
74

• When one photon penetrates the surface of
  cathode, all of its energy may be given
  completely to electron. If F is the energy
  necessary to remove an electron from the surface
  (F is called the work function and is a
  characteristic of the metal), the maximum kinetic
  energy of the electrons leaving the surface will
  be (h ? F) and the stopping potential V0 should
  be given by
• This equation is referred as the photoelectric
  effect equation.

75

• As can be seen from
• the slope of the line on the graph V0 versus ?
  should equal h/e.
• The minimum, or threshold, frequency for
  photoelectric effect, ?t and the corresponding
  threshold wavelength ?t are related to work
  function F by setting V0 0
• Photons of frequency lower than ?t ( and
  therefore having wavelength grater than ?t ) do
  not have enough energy to eject an electron from
  the metal.

76
For constant intensity Einsteins explanation of
the photoelectric effect indicates that the
magnitude of the stopping voltage should be
grater for f2 than f1, as observed, and that
there should be a threshold frequency ft below
which no photoelectrons were seen, also in
agreement with experiment.
77

• Millikans data for stopping potential versus
  frequency for the photoelectric effect. The data
  falls on a straight line of slope h/e, as
  predicted by Einstein. The intercept of the
  stopping potential axis is ?/e.

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Example Photoelectric Effect in Potassium

• The threshold wavelength of potassium is 558 nm.
  What is the work function for potassium? What is
  the stopping potential when light of wavelength
  400 nm is used?

80
Example Photoelectric Effect in Potassium

• The threshold wavelength of potassium is 558 nm.
  What is the work function for potassium? What is
  the stopping potential when light of wavelength
  400 nm is used?
• Solution

81
Example Photoelectric Effect in Potassium

• The threshold wavelength of potassium is 558 nm.
  What is the work function for potassium? What is
  the stopping potential when light of wavelength
  400 nm is used?

82
Photoelectric Effect for Sodium

• A sodium surface is illuminated with light of
  wavelength 3x10-7m. The work function for sodium
  is 2.28 eV. Find (a) the kinetic energy of the
  ejected photoelectrons and (b) the cutoff
  wavelength for sodium.

83
X Rays and the Compton Effect

• Future evidence of the correctness of the
  photon concept was given by Arthur Compton, who
  measured the scattering of x-rays by free
  electrons.
• The German physicist Wilhelm Röentgen
  discovered x-rays in 1895 when he was
  working with a cathode-ray tube. He found, that
  rays, originating from the point where the
  cathode rays (electrons) hit the glass tube, or a
  target within the tube, could pass through the
  materials opaque to light and activate a
  fluorescent screen or photographic film. He found
  that all materials were transparent to this rays
  to some degree, depending of the density of this
  materials.
• The slight diffraction of an x-ray beam after
  passing slit of a few thousands of a mm wide
  indicated their wavelength in other of 10-10m
  0.1nm.

84
Observation of X Rays

• Wilhelm Röntgen studied the effects of cathode
  rays passing through various materials. He
  noticed that a phosphorescent screen near the
  tube glowed during some of these experiments.
  These new rays were unaffected by magnetic fields
  and penetrated materials more than cathode rays.

Wilhelm Röntgen
He called them x-rays and deduced that they were
produced by the cathode rays bombarding the glass
walls of his vacuum tube.
85
Röntgens X-Ray Tube

• Röntgen constructed an x-ray tube by allowing
  cathode rays to impact the glass wall of the tube
  and produced x-rays. He used x-rays to make a
  shadowgram the bones of a hand on a
  phosphorescent screen.

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(a)
(b)
(a) Early x-ray tube and (b) typical of the
mid-twenties century x-ray tube design.
87
Diagram of the components of a modern x-ray tube.
Design technology has advanced enormously,
enabling very high operating voltages, beam
currents, and x-ray intensities, but the
essential elements of the tubes remain unchanged.
88
X-Ray Production Theory

• An energetic electron passing through matter
  will radiate photons and lose kinetic energy,
  called bremsstrahlung. Since momentum is
  conserved, the nucleus absorbs very little
  energy, and it can be ignored. The final energy
  of the electron is determined from the
  conservation of energy to be

89
X-Ray Production Experiment
Current passing through a filament produces
copious numbers of electrons by thermionic
emission. If one focuses these electrons by a
cathode structure into a beam and accelerates
them by potential differences of thousands of
volts until they impinge on a metal anode
surface, they produce x rays by bremsstrahlung as
they stop in the anode material.
90
Inverse Photoelectric Effect

• Conservation of energy requires that the electron
  kinetic energy equal the maximum photon energy
  (neglect the work function because its small
  compared to the electron potential energy). This
  yields the Duane-Hunt limit, first found
  experimentally. The photon wavelength depends
  only on the accelerating voltage and is the same
  for all targets.

91
Photons also have momentum!
Use our expression for the relativistic energy to
find the momentum of a photon, which has no mass
Alternatively
92
X-Rays Diffraction

• Experiment soon confirmed that x-rays are a form
  of electromagnetic radiation with wavelength of
  about 0.01nm to 0.10 nm.
• It was also known that atoms in crystals are
  arranged in regular arrays that are spaced by
  about same distances.
• In 1912, Laue suggested that since the
  wavelength of x-rays were on the same order
  of magnitude as the spacing of atoms in a
  crystal, the regular array of atoms in crystal
  might act as a three-dimensional grating for
  diffraction of x-rays.

93
X-Rays Diffraction

• W.L.Bragg, in 1912, proposed a simple and and
  convenient way of analyzing the diffraction of
  x-rays due to scattering from various sets of
  parallel planes of atoms, now called Bragg
  planes.
• Two sets of Bragg planes are illustrated for
  NaCl, which has a simple crystal structure called
  face-centered cubic.

94
A face-centered cubic crystal of NaCl showing two
sets of Bragg planes.
95

• Waves scattered at equal angels from atoms in
  two different planes will be in phase
  (constructive interference) if the difference in
  path length is an integer number of wavelength.
  This condition is satisfied if
• 2d sin ? m? where m an integer
• This equation called the Bragg condition.

96
Bragg scattering from two successive planes. The
waves from the two atoms shown have a path
difference of 2dSin?. They will be in phase if
the Bragg condition 2dSin? m? is met.
97
Reflection from Calcite

• If the spacing between certain planes in crystal
  of calcite is 0.314nm, find the grazing angles at
  which first and third order interference will
  occur for x-rays of wavelength 0.070nm.

98
Schematic sketch of the Laue experiment. The
crystal acts as a three-dimensional grating,
which diffracts the x-ray beam and produce a
regular array of spots, called a Laue pattern, on
a photographic plate.
99
Modern Laue-type x-ray diffraction pattern using
a niobium diboride crystal and 20-kV molybdenum
x-rays. General Electric Company.
100
Incident X-ray Beam

Scattered X-rays
101
Schematic diagram of Bragg crystal spectrometer.
A collimated x-ray beam is incident on a
crystal and scattered into an ionization chamber.
The crystal and ionization chamber can be rotated
to keep the angles of incidence and scattering
equal as both are varied. By measuring the
ionization in the chamber as a function of angle,
the spectrum of the x-rays can be determined
using the Bragg condition 2dSin? m?, where d is
the separation of the Bragg planes in the
crystal. If the wavelength ? is known, the
spacing d can be determined.
102
Two typical x-ray spectra are produced by
accelerating electrons through two voltages V and
bombarding a tungsten target. I(?) is the
intensity emitted with the wavelength interval d?
at each value of ?.
103
The spectrum consist of a series of sharp lines,
called the characteristic spectrum. The line
spectrum is a characteristic of target material
and varies from element to element.
104
The continuous spectrum has a sharp cutoff
wavelength ?m which is independent of the target
material but depends on the energy of the
bombarding electrons.
105

• If the voltage of the x-ray tube is V in volts,
  the cutoff wavelength was found empirically by
• It was pointed out by Einstein that x-ray
  production by electron bombardment was an
  inverse photoelectric effect and equation
• should be applied. The ?m simply correspond to a
  photon with the maximum energy of the electrons,
  i.e. the photon emitted when the electron losses
  all of its kinetic energy in a simple collision.

106

• Since the kinetic energy of the electron in the
  x-ray tube is 20,000 eV or larger, the work
  function F is negligible by comparison and
  equation
• becomes
• or
• Thus, the x-ray spectrum can be explained by
  Planks quantum hypothesis and ?m can be used to
  determine h/e.

107
Compton Effect
Schematic sketch of Compton apparatus. X-rays
from the tube strike the carbon bloke R and are
scattered into a Bragg-type crystal spectrometer.
In this diagram, the scattering angle is 300. The
beam was defined by slits S1 and S2. Although
the entire spectrum is being scattered by R, the
spectrometer scanned the region around Ka line
of molybdenum.
108
Derivation of Comptons Equation

• Let ?1 and ?2 be the wavelengths of the incident
  and scattered x rays, respectively. The
  corresponding momentum are
• and
• using f? c. Since Compton used the Ka line of
  molybdenum ( ? 0.0711 nm), the energy of the
  incident x ray (17.4 keV) is much greater than
  the binding energy of the valence electrons in
  the carbon scattering block (about 11 eV)
  therefore, the carbon electron can be considered
  to be free.

109
Conservation of momentum gives

• where pe is the momentum of the electron after
  the collision and ? is the scattering angle for
  the photon, measured as shown in Figure. The
  energy of the electron before the collision is
  simply its rest energy E0 mc2.

110

• After the collision, the energy of the electron
  is .
• Conservation of energy gives
• Transposing the term p2c and squaring we obtain
• or

111

• If we eliminate pe2 from the previous equations,
  we obtain
• Multiplying each term by and
  using , we obtain
• Comptons equation
• or

112
where is the energy of the incident
photon, is the energy of the scattered
photon, and Ke is the kinetic energy of the
recoiled electron.
113

• Because the electron may recoil at a speed
  comparable to that of light, we must use the
  relativistic expression. Therefore,
• where

and v is the speed of electron
Next, lets apply the law of conservation of
momentum to this collision, noting that the x and
y components of momentum are each conserved
independently.
114
Because the relativistic expression for the
momentum of the recoiling electron is
we obtain the following expressions
for the x and y components of linear momentum
115
Isolate the terms involving F in these equations,
square and add to eliminate F.
Solve for where we define
116

• Substitute into

Square each side
From this we get
117
Compton Wavelength

The increase in wavelengths is independent of the
wavelength ?1 of the incident photon. The
quantity has dimensions of length and is
called the Compton Wavelength. Its
value is
118
Compton Wavelength
Because ?2?1 is small it is difficult to observe
unless ?1 is so small that the fractional change
(?2?1)/?1 is appreciable. Compton used X-rays of
wavelength 71.1pm. The energy of a photon of this
wavelength is

Since this is much greater than the binding
energy of the valence electrons in most atoms,
these electrons can be considered to be
essentially free. Comptons measurements
confirmed the correctness of the photon concept.
119
Compton Scattering at 450

• X-rays wavelength ?00.200 000 nm are scattered
  from a block of material. The scattered x-rays
  are observed at an angle of 450 to the incident
  beam. Calculate the wavelength of the x-rays
  scattered at this angle.
• Find the fraction of energy lost by the photon
  in this collision.

120

• X-rays with a wavelength of 120.0 pm undergo
  Compton scattering. (a) Find the wavelengths of
  the photons scattered at angles of 30.00, 60.00,
  90.00, 120.00, 150.00, and 180.00. (b) Find the
  energy of the scattered electron in each case.
  (c) Which of the scattering angles provides the
  electron with the greatest energy? Explain
  weather you could answer this question without
  any calculations.

121

• X-rays with a wavelength of 120.0 pm undergo
  Compton scattering. (a) Find the wavelengths of
  the photons scattered at angles of 30.00, 60.00,
  90.00, 120.00, 150.00, and 180.00. (b) Find the
  energy of the scattered electron in each case.
  (c) Which of the scattering angles provides the
  electron with the greatest energy? Explain
  weather you could answer this question without
  any calculations.

(a) and (b) From
we calculate the wavelength of the scattered
photon. For example, at ? 30? we have
The electron carries off the energy the photon
loses
122

• X-rays with a wavelength of 120.0 pm undergo
  Compton scattering. (a) Find the wavelengths of
  the photons scattered at angles of 30.00, 60.00,
  90.00, 120.00, 150.00, and 180.00. (b) Find the
  energy of the scattered electron in each case.
  (c) Which of the scattering angles provides the
  electron with the greatest energy? Explain
  weather you could answer this question without
  any calculations.

The electron carries off the energy the photon
loses
123

• X-rays with a wavelength of 120.0 pm undergo
  Compton scattering. (a) Find the wavelengths of
  the photons scattered at angles of 30.00, 60.00,
  90.00, 120.00, 150.00, and 180.00. (b) Find the
  energy of the scattered electron in each case.
  (c) Which of the scattering angles provides the
  electron with the greatest energy? Explain
  weather you could answer this question without
  any calculations.

The other entries are computed similarly.
?, degrees 0 30 60 90
120 150 180 ?', pm 120.0
120.3 121.2 122.4 123.6 124.5
124.8 Ke, eV 0 27.9 104 205
305 376 402
(c) 180?. We could answer like this The photon
imparts the greatest momentum to the originally
stationary electron in a head-on collision. Here
the photon recoils straight back and the electron
has maximum kinetic energy.
124
The Minimum X-ray Wavelength

• Calculate the minimum x-rays wavelength produced
  when electrons are accelerated through a
  potential difference of 100 000V, a not-uncommon
  voltage for an x-ray tube.

125
Photoelectric Effect in Lithium

• Light of wavelength of 400nm is incident upon
  lithium (F 2.9eV). Calculate (a) the photon
  energy and (b) the stopping potential V0.
• What frequency of light is needed to produce
  electrons of kinetic energy 3eV from illumination
  from Li?

126

• An isolated copper sphere of radius 5.0 cm,
  initially uncharged, is illuminated by
  ultraviolet light of wavelength 200 nm. What
  charge does the photoelectric effect induce on
  the sphere? The work function of copper is 4.70
  ev.

127

• An isolated copper sphere of radius 5.0 cm,
  initially uncharged, is illuminated by
  ultraviolet light of wavelength 200 nm. What
  charge does the photoelectric effect induce on
  the sphere? The work function of copper is 4.70
  ev.

Ultraviolet photons will be absorbed to knock
electrons out of the sphere with maximum kinetic
energy
,
,
The sphere is left with positive charge and so
with positive potential relative to
at
As its potential approaches 1.51 V, no further
electrons will be able to escape, but will fall
back onto the sphere. Its charge is then given by
or