Quantization of Charge, Light, and
Energy
CHAPTER 3
Prelude to Quantum Theory
Max Karl Ernst Ludwig
Planck (1858-1947)
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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
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
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•
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1. Quantization of Electric Charge;
2. Blackbody Radiation;
3. The Photoelectric Effect
4. X-rays and the Compton Effect.
Three great quantization discoveries:
1. Quantization of Electrical Charge
2. Quantization of Light Energy
3. Quantization of energy of Oscillating
Mechanical System
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 Faraday’s 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.
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 – Avogadro’s number
e – minimum amount of charge, that was
called an electron
Discovery of the the Electron
Wilhelm Röntgen
(1845-1923)
J. J. Thomson
(1856-1940)
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.
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.
Discovery of electron: Thomson’s 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.
Thomson’s 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.
Thomson’s Experiment: e/m
Thomson’s method of measuring
the ratio of the electron’s charge to
mass was to send electrons through
a region containing a magnetic field
perpendicular to an electric field.
J. J. Thomson
Thomson’s 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 tube’s screen, e/m can be
determined.
FE=qE
_q
-
FB=qvB
FE=FB
qE = qvB
Crossed electric and magnetic fields. When a negative
particle moves to the right the particle experience a
downward magnetic force FB=qvB and an upward electric
force FE=qE. If these forces are balanced, the speed of the
particle is related to the field strengths by
v=E/B
Thomson’s Experiment
In his experiment Thomson adjusted E
that the particles were undeflected.
┴B
so
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.
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.
With the magnetic field turned off, the beam are deflected by an amount
y=y1+y2. 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 t1=x1/v0 , and the vertical component of velocity
when it leaves the plates is
qEy
qEy x1
vy  a yt1 
t1 
m
m v0
where Ey is the upward component of electric field. The deflection
y1 is:
2
qE
qE
y 2
y  x1 
2
1
1
1
 
y1 
a yt1 
t1 
2
2 m
2 m v 
 0
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 t2=x2/v0 , and the additional vertical
deflection is:
y2  v yt2
qE y x1 x2

m v0 v0
The total deflection at screen is therefore

 qE y  2  qE y 
 E y  x12 
q


y  y1  y2  1  2  x1   2  x1x2    2    x1x2 
2  mv 
 mv 
 v  2 
m



 0
 0
 0  
This equation cab be used to determine the charge tomass ratio q/m from measured deflection y.
Determination of
Electron Charge
• Millikan’s oil-drop
experiment
Robert Andrews Millikan
(1868 – 1953)
• Millikan was able
to show that
electrons had a
particular charge.
Calculation of the oil drop charge
• Millikan used an electric field to balance
gravity and suspend a charged oil drop:
V
F y  eE  e   m drop g
d
 e   m drop gd / V
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, , and the air
viscosity, h :
Drop
radius:
r  3 h vt / 2 g 
and
Thousands of experiments showed that
there is a basic quantized electron charge:
m drop  43  r 3 
e = 1.602 x 10-19 C
Example
Electrons pass undeflected through the plates of Thomson’s
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.
me=9.11 x 10-31kg;
q=e=-1.6 x 10-19C
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 24:25:26.
Schematic drawing of a mass spectrometer.
Positive ions from an ion source
are accelerated through a
potential difference ΔV 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
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 ΔV, the ions kinetic energy
when they enter the magnetic field
equals their loss in potential energy,
q|ΔV|:
1 2
mv  q V
2
F  ma
2
v
qvB  m
R
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 Newton’s 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.
mv
R
qB
The Mass Spectrometer
The speed v can be eliminate from equations
1 2
mv  q V
2
mv
R
qB
and
2
2
Rq B
v 
2
m
2
2
Substituting this for v2:
1 R2q 2 B2
m
 q V
2
2
m
Simplifying and solving for (m/q):
2
2
m B R

q 2 V
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 58:60).
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.
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.
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.
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.
Spectral distribution function R(λ) measured at different
temperatures. The R(λ) axis is in arbitrary units for comparison
only. Notice the range in λ of the visible spectrum. The Sun
emits radiation very close to that of a blackbody at 5800 K. λm
is indicated for the 5000-K and 6000-K curves.
Wien’s Displacement Law
 The spectral intensity R(λ, T) is the total power irradiated
per unit area per unit wavelength at a given temperature.
 Wien’s displacement law: The maximum of the spectrum
shifts to smaller wavelengths as the temperature is
increased.
Wien’s Displacement Law
λmT = 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 Wien’s law, we have
λ1T1 = λ2T2
(674 x 10-9m)(4300 K) = (420 x 10-9m)(T2)
T2=6900 K
 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.
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 λmax for a
blackbody at room temperature, T=300 K.
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 StefanBoltzman radiation Law:
The total power radiated increases with the temperature

P(T )   R( , T )d  T
4
0
with the constant σ experimentally measured to be 5.6705 ×
10−8 W / (m2 · K4).
Blackbody Radiation
P = σT4
 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 24=16.
 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.
The calculation of the
distribution function R(λ)
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:
c
R ( )  U
4
The
spectral
distribution of the power
proportional
to
the
spectral distribution of the
energy density in the
cavity.
 If u(λ)dλ is the fraction of
the energy per unit
volume in the cavity in the
range dλ, then u(λ) and
R(λ) are related by

1
R ( )  cu ( )
4
The energy density distribution function u(λ) can be calculated
from classical physics.
We can find the
number of modes of
oscillation
of
the
electromagnetic field
in the cavity with
wavelength λ in the
interval
dλ
and
multiply it by average
energy per mode.
The result is that the number of modes of oscillation per unit
volume, n(λ), is independent of the shape of cavity and is
given by:
n( )  8c
4
Rayleigh-Jeans Equation
The number of modes of oscillation per unit volume:
n( )  8c
4
According to the classical kinetic theory, the average energy
per mode of oscillation is kT, the same as for a onedimensional harmonic oscillator, where k is the Boltzman
constant.
Classical theory thus predicts for the energy density spectral
distribution function
u( )  kTn( )  8ckT
4
Rayleigh-Jeans Formula
This
prediction,
initially
derived by Lord Rayleigh
using the classical theories
of electromagnetism and
thermodynamics is called the
Rayleigh-Jeans Law:
u ( ) 
8ckT

4
It approaches the experimental data at longer wavelengths,
but it deviates badly at short wavelengths.
u ( ) 
8ckT
4
At short wavelength this law predicts that u(λ) becomes large,
approaching infinity as λ→0, whereas experiment shows that
the distribution actually approaches zero as λ→0.
This problem for small wavelengths became known as the
ultraviolet catastrophe and was one of the outstanding
exceptions that classical physics could not explain.
Planck’s Radiation Law
In 1900 the German physicist Max Plank by making some
unusual assumptions derived a function u(λ) 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:

f ( E )  Ae
E
kT
where A is a constant and f(E) is the fraction of oscillators with
energy E.
Maxwell-Boltzman distribution function

E
kT
f ( E )  Ae
The average energy is then found, as is any weighted
average:

E   Ef E dE   EAe
E
kT
dE
Planck made two modifications to the classical theory:
1. The oscillators (of electromagnetic origin) can only have
certain discrete energies,
En  nh
where n is an integer, v is the frequency, and h is called
Planck’s constant: h = 6.6261 × 10−34 J·s.
2. The oscillators can absorb or emit energy in discrete
multiples of the fundamental quantum of energy given by:
E  h
The Maxwell-Boltzman distribution than becomes:

f ( E )  Ae
E
kT

 Ae
nh
kT
where A is determined by normalization condition that
the sum of all fractions fn must be equal 1.
The normalization condition
f
n
e
A

nh
kT
1
The average energy of oscillation is then given by discretesum equivalent
E

E   En f n   En Ae
h
x
kT
To solve this equation let put
f
n
 A e
 nx
where y=e-x.
3
x
x 2
x 3
 A[e  e  (e )  (e )  ....] 
0
A(1  y  y  y  ....)  1
2
kT
This sum is a series expansion of (1-y)-1, so
(1-y)-1=1+y+y2+y3+… ,
then Σfn=A(1-y)-1=1 gives A=1-y and


E   En Ae
0
E
kT

 A nhe

nh
kT

 Ah  nenx
0
0
Note that
 ne
 nx
d
   e nx . But
dx
 nx
1
e

(
1

y
)

so we have
 ne
 nx
since
d
d
dy 
 nx
1
2 
   e   (1  y)  (1  y)     y(1  y ) 2
dx
dx
 dx 
dy d  x

(e )   e  x   y
dx dx
Plank’s Law
Multiplying this sum by hv and using A=(1-y), the
average energy is:
E  hA
 ne
 nx
hy he
 h (1  y) y(1  y) 

x
1 y 1 e
x
2
Multiplying the numerator and the denominator by
ex and substituting for x, we obtain:
E
h

e


h
kT

 1


hc

 hkT

 e  1




Plank’s Law
Multiplying this result by the number of oscillators
per unit volume in the interval dλ given by
n(λ)=8πcλ-4 (the number of modes of oscillation
per unit volume) we obtain the energy distribution
function for the radiation in cavity:
8h c 
u ( ) 
hc
 kT

e


1




2
5
This function is called Plank’s Law.
Plank’s Law
The value of Plank’s constant h can be
determined by fitting the function
8hc 
u ( ) 
hc
 kT

e


1




2
5
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 J·s = 4.136 x 10-15 eV·s
u ( ) 
8ckT
4
Comparison of Plank’s Law and the Rayleigh-Jeans Law
with experimental data at T=1600 K. The u(λ) axis is linear.
Plank’s Law
A dramatic example of an application of Planck’s 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).
The energy density spectral distribution of the cosmic
microwave background radiation. The solid line is Plank’s Law
with T=2.735 K. The measurements were made by the
Cosmic Back Ground Exploder (COBE) satellite.
Problem 1. Thermal Radiation from the Human Body.
The temperature of the skin is approximately 35°C. What
is the wavelength at which the peak occurs in the
radiation emitted from the skin?
Problem 2. The Quantized Oscillator. A 2-kg mass is
attached to a massless spring of force constant
k=25N/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?
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?
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 peakintensity wavelength λm? (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.
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 Maxwell’s 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.
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 α. A magnetic
field, indicated by the circular pole piece, could deflect the
particles to an electrometer connected to β, enabling the
establishment of the sign of their charge and their q/m ratio.
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 α.
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:
 Photocurrent i versus anode voltage V for light of frequency
f with two intensities I1 and I2 , where I2>I1 . 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.
 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.
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
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
Photo-electric effect
observations
Electron
kinetic
energy
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.
Photo-electric effect
observations
There was a threshold
frequency of the light,
below which no
photoelectrons were
ejected.
Electron
kinetic
energy
The existence of a threshold frequency is completely
inexplicable in classical theory.
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.
 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 hν.
Einstein’s Theory: Photons
Einstein suggested that the electro-magnetic radiation field
is quantized into particles called photons. Each photon has
the energy quantum:
E  h
where ν is the frequency of the light and h is Planck’s
constant.
Alternatively,
E 
where:
 h / 2
Electron kinetic energy
Einstein’s Theory
Conservation of energy
yields:
h  f  mv
1
2
2
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 electron’s energy can be reduced
by the emitter material, consider vmax (not v):
h  f  mv
1
2
2
max
 When one photon penetrates the surface of cathode, all of
its energy may be given completely to electron. If Φ is the
energy necessary to remove an electron from the surface
(Φ 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 ν – Φ) and the stopping
potential V0 should be given by
 m v2 
  h  
eV0  
 2  max
 This equation is referred as the photoelectric effect
equation.
 As can be seen from
 m v2 
  h  
eV0  
 2  max
the slope of the line on the graph V0 versus ν should equal h/e.
 The minimum, or threshold, frequency for photoelectric effect,
νt and the corresponding threshold wavelength λt are related
to work function Φ by setting V0 = 0 :
  h t 
hc
t
 Photons of frequency lower than νt ( and therefore having
wavelength grater than λt ) do not have enough energy to
eject an electron from the metal.
For constant intensity Einstein’s 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.
Millikan’s 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 –Ф/e.
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?
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:
1 2
eV0   m v   hf  
2
 max
hf 
V0  
e e
 hft hc


e
e et
hc 1240eV  nm


 2.22eV
t
558nm
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?
1240eV  nm
eV0    
 2.22eV

400nm
 3.10eV  2.22eV  0.88eV
hc
V0  0.88V
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.
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.
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.
Röntgen’s 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.
(a)
(b)
(a) Early x-ray tube and (b) typical of the mid-twenties
century x-ray tube design.
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.
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:
Ei
Ef
h
E f  Ei  h
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.
Inverse Photoelectric Effect
Conservation of energy requires that
the electron kinetic energy equal the
maximum photon energy (neglect
the work function because it’s 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.
eV0  h max 
hc
min
Photons also have momentum!
Use our expression for the
relativistic energy to find the
momentum of a photon, which has
no mass:
E  (mc )  p c
2
2 2
Alternatively:
2 2

E h h
p 

c
c 
h 2
p
 k
2 
X-Ray’s 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.
X-Ray’s 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.
A face-centered cubic crystal of NaCl showing
two sets of Bragg planes.
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 θ = mλ where
m = an integer
This equation called the Bragg condition.
Bragg scattering from two successive planes. The waves from
the two atoms shown have a path difference of 2dSinθ. They
will be in phase if the Bragg condition 2dSinθ = mλ is met.
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.
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.
Modern Laue-type x-ray
diffraction pattern using a
niobium diboride crystal
and 20-kV molybdenum
x-rays. [General Electric
Company.]
Incident
X-ray
Beam
Scattered
X-rays
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 2dSinθ = mλ, where d is the
separation of the Bragg planes in the crystal. If the wavelength λ
is known, the spacing d can be determined.
Two typical x-ray spectra are produced by accelerating
electrons through two voltages V and bombarding a tungsten
target. I(λ) is the intensity emitted with the wavelength interval
dλ at each value of λ.
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.
The continuous spectrum has a sharp cutoff wavelength λm
which is independent of the target material but depends on
the energy of the bombarding electrons.
If the voltage of the x-ray tube is V in volts, the cutoff
wavelength was found empirically by
1 .24  10 3
m 
nm
V
It was pointed out by Einstein that x-ray production by
electron bombardment was an inverse photoelectric effect and
equation
 mv
eV0  
 2
2

  hf  
 max
should be applied. The λm 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.
• Since the kinetic energy of the electron in the x-ray tube is
20,000 eV or larger, the work function Φ is negligible by
comparison and equation
 mv2 

eV0  
 hf  
 2 

max
becomes
eV  hf 
or
hc

hc 1240 eV  nm


 1.24 10 3 nm
eV
eV
Thus, the x-ray spectrum can be explained by Plank’s
quantum hypothesis and λm can be used to determine h/e.
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 Kα line of molybdenum.
Derivation of Compton’s Equation
Let λ1 and λ2 be the wavelengths of the incident and scattered
x rays, respectively. The corresponding momentum are
E1 hf1
h
P1 


c
c
1
and
E2
h
P2 

c
2
using fλ = c. Since Compton used the Kα line of molybdenum
( λ = 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.
Conservation of momentum gives
 

p1  p2  pe
2
 
2
2
pe  p1  p2  2 p1  p2  p12  p 22  2 p1 p2 cos
where pe is the momentum of the electron after the collision
and θ 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.
After the collision, the energy of the electron is
Conservation of energy gives
E 02  p e2 c 2 .
p1 c  E 0  p 2 c  E  p c
2
0
2 2
e
Transposing the term p2c and squaring we obtain
E  c ( p 1  p 2 )  2 cE 0 ( p 1  p 2 )  E  p c
2
0
2
2
2
0
or
2 E 0 ( p1  p 2 )
p  p  p  2 p1 p 2 
c
2
e
2
1
2
2
2 2
e
2 E 0 ( p1  p 2 )
p  p  p  2 p1 p 2 
c
2
If we eliminate p from the previous equations, we obtain
2
e
2
1
2
2
e
E 0 ( p1  p 2 )
 p1 p 2 (1  cos  )
c
hc
Multiplying each term by
p1 p2 E0
we obtain
and using   h ,
Compton’s equation:
or
hc (1  cos  ) hc (1  cos  )
 2  1 

E0
mc 2
 h 
   0  
  1  cos 
 m ec 
p
hc
1

hc
2
 Ke
hc
hc
where
1
is the energy of the incident photon,
2
is the
energy of the scattered photon, and Ke is the kinetic energy of
the recoiled electron.
Because the electron may recoil at a speed comparable to
that of light, we must use the relativistic expression.
Therefore,
hc
1

hc
2
 (  1)me c
2
where

1
2
v
1 2
c
and v is the speed of electron
Next, let’s apply the law of conservation of momentum to
this collision, noting that the x and y components of
momentum are each conserved independently.
E2
h
P2 

c
2
E1 hf1
h
P1 


c
c
1
Because the relativistic expression for the momentum of the
recoiling electron is pe  mev we obtain the following
expressions for the x and y components of linear momentum
x
y
com ponent:
com ponent:
h
1

0
h
2
h
2
cos  meu cosf
sin   meu sin f
x
y
com ponent:
com ponent:
h
1

0
h
2
h
2
cos  meu cosf
sin   meu sin f
Isolate the terms involving Φ in these equations, square and
add to eliminate Φ.
1
1 2cos 
2 2 2
h  2 2


me u

0  
 0  
2
u2
Solve for 2  b 2
c
 b c 
h2
b 2
me
where we define
1
1 2cos 
 2  2 






0
 0

Substitute into
hc
1

hc
2
 (  1)me c 2
 h  1 1 
b 

1 
        1
2 

m
c


b

c


 e  0

1 2
c2  b

c2
Square each side:
2hc  1 1  h
2
c 
    2
me  0   me
2
From this we get:
2
2
 1 1

 1
h
1 2cos 
2
     c   2   2  2 




m




0
 0

 e  0

 h 
   0  
 1  cos 
 mec 
Compton Wavelength
h
  2  1 
(1  cos )
mec
The increase in wavelengths is independent of the
wavelength λ1 of the incident photon. The quantity
h
has dimensions of length and is called the
me c
Compton Wavelength. Its value is
h
hc
1240eV  nm
12
C 



2
.
43

10
m  2.43pm
2
5
me c me c
5.1110 eV
Compton Wavelength
Because λ2–λ1 is small it is difficult to observe
unless λ1 is so small that the fractional change
(λ2–λ1)/λ1 is appreciable. Compton used X-rays of
wavelength 71.1pm. The energy of a photon of this
wavelength is
1240 eV  nm
E

 17.1keV

0.0711nm
hc
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. Compton’s
measurements confirmed the correctness of the
photon concept.
Compton Scattering at 450
X-rays wavelength λ0=0.200 000 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.
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.
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   h 1  cos 
mec
we calculate the wavelength of the scattered photon. For
example, at  = 30 we have
     120  10
12
6.626  1034
12

1

cos30.0


120.3

10
m


31
8
 9.11 10  3.00 10 
The electron carries off the energy the photon loses:
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.
     120  10
12
6.626  1034
12

1

cos30.0


120.3

10
m


31
8
 9.11 10  3.00 10 
The electron carries off the energy the photon loses:
hc 6.626  10 34 J  s 3  10 8 m  1
1 
Ke   

 27.9 eV


-19
-12
0   1.6  10 J/ eV  s 10 m  120 120.3 
hc
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
', pm
120.0 120.3
Ke, eV
0
27.9
60
121.2
104
90
122.4
205
120
123.6
305
150
124.5
376
180
124.8
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.
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.
Photoelectric Effect in Lithium
Light of wavelength of 400nm is incident upon lithium
(Φ = 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?
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.
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 K m ax  hf  f
6.626  10


J s 3.00  108 m s  1.00 eV

19
200  109 m
1.60

10

34
Kmax

  4.70 eV  1.51 eV
J
,
The sphere is left with positive charge and so with positive
potential relative to V  0 at r  
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
keQ
V
or
r
rV  5.00  10
Q

2
ke
m  1.51 N  m C 
8.99  109 N  m 2 C2
 8.41 1012 C