CATHODE RAYS
To produce a supply of
electrons, hot cathodes are
used. By heating a metal
such as tungsten to a high
temperature, electrons can
be ejected. This is called
thermoionic emission.
Before they were known to be
tiny particles carrying a small
charge e, beams of electrons
were called cathode rays
because they came from the
cathode inside the tube.
Properties of Cathode Rays:
- They travel from the cathode in a straight line,
- They cause certain surfaces to fluoresce.
- They posses kinetic energy.
- They can be deflected by an electric field.
- They can be deflected by a magnetic field.
- They produce X-rays on striking matter.
DISCOVERY OF THE
ELECTRON
Thomson in 1897 performed
an experiment in which he
measured the speed and the
charge-to-mass ratio (e/m) of
cathode rays. The value (e/m)
was always the same.
Cathode rays soon came to
be recognized as beam of
particles, which we now call
electrons.
Magnetic force = evB
Electric force = eE
14.1 Combine the equations that describe the deflection of an
electron by a magnetic field and by an electric field to give the
charge/mass ratio in Thomson’s experiment.
Deflection by B
Fmagnetic = Fcentripetal
Deflection by E
Fmagnetic = Felectric
Bev = eE
mv
Bev 
r
2
e
v

m Br
E
v
B
e
E
 2
m B r
19
e 1.6 10
11 C/kg

=
1.75x10
m 9.111031
MILLIKAN’S OIL DROP EXPERIMENT
Robert Millikan was awarded the Nobel Prize in 1911 for
determining the charge on the electron. The experiment
was simple enough. Oil was sprayed out of an atomizer.
The small oil drops would pick up some unknown
number of electrons due to friction between the oil and
the nozzle. These oil drops were sprayed between two
oppositely charged parallel plates. The electric field E
between the plates was adjusted until the drop was
suspended in midair. The downward pull of gravity was
then just balanced by the upward force due to the
electric field.
qE = mg
The mass of the droplet was determined by measuring
its terminal velocity in the absence of the electric field.
The experimental results showed that any charge seems
to be an integral multiple of 1.6 x 10-19 C means that
electric charge is quantized.
ROBERT MILLIKAN
14.2 An electron emitted from a hot cathode is accelerated by a
voltage of 1000 V.
a. Calculate the KE and the speed acquired by the electron.
V = 1000 V
q = 1.6x10-19 C
m = 9.11x10-31 kg
KE = W = qV
KE = (1.6x10-19) 1000
= 1.6x10-16 J
KE 
1
2
mv
2
2(16
. x1016 )
2KE
7 m/s

=
1.8x10
v
31
9
.
11
x
10
m
b. The electron enters a uniform magnetic field of 1x10-3 T. Find the
radius of the path.
B = 1x10-3 T
mv 2
Bev 
r
mv 9.11x1031 (18
. x107 )
r
= 0.10 m

3
19
Be
1x10 (16
. x10 )
c. If the electric field plates are 2x10-2 m apart, what is the voltage
between them?
d=
2x10-2
m
FM = FE
eBv = eE
E = Bv
=1x10-3(1.8x107)
= 1.8x104 V/m
V = Ed
= 1.8x104 (2x10-2)
= 360 V
QUANTUM THEORY
Planck postulated that
electromagnetic energy is absorbed
or emitted in discrete packets or
quanta. The energy of such photons
is proportional to the frequency
of the radiation:
E=hf
h is Planck’s constant
h = 6.63 xl0-34 J/Hz
The energy is usually given in
electronvolts: 1 eV = 1.6x10-19 J
Max Planck
(1858-1947)
1918 Nobel Prize
PHOTOELECTRIC EFFECT
In photoelectric emission, electrons are ejected from
metal surfaces when electromagnetic radiation
(photons) of high enough frequency falls on them.
Example: zinc exposed to UV (ultraviolet) rays and Xrays; sodium exposed to X-rays UV rays and all colors
except orange and red.
Einstein used Planck’s equation to
explain the photoelectric effect.
When light strikes a metallic surface
its energy is transferred to a single
electron, the electron might be
expected to leave the metal with the
same amount of energy. However, at
least an amount of energy is needed
to remove the electron from the
metal.
This is called the work function of
the surface and is given by:
Wo = h fo
fo is called the threshold frequency.
Photoelectric Emission:
1. Electrons are emitted only when the frequency of the
light is above some threshold value (fo), no matter how
intense the light.
2. The maximum kinetic energy of the emitted electrons
depends on the frequency of the light.
3. The photoelectrons are emitted almost at once when
the light strikes.
4. The minimum energy required to release an electron
from a certain metal is called the work function of the
metal Wo.
5. The number of photoelectrons emitted is directly
proportional to the intensity of the light.
P
The number of photons emitted per second =
hf
where P is the power of the source.
The stopping potential Vstop is the voltage needed to
stop the emission of electrons: K max
Vstop 
q
Blue light will eject electrons
but red light will not.
The kinetic energy maximum that the ejected electrons
have is given by:
K max = h f - h fo
Effect
Einstein’s Photoelectric
A graph of the maximum kinetic energy of the emitted
electrons vs. the frequency of the light can be used to
find the following:
KEMAX
fo
Wo
f
slope of the line: Planck's constant
x- intercept: threshold frequency
y-intercept: work function
14.3 A surface has a threshold wavelength of 0.65 μm. Calculate:
a. The threshold frequency
λo = 0.65x10-6 m
c = 3x108 m/s
fo 
c
o
3x108 = 4.6x1014 Hz

0.65x106
b. The work function in electronvolts
Wo = h fo
= 6.63x10-34(4.6x1014)
= 3.06x10-19 J
= 3.06x10-19/1.6x10-19
= 1.9 eV
c. The maximum speed of the electrons emitted by violet light of
wavelength 0.4 μm
λ = 0.4x10-6 m
8
(
3
x
10
)
34
-19 J
E  hf  h  6.63x10
=
5x10
(0.4 x106 )

c
Kmax = hf - Wo =
= 5x10-19- 3.06x10-19
= 2x10-19 J
2K
2(2 x1019 )

v
9.11x1031
m
1
K  mv 2
2
= 6.5x105 m/s
14.4 Light of wavelength 650 nm is required to cause electrons to
be ejected from the surface of a metal. What is the K of the ejected
electrons if the surface is bombarded with light of 450 nm?
K h
c

h
λo = 650x10-9 m
λ = 450x10-9 m
c
o
1 1 
 hc   
  o 
 6.63 10
34


1
1
(3 10 ) 

9
9 
(450

10
)
(650

10
)

= 1.36x10-19 J
8
PHOTON INTERACTIONS
Photons of light have energy and momentum but no
mass. Thus, to determine the momentum of a photon,
begin by considering the momentum of a particle with
mass: p = mv
For the mass of the photon, substitute its energy
equivalent from
E
E = mc2,
m 2
c
E
E
p  mv  c 
c
c2
Also c
 f
therefore:
and
E  hf
E
h
hf
p


c
f 
The momentum of a photon, then, is proportional to its
frequency and inversely proportional to its wavelength.
Since photons have momentum, their momentum is
conserved in collisions.
The Compton Effect uses this concept to explain the
scattering that occurs when a photon collides with an
electron at rest. As expected, the electron's increase in
kinetic energy is matched by a decrease in kinetic
energy of the photon. A decrease in KE of the photon
would mean a decrease in frequency and an increase in
its wavelength after the collision.
14.5 A photon of wavelength 0.47 nm strikes an electron at rest.
After the collision, the photon has a wavelength of 0.50 nm.
Calculate:
a. the frequency of the photon before the collision,
λo = 0.47x10-9 m
λf = 0.50x10-9 m
3 108
17 Hz

f 
=
6.38x10
9
0.47

10
o
c
b. the frequency of the photon after the collision
f 
c
f
3 10
17 Hz

9 = 6.0x10
0.50 10
8
c. the speed of the electron after the collision.
Kelectron = Ko(photon) - Kf
(photon)
1
2
me ve  hf o  hf f
2
ve 
2h  f o  f f
m

2 (6.63 1034 (6.4  6) 1017 
9.111031
= 7.6x106 m/s
WAVE NATURE OF MATTER
Just as light exhibits properties of both particles and
waves, particles such as electrons, protons, and
neutrons also exhibit wave properties. Thus the
wave-particle theory extends to matter as well as light.
In 1923, Louis de
Broglie suggested
that the wavelength of
a particle of mass m
traveling at speed v is
given by:
h

mv
Units: meters (m)
Louis de Broglie
(1892-1987)
In 1927, two Americans, Davisson and Germer, produced
diffraction patterns by scattering electron beams from
the surface of a metal crystal. The calculated wavelength
of the electron waves agreed with de Broglie’s
prediction. It was later shown that protons, and neutrons
as well as other particles exhibit wave properties as well
as particle properties.
14.6 What is the de Broglie wavelength of an electron that has a
kinetic energy of 100 eV?
K = 100 eV
= (100 eV)(1.6x10-19J/eV)
= 1.6x10-17 J
m = 9.11x10-31 kg
1
K  mv
2
17
2K
2
(
16
.
x
10
)
6 m/s
v
=
5.93x10

m
9.11x1031
34
h
6.63x10
-10 m

=
1.23x10

mv 9.11x1031 (5.93x106 )
2
PAIR PRODUCTION
A photon can actually create matter, such as the production of an
electron and a positron. A positron has the same mass as an
electron, but the opposite charge. In this process the photon
disappears when creating the pair.
This is an example of rest mass being created from pure energy
according to Einstein's equation: E = mc2. Notice that a photon
cannot create an electron alone since electric charge would not
then be conserved.
The inverse of pair production also occurs: if an electron collides
with a positron, the two annihilate each other and their energy,
including their mass, appears as electromagnetic energy of
photons.
Electron-positron annihilation is the basis for the type of medical
imaging known as PET.
27.7 a. What is the minimum energy of a photon that can produce
an electron-positron pair?
m = 9.11x10-31 kg
E = mc2
= 2(9.11x10-31)(3x108)2
= 1.64x10-13 J
= 1.02 MeV
b. What is the photon's wavelength?
hc (6.63 1034 )(3 108 )
 
13
1.64

10
E
E
hc

= 1.2x10-12 m
STRUCTURE AND PROPERTIES OF THE NUCLEUS
A nucleus can be considered to be made up of two
types of particles: protons and neutrons. These
particles also have wave properties.
Proton: The nucleus of the simplest atom, hydrogen.
Neutron: A particle found in the nucleus that is
electrically neutral and that has a mass almost identical
to the proton.
Nucleons: The term that refers to the two constituent
particles of a nucleus (protons and neutrons).
Isotopes: Nuclei that contain the same number of
protons but different numbers of neutrons.
Thomson’s Model for the Atom
J.J. Thomson’s plum
pudding model consists
of a sphere of positive
charge with electrons
embedded inside.
This model would explain
that most of the mass
was positive charge and
that the atom was
electrically neutral.
Positive
pudding
Electron
Thompson’s
plum pudding
The size of the atom
(10-10 m) prevented
direct confirmation.
Rutherford’s Experiment
The Thomson model was abandoned in
1911 when Rutherford bombarded a thin
metal foil with a stream of positively
charged alpha particles.
Rutherford Scattering Exp.
Alpha source
Gold foil
Screen
Most particles
pass right
through the
foil, but a few
are scattered
in a backward
direction.
The Nucleus of an Atom
If electrons were distributed uniformly, particles
would pass straight through an atom. Rutherford
proposed an atom that is open space with positive
charge concentrated in a very dense nucleus.
Alpha scattering
+
Gold foil Screen
-
Electrons must orbit at a distance in order
not to be attracted into the nucleus of atom.
The Bohr Atom
Energy levels, n
+
The Bohr atom
Bohr’s postulate: When an electron
changes from one orbit to another, it gains
or loses energy equal to the difference in
energy between initial and final levels.
Bohr’s Atom and Radiation
Emission
Absorption
When an electron drops to a
lower level, radiation is
emitted; when radiation is
absorbed, the electron
moves to a higher level.
Energy: hf = Ef - Ei
By combining the idea of energy levels with
classical theory, Bohr was able to predict
the radius of the hydrogen atom.
ENERGY LEVEL DIAGRAMS
Radiation is absorbed or released when an atom
changes from one stationary state to another. The
energy of the emitted or absorbed photon is said to be
quantized and is equal to the difference in the energy
between these two states:
E  E f  Eo
The transition on an energy level diagram is shown
below.
Absorption of a photon to promote the electron to a
higher energy level.
nF
no
n=5
n=4
n=3
n=2
n=1
The transition on an energy level diagram is shown
below.
Emission of a photon when the electron falls from a
higher to a lower energy level.
no
nF
n=
n=6
n=5
n=4
n=3
n=2
n=1
Atomic Spectra
In an emission spectrum, light is separated into
characteristic wavelengths.
Emission Spectrum
Gas
Absorption Spectrum
1
2
In an absorption spectrum, a gas absorbs certain
wavelengths, which identify the element.
14.8 A hypothetical atom has 4 energy levels: 0 eV, 1 eV, 3 eV and
6 eV.a. Draw an energy level diagram for this atom indicating the
quantum numbers and the energies associated with them.
b. Use arrows to show all of the possible transitions between
energy levels.
The 12 possible transitions are indicated by the arrows
showing either absorption or emission of a photon.
c. For which transition is the associated photon energy highest?
The highest possible photon energy is 6 eV, corresponding to
a transition between the n = 1 and n = 4 levels.
d. For which transition is the associated photon energy lowest?
The lowest photon energy is 1 eV, corresponding to a
transition between the n = 1 and n = 2 levels.
e. For which transition is the associated photon wavelength
longest?
The longest wavelength corresponds to the lowest energy
since
hc

E
The transition between n = 1 and n = 2 corresponds to the longest
wavelength.
f. For which transition is the associated photon wavelength
shortest?
The transition between n = 1 and n = 4 (highest energy)
corresponds to the shortest wavelength.
g. A photon incident on the hypothetical atom causes the electron
to make a transition from the n = 2 orbital to the n = 4 orbital. What
is the wavelength of the photon?
E4  E2  6 eV 1 eV  5 eV
hc 1240 eV  nm


 248 nm
E
5 eV
h. How many wavelengths of emitted radiation are possible when
the electron returns to the n = 2 state?
i. An electron moving with a speed of 1.25x106 m/s collides with the
hypothetical atom. Is the energy provided by the electron enough
to excite the atom to the n = 3 state? Is it enough for the atom to
reach the n = 4 state?
1 2 1
KEe  mv  (9.111031 kg)(1.25 106 m/s) 2  7.12 10 19 J
2
2
7.12 1019 J
KEe 
 4.45 eV
19
1.6 10 J/eV
The electron has enough energy to excite the atom to the
n = 3 state but not to the n = 4 since that will require 6 eV.
Atomic Number: The number of protons in a nucleus
(designated by the letter Z).
Atomic Mass Number: The total number of protons and
neutrons (designated by the letter A).
Z = Atomic number
N = Number of neutrons
A = Atomic mass
A=Z+N
BINDING ENERGY
The total mass of a stable nucleus is always less than
the sum of the masses of its constituent particles.
Mass Defect: The difference in the mass of a nucleus
and the sum of the masses of its constituent particles.
Nuclear Binding Energy: The
amount of energy that must
be put into a nucleus to
break it into its constituent
particles.
It is the energy equivalent of
the mass defect found by
using:
E = m c2
The most convenient unit of
energy to use is the
electronvolt (eV)
E = 931MeV
E = mc2
One atomic mass unit (amu) = 1.6605 x 10-27 kg
-------------------------------------------------------------------E = mc2
E = (1.6605 x 10-27 kg) (3 x 108 m/s)2
= 1.49 x 10-10 J
1.49 x 10-10 J / 1.6 x 10-19 J /eV = 9.31 x 108 eV
= 931 x 106 eV
= 931 MeV
one amu = 931 MeV
The Binding Energy of Helium
The larger the binding energy of
a nucleus, the more stable it is.
14.9 The mass of a proton is: 1.007825 u. The mass of a neutron is
1.008665 u. The mass of the nucleus of the radioactive hydrogen
isotope tritium H-3 is 3.016049 u.
a. What is the nuclear mass defect of this isotope? 3 H
1
1 proton + 2 neutrons
1(1.007825) + 2(1.008665) = 3.025155 u
Mass defect = 3.025155 - 3.016049
= 0.009106 u
b. What is the binding energy of tritium?
E = 0.009106 (931)
= 8.47 MeV
TRANSMUTATION
Transmutation is the changing of one element into
another via radioactive decay.
Transmuting Uranium into Neptunium
RADIOACTIVE DECAY
Alpha particles are 8,000 times
as heavy as beta particles.
Paper or clothing will block
alpha particles.
Beta particles require a few
sheets of aluminum foil.
Gamma radiation is extremely
dangerous, a thousand times
more potent than x-rays.
Alpha Particle
The particle is emitted in alpha decay. It is essentially a
helium nucleus. It contains two protons and two
neutrons. It has a charge of q = -2 and a mass of A = 4.
When a nuclei decays by emitting an alpha particle,
the number of protons is reduced by two and its mass
is reduced by four. Alpha particles are emitted by
very
large nuclei where the strong nuclear force is
insufficient to hold the nuclei together.
It is abbreviated:
α or 4
2
He
Alpha Particle Emission
Distribution of Energy in Alpha Emission
Beta Particle
The particle is emitted in beta decay. Beta particles are
negative electrons or positrons emitted by the nucleus.
It is not an orbital electron, but one created in the
nucleus by the decay of a neutron into a proton and an
electron. Beta particles are emitted by nuclei that have
too many neutrons relative to the number of protons.
n p e
1
0
1
1
0
1
and
1
1
p n e
1
0
0
1
Another particle called the neutrino is also emitted in
beta decay. It is abbreviated 
The weak nuclear force is involved in the production of a
beta particle in the nucleus.
Beta Particle (Electron) Emission
Beta Particle (Positron) Emission by Oxygen-15
A positron has the same mass as the
electron, but opposite charge.
Gamma Radiation
Radiation emitted in gamma decay. Gamma radiation is
composed of high-energy photons. It is emitted by
excited state nuclei. Gamma radiation has no charge and
no mass. It is abbreviated : 
14.10 Write the nuclear equation for the transmutation of radium226 into radon-222 by the emission of an alpha particle.
226
88
Ra Rn He
222
86
4
2
14.11 Write the nuclear equation for the transmutation of lead-209
into bismuth-209 by the emission of a beta particle.
209
82
Pb Bi  e
209
83
0
1
NUCLEAR ENERGY
In 1934, Enrico Fermi and Emilio Segré bombarded
uranium with neutrons, producing new radioactive
isotopes. In 1939, German scientists Otto Hahn and Fritz
Strassmann found that barium was produced by
bombarding uranium with neutrons. Lisa Meitner and
Otto Frisch proposed that the neutrons caused the
uranium to divide into two smaller nuclei, accompanied
by a tremendous release of energy.
Fission: A division of a nucleus into two or more smaller
daughter nuclei.
Fusion: Two or more nuclei combine to form a larger
nucleus. The sun produces its energy by nuclear fusion.
Fusion is the opposite of fission. Deuterium
must be moving extremely fast to fuse.
A Nuclear Reactor
The Reactor Vessel
The water in the reactor vessel
has three purposes:
- The water, being composed of
relatively light molecules, acts
as a moderator.
- Water also acts to remove
heat from fuel rods which
otherwise would melt.
- The heated water, converted
to steam, is then converted
into electrical energy.
Chain Reaction: Neutrons produced by the fission of one
nucleus induce the fission of other nuclei.
A chain reaction
occurs if more
than one neutron
goes on to cause
another fission.