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Grade 12
Unit 9
SCIENCE 1209
ATOMIC AND NUCLEAR PHYSICS
CONTENTS
I. QUANTUM THEORY . . . . . . . . . . . . . . . . . . . .
2
ELECTROMAGNETIC RADIATION . . . . . . . . . . . . . . . . . . . .
MATTER WAVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ATOMIC MODELS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ATOMIC SPECTRA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
BOHR MODEL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
8
11
12
14
II. NUCLEAR THEORY . . . . . . . . . . . . . . . . . . . .
22
BUILDING BLOCKS OF THE NUCLEUS . . . . . . . . . . . . . . .
PROPERTIES OF THE NUCLEUS . . . . . . . . . . . . . . . . . . . .
22
23
III. NUCLEAR REACTIONS . . . . . . . . . . . . . . . . .
30
NUCLEAR FISSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
FUSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
NUCLEAR REACTION APPLICATIONS . . . . . . . . . . . . . . . .
30
32
33
GLOSSARY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
FORMULAS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
41
Author:
Editor:
Illustrations:
Robert Hughson, M.S.
Alan Christopherson, M.S.
Jim Lahowetz
Alpha Omega Graphics
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ATOMIC AND NUCLEAR PHYSICS
You are about to begin the study of modern physics, the physics of very small objects. You will learn
more about how atoms and their subparts interact. The laws of physics that you have already learned will
be used, but you will notice that at times some new concepts will be considered.
OBJECTIVES
Read these objectives. The objectives tell you what you will be able to do when you have
successfully completed this LIFEPAC®.
When
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
you have finished this LIFEPAC, you should be able to:
Explain when wave theory is applicable and when quantum theory is applicable.
Calculate the energy of photoelectrons and X-rays.
Explain de Broglie waves.
State and apply the uncertainty principle.
Define emission spectra and absorption spectra.
Define line spectra and continuous spectra.
Describe the Bohr model of the hydrogen atom.
Define nucleon and isotopes.
Explain mass defect in terms of mass-energy equivalent.
Explain how the binding energy curve shows that fission and fusion can release energy.
Explain the concept of half-life.
Explain how alpha and beta decay can stabilize a nucleus.
Describe fission reaction and fusion reaction.
Explain a chain reaction and nuclear energy.
Survey the LIFEPAC. Ask yourself some questions about this study. Write your questions here.
The last page of this LIFEPAC contains formulas and constants that may be used for
problem-solving, both on activities and on tests.
1
I. QUANTUM THEORY
For many centuries, the primary concern of
physics was to describe the behavior of large
objects. Newtonian mechanics describes the
behavior of matter influenced by gravitational
fields or by electromagnetic interactions.
Newtonian physics had been successful; so
successful, in fact, that most physicists believed
they were approaching the ultimate description of
nature. They thought that in just a matter of time
they would be able to work out the details of a
perfect, ordered world. Many believed that the
future work of physics would be essentially just
measuring the last decimal place.
Problems with this neat and supposedly
complete theoretical description began to arise
about the end of the nineteenth century. More
precise experiments opened a whole new world to
the physicist who was beginning to investigate the
properties of very small particles. The behavior of
these microscopic systems could not be explained
in terms of the classical physics and Newtonian
mechanics, which had been the physicists’ stock
and trade up to this point.
In classical physics the physicist had explained
the behavior of objects in terms of direct and
compelling deductions from a well defined set of
experiments. Attempts to do the same from new
observations were not successful. The new physics,
or quantum physics, was too abstract to make
deductive explanations possible. The main reasons
were twofold: The basic constructs of this quantum
theory were removed from everyday experience, and
the microscopic particles could not be observed
directly.
SECTION OBJECTIVES
Review these objectives. When you have completed this section, you should be able to:
1. Explain when wave theory is applicable and when quantum theory is applicable.
2. Calculate the energy of photoelectrons and X-rays.
3. Explain de Broglie waves.
4. State and apply the uncertainty principle.
5. Define emission spectra and absorption spectra.
6. Define line spectra and continuous spectra.
7. Describe the Bohr model of the hydrogen atom.
VOCABULARY
Study these words to enhance your learning success in this section.
absorption spectrum
energy levels
proton
alpha particle
excited atom
quanta
continuous spectrum
ground state
quantum number
de Broglie wave
line spectrum
radiation
electron volt
photoelectric effect
uncertainty principle
emission spectrum
photon
X-ray
Note: All vocabulary words in this LIFEPAC appear in boldface print the first time they are used. If you are unsure of
the meaning when you are reading study the definitions given.
ELECTROMAGNETIC RADIATION
The wave theory of light was well recognized by
the end of the nineteenth century. Experimental
work was being conducted in which both light rays
and high speed electrons were allowed to strike a
metal surface. The results of these experiments
could not be explained by existing theory. As a
result a new theory was developed, called the
quantum theory. At the time, physicists did not
fully realize what was happening; but modern
physics had been born.
2
Photoelectric effect. During the late 1800s
several experiments disclosed that electrons were
emitted when light fell on a metal plate. This
phenomenon is the photoelectric effect. At first no
one was surprised: Light waves were known to carry
energy; some of this energy, therefore, could
somehow be absorbed by electrons to give them
enough kinetic energy to escape from the metal
surface.
The symbol v (the Greek nu) represents the
frequency of incident light. The value of h was
always the same value, but v0 changed for each
metal.
An explanation of this phenomenon was given
in 1905 by Albert Einstein. He used a result of
earlier work by the German physicist Max Planck,
who described the radiation emitted by a glowing
object. Planck could get his theory to work if he
assumed that the light was emitted in little “lumps”
of energy. He called these little bursts of energy
quanta and showed that they had energy of
LIGHT
E = hv,
where v is the frequency and h is a constant known
as Planck’s constant. The value of Planck’s constant
h is given as
h = 6.63
•
10-34 joule•sec.
AMMETER
Few physicists really believed in Planck’s quanta,
but Einstein pointed out that the photoelectric
effect was experimental verification of Planck’s
equation. He suggested that the photoelectric
equation be rewritten:
Soon some problems developed: First, the
energy of the photoelectrons was not related to the
intensity of the light. When the light was made
brighter, more electrons were emitted; but they all
had the same average energy as those emitted by
dim light. Second, the energy of the photoelectrons
was related to the frequency of the light. If the
frequency was below a certain value, no electrons
were emitted even if the light was intense.
Conversely, if the frequency was above the value,
electrons were emitted even in very dim light.
Photoelectrons emitted by any given frequency
were observed to have energy that ranged from
zero to some maximum value. If the frequency of
the light was increased, the maximum electron
energy increased. In other words, a faint blue light
produced electrons with higher energy than did a
bright red light even though the bright red light
produced more electrons. The relationship between
KEmax, the maximum photoelectron energy, and the
frequency of the incident light can be stated:
hv = KEmax + hv0,
and be stated:
quantum (incident photon) energy = maximum
electron energy + energy required to eject an
electron from the metal surface.
A large portion of the photoelectron population
will have less than maximum energy for the
following reasons:
1. The photon (the quantum of incident light)
may have shared its energy with more than
one electron.
2. The electron may have lost part of its
energy before it got out of the metal.
The energy of photons is so small that the joule
is cumbersome. A unit of energy suited to the
atomic scale is the electron volt. One electron volt
is the energy change realized by an electron
moving through a potential change of one volt.
Numerically,
maximum
photoelectron
energy= h (frequency - frequency0)
KEmax = h (v - v0).
1 ev = 1.6
3
•
10-19 joules
LlFEPACs 1204 and 1205 dealt with diffraction
and interference. The observed phenomena could be
explained in terms of the wave theory of light. Now
you have seen an argument that light behaves like a
series of packets of energy, called photons, or quanta,
each photon small enough to interact with a single
electron. The wave theory cannot explain the
photoelectric effect and the quantum theory cannot
explain interference and diffraction. Which is the
correct theory? Many times before in physics, one
theory has been replaced by a new theory; but this
apparent conflict was the first time that two
✍
different theories were needed to describe a
phenomenon. Developing a model for light is a
situation in which science must acknowledge that
nature is not an “open book” to man. These two
theories are complementary: One is correct in some
experiments, but the other theory must be used in
others.
Interestingly, light can exhibit either a wave or
a particle nature, but never exhibits both in any
single situation. Since we cannot reconcile the two
descriptive models, we have no choice but to accept
them both.
Answer these questions.
1.1
What is the photoelectric effect?
1.2
Why do photoelectrons have a maximum energy?
1.3
Why does faint light not appear as a series of flashes?
1.4
Why can the photoelectric effect not be explained without quantum theory?
✍
Complete these sentences.
1.5
The phenomenon that describes the emission of electrons from a metal surface when light
shines on it is the
.
1.6
The maximum energy of photoelectrons a.
with frequency and b.
with increasing intensity of the light. (increases, decreases, remains constant)
1.7
Photon energy is proportional to
1.8
Photons are also known as
✍
1.9
.
.
Solve these problems.
A yellow lamp emits light with a wavelength of 6
•
10-7 m.
a. Write the common value for the speed of light, in meters per second.
4
b. Calculate the frequency of the yellow light from the equation given in Science LIFEPAC 1204.
c. Use Planck’s equation to calculate the energy of a single photon with the given wavelength.
d. How many such photons are required to produce 10 joules?
1.10
A 10,000-watt radio station transmits at 880 kHz.
a. Determine the number of joules transmitted per second.
b. Calculate the energy of a single photon at the transmitted frequency.
c. Calculate the number of photons that are emitted per second.
1.11
Light with a wavelength of 5 • 10-7 m strikes a surface that requires 2 ev to eject an electron.
a. Calculate the frequency of the given wave.
b. Calculate the energy, in joules, of one incident photon at this frequency.
c. Calculate the energy, in electron volts, of one incident photon.
5
d. Calculate the maximum KE, in electron volts, of the emitted photoelectron.
1.12
Photoelectrons with a maximum speed of 7 • 105 m/sec are ejected from a surface in the presence
of light with a frequency of 8 • 1014 Hz.
a. If the mass of an electron is 9.1 • 10-31 kg, calculate the maximum kinetic energy of a single
electron, in joules.
b. Calculate, in joules, the energy of incident photons.
c. Calculate the energy required to eject photoelectrons from the surface.
X-rays. When photons of light were found to
give up their energy to electrons, the question was
asked: Can the energy of moving electrons be
converted into photons? Further investigation
disclosed that not only could electrons produce
photons, but that several years previously the
German physicist Roentgen had in fact produced
the effect. Roentgen had conducted experiments in
which he allowed high speed electrons to strike the
surface of metal. He had observed that the metal
target emitted some very penetrating radiation.
These Roentgen rays, or X-rays, as they are now
called, exposed photographic plates. Roentgen was
certain that they were not particles; they were
affected by neither electric nor magnetic fields. He
also found that as the energy of the electrons
increased, the X-rays became more penetrating.
The density of emitted X-rays increased when
the number of electrons was greater. In addition
Roentgen established that X-rays could exhibit
both interference and polarization effects. He
believed these phenomena were evidence that the
emissions were electromagnetic radiations.
TARGET ANODE
FILAMENT
ee-
V
X-RAYS
When a charged particle accelerates, it emits
electromagnetic radiation. X-rays are identified as
electromagnetic radiation emitted when a very fast
electron is slowed suddenly as it strikes the metal.
Most of the incident electrons probably lose their
energy too slowly for X-rays to be emitted. Some of
the electrons, however, lose much or all of their
energy in a single collision with an atom in the
target metal. This radiation is the inverse of the
photoelectric effect. In fact, the highest frequency
found in X-rays corresponds to the photon energy,
6
and all frequencies below the maximum X-ray
frequency also exist.
When the electron energy is above a certain
value, the emission curves possess strange, sharp
spikes. The spikes indicate that some frequencies
are emitted preferentially. These emissions
originate from rearrangement of the electron
structure of the target atom.
30,000 v
X-RAY
INTENSITY
15,000 v
5,000 v
1
2
3
4
5
FREQUENCY x 1018 cycles/sec
6
AN X-RAY SPECTRUM
✍
Complete these sentences.
1.13
When the number of electrons striking the anode of an X-ray tube increases, the
of the emitted X-rays increases.
1.14
When the speed of electrons striking the anode of an X-ray tube increases, the
of the emitted X-rays increases.
✍
1.15
Complete these activities.
An X-ray tube emits X-rays with a wavelength of 10-11 m.
a. Calculate the photon energy, in joules, that the emitted X-rays possess.
b. Calculate the energy, in electron volts, that the X-rays possess.
c. Determine the energy, in electron volts, possessed by the incident electrons.
d. Calculate the potential that must be applied across the X-ray tube to give each incident
electron its energy.
7
1.16
Calculate the highest frequency X-rays produced by 8
1.17
A television tube can accelerate electrons to 2
X-rays with the highest energy.
1.18
Calculate the energy, in electron volts, of X-rays that have a frequency of 1019 Hz.
•
•
104 ev electrons.
104 ev. Calculate the wavelength of emitted
MATTER WAVES
and the relationship
wavelength, and velocity,
X-rays are used to study the configuration of
atoms in crystals. Atoms in a crystal interact with
X-rays, causing the scattered waves to interfere
constructively in some regions and destructively in
others. The interference pattern thus produced is
characteristic of the spatial arrangement of the
atoms in the crystal. The exact pattern is
determined by the wavelength of the X-rays and
the distance between the atoms.
If, instead of X-rays, electrons are used, a
pattern is produced almost exactly the same as the
X-ray pattern. This experiment is known as the
Davisson-Germer experiment. The implication of
this experiment is that electrons behave as waves.
X-rays and their diffraction pattern are used to
measure the distance between atoms in crystals.
frequency,
λv = c
Substituting the second equation into the first gives
momentum =
/c = h/λ
hv
Solving this relationship for the wavelength gives
λ=
/momentum
h
for a photon. The new idea that de Broglie
suggested was that this formula is much more
general and that it applies to particles as well as to
waves. The expression for the momentum of a
particle is
De Broglie waves. By using crystals with
known atomic spacing, the experiment can be used
to determine the wavelength of the electron. These
wavelengths are called de Broglie wavelengths; and
the waves themselves are de Broglie waves, or
matter waves. The wavelength of the de Broglie
wave is inversely proportional to the momentum of
the particle (generally an electron) that is being
observed. De Broglie took the formula for the
momentum of a photon,
momentum =
between
momentum = mv.
Substituting this equation into the former
relationship gives the expression for the
wavelength of a particle:
λ=
/mv
h
Again the relationship between the wavelength and
the momentum involves the quantity h, Planck’s
constant.
/c
hv
8
The value of h is extremely small, which
explains why waves are not observed for large
objects. The momentum must be small compared to
h for the wavelength to be appreciable. Particles at
times have the properties of a wave; at other times
they exhibit the properties of a particle. As in the
case of electromagnetic radiation, the particle is
never observed to exhibit both particle
characteristics and wave characteristics in the
same interaction; and we cannot determine which
description is the correct one. This phenomenon is
sometimes called the wave particle duality, or just
duality. The reason we cannot observe this
property of duality is because the wavelengths are
so small. For instance, the wavelength of an
automobile traveling 60 miles an hour is on the
order of 10-38 feet. This wavelength is much much
smaller than the automobile. On the other hand,
the wavelength of an electron traveling about 10
million meters per second is about 10-8 meters. This
very small wavelength is still the magnitude of
atomic dimensions. Since the wave length is
comparable to the size of the electron, the
wavelength is observable.
ELECTRON
e-
N
IO
CT
E
R
DI
VE
A
W
PHOTON
INITIAL MOMENTUM
ELECTRON
OBSERVER
e-
PHOTON
FINAL MOMENTUM
OBSERVER
Even the position is not known with absolute
accuracy since, even with the best of instruments,
the position will be off by at least part of a
wavelength. Thus,
Uncertainty principle. One of the tenets of
quantum physics is an inherent limitation on our
capability for making observations. To measure
something, some way is needed to carry
information from the object to the observer. You
must be able to touch it, shine a light on it, or
perform some other manipulation for information
about the object’s size or its position to be carried
back to the observer. Observing a large object is
fairly easy. For instance, you can measure a table
with a ruler. When the object being observed is an
electron, the problem is different. About the only
way that information about the electron’s size or its
position can be gathered is to shine light on the
electron and to reflect it to the observer. The
difficulty is that the light, to reflect from the
electron, must strike the electron. In striking the
electron, some of the light energy is transferred to
the electron. The reflected light enables the
position of the electron to be determined precisely:
Project back along the path of light and determine
where the electron was when struck by the light.
However, when the electron was struck by the
photon, the energy given up by the light changed
the direction of the electron. The very act of
observing the electron changes its momentum.
Although you now know where the electron was at
that instant, you do not know in exactly what
direction it is now going.
∆x ⬇ λ
is a reasonable estimate of the uncertainty about
the position of the particle. The photon had a
momentum:
momentum = h/λ .
Although you cannot know exactly how much the
momentum of the electron will change, it is
reasonably the same magnitude as the photon
momentum; thus,
∆mv ⬇ h/λ .
Combining the two equations of uncertainty gives
the result:
∆x
•
∆mv ⬇ h.
The uncertainty in position times the uncertainty
in momentum is approximately equal to Planck’s
constant, which means that we can never hope to
measure
both
position
and
momentum
9
simultaneously with perfect accuracy. In 1927 the
German physicist Heisenberg first proposed this
result, which is usually referred to as the
Heisenberg uncertainty principle.
Uncertainty does not result from deficiencies in
measuring instruments; it is a constraint of nature.
Remember again that Planck’s constant is very
small; matter waves have significance for atomic
particles, but not for large, “Newtonian” objects.
the relationships can be verified with absolute
accuracy. Are these relationships, then, not true?
Unfortunately, we cannot make as positive a
statement as we might wish. Individual particles
may not behave according to the Newtonian laws of
classical physics; but for statistically large groups of
particles, these expressions become statements of
probability. Many many small particles obey
classical laws of physics on the average, but
individual particles may deviate from this average
behavior. Fortunately, everything that we
physically encounter is made up of such large
numbers of particles that their average behavior is
all that is observed. Causality is reasonably reliable
for macroscopic events, but we cannot expect it to
hold true in situations that we cannot experience
directly.
Causality. Previously we assumed that the
laws of physics were cause-and-effect relationships:
that a given cause always produces a specific effect.
The equations studied in mechanics give
relationships; generally, that force is proportional to
something else. The uncertainty principle, however,
shows that all the factors that go into an equation
cannot be measured with total precision. None of
✍
Complete these sentences.
divided
1.19
The wavelength of a moving particle is equal to a.
.
by b.
1.20
The uncertainty principle states that the product of uncertainty in the momentum and in the
.
position of a particle is approximately equal to
✍
Complete these activities
1.21
Calculate the de Broglie wavelength of a 5,000 kg truck traveling at 80 kph. (Hint: convert
everything to fundamental units.)
1.22
Calculate the de Broglie wavelength of an electron traveling at 107 m/sec (me = 9.1
1.23
Calculate the approximate momentum change in a particle of mass 1.7
initially at rest, whose position is located to within 10-4 m.
1.24
Calculate the uncertainty of the velocity of a particle confined to a space of 10-9 m if the
particle is:
a. an electron
b. a proton
10
•
•
10-31 kg).
10-27 kg (a proton),
ATOMIC MODELS
By the year 1900 the field of chemistry had
accumulated a large amount of evidence that
material consisted of atoms. Very little evidence
existed, however, about the nature of the atoms
themselves. J.J. Thomson had recently discovered the
electron; this discovery suggested that electrical
forces were of significant strength on the atomic scale.
of Thomson’s model. This experiment was conducted
by two men named Geiger and Marsden. They placed
in a lead box material that emits alpha particles.
(Alpha particles will be discussed in detail later. All
we have to know about them now is that they are
small, positively charged radiations emitted by some
natural substances such as radium.) A small hole
drilled in the box allowed a stream of alpha particles
to escape. This beam of alpha particles was directed
at a very thin metal foil. Geiger and Marsden found
that, just as the Thomson model predicted, most of
the alpha particles passed through the foil or were
deflected through relatively small angles. The
particles were observed as they struck a zinc
sulphide screen surrounding the foil. When alpha
particles struck the screen a small amount of light
was emitted.
Geiger and Marsden made one interesting
observation: Most of the alpha particles passed
through the screen, but a few were scattered through
rather large angles. In fact, some were scattered back
nearly 180°. Alpha particles are much larger than
electrons, nearly 8,000 times greater in mass; only
very strong forces could change the direction of the
alpha particles so drastically.
Thomson model. Thomson proposed what is
referred to as the Thomson model, sometimes called
the plum pudding model, of the atom. He suggested
that atoms were spheres of positively charged matter
and the electrons were embedded in this matter
much as plums are embedded in plum pudding. This
model was quite reasonable using the evidence
available at the beginning of this century. It provided
an explanation for most of the atomic behavior that
had been observed. If this model were the nature of
atoms, then one would expect that charged particles
as they moved through an atom would not be
appreciably affected. Electric fields in atoms would be
quite weak, because the positive charge in an atom
would be diffused by negative “plums,” and vice versa.
Shortly afterwards, however, another experiment
was done that caused some concern as to the validity
ZINC-SULFIDE SCREEN
ALPHA EMITTER
LIGHT FLASHES
LEAD BOX
THIN METAL FOIL
THE GEIGER-MARSDEN EXPERIMENTS
Rutherford model. To explain this large
angle scattering, the British physicist Ernest
Rutherford proposed a new model of the atom. His
theory was that the positive charge was not
distributed throughout the entire atom, but was
concentrated in a very very small portion near the
center. The atom was mostly empty space. The
negative electrons moved around in this empty
space; but they were very small, very light, and
could not have an appreciable effect upon the
heavy alpha particle. The center of the atom, the
nucleus, was very small and would effect only those
alpha particles that came very near; most of the
alpha particles traveled through the empty space.
Alpha particles that approached a nucleus,
however, would experience strong electric forces
11
and would be scattered through a large angle. If an
alpha particle were to strike an electron, it would
knock the electron out of the way; the nucleus,
however, being much heavier than the alpha
particle, caused the alpha particle to change its
✍
1.25
direction. Using this idea and the rules of
electrostatics, Rutherford was able to derive a
formula that explained the backward scattering of
alpha particles. As a result, Rutherford is credited
with the discovery of the dense atomic nucleus.
Complete this activity.
Using the description of the Thomson and Rutherford models of the atom, draw a labeled
diagram of each model that illustrates the differences between the two.
THOMSON MODEL
✍
1.26
RUTHERFORD MODEL
Prepare a report.
Some of the men involved in the development of early atomic physics worked in the same
laboratory. Look up the following men: Ernest Rutherford, J. J. Thomson, and Neils Bohr.
Write a five-page, double-spaced, typewritten report on the men and their contributions to
atomic theory. Submit your report for evaluation.
Score
Adult check
______________________
Initial
Date
ATOMIC SPECTRA
A spectrum is an array of the different
wavelengths of radiant energy emitted by a source.
Different sources produce different wavelengths
and exhibit one of the four different kinds of
spectra:
1.
2.
3.
4.
Emission spectra. Heating a block of metal
produces no noticeable change until its
temperature reaches about 1000°K, at which point
the block begins to glow a dull red. As heating
continues, the glow becomes bright red and then
orange. When the temperature of the block is about
1500°K, the spectrum contains red, orange, and
yellow. At 2,000°K the block appears yellow, and
the spectrum now includes green. When the
temperature reaches 3,000°K, the block glows
white hot; and all the colors from red through
violet are present in the spectrum along with
invisible infrared and ultraviolet radiations.
Continuous emission spectra,
Line emission spectra,
Continuous absorption spectra, or
Line absorption spectra.
12
INTENSITY
VISIBLE
ULTRAVIOLET
INFRARED
WAVELENGTH
RADIATION EMITTED
BY A
HEATED SOLID
When hot gas is the source of the radiation, quite a different spectrum results. Instead of a
continuous spectrum, a number of bright lines constitute the spectrum.
SLIT
HOT GAS
PRISM
LENS
ARRANGEMENT
LENS
TO
DEMONSTRATE
The most intense line spectra are obtained
from metallic arc lamps, such as a carbon arc, when
one of the carbon rods has been impregnated with
a metal like mercury or sodium. Continuous
spectra come from high temperature solids but line
spectra always come from an incandescent gas.
A
LINE EMISSION SPECTRUM
SCREEN
wavelengths will be missing. When light from an
incandescent bulb (hot solid filament) passes
through colored glass, the glass absorbs all
wavelengths except the color of the glass.
When a continuous-emission spectrum is
passed through gas below its temperature of
incandescence, certain discrete wavelengths are
removed. The resulting absorption spectrum has
dark lines in the positions of the missing
wavelengths.
Absorption spectra. When a continuous
emission spectrum passes through a solid and
the radiation is then analyzed, a wide band of
SLIT
SODIUM GAS
HOT GAS
LENS
LENS
LENS
PRISM
SCREEN
SODIUM
GAS BURNER
ARRANGEMENT
TO
DEMONSTRATE LINE ABSORPTION SPECTRUM
13
When the sodium metal is heated, the chamber
is filled with relatively cool sodium vapor; the
spectrum on the screen displays two dark lines in
the yellow region. Alkali metals are about the only
substances that produce lines in the visible region
of the spectrum, but practically all substances
produce lines in the ultraviolet.
✍
1.27
Define these terms.
spectrum
1.28
emission spectrum
1.29
absorption spectrum
1.30
continuous spectrum
1.31
line spectrum
Again we see that solids produce a continuous
spectrum and that gases produce a line
spectrum. The reason for the linear display is the
slit through which the spectrum is focused. If the
light is focused instead through a small hole, small
disks would replace the line.
BOHR MODEL
A new era in physics began in 1913 when Niels
Bohr proposed a new structure of the hydrogen
atom. This theory gave a satisfactory explanation
for the spectrum emitted by hydrogen and was
later successfully extended to other elements.
v
e+
e-
Hydrogen model. Hydrogen is the simplest
of all atoms; its nucleus is a single proton. Bohr
proposed a model with the following properties:
Postulate One:
Fe
a. The hydrogen atom contains one electron in
a circular orbit.
b. The hydrogen nucleus is assumed to be at
rest.
c. The centripetal force on the electron is
equal to the electrostatic force between a
unit positive charge and a unit negative
charge.
BOHR’S HYDROGEN MODEL
From Coulomb’s law,
Fe = k
/
ee 2
r
and from the centripetal force,
Fc =
14
mv2
/r
Here n is the principal quantum number and is
allowed to have only whole number values, 1, 2, 3,
4, and so on; the orbit sizes and atomic energy
levels are thereby fixed. Bohr calculated the value
of k to be h/2π. Yes, Planck’s constant again.
Now two equations describe the allowed orbits:
Since the centripetal force is an electrostatic force,
mv2
/r = k ee/r2
Up to this point the physics has been classical
physics, but here Bohr introduced his quantum
hypothesis.
mv2
2
/r = k e /r2 and mvr = nh/2π
Postulate Two:
a. The electron is forbidden all but a few
orbits around the nucleus.
b. The orbits that are allowed are definite
and discrete.
c. For an orbit to be valid, the preceding
equation must be satisfied, and a stringent
constraint must be placed on the angular
momentum of the electron: its angular
momentum must equal an integer times a
small constant:
m
v
r
k
e
n
h
π
angular momentum mvr = nk.
✍
1.32
is the electron mass (9.1 • 10-31 kg),
is the orbital velocity,
is the orbital radius,
is the electrostatic constant from Coulomb’s
2
law (9 • 109 N•m /coul2),
is the size of the charge on an electron or a
proton (1.6 • 10-19 coul),
is the principal quantum number,
is Planck’s constant (6.63 • 10-34 joule•sec.), and
is taken to be 3.14.
Complete this activity.
Solve these two equations for v, equate the results, and solve for r, the radius of a discrete orbit.
mv2
2
/r = ke /r2
Score
Adult check
mvr =
/2π
nh
______________________
Initial
Date
velocity of light. The high velocity and small orbit
results in a very large number of revolutions per
second, about 1015. This frequency is the same order
of magnitude as the frequency of visible light.
A consequence of Bohr’s work up to this point is
that the single electron in the hydrogen atom has
several orbits that it can occupy. If the electron
moves from one orbit it must go directly to another
of the allowed orbits. Each orbit must be an
integral multiple of the electron’s wavelength in
circumference.
Solving this equation for the first orbit (n=1)
gives a radius of 0.000,000,000,053 m.
A more convenient unit to use for measuring
atomic distances is the angstrom:
1 meter = 1010 angstroms
1 A = 10-10 m
Thus, we have a radius for the innermost orbit of
0.53 A and a diameter of 1.06 A. If this value is put
into the angular momentum equation, the velocity
of the electron is calculated to be about 1/137 the
15
If a gas is excited with an electrical discharge,
many electrons will go to higher orbits. These
excited atoms do not stay excited for long. The
electron is attracted to the nucleus and will soon
jump to a lower orbit. It may go all the way to the
lowest orbit, the ground state, in one jump; or it
may make several jumps on its way down. The
photons emitted in this process cause the gas to
glow in discrete wavelengths that can be discerned
with the use of a prism.
n=5
n=4
n=3
2
IONIZATION
n
n
n
n
=
=
=
=
∞
5
4
3
n=2
ev
0.0
-0.54
-0.85
-1.5
-3.4
The final assumption that Bohr made in his
model concerned these orbit changes.
Postulate Three:
a. No radiation is emitted by an electron while
it is in one of its allowed orbits.
b. Electromagnetic radiation is emitted only
when an electron jumps from one orbit to
another.
c. The frequency of this radiation is
determined by the energy difference
between the two orbits.
GROUND STATE
n=1
ENERGY LEVELS
FOR
-13.6
HYDROGEN
The success of Bohr’s theory is not that it gives
us the “right answer,” for we now know that the
theory is much too simplified. The model gave rise
to an equation that agrees exactly with
experimental results, and this equation is the
significant contribution.
hv = E1 - E2
E1 is the energy of the electron in the initial orbit
and E2 is its energy in the final orbit. Of course, hv
is the energy of the radiated photons.
Quantum numbers. A man named Stoner
worked with Bohr to extend the orbital model of
the hydrogen atom to other elements. The helium
atom has an atomic number Z = 2: Its nucleus
contains two protons, which are electrically
balanced by two orbiting electrons. Lithium, Z = 3,
has three protons and three electrons. The allowed
orbits are called shells and are the orbits for the
quantum number n = 1, 2, 3 and so on. To explain
all the experimental data, Bohr and Stoner had to
limit the number of electrons in each shell. The
maximum number of electrons per shell is 2n2.
Atomic excitation. Obviously, radiation is
emitted only when an electron goes from the higher
orbit to the lower energy orbit. As with a
gravitational system the electron loses potential
energy as it falls into inner orbits; hence the
photons are emitted as the electrons move to orbits
represented by smaller quantum numbers.
In a sample of hydrogen gas, most of the
electrons are in the n = 1 orbit. Some of the
electrons, however, are knocked into higher orbits
by collisions with other atoms or with other
electrons. As the temperature of the hydrogen
increases, this excitation is more likely to happen.
16
Quantum Number n
Maximum Number of Electrons
1
2
•
12 = 2
2
2
•
22 = 8
3
2
•
32 = 18
4
2
•
42 = 32
5
2
•
52 = 50
6
2
•
62 = 72
The shells are generally filled in order as Z
increases, but exceptions occur for some of the
heavier elements. For some reason the number of
electrons that fill a shell is a “preferred number.”
For example, mercury with Z = 80 has the following
configuration:
n = 1 shell
2 electrons
n = 2 shell
8 electrons
n = 3 shell
18 electrons
n = 4 shell
32 electrons
n = 5 shell
18 electrons
n = 6 shell
2 electrons
electrons enter the next shell before the incomplete
shell (n = 5 in this case) finishes filling.
The equation for orbit size is valid only for
hydrogen. As Z gets larger, the electrons are
attracted more strongly; consequently, the orbits
get smaller. Therefore, all other atoms are smaller
than the hydrogen model would predict. For
instance, the helium (Z = 2) atom is only about 70
percent the size of the hydrogen atom. Lithium (Z
= 3) is slightly larger than hydrogen; but since only
two electrons can be in the n = 1 shell, one must be
in the n = 2 shell. Beryllium (Z = 4) is again smaller
than hydrogen. In fact, lithium (Z = 3), sodium (Z =
11) and mercury (Z = 80) are all about the same
size. As a result the heaviest elements are not
much larger in diameter than the lightest
elements.
The n = 5 shell stops filling at eighteen, the
number for a filled n = 3 shell, and two electrons go
to the n = 6 shell even though the n = 5 shell has
room for thirty-two more electrons. Not many
✍
Complete these activities.
1.33
Calculate the orbital radii of the hydrogen atom for the first six principal quantum numbers.
1.34
Assume that the radius of the hydrogen nucleus is 1.4 • 10-15 meters. How much larger than
the nucleus is the entire hydrogen atom? (Calculate the atomic radius for n = 1.)
1.35
Assume that the radius of the sun is 865,000 miles. According to the atomic scale of
distances for hydrogen, considering the earth to be an orbiting electron, calculate the earth’s
orbital radius.
17
1.36
An atom that has six principal quantum numbers can yield several emission lines. An excited
electron that jumps from the n = 1 to the n = 2 can only drop back to the n = 1. But, an excited
electron that jumps from the n = 1 to the n = 6 can jump back 1, 2, 3, 4, or 5 orbits to drop
back to the n = 1.
a.
Write the number of possible electron shell transitions for which energy is radiated.
(Hint: if an electron has jumped from n = 1 to n = 6, it can have these transitions: n = 6
to n = 5, n = 6 to n = 4, n = 6 to n = 3, n = 6 to n = 2, n = 6 to n = 1. That's 5 of them – can
you find the rest? Remember, the electron didn't have to jump all the way to the n=6 level
to start out with.)
b.
Use the Energy Levels for Hydrogen Chart to calculate the wavelength corresponding to
each electron transition. Note that energy is given in electron volts.
TRANSITION
ENERGY
18
IN EV’S
EMITTED WAVELENGTHS
✍
Answer these questions.
1.37
What is a quantum number?
1.38
What are the Bohr postulates?
a.
b.
c.
1.39
Why does a neon sign light up?
1.40
What forces must be balanced in the Bohr model?
✍
Solve these problems.
1.41
Calculate the speed of an electron in the innermost orbit of a hydrogen atom.
1.42
If an electron spends 10-8 seconds in the n = 2 orbit, calculate how many revolutions it makes
around the nucleus.
Review the material in this section in preparation for the Self Test. This Self Test will check
your mastery of this particular section. The items missed on this Self Test will indicate specific
areas where restudy is needed for mastery.
19
SELF TEST 1
Match these items (each answer, 2 points).
1.01
small unit of energy
a. electron volt
1.02
lowest energy level
b. de Broglie wave
1.03
characterizes an electron orbit
c. ground state
1.04
smallest amount of light
d. uncertainty principle
1.05
wave associated with moving matter
e. quantum number
1.06
statement about limit of position measurement
f. continuous spectrum
1.07
light from a heated metal
g. photon
1.08
light from a heated gas
h. line spectrum
1.09
radiation from decelerated electrons
i. wave theory
1.010
6.63
j. X-ray
•
10
-34
joule
•
sec.
k. Planck’s constant
Complete these sentences (each answer, 3 points).
1.011
When a photon collides with an electron and is deflected, the photon’s
1.012
According to Einstein’s interpretation, the maximum photoelectron energy is equal to the
energy of the incident photon minus the energy required to
.
1.013
When the number of electrons striking the anode of an X-ray tube is increased, the
of the emitted X-rays increases.
1.014
When the speed of the electrons striking the anode is increased, the
the emitted X-rays increases.
1.015
The lowest energy level in an atom is called its
1.016
The energy of a photon is equal to a.
1.017
The wavelength of a moving particle is equal to a.
by b.
.
1.018
De Broglie’s hypothesis was verified when electrons scattered by certain crystals were
observed to experience the wave phenomenon called
.
1.019
The orbit of an electron in a hydrogen atom is always an integral number of
in circumference.
decreases.
of
state.
times b.
.
divided
Choose the correct answer (each answer, 2 points).
1.020
In the Rutherford model of the atom, the positive charge in an atom is
a.
concentrated at its center
b.
spread uniformly throughout its volume
c.
in the form of positive electrons at some distance from its center
d. readily deflected by an incident alpha particle
1.021
Modern physics theories indicate that
.
a. all particles exhibit wave behavior
b. only moving particles exhibit wave behavior
c. only charged particles exhibit wave behavior
d. only uncharged Particles exhibit wave behavior
20
.
1.022
1.023
1.024
1.025
Photoelectrons are emitted by a metal surface only when the light shining on the surface
exceeds a certain minimum
.
a. wavelength
c. velocity
b. frequency
d. charge
.
The transition in a hydrogen atom that absorbs the photon of highest frequency is
a. n = 1 to n = 2
c. n = 2 to n = 6
b. n = 2 to n = 1
d. n = 6 to n = 2
When light shines on a metal surface, the maximum energies of emitted electrons
a. vary with the intensity of the light
c. vary with the speed of the light
b. vary with the frequency of the light
d. are random
.
An electron can rotate around an atomic nucleus indefinitely without radiating energy if its
.
orbit
a. is a perfect circle
b. is sufficiently far from the nucleus
c. is less than a de Broglie wavelength in circumference
d. contains an integral number of de Broglie wavelengths
1.026
1.027
A neon sign does not produce
.
a. a line spectrum
c. an absorption spectrum
b. an emission spectrum
d. photons
The de Broglie wavelength of a particle is
.
a. proportional to its momentum
c. inversely proportional to its momentum
b. proportional to its energy
d. inversely proportional to its energy
Complete these items (each answer, 5 points).
1.028
If the electron of a hydrogen atom is in the ninth circular orbit of diameter 85.9 A, the
frequency of revolution is
.
1.029
Make a diagram of an arsenic atom (Z = 33) according to the Bohr-Stoner scheme of atomic
structure. Show orbits with numbers indicating electrons in each.
Score
Adult check
63
79
______________________
Initial
21
Date