The Elementary Mathematics of the Atom 1114578339, 9781114578333
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MATHEMATICS OF THE ATOM
IRVING ADLER ^
uthor of ’
%
*
yr
THE NEW MATHEMATICS, INSIDE THE NUCLEUS,
/
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"
Diagrams by Ruth Adler
^
V
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etc
)
TEMOTA $ 4.50
THE ELEMENTARY MATHEMATICS OF THE ATOM By IRVING
ADLER
Author of Inside the Nucleus,
The New Mathematics,
etc.
with diagrams by ruth adler
Man’s knowledge of the atom, ture,
structure,
its
its
behavior
its
na-
— is
ex-
pressed in the language of mathematics.
He who wants
truly to understand the
atom, the unlocking of whose secrets has
opened the way
forward
in the history of
must approach It
for the greatest leaps
it
technology,
through mathematics.
has been generally assumed that the
mathematics needed But
in this
is
highly advanced.
book Irving Adler shows
that
considerable knowledge of the atom can
be understood by any interested person
who
has had one year of high school
algebra.
Thus
it is
a
• Students
book
for
of high school physics
and
of the usual one-year introductory col-
(Continued on hack flap
1.
ts
mmf’i
honofits «ia UioHcy. material Sale of this
Digitized by the Internet Archive in
2017 with funding from
Kahle/Austin Foundation
https://archive.org/details/elementarymathemOOadle
The Elementary Mathematics of the Man’s knowledge ture, its
ematics.
behavior
—
is
of the
Atom atom
—
its
nature,
its
struc-
expressed in the language of math-
He who wants
truly to understand the atom,
the unlocking of whose secrets has opened the
way
for
the greatest leaps forward in the history of technology,
must approach it through mathematics. It has been generally assumed that the mathematics needed is highly advanced. But in this book Irving Adler shows that considerable knowledge of the atom can be understood by any interested person who has had one year of high school algebra.
Without resort to calculus, it achieves the following: Develops in some detail the molecular theory of matter and the periodic table of the elements (Chapter II). Assuming some well known relationships of motion, electricity, and light, derives others by elementary methods (Chapters III, IV, and V). Using these relationships, develops in detail the Bohr model of the atom with circular electronic orbits (Chapter VI).
After describing the basic concepts of the quantum-
mechanical model of the atom, shows how
its
principal
conclusions explain the periodic table of the elements
(Chapter VII).
*
«
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t
'V iMf
I
1
THE ELEMENTARY MATHEMATICS OF THE ATOM
Books by Irving Adler
COLOR IN YOUR LIFE DUST THE ELEMENTARY MATHEMATICS OF THE ATOM FIRE IN YOUR LIFE
HOT AND COLD HOW LIFE BEGAN INSIDE THE NUCLEUS MAGIC HOUSE OF NUMBERS MAN-MADE MOONS
MONKEY BUSINESS: Hoaxes in the Name of Science A NEW LOOK AT ARITHMETIC THE NEW MATHEMATICS PROBABILITY AND STATISTICS FOR EVERYMAN THE SECRET OF LIGHT SEEING THE EARTH FROM SPACE THE STARS: Stepping Stones into Space THE SUN AND ITS FAMILY THINKING MACHINES TIME IN YOUR LIFE TOOLS IN YOUR LIFE THE TOOLS OF SCIENCE WEATHER IN YOUR LIFE WHAT WE WANT OF OUR SCHOOLS THE REASON WHY BOOKS {with Ruth
Adler)
IRVING ADLER
THE ELEMENTARY MATHEMATICS OF THE ATOM With diagrams by Ruth Adler
The John Day Company
New
York
©
1965 by Irving Adler
must not be reproduced in any form without permission. Published by The John Day Company, Inc., 62 West 46th Street, New York 36, New York, and simultaneously in Canada by Longmans Canada Limited, Toronto. All rights reserved. This book, or parts thereof,
qL/73 Library of Congress Catalogue Card
MANUFACTURED
IN
Number; 64-20699
THE UNITED STATES OF AMERICA
Contents
I.
II.
III.
Mathematics and the Atom
The Atom
in
Chemistry
Motion, Electricity and Light
IV. Electricity in the
V.
VI. '
VII.
Atoms
Atom
of Light
9 15 51
75
95
The Hydrogen Atom
103
The Electron Gets a Permanent Wave
I2I
Index
143
1
j
}
•
.
«
i i
I
I
THE ELEMENTARY MATHEMATICS OF THE ATOM
>
•i J
4
Matheviatics and the Known
Unseen, but Well
NOBODY we have a know that
large
Atmn
has ever seen an atom. Nevertheless
amount
We
knowledge about atoms.
of
there are about one hundred different kinds of
atoms, and that they are the building blocks out of which all
chemical substances are made.
outward behavior,
in
We
can describe their
which they interact with each other
form molecules, or with particles of light that they
to
send out or receive. ture, built
We can also describe their inner struc-
out of a central core called the nucleus, and
planetary electrons that surround the nucleus. explain, too,
by
how
their
outward behavior
their inner structure. All this
in the
may
knowledge
is
is
We
can
determined formulated
language of mathematics.
Mathematics Enters the Picture In the study of the atom, as in any science, mathe-
matics plays a part in three ways.
mathematics
in
a science
is
The
first
role
of
purely descriptive. In the
course of his experiments a scientist measures certain quantities and observes relationships
among them. He 9
then describes the observed relationships by means of an appropriate
mathematical equation or inequality. As
examples we
two equations that we
cite
casion to use in Chapter
have oc-
shall
II.
Temperature measurements
a scientific laboratory
in
are ordinarily expressed in terms of the Celsius scale, on
which the freezing point of water point of water
The study
100°.
is
is
0° and the boiling
of the behavior of heat
engines led to the introduction of another scale as the Kelvin scale.
point of water
The
the Kelvin scale the freezing
273°, and 0°
is
may
temperature that attained.
On
known
is
the absolute
minimum
be approached but can never be
relationship between a Celsius temperature
Tc and the equivalent Kelvin temperature Tk
is
given
by the equation
Tk = Tc
(1)
When
a sample of gas
can measure
its
volume,
its
is
+
273.
enclosed in a container w^e
pressure,
and
its
temperature.
we multiply the number of units in the volume by the number of units in the pressure, and then divide the product by the number of degrees Kelvin in the temperature, a definite number is obtained. Now suppose that If
the sample of gas
is
compressed or expanded or heated
or cooled, so that
its
volume, pressure and temperature
same computation is made with the new volume, pressure, and Kelvin temperature, the numare altered. If the
ber obtained tained before,
is if
found to be about equal to the one obthe pressure in both cases
is
low. This
observation shows that the volume, pressure, and temperature of a given sample of gas are not entirely inde-
pendent, but are related to each other. 10
If
P
is
the pres-
sure,
the relationship
K
is
a fixed
the Kelvin temperature,
is
expressed by the approximate equation
is
^=
(2)
where
T
the volume, and
T" is
K,
.
PV = KT
or
number.
The second role of mathematics in science is predictive. By using mathematical techniques to explore the logical implications of known relationships, we can obtain new relationships from old ones, and we can compute quantities
them
instead of measuring
For example,
directly.
suppose the original values of the pressure, volume and Kelvin temperature of a gas sample are Pi, Fi, and T\
and suppose that altered values for the same sample are P 2 TT, and T 2 respectively. Then, by equa-
respectively,
,
tion (2),
we have PiV,
ing these two equations,
we get
PiFi P2V2
..X ^
= KT,,
^
and
By
2.
Divid-
T, T2
V and T
to the old
we know the volume of given temperature and pressure, we can com-
using equation (3),
the gas at a
= KT
the proportion
which relates the new values of P, ones.
P2 F 2
if
pute what the volume would be at another temperature
and pressure.
We
shall use
it
for this
purpose on page 36.
Mathematics Makes the Picture
The It
third role of
assumes
this
mathematics
role
when
in science is explanatory.
the
scientist
mathematical model of the phenomenon he
constructs is
a
studying.
The mathematical model is a set of assumptions that he makes for the purpose of explaining the observed rela11
The observed
tionships.
be explained
if
relationships are considered to
they are derivable as logical consequences
made
of the assumptions
in
the model.
If,
in addition,
other observable relationships that are implied by the
model are
then the model
verified,
substantially true picture of the
An
vestigated.
exami:)le of such a
theory of matter.
The
phenomenon being model is
called molecules; that each molecule
of a molecule
pose
it,
the
arranged
is
in-
the molecular
basic assumptions of this theory
are that every chemical substance
atoms held together by
considered to be a
is
made up
is
of units
an assemblage of
electrical forces; that the
nature
depends on the kind of atoms that com-
number
in the
of each kind,
and the way they are
molecule; and that
all
molecules of the
same substance have the same composition, while molecules of different substances have different compositions. As it is worked out in detail, the theory specifies the actual composition of each kind of molecule. It tells us, for
example, that a water molecule consists of two atoms of
hydrogen joined to one atom of oxygen, that an oxygen molecule
we breathe
in the air
consists of
two atoms of
oxygen, and so on.
Models
of the
Atom
In order to explain the behavior of atoms, physicists
have constructed structure.
The
several
first
different
of these that
models of atomic
was
fairly successful
was the Bohr model of the atom. In this model, Bohr assumed that each planetary electron in an atom revolves about the nucleus in a circular orbit. Sornmerfeld con12
structed a refinement of this model that was in better
by assuming that electron orbits are elliptical. The Bohr-Sommerfeld model has since been replaced by the quantum-mechanical model constructed by Heisenberg, Born, Jordan, DeBroglie, Schrddinger and agreement with the
facts
Dirac, in which the idea that an electron orbit
is
moves
in
an
discarded altogether.
The Scope
of this
Book
Both elementary and advanced mathematical techniques were used to construct the molecular theory of matter and the models of atomic structure and to deduce their implications. Fortunately, significant parts of the
theories and their consequences can be developed
by
ele-
book we examine in detail aspects of the theories that require no more knowledge of mathematics than one year of high school algebra.
mentary methods
alone. In this
Wherever we have to use a result that can be derived only by using advanced methods, we shall state the result without proof. In spite of this self-imposed limitation on the techniques
we
use,
we
shall
be able to accomplish the
some basic laws of chemistry, we shall develop in some detail the molecular theory of matter, and its crowning achievement in chemfollowing
istry,
goals;
1)
Relying
on
the periodic table of the elements. (Chapter II)
Assuming some well-known relationships of motion, electricity, and light, we shall derive some others by elementary methods. (Chapters III, IV and V) 3) Using these relationships, we shall develop in detail the Bohr model of the atom with circular electronic orbits. (Chap2)
13
ter
VI) 4) After describing
in
simple terms and relating
to a simple picture the basic concepts used in the
tum-mechanical model of the atom, we
shall
quan-
show how
its
principal conclusions serve to explain the periodic table of the elements.
14
(Chapter VII)
The Atom in Chemistry Atomic Theories, Old and
AS
New
long ago as 420 B.C., the Greek philosopher
Democritus of Abdera expounded the view that
all
matter
consists of combinations of small unit particles called
atoms. At that time this view was only an inspired guess.
was purely speculative, and not related to any particular body of fact. The modern atomic theory of matter, initiated by John Dalton in 1805, has a different character. It is based on a multitude of facts uncovered by the sciences of chemistry and physics, and it successfully exIt
plains these facts. In this chapter
we
outline the chain of
thought, based on the facts of chemistry, that led to the
formulation and elaboration of Dalton’s Atomic Theory.
Mixtures vs Pure Chemicals All matter
is
made up
of chemicals.
Some samples
of
matter are pure chemicals, each of which has uniform properties other.
by which
it
can be distinguished from every
Other samples of matter are mixtures of these pure
chemicals.
The
battery
a pure chemical. So are the oxygen admin-
is
distilled
water you add to an automobile
15
a hospital, and the carbon deposited
isterecl
from tanks
as soot
on a cold spoon held
other hand,
is
in
in a
candle flame. Air, on the
a mixture of the pure chemicals nitrogen
and oxygen with small amounts of carbon dioxide and water vapor, and traces of some others. The science of chemistry
is
chiefly concerned with the properties of pure
chemicals that are expressed in the relationships of the chemicals to each other.
Chemical Reactions
The
central fact of chemistry
is
that
some chemicals
can be changed into others in chemical reactions. In one
kind of reaction, two or more chemicals combine to form another. For example,
if
hydrogen gas
is
burned
in air,
the hydrogen combines with oxygen from the air to form
water vapor. In a second kind of reaction, the opposite process takes place, and a single chemical to
is
decomposed
produce two or more others. For example, water can
be decomposed by intense heat or by an electric current to produce tion
hydrogen and oxygen. Reactions of combina-
and decomposition show that some chemicals are
related to each other as parts
and wholes. Thus, hydro-
gen and oxygen are the parts out of which water
Water
is
is
made.
one possible whole that can be obtained by com-
bining hydrogen and oxygen in a certain proportion.
drogen peroxide, used as an antiseptic, ent whole
made from
different proportion.
A
the
is
another
same parts combined
Hy-
differ-
in
a
third type of reaction consists of
a reshuffling of parts whereby
some combinations are
broken up and new combinations are formed from their parts.
16
Compounds
vs Elements
While water can be decomposed into hydrogen and oxygen, neither hydrogen nor oxygen can be decomposed into other chemicals. Chemicals that can be decomposed are called compounds. Those, like hydrogen
and oxygen,
The elecompounds
that cannot be decomposed, are called elements.
ments are the chemical parts out of which all are made. There are hundreds of thousands of different chemical compounds. There are only about one hundred different elements from which all these compounds are made. Each of the elements cial
symbol, such as
carbon, etc.
A
list
H
for
is
usually designated by a spe-
hydrogen,
0
for oxygen,
C
for
of the chemical elements appears in the
table on page 48.
Five Fundamental
Laws
Near the end of the eighteenth century a decisive change was introduced in the way in which chemical reactions were studied. Whereas chemists in the past had observed which chemicals entered into a reaction and which chemicals emerged from
it,
they began then to
measure how much of each chemical was involved in the reaction. By weighing the chemicals they determined their masses, because the number of grams in the mass of a body
is
equal to the number of grams in
its
weight at
sea level. In the case of gases, they also measured their
When
became clear from the experiments of Boyle, Charles, and Gay-Lussac that the volume of a gas depends on its pressure and temperature, careful volumes.
it
17
measurements were made
An examination
ture, too.
that entered into and
of the quantities of chemicals
came out
significant regularities that
and tempera-
of the pressure
of reactions revealed
were
summed up
some
in five basic
laws of chemistry: I.
The is
The Law total
of Conservation of Mass. (Lavoisier, 1774)
mass
of the chemicals that enter into a reaction
equal to the total mass of those that are produced by
For example when
grams
gram
of hydrogen combines with 8
oxygen to produce water, the mass of the water
of
produced II.
1
it.
is
9 grams.
The Law
The elements
of Constant Proportions. (Proust, 1797)
that combine to form a given
are always the same,
and the masses
of the
compound
combining
ele-
ments always have a fixed ratio that is characteristic of the combination. For example, the only elements that can
combine
to
form water are hydrogen and oxygen, and the masses
ratio of their
is
always 1:8. That
is,
in the
forma-
of oxygen, 2
gram of hydrogen combines with 8 grams grams of hydrogen combine with 16 grams of
oxygen,
In general,
tion of water,
etc.
1
if
x grams of hydrogen combine y grams of water,
with y grams of oxygen to form x
X
then X and y satisfy
the proportion L
y
The Law A given amount III.
=
1
8
of Multiple Proportions. (Dalton, 1804) of
an element may combine with
differ-
ent masses of a second element to form different com-
pounds. But then the masses of the secoyid element that enter into such combinations are always whole number multiples of one particular mass. For example, 16 grams of oxygen can
18
combine with either
1
gram
of
hydrogen to
form hydrogen peroxide, or 2 grams of hydrogen to form water. Notice that the masses of hydrogen that may combine with 10 grams of oxygen are whole
number multiples
gram =1X1 gram; 2 grams = 2X1 gram.) Similarly, 3 grams of carbon can combine with either 4 grams of oxygen to form carbon monoxide, a of
1
gram.
(1
automobile exhaust fumes, or 8 grams of oxygen to form carbon dioxide, the gas that is dissolved deadly gas
in
in
soda water. Notice that the masses of oxygen that
may
combine with 3 grams of carbon are whole number multiples of 4 grams. (4 grams = 1 X 4 grams; 8 grams =
2X4 grams.) A
particularly impressive example
is
given
by the combinations that nitrogen may form with oxygen. Seven grams of nitrogen can combine with either 4 grams of oxygen to form nitrous oxide, or 8 grams of oxygen to form nitric oxide, or 12 grams of oxygen to form nitrous anhydride, or 16 grams of oxygen to form nitrogen dioxide, or 20 grams of oxygen to form nitric anhydride. Notice that the
masses of oxygen that
may combine
with 7
grams of nitrogen are whole number multiples of 4 grams. grams; 8 grams = 2X4 grams; 12 (4 grams grams = grams; 20 grams; 16 grams =
=1X4
3X4
grams
4X4
= 5X4 grams.)
IV.
The Law
of
Equivalent Proportions.
(Richter,
1791) Ij two elements each react with a third element,
and
masses of the two that react with a given mass of the third element will react also react with each other, the
with each other, or simple
masses of the two example,
1
gram
77iultiples or fractions of these
ele77ients will react
of
with each other. For
hydrogen and 35.5 grams of chlorine,
each of which are capable of reacting with 8 grams of
oxygen
to
form water and chlorine monoxide respectively, 19
combine with each other
form hydrogen chloride.
to
On
the other hand, while 6 grams of carbon react with 8
form carbon monoxide, only 3 grams of carbon, or | of 6 grams of carbon, react with 35.5 grams of chlorine to form carbon tetrachloride.
grams of oxygen
to
In order to standardize the comparison of reacting
masses of different elements, the mass of
it is
customary
measure
to
any element that reacts with exactly 8 mass
The masses determined in this way are called “equivalent weights.” An element may have more than one equivalent weight, because, as we have seen, it units of oxygen.
may combine
with oxygen
in several different ratios
by
weight to form different compounds. Thus, the equivalent weight of carbon
in
carbon monoxide
is 6,
lent weight of carbon in carbon dioxide
but the equiva-
is 3.
Law of Constant Ratios by Volume. (Gay-Lussac, 1808) When gases enter into a reaction or are produced by V.
it,
the ratio of their volumes, measured under the
same
conditions of temperature and pressure, can be expressed as a ratio of small whole numbers. For example,
when
hydrogen combines with oxygen to form water vapor, the
volumes of the hydrogen and oxygen
(at fixed
tempera-
ture and pressure) have the ratio 2:1; the volumes of the
oxygen and water vapor have the ratio
Or, putting
it
another way, 2 volumes of hydrogen combine with
1
1 :2.
volume of oxygen to form 2 volumes of water vapor. Here are some other combining ratios by volume in reactions for which we have already given the combining ratios by mass: 2 volumes of nitrogen combine with
1
volume
of oxy-
gen to form 2 volumes of nitrous oxide; 2 volumes of nitrogen combine with 2 volumes of oxygen to form 4
20
volumes of
volumes of nitrogen combine with 3 volumes of oxygen to form 2 volumes of nitrous anhydride; 2 volumes of nitrogen combine with 4 volumes of
oxygen
umes
nitric oxide; 2
to
form 4 volumes of nitrogen dioxide; 2 vol-
of nitrogen
combine with 5 volumes
form 2 volumes of
of
oxygen to
nitric anhydride.
Dalton's Atomic Theory In 1805 John Dalton formulated his atomic theory of
matter.
The theory
explain
why
is
a mathematical
model designed
to
chemical reactions obey the five laws de-
was guided toward the ideas embodied in the theory by a significant clue that is inherent in the Law of Multiple Proportions. As an example of this law, we have cited the fact that while 7 grams of nitrogen can combine with several different amounts of oxygen, these amounts are not arbitrary, but are all multiples of 4 grams. Notice that there is a smallest amount of oxygen with which 7 grams of nitrogen can combine, namely 4 grams, and that increases above this amount can occur scribed above. Dalton
only in increments that are equal to significance of this clue, let us
it.
To understand
examine
first
the
a familiar
analogy. Suppose you are making a necklace out of ribbon. Since a
roll of
ribbon
is
continuous, you
may
choose
any arbitrary length for the necklace, and cut this length of ribbon from the roll to make the necklace. On the other
A
length of ribbon
is
continuous
21
you are making the necklace of uniform beads placed side by side, you cannot choose any arbitrary length. There is a smallest length your string of beads hand,
may
if
have, namely the diameter of one bead, and increases
above
can take place only in increments that
this length
A are equal to
string of
beads consists
The amount
it.
of discrete units
of
oxygen with which 7
grams of nitrogen can combine resembles
in this respect
the length of a string of beads, rather than the length of a piece of ribbon. This resemblance suggests that, just as a string of beads
not continuous, but
is
crete units, a quantity of
but
is
made up
oxygen
is
made up
of dis-
also not continuous, is
the fundamental
Law
of Multiple Pro-
of discrete units. This
idea that Dalton derived from the
is
portions and elaborated in his Atomic Theory. Dalton’s
theory consists essentially of the following assumptions:
1.
Any sample
free state (that
compound),
is
compound, or of an element in the not combined with other elements in a
of a
is,
an assemblage of discrete units called mole-
cules. 2.
There
is
a smallest unit of any element that
may
enter into chemical combinations. This smallest unit
Atoms
is
same element are all alike. Atoms of different elements have different masses and differ in their chemical and physical behavior. 3. Every molecule is a combination of a definite numcalled an atom.
of the
ber of particular kinds of atoms. Molecules of the
compound have 22
the
same composition. The mass
same of a
molecule
the
is
sum
of the masses of the
atoms that
it
contains. 4.
In chemical reactions, atoms are combined, sepa-
rated, or reshuffled, but never created or destroyed.
Chemical Formulas and Equations According to assumption
3,
the composition of a mole-
by stating which atoms are in it, and how many of each atom it contains. This is done by means of a molecular formula in which each kind of atom in the cule
is
specified
molecule the
represented by the appropriate symbol, and
is
number
of
atoms
of each kind in the molecule
is
represented by a subscript of the symbol. For example, as
we
hydrogen contains 2
shall see later, a molecule of
atoms of hydrogen, so the formula for a molecule of hydrogen is H 2 A molecule of oxygen contains 2 atoms of oxygen, so the formula for a molecule of oxygen is O 2 A molecule of water contains 2 atoms of hydrogen and .
.
1
atom
water
is
symbol
A
of oxygen,
H
2
so
the formula for a molecule of
O, where the absence of a subscript for the
0 means that it is understood to be
chemical reaction
is
described by
1.
means
of
an equa-
tion in which the molecules that enter the reaction are listed
on one
and the molecules produced by the on the other side. Where more than one
side,
reaction are listed
molecule of a kind occurs, the number of such molecules is
indicated by a coefficient written to the left of the
molecular formula. Thus, while of hydrogen, 2
arrow
in the
H
2
H
2
means one molecule
means two molecules
of hydrogen.
An
equation shows the direction in which the
reaction progresses. According to assumption
4,
the
23
num-
ber of atoms of each kind that occur on one side of an
equation must he the same as the number of atoms of
each kind that occur on the other side of the equation. However, the grouping of the atoms in molecules changes as a result of the reaction. For example, suppose
we want
which hydrogen
to write the equation for the reaction in
combines with oxygen to form water. At
least
one oxygen
molecule must participate in the reaction. However, a single
oxygen molecule,
To form water with
O2
contains two oxygen atoms.
,
we must provide
these oxygen atoms,
two hydrogen atoms
for
each oxygen atom, since the
H
That is, we need four hydrogen atoms. This can be provided by two molecules of hydrogen whose molecular formula is H 2 But four atoms of hydrogen and two atoms of oxygen provide enough raw material to make two molecules of water. formula for a water molecule
is
2
O.
.
Consequently the formula
2H2
-f
for the reaction is
^ 2H2O.
O2
Notice that the number of hydrogen atoms on each side of the equation
each side
is 4,
and the number
oxygen atoms on
is 2.
Explaining the Basic If
of
Laws
Dalton’s Atomic Theory
stitution of matter, five basic
is
we should be
a good model of the conable to derive from
laws governing chemical reactions.
it
We now
the
pro-
ceed to derive them, one at a time. I.
Under the assumptions
of Dalton’s theory, the
that enter into a reaction are the
come out 24
of
it.
Only
their
same
grouping
as the is
atoms
atoms that
altered.
Conse-
quently the total mass of the atoms that enter into the reaction
come out action
sum
is
same as the total mass of the atoms that it. But the mass of any molecule in the re-
the
is
of
the
sum
of the masses of
atoms. Therefore the
of the masses of the molecules that enter into the
reaction
Law
sum
equal to the
is
of the masses of the mole-
by the
cules that are produced
the
its
of Conservation of
reaction. This establishes
Mass
as a consequence of
Dalton’s theory. For example, consider the reaction whose
equation
is
2H2
+
02
^ 2H2O.
Let iuh represent the mass of a hydrogen atom, and
mo
mass
re])resent the
of a
molecule
2mH
+
is
mo.
2???o,
The
total
equation
mass
is
is
2mH, the mass of an oxygen
and the mass of a water molecule
total
side of the equation
The
Then the mass
of an oxygen atom.
hydrogen molecule
let
mass is
of the molecules
+ 2mo,
2(2???//)
on the
+
or
is
left
2mo.
of the molecules on the right side of the
2{2inH -f mo)
=
+ 2mo.
Suppose we use any arbitrary mass of oxygen, and combine it with as much hydrogen as is necessary in order 11.
to
form water from
all
the oxygen and hydrogen used.
According to Dalton’s theory, the oxygen of
is
an assemblage
oxygen molecules. Let the number of molecules
assemblage be
n.
in this
Since the equation for the reaction
shows that every oxygen molecule combines with two hydrogen molecules to form water, then n oxygen molecules
combine with 2n hydrogen molecules
The mass mass
2???//,
of is
to
form water.
2n hydrogen molecules, each of which has 2??(2?n//), or 4:nmn.
The mass
molecules, each of which has mass 2mo,
is
of
n oxygen
n(27?ro), or
25
n
2nvio. of
Then
the ratio of the mass of hydrogen to the mass
oxygen with which
_
AnrriH
2nmo That
it
combines
to
form water
^
_ 2m ^ ^ 2 m// ^ 2n mo {mo)2n mo
(2y??//)2n
ent of the
number n
lishes the
Law
of
^
mo
2mn:mo, independ-
the ratio has the fixed value
is,
is
oxygen molecules used. This estab-
of Constant Proportions for this reaction.
A similar argument establishes it for every reaction. Suppose we take an arbitrary mass of nitrogen and combine it with as much oxygen as is needed to form in III.
turn each of the five compounds of nitrogen and oxygen listed
be,
it
on page
Whatever each amount
consists, according to the
number amount
of
number multiple
a whole
is
Law
of nitrogen
A
mass
of each
of mo. This estab-
argument establishes two elements. To be more
similar
combinations of
specific in the case of
we must
may
of Multiple Proportions for combinations
and oxygen.
for all other
oxygen
Dalton theory, of a whole
of oxygen atoms. Consequently the
lishes the
it
10.
compounds
of nitrogen
and oxygen,
use the molecular formulas for them.
lecular formula for nitrogen
is
X The 2
.
The mo-
molecular formu-
compounds of nitrogen and oxygen listed on page 19 are N 2 O, XO, XXOs, X"02, and X’205 respectively. The minimum numbers of molecules of nitrogen and oxygen that are needed to form each of these comlas for the five
pounds are indicated
in the
following equations:
+02 N +02
2X2 2
+ X + 2X2 + 2X2
2
26
= 2N2O, = 2NO, 3O2 = 2X2O3, 2 O = 2 XO 5O2 = 2X2O5. 2
2,
amount
In order to show the same
of nitrogen in each
equation, multii^ly the second and fourth equations by
2,
Then we have,
N O = 2N + 2O = 2N2 + 3O2 = 2
2
2
21^20,
2
2
4N0,
2^2 2N2
~h
+
2N2O3,
= 4NO2, 5O2 = 2N2O5.
4O2
According to these equations, 2 nitrogen molecules can
combine with
1
oxygen molecule, or 2 oxygen molecules,
or 3 oxygen molecules, or 4 oxygen molecules, or 5 oxygen
The masses
molecules.
oxygen are 2mo, respectively.
2(2//zo),
Then
amounts of 3(2?no), 4(2mo), and b{277io)
of these five different
the masses of oxygen per nitrogen
molecule are half of these amounts, namely,
7110,
2??7o,
3^0, 47 ??o, and bmo. Then for any given mass of nitrogen that contains
71
nitrogen molecules, the masses of oxygen
that can combine with
and 717)70.
D7}7no, all of
That
is,
it
are
27i77io,
3n?^o,
4nmo
which are whole number multiples of
the masses conform to the
Law
of Multiple
Proportions. IV. According to
the Dalton theory and molecular
formulas that are based on bine with
1
atom
of
oxygen
it,
to
2
atoms
of
hydrogen com-
form a molecule of water
;
and 2 atoms of chlorine combine with 1 atom of oxygen to form a molecule of chlorine monoxide. That is, 2 atoms of hydrogen and 2 atoms of chlorine combine with the same
amount
mass of a hydrogen atom is tiih, and the mass of a chlorine atom is 777ci, it means that a mass of hydrogen equal to 2m// and a mass of chlorine equal to 2mcj combine with the same amount of oxygen. of oxygen. If the
27
According to the Dalton theory, bines with
1
atom
of chlorine to
drogen chloride (HCl). That to
iriH
form a molecule of hy-
a mass of hydrogen equal
amounts we
see that a
mass
of
mci.
Dou-
hydrogen equal
combines with a mass of chlorine equal to 2mcu
to 2w//
the Dalton theory and molecular formulas that are
based on rine that
explain
it
why
the masses of hydrogen and chlo-
combine with the same amount of oxygen
combine with each other; that Equivalent Proportions as
To
hydrogen com-
of
combines with a mass of chlorine equal to
bling these
Thus
is,
atom
1
Law
derive the
it
is,
Law
they explain the
manifests
also
of
itself in this case.
of Equivalent Proportions in general
form as a consequence of Dalton’s theory, we argue as
C be three elements such combine with C and also combine with
A
fol-
lows: Let A, B, and
that
B
each other.
each
Let
rriA,
7nn,
and
and
masses of a single atom of each
ync be the
Suppose that under the which A combines with C,
of these elements, respectively.
Dalton theory the reactions
B combines
with C, and
A
in
combines with
B
are given
by
the following equations: 1)
xX
2)
zB
3)
rA
In order to have the of the
first
two
equation 2) by
wxX
-
A,C
-f-
wQ
-
B,C
T
sB -
A3j
same number
of
atoms
of
reactions, multiply equation 1)
y.
+ wyC
yC
Then we
C
in
by
each
t/;
and
get
wkxCy,
yzB
+ wyC
—^
t/B^C^,.
That is, wx atoms of A and yz atoms of B combine with the same amount of C {wy atoms of C) or a mass wxvia of atoms of A and a mass yzmn of atoms of B combine ;
28
C
with the same amount of
mass
(a
of
C). Multiplying equation 3) by wxyz,
ivxyzrA
Then
a
mass
a mass of
of
+
wxyzsB
A atoms
B atoms
wxyzArBg.
= wxs(yzmn)
.
That
=
the
is,
that react with each other are simple
multiples of the masses of
how molecules
get
equal to ivxyzsmn. But wxyzriiiA
A and B
same amount of C. V. Assumptions
we
equal to ivxyzrtn.A combines with
yzr{wxmA), and wxyzsmn masses of
wynic of atoms of
1
A
and
B
that react with the
on page 22 say nothing about
to 4
are dispersed in a gas, so they provide no
clue to the volumes that different gases
sequently assumptions explanation for the
Law
Avogadro
gap
filled this
to 4 alone
1
of
may
occupy. Con-
cannot provide an
Constant Ratios by Volume.
in the
Dalton theory
by Avo-
in 1811
adding another assumption, usually referred to as gadro’ s hypothesis: 5.
Equal volumes
of all gases at the
same temperature
and pressure contain equal numbers of molecules.
To show how assumptions 1 to 5 together successfully explain the Law of Constant Ratios by Volume, let us consider as an example the reaction in which hydrogen
combines with oxygen
form water. Suppose that the
to
equation for the reaction
2
is
molecules of hydrogen and
1
H + 02^ 2 H 2
2 O,
that
is,
2
molecule of oxygen produce
by any positive integer hydrogen and n molecules
2 molecules of water. Multiplying n,
we
of
oxygen produce 2n molecules of water.
see that
2n molecules of
that, at a given
Now
suppose
temperature and pressure, n molecules of
oxygen occupy a particular volume. Then according to 29
i
Avogadro’s hypothesis, n molecules of hydrogen occupy the same volume, and n molecules of water vapor occupy the
same volume. Consequently 2n molecules
of either
occupy double that volume. This implies that 2 volumes of hydrogen combine with 1 volume of oxygen to
will
produce 2 volumes of vapor. Thus, the Ratios by Volume, as
manifests
it
Law
of Constant
itself in this reaction, is
a consequence of Dalton’s theory and Avogadro’s hypothesis.
To
derive the law in general form,
we observe
first
an
immediate consequence of Avogadro’s hypothesis. Suppose that at a given temperature and pressure, a unit of
volume
n molecules. Then, by Avogadro’s
of gas contains
hypothesis, a unit volume of any gas at the
perature and pressure contains n molecules.
ume
Vi of gas contains
nV
same tem-
Then
a vol-
molecules, and a volume
V2
nVo molecules. Then the ratio of the nummolecules contained in volumes V\ and V 2 respec-
of gas contains
bers of tively
is
nVi ^ 7i 72* nV,
That
same temperature and pressure, the ratio oj the volumes of two gases is the same as the ratio of the numbers of molecules they contain. Now, suppose that, in a reaction, x molecules of A comis,
at the
bine with y molecules of B to form z molecules of C. Then the ratio of the numbers of molecules of A, B and C in the reaction
is
x:y:z. Consequently, at a fixed tempera-
ture and pressure, the ratio of the volumes of A,
C
in the reaction is also x:y:z.
lishes the
Law
of
B and
This observation estab-
Constant Ratios by Volume as a conse-
quence of Dalton’s theory and Avogadro’s hypothesis. 30
Three Questions Assumptions
1
to 5 give only the general content of the
atomic theory of matter. Since we know now that they successfully explain the five basic laws of chemistry, it is
worthwhile to develop the theory
in detail.
In working
out the details of the theory, three fundamental questions
must be answered: 1) Given any particular compound or element, what is the mass of one of its molecules? 2) Given any element, what is the mass of one of its atoms? 3) Assuming that we know from the experiments of the chemists what elements are combined in a compound,
how many atoms of each element are found in a molecule of the compound? That is, what is the compound’s molecular formula? We shall show now how facts determined by experiment and interpreted
in the light of
the theory
provide answers to these questions.
Weighing Molecules
The
job of measuring the mass of a molecule
out in two steps. The
first
step
is
is
carried
to obtain the relative
mass of the molecule, by comparing it with the mass of an oxygen molecule. The second step is to obtain the absolute mass by expressing it in grams.
Atomic Mass Unit In order to express the relative mass of a molecule, chemists use a special scale of masses in which the mass of an oxygen molecule
is
arbitrarily assigned the value 32.
31
Then
the unit of this scale
of the
is
mass
oxygen
of an
mass unit and is abbreviated as amu. Consequently, by definition of amu, the mass of an oxygen molecule is 32 amu. If we can find out how heavy a molecule is compared to a molecule of oxygen, we can express its mass as a number of amu, and molecule. This unit
vice versa. Thus,
if
is
called the atomic
the mass of a given molecule
is
J the
mass of an oxygen molecule, then it is J (32 amu) = 24 amu. Conversely, if the mass of a molecule is 24 amu, then it is = | times as heavy as an oxygen molecule.
The number of amu in the mass of a molecule of an element or a compound is called the molecular weight of the element or compound. The number of amu in the mass of an atom of an element
is
called the atomic weight
of the element.
Measuring Molecular Weight
We
shall consider here only the
problem of measuring
the molecular weight of a chemical that can be observed the gaseous state.
in
The method
based on a simple rule that
is
measurement is derived from Avo-
of
easily
gadro’s hypothesis.
Let us consider two different gases which
and
II.
we
Let wi be the molecular weight of gas
the mass of one of
its
I,
that
is,
molecules, expressed in amu. Let
wii be the molecular weight of gas 11.
of these gases at the
shall call I
Take equal volumes
same temperature and
pressure.
By
Avogadro’s hypothesis, these equal volumes contain the
same number be
n.
umes
Let
Mi and
of gases
32
of molecules. Let this
I
Mu
and
II,
number
of molecules
be the masses of these equal volexpressed in amu.
Then Mi
= nwi,
and Mii
= nwii.
The
ratio of these masses
is
Mr.Mn —
nwr.nwii, which reduces to lor.wn. In short,
Mi _
^
II
wii
M
the ratio of the iiiasses of equal volumes of two gases at the same temperature and pressure is the same as
That
is,
the ratio of their molecular weights.
masses of two substances is the same as the ratio of their weights. If the weights in grains of the eQual volumes of gases I and II that we are considering are Ih/ and Wn respectively, then
However, the
ratio of the
Wn
Mn If
we combine equations
(4)
Wi
and
_
(5),
we
get
ITj
Wii
Wii
This gives us a simple rule for comparing the molecular weights of two gases: Simply compare the weights of equal volumes of the two gases at the same conditions of temperature and pressure. If one of the gases is oxygen,
comparison gives the molecular weight of the other amu. For example, suppose ga«? I is hydrogen and gas
this in
found by experiment that the ratio of the weights of equal volumes of hydrogen and oxygen at the same temperature and pressure is about 1:16. ConseII
is
oxygen. It
is
quently
— = h’ wu
or
wi
= ^wn,
16
Since
wn
is
the
number
of
approximately.
16
amu
in the
molecular weight of
33
oxygen, which
amu
in the
is
32,
we have
that Wi (the
molecular weight of hydrogen)
number
= tV
(32)
of
= 2,
approximately.
Standard Temperature, Pressure, and Volume In the procedure described in the preceding paragraph, the experimenter
is
free to choose
and any volume
pressure,
for the
any temperature, any
two
gases, provided that
he chooses the same temperature, pressure and volume for
both gases. However, tations, chemists
in
order to standardize the compu-
have agreed
to relate all
measurements and vol-
to a particular standard temperature, pressure
ume. As standard temperature they use 0° Celsius, and as standard pressure they use a pressure of 760 milli-
meters (that
is,
the pressure that will support a column
of 760 millimeters of mercury).
The standard volume
Take enough oxygen gas so that the number of grams in its weight is the same as the number of amu in its molecular weight. Then see how large a volume it fills at the standard temperature and
is
defined in this way:
pressure. Since the molecular weight of
the standard volume
is
oxygen
is
32 amu,
the volume at the standard tem-
perature and pressure of 32 grams of oxygen gas. This
volume
To ume,
found by experiment to be 22.4
is
liters.
see the significance of this choice of standard vollet
us observe the consequences of using
it.
In the
we found that if IF/ grams and Wn the weights at the same temperature and pres-
preceding paragraph,
grams are
volumes of two gases whose molecular weights are Wi and ivn respectively, then
sure of equal
34
Now
assume that gas
II
oxygen, and that
is
we
are using
the standard temjDerature and pressure, and the standard
volume
number
of
amu
in
liters of
Substituting these values,
number
of
grams
=
IT/.
oxygen
= 32,
^ Wr
32’
32
That
molecular weight of a gas
grams
the
we obtain
wi
Consequently Wi
Wn =
and wn = the the molecular weight of oxygen = 32.
weight of 22.4
in the
Then
of 22.4 liters.
is
is,
number of amu in the same as the number of
the
the
weight of 224 liters of the gas at the standard temperature and pressure. For example, if we weigh in the
22.4 liters of hydrogen gas at the standard temperature
and pressure, the weight
is
found to be 2.016 grams.
Therefore the molecular weight of hydrogen If
we weigh
is
2.016 amu.
22.4 liters of carbon dioxide at the standard
temperature and pressure, the weight
is
found to be about
44 grams. Therefore the molecular weight of carbon oxide
is
di-
about 44 amu.
Using Any Temperature, Pressure and Volume In practice
it is
not necessary to use a standard volume
of a gas at the standard temperature
and pressure
order to measure
suffices to
any sample sure.
its
molecular weight. It
in
weigh
any temperature and any presThen from the known weight and volume of the of the gas at
gas at these conditions the molecular weight can be calculated.
The
calculation
is
carried out in three steps:
35
Step
=
Use the formula Tk
1.
To
273 to convert the
Celsius temperature to a Kelvin temperature. (See page 10 )
Step
II.
Use the formula PiVi P2U2
^
T,
T2
what the volume of the sample would be the standard temperature and pressure. (See page 11) to calculate
Step
III.
The weight
and pressure
is
at
of a gas at constant temperature
proportional to
its
volume. This fact
is
expressed in the formula
W
^ V
W' Use
this
V''
formula to compute the weight of 22.4
liters of
the gas at the standard temperature and pressure.
For example, suppose we have a sample of carbon
monoxide whose weight is .35 grams, and we measurement that its volume is .30 liters when perature
Step
I.
is
24° Celsius and
pressure
its
II.
Tc
Tk = Tc-h
273.
Tk = 24
273
+
its
is
= 760),
ture
= 24. =
297.
Find the volume Vi that the sample would
if its
and pressure 36
tem-
equiva-
occupy at standard temperature and pressure {Ti Pi
by
765 millimeters.
Find the Kelvin temperature Tk that
lent to the Celsius temperature
Step
is
find
= .30 liters = 297, P = 765.
volume of
T2
is
IT
2
= 273,
at a tempera-
(760)
^ 273
V'l
297'
(765) (.30)
^
..
(273) (765) (.30)
'
Step
III.
Find the weight IF
and pressure weigh
.35
standard temperature
at
of 22.4 liters of carbon
grams. Use ir
U
= 22.4,
=
.35,
=
28.
F'
if
.28 liters
= .28.
U'‘
^
22.4 .28'
.35
ir
11^'
monoxide
^
W'
W
.28 liters.
(297) (760)
= .28
That
is,
a standard
volume
of 22.4 liters of carbon
mon-
oxide at standard conditions of temperature and pressure
weighs 28 grams. Therefore the molecular weight of car-
bon monoxide
is
28 amu.
Absolute Mass of a Molecule
The molecular weight of a molecule is the number of amu in its mass. To convert from a mass expressed in amu to a mass expressed in grams it is necessary to know the number of grams there are in one amu. To determine this number, let us think for a moment about the standard volume (22.4 liters) of a gas at standard conditions of temperature
molecules
it
and pressure. Let No be the number
of
contains. According to Avogadro’s hypothesis
37
it
is
number
the same
known
for all gases.
We w
already
know
is
that
w grams, liters of the gas is w grams.
X
we
Dividing by
(1
We shall
value found
is
w
Consequently
amu) = w grams.
amu =
1
gram.
which the value of No can be
in
see one of these
No
is
and therefore the mass of 22.4
find that
There are several ways found.
is
of 22.4 liters of the gas at stand-
is
(Now)
suppose that the
the molecular weight of a gas
if
amu, then the weight
ard conditions
is
amu. Then the mass of No(w amu) = (Now) X (1 amu).
molecular weight of the gas 22.4 liters of the gas
Now
number.
as Avogadro's
The number No
= about 6 X
ways 10"^,
in
that
Chapter IV. The is,
6 followed
by
23 zeros, or 600 thousand million million million. Conse-
quently
1
No
gram
= =
-7-
1.66
X
10^'^^
10“^'^
using
10~‘^
understood to
is
gram =
24 places to the
Now
-r-
10^"^)
gram
gram,
mean
turn, can be accomplished
1
(1.66
the customary notation, multiplying by
where,
is
10
1
1
amu
left.
That
dividing by
10'^,
and
this, in
by moving the decimal point
is,
amu = .00000000000000000000000166 gram. that
we know
the
number
of
grams
in
one amu,
easy to find the mass, in grams, of any molecule,
38
if
it
we
know
its
mass
gram
=
gram. The mass of a hydrogen molecule
is
gen
molecule
5.31
X
10““’^
about
2
For example, the mass of an oxy-
in ainu.
amu
is
amu
32
=2X
1.66
= 32 X X
1.66
10~^^
X
10““^
= 3.32 X
gram
10“^^
gram. The mass of a molecule of carbon dioxide 44
amu
= 44 X
X
1.66
10““^
gram
= 7.30 X
is
gram.
Weighing Atoms It is
found by experiment that
gas combines with
1
volume
1
volume
of
hydrogen
of chlorine gas to produce
same tempera-
2 volumes of hydrochloric acid gas at the
ture and pressure. Since 2 volumes of gas contain twice as
many
molecules as
1
volume
ture and pressure, this
of gas at the
same tempera-
means that each molecule
of hy-
drogen gas combines with one molecule of chlorine gas to
produce two molecules of hydrochloric
acid. Since each
molecule of hydrochloric acid contains at least one atom of hydrogen, a molecule of tain at least
molecule of
hydrogen must therefore con-
two atoms of hydrogen. Thus we see that a an element may contain more than one atom
of the element. Consequently the atomic weight of an
element need not be the same as the molecular weight of the element. If a molecule of an element contains two
atoms, the atomic weight
is
one half of the molecular
weight. If a molecule of an element contains three atoms, the atomic weight
and the
is
so on. This fact
way
one third of the molecular weight, is
the chief obstacle that stands in
of determining the atomic weight of an element.
In I860 Cannizzaro showed that there
bypassing this obstacle.
He
is
a simple
way
of
pointed out that although a
39
:
molecule of an element
may
contain more than one atom
some compounds of the element whose molecules contain only one atom of the element. The mass contributed by the element to the molecular weight of such a compound would be the of the element,
it is
likely that there are
atomic weight of the element. Since no compound of an than one atom of the element, the
element contains
less
atomic weight
easily recognized as the smallest
is
mass
contributed by the element to the molecular weight of
any of its compounds. For example, chemical analysis shows that chloroform, carbon tetrachloride, hydrogen chloride, sulfur monochloride,
and ethyl chloride are
all
compounds
ment chlorine. The molecular weights are shown in the table below
Compound
compounds
Molecular Weight
amu 154 amu 36.5 amu 135 amu 64.5 amu
Chloroform
119
Carbon tetrachloride Hydrogen chloride Sulfur monochloride
Ethyl chloride
By
of these
of the ele-
quantitative chemical analysis
it is
possible to find
out the percentage of chlorine in each of these compounds.
These percentages are found
to
be 89.10%, 92.19%,
97.24%, 52.51%, and 54.96% respectively. Consequently the mass of chlorine in a chloroform molecule
is
89.10%
amu; the mass of chlorine in a carbon tetrachloride molecule is 92.19% of 154 amu; etc. The masses of chlorine computed in this way are shown in this table: of 119
40
Compound
Mass
Chloroform
of chlorine in
Carbon tetrachloride Hydrogen chloride
89.10% 92.19% 97.24%
Sulfur monochloride
52.5,1% of
Ethyl chloride
54.96%
one molecule
amu = 106 amu 154 amu = 142 amu 36.5 amu = 35.5 amu 15 amu = 70.9 amu 64.5 amu = 35.4 amu
of 119
of of
of
Notice that the smallest mass of chlorine in any of these
compounds
is
about 35.4 amu. Then this must be the
approximate mass of one chlorme atom. Moreover, we can tell from the table how many atoms of chlorine there are in a molecule of each of the
chloroform molecule contains 106
compounds
=3 X
amu
listed.
A
amu
of
35.4
chlorine.
Therefore a chloroform molecule contains 3
atoms of
chlorine.
tains 142
A carbon tetrachloride molecule conamu = 4 X 35.4 amu of chlorine. Therefore a
carbon tetrachloride molecule contains 4 atoms of chlorine. Similarly,
atom
1
atom contains
2
of chlorine, a sulfur monochloride
atoms of tains
a hydrogen chloride molecule contains
1
chlorine,
atom
and an ethyl chloride molecule con-
of chlorine.
The molecular weight the atomic weight
is
of chlorine gas
about 35.4 amu,
is
it
71.0
follows that a
chlorine molecule contains 2 atoms of chlorine.
methods, using the principle of Cannizzaro, that the atomic weight of oxygen
molecular weight of oxygen
is
is
16
amu. Since
By it
is
similar
found
amu. Since the
32 amu, a molecule of
oxygen contains two atoms of oxygen. The atomic weight of hydrogen lar
is
found to be about
weight of hydrogen
is
2
1
amu. Since the molecu-
amu, a molecule of hydrogen
contains two hydrogen atoms. 41
Atomic Weight via
Heat
Specific
Cannizzaro’s method can be used to find the atomic
weight of any elements whose compounds are easily
compounds
vaporized. For those elements whose
method
easily vaporized, a different
measuring the state.
The
calories of heat
gram
used, based on
element
specific heat of the
specific
is
heat of a substance
are not
is
in the solid
the
number
of
needed to raise the temperature of one
of the substance one degree Celsius. In 1819,
Du-
long and Petit found that for elements whose atomic
weights were known, the product of the atomic weight
and the
specific heat of the solid
namely
6.3.
Assuming that
element
is
a constant,
this rule applies to all ele-
ments, the atomic weight of an element can be computed
from the formula atomic weight
=
6.3 specific heat
For example, the
specific heat of iron
is
0.113 calories per
gram. Therefore the atomic weight of iron 0.113
is
about 6.3
-J-
= 56.
Molecular Formulas
Once we know the molecular weight atomic weight of each element that
compound, the contains, and the
of a
it
combining ratios by weight of the elements in the compound, it is easy to figure out the molecular formula for the compound. For example, chemical analysis of the
compound 42
called carbon tetrachloride
shows that
it is
a
compound
of carbon (C)
analysis shows that of the
compound
is
and chlorine (Cl). Quantitative
92.19%
of the
contributed by chlorine.
namely 100%
of the mass,
mass of each molecule
The balance
— 92.19% = 7.81%
is
con-
by carbon. The molecular weight of carbon tetrachloride is 154 amu. Therefore the mass of chlorine in a tributed
carbon tetrachloride molecule
92.19% in
of 154
is,
as
we have already
amu, or 142 amu; and the mass
a carbon tetrachloride molecule
is
7.81%
seen,
of carbon
of 154
amu,
amu. By the method of Cannizzaro it is found that a chlorine atom weighs about 35.4 amu, and a carbon atom weighs about 12 amu. Since 142 is about 4 X 35.4, and
or 12
12
is
1
X
12,
it
follows that each molecule of carbon
tetrachloride contains 4 atoms of chlorine and
1
atom
of
carbon. Consequently the molecular formula for carbon tetrachloride
is
CCb.
The molecular formula
for
water can be determined
in
same way. Chemical analysis of water shows that it is a compound of hydrogen (H) and oxygen (0). Quantitative analysis shows that oxygen contributes about 89% of the mass of water, while hydrogen contributes about the
11%
of the mass.
amu,
The molecular weight
of water
is
18
mass that oxygen contributes to a single molecule of water is 89% of 18 amu, or about 16 amu, while the mass that hydrogen contributes to a single molecule of water is 11% of 18 amu, or about 2 amu. Since the atomic weight of oxygen is 16, and the atomic weight of hydrogen is about 1, it follows that a molecule of water contains 2 atoms of hydrogen and 1 atom of so the
oxygen. Consequently the molecular formula for water is
H
2
O.
We
have already seen on page 41 that a molecule 43
of
chlorine gas contains 2 atoms of chlorine, a molecule of
oxygen gas contains 2 atoms of oxygen, and a molecule of hydrogen gas contains 2 atoms of hydrogen. Consequently the molecular formulas for chlorine, oxygen and
hydrogen are CI 2
The Valence
,
and
H
respectively.
2
Element
of an
An important atoms
O2
,
property of an element
of other elements with
which
it
number
of
can combine.
A
is
the
numerical measure of this property, called the valence of the element,
introduced in the following way:
is
valence of hydrogen
is
defined to be
1,
The
and the valence
any other element is defined to be the number of atoms of hydrogen with which an atom of the element combines. For example, since an oxygen atom combines with of
2 hydrogen atoms to form a water molecule, the valence of 1
oxygen
is
Since a chlorine atom combines with
2.
hydrogen atom to form a hydrogen chloride molecule,
the valence of chlorine
is 1.
Since a carbon atom combines
with 4 hydrogen atoms to form a molecule of methane
whose molecular formula
is
CH4, the valence of carbon
is 4.
When
a
compound contains only two elements, the
numbers of atoms
compound obey
of each element in a molecule of the
this rule:
the
number
atoms of one
of
element times the valence of that element equals the
number
atoms of the other element times the valence of that element. For example, the molecular formula for of
carbon dioxide cule of
1X4
CO 2
,
equals
44
is
CO
2.
There
is 1
carbon atom
and the valence of carbon 4.
is 4.
There are 2 oxygen atoms
in
a mole-
The product
in a
molecule
CO 2 and the X 2 is also 4.
of
2
valence of oxygen
,
If the
valence,
is
atomic weight of an element
we
get
its
is
divided by
its
equivalent weight. For example, the
atomic weight of oxygen
is
equivalent weight
2
is
The product
2.
16
16, its
= 8.
valence
is 2,
its
(See page 20)
Some elements have more than one
valence. For ex-
ample, the element iron (Fe) has the valence 2
compounds and the valence
and
3 in other
in
some
compounds.
Families of Elements
There are some elements that resemble each other their physical
and chemical properties. They tend
to
in
com-
bine with the same elements to form similar compounds,
and they have the same valences
in these
elements that resemble each other are in
compounds. The
classified together
groups or families of elements. For example, lithium
(Li),
sodium (Xa), and potassium (K) are members of
the family of alkali metals. Fluorine (F), chlorine (Cl),
bromine (Br), and iodine
(I), are
members
of the halogen
family of non-metals. Helium (He), neon (Ne), argon (Ar), and krypton (Kr) are
noble gases.
The
alkali metals
active chemically, that
is,
members
of the family of
and the halogens are very
they react easily with
many
The noble gases, on chemically. They form
other elements to form compounds. the other hand, are very inactive
almost no compounds at
The Periodic Table
all.
of the
Elements
Meyer and Mendeleyev, working independmade an important discovery about the chemical
In 1869 ently,
45
elements.
They found
that
if
the elements are listed in
order of increasing atomic weight, elements that are in the
same family occur
this respect the list of
at regular intervals in the
elements resembles a
days of a month. The days of a month are families
month
list
In
of the
classified into
by the day of the week on which each day of the falls.
Thus, some of the days of the month are
Sundays, some are Mondays, and so on.
month
list.
the days of a
If
are listed in order of increasing date, days that
belong to the same family occur at regular intervals in the list.
For example,
if
month
the 1st of the
are the 8th, the 15th, the 22nd, this property of the list of
a Sunday, so
is
and the 29th. To describe
days of a month we say that
the days of the week occur periodically in the
cause of this periodic property of the
days of a month
in a table in
M
A
page
T
we can
W
T
F
S
1
2
3
4
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
calendar
is
a periodic table of the days of a month.
designating the days of the
month
increase from left to
right in each line,
and from top
the days that
on the same weekday
fall
the
1964
5
of a
list
Be-
which, while the numbers
January
S
list
list.
to
bottom fall
line
by
line,
under each
other in the same vertical column. This arrangement,
46
used in the calendar,
is
called a periodic table. Similarly
the chemical elements can be arranged in a periodic table, in which, while the
crease from
bottom family
line fall
A modern
left
by
atomic weights of the elements
and from top
to right in each line,
line,
in-
to
the elements that belong to the same
under each other
in the
same
vertical column.
version of the periodic table of the elements
shown on page 48. There were three imperfections
is
in the original periodic
table of the elements. 1) In order to keep elements of the
same family in the same vertical column it was necessary to leave some gaps in the table. For example, there was a gap in the 21st place, right after calcium. 2) To keep the families in vertical columns it was sometimes necessary to depart from the order of increasing atomic weight.
For example, the cobalt atom has a higher atomic weight than the nickel atom, but
atom
in the table. 3)
the table at
all,
it is
listed before the nickel
Some elements
but had to be
didn’t
fit
well into
listed separately.
This was
true of the entire family of elements called the rare earths, or the lanthanide series.
The
discovery of the periodic table raised two very
challenging questions:
What
1)
accounts for the near-
perfection of the table, as a result of which chemical families occur periodically in the
lated
by atomic weight?
2)
What
list
of elements tabu-
accounts for the imper-
fections in the table? Part of the second question
swered immediately by Mendeleyev.
He
was an-
guessed that the
gaps in the table represented elements that had not yet been discovered. As these unknown elements were covered one by one, the gaps were for example,
is
now
filled
filled.
The
dis-
21st place,
by the element scandium. The 47
Ne
o
He
2
Kr
Ar 20.183
4.003
Xe
Lu
18
174.99
(222)
83.80
10
Lw
Rn 131.30
39.944
36
54
86
103
71
(257)
No Ni
Yb
Pt
Pd
173.04 195.09
78
46
(254)
Md
Rh
Co
102
70
106.4
58.71
28
Ir
VIII
Tm
102.91
192.2
58.94
27
45
77
Ru
Fe
168.94
Os Er
190.2
76
44
167.27
1(X)
68
Br
Cl
F 9
35.457
I
79.916
At
Ho
126.91
(253)
Es
(210)
19.00
17
(256)
Fm
101.10
55.85
26
101
69
35
88)
164.94
85
53
(254)
67
99
Dy
Cf
number
VII page
Mn
Re
Tc
186.22
54.94
25
43
(98)
75
mass
162.51
(see
(251)
0
16.000
S
Te
Se 34
Tb
Bk
show
(210)
78.96
8
98
Po 127.61
32.066
16
66
84
52
isotope
158.93
(247)
97
65
VI Elements
Mo
Cr 52.01
W
pareotbeses
known 183.86
95.95
24
74
42
Gd
Cm 157.26
(247)
96
64 the
io
N 7
of
>
14.008
P
As
121.76
30.975
209.00
74.91
15
33
stable
Bi
Sb 51
Eu
83
Am most Numbers
152.0
Nb
V
Table
(243)
95
63
of
Ta 180.95
50.95
92.91
73
41
23
Pu
Sm 150.35
(242)
62 Periodic
C
Ge
Si
207.21
118.70
28.09
6
Pb
Sn
12.011
94
72.60
50
32
14
82
Pm
IV
Np (147)
61 Ti
Symbol Atomic
(237)
93
Weight
Hf
Zr
178.50
91.22
47.90
40
22
72
Nd 144.27
60
B 5
10.82
A1 13
Ga
238.07
92
Atomic
Tl
In
umber itomic
114.82
26.98
U
204.39
69.72
31
81
49
Pa
Pr
III
140.92 series*
44.96
89-103
57-71
88.92
39
21
91
59
Y
Sc
(231)
series**
Ac
La
Ce
Th 140.13
Cd
Zn
232.05
90
58
Hg
112.41
200.61
65.38
48
30
80
NH NN
Ac
La 138.92
(227)
Ca
Mg
Be
12
89
226.05
87.63
40.08
20
38
56
Cu
Ag
88
Au 107.880
63.54
197.0
29
H 1
57
Ra 137.36
24.32
9.013
4
Ba
Sr
79
47
Lanthanide
series:
1.008
Na Li
3
48
22.991
6.940
11
K 19
Rb
Cs
39.100
Actinide
Fr 132.91
(223)
85.48
37
55
87
*
**
series:
answers to the
first
question and the rest of the second
question are supplied by the modern theory of atomic structure. In the remainder of this
development of
how and
this theory. In
book we outline the
Chapter VII we
shall see
the theory successfully explains the periodic table its
imperfections.
The number
of the place that an element occupies in
the periodic table ally designated
helium,
Z
= 2,
is
number and is usuFor hydrogen, Z = 1, for 3, and so on. The signifi-
called its atomic
by the
letter Z.
for lithium
Z
=
cance of the atomic number in the theory of atomic structure
is
discussed in Chapter IV.
49
f
Ill
Motion, Electricity THE cists
and Light
chemists discovered the atom, but the physi-
analysed
its
structure. In this chapter
we present
some physical concepts and relationships that play a part in the analysis. All measurements used will be expressed in the
centimeter-gram-second system of measurements.
It will
be understood then, w^hen units are not mentioned
explicitly,
that distances are measured in centimeters,
masses are measured
in
grams, time
is
measured
onds, forces are measured in dynes, energy
is
in sec-
measured
in
ergs, etc.
Velocity
A
w^eather report on wind conditions always gives the
may
speed of the wind and
its direction. It
ample, that the wind
from the west with a speed of 10
is
say, for ex-
The combination of speed and direction of motion is known as velocity. A velocity can be represented graphically by an arrow. The direction in which the armiles per hour.
row points shows the direction of motion, and the length of the arrow shows the speed of the motion. If a
body moves through a distance
s in
a time interval 51
t,
its
average speed v in that interval
is
given by the
formula
the speed does not change during the interval, this
If
formula also gives the exact speed at each instant
in the
interval. If a
body, after moving for a length of time
move
to
for
t,
continues
an additional small amount of time,
it is
tomary
to represent this small increment in time
symbol
At, read as “delta
“the change in t” If
moved
in the
time
s t,
to
The symbol As is mean “the change
by the
and understood to mean
represents the distance the body has it is
customary to write As
for the
moves during the time interval read as “delta s,” and is understood
additional distance that At.
t,’’
cus-
it
By formula
in s.”
speed during the time interval At
V
(8), the
average
is
= As At'
If
we multiply both
side of equation (9)
As =
(10)
by
At,
we
get
v{At).
Momentum
A moving
body tends
to keep
moving. There
ure of this persistence of motion called the of the body. If a
momentum M,
body with mass
M
11 )
52
=
mv.
a meas-
momentum v,
and
by the
for-
has a speed
these quantities are related
mula (
m
is
Changing Velocity
The
velocity of a
moving body may change
may change, the both may change. If
ways: the speed of motion
motion
may
changing,
it
change, or is
make an
possible to
at a particular instant in this
in three
direction of
the speed
is
estimate of the speed
way:
find the distance As
moves during a small interval of time At starting with that instant. Then equation (9) gives an approximate value of the speed at that instant. The that the body
smaller the interval of time At
mation
The
is.
is,
the better the approxi-
exact value of the speed at that instant
is
the limit approached by these approximate values as
At approaches zero.
A
change
in velocity, like the velocity itself,
has both
and can be represented by an arrow. When a velocity and a change in the velocity are known, the new velocity that results can be found a
magnitude and a
direction,
graphically by following this simple rule to represent the original velocity.
From
:
Draw an arrow the head of this
arrow draw an arrow to represent the change of velocity.
Then
the
new
velocity
from the
tail of
arrow, as
shown
shows how velocity
the
is
first
in the
to find the
represented by an arrow drawn
arrow to the head of the second
diagram below. The diagram change
in velocity, if
both the old
and the new velocity are known: Draw the
new
also
velocity
53
ar-
:
rows that represent the old and the new velocities so that
same point. Then the change in velocity is represented by the arrow drawn from the head of the old velocity to the head of the new velocity. Let V be the speed of a moving body (the magnitude of its velocity), and let be the magnitude of the their tails are at the
change
in velocity that takes place in a
The average
rate of change of the velocity
average acceleration
a,
and
is
is
AL
called the
given by the formula
Av
a
12 ) (
Xi'
If the acceleration
this
time interval
does not change during the interval,
formula also gives the exact acceleration at each
in-
stant in the interval. If the acceleration does change, the
formula gives an approximation of the acceleration at a particular instant
if
we
use as At a small interval that
The smaller the interval is, the approximation is. The exact value of the accel-
begins with that instant. better the
eration at that instant
is
the limit approached by these
approximate values as At approaches
zero. If
we multiply
both sides of equation (12) by At, we get a formula for the approximate magnitude of the change in velocity,
Av, during a small interval of time At, in terms of the acceleration a at the start of the interval
=
a{At).
a push or a pull.
When
Av
(13)
Force
A that
force is '
is
free to
54
move,
it
a force acts on a
body
changes the motion of the body.
The magnitude
of the force F, the
mass
and the acceleration a caused by the
m
of the body,
by
force are related
the equation
F =
(14) If
the values of
F and
then the value of a
is
ma.
ni are
unchanging, or constant,
constant.
Electrical Charges
There are two kinds of
negative. Electrical charges
may
be measured
of units called electrostatic units (esu).
a charge
is
and
a
it is
of
2
—3
+
of
in
is
a,
6,
The measure
and another charge b b,
=—
1.
If a
2,
of
is
charge of
0,
placed on is
a charge
if
the total charge becomes
the total charge becomes 6
charge on a body
is
where the addition
elementary algebra. For example,
added to a charge of
(—3)
terms
number if the charge is positive, negative number if the charge is negative. If
the total charge becomes a-\-
done as
in
a positive
a body has a charge it,
and
electrical charges, positive
we say
—6
+
is
added to a charge
(—6)
= 0. When
that the body
is
the
elec-
removed from a body, the charge that remains can be calculated by subtraction. For example, if a charge of —3 is removed from a body that is electrically neutral, the remaining charge trically neutral.
If
isO- (-3) =3. Any two electrical If the
them
a charge
is
charges exert a force on each other.
charges are of the same kind, the force between
is
a force of repulsion. If the charges are of opposite
55
kinds, the force
charges
is
is
a force of attraction. If each of two
concentrated at a separate point,
between the points
r,
is
(
between them
is
the distance
and the charges, measured
in
and q respectively, then the given by the formula
Q
electrostatic units, are
force
if
15 )
If
an
moves
electrical charge
across a magnetic
field,
the field exerts a force on the charge, tending to deflect
from
its
The
path.
an arrow.
If
field
may
the initial velocity of the
perpendicular to the is
strength
field,
be represented by
moving charge
is
the force exerted on the charge
perpendicular to both the magnetic
locity of the charge.
it
Then
the charge
field
and the ve-
moves
in a circle
V
whose plane
perpendicular to the magnetic
is
strength of the magnetic field
the charge
is e,
and
its
speed
is v,
the force exerted on the charge
(
is
H,
if
field. If
the
the magnitude of
then the magnitude of
is
16 )
where
c is the
speed of
toward the center of the 56
light.
The
circle.
direction of the force
is
Work
A
force
capable of doing work. It does work when
is
pushes an object from one place to another.
F
magnitude
force with
is
done
is
defined
If
the force
close to
is
=
work
of
F(As).
not constant but
some value F
amount
dis-
by the equation
W
(17)
constant
pushes an object through a
tance As in the direction of the force, the
W that
If a
it
as
it
its
magnitude
always
is
acts through the distance As,
then equation (17) gives an approximate value of the
work done.
we can way:
If
the force changes considerably as
it acts,
work done
in this
Divide the distance As into n segments
(As)i,
get an approximate value of the
(As) 2, ...
,
(As)w.
Make
the lengths of the segments small
enough so that the force does not change much as the body moves through one of these segments. Let Fi be one of the values of the force as (As)i, let Fo be
it
acts through the distance
one of the values of the force as
through the distance (As) 2,
etc.
it
acts
Then an approximate
value of the work done as the force acts through the total distance
given by
=
(18)
We
is
Fi(As)i
-t-
F^iAsh
+
•
•
•
+
F„(As)n.
can get better and better approximations to the value
work done by dividing the distance into more and more segments with smaller and smaller lengths. The exact value TT^ of the work done is the limit approached by of the
these approximations as the
number
creased to infinity while their
of segments
maximum
length
creased to zero.
57
is is
in-
de-
,
Kinetic Energy
A moving body work that
can do by virtue of
it
kinetic energy. If a
The
also capable of doing work.
is
body that
is
motion
its
is
called its
initially at rest is
pushed
energy of the work done by
into motion
by a
the force
transformed into the kinetic energy of the
body.
We
is
force, the
can use this fact to derive a formula for kinetic
F
energy. Suppose that a constant force
with mass m, pushes
it
acting on a
through a distance As
body
an
in
in-
terval of time A^. Let the acceleration imparted to the
body be
a.
Then
the value of a
the body’s initial speed
Then
By
is 0,
is
constant.
and that
Assume
that
speed
is v.
its final
the magnitude of the change in velocity, At;
equation
work done, IF
(17), the
= /^(As).
=
v.
Let v
bar”) be the average speed of the body during
(read as
AL By equation
the interval
(10) on page 52, As
= v{At),
and by equation (14) on page 55, F = ma. Substituting these values of As and F into equation ( 17) we get ,
W
(19)
By equation case equals V,
v.
=
is,
=
mv[a(At)].
(13) on page 54, a{At)
—
Av, which in this
Moreover, since the value of a
the average speed,
speed, that
mav{At)
D
=
tions in equation (19),
W
(20)
+
^)
we
=
=
Iv.
Making
V
is
speed,
these substitu-
get
m{\v)v
=
^mv^.
Since the kinetic energy stored in the body is
constant,
the average of the initial and final
is
i(0
is
when
its
equal to the work that was done to give
we have,
finally,
Kinetic energy
(21)
58
=
^mv^.
speed
it
that
Energy
Potential
An
electrical
charge that charge
is
charge
near
it.
tends to push or pull any other
Q
To
describe this fact
surrounded by a
is
we say
that the
field of electrostatic force.
Any
charge q that is in this field of force is subjected to either a push or a pull. Since the push or pull can make it move,
moves,
can do work, the charge q is capable of doing work because of its position in the electrostatic and, as
it
field of force.
it
The work
that
it
can do by virtue of
position in the electrostatic field of force (electrostatic) potential energxy. Specifically,
is if
its
called its
g
is
at a
from Q, we define the potexitial energxy of q to be the work done by the force exerted by Q on q sls q moves away to an infinite distance from Q. To obtain a distance
r
formula for the potential energy of culate the
work done
as q
q,
we
shall first cal-
moves from the distance
We
greater but finite distance R.
r to
a
shall designate this
work by the symbol W{r, R). Then the potential energy of q at a distance r from Q will be the limit of W{r, J?) as becomes infinite. The electrostatic force between Q and q when the distance between them is r is given by formula (15), namely,
amount
of
f =
9£. r2
In the special case where
Q=
q
=
1,
the formula becomes
Notice that the force in the general formula
Qq
is
times the force that occurs in this special case.
59
simply
To
de-
formula for the potential energy, we shall derive
rive the
Then
for the special case only.
it first
the general formula
can be obtained easily from the special case by simply multiplying the force, wherever
occurs in the computa-
it
by Qq. For simplicity, we shall assume that the charge q moves away from Q in a straight line. In the diagram tion,
below, points on the straight line are designated by their distances from Q. Thus, the point labeled 0
where
Q
located.
is
from Q. distance R from Q. distance
r
The point labeled r The point labeled R
is
the point
is
the point at a
is
the point at a
The charge q moves Position of
from
charge
here
'
Q
... .
.
to
here
.
t
’
»
0
R
Assume Q acting on
it
= q=\.
As q moves from r to R, the force changes from 1/r^ to \/R^. To calculate the
work done, we follow the procedure outlined on page
We
shall divide the distance
between
r
and
R
57.
into
n
equal segments, and use formula (18) to find an approxi-
mation
Wn
work done. Then we shall of the work done by finding
to the value of the
find the exact value
Wn
W{r,
J?)
n becomes infinite. Let us begin by choosing n = 1 to get a very crude approximation, Tri = Fi(As)i, where Fi is a value that F takes on as the charge moves from point r to R, and where (As)i is the distance from r to R. The distance the limit of
60
as
(A5)i
equal to
is
R—
r.
The
force
any value between \/r (the value (the value of
We
at 7?).
shall
of
make
may F at
be chosen as
and
r),
1/7?“
a choice of F\ that
will
enable us to put the formula for lib in a simple form.
We
note
first
multiply both
r
.
The
length of
length of the chord
SR
QP
is v.
is
equal to the length of the arc QP, which these substitutions,
we
approximately is
As.
Making
_ As r
V
=
equation (10) on page 52, As
substitution,
length
get the approximate equation At>
By
The
v(At).
Making
this
we
ob-
we get ^ (AO. r
V
Solving this equation for Av,
Av =
we get,
—
(At),
r
Then, dividing by
A^,
we
get
Av _ At This
is
r
only an approximate equation, because
by using an approximate value for the length of the chord QP. However, it becomes more and more exact as At approaches zero. But the limiting value of tained
Av —
it
as At approaches zero
acceleration
(
a.
is
precisely
what we
call
(See page 54) Therefore
34 ) 67
the
Wherever there is an acceleration that is not zero, there must be a force that accounts for the acceleration. This force, which keeps the body moving in a circle, is called the centripetal force. The magnitude of the force is given by equation (14): F = ma. If we substitute for a the value given by equation (34), we get this formula for the centripetal force:
(35) r
Momentum
Angular
A spinning body tends to keep spinning. There is a measure of this persistence of spinning motion called angular
momentum. When
body moves along the circumference of a circle, it is as though the circle were spinning around its center and carrying the body with it as it spins. Consequently there
is
a
circular motion. If the
radius of the circle
momentum associated with body has momentum M, and the the angular momentum p is given
an angular
is r,
by the formula p
(36)
= Mr.
M=
According to equation (11), mv. Substituting this value of into equation (36), we get
M
p
(37)
Light
mvr.
Waves
Light in
=
is
waves. 68
a form of energy that
is
radiated through space
The waves may be represented diagrammati-
by a sinuous line like the one shown below. The distance between consecutive crests in a wave is called a wavelength, and is represented by the Greek letter X cally
(lambda). Sunlight
a mixture of light of different
is
In a rainbow, the colors are separated and ar-
colors.
ranged side by
side.
Each
The wavelength
a different wavelength. single color
distinct color in a
(monochromatic
the following way:
A beam
verging from a light source S a lens
L and then two narrow
is
it
monochromatic
light di-
allowed to pass through
slits
a small distance d on a screen R.
X of light of a
can be measured in
light)
of
rainbow has
that are separated by
The
lens
is
placed so that
bends the diverging rays of light and makes them paral-
On
lel.
the other side of the screen each
slit
becomes a
separate source of diverging rays of light. These rays of light are this
caught on a second screen T. (See diagram
way each
illuminated point on the screen
tw^o rays of light,
one from each
slit.
T
I)
In
receives
These two rays may
have traveled along paths of different lengths, as shown in
diagram
lengths in the crest,
is
II.
nX,
Where
where n
two rays are
the difference between the path is
an integer, then the wave-trains
in step.
That
is,
crest coincides with
and trough coincides with trough. Then the waves
reenforce each other, and the point where they is
brightly illuminated.
path lengths
is
are out of step.
Where
fall
on
T
the difference between the
(n -b i)X, the wave-trains in the two rays
That
is,
crest coincides with trough
69
and
trough coincides with
crest.
Then
the waves interfere with
each other, or cancel each other, and the point where they fall
on
T
is
not illuminated at
series of alternating light
all.
As a
result, there is a
and dark bands on the screen T.
These bands are called interference bands. The as though the screen
R
splits the
beam
effect
that crosses
R
it
is
into
T
II
S
many
separate beams, each turned through a different
angle from the direction of the original beam. Diagram II
above shows one of these beams turned through an angle
0,
70
to
form a bright band at
P
on screen T. The
sides of angle sides of
0,
ABC
are respectively perpendicular to the
and therefore angle
ABC = 7?X/d,
where
7?X is
ABC =
The
slits
=
For
\.
this case
Multiplying both sides of this equation by \
=d
sin 6. Since
d,
beam
is
= X/d.
sin
d
we
find that
both d and 6 can be measured, this
formula allows us to compute the wavelength
A
on their way
smallest angle 6 through which a
turned occurs when n
= sin
sin 6
the difference in length of the
paths followed by the rays from the two to P.
Then
6.
similar separation of a
beam
of
X.
monochromatic
light
many divergent beams can be effected if the screen R has many slits instead of only two. Such a screen with many slits is called a diffraction grating. If a mixture of into
colors
is
passed through a diffraction grating, the smallest
angle through which each different color
with to
its
turned varies
wavelength. As a result, the colors are separated
form a spectrum on the screen
light
is
T where monochromatic
forms only a single bright band. The wavelength of
each color in the spectrum can be calculated from the
formula X
=d
sin 6,
where 6
the angle through which
is
that particular color was turned.
Light waves travel at a speed of 3 per second. This speed c.
the
number
of
10^^ centimeters
usually represented by the sym-
When monochromatic
bol
is
is
X
moves through
light
waves that pass a
fixed point in a second
called the frequency of the wave.
the frequency by the symbol
/.
space,
If /
We
shall designate
waves pass a point
wave has length X, then /X is the tance that the wave advances in a second. But the a second, and each
tance
and (38)
it
advances per second
c are related
is its
speed
c.
Therefore
by the equation /X
=
c.
71
in
disdis/, X,
Monochromatic either
its
light
can
be identified by specifying
its
frequency. It can also be iden-
wavelength or
wave number, the number of waves in 1 centimeter, usually designated by the Greek letter v (nu). Since the length of one wave is X, the number of waves in 1 centimeter is 1/X. P’rom equation (38) we find that l/X = //c. Consequently we have this formula for the wave number: by
tified
its
(39)
i;
=
= 4 Ac
i
or
cv
=
f.
Electromagnetic Waves
An
oscillating electric current produces electromagnetic
waves that radiate through space with the speed of light. Radio waves are examples of these electromagnetic waves. It is
now understood
radiation, differing
that light
is
also electromagnetic
from radio waves
in w^avelength
and
frequency. Electromagnetic waves from the shortest to wavelength
III 000(1)0
2
ooo
00
CM
I
(wavelength
is in
centimeters)
^
VO
the longest ones
known
are
now
classified into families in
order of increasing wavelength as follows:
gamma
rays,
X
rays, ultraviolet rays, visible light, infrared rays, micro-
waves, and radio waves. radiation
is
shown
The spectrum
in the
of electromagnetic
diagram on page
72.
73
IV Electricity in the
Electrons in the
Atom
A TELEVISION of a cathode ray tube, first
tube
Atom
picture tube
is
a
whose interesting
studied over a hundred years ago. is
a glass tube enclosing a gas
modern version pro})erties
A
were
cathode ray
whose pressure has
been reduced to below one thousandth of a millimeter.
There are two electrodes
in the
high voltage
is
The negative
the cathode.
The
attached.
If the voltage is
tube to which a source of
positive electrode
is
high enough, rays
rays emanate from the cathode.
electrode
is
called
called the anode.
known
When
as cathode
the rays strike
the glass wall of the tube opposite the cathode, the glass
glows with a fluorescent
Laboratory studies of
light.
cathode rays showed that they have the following properties:
I)
the rays normally travel in straight lines; 2)
they can be deflected by electrostatic fields; 3) their
behavior
is
fields or
independent of the chemical
composition of the cathode or the gas that in the tube. Properties 1)
magnetic
and
is
enclosed
2) can be explained
by
the assumption that the rays are streams of small charged particles.
by an
The
direction in which the rays are deflected
electrostatic or a
magnetic
field
requires that the
75
charges on the particles be negative. Property 3) indiall
matter.
now known assumption made to explain
as elec-
cates that these particles are constituents of
The
particles in a cathode ray are
So the basic
trons.
havior of cathode rays
may
be expressed
the be-
words:
in these
Every atom contains small negatively charged
particles
called electrons.
—e
Let
be the charge on a single electron. Let
m be
the
mass of the electron. We outline now the procedure by which these quantities have been measured.
Measuring e/m
The
first
step toward measuring e and
m
is
to
measure
was first done by Thomson in 1894. There are many ways of determining the ratio e/m. We shall outline one method in which the mathematical their ratio. This
reasoning
Step
I:
is
easy to follow.
In the next diagram, the horizontal line indi-
cates the path of
some
electrons in a cathode ray tube.
Two
wire grids are placed across the path at right angles
to
The
it.
grids are connected to a voltage source so that
the electrons pass through the negative grid before reaching the positive grid. Lender these conditions the voltage
between the grids accelerates the electrons as they pass.
moving slowly when
Suppose an electron
is
the grids so that
kinetic energy
its
grids boost the electron’s speed to a
is
76
The
approaches
almost zero:
magnitude
equation (20) on page 58 they give equal to imv‘^.
it
it
source of this energy
v,
If
the
then by
a kinetic energy is
the loss in po-
tential energy experienced
from the negative tial
by the electron when
to the positive grid.
energy per unit charge
is
ence, usually expressed as the
the charge on the electron
ergy
is
—eV. Since
is
The
loss in
poten-
called the potential differ-
number
—e,
of volts V. Since
its loss in
potential en-
the potential energy lost
the kinetic energy gained,
moves
it
is
equal to
we have
= -eV.
(40)
Multiplying by 2 and dividing by
e,
we
get
-v^=-2V.
(41)
e
Step
II.
After the electrons acquire the speed
stant magnetic field of strength
H
angles to the path of the electrons. field exerts a force
move
in a circle.
By
is
v,
a con-
placed at right
Then the magnetic
on each electron that compels
it
to
equation (16) on page 56, the force 77
H
V
directed to the center of the circle
e- H.
is
But
this
is
c
the centripetal force, which, according to equation (35) is
equal to mir/r, where
r
the radius of the
is
circle.
Equating these two expressions, we get V jT
e-
(42)
H —
—
mv' r
c
Dividing by ev and multiplying by
r,
we
get
m V = Hr — —
(43)
e
c
Dividing equation (41) by equation (43) we get
Since
-2cV Hr
V
(44)
c,
the speed of light,
is
known, and
T',
H, and
r
can
be measured in the experiment, equation (44) allows us to calculate the speed acquired
By equation
(43), 7n./e
=
by the electrons
(Hr/c)
V the value given in equation (44),
m — = Hr e
78
c
V
v.
we get
in
Step
1.
Substituting for
771 -
Hr
-2cV Hr Hr -2cV
• •
e
c
771
_
e
c
m _ 7 “ -2cW' Inverting both sides of the last equation,
we
get
^ -2cW
(45) 771
Since
c,
V,
H and r are all known, equation
(45) allows us
to calculate the ratio e/771.
Mass Varies with Speed Repeated experiments show that the ratio e/771
the one described above
like is
not constant but depends on
the speed of the electron. This variability of e/ 77 i
is
ac-
counted for by the theory of relativity which shows that the mass of a body depends on the observer. If rest,
and
m
is
7710 is
its
its
speed with respect to
the mass of an electron
mass when
it
then, according to the theory of
when
it is
at
moves with a speed v, relativity, m and 771q are
related as follows:
Substituting this value of solving for e/mo,
we
(45),
and then
finally get
e
(47)
m into equation -2cW
mo 79
from which the value of the ratio e/mo can be calculated. It is
found that
—
(48)
=
5.3
mo
X
10^^
esu per gram.
Measuring e In 1910, the magnitude of the electrostatic charge on an
measured
electron w^as
formed by Millikan. air
A
in
an ingenious experiment per-
tiny
oil
droplet was placed in the
space between the plates of a condenser, as shown in
The
the diagram.
voltage between the plates was adjusted
+
Mg
SO that the
upward push on the droplet exerted by the
electrostatic field
downward
O
between the plates just balanced the
pull of the weight of the droplet, so that the
droplet remained stationary. If let,
and g
downward
is
M
is
the mass of the drop-
the acceleration due to gravity, then the
pull, the
equation (14),
is
weight of the droplet, according to
given by
Mg.
If
E
is
the force the elec-
would exert on a unit charge, and q is the charge on the droplet, the upward push on the droplet is trostatic field
qE = Mg, and therefore q = Mg/E. The quantity E, known as the field strength between the plates, can be measured. The mass can be qE.
When
the forces balance,
M
computed from the speed with which the 80
oil
droplet falls
when
the electric current
ation due to pjravity,
is
turned
And
off.
g,
the acceler-
known. Therefore q can be computed. Repeated measurements on many droplets showed that there
is
is
a smallest value e that q
may
have, and that
other values q may have are whole number multiples of e. Assuming that the smallest possible charge e is the all
magnitude of the charge of a
ment
gives us the value of e
(49)
=
e.
4.80
single electron, the experi-
It is
X
found that
10“^® esu.
Computing mo Substituting this value of e into equation (48),
we
find
the value of the mass of an electron at rest: viq
(50)
=
(4.80
=
9.11
X 10“^°) (5.3 X X 10”^® gram.
Positive Charges in the
10^^)
gram
Atom
In 1886 Goldstein used a cathode ray tube in which holes or canals had been drilled through the cathode.
He
found that while the negative cathode rays were flowing
from the cathode (toward the anode), other positive rays were flowing through the canals of the cathode posite direction. These are
known
in the op-
as canal rays. In other
was found, too, that positive rays sometimes flowed from the anode of a cathode ray tube. These are known as anode rays. The existence of these positive rays can be explained by assuming, as Thomson did, that experiments
it
there are positive charges as well as negatively charged electrons in every atom. Ordinarily, every positive charge of
magnitude
e in
an atom
is
balanced by a negative 81
charge of
Under is
—e
on an electron
these conditions, as
we saw on page
atom
55, the
However, a neutral atom can accharge in several ways. For example,
one or more electrons are added to
it
it. it
acquires a nega-
one or more electrons are removed from
tive charge. If it.
atom, and vice versa.
electrically neutral.
quire an electrostatic if
in the
acquires a positive charge.
atoms that has a charge producing ions
is
is
called
An atom or an ion. The
called ionization.
cluster of
process of
Using these concepts,
the canal rays and the anode rays are explained in this
way: As the electrons of the cathode rays stream through the tube they occasionally collide with molecules of the
gas in the tube. Sometimes a collision to tear
is
violent enough
an electron out of a molecule and ionize
resulting positive ion
cathode.
Some
is
it.
attracted toward the negative
of the ions hit the cathode
Others pass through the canals
in the
and
stick to
may
tear electrons out of
in the anode. Positive ions torn
it.
cathode and form
bom-
the canal rays. Electrons in the cathode rays also
bard the anode. Here they
The
atoms
out of the anode become
the source of anode rays.
There are various ways of measuring the charge and the
The charge is always found to be an integral multiple of e. The mass is always found to agree with the known chemical composition of the ion. The ion mass of an
ion.
with smallest mass and smallest positive charge
hydrogen
ion,
with a charge of
is
a
e.
Electrolytes
Additional evidence, produced by both chemists and physicists, supported the
82
assumption that there are
elec-
the atom. There are some chemical
trical particles inside
compounds, which, when dissolved
in water,
produce solu-
Such com-
tions capable of conducting an electric current.
pounds are known as chloride
is
electrolytes.
an electrolyte.
If
For example, hydrogen
two electrodes connected
to
a source of direct current are inserted into a solution of
hydrogen chloride, a current flows through the solution. Moreover, while the current
flows, the
hydrogen chloride
decomposed by it into hydrogen gas and chlorine gas. The hydrogen gas appears in bubbles on the cathode, and is
The
the chlorine gas appears in bubbles on the anode. process of decomposing an electrolyte by electric current
is
of an
called electrolysis. In the electrolysis
of copper bromide, metallic copper
cathode, while bromine
To
means
is
deposited on the
is
deposited on the anode.
explain the behavior of electrolytes, Arrhenius pro-
posed in 1887 the theory that an electrolyte in solution dissociates into both positive
up enough
molecules.
ions.
The
posi-
toward the cathode where they
tive ions are attracted
pick
and negative
electrons to
The negative
become neutral atoms
or
ions are attracted toward the
anode where they give up enough electrons to become neutral atoms or molecules.
The charge on an
ion
counted for by a shortage of electrons in a positive
and an excess of electrons
in a
negative ion.
electron
chlorine
is
A
hydrogen atom that
denoted by
atom that has one
by Cl~ and
also has a valence of
has a shortage of two electrons
has a valence of
is
is
1.
A
many
is
the
short one
and has a valence of electron too
ion,
The number
of electrons short or in excess in an atomic ion
valence of the ion.
ac-
is
is
1.
A
denoted
copper atom that
denoted by Cu^'*’, and
2.
83
Gram-atom In experiments with electrolysis Faraday discovered a relationship between the quantity of charge transported
through an electrolytic solution and the quantity of an
element deposited on an electrode. Before stating what this relationship
it is
is,
necessary to introduce the con-
cept of a gram-atom of an element. It cept of standard
is
related to the con-
volume described on page
standard volume of 22.4
of a
liters
The
34.
compound
in
the
gaseous state has the property that, at standard tempera-
number
ture and pressure, the
that
contains
it
of
grams of the compound
number of amu in the the compound. This same number of
equal to the
is
molecular weight of
grams of the compound, whether it is in the gaseous, liquid or solid state, is called a gram-molecule of the compound. Analogously, that weight of an element that contains as
many grams
of the element as there are
the atomic weight of the element
is
called a
of
hydrogen
is
1.008
gen. Since the atomic weight of chlorine
gram-atom
We
of chlorine
is
in
gram-atom.
For example, since the atomic weight of hydrogen
amu, a gram-atom
amu is
1.008
grams of hydrois
35.457 amu, a
35.457 grams of chlorine.
have already seen that a standard volume of 22.4
liters of a
gas at standard temperature and pressure al-
ways contains the same number of molecules, namely. No molecules, where No is Avogadro’s number. It follows,
compound contains No now show that, similarly,
then, that a gram-molecule of a
molecules of the compound.
We
a gram-atom of an element contains
No
atoms.
Let the atomic weight of an element be
84
w
amu. Then
the weight of a gram-atom of the element
is
w
grams. Let
N be the number of atoms in a gram-atom of the element. Since each atom weighs w amu, then N atoms weigh Nw amu. Consequently, w grams = Nw amu = Niv (1 amu). We saw on page 38 that amu = l/No gram. 1
Substituting this value for
w grams =
1
amu, we
find that
Niv
we find that N grams = No grams, that is, that N = No. The number of atoms in a gram-atom of an element is Avogadro's numMultiplying by iVo and dividing by
iv,
ber.
Law
Faraday’s
Faraday discovered of charge
mass
is
in
1833 that
when
a fixed
passed through an electrolytic solution, the
of an element deposited on an electrode
tional
to
amount
the equivalent weight, which
is
is
propor-
the atomic
weight divided by the valence. In particular there certain definite
amount
is
a
of charge needed to deposit a
weight of an element equal to a gram-atom divided by the valence. This definite constant. It
is
amount
denoted by
F and
is
known
as Faraday’s
has the value 2.90
X
esu.
The
fact observed
by Faraday
is
easily explained
basis of the ionic theory of electrolytes.
on the
Suppose an
ele-
ment deposited on the cathode during electrolysis has atomic weight w. Then a gram-atom of the element contains IV grams of the element, and, as we have seen, consists of A’o
atoms. Suppose a positive ion of the element
85
.
lacks
ion
n
electrons.
is n.
w — grams n
Since
Then
w grams
the valence of the element in this
No
of the element contains
of the element contain
No/n atoms. To
atoms,
convert
one ion into a neutral atom deposited on the cathode, n electrons are needed (one for each electron
convert iVo/n ions into
X
(No/n)
n
the same
lacks).
it
number
electrons are needed. So,
No
of
To
atoms,
electrons are
needed to deposit a weight of w/n grams of an element
whose valence is 7i. Then the charge F measured by Faraday is the magnitude of the charge of No electrons, that is,
F =
(51)
where is
eNo,
magnitude of the charge of one a constant because e and No are constants. e is the
Avogadro's If
we
electron.
F
Number
divide both sides of equation (51) by
No =
(52)
e,
we
get
-> e
from which Avogadro’s number can be computed. Since F = 2.9 X 10^^ esu, and e = 4.8 X 10“^^ esu, (53)
No =
^ iQ-io
=
6
X
approximately.
Radioactivity
More evidence
that there are electrical particles in the
atom was provided by the phenomenon of radioactivity Atoms of uranium, radium and thorium spontaneously 86
release rays.
Three
distinct kinds of rays
and given the names alpha rays.
The alpha
each of which
were
rays, beta rays,
and
gamma
rays were found to be streams of particles
is
a helium ion, with a charge of 2e. These
The beta
particles are called alpha particles.
found to be streams of particles each of which tron.
identified,
The gamma
rays were is
an elec-
rays were found to be electromagnetic
radiation of very high frequency.
When
an atom of a
radioactive element releases either an alpha particle or a
beta particle the atom
is
transformed into an atom of a
different element.
A
Building Block for Atoms
The atomic weights
of
some elements are very nearly
whole numbers of amu. This partial
list
is
shown, for example,
in the
of elements printed below:
Element
Atomic weight (number of amu)
Hydrogen Helium
4.003
Beryllium
9.013
1.008
Carbon
12.010
N itrogen
14.008
Oxygen
16.000
Fluorine
19.00
This fact suggested to Prout
in
1815 the idea that the
smallest atom, hydrogen, with atomic weight of about
1
amu, might be the building block out of which all other atoms are made. According to Front’s hypothesis, a he87
amu
lium atom has an atomic weight of about 4 it
contains 4 hydrogen atoms; a carbon
atomic weight of about 12
amu
because
atom has an it
contains 12
Front’s hypothesis was not widely
hydrogen atoms;
etc.
supported at
because there are
first
because
many
elements whose
atomic weights are not close to a whole number of amu.
For example, the atomic weight of chlorine
However, a
later discovery
modern theory
35.457.
gave strong support to Front’s
hypothesis, and in a modified form
the
is
it
has become part of
of the atom.
Chemical Twins
when
was discovered by Thomson, Aston and others that the atoms of some Front’s hypothesis was revived
elements are not
all identical.
it
Chlorine, for example,
is
a
mixture of two different kinds of atoms that are the same in
chemical behavior but have different atomic
their
Moreover the atomic weights of these different kinds of chlorine atoms are almost whole numbers of amu, namely 35 amu and 37 amu. Atoms that are chemically the same but have different weights are called isotopes. By now, every element has been found to have two or more isotopes, and the atomic weight of every isotope of
weights.
an element has been found to be nearly a whole number of
amu.
It is clear
then that the whole number nearest
to the atomic weight of an isotope of
an element must
have some physical meaning. The whole number nearest to the atomic weight of an isotope is called its mass number,
and
is
designated by the symbol A.
Since every element
is
a mixture of isotopes, the atomic
weight of an element depends on
88
how much
of each iso-
tope
in the
is
atoms have an atomic weight ing
25%
75%
mixture. For example,
amu, and the remainhave an atomic weight of 37 amu. Consequently of 35
the mixture has an atomic weight of (.75 (.25
X
of all chlorine
37 amu)
X
35
amu)
+
= 35.5 amu, approximately. All fractional
atomic weights can be explained
Tlie Nucleus of the
in
the same way.
Atom
In 1911 Rutherford tried to get information about the structure of atoms by throwing small fast-moving parti-
Fluorescent
Gold
\ '
foil
Radium
Schematic diagram of the experiment
in
which
Rutherford discovered the nucleus cles at
them. The atoms he bombarded were atoms of
gold in a thin sheet of gold
them were alpha
foil.
particles released
integration of radium.
Some
Others were deflected as
The
if
particles
he threw at
by the radioactive
dis-
of the alpha particles passed
right through the gold foil as
positive charge.
The
if
there were nothing there.
they had passed close to a
detailed results of the experiment
could be explained by the assumption that each atom consists of
a central core or nucleus with a positive charge,
89
surrounded by enough electrons to make the total charge
on the atom equal
to zero.
Deflection of alpha particles by positive charge of nucleus
From cles
the paths followed by the deflected alpha parti-
Rutherford calculated the deflecting force exerted by
The force atomic number
the nucleus on an alpha particle at a distance
where Z
turned out to be
and
of the atom,
equation (49).
By
e
is
\{
Q-=
charge 2e,
amount
the
of charge given
equation (15) on page 56, this
cisely the force exerted q,
the
is
r.
by
a charge
Q
is
by
pre-
on another charge
and q — 2e. Since an alpha particle has Rutherford concluded that the charge Ze must
7je,
be the charge on the nucleus.
To balance
this charge,
must be Z electrons surrounding the nucleus. Later experiments by Moseley, in which he examined the X rays produced when different elements were bombarded by cathode rays, confirmed Rutherford’s discovery that the atomic number Z of an atom is a measure of the there
charge on
90
its
nucleus, using e as the unit of charge. This
discovery casts
new
light
on the order of the elements
the periodic table of the elements.
ment
in the table
of an
atom
The
in
place of an ele-
depends on the charge on the nucleus
of the element, -rather than its atomic weight.
Particles in the Nucleus
Nearly
all
the mass of an
atom
is
in the nucleus of the
atom. Since the mass of an isotope of an atom
number a whole number a whole
is
nearly
amu, the mass of its nucleus is nearly of amu. This fact leads to a modification of Trout’s hypothesis. Trout assumed that an atom whose mass number is A contains A hydrogen atoms, each with a mass of 1 amu. The modified version of Trout’s hyof
pothesis asserts instead that the nucleus of an
atom with
mass number A contains A hydrogen nuclei, each with a mass of 1 amu. The hydrogen nucleus with mass 1 amu is called a 'proton. The charge on a proton is e. Consequently A protons have a total charge of Ae. However, the charge on a nucleus is Ze, and for all atoms except the hydrogen atom with mass number 1 the atomic number Z is less than the mass number A. For example, in the most common isotope of helium, while the mass number is 4, the atomic number is only 2. This fact makes it necessary to modify Trout’s hypothesis even further.
The Proton-Electron Theory
of the Nucleus
The discrepancy between the mass number A and the atomic number Z of an atom could be accounted for if the nucleus contained not only protons, but also electrons 91
that balanced the charge of
ample,
if
we assume
of the protons.
For ex-
that a helium nucleus contains 4
protons and 2 electrons, so its atomic
some
its
charge would be 4e
number would be
electrons contribute a negligible
2.
—
2c
= 2c,
Moreover, since the 2
amount
of
mass
to the
mass would be about 4 amu (the mass of the 4 protons), and hence its mass number would be 4. In general, if an atom has mass number A and atomic number Z, we could account for the discrepancy between A nucleus,
its
and Z by assuming that the nucleus contains A protons and enough electrons
to neutralize the charge of all
but
Z
of the protons.
These considerations
a nucleus with
mass number A and atomic number Z con-
sists of
A
protons and
A
led to the theory that
— Z electrons.
The theory broke down, however, when
it
was discov-
ered that the nuclei of atoms are spinning like tops and
hence have angular momentum.
The
proton-electron
theory of the nucleus turned out to be inconsistent with
measurements made of nuclear angular momentum.*
Discovery of the Neutron
A
different
came ments
and better version of Front’s hypothesis be-
possible with the discovery of the neutron. Experiin
which alpha particles were thrown at atoms of
beryllium or boron caused the bombarded atoms to eject
was thought at first that However, in 1932, Chadwick
electrically neutral particles. It
these were
gamma
rays.
showed that they were *
For
particles that
details, see Inside the
Day Company, New York,
92
1963.
had the same mass
Nucleus, by the same author,
The John
:
as protons,
namely about
1
amu
each. Because they are
electrically neutral, these particles are
The new and this
form:
number
If a
known
final version of Front’s
hypothesis takes
nucleus has mass number
Z
Z, the nucleus consists of
as neutrons.
A and
protons and
atomic
A
—Z
The Z protons account for the atomic number Z. The total number of j^articles of mass 1 amu is A, and this accounts for the mass number A. It is found, too, that this theory fits the known facts about nuclear angular momentum. neutrons.
A Model The
Atom
of the
and theories outlined
facts
culminated
preceding pages
in the
in the construction of the following theoretical
model of an atom
Every atom
tary electrons. If the
atomic number
its
by planethe atom is A and
consists of a nucleus surrounded
is
mass number
of
Z, the nucleus contains
A
particles
amu each. Z of them are protons, and the remaining A — Z oi them are neutrons. The charge on the nucleus is Ze, the charge of the Z protons. The number of of
mass about
1
electrons that surround the nucleus
number This
is
the same as the
of protons that are in the nucleus. is
only a crude model, because while
how many
it tells
electrons surround a nucleus in an atom,
us it
how
these electrons are arranged. In the
chapters that follow
we show how the model has been
does not
tell
us
refined in order to give precise information about the
arrangement of the electrons around the nucleus.
93
I
i
!
J
1
V Atoms of Light THE
next steps in the development of the theory
of atomic structure tion of
were based on studying the interac-
atoms and electromagnetic radiation. The form
that the theory was compelled to take was determined in large part
by the discovery that electromagnetic radianot continuous, but
tion, like matter, is
discrete units. In this chapter
we
made up
is
of
outline the experimental
data and the theoretical ideas that converged toward this discovery.
Black Body Radiation In the theory of heat, a body that absorbs
tromagnetic radiant energy that
falls
on
black body. Every body whose temperature
all
the elec-
is
called a
it is
above zero
degrees Kelvin radiates energy into space. In the case of a black body, the tion
among
amount
of radiation
different frequencies
perature of the body.
A
black body: Radiation that
falls
outside enters the cavity and
between the walls
of the
it
is
its
distribu-
depend only on the tem-
box that
small opening into the cavity
and
is
closed except for one
encloses behaves like a
on the opening from the reflected
box until
it is
back and forth completely ab95
sorbed. Consequently any radiation that emerges from
the opening has the characteristics of black body radia-
With the help
thermodynamic theory and the electromagnetic theory of light, two laws governing black body radiation can be derived. One, known as Stefan’s law, asserts that if the Kelvin temperature of the body is T, the total amount of energy the body radiates is protion.
portional to T^.
The
of
other law,
known
as Wien’s law,
asserts that the formula for Uf, the rate per unit of fre-
quency at which
this total
is
shared
among
different fre-
quencies, takes the form
(54)
stands for a function of is,
f
that
a variable whose value depends on the value of the
this function
Planck’s
is.
Quantum Hypothesis
In 1900 Planck undertook to identify the function that occurs in Wien’s law.
To do
so,
F
he had to make some
assumptions about the mechanism that produces the radiation from a black body. According to thermody-
namic theory, the function F should be the same no matter what the mechanism
is,
so he naturally used the
simplest assumptions that he could. oscillator that electric
He assumed
produces radiation with frequency
charge vibrating with frequency
96
that the
/.
On
/ is
an
the basis of
assumption he could show that Uf is related to E, the average energy of the oscillator, by the formula this
(55)
=
Uf
^
E.
assumed that the energy radiated by divisible into arbitrarily small amounts,
addition, he
If, in
the oscillator
is
he could show that
E
is
related to the temperature
T by
the formula,
E =
(56)
where k
is
known
a constant
Substituting this value of
(57)
kT,
Uf
E
as Boltzmann’s constant.
into equation (55) gives
^
=
kT.
Multiplying the right hand side of this equation by j
(which in the
is
equal to
1),
we
find that
can also be written
it
form Uf
(58)
-M'l. = k I y =P X f
Sirk ,3
//
Comparing equation (58) with equation this result agrees with
is
a function of
imental
fact.
frequency
(54),
we
see that
Wien’s law, because Swk/
However,
it
does not agree with exper-
Equation (57) implies that the higher the
/ is,
the higher the value of Uf
is.
The
fact
is,
however, that as we examine higher and higher frequencies,
Uf at first increases to a
maximum, and then
creases again.
97
de-
Since the assumptions he used led to a formula that is
wrong, Planck re-examined his assumptions to see
what changes he should make that
is
right.
One
He
is
formula
was that the energy
of the assumptions
radiated by the oscillator
amounts.
in order to derive a
divisible into arbitrarily small
replaced this assumption by the opposite
assumption that the energy
is
radiated only in whole
number multiples of an indivisible small amount called a quantum. Under this assumption, if the quantum of energy is represented by the symbol e (epsilon), he showed that equation (56) is replaced by
E =
(59)
where of
e is
approximately 2.718. Substituting this value
E into equation
(55) gives
U;
(60)
1
=
-^X -
1
According to Wien’s law, the temperature into the formula only in the ratio tion
f
T
So, to
(60) have the form prescribed
should enter
make equa-
by equation
(54),
Planck made the assumption that €
(61)
where
/i
is
=
hf,
a constant. Substituting this value of
equation (60) yields the formula
Uf
(62)
=
98
1
e
into
in
which
Uf
is
seen to be the product of
f
and a function
of j/T, as prescribed
by Wien’s law. Equation (62) is known as Planck's law. It is fully confirmed by the facts of experiment. The constant h is known as Planck's constant, and has been found to have the value 6.62 X 10“^^ erg second.
The Photon Planck merely assumed that the quantum of energy hf
was the smallest amount
with frequency
of electromagnetic energy
that can be radiated. In 1905, Einstein
/
quantum idea one step further by assuming quantum hf was the smallest amount of electro-
carried the
that the
magnetic energy of frequency fact a
beam
/
that can exist,
of light of frequency /
visible corpuscles each of
is
and that
in
a stream of indi-
which contains an amount of
These corpuscles are called photons.
energy equal to
hf.
By making
assumption he was able to explain the
this
hitherto unexplained details of the photoelectric effect. If ultraviolet light is
kept
in a
vacuum
allowed to
move
to
on a metal surface
tube, the light knocks electrons out of
the surface. If a voltage
trons
fall
is
applied to the tube, the elec-
produce an
electric
current.
possible to measure the current, which
the to
number
of electrons
knocked
is
out. It
It
is
then
proportional to is
also possible
measure the speed of the electrons when they are
ejected from the metal surface. It turns out that increas-
ing the intensity of the light increases the
number
of elec-
trons knocked out, but does not increase the speed of the electrons.
In fact the kinetic energy of each electron
knocked out depends only on the frequency
of the light,
99
and
is
given by the formula hf
—
A, where
a constant
.4 is
whose value depends on the metal that was
irradiated.
beam
of light of
If
we assume
frequency then
/ is
these
as Einstein did that a
a stream of photons each with energy hf, are
facts
easily
An
explained.
knocked out of the surface by the
light
when
electron a
photon of
light collides with the electron. In this collision, the
ton delivers
of its energy, an
all
the electron. Part of this energy,
pho-
amount equal to hf, to an amount equal to A,
used up to overcome the force holding the electron in
is
The
place in the metal. kinetic energy.
its is
is
increased,
balance, hf
When
—
A, gives the electron
the intensity of the
more photons
beam
of light
and more
strike the metal,
elec-
trons are knocked out.
The theory
that light
made up
is
firmed in an experiment with ton in 1922.
X
photons
of
is
con-
X rays performed by Comp-
rays passed through a block of paraffin
are scattered by the electrons in the paraffin. According
wave theory of light, if radiation with frequency / is scattered in any direction, the scattered radiation should also have frequency /. However, Compton’s experiment showed that if the radiation is turned aside to the
through an angle that
is less
tion has a frequency that easily explained
than 90°, the scattered radia-
is
lower than
by the photon theory.
This result
/.
When
with energy hf collides with an electron,
it
of its energy on to the electron. Therefore the
leaves the scene of the collision has an J}f'
h,
that
we
is
less
than
find that
scattered photon
photon.
100
hf.
/'