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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,

/

I

"

Diagrams by Ruth Adler

^

V

\\

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).

*

«

-I I

w

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.

/'