quantum mechanics part - university of colorado boulder€¦ · quantum mechanics chapter 3 atoms...

QUANTUM MECHANICS

Chapter 3 AtomsChapter 4 Quantization of LightChapter 5 Quantization of Atomic Energy LevelsChapter 6 Matter WavesChapter 7 The Schrödinger Equation in One DimensionChapter 8 The Three-Dimensional Schrödinger EquationChapter 9 Electron SpinChapter 10 Multielectron Atoms: The Pauli Principle and the Periodic TableChapter 11 Atomic Transitions and Radiation

In Part II (Chapters 3 to 11) we describe the second of the two theories thattransformed twentieth-century physics — quantum theory. Just as relativitycan be roughly characterized as the study of phenomena involving high speeds

so quantum theory can be described as the study of phenomenainvolving small objects — generally of atomic size or smaller.

We begin our account of quantum theory in Chapter 3, with a descriptivesurvey of the main properties of atoms (their size, mass, constituents, and so on).In Chapters 4 and 5 we describe some puzzling properties of microscopic sys-tems that began to emerge in the late nineteenth century — some properties oflight in Chapter 4, and of atoms in Chapter 5.All of these puzzles pointed up theneed for a new mechanics — quantum mechanics, as we now say — to replaceclassical mechanics in the treatment of microscopic systems. In Chapters 6 to 9we describe the basic ideas of the new quantum mechanics, which began to de-velop around 1900 and was nearly complete by 1930. In particular, we introducethe Schrödinger equation, which is the basic equation of quantum mechanics,just as Newton’s second law is the basic equation of Newtonian mechanics.

Armed with the Schrödinger equation, we can calculate most of the im-portant properties of the simplest of all atoms, the hydrogen atom (with its oneelectron), as we describe in Chapter 8. In Chapter 9 we introduce one moreimportant idea of the new quantum mechanics — the electron’s spin angularmomentum. Then in Chapters 10 and 11 we are ready to apply all these ideasto the general multielectron atom. We will see that quantum theory gives a re-markably complete account of all the 100 or so known different atoms, andthat, in principle at least, it explains all of chemistry.

Although quantum mechanics was first developed to explain the propertiesof atoms, it has also been applied with extraordinary success to other systems,some larger than atoms, such as molecules and solids, and some smaller, such assubatomic particles.We will describe these other applications in Parts III and IV.

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PARTI I

85

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86

C h a p t e r 3Atoms

3.1 Introduction3.2 Elements, Atoms, and Molecules3.3 Electrons, Protons, and Neutrons3.4 Some Atomic Parameters3.5 The Atomic Mass Unit3.6 Avogadro’s Number and the Mole3.7 Kinetic Theory3.8 The Mean Free Path and Diffusion3.9 Brownian Motion3.10 Thomson’s Discovery of the Electron�

3.11 Millikan’s Oil-Drop Experiment�

3.12 Rutherford and the Nuclear Atom�

3.13 Derivation of Rutherford’s Formula�

Problems for Chapter 3�These sections can be omitted without serious loss of continuity.

3.1 Introduction

Quantum physics is primarily the physics of microscopic systems. Historically,the most important such system was the atom, and even today the greatest tri-umph of quantum theory is the complete and accurate account that it gives usof atomic properties. For the next several chapters we will be discussing atoms,and in this chapter, therefore, we give a brief description of the atom and itsconstituents, the electron, proton, and neutron, together with some of theevidence for these entities.

3.2 Elements, Atoms, and Molecules

The concept of the atom arose in scholars’ search to identify the basic con-stituents of matter. More than 2000 years ago Greek philosophers had recog-nized that this search requires answers to two questions: First, among thethousands of different substances we find around us — sand, air, water, soil,gold, diamonds — how many are the basic substances, or elements, from whichall the rest are formed? Second, if one took a sample of one of these elementsand subdivided it over and over again, could this process of repeated subdivi-sion go on forever, or would one eventually arrive at some smallest, indivisibleunit, or atom (from a Greek word meaning “indivisible”)? With almost no ex-perimental data, the ancient Greeks could not find satisfactory answers tothese questions; but this in no way lessens their achievement in identifying theright questions to ask.

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John Dalton(1766–1844, English chemist)

The son of an English weaver, Dal-ton left school at age eleven. Ayear later he became a school-teacher, and this led to his interestin science. In his book, New Systemof Chemical Philosophy, Dalton ar-gued that the chemists’ law of def-inite proportions is evidence forthe existence of atoms. Using thisidea, he was the first to draw up atable of relative atomic masses. Hehad many other interests, includingmeteorology and color blindness.(He was himself color-blind —which can’t have helped his chemi-cal research.) He was a Quakerand extremely modest. In 1810 hedeclined membership in the RoyalSociety, although he was subse-quently elected without his priorconsent.

Section 3.2 • Elements, Atoms, and Molecules 87

TABLE 3.1A few of the elements with their chemical symbols. For a complete list of the elements,see the periodic table inside the back cover or the alphabetical list in Appendix C.

Hydrogen (H) Oxygen (O) Iron (Fe)Helium (He) Sodium (Na) Nitrogen (N)Carbon (C) Aluminum (Al) Lead (Pb)Copper (Cu) Chlorine (Cl) Uranium (U)

Serious experimental efforts to identify the elements began in the eigh-teenth century with the work of Lavoisier, Priestley, and other chemists. By theend of the nineteenth century, about 80 of the elements had been correctlyidentified, including all of the examples listed in Table 3.1.Today we know thatthere are 90 elements that occur naturally on earth. All of the matter on earthis made up from these 90 elements, occasionally in the form of a pure element,but usually as a chemical compound of two or more elements, or as a mixtureof such compounds.

*A few elements with half-lives much less than the earth’s age are observed to occurnaturally, but this is because some natural nuclear process (radioactivity or cosmic-raycollisions, for example) is creating a fresh supply all the time.†In stating Dalton’s arguments we have replaced his terminology and measured masseswith their modern counterparts. In particular, Dalton had measured the ratio of the Cand O masses to be not as we now know it.3 : 4,5 : 7,

In addition to the 90 naturally occurring elements, there are about twodozen elements that can be created artificially in nuclear reactions.All of theseartificial elements are unstable and disintegrate with half-lives much less thanthe age of the earth; this means that even if any of them were present when theearth was formed, they have long since decayed and are not found naturally inappreciable amounts.*

The evidence that the elements are composed of characteristic smallestunits, or atoms, began to emerge about the year 1800. Chemists discovered thelaw of definite proportions:When two elements combine to form a pure chem-ical compound, they always combine in a definite proportion by mass. For ex-ample, when carbon (C) and oxygen (O) combine to form carbon monoxide(CO), they do so in the proportion three grams of C combine with four ofO to form seven grams of CO.

If we were to add some extra carbon, we would not get any additional CO; rather,we would get the same 7 g of CO, with all the extra carbon remaining unreacted.

The law of definite proportions was correctly interpreted by the Englishchemist John Dalton as evidence for the existence of atoms. Dalton arguedthat if we assume that carbon and oxygen are composed of atoms whose mass-es are in the ratio† and if CO is the result of an exact pairing of theseatoms (one atom of C paired with each atom of O), the law of definite propor-tions would immediately follow: The total masses of C and O that combine toform CO would be in the same ratio as the masses of the individual atoms,namely If we add extra C (without any additional O), the extra C atomshave no O atoms with which to pair; thus we get no additional CO and theextra C remains unreacted.

3 : 4.

3 : 4,

(3 g of C) + (4 g of O) : (7 g of CO)

3 : 4;

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88 Chapter 3 • Atoms

We now know that this interpretation of the law of definite proportionswas exactly correct. However, its general acceptance took 60 years or more,mainly because the situation was considerably more complicated than our sin-gle example suggests. For instance, carbon and oxygen can also combine in theratio to form carbon dioxide, Dalton was aware of this particularcomplication and argued (correctly) that the carbon atom must be able tocombine with one O atom (to form CO) or two O atoms (to form ). Nev-ertheless, this kind of ambiguity clouded the issue for many years.

A stable group of atoms, such as CO or is called a molecule. A fewexamples of molecules are listed in Table 3.2, where we follow the conventionthat a subscript on any symbol indicates the number of atoms of that kind (andsingle atoms carry no subscript — thus has two H atoms and one Oatom). As the list shows, a molecule may contain two or more different atoms,for example, the CO and molecules, in which case we say that the twoor more elements have combined to form a compound, of which the moleculeis the smallest unit. Molecules can also contain atoms of just one kind, for ex-ample, In this case, we do not speak of a new compound; we say simplythat the element normally occurs (or sometimes occurs) not as separate atoms,but as groups of atoms clustered together, that is, as molecules.* Molecules cancontain small numbers of atoms, like the examples in Table 3.2; but certainorganic molecules, such as proteins, can contain tens of thousands of atoms.

O2.

C6H12O6

H2O

CO2,

CO2

CO2.3 : 8

*This possibility was probably the source of greatest confusion for the nineteenth-century chemists; if one wrongly identified the molecule as an atom, then one’s in-terpretation of all other molecules containing oxygen — CO, and so on — wouldalso be incorrect.

CO2,O2

Although most of the credit for establishing the existence of atoms andmolecules goes to the chemists, a second strand of evidence came from the ki-netic theory of gases — which we would regard today as a part of physics, andwhich we discuss in sections 3.7 to 3.9.This theory assumes, correctly, that a gasconsists of many tiny molecules in rapid motion. By applying Newton’s laws ofmotion to the molecules, one could explain several important properties ofgases (Boyle’s law, viscosity, Brownian motion, and more). These successesgave strong support to the atomic and molecular hypotheses. In addition, ki-netic theory, unlike the chemical line of reasoning, gave information on the ac-tual size and mass of molecules. (The law of definite proportions implied thatthe C and O atoms have masses in the proportion but gave no clue as tothe actual magnitude of either.) Kinetic theory gave, for example, an expres-sion for the viscosity of a gas in terms of the size of the individual molecules.Thus, measurement of viscosity allowed one to determine the actual size of themolecules — a method first exploited by Loschmidt in 1885.

By the beginning of the twentieth century it was fairly generally accept-ed that all matter was made up of elements, the smallest units of which were

3 : 4

TABLE 3.2A few simple molecules.

Carbon monoxide, CO Oxygen, Glucose,Water, Nitrogen, Ethyl alcohol,Ammonia, Sulfur, Urea, CON2H4S8NH3

C2H6ON2H2OC6H12O6O2

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Joseph JohnThomson(1856–1940, English)

As head of the famous CavendishLaboratory in Cambridge for 35years, J. J. Thomson was one of themost influential figures in the de-velopment of modern physics.Seven of his students and assistantswent on to win the Nobel Prize,and he won it himself in 1906 forhis discovery of the electron.

Section 3.3 • Electrons, Protons, and Neutrons 89

atoms. Atoms could group together into molecules, the formation of which ex-plained the chemical compounds. The relative masses of many atoms and mol-ecules were known quite accurately, and there were already reasonablyreliable estimates of their actual masses.

3.3 Electrons, Protons, and Neutrons

Our story so far has the atom — true to its name — as the indivisible smallestunit of matter, and until the late nineteenth century there was, in fact, no directevidence that atoms could be subdivided. The discovery of the first subatomicparticle, the negatively charged electron, is generally attributed to J. J.Thomson (1897), whose experiments we describe briefly in Section 3.10. Onthe basis of his experiments,Thomson argued that electrons must be containedinside atoms, and hence that the atom is in fact divisible.

Fourteen years later (1911) Ernest Rutherford argued convincingly forthe now familiar picture of the atom as a tiny planetary system, in which thenegative electrons move far outside a positive nucleus. Rutherford’s conclu-sion was based on an experiment that was the forerunner of many modern ex-periments in atomic and subatomic physics. In these experiments, calledscattering experiments, one fires a subatomic projectile, such as an electron, atan atom or nucleus. By observing how the projectile is deflected, or scattered,one can deduce the properties of the target atom or nucleus. In the Rutherfordexperiment, the projectiles were alpha particles — positively charged, sub-atomic particles ejected by certain radioactive substances.When these were di-rected at a thin metal foil, Rutherford found that almost all of them passedstraight through, but that a few were deflected through large angles. In termsof his planetary model, Rutherford argued that the great majority of alphaparticles never came close to any nuclei and encountered only a few electrons,which were too light to deflect them appreciably. On the other hand, a few ofthe alpha particles would pass close to a nucleus and would be deflected by thestrong electrostatic force between the alpha particle and the nucleus. These,Rutherford argued, were the alpha particles that scattered through large an-gles. As we will describe in Section 3.12, Rutherford used his model to predictthe number of alpha particles that should be scattered as a function of scatter-ing angle and energy. The beautiful agreement between Rutherford’s predic-tions and the experimental observations was strong evidence for the nuclearmodel of the atom, with its electrons far outside a tiny heavy nucleus.

Within another eight years (1919) Rutherford had shown that the atom-ic nucleus can itself be subdivided, by establishing that nuclear collisions canbreak up a nitrogen nucleus. He identified one of the ejected particles as a hy-drogen nucleus, for which he proposed the new name proton (“first one”) inhonor of its role in other nuclei. In 1932 Chadwick showed that nuclei containa second kind of particle, the neutral neutron. (The experiments that identifiedthe proton and neutron as constituents of nuclei will be described inChapter 17.) With this discovery, the modern picture of the constituents ofatoms was complete. Every atom contains a definite number of electrons, eachwith charge in orbit around a central nucleus; and nuclei consist of twokinds of nuclear particles, or nucleons, the proton, with charge and theuncharged neutron.

The constituents of five common atoms are shown in Table 3.3. In everycase the number of electrons is equal to the number of protons, reflecting thatthe atom is neutral (in its normal state). Hydrogen is the only atom that has no

+e,-e,

1a2

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TABLE 3.3The constituents of five representative atoms.

Nucleons

Atom Electrons Protons Neutrons

Hydrogen, H 1 1 0Helium, He 2 2 2Carbon, C 6 6 6Iron, Fe 26 26 30Uranium, U 92 92 146

*As we will discuss in Chapter 18, it has been established that certain “sub-subatomic”particles, called quarks, have charge or However, quarks are never foundin isolation; they always occur as a member of a bound collection of quarks with totalcharge 0 or an integer multiple of e.

;2e>3.;e>3

neutrons. In all other atoms the numbers of neutrons and protons are roughlyequal. (We will see the reason for this in Chapter 16.) In many lighter atoms(helium or carbon, for example) the two numbers are exactly the same, whilein most medium atoms the number of neutrons is a little larger; in the heaviestatoms there are about 50% more neutrons than protons.

In addition to their role as the building blocks of atoms, the electron andproton also define the smallest observed unit of charge. Their charges areexactly equal and opposite, with magnitude e,

where “C” stands for coulomb. No charge smaller than e has been detected,and all known observed charges are integral multiples of e,*

Because e is so small, typical macroscopic charges are very large multiples of e,and the restriction to integral multiples is usually unimportant. On the atomiclevel, the existence of a smallest unit of charge is obviously very important.

By 1932 it appeared that all matter was made from just three subatomicparticles, the electron, proton, and neutron. This picture of matter was a dis-tinct simplification compared with its predecessor, with its 100 or so elements,each with a characteristic atom. As we will describe in Chapter 18, we nowknow that at least some of the subatomic particles themselves have an internalstructure, being made of sub subatomic particles called quarks. However, forthe purposes of atomic physics, and much of nuclear physics, the picture ofmatter as made of electrons, protons, and neutrons seems to be quite suffi-cient, and this is where we will stop the story for now.

3.4 Some Atomic Parameters

The distribution of electrons in their atomic orbits, and of protons and neu-trons inside the nucleus, are two of the major concerns of quantum theory, aswe will describe in several later chapters. Indeed, they are still the subjects of

0, ;e, ;2e, ;3e, Á

qp = -qe = e = 1.60 * 10-19 C

Ernest Rutherford(1871–1937, British)

Rutherford grew up, one of twelvechildren,on a farm in New Zealand.In 1895 he won a scholarship toCambridge, where he was a stu-dent of J. J. Thomson. After highlyproductive periods at McGill Uni-versity in Montreal, Canada, and atManchester University in England,he succeeded J. J. Thomson asCavendish Professor at Cambridgein 1919. In 1908 he won the NobelPrize in chemistry for showing thatradioactivity transforms one ele-ment into another. Within anotherfour years, he had established theexistence of the atomic nucleus,and in 1917 he identified the firstman-made nuclear reaction andproved that the proton is one ofthe constituents of nuclei. Ruther-ford’s booming voice was known todisturb delicate experiments, andhe is shown here (on the right)with his assistant Jack Ratcliffeunder a sign that says “TALKSOFTLY PLEASE.”

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Section 3.4 • Some Atomic Parameters 91

current research. Nevertheless, a surprising number of atomic and nuclearproperties can be understood just from an approximate knowledge of a fewparameters, such as the size of the electron orbits and the masses of the elec-tron, proton, and neutron. In this section we discuss some of these parameters,with which you should become familiar.

The size of an atom is not a precisely defined quantity, but can be char-acterized roughly as the radius of the outermost electron’s orbit. This radiusvaries surprisingly little among the atoms, ranging from about 0.05 nm(helium) to about 0.3 nm (cesium, for example). In the majority of atoms it isbetween 0.1 and 0.2 nm. Thus for all atoms we can say that

(3.1)

The radii of nuclei are all much smaller than atomic radii and are usuallymeasured in terms of the femtometer (fm) or fermi, defined as

(3.2)

Nuclear radii increase steadily from about 1 fm for the lightest nucleus, hydro-gen (which is a single proton, of course), to about 8 fm for the heaviest nuclei(for example, uranium). Thus we can say that

(3.3)

The single most important thing about the two sizes (3.1) and (3.3) istheir great difference. The atomic radius is at least times larger than thenuclear radius. Thus, if we made a scale model in which the nucleus was repre-sented by a pea, the atom would be the size of a football stadium. Now, it turnsout that in chemical reactions atoms approach one another only close enoughfor their outer electron orbits to overlap.This means that the chemical proper-ties of atoms are determined almost entirely by the distribution of electrons,and are substantially independent of what is happening inside the nucleus.Further, quantum mechanics predicts (as we will find in Chapter 10) that thedistribution of electrons is almost completely determined just by the numberof electrons. Thus the chemical properties of an atom are determined almostentirely by the number of its electrons. This important number is called theatomic number and is denoted by the letter Z.

(3.4)

That Z is also the number of protons in the nucleus follows because the num-bers of electrons and protons are equal in a neutral atom.

We should mention that many atoms can gain or lose a few electrons fair-ly easily. When this happens, we say that the atom is ionized and refer to thecharged atom as an ion. (For example, the ion is a calcium atom that haslost two electrons; the ion is a chlorine atom that has gained one electron.)Since the number of electrons in an atom can vary in this way, one must speci-fy that Z is the number of electrons in the neutral atom, as was done in (3.4).

Cl-Ca2+

= number of protons in nucleus.

= number of electrons in neutral atom

atomic number, Z, of an atom

104

nuclear radius L a few fm = a few * 10-15 m

1 fm = 10-15 m

'0.1 nm = 10-10 m

atomic radius L radius of outer electron orbits

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Since chemical properties are determined mainly by the atomic numberZ, one might expect that each chemical element could be identified by theatomic number of its atoms, and this proves to be so. All the atoms of a givenelement have the same atomic number Z, and, conversely, for every number Zbetween 1 and 109 (the highest value to be officially recognized with a name),there is exactly one element. Table 3.4 lists a few elements with their atomicnumbers. We see that the hydrogen atom has one electron heliumtwo carbon six, oxygen eight, and so on through uranium with 92, andon to the artificial Meitnerium with 109. An alphabetical list of all the ele-ments can be found in Appendix C; a table of their properties — the periodictable — is inside the back cover.

1Z = 22, 1Z = 12,

The mass of an atom depends on the masses of the electron, proton, andneutron, which are as follows:

electron:

proton: (3.5)

neutron:

Within 1 part in 1000, the proton and neutron masses are equal, and, by com-parison, the electron mass is negligible (one part in 2000).Thus we can say that

(3.6)

where denotes the mass of the hydrogen atom (a proton plus an electron).Formany purposes it is sufficient to approximate the mass (3.6) as roughly

The result (3.6) makes the approximate calculation of atomic masses ex-tremely simple, since one has only to count the total number of nucleons (thatis, the number of protons plus the number of neutrons). For example, the he-lium atom has two protons and two neutrons and so is four times as massiveas hydrogen:

the carbon atom has six protons and six neutrons and so

mC L 12mH

mHe L 4mH

1 GeV>c2.mH

mp L mn L mH L 940 MeV>c2

mn = 939.6 MeV>c2 = 1.675 * 10-27 kg

mp = 938.3 MeV>c2 = 1.673 * 10-27 kg

me = 0.511 MeV>c2 = 9.11 * 10-31 kg

TABLE 3.4A few elements listed by atomic number, Z.

1 Hydrogen 6 Carbon 82 Lead2 Helium 7 Nitrogen3 Lithium 8 Oxygen 92 Uranium4 Beryllium5 Boron 26 Iron 109 Meitnerium

oo

o

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Section 3.4 • Some Atomic Parameters 93

Since its number of nucleons determines an atom’s mass, this number is calledthe mass number of the atom. It is denoted by A:

(3.7)

An atom with mass number A has mass approximately equal to It often happens that two atoms with the same atomic number Z (and

hence the same numbers of electrons and protons) have different numbers ofneutrons in their nuclei. Such atoms are said to be isotopes of one another.Since the two isotopes have the same number of electrons, they have almostidentical chemical properties and so belong to the same chemical element. Butsince they have different numbers of neutrons, their mass numbers, and hencemasses, are different. For example, the commonest carbon atom has 6 protonsand 6 neutrons in its nucleus, but there is also an isotope with 6 protons but 7neutrons.These two atoms have the same chemical properties, but have differ-ent masses, about and respectively. To distinguish isotopes, wesometimes write the mass number A as a superscript before the chemical sym-bol. Thus the two isotopes of carbon just mentioned are denoted and (usually read as “carbon 12” and “carbon 13”).

While some elements have only one type of stable atom, the majorityhave two or more stable isotopes.The maximum number belongs to tin with 10stable isotopes, and on average there are about 2.5 stable isotopes for each el-ement. Table 3.5 lists all of the stable isotopes of four representative elements.

13C12C

13mH,12mH

AmH.

= (number of protons) + (number of neutrons)

mass number, A, of an atom = number of nucleons in atom

Since the chemical properties of isotopes are so similar, the proportionof isotopes that occur in nature does not change in normal chemical processes.This means that the atomic mass of an element, as measured by chemists, is theweighted average of the masses of its various natural isotopes. This explainswhy chemical atomic masses are not always close to integer multiples of For example, we see from Table 3.5 that natural chlorine is the isotope and the isotope Thus the chemical atomic mass of Cl is the weighted av-erage This complication caused some confusion in the historical de-velopment of atomic theory. The English physician Prout had pointed out asearly as 1815 that the masses of atoms appeared to be integral multiples of

suggesting that all atoms were made from hydrogen atoms. As moreatomic masses were measured, examples of nonintegral masses (such as chlo-rine) were found, and Prout’s hypothesis was rejected. Not until a hundredyears later was it seen to be very nearly correct.

mH,

35.5mH.

37Cl.14

35Cl34

mH.

TABLE 3.5The stable isotopes of four elements. The percent of each element that occurs naturally in each isotope isshown in parentheses. A complete list, including all stable isotopes of all the elements and their abundances,is given in Appendix D.

Element Atomic Number Isotopes

Carbon 6

Magnesium 12

Chlorine 17

Iron 26 54Fe(5.8%), 56Fe(91.7%), 57Fe(2.2%), 58Fe(0.3%)

35Cl(75.8%), 37Cl(24.2%)

14Mg(79.0%), 21Mg(10.0%) 26Mg(11.0%)

12C(98.9%), 13C(1.1%)

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*There is also a correction for the binding energy of the electrons, but this is muchsmaller (1 part in 10,000 at the most) and is almost always completely negligible.

3.5 The Atomic Mass Unit

Since all atomic masses are approximately integer multiples of the hydrogenatom’s mass, it would be natural to use a mass scale with as the unit ofmass, so that all atomic masses would be close to integers. In fact, this is ap-proximately (although not exactly) how the atomic mass scale is defined, andthe so-called atomic mass unit, or u, is to a good approximation just

To understand the exact definition of the atomic mass unit, we must ex-amine why atomic masses are only approximately integral multiples of Let us consider an atom with atomic number Z and mass number A.The num-ber of neutrons we denote by N, so that

If we ignore for a moment the requirements of relativity, the mass ofour atom would be just the sum of the masses of Z electrons, Z protons, andN neutrons.

(3.8)

To the extent that we can write this as

(3.9)

or

(3.10)

In fact, of course, and are not exactly the same, the difference being oforder 1 part in 1000.Thus the step from (3.8) to (3.9) is only an approximation,and our conclusion (3.10) can be in error by about 1 part in 1000.

There is a second, and more important, reason why (3.10) is only an ap-proximation. We have learned from relativity that when bodies come togetherto form a stable bound system, the bound system has less mass than its sepa-rate constituents by an amount

(3.11)

where B is the binding energy, or energy required to pull the bound systemapart into its separate pieces. For example, the helium nucleus

has a binding energy of about 28 MeV; thus the heli-um atom has less mass than the nearly predicted by(3.10). This is a correction of about 7 parts in 1000 and is appreciably more im-portant than the correction discussed previously.*

For all atoms (except hydrogen itself) there is a similar adjustment dueto the nuclear binding energy. Further, it turns out that in almost all atoms thecorrection is in the same proportion, namely 7 or 8 parts in 1000. (We will see

4000 MeV>c228 MeV>c2(2 protons + 2 neutrons)

¢m =B

c2

mnmH

m L AmH

m L ZmH + NmH

mn L mH

m = Zme + Zmp + Nmn = ZmH + Nmn

A = Z + N

mH.

mH.

mHmH,

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Example 3.1

Using the natural abundances in Table 3.5, find the chemical atomic mass ofmagnesium in u to three significant figures.

The required mass is the weighted average of the masses of the isotopesas listed in Table 3.5. To three significant figures we can

use the approximation (3.13) to give

(3.14)

in agreement with the observed value given inside the back cover. On thoserare occasions when one needs greater accuracy, the more precise atomicmasses given in Appendix D can be used.

= 310.79 * 242 + 10.10 * 252 + 10.11 * 2624 u = 24.3 u

average atomic mass of Mg

24Mg, 25Mg, 26Mg,

Section 3.5 • The Atomic Mass Unit 95

why the correction is always about the same in Chapter 16.) The only signifi-cant exception to this statement is the hydrogen atom itself, which has no nu-clear binding energy and hence no correction. Thus if we want a mass scale onwhich most atomic masses are as close as possible to integers, the mass of theH atom is certainly not the best choice for the unit. If we chose instead or

and defined the unit as the mass of or the mass of all atoms(except hydrogen) would have masses closer to integral multiples of the atom-ic mass unit.Which atom we actually choose is a matter of convenience, and byinternational agreement we use defining the u as follows:*

(3.12)

With the definition (3.12) we can write the mass of an atom whose massnumber is A as

(3.13)

For most atoms this approximation is good to 1 part in 1000, and even in theworst case, hydrogen, it is better than 1 part in 100. Both of these claims areillustrated by the examples in Table 3.6.

(mass of atom with mass number A) L A u

= 1.661 * 10-12 kg = 931.5 MeV>c2

1 atomic mass unit = 1 u = 112 (mass of one neutral 12C atom)

12C,

12C,112

4He14

12C

4He

*The atomic mass unit defined here, sometimes called the unified mass unit, replacestwo older definitions, one based on and the other on the natural mixture of thethree oxygen isotopes.

16O

TABLE 3.6Atomic masses in u are always close to an integer.

1.008 34.9694.003 55.939

12 exactly 207.97715.995 238.049238U16O

208Pb12C

56Fe2He

35Cl1H

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3.6 Avogadro’s Number and the Mole

From a fundamental point of view the natural way to measure a quantity ofmatter is to count the number of molecules (or atoms). In practice, a conve-nient amount of matter usually has a very large number of molecules, and it issometimes better to have a larger unit than the individual molecule. The usualchoice for that larger unit is the mole, which is defined as follows.

We first define Avogadro’s number, as the number of atoms in 12grams of

(3.15)

In this definition there is nothing sacred or fundamental about 12 grams; it issimply a reasonable macroscopic amount. (There is something convenient aboutthe number 12 when used in conjunction with as we will see directly.) Since

atoms of (each of mass 12 u) have total mass 12 g we see that

Canceling the 12 and dividing, we see that

(3.16)

That is, is just the number of u in a gram.We now define a mole of objects (carbon atoms, water molecules, physi-

cists) as objects.Thus a mole of carbon is carbon atoms, a mole of wateris water molecules (and a mole of physicists is physicists). Since a molealways contains the same number of objects, the mass of a mole is proportion-al to the mass of the object concerned. From its definition, a mole of hasmass exactly 12 grams. Thus a mole of has mass approximately 4 grams(more exactly 4.003 grams); a mole of hydrogen atoms, about 1 gram; a mole of

molecules, about 2 grams; a mole of water, about 18 grams (since the molecule has mass ); and so on. The official abbreviation for themole is “mol.”

In chemistry the masses of atoms and molecules are usually specified bygiving, not the mass of an individual atom or molecule, but the mass of a mole.Thus, instead of saying that

one says that

Masses expressed in this way, in grams per mole, are often misleadinglycalled atomic or molecular weights. Evidently, the mass of an atom or mole-cule in u has the same numerical value as its atomic or molecular weight ingrams per mole.

(mass of 1 mole of 12C) = 12 grams

(mass of 12C atom) = 12 u

2 + 16 = 18 uH2OH2

4He

12C

NANA

NANA

NA

NA =1 gram

1 u

NA * 112 u2 = 12 grams

12CNA

12C,

= (number of atoms in 12 grams of 12C) = 6.022 * 1023

Avogadro’s number, NA

12C.NA,

TAYL03-085-124.I 12/9/02 3:00 PM Page 96

Section 3.7 • Kinetic Theory 97

*This kind of microscope and others capable of resolving atoms are further describedin Chapter 14.

The mole is not a fundamental unit. It is an arbitrary unit, which is con-venient for people who deal with large numbers of atoms and molecules. (Itcan be compared with the “dozen,” another arbitrary unit, which is convenientfor people who deal with large numbers of eggs or doughnuts.) A purely theo-retical physicist has little reason to work with either moles or Avogadro’s num-ber, but for most of us, it is important to be able to translate between thefundamental language of molecules and the practical language of moles.

To conclude this brief discussion of the mole, we should mention that themole is officially considered to have an independent dimension called the“amount of substance” (in the same way that meters have the dimensioncalled “length”). For this reason Avogadro’s number,

is often called Avogadro’s constant, since it is viewed as a dimensional quantitywith the units (or ).

3.7 Kinetic Theory

The description of the macroscopic properties of gases in terms of the micro-scopic behavior of atoms is called kinetic theory. In 1900 the notion that theworld is made of very tiny particles, atoms, was widely, but not fully, acceptedby the scientific community. As the twentieth century dawned, a vocal minori-ty of competent scientists refused to accept the concept of atoms as fact be-cause there was no direct, irrefutable evidence of their existence. Atoms weremuch too small to see directly; it was not until 1953 that individual atoms werefirst imaged with the newly invented field-ion microscope.* The evidence forthe existence of atoms from chemistry, though viewed as compelling by manyscientists, was indirect and not considered conclusive by all. Only when the in-creasingly precise predictions of kinetic theory were borne out by experimentdid the scientific community completely embrace the atomic hypothesis. Thelast skeptics were finally convinced when experiment verified the theory ofBrownian motion (described in Sec. 3.9) — a theory developed by a young, un-known, scientist named Albert Einstein.

As our first application of kinetic theory, we show that the ideal gas lawan experimentally discovered relation, can be derived from con-

siderations of the motion of atoms. We will also find that the microscopic viewof the world afforded by kinetic theory provides a very intuitive and satisfacto-ry explanation of the meaning of temperature, a concept that was poorly un-derstood before kinetic theory. Thus, the kinetic theory derivation of the idealgas law was an important step in the path to the modern atomic view of matter.

Chemists usually write the ideal gas law as

(3.17)

where n is the number of moles of the gas, the constant iscalled the universal gas constant, and p, V, and T are the pressure, volume, and

R = 8.314 J>K # mole

pV = nRT

1pV = nRT2,

mol-1mole-1

NA = 6.022 * 1023 objects>mole

TAYL03-085-124.I 12/9/02 3:00 PM Page 97

L

Vfy

Vfx

Viy

Vix

X

Y

FIGURE 3.1A gas molecule bounces off the wallof a container.


temperature of the gas. Physicists usually think in terms of the number N ofmolecules, rather than the number n of moles, so they replace n by and rewrite (3.17) as or

(3.18)

where the constant is called Boltzmann’s constant.

(3.19)

In 1900 the value of the universal gas constant was well established bycomparing (3.17) with experiment. However, Avogadro’s number was notwell known. Estimates of its value varied by a factor of 100 or more. Basically,there was no good way known to count atoms. The inability of experimental-ists to determine Avogadro’s number contributed to skeptics’ argumentsagainst the atomic hypothesis.

We now derive the ideal gas law in the form by consideringthe motion of atoms. We consider a gas of N atoms or molecules, each withmass m, in thermal equilibrium enclosed in a rectangular container of length Land cross-sectional area A. A system is in thermal equilibrium when all partsof the system have the same average energy. We assume that the gas is so di-lute and the interactions among the molecules are so weak that the gas can beconsidered ideal, that is, the molecules are noninteracting. Our goal is to de-rive an expression for the pressure that the gas exerts on the walls of the con-tainer, in terms of the microscopic properties of the molecules.

We orient a coordinate system so that the right and left walls of thecontainer are perpendicular to the x axis as shown in Fig. 3.1. Consider now aparticular molecule labeled i moving toward the right wall with an xcomponent of velocity If the wall is smooth and rigid, the molecule willbounce off the wall like a light ray reflecting from a mirror, and after the colli-sion, the new velocity will have x component while the y and zcomponents will be unchanged. The change in momentum of the molecule isthen The time between collisions ofmolecule i with the right side of the container is the time for the molecule tobounce back and forth once between the right and left walls, so Recall that we are assuming a gas of non-interacting atoms, so that each mole-cule bounces between the right and left walls without interacting with theother molecules. Except for extremely dilute gases, this assumption ofcomplete non-interaction is unfounded, but, as we will discuss below, collisionsamong the atoms do not affect our final results. The force from this one mole-cule on the right wall is a series of brief impulses, but the time-averaged forcedue to this molecule is the total change in momentum between collisionsdivided by the total time, (recall that )

Fxi =ƒ ¢pxi ƒ

¢ti=

ƒ2mvxi ƒ¢ti

=mv2

xi

L

F = ¢p>¢t

¢ti = 2L>vxi .

¢ti¢pxi = pxfinal - px

initial = -2mvxi .

-vxi

+vxi .

pV = NkT

NA

kB =R

NA= 1.38 * 10-23 J>K = 8.62 * 10-5 eV>K

kB

pV = Nk B T

pV = 1N>NA2RTn = N>NA

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Section 3.7 • Kinetic Theory 99

The total pressure p on the wall due to all the molecules in the con-tainer is obtained by summing over all N molecules and dividing by the areaA of the wall.

(3.20)

In the last step we have replaced AL with the volume V of the box, and wehave multiplied and divided by N. The last expression is theaverage value of (The brackets represent an average over all mole-cules). Equation (3.20) now becomes

(3.21)

Finally, we relate to the average squared speed of a moleculeBecause there is no preferred direction for the motion of the mole-

cules, we must have

(3.22)

The square of the speed of a molecule is Therefore,

(3.23)

and equation (3.21) becomes

(3.24)

a result obtained by the Swiss mathematician Daniel Bernoulli at the amaz-ingly early date of 1760.

Comparing this expression with the empirical relation

we see that our derived expression matches the ideal gas law if we assume thatthe average kinetic energy of a molecule is related to theabsolute temperature T by

(3.25)

Our derivation has not only predicted the form of the ideal gas law but hasalso shown the physical meaning of temperature. Equation (3.25) says thattemperature is proportional to the average kinetic energy per molecule. For agas, “hotter” means “more kinetic energy per molecule, faster moving mole-cules.” In Chapter 15 we will give a more general definition of temperature,one that applies to solids as well as gases, but the basic idea remains the same:Temperature is a measure of the energy per atom.

12 m8v29 = 3

2 kT

8KE9 = 12 m8v29

pV = NkT

pV = Nm 13 8v29

8v29 = 8vx29 + 8vy

29 + 8vz29 = 38vx

29v2 = vx

2 + vy2 + vz

2 .

8vx29 = 8vy

29 = 8vz29

8v29. 8vx29

p =Nm8vx

29V

8 Á 9vx2 .

a

v2xi>N = 8vx

29

p =a

i Fxi

A=

m a

i v2

xi

AL=

Nm

V a

i v2

xi

N

TAYL03-085-124.I 12/9/02 3:00 PM Page 99


2R

Area � � � �(2R)2

FIGURE 3.2Two molecules will collide if theircenters are closer than where Ris the radius of either molecule.

2R,

Area � �

FIGURE 3.3A molecule in gas collides with itsneighbors, sweeping out a tube withcross-sectional area The meanfree path is the average length ofthe separate straight sections.

ls.

3.8 The Mean Free Path and Diffusion

In this section we examine some further features of the kinetic theory ofgases. In particular, we look at the consequences of the fact that gases aremade of molecules that have a nonzero size. In the derivation of the ideal gaslaw, we made the assumption that molecules collide only with the walls ofthe container and not with each other. In fact, because of their nonzero size,molecules do collide with each other, but these intermolecule collisions*alter neither the average speed of the molecules nor the validity of equation(3.21), so the ideal gas law remains valid in the presence of collisions. [An al-ternate derivation of (3.21), in the presence of collisions, is explored inProblem 3.36.].

The average distance that a molecule travels between collisions withother molecules is called the mean free path. As one might expect, the meanfree path depends on both the size and density of the gas molecules; collisionsare more frequent if the molecules are larger and if there are more of them. Ifwe approximate the molecules as spheres of radius R, two molecules will over-lap and hence collide if their centers come within of each other (seeFig. 3.2).The collision cross section is defined as the collision area that a molecule presents as it moves through space. We now show that themean free path the collision cross section and the number density n ofmolecules ( per volume†) are related by the simple equation

(3.26)

Consider a molecule that undergoes a large number N of collisions andin so doing moves along a zigzag path of total length L (see Fig. 3.3). Since isdefined as the mean distance traveled between collisions, we have Asthe molecule travels the distance L, it sweeps through a volume of length Land cross-sectional area colliding with the N molecules it encounters whosecenters are within this volume Assuming that the numberdensity in this zigzag tube-shaped volume is the same as the number den-sity n in the sample as a whole, we have

which gives the desired result (3.26). This derivation is not rigorously correctbecause we have not taken proper account of the fact that the other moleculesare moving. A more careful calculation yields

Example 3.2

Given that the mean radius of an air molecule (either or ) is aboutwhat is the approximate mean free path of the molecules in air

at atmospheric pressure and room temperature? 11 atm L 1.01 * 105 Pa2lR = 0.15 nm,N2O2

l = 1> A22ns B .

n =N

V=

N

Nls=

1ls

N>V V = Ls = Nls.s,

L = Nl.l

l L1

ns

n = numbers,l,

s = p12R222R

*We are assuming so-called hard-sphere collisions, that is, a short-range repulsiveinteraction only, no long-range interaction.†Note well that, here, n is not the number of moles, as in (3.17). It is the number ofmolecules per unit volume.

TAYL03-085-124.I 12/9/02 3:00 PM Page 100

Section 3.8 • The Mean Free Path and Diffusion 101

To use the formula we must compute the number densityn and the cross section From the ideal gas law (3.18), we have

The collision cross section is

So we have

This is about 40 times greater than the average distance between nearest-neighbor molecules in the gas.

The convoluted zigzag motion of an individual molecule as it collideswith its neighbors is an example of diffusive motion or diffusion. We will shownext that with diffusion the average magnitude of the displacement D attainedby a molecule in a time t is proportional to

This is to be contrasted with ordinary ballistic motion, which is the straight-line, constant-velocity motion of a projectile in the absence of friction or grav-ity. For ballistic motion, with speed , the net displacement is that is,displacement is proportional to time,

Compared to a ballistic projectile, a diffuser with the same speed does notget very far.

Diffusion is also called random-walk motion because the motion is thatof a bewildered walker who wanders about, making frequent, random changesin direction. The standard random-walk problem is usually formulated likethis: A walker moving in 1, 2, or 3 dimensions, takes a series of steps labeled

The step is described by a vector which has length d, anda random direction. After N steps, the net displacement of the walker, relativeto his starting point, is

(3.27)D = aN

i = 1 di

di ,ithi = 1, 2, Á N.

Dbal r t

D = vt;v

Ddif r 2t

t1>2,

l L1

ns=

1

12.45 * 10252 * 12.8 * 10-192 = 1.4 * 10-7 m = 140 nm

s = p12R22 = 4p10.15 * 10-9 m22 = 2.8 * 10-19 m2

n =N

V=

p

kT=

1.01 * 105 Pa

11.38 * 10-23 J>K2 * 1300 K2 = 2.45 * 1025 m-3

s.l L 1>1ns2,

TAYL03-085-124.I 12/9/02 3:00 PM Page 101


In the case of a gas, the walker is a molecule moving in 3 dimensions, takingsteps of average length the mean free path (see Fig. 3.4).

We seek the average distance of the walker from the origin after N steps.The word “average” must be carefully defined since there is more than onekind of average. We will use the root-mean-square average distance, which isdefined as the square root of the average of the distance squared,

(3.28)

Here the brackets indicate an average over all possible N-step randomwalks, often called the ensemble average. The rms average is usually chosenover other possible averages (such as the average of the absolute value of thedisplacement, ) because it is much easier to work with mathematically.

We now show that after N steps, each of length d, the rms average dis-tance of a random walker from the starting point is

(3.29)

We begin by noting that the magnitude of the vector D is The dot product of D with itself is

(3.30)

We now take the ensemble average, the average over all possible N-stepwalks, and obtain

(3.31)

Since the direction of is random, the quantity is positive asoften as it is negative, and so averages to zero. Taking the squareroot of (3.31), we have our desired result48D29 = Drms = 2Nd

1a

di# dj2

di# dj1i Z j2di

0(')'*

8D29 = 8Nd29 + ha

i Z j di

# dji = Nd2

Nd2(')'*

D # D = D2 = aaN

i = 1 dib # aa

N

j = 1 djb = a

N

i = 1 di

2 + a

i Z j di

# dj

D = ƒD ƒ = 2D # D .

Drms = 2Nd

8 ƒD ƒ9

8 Á 9Drms = 48D29

l,

D

Start

Finish

d1

d2

dN

diFIGURE 3.4Random-walk motion. A randomwalker takes N steps (labeled

) whereindividual steps have a randomdirection and, perhaps, random-steplength; the resulting netdisplacement is the vector D.

d1 , d2 , Á di , Á dN

TAYL03-085-124.I 12/9/02 3:00 PM Page 102

Section 3.8 • The Mean Free Path and Diffusion 103

Example 3.3

An aimless physics student performs a random walk by taking 1-meter longsteps, at a rate of one step each second, with each step in a completely ran-dom direction. How long does it take, on average, for the student to wanderto a point 1 kilometer from his starting point?

Using equation (3.29), with step length and final displacementwe have for the number of steps N,

That is, on average, the student needs to take one million steps in order to getone kilometer away from his starting point. At 1 second per step, this walkwill take nearly 12 days.

We can now derive a formula for how far a molecule has diffused in a gasafter a time t. This is a random-walk problem in which the step length is themean free path of the molecule. The rms average speed of our molecularrandom walker is, according to (3.25),

(3.32)

We can define a collision time, as the mean time between collisions(that is, between steps) by the equation

(3.33)

The time t required to make N steps is then We can now rewrite (3.29)in terms of the time t of the walk and other microscopic variables

Finally, using (3.26) and (3.32), we have

(3.34)

This is a very useful formula for a present-day experimental physicist.However, to a physicist working in the late 1800s, it was interesting, but not es-pecially useful, since none of the quantities m, or could be measuredexperimentally. (Recall that at that time there was not even any good way tomeasure Avogadro’s number.) However, in 1905 Einstein derived a similarformula for the case of larger particles, particles suspended in air or water andlarge enough to be seen and measured in a microscope.

Drmss,

Drms L11ns

a3kTmb1>42t

Drms = 2Nl = A tt

l = Av rms t

l l = 2lvrms 2t

t = Nt.

vrms =l

t

t,

vrms = A3kTm

l

N = ¢Drms

d≤2

= a10001b2

= 106

Drms = 1000 m,d = 1 m

TAYL03-085-124.I 12/9/02 3:00 PM Page 103

Example 3.4

A bottle of ammonia is opened in a closed room where the air is per-fectly still. Estimate how long it will take for the evaporating ammonia mole-cules to diffuse from one side of the room to the other, a distance of 5 m. Themean radius of an ammonia molecule is about

The size of the ammonia molecule is very nearly the same as that of anor molecule, so we can use the values of the collision cross section,

and n, the number density of molecules, from Example 3.2 above. The massof an ammonia molecule is

Solving equation (3.34) for the time t, we have

(3.35)

This long time seems to contradict everyday experience. If you open a bottleof ammonia, the smell will usually permeate the room in minutes or perhaps,at most, an hour. That is because, under normal circumstances, the mixing ofair (indoors or out) is accomplished primarily by convection, not diffusion.The point here is that diffusion is very slow compared to convection; if youwant to be sure that gases or liquids are mixed, you must stir them.

3.9 Brownian Motion

In 1828 a Scottish botanist, Robert Brown, discovered that tiny pollen grains,when suspended in water and viewed under a microscope, exhibited an irregu-lar jiggling motion, which was later dubbed Brownian motion (see Fig. 3.5).Brown initially thought that this motion was due to some “life force”; howev-er, he quickly established that tiny particles of any composition, whether or-ganic or inorganic, suspended in any fluid, whether liquid or gas, also exhibitedthis erratic motion. In the decades that followed, Brownian motion was care-fully studied and a variety of explanations were suggested. In the late 1880s, itwas proposed that Brownian motion is caused by the random motion of themolecules in the surrounding fluid. A small suspended particle experiences acontinuous, irregular pounding from all sides because of collisions with themolecules of the fluid. This explanation, which presumed the existence ofatoms, was controversial; many thought Brownian motion was due to convec-tion currents or vibrations transmitted through the fluid — explanations thatdid not require the existence of atoms.

In 1905 the twenty-six-year-old Einstein, having failed to procure an aca-demic position, was supporting himself and his young family by working in aSwiss Patent Office as a “technical expert, third-class.” Though not very presti-gious, this job brought him financial security and allowed him enough leisure

L 3 * 105 s L 3 days

= 52 * 12.5 * 10252 * 12.8 * 10-192 * ¢ 2.8 * 10-26

3 * 11.38 * 10-232 * 13002 ≤1>2

t L 1Drms22 nsa m

3kTb1>2

= 2.8 * 10-26 kg

m = (mass of 1 mole)>(Avogadro’s number) = 117 g2>16 * 10232

s,N2O2

R = 0.15 nm.

(NH3)


TAYL03-085-124.I 12/9/02 3:00 PM Page 104

Section 3.9 • Brownian Motion 105

FIGURE 3.5The positions of a Brownianparticle recorded at regularlyspaced time intervals, from thebook Atoms by Jean Perrin.

*The Brownian particle is assumed to be much larger than an atom, but small enoughto remain suspended in the fluid. The Brownian motion of larger-than-atom-sized par-ticles is similar to, but not the same as the random-walk motion of individual atoms; theBrownian motion equation (3.36) cannot be derived from the atom motion equation(3.34).

time to pursue his own studies of physics. In that year, working in his sparetime, Einstein wrote six history-making papers — a creative outburst rivaledonly by the work of the young Isaac Newton, two and a half centuries before.The six articles, all published in the German scientific journal Annalen derPhysik (Annals of Physics) consisted of a paper on the photoelectric effect(Section 4.3), for which he would receive the Nobel Prize in 1922; two paperson special relativity, the second of which contained the formula apaper on the viscosity of dilute solutions, which turned out to have enormouspractical importance and which is the Einstein paper most cited in the scientif-ic literature; and two papers on the theory of Brownian motion.

Einstein was able to derive an equation for the motion of a Brownianparticle, an equation similar to (3.34), but one containing quantities that couldbe measured experimentally. Einstein showed that the average displacement

in a time t of a spherical particle* of radius a, suspended in a fluid withviscosity is given by

(3.36)

(The viscosity of a fluid is a measure of its resistance to flow: thicker, moresyrupy fluids have higher viscosity.)

In a series of painstaking measurements, the French experimentalistJean-Baptiste Perrin verified Eq. (3.36) and used it to determine experimen-tally Boltzmann’s constant and hence, Avogadro’s number Perrin also determined experimentally using two other methods: Onemethod relied on a theory of the rotation of Brownian particles, a theory alsodue to Einstein; the other involved determining the distribution of suspended

NA

NA = R>kB.kB,

Drms = A kT

3pha 2t

h,Drms

E = mc2;

TAYL03-085-124.I 12/9/02 3:00 PM Page 105

Jean BaptistePerrin(1870–1942, French)

Perrin was professor of chemistryat the University of Paris,but is bestremembered for two importantcontributions to modern physics.His investigations of cathode rays,which he showed conclusively tocarry negative charge, preceded bya couple of years Thomson’s proofthat these “rays” were actually thenegative particles that we now callelectrons. In 1913 Perrin publishedthe results of his extensive studiesof Brownian motion, which yieldeda precise value for Avogadro’s num-ber, provided convincing directproof of the atomic hypothesis, andearned him the Nobel Prize forphysics in 1926.


*Not quite all. For instance, Ernst Mach (of Mach’s number) died in 1916, still refusingto accept the reality of atoms.†You can find an authoritative but highly readable history of atomic theory from about600 B.C. to 1960 in The World of the Atom by H. A. Boorse and L. Motz, New York:Basic Books, 1966.

particles in a gravitational field.All three methods produced the same value ofwithin a few percent. It was this concordance of different measurements of

Avogadro’s number that convinced almost all* the remaining skeptics of theatomic hypothesis.

3.10 Thomson’s Discovery of the Electron�

�In the last four sections of this chapter, we describe three of the pivotal experimentsof modern physics: the discovery of the electron, the measurement of its charge, and thediscovery of the atomic nucleus. We believe that every well-educated scientist shouldknow some of the details of these experiments. Nevertheless, we will not be using thismaterial again, so if you are pressed for time, you could omit these sections without lossof continuity.

The main purpose of this chapter has been to introduce the principal charac-ters in the story of atomic physics — atoms and molecules, electrons, protons,and neutrons — and to describe one of the first successes of atomic theory, thekinetic theory of gases. This early history, culminating in the acceptance of theatomic hypothesis by the scientific community, is not regarded as part of“modern physics,” and in this book we will content ourselves with the brief de-scription already given.† On the other hand, the more recent history of atomicphysics is very much a part of modern physics, and from time to time we willgive more detailed descriptions of some of its experimental highlights. In par-ticular, we conclude this chapter by describing three key experiments: the ex-periments of J. J.Thomson and Millikan, which together identified the electronand its principal properties, and the so-called Rutherford scattering experi-ment, which established the existnce of the atomic nucleus.

The discovery of the electron is generally attributed to J. J. Thomson(1897) for a series of experiments in which he showed that “cathode rays”were in fact a stream of the negative particles that we now call electrons. Cath-ode rays, which had been discovered some 30 years earlier, were the “rays”emitted from the cathode, or negative electrode, of a cathode ray tube — asealed glass tube containing two electrodes and low-pressure gas (Fig. 3.6).When a large potential is applied between the cathode and anode, some of thegas atoms ionize and an electric discharge occurs. Positive ions hitting the cath-ode eject electrons, which are then accelerated toward the anode. With thearrangement shown in Fig. 3.6, some of the electrons pass through the hole inthe anode and coast on to the far end of the tube. At certain pressures the rayscan be seen by the glow that they produce in the gas, while at lower pressuresthey produce a fluorescent patch where they strike the end of the tube (as in astandard television tube).

It had been shown by William Crookes (1879) and others that cathoderays normally travel in straight lines and that they carry momentum. (Whendirected at the mica vanes of a tiny “windmill,” they caused the vanes to ro-tate.) Crookes had also shown that the rays can be bent by a magnetic field,the direction of deflection being what would be expected for negative charges.All of this suggested that the rays were actually material particles carrying

NA

TAYL03-085-124.I 12/9/02 3:00 PM Page 106

Section 3.10 • Thomson’s Discovery of the Electron 107

Cathode

Cathode ray

Anode

Deflector plates

�

FIGURE 3.6Thomson’s cathode ray tube. Whenopposite charges were placed onthe deflector plates, the “rays” weredeflected through an angle asshown.

u

negative charge. Unluckily, attempts to deflect the rays by a transverse electricfield had failed, and this failure had led to the suggestion that cathode rayswere not particles at all, but were instead some totally new phenomenon —possibly some kind of disturbance of the ether.

J. J. Thomson carried out a series of experiments that settled these ques-tions beyond any reasonable doubt. He showed that if the rays were deflectedinto an insulated metal cup, the cup became negatively charged; and that assoon as the rays were deflected away from the cup, the charging stopped. Heshowed, by putting opposite charges on the two deflecting plates in Fig. 3.6,that cathode rays were deflected by a transverse electric field. (He was alsoable to explain the earlier failures to observe this effect. Previous experi-menters had been unable to achieve as good a vacuum as Thomson’s. The re-maining gas in the tube was ionized by the cathode rays, and the ions wereattracted to the deflecting plates, neutralizing the charge on the plates andcanceling the electric field.) After further experiments with magnetic deflec-tion of cathode rays, Thomson concluded, “I can see no escape from the con-clusion that they are charges of negative electricity carried by particles ofmatter.” Accepting this hypothesis, he next measured some of the particles’properties. By making each measurement in several ways, he was able todemonstrate the consistency of his measurements and to add weight to hisidentification of cathode rays as negatively charged particles.

Thomson measured the speed of the cathode rays by applying electricand magnetic fields. The force on each electron in fields E and B is the wellknown Lorentz force

(3.37)

(where v is the electron’s velocity). Using an E field alone, he measured thedeflection of the electrons. After removing E and switching on a transversemagnetic field B, he adjusted B until the magnetic deflection was equal to theprevious electric deflection. Under these conditions or since v andB were perpendicular,

(3.38)

(In Thomson’s experiments was of order Given his experimental uncer-tainties, this means that his experiments can be analyzed nonrelativistically.)By measuring the heating of a solid body onto which he directed the electrons,he got a second estimate of , which agreed with the value given by (3.38)considering his fairly large uncertainties. (See Problem 3.43.)

v

0.1c.v

v =E

B

E = v * B,

F = -e1E + v * B2

TAYL03-085-124.I 12/9/02 3:00 PM Page 107


Knowing the electrons’ speed he could next find their “mass-to-charge”ratio, For example, as we saw in Chapter 2 — Eq. (2.48) — a B fieldcauses electrons to move in a circular path of radius

(3.39)

(In discussing relativity, we used for the relative speed of two frames and ufor the speed of a particle. Here we have reverted to the more standard forthe electron’s speed.) Thus, measurement of R (plus knowledge of and B),gave the ratio Similarly, you can show (Problem 3.40) that an E fieldalone deflects the electrons through an angle

(3.40)

where l denotes the length of the plates that produce E. Thus, by measuring thedeflection produced by a known E field, he could again determine the ratio

Even though Thomson used several different gases in his tube and differ-ent metals for his cathode, he always found the same value for (within hisexperimental uncertainties). From this observation, he argued correctly thatthere was apparently just one kind of electron, which must be contained in allatoms. He could also compare his value of for electrons with the knownvalues of for ionized atoms. (The mass to charge ratio for ionized atomshad been known for some time from experiments in electrolysis — the conduc-tion of currents through liquids by transport of ionized atoms.) Thomson foundthat even for the lightest atom (hydrogen) the value of was about 2000times greater than its value for the electron.The smallness of for electronshad to be due to the smallness of their mass or the largeness of their charge, orsome combination of both.Thomson argued (again correctly, as we now know)that it was probably due to the smallness of the electrons’ mass.

3.11 Millikan’s Oil Drop Experiment�

A frustrating shortcoming of Thomson’s measurements was that they al-lowed him to calculate the ratio but not m or e separately. The reasonfor this is easy to see: Newton’s second law for an electron in E and B fieldsstates that

Evidently, any calculation of the electron’s motion is bound to involve just theratio and not m or e separately. One way around this difficulty was tostudy the motion of some larger body, such as a droplet of water, whose massM could be measured and which had been charged by the gain or loss of anelectron. In this way one could measure the ratio Knowing the mass M ofthe droplet, one could find e, and thence the mass m of the electron from theknown value of

The need to measure e (or m) separately was recognized by Thomson andhis associates, who quickly set about measuring e using drops of water. Howev-er, the method proved difficult (for example, the water drops evaporatedrapidly), and Thomson could not reduce his uncertainties below about 50%.

m>e.

M>e.

m>e

ma = -e1E + v * B2

m>e,

m>em>e

m>e m>e

m>em>e.u

u =eEl

mv2

m>e.v

vv

R =mv

eB

m>e.

TAYL03-085-124.I 12/9/02 3:00 PM Page 108

Section 3.11 • Millikan’s Oil Drop Experiment 109

FIGURE 3.7Millikan’s oil-drop experiment. Thepotential difference between theplates is adjustable, both inmagnitude and direction.

In an effort to avoid the problem of evaporation, the American physicistRobert Millikan tried using oil drops and quickly developed an extremely ef-fective technique. His apparatus is sketched in Fig. 3.7. Droplets of oil from afine spray were allowed to drift into the space between two horizontal plates.The plates were connected to an adjustable voltage, which produced a verticalelectric field E between the plates. With the field off, the droplets all drifteddownward and quickly acquired their terminal speed, with their weight bal-anced by the viscous drag of the air. When the E field was switched on, Mil-likan found that some of the drops moved down more rapidly, while othersstarted moving upward. This showed that the drops had already acquired elec-tric charges of both signs, presumably as a result of friction in the sprayer. Themethod by which Millikan measured these charges was ingenious and intri-cate, but the essential point can be understood from the following simplifiedaccount. (For some more details, see Problem 3.45.)

By adjusting the electric field E between the plates, Millikan could holdany chosen droplet stationary. When this was the case, the upward electricforce must exactly have balanced the downward force of gravity:

(3.41)

where q was the charge on the droplet. Since E and g were known, it remainedonly to find M in order to determine the charge q.

To measure M, Millikan switched off the E field and observed the termi-nal speed with which the droplet fell. (This speed was very small and so eas-ily measured.) From classical fluid mechanics, it was known that the terminalspeed of a small sphere acted on by a constant force F is

(3.42)

where r is the radius of the sphere and the viscosity of the gas in which itmoves. In the present case F is just the weight of the droplet,

(3.43)

where is the density of the oil. Substitution of (3.43) in (3.42) gives the speedof the falling droplet as

(3.44)

Since g, and were known, measurement of allowed Millikan to calculatethe droplet’s radius r and thence, from (3.43), its mass M. Using (3.41), hecould then find the charge q.

From time to time it was found that a balanced droplet would suddenlymove up or down, indicating that it had picked up an extra ion from the air.

vhr,

v =2r2

rg

9h

r

F = Mg = 43 pr3

rg

h

v =F

6prh

v

qE = Mg

TAYL03-085-124.I 12/9/02 3:00 PM Page 109


Millikan quickly learned to change the charge of the droplet at will by ionizingthe air (by passing X-rays through the apparatus, for example). In this way hecould measure not only the initial charge on the droplet but also the change inthe charge as it acquired extra positive and negative ions.

In the course of several years, starting in 1906, Millikan observed thou-sands of droplets, sometimes watching a single droplet for several hours as itchanged its charge a score or more times. He used several different liquids forhis droplets and various procedures for changing the droplets’ charges. Inevery case the original charge q and all subsequent changes in q were found tobe integer multiples (positive and negative) of a single basic charge,

That the basic unit of charge was the same as the electron’s charge waschecked by pumping most of the air out of the apparatus. With few air mole-cules, there could be few ions for the droplet to pick up. Nevertheless, Millikanfound that X-rays could still change the charge on the droplets, but only in thedirection of increasing positive charge. He interpreted this, correctly, to meanthat X-rays were knocking electrons out of the droplet. Since the changes incharge were still multiples of e (including sometimes e itself), it was clear thatthe electron’s charge was itself equal to the unit of charge e (or rather tobe precise).

It is worth emphasizing the double importance of Millikan’s beautifulexperiments. First, he had measured the charge of the electron, Com-bined with the measurement of by Thomson and others, this also deter-mined the mass of the electron (as about 1/2000 of the mass of the hydrogenatom, as Thomson had guessed). Second, and possibly even more important,he had established that all charges, positive and negative, come in multiples ofthe same basic unit e.

3.12 Rutherford and the Nuclear Atom�

As soon as the electron had been identified by Thomson, physicists began con-sidering its role in the structure of atoms. In particular, Kelvin and Thomsondeveloped a model of the atom (usually called the Thomson model), in whichthe electrons were supposed to be embedded in a uniform sphere of positivecharge, somewhat like the berries in a blueberry muffin. The nature of the pos-itive charge was unknown, but the small mass of the electron suggested thatwhatever carried the positive charge must account for most of the atom’s mass.

To investigate these ideas, it was necessary to find an experimental probeof the atom, and the energetic charged particles ejected by radioactive sub-stances proved suitable for this purpose. Especially suitable was the alpha par-ticle, which is emitted by many radioactive substances and which Rutherfordhad identified (1909) as a positively ionized helium atom. (In fact, it is a heliumatom that has lost both of its electrons; that is, it is a helium nucleus.) It wasfound that if alpha particles were directed at a thin layer of matter, such as ametal foil, the great majority passed almost straight through, suffering onlysmall deflections. It seemed clear that the alpha particles must be passingthrough the atoms in their path and that the deflections must be caused by theelectric fields inside the atoms. All of this was consistent with the Thomsonmodel, which necessarily predicted only small deflections, since the electrons

m>e q = -e.

-e,

e = 1.6 * 10-19 C

TAYL03-085-124.I 12/9/02 3:00 PM Page 110

Section 3.12 • Rutherford and the Nuclear Atom 111

Shield

Foil

Microscope

ZnS screen

�

Radioactivesource

FIGURE 3.8The “Rutherford scattering”experiment of Geiger and Marsden.Alpha particles from a radioactivesource pass through a narrowopening in a thick metal shield andimpinge on the thin foil. Thenumber scattered through the angle

is counted by observing thescintillations they cause on the zincsulfide screen.

u

are too light and the field of the uniform positive charge too small to producelarge changes in the velocity of a massive alpha particle.

However, Ernest Rutherford and his two assistants, Hans Geiger andErnest Marsden, found (around 1910) that even with the thinnest of metalfoils (of order ) a few alpha particles were deflected through very largeangles — 90° and even more. Since a single encounter with Thomson’s atomcould not possibly cause such a deflection, it was necessary to assume (if onewished to retain the Thomson model) that these large deflections were the re-sult of many encounters, each causing a small deflection. But Rutherford wasable to show that the probability of such multiple encounters was far too smallto explain the observations.

Rutherford argued that the atom must contain electric fields far greaterthan predicted by the Thomson model, and large enough to produce the ob-served large deflections in a single encounter. To account for these enormouselectric fields he proposed his famous nuclear atom, with the positive chargeconcentrated in a tiny massive nucleus. (Initially, he did not exclude the possi-bility of a negative nucleus with positive charges outside, but it was quicklyestablished that the nucleus was positive.)

According to Rutherford’s model, the majority of alpha particles goingthrough a thin foil would not pass close enough to any nuclei to be appreciablydeflected. (Collisions with atomic electrons would not cause significant deflec-tions because the elecron is so much lighter than the alpha particles.) On theother hand, a few would come close to a nucleus, and these would be the onesscattered through a large angle. A simplifying feature of the model is that thelarge deflections occur close to the nucleus, well inside the electron orbits, andare unaffected by the electrons.That is, one can analyze the large deflections interms of the Coulomb force of the nucleus, ignoring all the atomic electrons.

Rutherford was able to calculate the trajectory of an alpha particle in theCoulomb field of a nucleus and hence to predict the number of deflectionsthrough different angles; he also predicted how this number should vary withthe particle’s energy, the foil’s thickness, and other variables.These predictionswere published in 1911, and all were subsequently verified by Geiger andMarsden (1913), whose results provided the most convincing support for thenuclear atom.

The experiment of Geiger and Marsden is shown schematically inFig. 3.8. A narrow stream of alpha particles from a radioactive source was di-rected toward a thin metal foil. The number of particles deflected through anangle was observed with the help of a zinc sulfide screen, which gave off avisible scintillation when hit by an alpha particle; these scintillations could beobserved through a microscope and counted.

To calculate the number of deflections expected on the basis of Ruther-ford’s model of the atom, let us consider a single alpha particle of mass m,charge and energy E. Since we are interested in large deflections q = 2e,

u

1 mm

TAYL03-085-124.I 12/9/02 3:00 PM Page 111


�

��

Nucleus

pi

pf

F

Impact parameter b

x

FIGURE 3.9Trajectory of the alpha particle inthe Coulomb field of a nucleus. Theinitial and final momenta are labeled

and the direction labeled x isused as x axis in Section 3.13.

pf ;pi

*The Coulomb force constant k is often written in the form where iscalled the permittivity of the vacuum.

eok = 1>14peo2,

( more than a couple of degrees, say) we suppose that the particle passes rea-sonably close to the nucleus of an atom and well inside the atomic electrons,whose presence we can therefore ignore. We denote the nuclear charge by

and, for simplicity, suppose the nucleus to be fixed. (This is a good ap-proximation since most nuclei are much heavier than the alpha particle.) Theonly force acting on the alpha particle is the Coulomb repulsion of the nucleus,

(3.45)

where k is the Coulomb force constant* Under theinfluence of this inverse-square force the alpha particle follows a hyperbolicpath as shown in Fig. 3.9.

We can characterize the path followed by any particular alpha particle byits impact parameter b, defined as the perpendicular distance from the nucleusto the alpha particle’s original line of approach (Fig. 3.9). Our first task is to re-late the angle of deflection to the impact parameter b. We defer this some-what tedious exercise in mechanics to Section 3.13, where we will prove that

(3.46)

where E denotes the energy of the incident alpha particles. Notice that a largescattering angle corresponds to a small impact parameter b, and vice versa,just as one would expect.

Let us now focus on a particular value of the impact parameter b and thecorresponding angle All particles whose impact parameter is less than b willbe deflected by more than [Fig. 3.10(a)].Thus all those particles that impingeon a circle of radius b (and area ) are scattered by or more. If the originalbeam of particles has cross-sectional area A [Fig. 3.10(b)], the proportion ofparticles scattered by or more is If the total number of particles is N,the number scattered through or more by any one atom in the foil is

(3.47)(number scattered through u or more by one atom) = N pb2

A

u

pb2>A.u

upb2u

u.

u

b =Zke2

E tan1u>22

u

k = 8.99 * 109 N # m2>C2.

F =kqQ

r2 =2Zke2

r2

Q = Ze

u

TAYL03-085-124.I 12/9/02 3:00 PM Page 112


Area ofbeam, A

Thickness t

b

(a) (b)

�

Area �b2

FIGURE 3.10(a) A particle with impactparameter b is deflected by an angle

all those particles that impingeon the circle of area aredeflected by more than (b) Thecross-sectional area of the wholebeam is A; the volume of targetintersected by the beam is At.

u.pb2

u;

This must now be multiplied by the number of target atoms that the beam ofalpha particles encounters. If the target foil has thickness t and contains natoms in unit volume, the number of atoms that the beam meets is[Fig. 3.10(b)]

(3.48)

Combining (3.47) and (3.48), we find that the total number of alpha particlesscattered through or more is

(3.49)

By differentiating (3.49) with respect to we can find the number ofparticles emerging between and and finally, by elementary geometrywe can find the number that hit unit area on the zinc sulfide screen at angle and distance s from the foil. We again defer the details of the calculation toSection 3.13. The result is that

(3.50)

This important result is called the Rutherford formula.Some features of the Rutherford formula would be expected of almost

any reasonable atomic model. For example, it is almost inevitable that should be proportional to N, the original number of alpha particles, and in-versely proportional to the square of the distance to the detector. On theother hand, several features are specific to Rutherford’s assumption that eachlarge angle deflection results from an encounter with the tiny but massivecharged nucleus of a single target atom. Among the features specific toRutherford’s model are

1. is proportional to the thickness t of the target foil.2. is proportional to the nuclear charge squared.3. is inversely proportional to the incident energy squared.4. is inversely proportional to the fourth power of sin1u>22.nsc1u2

E2,nsc1u2Z2,nsc1u2

nsc1u2

s2,

nsc1u2

=Nnt

4s2# ¢Zke2

E≤2 # 1

sin41u>22

nsc1u2 = number of particles per unit area at u

u

u + du;u

u,

Nsc1u or more2 =Npb2

A* nAt = pNntb2

u

(number of target atoms encountered) = nAt

TAYL03-085-124.I 12/9/02 3:00 PM Page 113

Geiger and Marsden were able to check each of these predictions separatelyand found excellent agreement in all cases. Their observations of item (4), thevariation the scattering with the angle are plotted in Fig. 3.11, which showstheir measurements taken at 14 different angles* and fitted to a curve of theform These and their other equally beautiful results establishedRutherford’s nuclear atom beyond reasonable doubt and paved the way forthe modern quantum theory of the atom.

One final consequence of Rutherford’s analysis deserves mention. TheRutherford formula (3.50) was derived by assuming that the force of thenucleus on the alpha particles is the Coulomb force

(3.51)

This is true provided that the alpha particles remain outside the nucleus at alltimes. If the alpha particles penetrated the nucleus, the force would not begiven by (3.51) and the Rutherford formula would presumably not hold. Thusthe fact that Geiger and Marsden found the Rutherford formula to be correctin all cases indicated that all alpha particles were deflected before they couldpenetrate the target nuclei. Since it was easy to calculate the minimum dis-tance of the alpha particle from the center of any nucleus, this gave an upperlimit on the nuclear radius, as we see in the following example.

Example 3.5

When 7.7-MeV alpha particles are fired at a gold foil the Ruther-ford formula agrees with the observations at all angles. Use this fact to obtainan upper limit on the radius R of the gold nucleus.

Since the Rutherford formula holds at all angles, none of the alpha par-ticles penetrate inside the nucleus; that is, for all trajectories the distance rfrom the alpha particle to the nuclear center is always greater than R

(3.52)

(See Fig. 3.12.) Now, the minimum value of r occurs for the case of a head-oncollision, in which the alpha particle comes instantaneously to rest. At this

r 7 R

1Z = 792,

F =kqQ

r2

K>sin41u>22.u,


1

102

104

106

0 �60 120

Scat

teri

ng r

ate

FIGURE 3.11Geiger and Marsden measured theflux of scattered alpha particles at14 different angles. Theirmeasurements fit Rutherford’spredicted behaviorbeautifully. Notice the vertical scaleis logarithmic and themeasurements span an amazing 5orders of magnitude.

1>sin4(u>2)

*Data is taken from Geiger and Marsden, Philosophical Magazine, vol. 25, pp. 604–623,(1913) — a surprisingly readable paper.

TAYL03-085-124.I 12/9/02 3:00 PM Page 114


Nucleus of radius R

r � R

rmin

FIGURE 3.12If none of the alpha particlespenetrates into the nucleus,for all points on all orbits. Theclosest approach occurs in ahead-on collision.

rmin

r 7 R

point its kinetic energy is zero and its potential energyis equal to its total energy; that is, Thus

(3.53)

whence

(3.54)

Since for all orbits, it follows that and hence

(3.55)

We have used the sign to emphasize that this is only an approximate re-sult. In the first place, the nuclear radius is itself only an approximate notion.Second, we are ignoring the size of the alpha particle. [We could improve on(3.55) by replacing R by ] Finally, it is possible that non-Coulombic,nuclear forces may have an appreciable effect even a little before the alphaparticle penetrates the nucleus.

Before we substitute numbers into the inequality (3.55), it is convenientto note that the Coulomb constant, almost alwaysappears in atomic and nuclear calculations in the combination Since

is an energy, has the dimensions of energy length and is conve-niently expressed in the units or, what is the same thing,To this end, we detach one factor of e and write

Now the unit is the same as a volt. Multiplied by e, this givesan electron volt; therefore, or

(3.56)

Returning to the inequality (3.55), we find that

R f2Zke2

E=

2 * 79 * 11.44 MeV # fm27.7 MeV

ke2 = 1.44 eV # nm = 1.44 MeV # fm

ke2 = 1.44 * 10-9 eVN # m>C = J>C = 1.44 * 10-9 e # N # m2>C

ke2 = 18.99 * 109 N # m2>C22 * 11.60 * 10-19 C2 * e

MeV # fm.eV # nm*ke2ke2>r ke2.

k = 9 * 109 N # m2>C2,

R + Ra .

f

R f2Zke2

E

R 6 rminR 6 r

rmin =2Zke2

E

2Zke2

rmin= E

U = E = 7.7 MeV.U = 2Zke2>rmin ,1K = 02

TAYL03-085-124.I 12/9/02 3:00 PM Page 115


��

�

/2

�i�f

ppf

pf

pi

FIGURE 3.13Geometry of the initial and finalmomenta in Rutherford scattering.

or

(3.57)

That is, the fact that Geiger and Marsden found the Rutherford formula validfor 7.7 MeV alpha particles on gold implied that the gold nucleus has radiusless than 30 fm. Since the radius of the gold nucleus is in fact about 8 fm, thiswas perfectly correct.

If the incident energy E were steadily increased, some of the alpha par-ticles would eventually penetrate the target nucleus, and the Rutherford for-mula would cease to hold, first at (that is, for head-on collisions),then as the energy increased, at smaller angles as well. From (3.55) we seethat the energy at which the Rutherford formula first breaks down is

(3.58)

For gold this energy is about 30 MeV, an energy that was not available toRutherford until much later. However, for aluminum, with Z only 13, it isabout 6 MeV (Problem 3.49), and Rutherford was able to detect somedeparture from his formula at large angles.

3.13 Derivation of Rutherford’s Formula�

In deriving the Rutherford formula (3.50), we omitted two slightly tedious cal-culations. First, the relation (3.46) between impact parameter b and scatteringangle If we call the initial and final momenta of the alpha particle and the angle between and is the scattering angle as shown in Fig. 3.13.Since the alpha particle’s energy is unchanged, and are equal in magni-tude and the triangle on the left of Fig. 3.13 is isosceles. If we denote thechange in momentum by it is clear from the constructionshown that

(3.59)

Now, we know from Newton’s second law that

¢p = Lq

-q F dt

1dp>dt = F2

¢p = 2pi sin u

2

¢p = pf - pi ,

pfpi

u,pfpi ,pf ,piu:

E L2Zke2

R

u = 180°

R f 30 fm

TAYL03-085-124.I 12/9/02 3:00 PM Page 116

Section 3.13 • Derivation of Rutherford’s Formula 117

To evaluate this integral, we choose our x axis in the direction of Thisdirection was indicated in Fig. 3.9, where we labeled the polar angle of thealpha particle as With these notations

(3.60)

This integral can be easily evaluated by a trick that exploits conservation ofangular momentum. (Since the force on the alpha particle is radially outwardfrom the nucleus, the angular momentum L about the nucleus is constant.)Long before the collision whereas we know that at all times

Equating these two expressions for L, we find that

Thus we can rewrite (3.60) as

(3.61)

Comparing Figs. 3.9 and 3.13, we see that and that Therefore, and Thus (3.61) becomes

(3.62)

Equating the two expressions (3.59) and (3.61) for we see that

which we can solve for b:

(3.63)

since This completes the proof of the relation (3.46).E = p2i >2m.

b =Zke2

E tan1u>22

2pi sin u

2=

4Zke2 m

bpi cos u

2

¢p,

¢p =4Zke2

m

bpi cos u

2

sin ff = cos1u>22.ff = 90° - u>2 u + 2ff = 180°.ff = -fi

=2Zke2

m

bpi 1sin ff - sin fi2

=2Zke2

m

bpi Lf

f

fi

cos f df

= 2Zke2 L

cos f

r2# mr2

bpi df

¢p = 2Zke2 L

cos f

r2# dtdf

df

df

dt=

bpi

mr2

L = mr2 v = mr2

df

dt

L = bpi ,

¢p = Lq

-q Fx dt = L

q

-q 2Zke2

r2 cos f dt

f.

¢p.

TAYL03-085-124.I 12/9/02 3:00 PM Page 117


s d�

�

s sin �d�s

FIGURE 3.14Particles whose deflection isbetween and pass throughthe ring-shaped surface shown.

u + duu

The other step that was omitted in Section 3.12 was the derivation of thefinal result, the Rutherford formula (3.50), from Equation (3.49),

Substituting (3.63), we can rewrite the latter as

(3.64)

This gives the number of alpha particles deflected through or more. Thenumber that emerge between and is found by differentiating (3.64)to give (see Problem 3.51)

(3.65)

Now, at a distance s from the target (where the detector is placed) theparticles emerging between and are distributed uniformly over thering-shaped surface shown in Fig. 3.14. This ring has area

(3.66)

To find the number of particles per unit area at distance s, we must divide thenumber (3.65) by this area, to give

where we have used the identity This completesour derivation of the Rutherford formula.

sin1u2 = 2 sin1u>22 cos1u>22.

=Nnt

4s2# ¢Zke2

E≤2 # 1

sin41u>22

nsc1u2 =Nsc1u to u + du2

2ps2 sin udu

area of ring = 12ps sin u2 * 1s du2

u + duu

Nsc1u to u + du2 = pNnt¢Zke2

E≤2

cos1u>22sin31u>22 du.

u + duu

u

Nsc1u or more2 = pNnt¢ Zke2

E tan1u>22 ≤2

Nsc1u or more2 = pNntb2

CHECKLIST FOR CHAPTER 3CONCEPT DETAILS

Elements and compounds, atoms and Sec. 3.2molecules

Electrons, nuclei, protons, and neutrons Building blocks of atoms (Sec. 3.3)

Rutherford scattering Scattering of alpha particles off a metal foil; first evidence for the nuclear atom (Secs. 3.3, 3.12 and 3.13)

TAYL03-085-124.I 12/9/02 3:00 PM Page 118

Problems for Chapter 3 119

Nucleon Name for a proton or neutron

Unit of charge e

Atomic radii (3.1)

Nuclear radii (3.3)

Atomic number of an atom Z of electrons in neutral atom)of protons in nucleus) (3.4)

Ions Atoms that have lost or gained one or more electrons

Relative masses of e, p, and n (3.5)

Mass number A(3.7)

Isotopes Two or more atoms with the same atomic number Z butdifferent neutron number. Same chemical properties, butdifferent masses

Atomic mass unit, u

(3.12)

Avogadro’s constant, objects/mole (3.15)

Mole (mol) objects

Ideal gas law (3.17) and (3.18)derivation from kinetic theory (Sec. 3.7)

Universal gas constant, R

Boltzmann’s constant,(3.19)

Mean free path, Mean distance between collisions (Sec. 3.8)

Diffusive motion Random-walk motion. Displacement (Sec. 3.8)

Brownian motion Jiggling of small particles due to impacts of atoms(Sec. 3.9)

Thomson’s discovery of the electron� Sec. 3.10

Millikan’s oil-drop experiment� Measurement of electron charge (Sec. 3.11)

Rutherford’s scattering experiment� Established nuclear model of atom (Secs. 3.12 and 3.13)

Ddif r 1t

l

kB = R>N = 1.38 * 10-23 J>K = 8.62 * 10-5 eV>KkB

R = 8.31 J>K # mole

pV = nRT = Nk B T

1 mole = NA

NA = 6.022 * 1023NA

1 u =1

12 (mass of neutral atom of 12C) L (mass of 1H atom)

A = number of nucleons (protons + neutrons) in nucleus

mp L mn L 2000 me

= (number Z = (number

L a few fm = a few * 10-15 m

L0.1 nm = 10-10 m

e = -(electron charge) = (proton charge)

PROBLEMS FOR CHAPTER 3SECTION 3.2 (Elements, Atoms, and Molecules)3.1 • One of the triumphs of nineteenth-century science

was the discovery by Mendeleev of the periodic tableof the elements, which brought order to the confusingmultitude of known elements, as described inChapter 10. A crucial step in this discovery was to listthe elements in the correct order — that is, as we nowknow, in order of increasing atomic number (whichwe define in Section 3.4, as the number of electrons inthe neutral atom).At the time, atomic number was anunknown concept, and chemists listed the elements in

order of atomic mass. Fortunately, these two order-ings are very nearly the same. Use the periodic tableinside the back cover of the book (where both para-meters are listed) to find out how many times the twoorders are different. The exception involving argonand potassium was a stumbling block in the develop-ment of the periodic table since it was found neces-sary to reverse the order of these two (when orderedby mass) to get the periodic table to make sense. Wewill see in Chapter 16 why it is that atomic numberand atomic mass increase almost perfectly in step.

TAYL03-085-124.I 12/9/02 3:00 PM Page 119


3.2 • The elements are generally listed in order of their“atomic number” Z (which is just the number of elec-trons in a neutral atom of the element). Hydrogen isfirst with helium next with and so on.There are 90 elements that occur naturally on earth,but they are not just numbers 1 through 90. What arethey? What are the atomic numbers of the known ar-tificial elements? (You can find this information inthe periodic table inside the back cover.)

3.3 • With what mass of oxygen would 2 g of hydrogencombine to form water

[HINT: All you need to know is the ratio of the masses of theatoms concerned; you can find these masses, listed in atom-ic mass units, inside the back cover.]

3.4 • With what mass of carbon would 1 g of hydrogencombine to form acetylene, (Read the hint toProblem 3.3.)

3.5 • With what mass of nitrogen would 1 g of hydrogencombine to form ammonia (Read the hint toProblem 3.3.)

3.6 • (a) With what mass of oxygen would 1 g of carboncombine to form carbon monoxide CO? (b) Withwhat mass of oxygen would 1 g of carbon combine toform carbon dioxide (In practice, when oxygenand carbon combine — as in a car engine — both COand are formed in proportions that depend onconditions such as the temperature.)

SECTION 3.4 (Some Atomic Parameters)

3.7 • Make a table showing the numbers of electrons,protons, neutrons, and nucleons in the most commonform of the following atoms: He, Li, Na, Ca. (The nec-essary information is in Appendix D.)

3.8 • Tabulate the numbers of electrons, protons, neu-trons, and nucleons in the most common form of thefollowing neutral atoms: hydrogen, carbon, nitrogen,oxygen, aluminum, iron, lead. (The necessary infor-mation can be found in Appendix D.) On the basis ofthis information, can you suggest why lead is muchdenser than aluminum?

3.9 • Make a table similar to Table 3.5 showing all thestable isotopes for hydrogen, helium, oxygen, and alu-minum, inclu their percentage abundances. (Use theinformation in Appendix D.)

3.10 • Use Appendix D to find four elements that haveonly one type of stable atom (that is, an atom with nostable isotopes).

3.11 • Two atoms that have the same mass number A butdifferent atomic numbers Z are called isobars. (Iso-bar means “of equal mass.”) For example, and are isobars. Use the data in Appendix D to find threeexamples of pairs of isobars. Find an example of atriplet of isobars (that is, three different atoms withthe same mass number).

3H3He

CO2

CO 2 ?

(NH3)?

C2H 2 ?

(H2O)?

Z = 2,Z = 1,

3.12 •• It is found that the radius R of any nucleus is givenapproximately by

where A is the mass number of the nucleus and isa constant whose value depends a little on how R isdefined, but is about 1.1 fm. (a) What are the radii ofthe nuclei of helium, carbon, iron, lead, and lawrenci-um? (b) How does the volume of a nucleus (assumedspherical) depend on A? What does your answer tellyou about the average density of nuclei?

SECTION 3.5 (The Atomic Mass Unit)

3.13 • The mass of a carbon 12 atom is (with an uncertainty of 1 in the final digit). Calculatethe value of the atomic mass unit to six significant fig-ures in kg and in

3.14 • Find the mass of a nucleus in u correct to five sig-nificant figures. (The mass of an electron is 0.000549 u.Ignore the binding energy of the electrons.)

3.15 • Using the natural abundances listed in Table 3.5 (orAppendix D) calculate the average atomic masses ofnaturally occurring carbon, chlorine, and iron. (Youranswers should be in u and correct to three signifi-cant figures.)

3.16 • What is the mass (correct to the nearest u) of the fol-lowing molecules:(a) water, (b) laughing gas,(c) ozone, (d) glucose, (e) ammonia,

(f) limestone,

3.17 •• For the great majority of atoms, it is found that theatomic mass (in atomic mass units) is close to twicethe atomic number (an observation we will explain inChapter 16). (a) Using the data in the periodic tableinside the back cover, verify this observation by com-puting the ratio

for the elements with Commenton your results. (b) In terms of the mass number A,your results should show that A is close to (or insome cases a bit more than) What does thismean for the neutron number N in relation to theproton number Z?

3.18 •• (a) Calculate the mass of the nucleus in atom-ic mass units by subtracting the mass of two electronsfrom that of a neutral 4 He atom. (See Appendix D,and give four decimal places.) (b) The procedure sug-gested for part (a) is theoretically incorrect becausewe have neglected the binding energy of the elec-trons. Given that the energy needed to pull both elec-trons far away from the nucleus is 80 eV, is the correctanswer larger or smaller than implied by part (a), andby how much (in u)? In stating the mass of the nucle-us, about how many significant figures can you givebefore this effect would show up?

4He

2Z.

Z = 10, 20, Á , 90.

(atomic mass in u)>(atomic number)

CaCO3.NH3;C6H12O6;O3;

N2O;H2O;

12C

MeV>c2.

1.992648 * 10-26 kg

R0

R = R 0 A1>3

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SECTION 3.6 (Avogadro’s Number and the Mole)

3.19 • How many moles are in (a) 1 g of carbon? (b) 1 gof hydrogen molecules? (c) 10 g of water? (d) 1ounce of gold?

3.20 • How many moles are in (a) 10 g of NaCl? (b) 2 kgof (c) of Hg (density )? (d) 1pound of sugar

3.21 • A mole of molecules has mass 32 g (that is, themolecular “weight” of is 32 g/mole). What isthe mass of a single molecule (in u and in grams)?

3.22 • What is the total charge of 1 mole of ions? (Theion is a chlorine atom that has acquired one extra

electron.) This important quantity is called the fara-day; it is easily measured since it is the charge thatmust pass to release 1 mole of chlorine in electrolysisof NaCl solution, (or, more generally, 1 mole of anymonovalent ion in electrolysis of an appropriate solu-tion). Once the electron charge was known, knowl-edge of the faraday let one calculate Avogadro’snumber.

3.23 • (a) How many atomic mass units are there in onepound mass? (b) How many in one gram? Commenton your answer to part (b).

3.24 •• How many C atoms are there in (a) 1 g of (b) 1 mole (c) 1 kmole of In each casewhat fraction of the atoms are C atoms, and whatfraction of the mass is C?

SECTION 3.7 Kinetic Theory

3.25 • Show that the number density (number per vol-ume) of molecules in an ideal gas is

3.26 • What is the edge length of a cube that contains 1billion molecules of a gas at standard tempera-ture and pressure

3.27 • At what pressure would the density of He gas atequal the density of ordinary water? Give

your answer in atmospheres.

3.28 •• Consider an ideal gas at standard temperatureand pressure

Imagine that the molecules of the gas areequally spaced so that every molecule is at the centerof a small cube. (a) What is the edge length of thiscube, which is the average distance between nearest-neighbor molecules in the gas? (b) Qualitatively, howdoes this mean molecular separation distances com-pare with the diameter of a small molecule such as

3.29 ••• (a) Consider an ideal gas at temperature T withmolecules of mass m and number density (number pervolume) n. Show that the number of molecules thatstrike a surface of area A in a time is given roughly by To simplify this calculation,consider a wall perpendicular to the x-axis and make1nA ¢t>222kT>m .

¢t

O 2 ?

105 N>m22. p = 1 atm = 1.01 * 1STP: T = 273 K,

31 atm = 1.01 * 105 N>m24.T = 300 K

1.01 * 105 N>m22. p = 1 atm = 1STP: T = 273 K,11092

p>kT.

C2H 6 ?CH 4 ?

CO 2 ?

Cl-Cl-

O2

O2

O2

(C12H22O11)?13.6 g>cm310 cm3NH

3 ?

the radical assumption that the x-components of thevelocities of all molecules have the same magnitude;that is, assume that all molecules have the same with half the molecules moving to the right and theother half moving to the left. (b) For oxygen at stan-dard temperature and pressure

what is the average

number of particles that strike a surface in 1 s?

SECTION 3.8 The Mean Free Path and Diffusion

3.30 • Consider a large container of oxygen gas at STPAs-

sume a molecular radius of 0.15 nm. (a) How longdoes it take for a molecule to diffuse to a point 1 mfrom its starting point? (b) How long would it takefor the molecule to diffuse the same distance at apressure of 0.01 atm? (c) Solve for the pressure atwhich a molecule will diffuse a displacement of mag-nitude d in a time t.

3.31 • (a) Show that for a gas, the mean free path between collisions is related to the mean distancebetween nearest neighbors r by the approximaterelation where is the collision cross-section. (b) Given that the molecular radius of a gasmolecule such as or is about 0.15 nm,estimate the value of r and for air at STP(standard temperature and pressure,

). (c) Why, qualita-tively, is for dilute gases?

3.32 • The Knudsen regime is the pressure regime inwhich the mean free path is comparable to the di-mensions of the container. (a) Given the container di-ameter d, the molecule’s collision cross section andthe temperature T, derive an expression for the pres-sure p at which the Knudsen regime occurs. (b) Whatis this pressure (in atmospheres) for a gas at roomtemperature (about 293 K) in a container with diam-eter 20 cm? Assume a molecular radius of 0.15 nm. Agood mechanical pump can achieve a pressure of atm; can such a pump produce Knudsen pressures in

a 20-cm diameter container?

3.33 •• The best vacuum that can be readily achieved inthe lab is called “ultra-high vacuum” or UHV and is

torr At that pressureand at room temperature (a) how manyair molecules are present in (b) what is themean free path of the molecules? Assume a molec-ular radius of 0.15 nm.

3.34 •• An aimless physics student, wandering around ona flat plane, takes a 1-m step in a random directioneach second. (a) After one year of continuous ran-dom walking, what is the student’s expected distancefrom his starting point? (b) If the student wanderedin 3D space, rather than in a plane, but still took 1-msteps each second in random directions, would his ex-pected distance from the origin be greater, less, or thesame as before. Explain.

l1 cm3,

1T = 293 K2,(mm Hg) L 10-13 atm.10-10

10-4

s,

l W r,p = 1.00 atm = 1.01 * 105 N>m2

T = 273 K,l

CO2O2, N2,

sl L r1r2>s2

l

p = 1 atm = 1.01 * 105 N>m22.1T = 273 K,

1 cm2

p = 1 atm = 1.01 * 105 N>m22, 1STP: T = 273 K,

ƒvx ƒ ,

TAYL03-085-124.I 12/9/02 3:00 PM Page 121


y

xv E

�

l

FIGURE 3.15(Problem 3.40)

3.35 •• A gambler has $100 and makes a series of $1 betsthat he can correctly guess the outcome of a coin flip— heads or tails. (The coin and the game are honest,so for every coin flip, there is a 50% chance that thegambler will guess correctly.) The gambler will quitwhen he has either won an additional $100 or has lostall his money. About how many times will the gam-bler bet before quitting?

[HINT: This situation is like a random walk in one dimension.]

3.36 ••• Derive the ideal gas law without the assumptionof no intermolecule collisions. To simplify this calcu-lation, make the radical assumption that thex-components of the velocities of all molecules havethe same magnitude; that is, assume that all moleculeshave the same with half the molecules moving tothe right and the other half moving to the left; makesimilar assumptions for y- and z-directions. Considera gas of molecules with number density n in a cubicalcontainer, and consider only those molecules that arewithin a short distance of one wall of the contain-er where the mean free path. At any in-stant, half of those molecules are heading toward thewall and will collide with the wall in an average time

where Proceed to compute the av-erage force on the the wall due to molecular colli-sions, and then deduce the ideal gas law.

SECTION 3.9 Brownian Motion

3.37 •• A Brownian particle is observed to diffuse in 60 s, on average. (Diffusion distance means magni-tude of the net displacement.) A second particle withtwice the diameter and ten times the mass of the firstis observed. How far, on average, will the second par-ticle diffuse in 60 s at the same temperature?

3.38 •• A Brownian particle is observed to diffuse a dis-tance of in 20 s, on average. How far, on aver-age, will the particle diffuse in 200 s?

3.39 •• A good microscope can readily resolve objectsthat are in diameter ( is about twice thewavelength of visible light). The unaided human eyecan resolve objects no smaller than roughly

Suppose a diameter Brown-ian particle is observed (with a microscope) to dif-fuse an average distance of 5 times its diameter

in 20 s. Under the same conditions of temper-ature and air viscosity, another Brownian particle of

diameter is observed with the unaided eye.How long will this particle take to diffuse an averagedistance of 5 times its diameter? Why was Brownianmotion not discovered until after the invention of themicroscope?

SECTION 3.10 (Thomson’s Discovery of the Electron)

3.40 •• In Thomson’s experiment electrons travel with ve-locity in the x direction. They enter a uniform elec-tric field E, which points in the y-direction and has

v

100 mm

(5 mm)

1-mm0.1 mm = 100 mm.

1 mm1 mm

10 mm

100 mm

¢x>¢t = ƒvx ƒ .¢t,

¢x V l,¢x

ƒvx ƒ ,

total width l (Fig. 3.15). Find the time for an electronto cross the field and the y-component of its velocitywhen it leaves the field. Hence show that its velocityis deflected through an angle (provid-ed that is small). Assume that the electrons arenonrelativistic, as was the case for Thomson.

uu L eEl>1mv22

3.41 •• Suppose that the electrons in Thomson’s experi-ment enter a uniform magnetic field B, which is in thez-direction (with axes defined as in Fig. 3.15) and hastotal width l. Show that they are deflected through anangle (provided that is small).Assume that the electrons are nonrelativistic.

3.42 •• The formulas for the deflection of an electron inan electric or magnetic field (Problems 3.40 and 3.41)are quite similar. Explain in words why both formulashave factors of in the denominator and why themagnetic deflection has just one power of wherethe electric has two.

3.43 •• Thomson directed his cathode rays at a metal bodyand measured the total charge Q that it acquired andits rise in temperature From and the body’sknown thermal capacity, he could find the heat givento the body. Show that this heat should be

where and e are the electron’s mass, speed (non-relativistic), and charge. Combine this with the resultof Problem 3.41 to give expressions for and interms of measured quantities.

3.44 •• In one version of Thomson’s measurement of ,nonrelativistic electrons are accelerated through ameasured potential difference This means thattheir kinetic energy is They are then passedinto a known magnetic field B and the radius

of their circular orbit is measured.Eliminate the speed u from these two equations, andderive an expression for in terms of the mea-sured quantities and R.

SECTION 3.11 (Millikan’s Oil-Drop Experiment)

3.45 ••• In order to find the electron’s charge e, Millikanneeded to know the mass M of his oil drops, and thiswas actually the source of his greatest uncertainty indetermining e. (His value of e was about 0.5% low asa result.) However, to show that all charges are multi-ples of some basic unit charge, it was not necessary to

V0 , B,m>e

R = mv>1eB2K = eV0 .

V0 .

m>em>ev

m, v,

heat =Qmv2

2e

¢T¢T.

vv

uu L eBl>1mv2

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Table 3.7

A series of measurements of and the times for a single dropletto travel down and up a fixed distance l. All times are in seconds.(Problem 3.45)

15.2 15.0 15.1 15.0 14.9 15.1 15.1 15.0 15.2 15.26.4 6.3 6.1 24.4 24.2 3.7 3.6 1.8 2.0 1.9tu :

td :

tu ,td

Table 3.8

Additional data for the oil-drop experiment.

(distance traveled by the droplet)(density of oil)

(acceleration of gravity)

(viscosity of air, adjusted*)

(electric field)

* For very small droplets, there is a small correction to the formula (3.42) forthe terminal velocity. For the experiment reported here, this correctionamounts to a 12% reduction in the viscosity. We have included this correc-tion in the value given.

E = 1.21 * 105 N>C h = 1.60 * 10-5 N # s>m2

g = 9.80 m>s2

r = 839 kg>m3 l = 8.3 * 10-4 m

know M, which cancels out of the charge ratios. Thefollowing problem illustrates this point and givessome more details of Millikan’s experiment.

By switching the E field off and on, Millikan couldtime an oil drop as it fell and rose through a mea-sured distance l (falling under the influence of gravi-ty, rising under that of E and gravity). The downwardand upward speeds are given by (3.42) as

(3.67)

Both speeds were measured in the form where t is the time for the droplet to traverse l.

(a) By adding the two equations (3.67) and rear-ranging, show that

where the quantity was constantas long as Millikan watched a single droplet anddid not vary E.

From (3.68) we see that the charge q is pro-portional to the quantity Thus ifit is true that q is always an integer multiple of e,it must also be true that is alwaysan integer multiple of some fixed quantity. Table3.7 shows a series of timings for a single droplet(made with a commercial version of the Millikanexperiment used in a teaching laboratory). That

changes abruptly from time to time indicatesthat the charge on the droplet has changed as de-scribed in Section 3.11. (The times should the-oretically all be the same, of course. The smallvariations give you a good indication of the un-certainties in all timings. For you should usethe average of all measurements of )td .

td

td

tu

11>td2 + 11>tu211>td2 + 11>tu2.

K = E>16prhl2

1td

+1tu

= a E

6prhlbq = Kq

v = l>t,

vd =Mg

6prh and vu =

qE - Mg

6prh

SECTION 3.12 (Rutherford and the Nuclear Atom)

3.46 • A student doing the Rutherford scattering experi-ment arranges matters so that she gets 80 counts/minat a scattering angle of If she now moves herdetector around to keeping it at the samedistance from the target, how many counts would sheexpect to observe in a minute? (This illustrates anawkward feature of the experiment, especially be-fore the days of automatic counters. An arrangementthat gives a reasonable counting rate at small givesfar too few counts at large and one that gives a rea-sonable rate for large will overwhelm the counterat small )

3.47 •• The Rutherford model of the atom could explainthe large-angle scattering of alpha particles becauseit led to very large electric fields compared to theThomson model. To see this, note that in the Thom-son model the positive charge of an atom was uni-formly distributed through a sphere of the same sizeas the atom itself. According to this model, whatwould be the maximum E field (in volts/meter) pro-duced by the positive charge in a gold atom (atomic )? What is the correspond-ing maximum field in Rutherford’s model of thegold atom, with the positive charge confined to asphere of radius about 8 fm? (In the Thomson modelthe actual field would be even less because of theelectrons. Note that since it fol-lows that )

3.48 •• Consider several different alpha particles ap-proaching a nucleus, with various different impact pa-rameters, b, but all with the same energy, E. Prove thatan alpha particle that approaches the nucleus head-on

gets closer to the nucleus than any other.

3.49 •• (a) If the Rutherford formula is found to be cor-rect at all angles when 15-MeV alpha particles arefired at a silver foil what can you say aboutthe radius of the silver nucleus? (b) Aluminum hasatomic number and a nuclear radius about

If one were to bombard an aluminumfoil with alpha particles and slowly increase their en-ergy, at about what energy would you expect theRutherford formula to break down? You can makethe estimate (3.58) a bit more realistic by taking R to

RAl L 4 fm.Z = 13

1Z = 472,

1b = 02

ke = 1.44 V # nm.ke2 = 1.44 eV # nm,

radius L 0.18 nmZ = 79,

u.uu,

u

u = 150°,u = 10°.

(b) Calculate and show that (withinthe uncertainties) this quantity is always an inte-ger multiple of one fixed number, and hence thatthe charge is always a multiple of one fixed charge.

(c) Use Equation (3.44) for and the additionaldata in Table 3.8 to find the radius r of the droplet.

(d) Finally, use (3.68) to find the four differentcharges q on the droplet. What is the best esti-mate of e based on this experiment?

vd

11>td2 + 11>tu2

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be where is the alpha particle’sradius (about 2 fm).

3.50 •• In a student version of the Rutherford experimentis used as a source of 5.2-MeV alpha particles,

which are directed at a gold foil (thickness ) at arate of particles per minute. The scattered parti-cles are detected on a screen of area at a dis-tance of 12 cm. Use the Rutherford formula (3.50) topredict the number of alpha particles observed in 10minutes at and 45°, given the followingdata:

(number of incident particles in 10 min)(density of gold)

(mass of gold atom)(thickness of foil)t = 2 * 10-6 m

mAu = 197 ur = 19.3 g>cm3N = 106

u = 15°, 30°,

1 cm2105

2 mm

210Po

RHeRAl + RHe, (atomic number of gold)(energy of alpha particles)

[Note that the number density n of gold nuclei isand don’t forget that ]

SECTION 3.13 (Derivation of Rutherford’s Formula)

3.51 • (a) If ( or more) denotes the number of alphaparticles scattered by or more, show that the num-ber whose deflection is between and is

(b) Differentiate the expression (3.64) for and verify the result (3.65).1u or more2 Nsc

Nsc1u to u + du2 = -

dNsc1u or more2du

du

u + duuu

uNsc

ke2 = 1.44 MeV # fm.r>mAu,

E = 5.2 MeVZ = 79

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