Chapter 6 the Rutherford-Bohr Model of the Atom

8/2/2019 Chapter 6 the Rutherford-Bohr Model of the Atom

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H A T E

modelf the atom, based on the work of,andBohr,shows the electrons circu-the nucleus like planets circulating

Siln. Itcanbea useful model for some

purposes, but we shall see in Chapters 7 and 8that it is not a correct picture of the structure ofthe atom.

R

T H ER U T H E R F O R 0D O H R M O O E ~o ' F T H E A T O M


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Our goal in this chapter is to understand some of the details of atomicstructure based on experimental studies of atoms. In particular, wetwo types of experiments that are important in the development oftheory of atomic structure: the scattering of charged particles by atoms,which tells us about the distribution of electric charge in atoms, andemission or absorption of radiation by atoms, which tells us aboutexcited states.

We use the information obtained from these experiments to develop anatomic model, which helps us understand and explain the properties 0 1atoms. A model is usually an oversimplified picture of a more cOlnpllex.system, which provides some insight into its operation but may not b esufficiently detailed to explain all of its properties.

In this chapter, we discuss the experiments that led to the Rutherford-Bohr model (known simply the Bohr model), which is based on the familiar"planetary" structure in which the electrons orbit about the nucleus likeplanets about the Sun. Even though this model is not stricly valid fromstandpoint of wave mechanics, it does help us understand manyproperties, especially the excited states of the simplest atom, hydrogen, InChapter 7 , we show how wave mechanics changes our picture of the hydrogen atom, and in Chapter 8 we consider the structure of more compicated atoms.

174 Chapter 6 The Rutherford-Bohr Model of the Atom

6.1 BASIC PROPERTIES OF ATOMSBefore we begin to construct a model of the atom, let us summarizeof the basic properties of atoms.1. Atoms are very small, about 0.1 nm (0.1 X 10-9 m) in radius. Thus an)

effort to "see" an atom using visible light (A : : : = 500 nm) is hopeless owin!to diffraction effects. We can make a crude estimate of the maximum'of an atom in the following way. Consider a cube of elementalfor example iron. Iron has a density of about 8 g/cm' and a molarof 56 g. One mole of iron (56 g) contains Avogadro's number ofabout 6 X 1023. Thus 6 X 1023 atoms occupy about 7 ern" and so 1occupies about 10-23 em". Ifwe assume the atoms of a solid aretogether in the most efficient possible way, like hard spheres inthen the diameter of one atom is about ~1 O 23 cn r ' : : : = 2 X 10-8 em0.2 nm.

2. Atoms are stable-they do not spontaneously break apart intopieces or collapse; therefore the internal forces that hold thetogether must be in equilibrium. This immediately tells us that thethat pull the parts of an atom together must be opposed in someotherwise atoms would collapse.

3. Atoms contain negatively charged electrons, but areneutral. If we disturb an atom or collection of atoms with


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atomicdiscussof ouratoms,d thet their

6.2 The Thomson Model 17 5

force,electrons are emitted. We learn this fact from studying the Comp-ton effect and the photoelectric effect. We also learned in Chapter 4that even though electrons were emitted from the nuclei of atoms incertain radioactive decay processes, they don't "exist" in those nucleibut are manufactured there by some process. Electrons were excludedfromthe nucleus based on the uncertainty principle, which forbids emit-ted electrons of the energies observed in the laboratory from existinginthe nucleus. The uncertainty principle places no such restriction onatoms (see Problem 1), so we assume the electron to be present inatoms.We can also easily observe that bulk matter is electrically neutral,andwe assume that this is likewise a property of the atoms. Experimentswith beams of individual atoms support this assumption. From theseexperimental facts we deduce that an atom with Z negatively chargedelectrons must also contain a net positive charge of Ze.4.Atoms emit and absorb electromagnetic radiation. This radiation maytake many forms-visible light ( A ~ 500 nm), X rays ( A ~ 1 nm),ultraviolet rays (A ~ 10 nm), infrared rays ( A ~ 0.1 /Lm), and so forth.In fact it is from observation of these emitted and absorbed radiationsthatwe learn most of what we know about atoms. In a typical emissionmeasurement, an electric current is passed through a tube containing asmallsample of the gas phase of the element under study, and radiationisemitted when an excited atom returns to its ground state. The wave-lengthsof the many different emitted radiations can be measured withgreat precision, such as with a diffraction grating in the case of visiblelight.The absorption wavelengths can be measured by passing a beamofwhite light through a sample of the gas and noting which colors areremoved from the white light by absorption in the gas. One particularlycuriousfeature of the atomic radiations is that atoms don't always emitandabsorb radiations at the same wavelengths-some wavelengths pres-ent in the emission experiment do not also appear in the absorptionexperiment. Any successful theory of atomic structure must be able toaccount for these emission and absorption wavelengths.

*6.2 THE THOMSON MODEL .A n earlymodel of the structure of the atom was proposed by J. J. Thomson,whowasknown for his previous identification of the electron and measure-mentof its charge-to-mass ratio elm. The Thomson model incorporatesmanyof the known properties of atoms: size, mass, number of electrons,andelectrical neutrality. In this model, an atom contains Z electrons whichareimbedded in a uniform sphere of positive charge (Figure 6.1). The totalpositivecharge of the sphere is Ze, the mass of the sphere is essentiallyThisisan optional section that may be skipped without loss of continuity.


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Electrons the mass of the atom (the electrons don't contribute significantly tototal mass), and the radius R of the sphere is the radius of the(This model is sometimes known as the "plum-pudding" model, sinceelectrons are distributed throughout the atom like raisins in apudding.)The force on an electron at a distance r from the center of a U1J.lLVlllII1.charged sphere of radius R can be computed using Gauss' law (seelem 2):

."'"...--- . . . . . .,/ "1/ .',,'. \

~ .' I,_-R r I\ / I, . /X . . . . . . . :

FIGURE 6.1 The Thomsonmodel of the atom. Z tiny electronsare imbedded in a uniform sphere ofpositivecharge Ze and radiusR. Animaginaryspherical surfaceof radiusr contains a fraction ,3/R3 of thecharge.

y

FIGURE 6.2 A positivelychargedalphaparticleisdeflectedbyan angle()as it passes through a Thomson-model atom. The coordinates r and4 > locate the alpha particlewhileit isinsidethe atom.

. . .This linear restoring force permits the electrons to oscillate aboutequilibrium positions just like a mass on a spring subject to therestoring force F = kx.We therefore expect the electrons in the Thomson atom to lJ~Lllall __about their equilibrium positions with a frequency v = (211)-1 Vk];;i,k is the constant defined by Equation 6.1. Since an oscillatingcharge radiates electromagnetic waves whose frequency is identical tooscillation frequency, we might expect, based on the Thomson model,the radiation emitted by atoms would show this characteristic +.. ""' ' ' ' 'n,.,,This turns out not to be true (see Problem 3); the calculated frequenciesnot correspond to the frequencies observed for radiation emitted byThe most serious failure of the Thomson model arises from theof charged particles by atoms. Consider the passage of a singlecharged particle through the atom. The particle is deflected somewhatits original trajectory owing to the electrical forces exerted on theby the atom. These forces are (1) a repulsive force due to thecharge of the atom, and (2) an attractive force due to the negativelyelectrons. We assume that the mass of the deflected particle is bothgreater than the mass of an electron and yet much less than the massthe atom. In the encounter between the projectile and an electron,forces exerted on each by the other are equal and opposite (bythird law), and so the principal victim of the encounter is the muchmassive electron; the effect on the projectile is negligible. (Imaginea bowling ball through a field of Ping-Pong ballsl) We thus needconsider the positively charged atom as a cause of the deflection ofparticle. By the same argument, we neglect any possible motion ofmore massive atom caused by the passage of the projectile. Ourthen, is the scattering of a positively charged projectile by thepositively charged massive part of the atom.Figure 6.2 shows a representation of the deflection of the nrotecnewhich is moving at speed u (we assume u c and use UUJIU"Wlmechanics, so K = !mu 2) along a line that would pass a distance bthe center of the atom if the projectile were not deflected. (Theb is called the impact parameter.) The electrical repulsion causes adeflection, and so the particle leaves the atom moving in a slightlydirection, at an angle (J with respect to the original direction.


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(6.1)ut theire linear

ass ofon, theh lessrollingd onlyof theof thenary

fromtancesmall

Wecan calculate the deflection angle (}by considering the impulse re-ceivedby the projectile, which gives it some momentum in the y directionflpy = f r,dt (6.2)

Atan arbitrary point along the trajectory,Fy = F cos c p

Theprojectile, which has a charge q = ze, experiences a force F given byanalogywith Equation 6.1:1 zZe2F=----r= zkr4 1 T e o R 3 (6.3)

wherek is the same constant defined by Equation 6.1. Since cos c p = =b l r , wehave

flpy = = f zkr ~ . dt = zkb f dt=kbT (6.4)

whereT is the total time it takes for the projectile to travel through theatom,which is the total distance traveled in the atom divided by the averagespeed.Since the deflection is small, the path can be approximated as astraightline, as shown in Figure 6.3, and the average speed is very nearlyequalto v, so(6.5)

andEquation 6.4 then givesflpy = = 2zkb VR2 - b2

V (6.6)Assumingthat the change in Px is negligible, we have

tan (}=py = = flpyPx P (6.7)

andif e is small, tan (}= = (},so( } = = flpy = 2zkb VR2 _ b2p mu? (6.8)

The value of the scattering angle e depends on the value of the impactparameter b. When b = 0, the projectile is undeflected ( } = 0) becausethenet force on the projectile is zero. When b = R, the projectile is againundeflected,because we assume the force acts only when the projectile iswithinthe atom. We can find a representative average value of (}correspond-ingto an average value of b, for which we choose the intermediate valueb = R12 . In this case, Equation 6.8 gives

( } avg = ~~:: (6.9)

6.2 The Thomson Model 177

FIGURE 6.3 An approximate ge-ometry of the deflection of an alphaparticle by a Thomson atom. Thescattering angle, whose maximumvalue is about 0.010, has beengreatly exaggerated.

Scattering angle forThomson atom


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I n ci de n t b eam

FIGURE 6.4 A typical scatteringexperiment. An incident beam isscat-tered by a thin foil; scattered particlesare observed at all possible values of8 in the laboratory.

FIGURE 6.5 A microscopic rep-resentation of the scattering. Someindividual scatterings tend to increase8, while others tend to decrease 8.

AntMarsdeshowethan 9valuein this

Equation 6 .9 gives an estimate of the average scattering angle for a Thomson atom.

E X H M P L E G . lUsing the Thomson model with R =0.1 nm (a typical atomic radius),calculate the average deflection angle per collision when 5 Me Y alphaparticles (z = 2) are scattered from gold (Z = 79). It wlife

tiSSS H U T I O NThe apropo:uniforan exatom.the la

The quantity zkR2 can be computed to bezkR2 =2 (~) R2 =(2)(79) 1.44 eY'nm ~ 2.3 keY4 1 T o R 3 0.1 nm

(Recall that e 2/ 4 1 T 0 = 1.44 eYnm.) Also, mo? = 2K = 10 MeY, so- ! 3 2.3 keY - 2 10-4 d - 0 01{}avg= '{ 4 10MeY= X ra - .

Figure 6.4 shows the geometry for a typical scattering experiment. Abeam of particles is incident on a thin foil. Instead of the scattering of a singleprojectile by a single atom, we observe the scattering of many projectiles b ymany atoms and we have no control over the impact parameter b for eachscattering. Figure 6.5 shows an enlarged cross-section of the scattering foil.In passing through the foil, a projectile is scattered many times. For atypical foil thickness of 1 /Lm (10-6 m), the projectile is scattered by about104 atoms, each of which deflects the projectile by an angle whose averagevalue is {}avg.The total scattering angle for any particular projectile isdetermined by statistical considerations, since some of the individual scat-terings move the projectile toward larger scattering angles and some towardsmaller angles, as shown in Figure 6.5. This is an example of a "randomwalk" problem-for N scatterings, the average net scattering angle e isrelated to the average individual scattering angle by

{}=vN {}avg ( 6 . 1 0 )For N = 104 and {}avg= 0.01, we expect to see an average net scatteringangle of about 1, which is consistent with observations.

The Thomson model for scattering fails when we examine the probabilityfor scattering at large angles. If each individual scattering deflects theprojectile through an angle of around 0.01, then to observe projectilesscattered through a total angle greater than 90, we must have about 104successive scatterings, all of which push the projectile toward larger angles.Since the probabilities of individual scatterings toward either larger orsmaller angles are equal, the probability of having 104 successive scatteringstoward larger angles, like the probability of finding 104 successive heads intossing a coin, is about (112)10.000=10-3000.

In anthe Ilargethereat itsing gsive

(Wefullnoton tthannonK o

Ascattpro

* SeSolid


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om-

Alebych.r auteis

is

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6.3 The Rutherford Nuclear Atom 179

6.3 THE RUTHERFORD NUCLEAR ATOM

An experiment of this sort was performed by Hans Geiger and ErnestMarsdenin the laboratory of Ernest Rutherford in 1910. Their resultsshowedthat the probability of an alpha particle scattering at angles greaterthan900was about 10-4. This remarkable discrepancy between the expectedvalue(10-3000) and the observed value (10-4) was described by Rutherfordinth~sway:

It was quite the most incredible event that ever happened to me in mylife. I t was as incredible as if you fired a I5-inch shell at a piece oftissuepaper and it came back and hit you.Theanalysis of the results of such scattering experiments led Rutherford toproposethat the mass and positive charge of the atom were not distributeduniformlyover the volume of the atom, but instead were concentrated inanextremely small region, about 10-14 m in diameter, at the center of theatom.In the next section we will see how this proposal is consistent withthelarge-angle scattering results.

In analyzing the scattering of alpha particles, Rutherford concluded thatthemost likely wayan alpha particle (m =4 u) can be deflected throughlargeangles is by a single collision with a more massive object. Rutherfordthereforeproposed that the charge and mass of the atom were concentratedatitscenter, in a region called the nucleus. Figure 6.6 illustrates the scatter-inggeometry in this case. The projectile, of charge ze , experiences a repul-siveforce due to the positively charged nucleus:F = (ze)(Ze)4m;or2 (6.11)

(Weassume that the projectile is always outside the nucleus, so it feels thefullnuclear charge Ze.) The atomic electrons, with their small mass, donotappreciably affect the path of the projectile and we neglect their effectonthe scattering. We also assume that the nucleus is so much more massivethanthe projectile that it does not move during the scattering process; sincenorecoilmotion is given to the nucleus, the initial and final kinetic energiesK of the projectile are equal.As Figure 6.6 shows, for each impact parameter b, there is a certainscatteringangle 8, and we need the relationship between band 8. Theprojectilecan be shown" to follow a hyperbolic path; in polar coordinates

'See, for example, R. M. Eisberg and R. Resnick, Quantum Physics of Atoms, Molecules,Solids,Nuclei, and Particles, 2nd ed. (New York, Wiley, 1985).

Ernest Rutherford (1871-1937, En-gland). Founder of nuclear physics,he is known for his pioneering workon alpha-particle scattering and ra-dioactive decays. His inspiring lead-ership influenced a generation ofBritish nuclear and atomic scientists.


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Rutherford scattering angle

FIGURE 6.8 Scattering geometryfor many atoms. For impact parame-ter b, the scattering angle is e . If theparticle enters the atom within thedisc of area 1 T b 2, its scattering anglewill be larger than e .

~\\\\\ _ _ \ . r\ --- -..----.....--\~--T ()

... / b _j______________ t. _FIGURE 6.6 Scattering by a nu-clear atom. The path of the scatteredparticle is a hyperbola. Smaller im-pact parameters give larger scatter-ing angles. FIGURE 6.7 The hyperbolic tra-jectory of a scattered particle.

(b )

rand c f > , the equation of the hyperbola is1 1. zZe2-;.= b sm c f> + 8 1 T B O b 2 K (cos c f > - 1)

As shown in Figure 6.7, the initial position of the particle is c f> = 0,r ~ 00, and the final position is c f> = 1 T - 8, r ~ 00. Using the coordinatesat the final position, Equation 6.12 reduces to

( 6 . 1 2 )

zZe2 zZ e2b= - 8 Kcoti8 =2K-4-coti81 T B O 1 T B O ( 6 . 1 3 )(This result is written in this form so that e 2 / 4 1 T B o = 1.44 eVnm canbe easily inserted.) A projectile that approaches the nucleus with impactparameter b will be scattered at an angle 8; projectiles approaching withsmaller values of b will be scattered through larger angles, as shown i nFigure 6.6.We divide our study of the scattering of charged projectiles by nuclei(which is commonly called Rutherford scattering) into three parts: ( 1) c alc u-lation of the fraction of projectiles scattered at angles greater than somevalue of 8, (2) the Rutherford scattering formula and its experimentalverification, and (3) the closest approach of a projectile to the nucleus.

1. The Fraction of Projectiles Scattered at Angles Greater t h a n9. From Figure 6.6 we see immediately that every projectile with impactparameters less than a given value of b will be scattered at angles greaterthan its corresponding 8.What isthe chance of a projectile having an impactparameter less than a given value of b? Suppose the foil were one atomthick-a single layer of atoms packed tightly together, as in Figure 6 . 8 ,Each atom is represented by a circular disc, of area 1 T R z . If the foil containsN atoms, its total area is N 1 T R z . For scattering at angles greater than 0 , theimpact parameter must fall between zero and b-that is, the projectilemust approach the atom within a circular disc of area 1 T b z . If the projectilesare spread uniformly over the area of the disc, then the fraction ofprojectilesthat fall within that area is just 1 T b z / 1 T R z .A real scattering foil may be thousands or tens of thousands of atomsthick. Let t be the thickness of the foil and A its area, and let p and M bethe density and molar mass of the material of which the foil is made. Thevolume of the foil is then At, its mass is pAt, the number of moles is

so


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p A r l M , and the number of atoms or nuclei per unit volume is(6.14)

whereNA is Avogadro's number (the number of atoms per mole). As seenb y an incident projectile, the number of nuclei per unit area is nt =N A p t I M ; that is, on the average, each nucleus contributes an area( N A P t I M t 1 to the field of view of the projectile. For scattering at anglesgreaterthan 8, it must once again be true that the projectile must fall withinanarea rr b 2 of the center of an atom; the fraction scattered at angles greaterthane is just the fraction that approaches an atom within the area nb':

f< b = f> o = ntnb? (6.15) Fraction scattered at anglesgreater than ()assumingthat the incident particles are spread uniformly over the area ofthefoil.

f l ~ M n E 6 . 2A goldfoil ( p = 19.3 g/crrr', M = 197 g/mole) has a thickness of 2.0 X 10-4e m . It is u sed to scatter alpha particles of kinetic energy 8.0 MeV. (a ) Whatfraction of the alpha particles is scattered at angles greater than 90?(b ) What fraction of the alpha particles is scattered at angles between 90and45?~ H ~ T I O N( a ) Fo r this case the number of nuclei per unit volume can be computed as

N A P (6.02 X 1023atoms/mole)(19.3 g/cm ')n = - = - '- -- - -- - -- - --"- -'- - -"'- - -- - '-M 197 g/mole= 5.9 X 1022atoms/em-= 5.9 X 1028atoms/m '

Forscattering at 90, the impact parameter b can be found from Equa-tion6.13:

b- (2)(79) ( ) 0_ -14.- 2(8.0 X 106eV) 1.44eYnm cot45 -1.4 X 1~ mso rr b 2 = 6.4 X 10-28 m2/nucleus and we then have

1>90 ' = (5.9 X 1028nuclei/m3)(2.0 X 10-6 m)(6.4 X 10-28m2/nucleus)=7.5 X 10-5

( b ) Repeating the calculation for 8 = 45, we findb - (2)(79) ( ) _ -14-2(8.0X106eY) 1.44eYnm cot22.5 -3.4x10 m


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and f>45" = 4.4 X 10-4. If a total fraction of 4.4 X 10-4 is scattered at anglesgreater than 45, and of that, 7.5 X 10-5 is scattered at angles greater thandb 90, the fraction scattered between 45 and 90 must bet

FIGURE 6.9 Particles _ enteringthe ring between band b + db aredistributed uniformlyalonga ring ofangularwidth dO . A detector is at adistance r from the scatteringfoil.

Rutherford scattering formula

4.4 X 10-4 - 7.5 X 10-5 = 3.6 X 10-4

2. The Rutherford Scattering Formula and Its ExperimentalVerification. In order to find the probability that a projectile be scatteredinto a small angular range at e (between e and e + de), we require thaithe impact parameter lie within a small range of values db at b (see Figure6.9). The fraction, df, is thendf = nt(21Tb db)

from Equation 6.15. Differentiating Equation 6.13 we find db in terms o fde:

zZ e2 ( 2 1 )(1 d )db =- -- - csc " 2 e "2 e2K 41T80 ( 6 . 1 6 )and so

( Z Z ) 2 ( e 2 ) 2Idfl = tint 2K 41T80 csc2!ecot!ede ( 6 . 1 7 )(This minus sign in Equation 6.16 is not important-it just tells us t h a t Pincreases as b decreases.) Suppose we place a detector for the scatteredprojectiles at the angle e a distance r from the nucleus. The probability fora projectile to be scattered into the detector depends on df, which givesthe probability for scattered particles to pass through the ring of radiusr sin e and width r de. The area of the ring is dA = (21Trsin e)r de. In orderto calculate the rate at which projectiles are scattered into the detector wemust know the probability per unit area for scattering into the ring. Thisis IdflldA, which we call N(e), and, after some manipulation, we find:

N(e) =4~t2( ~ ~ r ( 4 : : J 2i)!eThis is the Rutherford scattering formula.In Rutherford's laboratory, Geiger and Marsden tested the predictionsof this formula in a remarkable series of experiments requiring great careand experimental skill. They used alpha particles (z = 2) to scatter fromnuclei in a variety of thin metal foils. In those days before electronicrecording and processing equipment was available, Geiger and Marsdenobserved and recorded the alpha particles by counting the scintillations(flashes of light) produced when the alpha particles struck a zinc sulfidescreen. A schematic view of their apparatus is shown in Figure 6.10. In a U ,four predictions of the Rutherford scattering formula were tested:(a ) N(e) e x : t. With a source of 8-MeV alpha particles from radioactive

decay, Geiger and Marsden used scattering foils of varying thicknesses ,while keeping the scattering angle e fixed at about 25. Their results aresummarized in Figure 6.11, and the linear dependence of N(e) on t is

( 6 . 1 8 )

proje


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apparent. This is also evidence that, even at this moderate scattering angle,singlescattering is much more important than multiple scattering. (In arandomstatistical theory of multiple scattering, the probability for scatter-ingat a large angle would be proportional to the square root of the numberofsingle scatterings, and we would expect N(e) e x : t1/2 Figure 6.11 shows [ B B .clearlythat this is not true.) ,----1"This result emphasizes a significant difference between scattering by a

Thomsonmodel atom and a Rutherford nuclear atom: In the Thomsonmodel,the projectile is scattered by every atom along its path as it passesthrough the foil (see Figure 6.5), while in the Rutherford nuclear modelthenucleus is so tiny that the chance of even a single significant encounterissmalland the chance of encountering more than one nucleus is negligible.

(b ) N (e ) e x : Z2 . In this experiment, Geiger and Marsden used a varietyofdifferent scattering materials, of approximately (but not exactly) thesamethickness. This proportionality is therefore much more difficult totest than the previous one, since it involves the comparison of differentthicknessesof different materials. However, as shown in Figure 6.12, theresultsare consistent with the proportionality of N(e) to Z2 .(c ) N(e) e x: K-2. In order to test this prediction of the Rutherford scatter-

ingformula, Geiger and Marsden kept the thickness of the scattering foilconstant and varied the speed of the alpha particles. They accomplishedthisbyslowing down the alpha particles emitted from the radioactive source,usingthin sheets of mica. From independent measurements they knew theeffectof different thicknesses of mica on the velocity of the alpha particles.The results of the experiment are shown in Figure 6.13; once again we seeexcellentagreement with the expected relationship.( d ) N(e) e x: sin" !e . This dependence of N on e is perhaps the most

important and distinctive feature of the Rutherford scattering formula.It a lso produces the largest variation in N over the range accessible byexperiment. In the previous tests, N varies by perhaps an order of magni-tude;in this case N varies by about five orders of magnitude from thesmallerto the larger angles. Geiger and Marsden used a gold foil and variede from 5 to 150, to obtain the relationship between Nand e plotted inFigure6.14. The agreement with the Rutherford formula is again very good.Thus all predictions of the Rutherford scattering formula were confirmedb y experiment, and the "nuclear atom" was verified.

3 . The Closest Approach of a Projectile to the Nucleus. A positivelychargedprojectile slows down as it approaches a nucleus, exchanging partof its initial kinetic energy for the electrostatic potential energy due to thenuclear repulsion. The closer the projectile gets to the nucleus, the morepotential energy it gains, since

U= _1_zZe247TBO r

The maximum potential energy, and thus the minimum kinetic energy,occursat the minimum value of r. We assume that U=0 when the projectileis far from the nucleus, where it has total energy E = K = !mv2 As theprojectile approaches the nucleus, K decreases and U increases, but

FFIGURE 6.10 Schematic diagramof alpha-particle scattering experi-ment. A radioactive source of alphaparticles is in a shield with a smallhole. Alpha particles strike the foilF and are scattered into the angularrange dB . Each time a scattered parti-cle strikes the screen S a flash of lightis emitted and observed with themovable microscope M.

Silver

2 lE:::JZFo il t h ic k nes s

FIGURE 6.11 The dependence ofscattering rate on foil thickness forthree different scattering foils.

FIGURE 6.12 The dependence ofscattering rate on the nuclear chargeZ for foils of different materials. Thedata are plotted against Z2.


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1000 ---.--1-- '-----_._---_.500 _.-1--- -----\ --- ---

-0 \! ! : ' 200Q) Slope =12 \ 1 \ 1OJUIf)100Q ; --.CJE 50: : :JZ10 , - - - - - - - - , - - - - - - , - - - , - - - , - - 1 - - , - - I- - - - ' - - - - ' - - ' - - - - - - - - '0.1 0.2 0.5 1

Re la tiv e k in e tic e n er gyo f a lp h a p ar tic le s

FIGURE 6.13 The dependence ofscattering rate on the kinetic energyof the incident alpha particles forscattering by a single foil. Note thelog-log scale; the slope of -2 showsthat log N DC -2 log K, or N DC K-2,as expected from the Rutherfordformula.

Distance of closest approach

(See Figure 6.15.)Angular momentum is also conserved. Far from the nucleus, themomentum L is mvb, and at rmin, the angular momentum is mVminrmin,

u + K remains constant. At the distance rmin, the speed is Vmin and:_ 1 2 1 zZe2 _ 1 2E --2mVmin +-4---- 2-mv1TBO rmin

orbVmin =-v,rmin

Combining Equations 6.19 and 6.20, we find!mv2 =!m (b2v2) + _1_zZe22 2 r~in 41TBo r min ( 6 . 2 1 )

This expression can be solved for the value of rmin'Notice that the kinetic energy of the projectile is not zero at r min, u n l e s sb = O. (See Figure 6.15.) In this case, the projectile would lose all of i t skinetic energy, and thus get closest to the nucleus. At this point its distancefrom the nucleus is d, the distance of closest approach. We find this distanceby solving Equation 6.21 for rmin when b = 0, and obtain

d = _ _ ! _ . , zZe241TBO K

!\ ---_.

\.--~ !----- -----,--- - - -_ .. , - - _ . _ . . . . _.\ ~~ in4~-,

~


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angularin, SO:

(6.20)

(6.21)

unlessof itsance

6.22)

(6.19)

6.4 LINE SPECTRA -

f U M P L E G . 3Findthe distance of closest approach of an 8.0-MeV alpha particle incidentona gold foil.~H~ IO N

zZe21 1d = 4 1 T e o K = (2)(79)(1.44 eVnm) 8.0 X 106 eV= 28 X 10-6nm= 2.8 X 10-14 m

Althoughthis distance is very small (much less than an atomic radius, forexample)it is larger than the nuclear radius of gold (about 7 X 10-15 m).Thusthe projectile is always outside of the nuclear charge distribution, andtheRutherford scattering law, which was derived assuming the projectileto remain outside the nucleus, correctly describes the scattering. If weincreasethe kinetic energy of the projectile, or decrease the electrostaticrepulsionby using a target nucleus with low Z, this may not be the case.Undercertain circumstances, the distance of closest approach can be lessthanthe nuclear radius. When this happens, the projectile no longer feelsthefullnuclear charge, and the Rutherford scattering law no longer holds.In f act, as we discuss in Chapter 12, this gives us a convenient way ofmeasuringthe size of the nucleus.

Theradiation from atoms can be classified into continuous spectra anddiscreteor line spectra. In a continuous spectrum, all wavelengths fromsomeminimum, perhaps 0, to some maximum, perhaps approaching 00, areemitted. The radiation from a hot glowing object is an example of thiscategory.White light is a mixture of all of the different colors of visiblelight;an object that glows white hot is emitting light at all wavelengths ofthevisible spectrum. If,on the other hand, we force an electric dischargein a tube containing a small amount of the gas or vapor of a certainelement,such as mercury, sodium, or neon, light is emitted at a few discretewavelengths and not at any others. Examples of such "line" spectra areshownin Figure 6.16 and Color Plate 4. The strong 436 nm (blue) and 546nm(green) lines in the mercury spectrum give mercury-vapor street lightstheirblue-green tint; the strong yellow line at 590 nm in the sodium spectrum(whichis actually a doublet-two very closely spaced lines) gives sodium-vaporstreet lights a softer, yellowish color. The intense red lines of neonareresponsible for the red color of "neon signs."

6.4 Line Spectra 185

u= zZe2 _1_4nEo " m m1 2K = 2 mV min

FIGURE 6.15 Closest approachof the projectile to the nucleus.


14/33


Vaportubev

I I I I HgU ltravio le t V is ible

I I I I I Na200 300 400 500 600 700

Wa ve le ng th (n rn )FIGURE 6.16 Apparatus for observing line spectra. Light isemit-ted when an electric discharge iscreated in a tube containing a vaporof an element. The light passes through a dispersive medium, suchas a prism or a diffraction grating, which displays the individualcomponent wavelengths at different positions. Sample line spectraare shown for mercury and sodium in the visible and near ultraviolet.

Another possible experiment is to pass a beam of white light, containingall wavelengths, through a sample of a gas. When we do so, we find thatcertain wavelengths have been absorbed from the light, and again a linespectrum results. These wavelengths correspond to many (but not all) o fthe wavelengths seen in the emission spectrum. Examples are showninFigure 6.17.In general, the interpretation of line spectra is very difficult in complexatoms, and so we will deal for now with the line spectra of the simplestatom, hydrogen. Regularities appear in both the emission and absorptionspectra, as shown in Figure 6.18. Notice that, as with the mercury andsodium spectra, some lines present in the emission spectrum are missingfrom the absorption spectrum.In our discussion of blackbody radiation, we found an example of the"reverse scientific method," in which, in the absence of a theory thatexplains the data, we try to find a function that fits the data, and then try


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6.4 Line Spectra 187

Blue

White Gasl ight cel lsource

f\IU SlitHg

Visible

I I I I I INa

2 0 0 300 400 500 600 700Wa ve le ng th (n rn )

FIGURE 6.17 Apparatus for observing absorption spectra. Alightsource produces a continuous range of wavelengths, some ofwhichare absorbed by a gaseous element. The light is dispersed, asinFigure 6.16. The result is a continuous "rainbow" spectrum, withd a r k lines at wavelengths where the light was absorbed by the gas.

to find a theory that explains the derived function. Johannes Balmer, aSwissschoolteacher, noticed that the wavelengths of the group of emissionlines of hydrogen in the visible region could be very accurately calculatedf r o m the formula

n2A= 364.5~4n - (6.23) Balmer formulawhere" is in units of nm and where n can take integer values beginningwith3. For example, for n = 3, A = 656.1 nm. This formula is now knownastheBalmer formula and the series of lines that it fits is called the Balmerseries.The wavelength 364.5 nm, corresponding to n -i> 00, is called thes e r i e s limit. Itwas soon discovered that all of the groupings of lines in thehydrogen spectrum could be fit with a similar formula of the form

n2A =Alimit -2--2n - no (6.24)where "limit is the wavelength of the appropriate series limit and where ntakesinteger values beginning with no + 1. (For the Balmer series, no =


16/33


90nm 100 nm 110 nm 120 nmI

i 1 1 1 1 1 1 II

I

I I

Lyman(ultraviolet)4 00 nm 500 nm 600 nm

i 1 1 1 1 1 ' I I II I

Balmer(visible)

0.5.um 1.0.um 1.5.um z . o u mI

i I I I l l l l II I

Paschen(infrared)1.0.um 2.0.um 3.0.um 4.0.um

I

i 1 1 1 1 1 1 1 ' I II

'IBrackett(infrared)

2.0.um 4.0.um 6.0.um 8.0.um

, i 1 1 1 1 1 1 1 II

I

I I

Pfund(infrared)FIGURE 6.18 Emission and absorption spectral series of hydro-gen. Note the regularities in the spacing of the spectral lines. Thelines get closer together as the limit of each series (dashed line)is approached. Only the Lyman series appears in the absorptionspectrum; all series are present in the emission spectrum.2.) The other series are today known as Lyman (n o = 1), Paschen (n o =3), Brackett (n o = 4), and Pfund (n o = 5).Another interesting property of the hydrogen wavelengths issummarizedRitz combination principle in the Ritz combination principle. If we convert the hydrogen emissionwavelengths to frequencies, we find the curious property that certain pairsof frequencies added together give other frequencies which appear inthe spectrum.Any successful model of the hydrogen atom must be able to explain th eoccurrence of these interesting arithmetic regularities in the emissionspectra.E X H M P L E 6 . 4The series limit of the Paschen series (n o = 3) is 820.1 nm. What are th ethree longest wavelengths of the Paschen series?S O L U T I O NFrom Equation 6.24, n2A=820.1~32n - (n=4,5,6, ... )


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o =

r in

he

6.5 THE BOHR MODEL

Thethree longest wavelengths are:n =4: 42A = 820.1 42 _ 32 = 1875 nm

52A=820.1 52 _ 32=1281 nm62A=820.1 62 _ 32 = 1094 nm

n =5:

n =6:Thesetransitions are in the infrared region of the electromagnetic spectrum.

B ~ M P ~ E 6 . SShowthat the longest wavelength of the Balmer series and the longest twowavelengths of the Lyman series satisfy the Ritz combination principle.Forthe Lyman series, Alimit =91.13 nm.~ H U T I O NThelongest wavelength of the Balmer series was found previously to be656.1 nm, using Equation 6.23 with n = 3. Converting this to a frequency,wefind v = 4.57 X 1014 Hz. Using Equation 6.24 for ri = 2 and 3 withn o = 1, we find the longest two wavelengths of the Lyman series to be121.5 nm and 102.5 nm, corresponding to frequencies of 24.67 x 1014 Hzand29.24 X 1014 Hz. Adding the smallest frequency of the Lyman seriesto the smallest frequency of the Balmer series gives the next smallestLymanfrequency:

24.67 X 1014 Hz + 4.57 X 1014 Hz = 29.24 X 1014 Hzdemonstrating the Ritz combination principle.

Following Rutherford's proposal that the mass and positive charge areconcentrated in a very small region at the center of the atom, the Danishphysicist Niels Bohr in 1913 suggested that the atom was in fact like aminiature planetary system, with the electrons circulating about the nucleuslikeplanets circulating about the Sun. The atom thus doesn't collapse underthe influence of the electrostatic Coulomb attraction between nucleus andelectrons for the same reason that the solar system doesn't collapse underthe influence of the gravitational attraction between Sun and planets. Inboth cases, the attractive force provides the centripetal acceleration neces-sary to maintain the orbital motion.We consider for simplicity the hydrogen atom, with a single electroncirculating about a nucleus with a single positive charge, as in Figure 6.19.

6.5 The Bohr Model 189

Niels Bohr (1885-1962, Denmark).He developed a successful theory ofthe radiation spectrum of atomic hy-drogen and also contributed the con-cepts of stationary states and comple-mentarity to quantum mechanics.Later he developed a successful the-ory of nuclear fission. The instituteof theoretical physics he founded inCopenhagen attracts scholars fromthroughout the world.


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19 0 Chapter 6 The Rutherford-Bohr M odel of the A tom

~

v -e ...." . . . ." . . . ./ '

II F "I \I \I \I II \I \

I \\ ~+Ze \I\ I\ I1\ -: /\( /

, I, /-, /. . . . ". . . . - -.. . - -------FIGURE 6.19 The Bohr model ofthe atom (Z = 1 for hydrogen).

Quantization of angularmomentum

Hydrogen atom radii

The radius of the circular orbit is r, and the electron (of mass m) moveswith constant tangential speed u. The attractive Coulomb force providesthe centripetal acceleration u 2/ r, soF _ 1 qsqz _ 1 e2 _ mo? ( )- 417 1::07-7TBO2 - -r- 6 . 2 5

Manipulating this equation, we can find the kinetic energy of the electron(we are assuming the nucleus to remain at rest-more about this later)1 1 e2K=-mu2=---2 87TBOr

The potential energy of the electron-nucleus system is the Coulomb patential energy:1 e2U=---47TBOr

The total energy E =K + U is obtained by adding Equations 6.26 and 6 . 2 7 :1 e2E=---87TBOr

We have ignored one serious difficulty with this model thus far. Classicalphysics requires that an accelerated electric charge, such as our orbitingelectron, must continuously radiate electromagnetic energy. As it radiatesthis energy, its total energy decreases, the electron spirals in toward thenucleus and the atom collapses. To overcome this difficulty, Bohr madeabold and daring hypothesis-he proposed that there are certain specialstates of motion, called s ta tio na ry s ta te s, in which the electron may existwithout radiating electromagnetic energy. In these states, according toBohr,the angular momentum L of the electron takes values that are integermultiples of h. In stationary states, the angular momentum of the electronmay have magnitude h, 2 h, 3h, ... , but never such values as 2 .5h or 3.lfl.This is calJed the q uan tiza tion of an gula r m om entu m.In a circular orbit, the position vector r that locates the electron relativetc? the nucleus is always perpendicular to its linear momentum p . Theangular momentum, which is defined as L = r X p, has magnitude L =rp = mur when r is perpendicular to p. Thus Bohr's postulate is

mu r = nhwhere n is an integer (n = 1, 2, 3, ... ). We can use this expression withEquation 6.26 for the kinetic energy

1 1 (nh)2 1 e22mu2 = 2m mr = 87TBO--- ;to find a series of allowed values of the radius r:

47TBOh2r = -- - n2 = a n211 me: 0where the Bohr radius ao is defined as

47TBoh2ao =--2- =0.0529 nmme

( 6 . 2 6 )

( 6 . 2 7 )

( 6 . 2 8 )

( 6 . 2 9 )

( 6 . 3 0 )

or


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movesdes

(6.25)ctron

later)(6.26)poten-

(6.27)6.27:(6.28)

iatesd theade acialexisteger3.1h.tiveTheL =

with

Thisimportant result isvery different from what we expect from classicalphysics.A satellite may be placed into Earth orbit at any desired radiusbyboosting it to the appropriate altitude and then supplying the propertangential speed. This is not true for an electron's orbit-only certain radiiareallowed by the Bohr model. The radius of the electron's orbit may bea Q , 4ao, 9ao, 16ao, and so forth, but never 3ao or 5.3ao..Our expression for r may be combined with Equation 6.28 to give

me' 1E =- -n 32rrzdhz nZWehave added a subscript n to the energy to serve as an index to identifythe energy levels. The constants may be evaluated, yielding

(6.33)

E = -13.6 eV11 nZ (6.34)

Theenergy levels are indicated schematically in Figure 6.20. The electron'senergyis quantized-only certain energy values are possible. In its lowestlevel,with n = 1, the electron has energy E1 = -13.6 eV and orbits witharadius of rl = 0.0529 nm. This state is the ground state. The higher states(n = 2 with Ez = -3.4 eV, n = 3 with E3 = -1.5 eV, etc.) are theexcited states.When the electron and nucleus are separated by an infinite distance,corresponding to n = 00, we have E = O . We might construct a hydrogenatomby beginning with the electron and nucleus separated by an infinitedistanceand then bring the electron closer to the nucleus until it is in orbitin a particular state n. Since this state has less energy than the E = 0 withwhichwe began, we have released an amount of energy equal to I E , , ! .Conversely, if we have an electron in a state n, we can "take the atomapart" by supplying an energy I En l . This energy is known as the bindingenergy of the state n. If we supply more energy than I E " I to the electron,the excess will appear as kinetic energy of the now free electron.The excitation energy of an excited state n is the energy above the groundstate, En - E 1 Thus the first excited state (n = 2) has excitation energy

!1 E = Ez - E1 = -3.4 eV - (-13.6 eV) = 10.2 eVthe second excited state has excitation energy 12.1 eV, and so forth.We previously discussed the emission and absorption spectra of atomichydrogen, and our discussion of the Bohr model is not complete withoutanunderstanding of the origin of these spectra. Bohr postulated that, eventhoughthe electron doesn't radiate when it remains in any particular station-arys tate, it can emit radiation when it moves to a lower energy level. Inthislower level, the electron has less energy than in the original level, andthe energy difference appears as a quantum of radiation whose energy hvisequal to the energy difference between the levels. That is, if the electronjumpsfrom n =n1 to n =n2, as in Figure 6.21,a photon appears with energyh u = En, - E" 2

or,using Equation 6.33 for the energies,(6.35)

(6.36)

6.5 The Bohr Model 19 1

Hydrogen atom energy levels

n = 00 " "E~~~~~1 7 .= 4 : :11=3------ ==04 =-D.8 eV3 =-1.5 eV2=-34V1=2------

11=1------ 1=-13.6 eVFIGURE 6.20 The energy levelsof atomic hydrogen.

~~)----_'!_::1/ . . . .

// . . . . . . . - - . . . . - - - . . . . . . . . . "1./ ,,\1./ "\I / n=n2'\ \I I \ \I I \ \, , I I, I , '\ \ I '\ \ I I\ \ I I\" / I\" ./ /\ ' . . . . . . . _,/ /, /, /

. . . . /


20/33


The wavelength of the emitted radiation isA = . = 6 4 1 T2S5 h3c ( nin~ )

II me" ni - n~ ( 6 . 3 7 )The inverse of the numerical coefficient in Equation 6.37 is called theRydberg constant R",

me4R =-::-:--;;--;;:-::-:;-cc 6 4 1 T 3s 1 l h 3cThe presently accepted numerical value is

R ", = 1.097 X 107 m

( 6 . 3 8 )

E x n M P L E G . GFind the wavelengths of the transitions from nl = 3 to nz = 2 and fromni = 4 to n2 = 2.S H U T I O NEquation 6.37 gives

1 ( 3222 )A = 1.097 X 107 m I 32 _ 22 = 656.1 nmand

These wavelengths are remarkably close to the values of the two longestwavelengths of the Balmer series (Figure 6.18). In fact, Equation 6.37 gives

A = (364.5 nm) ( n t ~4 )for the wavelength of transitions from any state nl to n2=2. This is identicalwith Equation 6.23 for the Balmer series. Thus we see that the radiationsidentified as the Balmer series correspond to transitions from higher levelsto the n = 2 level. Similar identifications can be made for other series ofradiations, as shown in Figure 6.22.The Bohr formulas also explain the Ritz combination principle, accordingto which certain frequencies in the emission spectrum can be summed togive other frequencies. Let us consider a transition from a state n3 to astate n2, which is followed by a transition from nz to nl . Equation 6.36 c a nbe used for this case to give .11"3->"2 = cR", ( ~ - ~)n3 n2

in

Thm

whitram

TheprimbyeneeneovetheTand

foun(lesspeFig


21/33

om

lsof

a

the

6.5 The Bohr Model 193

P h o t o n L ! " ! LO C '! 01 r - - . .-4 0 0 LOw a v e l e n g t h .-4 N r-, ._f r r: i r- t u:i u:i - s c :: i . .. ._fN0 01 01 01 01 LO 00 (Y) .-4


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1 Ze2F=--4 1 T e o r2

found in the absorption spectrum of hydrogen. A hydrogen atom in itsground state can absorb radiation of 10.20 eV and reach the first excitedstate, or of 12.09 eV and reach the second excited state, and so forth. Ahydrogen atom cannot absorb a photon of energy 1.89 eV (the first lineofthe Balmer series), because the atom is originally not in the n = 2 level.The Balmer series is therefore not found in the absorption spectrum.The Bohr theory for hydrogen can be used for any atom with a singleelectron, even though the nuclear charge may be greater than 1. For exampIe, we can calculate the energy levels of singly ionized helium (heliumwith one electron removed), doubly ionized lithium, etc. The nuclear electriccharge enters the Bohr theory in only one place-in the expression for theelectrostatic force between nucleus and electron, Equation 6.25. For anucleus of charge Ze, the Coulomb force acting on the electron is


Radii and energies oj one-electron atoms

( 6 . 4 0 )

That is, where we had e2 previously, we now have Ze2. Making the samesubstitution in the final results, we can find the allowed radii:

( 6 . 4 1 )and the energies become

( 6 . 4 2 )

The orbits in the higher-Z atoms are closer to the nucleus and havelarger (negative) energies; that is, the electron is more tightly bound tothe nucleus.

E x n M P ~ E 6 . 7Calculate the two longest wavelengths of the Balmer series of triply ionizedberyllium (Z = 4).S O L U T I O NSince the radiations of the Balmer series end with the n = 2 level, the twolongest wavelengths are the radiations corresponding to n = 3 ~ n = 2and n = 4 ~ n = 2. The energies of the radiations and their correspondingwavelengths are

E3 - E2 = -(13.6 eV)(42) G - 1 ) = 30.2 eV, =hc = 1240eV'nm =41 0/l E 30.2 eV . nm


23/33

me

to

2

6.6 The Franck-Hertz Experiment 195

E4 - E2 = -(13.6 eV)(42) C ~ - ~ ) =40.8 eVA= he = 1240 eV .nm = 30 4E 40.8eV . nm

These"adiations are in the ultraviolet region.

6.6 THE FRANCK-HERTZ EXPERIMENTLetus imagine the following experiment, performed with the apparatusshownschematically in Figure 6.23. A filament heats the cathode, whichthenemits electrons. These electrons are accelerated toward the grid bythepotential difference V, which we control. Electrons pass through thegridand reach the plate if V exceeds Vo , a small retarding voltage betweenthegrid and the plate. The current of electrons reaching the plate is mea-suredusing the ammeter A.Nowsuppose the tube isfilled with atomic hydrogen gas at a low pressure.A s the voltage is increased from zero, more and more electrons reach theplate,and the current rises accordingly. The electrons inside the tube maymakecollisions with atoms of hydrogen, but lose no energy in these colli-sions-the collisions are perfectly elastic. The only way the electron cangiveup energy in a collision is if the electron has enough energy to causethehydrogen atom to make a transition to an excited state. Thus, whentheenergy of the electrons reaches and barely exceeds 10.2 eV (or whenthe voltage reaches 10.2 V), the electrons can make inelastic collisions,leaving10.2 eV of energy with the atom (now in the n = 2 level), and theoriginalelectron moves off with very little energy. If it should pass throughthe grid, the electron might not have sufficient energy to overcome the

= : t FCI~'~~

p

FIGURE 6.23 Franck-Hertz apparatus. Electrons leave the cath-ode C, are accelerated by the voltage V toward the grid G, andreach the plate P where they are recorded on the ammeter A.


24/33


*6.7 THE CORRESPONDENCE PRINCIPLE

- - - - - - - T - -i--I 14.7 Vi 9.8V

C 4.9V~:;u

5 10 15VoltageFIGURE 6.24 Result of Franck-Hertz experiment usingmercury va-por.The current drops at voltagesof4.9 V, 9.8 V (= 2 X 4.9 V), 14.7 V(= 3 X 4.9 V).

small retarding potential and reach the plate. Thus when V = 10.2 V, adrop in the current is observed. As V is increased further, we begin to seethe effects of multiple collisions. That is, when V = 20.4 V, an electroncan make an inelastic collision, leaving the atom in the n = 2 state. Theelectron loses 10.2 eV of energy in this process, and so it moves off afterthe collision with a remaining 10.2 eV of energy, which is sufficient to excitea second hydrogen atom in an inelastic collision. Thus, if a drop in thecurrent is observed at V, similar drops are observed at 2V, 3V, ....This experiment should thus give rather direct evidence for the existenceof atomic excited states. Unfortunately, it is not easy to do this experimenlwith hydrogen, because hydrogen occurs naturally in the molecular formH2, rather than in atomic form. The molecules can absorb energy inavariety of ways, which would confuse the interpretation of the experiment.A similar experiment was done in 1914 by James Franck and Gustav Hertz,using a tube filled with mercury vapor. Their results are shown in Figure6.24, which shows clear evidence for an excited state at 4.9 eV; wheneverthe voltage is a multiple of 4.9 V, a drop in the current appears. Coinciden-tally, the emission spectrum of mercury shows an intense ultraviolet lineof wavelength 254 nm, which corresponds to an energy of 4.9 eV; thisresults from a transition between the same 4.9-eV excited state and theground state. The Franck-Hertz experiment showed that an electron musthave a certain minimum energy to make an inelastic collision with an atom;we now interpret that minimum energy as the energy of an excited stateof the atom. Franck and Hertz were awarded the 1925 Nobel prize inphysics for this work.

We have seen how Bohr's model permits calculations of transition wave-lengths in atomic hydrogen that are in excellent agreement with the wave-lengths observed in the emission and absorption spectra. However, in orderto obtain this agreement, Bohr had to introduce two postulates that areradical departures from classical physics. In particular, an acceleratedcharged particle radiates electromagnetic energy according to classical p h y s -ics, but an electron in Bohr's atomic model, accelerated as it moves inacircular orbit, does not radiate (unless it jumps to another orbit). Here wehave a very different case than we did in our study of special relativity.You will recall, for example, that relativity gives us one expression for thekinetic energy, K = E - Eo, and classical physics gives us another, K =imu2; however, we showed that E - Eo reduces to imu2 when u c. Thusthese two expressions are really not very different-one is merely a specialcase of the other. The dilemma associated with the accelerated electron isnot simply a matter of atomic physics (as an example of quantum physics)being a special case of classical physics. Either the accelerated charge* This is an optional section that may be skipped without loss of continuity.


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6.7 The Correspondence Principle 19 7

radiates,or it doesn't! Bohr's solution to this serious dilemma was toproposethe correspondence principle, which states thatQuantum theory must agree with classical theory in the limit in whichclassicaltheory is known to agree with experiment,

orequivalentlyQuantum theory must agree with classical theory in the limit of largequantum numbers.Let us see how we can apply this principle to the Bohr atom. Accordingto classical physics, an electric charge moving in a circle radiates at afrequencyequal to its frequency of rotation. For an atomic orbit, the periodofrevolution is the distance traveled in one orbit, Znr, divided by the orbitalspeedv = V2Klm, where K is the kinetic energy:

T = 21fr V161f3eomr3V2Klm e (6.43)

(6.44)

wherewe use Equation 6.26 for the kinetic energy. The frequency v is theinverseof the period:1 ev =- = ---;:.=::;;===:;:T V161f3eomr3

UsingEquation 6.31 for the allowed orbits, we findme4 1v =n 321f3e~n3 n3 (6.45)

A "classical" electron moving in an orbit of radius r n would radiate at thisfrequencyvnIf we made the radius of the Bohr atom so large that it went from a

quantum-sized object (10-10 m) to a laboratory-sized object (10-3 m), theatomshould behave classically. Since the radius increases with increasingn like n2, this classical behavior should occur for n in the range 103-104Let us then calculate the frequency of the radiation emitted by such anatomwhen the electron drops from the orbit n to the orbit n - 1.AccordingtoEquation 6.36, the frequency is

v = 64;!:~n3 (n ~ 1)2 - ~ 2 )me4 2n-1

641f3e~n3 n2(n - 1) 2 (6.46),If n is very large, then we can approximate n - 1 by nand 2n - 1 by

2 n , which gives(6.47)


26/33

This is identical with Equation 6.45 for the "classical" frequency. The"classical" electron spirals slowly in toward the nucleus, radiating at thefrequency given by Equation 6.45, while the "quantum" electron jumpsfrom the orbit n to the orbit n - 1 and then to the orbit n - 2, and soforth, radiating at the frequency given by the identical Equation 6.47. (Whenthe circular orbits are very large, this jumping from one circular orbit tothe next smaller one looks very much like a spiral, as in Figure 6.25.)In the region of large n, where classical and quantum physics overlap,the classical and quantum expressions for the radiation frequencies areidentical. This is an example of an application of Bohr's correspondenceprinciple. The applications of the correspondence principle go far beyondthe Bohr atom, and this principle is important in understanding how weget from the domain in which the laws of classical physics are valid to thedomain in which the laws of quantum physics are valid.


6.8 DEFICIENCIES OF THE BOHR MODEL

FIGURE 6.25 (Top) A largequantum atom. Photons are emittedin discrete transitionsas the electronjumps to lower states. (Bottom) Aclassical atom. Photons are emittedcontinuously by the acceleratedelectron.

The Bohr model gives us a picture of how electrons move about the nucleus,and many of our attempts to explain the behavior of atoms refer to thispicture, even though it is not strictly correct. It is remarkable (perhapseven fortunate or coincidental) that the successes of the model, with itsnew concepts of discrete energy levels and stationary states, occurred afulldecade before de Broglie's work and the birth of wave mechanics.In a way, the good agreement between our calculated values of tbeBalmer series wavelengths and the experimental values is fortuitous, be-cause we made two errors. The first error is related to the finite massofthe nucleus and results from our neglect of the motion of the nucleus. Tbeelectron does not orbit about the nucleus, but instead the electron and thenucleus both orbit about their common center of mass. The kinetic energyshould thus include an additional term due to the nuclear motion, and theeffect of this inclusion is to reduce the Rydberg constant from the valueR o o (so called because it would be correct if the nuclear mass were infinite)to the value R = R,,/(l + mIM), where M is the mass of the nucleus. Tbiseffect tends to increase the calculated wavelengths by a factor of about1.00055. The second mistake we made was in converting the frequenciesinto wavelengths. The frequency of the emitted photon is directly relatedto the energy difference of the atomic levels; to find the wavelength, weuse the expression c = AV, which is correct only in vacuum. Since theexperiment is usually performed in laboratories in air, the correct expressionis AV =Vai" where Vair is the speed of light in air. This effect tends todecrease the calculated wavelengths by a factor of about 1.00029. Thus ourtwo mistakes offset one another to some extent.In spite of its successes, the Bohr model is at best an incomplete model.It is useful only for atoms that contain one electron (hydrogen, singlyionized helium, doubly ionized lithium, etc.), but not for atoms with twoor more electrons, since we have considered only the force between electronand nucleus, and not the force between the electrons themselves. Further-


27/33

to

re

e

Suggestions For Further Reading 19 9

more, fwe look very carefully at the emission spectrum, we find that manylinesare in fact not single lines, but very closely spaced combinations oftwoor more lines; the Bohr model is unable to account for these doubletsofspectral lines. The model is also limited in its usefulness as a basis fromwhicho calculate other properties of the atom; although we can accuratelycalculatethe energies of the spectral lines, we cannot calculate their intensi-ties.For example, how often will an electron in the n =3 state jump directlytothe n =: 1state, emitting the corresponding photon, and how often willit jump first to the n = 2 state and then to the n = 1 state, emitting twophotons?A complete theory should provide a way to calculate this property.A serious defect in the Bohr model is that it gives incorrect predictions

fortheangular momentum of the electron. For the ground state of hydrogen(n = 1), the Bohr theory gives L = h, while experiment clearly showsL = O .A more serious deficiency of the model is that it completely violates the

uncertainty relationship. (In Bohr's defense, remember that this was adecadebefore the introduction of wave mechanics, with its accompanyingideasof uncertainty.) The uncertainty relationship dx dpx ~ h is valid foranydirection in space. Ifwe choose the radial direction, then dr ts p, ~ h.Foran electron moving in a circular orbit, we know the value of r exactly,andthus A r = O . If it is moving in a circle we also know P I ' exactly (in factit isexactly zero), and so dpr = O . This simultaneous exact knowledge ofr andP I' violates the uncertainty principle.We do not wish, however, to discard the model completely. The Bohr

modelgives a mental picture of the structure of an atom that sometimescanbe useful. There are many atomic properties, especially those associatedwithmagnetism, which can be understood on the basis of Bohr orbits.Whenwe treat the problem correctly in the next chapter, we find that theenergylevels of hydrogen calculated by solving the Schrodinger equationare in fact identical with those of the Bohr model.In this chapter we considered two problems of atomic physics-

Rutherford scattering and the hydrogen emission spectrum. Both can beunderstood based on calculations done without reference to wave mechan-ics,even though we expect wave mechanics to be an important considerationon the atomic scale. In fact, these two examples are unique in that the"correct" wave-mechanical calculations of the Rutherford scattering for-mulaand the hydrogen emission wavelengths give the same results as ourclassicalcalculation! This was not the case with many other phenomenawehave studied, including blackbody radiation, Compton scattering, andtbephotoelectric effect. It is interesting to speculate on the developmentofphysics in the era of Bohr and Rutherford if their classical calculationsbadnot yielded correct results.

SUGGESTIONS FOR FURTHER READINGA discussionof many of the basic properties of atoms may be found in:M.R. Wehr, J. A. Richards, and T. W. Adair, Physics of the Atom (Reading,Addison-Wesley, 1978).


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For a historical perspective on the development of atomic theory, see:H. A. Boorse and L. Motz, editors, The World of the Atom (New York,Books, 1966).G. K. T. Conn and H. D. Turner, The Evolution of the Nuclear Atom (London,

IIiffe Books, 1965).F. Friedman and L. Sartori, The Classical Atom (Reading, Addison-Wesley, 1 9 6 5 ) .

For more details on the history of the Thomson and Bohr models, see:J. L. Heilbron, "J. J. Thomson and the Bohr Atom," Physics Today, April 1 9 7 7 ,p.23.J. L. Heilbron, "Bohr's First Theories of the Atom," Physics Today, October 1 9 8 5 ,

p.28.For a popular summary of Rutherford's work, see:E. N. da C. Andrade, "The Birth of the Nuclear Atom," Scientific American 195,93 (November 1956).The early papers on the Rutherford model and its experimental confirmation illustrate the difficulty of the experiments and the care and abilities of the experimenters,They are easily readable and require no mathematics beyond the present level.E. Rutherford, Philosophical Magazine 21, 669 (1911).H. Geiger, Proceedings of the Royal Society of London AS3, 492 (1910).H. Geiger and E. Marsden, Philosophical Magazine 25, 604 (1913).

QUESTIONS1. Does the Thomson model fail at large scattering angles or at small scatteringangles? Why?2. What principles of physics would be violated if we scattered a beam of a l p h a

particles with a single impact parameter from a single target atom at rest?3. Could we use the Rutherford scattering formula to analyze the scattering o f:(a) Protons incident on iron?

(b) Alpha particles incident on lithium (Z = 3)?(c) Silver nuclei incident on gold?(d) Hydrogen atoms incident on gold?(e) Electrons incident on gold?4. What determines the angular range de in the alpha-particle scattering e x p e r i -ment (Figure 6.10)?5. Why didn't Bohr use the concept of de Broglie waves in his theory?6. In which Bohr orbit does the electron have the largest velocity? Are w ejustified in treating the electron nonrelativistically in that case?7. How does an electron in hydrogen get from r = 4ao to r = ao without beinganywhere in between?


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77,

1 1 .

12 .

13 .

a:

Problems 201

8. How is the quantization of the energy in the hydrogen atom similar to thequantization of the systems discussed in Chapter 5? How is it different? Dothe quantizations originate from similar causes?

9 . In a Bohr atom, an electron jumps from state nl, with angular momentumn,f!, to state nz, with angular momentum n2n. How can an isolated systemchange its angular momentum? (In classical physics, a change in angular mo-mentum requires an external torque.) Can the photon carry away the differenceinangular momentum? Estimate the maximum angular momentum, relativeto the center of the atom, which the photon can have. Does this, suggestanother failure of the Bohr model?

10. Theproduct Enrn for the hydrogen atom is (1) independent of Planck's constantand (2) independent of the quantum number n. Does this observation haveanysignificance? Is this a classical or a quantum effect?(a) How does a Bohr atom violate the ~x ~p uncertainty relationship?(b) How does a Bohr atom violate the ~E ~t uncertainty relationship? (Whatis ! : : . . E ? What does this imply about ~t?What do you conclude about transitionsbetween levels"]List the assumptions made in deriving the Bohr theory. Which of these are aresultof neglecting small quantities? Which of these violate basic principlesofrelativity or quantum physics?List the assumptions made in deriving the Rutherford scattering formula.Which of these are a result of neglecting small quantities? Which of theseviolate basic principles of relativity or quantum physics?

14 . In both the Rutherford theory and the Bohr theory, we used the classicalexpression for the kinetic energy. Estimate the velocity of an electron in theBohr atom and of an alpha particle in a typical scattering experiment, anddecideif the use of the classical formula isjustified.

1 5 . In both the Rutherford theory and the Bohr theory, we neglected any waveproperties of the particles. Estimate the de Broglie wavelength of an electronin a Bohr atom and compare it with the size of the atom. Estimate the deBrogliewavelength of an alpha particle and compare it with the size of thenucleus.Is the wave behavior ~pected to be important in either case?

16. Whyare the decreases in currennin the Franck-Hertz experiment not sharp?17 . Asindicated by the Franck-Her tz experiment, the first excited state of mercuryisat an energy of 4.9 eV. Do you expect mercury to show absorption lines inthevisible spectrum?18 , Is the correspondence principle a necessary part of quantum physics or is it

merelyan accidental agreement of two formulas? Where do we draw the linebetween the world of quantum physics and the world of classical, nonquan-tumphysics?

PROBLEMS1 . Electrons in atoms are known to have energies in the range of a few eV. Showthat the uncertainty principle allows electrons of this energy to be confinedin a region the size of an atom (0.1 nm).


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2. Consider an electron in Figure 6.1 subject to the electric field of the sphereof positive charge Ze in which it is embedded. (a) Using Gauss' Law, sho~that the electric field on the electron due to the positive charge is 14.

1 ZeE=4 R3r1 T c o

(b) For this electric field, show that the force on the electron is given b )Equation 6.1. 15.

3. (a) Compute the oscillation frequency of the electron and the expected absorption or emission wavelength in a Thomson-model hydrogen atom. Use R =0.053 nm. Compare with the observed wavelength of the strongest emissionand absorption line in hydrogen, 122 nm. (b) Repeat for sodium (Z = 1 1 ) .Use R = 0.18 nm. Compare with the observed wavelength, 590 nm. 16.4. Use Equation 6.8 to find the maximum scattering angle for the Thomson atomand the impact parameter for which it occurs. 17.5. Consider the Thomson model for an atom with 2 electrons. Let the electronbe located along a diameter on opposite sides of the center of the sphere,each a distance x from the center. (a) Show that the configuration is stable if 18.x = R12. (b) Try to construct similar stable configurations for atoms with],4, 5, and 6 electrons.6. Alpha particles of kinetic energy 5.00 MeV are scattered at 90 by a gold foil. 19.(a) What is the impact parameter? (b) What is the minimum distance betweenalpha particles and gold nucleus? (c) Find the kinetic and potential energies 20. Aat that minimum distance. rn

7. How much kinetic energy must an alpha particle have before its distance of grclosest approach to a gold nucleus is equal to the nuclear radius (7.0 X 10-15 m )l di

8. What is the distance of closest approach when alpha particles of kinetic energy er6.0 MeV are scattered by a thin copper foil? 21. C

9. Protons of energy 5.0 MeV are incident on a silver foil of thickness in4.0 X 10-6 m. What fraction of the incident protons is scattered at angles: 22. A(a) Greater than 90? (b) Greater than 10? (c) Between 5 and 10? (d) Less lethan 50? 23. T10. Protons are incident on a copper foil 12 /-Lmthick. (a) What should the proton akinetic energy be in order that the distance of closest approach equal the nnuclear radius (5.0 frn)? (b) If the proton energy were 7.5 MeV, what is the iimpact parameter for scattering at 120? (c) What is the minimum distance 24. Ubetween proton and nucleus for this case? (d) What fraction of the protonsis scattered beyond 120? a

11. Alpha particles of kinetic energy K are scattered either from a gold foil or a tsilver foil of identical thickness. What is the ratio of the number of particlesscattered at angles greater than 90 by the gold foil to the same number for 25. ~the silver foil?

12. The maximum kinetic energy given to the target nucleus will occur in a h e a d -on collision with b = O. (Why?) Evaluate the maximum kinetic energy given 26.to the target nucleus when 8.0 MeV alpha particles are incident on a gold foil.Are we justified in neglecting this energy? 27.

13. The maximum kinetic energy that an alpha particle can transmit to an e l e c t r o noccurs during a head-on collision. Compute the kinetic energy lost by an alpha



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Problems 20 3

particle of kinetic energy 8.0 Me V in a head-on collision with an electron atrest. Are we justified in neglecting this energy in the Rutherford theory?

1 4 . Alpha particles of energy 9 .6 MeV are incident on a silver foil of thickness7. 0 /Lm. For a certain value of the impact parameter, the alpha particles loseexactly half their incident kinetic energy when they reach their minimumseparation from the nucleus. Find the minimum separation, the impact parame-ter, and the scattering angle.

15. Alpha particles of kinetic energy 6.0 Me V are incident at a rate of 3.0 X 107per second on a gold foil a t thickness 3.0 X 10-6 m. A circular detector ofdiameter 1.0 em is placed 12 ern from the foil at an angle of 30 with thedirection ofthe incident alpha particles. At what rate does the detector measurescattered alpha particles?

1 6 . In the n = 3 state of hydrogen, find the electron's velocity, kinetic energy,and potential energy.

II Use the Bohr theory to find the series wavelength limits of the Lyman andPaschen series of hydrogen.

1 8 . Show that the speed of an electron in the nth Bohr orbit of hydrogen isac/n , where a is the fine structure constant. What would be the speed in ahydrogenlike atom with a nuclear charge of Ze?

1 9. An electron is in the n = 5 state of hydrogen. To what states can the electronmake transitions, and what are the energies of the emitted radiations?2 0 . A hydrogen atom is in the n = 6 state. (a) Counting all possible paths, how

many different photon energies can be emitted if the atom ends up in theground state? (b) Suppose only Sn =1transitions were allowed. How manydifferent photon energies would be emitted? (c) How many different photonenergies would occur in a Thomson-model hydrogen atom?

11 Continue Figure 6.22, showing the transitions of the Paschen series and comput-ing their energies and wavelengths.

2 2 . A collection of hydrogen atoms in the ground state is illuminated with ultravio-let light of wavelength 59.0 nm. Find the kinetic energy of the emitted electrons.

2 3 . The ionization energy is the energy required to remove an electron from anatom. Find the ionization energy of: (a) The n = 3 level of hydrogen. (b) Then =2 level of He+ (singly ionized helium). (c) The n =4 level of Li"" (doublyionized lithium).

2 4 . Use the Bohr formula to find the energy differences E(nJ ~ n2 ) =En, - En2and show that (a) E(4 ~ 2) = E(4 ~ 3) + E(3 ~ 2); (b) E(4 ~ 1) =(4 ~ 2) + E(2 ~ 1). (c) Interpret these results based on the Ritz combina-tion principle.

2 5 . What is the difference in wavelength between the first line of the Balmerseries in ordinary hydrogen (M = = 1.01 u) and in "heavy" hydrogen (M = =2.01 u)?

26. Find the shortest and the longest wavelengths of the Lyman series of singlyionized helium.

27. Draw an energy-level diagram showing the lowest four levels of singly ionizedhelium. Show all possible transitions from the levels and label each transitionwith its wavelength.


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28. An electron is in the n = 8 level of ionized helium. (a) Find the three longeswavelengths that are emitted when the electron makes a transition from then = 8 level to a lower level. (b) Find the shortest wavelength that canbeemitted. (c) Find the three longest wavelengths at which the electron in then = 8 level will absorb a photon and move to a higher state, if we couldsomehow keep it in that level long enough to absorb. (d) Find the shorteslwavelength that can be absorbed.29. The lifetimes of the levels in a hydrogen atom are of the order of 10-8 s. Findthe energy uncertainty of the first excited state and compare it with the energy

of the state.30. The Handbook of Chemistry and Physics lists the following emission w a v e -lengths (in nm) for ionized helium:

251.1273.4320.3468.6541.1656.0

1012.41162.61863.723.7324.3025.63

30.38102.5108.5121.5164.0238.5

Using the same values of no as the hydrogen spectrum, group these spectrallines into series, showing the index n for each line identified. Give the serieslimit of each series. In what region of the electromagnetic spectrum is eachseries located?

31. The Handbook of Chemistry and Physics lists the following emission w a v e -lengths (in nm) for doubly ionized lithium: 11.39, 13.50, 54.00, 72.91. Identifythese lines with the proper spectral series (as in hydrogen), giving the indexn for each line and the series limit for each series.32. When an atom emits a photon in a transition from a state of energy EI toa

state of energy E2, the photon energy is not precisely equal to EI - 2 'Conservation of momentum requires that the atom must recoil, and so someenergy must go into recoil kinetic energy KR. Show that KR ;:(E1 - E2)2 /2Mc2 where M is the mass of the atom. Evaluate this recoil energyfor the n = 2 to n = 1 transition of hydrogen.33. A long time ago, in a galaxy far, far away, electric charge had not yet been

invented, and atoms were held together by gravitational forces. Compute theBohr radius and the n = 2 to n = 1 transition energy in a gravitationallybound hydrogen atom.

34. (a) Find an expression for the Bohr radius ao in terms of the fine structureconstant a (see Chapter 1), the rest energy of the electron, and the constanthe. (b) Do the same for the hydrogen ground-state energy E1 .

35. In a muonic atom, the electron is replaced by a negatively charged particlecalled the muon. The muon mass is 207 times the electron mass. (a) What isthe shortest wavelength of the Lyman series in a muonic hydrogen atom? Inwhat region of the electromagnetic spectrum does this belong? (b) How largeis the correction for the finite nuclear mass in this case? (See the discussionat the beginning of Section 6.8.)36. What is the radius of the first Bohr orbit of a muonic lead atom (Z = 8 2 ) 1

Compare with the nuclear radius of about 7 fm.37. An alternative development of the Bohr theory begins by assuming that thestationary states are those for which the circumference of the orbit is anintegral number of de Broglie wavelengths. (a) Show that this condition leads

to stangives t38. Showa trans~(mc2139. Supposgo to bsoon t

40. Compphoto](c) n

41. A hYa groufor thdecrea42. The f

photoexper43. CUPS44. CUPS


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n ben the

isIn

Problems 20 5

to standing de Broglie waves around the orbit. (b) Show that this conditiongivesthe angular momentum condition, Equation 6.29, used in the Bohr theory.

3 8 . Show that the energy of the photon emitted when a hydrogen atom makesa transition from state n to state n - 1 is, when n is very large, ilE = =rJ(mc2/n 3) where ll' is the fine structure constant.

39. Suppose all of the excited levels of hydrogen had lifetimes of 10-8 s. As wego to higher and higher excited states, they get closer and closer together, andsoon they are so close in energy that the energy uncertainty of each statebecomes as large as the energy spacing between states, and we can no longerresolve individual states. Use the result of Problem 38 for the energy spacingand find the value of n for which this occurs. What is the radius of such an atom?

4 0 . Compare the frequency of revolution of an electron with the frequency of thephotons emitted in transitions from n to n - 1 for (a) n = 10; (b) n = 100;(c ) n = 1000; (d) n = 10,000.

41. A hypothetical atom has only two excited states, at 4.0 and 7.0 eV, and hasa ground-state ionization energy of 9.0 eY. If we used a vapor of such atomsfor the Franck-Hertz experiment, for what voltages would we expect to seedecreases in the current? List all voltages up to 20 V.

4 2 . The first excited state of sodium decays to the ground state by emitting aphoton of wavelength 590 nm. If sodium vapor is used for the Franck-Hertzexperiment, at what voltage will the first current drop be recorded?

4 3 . CUPS Exercise 2 . 8 .44. CUPS Exercise 2.9.

Chapter 6 the Rutherford-Bohr Model of the Atom

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Transcript of Chapter 6 the Rutherford-Bohr Model of the Atom