nuclear atom

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Structure of the Atom



▪ The Atomic Models of Thomson andRutherford

▪ Rutherford Scattering

▪ The Classic Atomic Model

▪ The Bohr Model of the Hydrogen Atom

▪ Successes & Failures of the Bohr Model

▪ Characteristic X-Ray Spectra and Atomic

Number

▪ Atomic Excitation by Electrons

CHAPTER 4


The opposite of a correct statement is a false statement. But the opposite of a profound

truth may well be another profound truth.

An expert is a person who has made all the mistakes that can be made in a very narrow

field.

Never express yourself more clearly than you are able to think.

Prediction is very difficult, especially about the future. - Niels Bohr

Niels Bohr (1885-1962)



History

• 450 BC, Democritus – The idea that matter is composed of

tiny particles, or atoms.

• XVII-th century, Pierre Cassendi, Robert Hook – explained

states of matter and transactions between them with a model

of tiny indestructible solid objects.

• 1811 – Avogadro’s hypothesis that all gases at giventemperature contain the same number of molecules per unit

volume.

• 1900 – Kinetic theory of gases.

Consequence – Great three quantization discoveries of XXcentury: (1) electric charge: (2) light energy; (3) energy of

oscillating mechanical systems.



Historical Developments in Modern Physics • 1895 – Discovery of x-rays by Wilhelm Röntgen.

• 1896 – Discovery of radioactivity of uranium by HenriBecquerel

• 1897 – Discovery of electron by J.J.Thomson

• 1900 – Derivation of black-body radiation formula by MaxPlank.

• 1905 – Development of special relativity by Albert Einstein,and interpretation of the photoelectric effect.

• 1911 – Determination of electron charge by Robert Millikan.

• 1911 – Proposal of the atomic nucleus by Ernest Rutherford .

• 1913 – Development of atomic theory by Niels Bohr .• 1915 – Development of general relativity by Albert Einstein.

• 1924+ - Development of Quantum Mechanics by deBroglie,Pauli, Schrödinger, Born, Heisenberg, Dirac,….



There are 112 chemical elements that have

been discovered, and there are a couple ofadditional chemical elements that recently have

been reported.

Flerovium is the radioactive chemical element

with the symbol Fl and atomic number 114. The

element is named after Russian physicist GeorgyFlerov , the founder of the Joint Institute for Nuclear

Research in Dubna, Russia, where the element was

discovered.

Georgi Flerov (1913-1990)

The Structure of Atoms

The name was adopted by IUPAC on May 30, 2012. About 80 decays of atoms of

flerovium have been observed to date. All decays have been assigned to the fiveneighbouring isotopes with mass numbers 285–289. The longest-lived isotope

currently known is 289Fl with a half-life of ~2.6 s, although there is evidence for a

nuclear isomer, 289bFl, with a half-life of ~66 s, that would be one of the longest-

lived nuclei in the superheavy element region.



The Structure of Atoms

Each element is characterized by atom thatcontains a number of protons Z , and equal number

of electrons, and a number of neutrons N . The

number of protons Z is called the atomic number .The lightest atom, hydrogen (H ), has Z=1; the

next lightest atom, helium (He), has Z=2 ; the third

lightest, lithium (Li ), has Z=3 and so forth.



The Nuclear Atoms

Nearly all the mass of the atom is concentrated

in a tiny nucleus which contains the protons andneutrons.

Typically, the nuclear radius is approximately

from 1 fm to 10 fm (1fm = 10

-15

m). The distancebetween the nucleus and the electrons is

approximately 0.1 nm=100,000fm. This distance

determines the size of the atom.



Nuclear Structure

An atom consists of an extremely small, positivelycharged nucleus surrounded by a cloud of negativelycharged electrons. Although typically the nucleus is lessthan one ten-thousandth the size of the atom, thenucleus contains more than 99.9% of the mass of theatom!



The number of protons in the

nucleus, Z , is called the atomic

number . This determines whatchemical element the atom is. The

number of neutrons in the nucleus

is denoted by N . The atomic mass

of the nucleus, A, is equal to Z + N .

A given element can have

many different isotopes, which

differ from one another by the

number of neutrons contained in

the nuclei. In a neutral atom, the

number of electrons orbiting the

nucleus equals the number of

protons in the nucleus.




Evidence in 1900 indicated that

the atom was not a fundamental unit:

1. There seemed to be too many kinds

of atoms, each belonging to a distinct chemical

element (way more than earth, air, water, and fire!).

2. Atoms and electromagnetic phenomena wereintimately related (magnetic materials; insulators vs.

conductors; different emission spectra).

3. Elements combine with some elements but not with

others, a characteristic that hinted at an internal atomicstructure (valence).

4. The discoveries of radioactivity, x-rays, and the

electron (all seemed to involve atoms breaking apart in

some way).



The Nuclear Atoms

We will begin our study of atoms bydiscussing some early models, developed in

beginning of 20 century to explain the spectra

emitted by hydrogen atoms.



Atomic Spectra

By the beginning of the 20th century a large body of datahas been collected on the emission of light by atoms of

individual elements in a flame or in a gas exited by electricaldischarge.

Diagram of the spectrometer



Atomic Spectra

Light from the source passed through a narrow slit before

falling on the prism. The purpose of this slit is to ensure that allthe incident light strikes the prism face at the same angle so that

the dispersion by the prism caused the various frequencies that

may be present to strike the screen at different places with

minimum overlap.



The source emits only two wavelengths, λ2>λ

1. The source is

located at the focal point of the lens so that parallel light passes

through the narrow slit, projecting a narrow line onto the face ofthe prism. Ordinary dispersion in the prism bends the shorter

wavelength through the lager total angel, separating the two

wavelength at the screen.



In this arrangement each wavelength appears as a narrow line,

which is the image of the slit. Such a spectrum was dubbed aline spectrum for that reason. Prisms have been almost entirely

replaced in modern spectroscopes by diffraction gratings, which

have much higher resolving power.



When viewed through the spectroscope,

the characteristic radiation, emitted byatoms of individual elements in flame or ingas exited by electrical charge, appears asa set of discrete lines, each of a particular

color or wavelength.

The positions and intensities of the lines

are a characteristic of the element. Thewavelength of these lines could bedetermined with great precision.



Emission line spectrum of hydrogen in the visible and

near ultraviolet. The lines appear dark because thespectrum was photographed. The names of the first fivelines are shown. As is the point beyond which no linesappear, H

∞ called the limits of the series.



Atomic Spectra

In 1885 a Swiss schoolteacher, Johann Balmer,

found that the wavelengths of the lines in the visiblespectrum of hydrogen can be represented by formula

Balmer suggested that this might be a specialcase of more general expression that would be

applicable to the spectra of other elements.



Atomic Spectra

Such an expression, found by J.R.Rydberg

and W. Ritz and known as the Rydberg-Ritzformula, gives the reciprocal wavelengths as:

where m and n are integers with n>m, and R is

the Rydberg constant .



Atomic Spectra

The Rydberg constant is the same for all

spectral series.

For hydrogen the R H

= 1.096776 x 10 7 m-1.

For very heavy elements R approaches the

value R ∞ = 1.097373 x 10 7 m-1. Such empirical expressions were successful

in predicting other spectra, such as other

hydrogen lines outside the visible spectrum.



Atomic Spectra

So, the hydrogen Balmer series wavelength are

those given by Rydberg equation

with m=2 and n=3,4,5,…

Other series of hydrogen spectral lines were foundfor m=1 (by Lyman) and m=3 (by Paschen).



Hydrogen Spectral Series

Compute the wavelengths of the first lines

of the Lyman, Balmer , and Paschen series.



Emission line spectrum of hydrogen in the visible and

near ultraviolet. The lines appear dark because thespectrum was photographed. The names of the first fivelines are shown, as is the point beyond which no linesappear, H

∞ called the limits of the series.



The Limits of Series

Find the predicted by Rydberg-Ritz

formula for Lyman, Balmer, and Paschenseries.



A portion of the emission spectrum of sodium. The two

very close bright lines at 589 nm are the D1 and D

2 lines.

They are the principal radiation from sodium street

lighting.



A portion of emission spectrum of mercury.



Part of the dark line (absorption) spectrum of sodium.White light shining through sodium vapor is absorbed at

certain wavelength, resulting in no exposure of the film atthose points. Note that frequency increases toward theright , wavelength toward the left in the spectra shown.



Nuclear Models

Many attempts were made to construct a

model of the atom that yielded the Balmer and

Rydberg-Ritz formulas.

It was known that an atom was about 10-10m

in diameter, that it contained electrons muchlighter than the atom, and that it was electrically

neutral.

The most popular model was that of J.J.

Thomson, already quite successful in explainingchemical reactions.



Knowledge of atoms in 1900

Electrons (discovered in1897) carried the negative

charge.

Electrons were very light,

even compared to the atom.Protons had not yet been

discovered, but clearly

positive charge had to be

present to achieve chargeneutrality.



In Thomson’s view, when the atom was heated, the

electrons could vibrate about their equilibrium positions,thus producing electromagnetic radiation.

Unfortunately, Thomson couldn’t explain spectra with this

model.

Thomson’s Atomic Model

Thomson’s “plum-pudding”

model of the atom had the

positive charges spread

uniformly throughout a sphere

the size of the atom, with

electrons embedded in theuniform background.



The difficulty with all such models was that

electrostatic forces alone cannot produce stable

equilibrium. Thus the charges were required to move

and, if they stayed within the atom, to accelerate.However, the acceleration would result in continuous

radiation, which is not observed.

Thomson was unable to obtain from his model a set of

frequencies that corresponded with the frequencies of

observed spectra.

The Thomson model of the atom was replaced by

one based on results of a set of experimentsconducted by Ernest Rutherford and his student H.W.

Geiger .

E i t f G i d M d



Experiments of Geiger and Marsden

Rutherford, Geiger, andMarsden conceived a new

technique for investigating

the structure of matter by

scattering α particles from

atoms.



Rutherford was investigating radioactivity

and had shown that the radiations from uranium

consist of at least two types, which he labeled α and β.

He showed, by an experiment similar to that

of Thompson, that q /m for the α - particles was

half that of the proton.Suspecting that the α particles were double

ionized helium, Rutherford in his classical

experiment let a radioactive substance decay

and then, by spectroscopy, detected the spectra

line of ordinary helium.

Beta decay



Beta decay Beta decay occurs when the neutron to proton ratio is too

great in the nucleus and causes instability. In basic beta

decay, a neutron is turned into a proton and an electron.The electron is then emitted. Here's a diagram of beta

decay with hydrogen-3:

Beta Decay of Hydrogen-3 to Helium-3.

Alpha Decay



Alpha DecayThe reason alpha decay occurs is because the nucleus has too many

protons which cause excessive repulsion. In an attempt to reduce the

repulsion, a Helium nucleus is emitted. The way it works is that the Helium

nuclei are in constant collision with the walls of the nucleus and because

of its energy and mass, there exists a nonzero probability of transmission.

That is, an alpha particle (Helium nucleus) will tunnel out of the nucleus.

Here is an example of alpha emission with americium-241:

Alpha Decay of Americium-241 to Neptunium-237

G D



Gamma DecayGamma decay occurs because the nucleus is at too high

an energy. The nucleus falls down to a lower energy state

and, in the process, emits a high energy photon known as agamma particle. Here's a diagram of gamma decay with

helium-3:

Gamma Decay of Helium-3



Rutherford was investigating radioactivity

and had shown that the radiations from uranium

consist of at least two types, which he labeled α and β.

He showed, by an experiment similar to that

of Thompson, that q /m for the α - particles was

half that of the proton.Suspecting that the α particles were double

ionized helium, Rutherford in his classical

experiment let a radioactive substance decay

and then, by spectroscopy, detected the spectra

line of ordinary helium.



Schematic diagram of the Rutherford apparatus. The beam

of α - particles is defined by the small hole D in the shieldsurrounding the radioactive source R of 214Bi . The α beamstrikes an ultra thin gold foil F , and α particles are individuallyscattered through various angels θ . The experiment consistedof counting the number of scintillations on the screen S as a

function of θ .



A diagram of the original apparatus as it appear

in Geiger’s paper describing the results.



An α -particle by such an atom (Thompson model) would have

a scattering angle θ much smaller than 10 . In the Rutherford’s

scattering experiment most of the α -particles were eitherundeflected, or deflected through very small angles of the

order 10 , however, a few α -particles were deflected through

angles of 90 0 and more.



An α -particle by such an atom (Thompson model) would have

a scattering angle θ much smaller than 10 . In the Rutherford’s

scattering experiment a few α -particles were deflected through

angles of 90 0 and more.



Experiments of Geiger and Marsden

Geiger showed that many α particles were scattered from

thin gold-leaf targets at backward angles greater than 90°.

B f Aft



Electrons

can’t back-

scatter α

particles.

Calculate the maximum scattering angle - corresponding to the

maximum momentum change.

It can be shown that the maximummomentum transfer to the α particle is:

Determine θ max

by letting

Δ pmax

be perpendicular

to the direction of motion:

Before After

too small!



If an α particle is scattered by N electrons:

Try multiple scattering from electrons

The distance between atoms, d = n-1/3

, is:

N = the number of atoms across the thin gold layer , t = 6 × 10−7 m:

still too small!

n =

N = t / d



If the atom consisted of a positively charged

sphere of radius 10 -10 m, containing electrons as in

the Thomson model, only a very small scatteringdeflection angle could be observed.

Such model could not possibly account for the

large angles scattering. The unexpected large

angles α -particles scattering was described byRutherford with these words:

It was quite incredible event that ever

happened to me in my life. It was as incredibleas if you fired a 15-inch shell at a piece of tissue

paper and it came back and hit you.



even if the α

particle is scattered from all 79electrons in each atom of gold.

Experimental results were not

consistent with Thomson’s atomic

model.

Rutherford proposed that an atom

has a positively charged core

(nucleus!) surrounded by the

negative electrons.

Geiger and Marsden confirmed theidea in 1913.

Rutherford’s Atomic Model

Ernest Rutherford

(1871-1937)



Rutherford concluded that the large angle

scattering could result only from a single encounter

of the α particle with a massive charge with volume

much smaller than the whole atom.

Assuming this “nucleus” to be a point charge,

he calculated the expected angular distribution forthe scattered α particles.

His predictions on the dependence of scatteringprobability on angle, nuclear charge and kinetic

energy were completely verified in experiments.



Rutherford Scattering geometry. The nucleus is assumed to

be a point charge Q at the origin O . At any distance r the α particle experiences a repulsive force kq

α Q/r 2 . The α particle

travel along a hyperbolic path that is initially parallel to line

OA a distance b from it and finally parallel to OB, which

makes an angle θ with OA.

Rutherford Scattering



There’s a relationship

between the impact

parameter b and thescattering angle θ .

Rutherford Scattering

When b is small,

r is small.

the Coulomb force is large.

θ can be large and the particle can be

repelled backward.

where

cot(θ /2)

θ π 0



Any particle inside

the circle of area π

b0

2 will be similarly (or

more) scattered.

Rutherford

Scattering

The cross section σ = π b2 is related to

the probability for a particle being

scattered by a nucleus (t = foil thickness):

The fraction of incident

particles scattered is:



Force on a point charge versus distance r from the center

of a uniformly charged sphere of radius R . Outside thesphere the force is proportional to Q/r 2 , inside the sphere

the force is proportional to q I /r 2 = Qr/R 3, where q I = Q(r/R)3

is the charge within a sphere of radius r . The maximum

force occurs at r =R



Two α particles with equal kinetic energiesapproach the positive charge Q = +Ze withimpact parameters b

1 and b

2 , where b

1<b

2 .

According to equation for impact parameter inthis case θ

1

> θ 2

.

The path of α particle can be shown to be a



The path of α particle can be shown to be a

hyperbola, and the scattering angle θ can be relate to

the impact parameter b from the laws of classical

mechanics.

The quantity πb2 , which has the dimension of the

area , is called the cross section σ for scattering.



The cross section σ

is thus defined as the

number of particles

scattered per nucleusper unit time divided

by the incident

intensity.

The total number of nuclei of foil atoms in the area coveredby the beam is nAt , where n is the number of foil atoms per

unit volume. A is the area of the beam, and t is the thickness

of the foil.



The total number of particles scattered per second isobtained by multiplying πb2 I

0 by the number of nuclei in the

scattering foil. Let n be the number of nuclei per unit volume:

For a foil of thickness t , the total number of nuclei as “seen”

by the beam is nAt , where A is the area of the beam. The totalnumber scattered per second through angles grater than θ isthus πb2 I

0 ntA. If divide this by the number of α particles

incident per second I 0 A we get the fraction of α particles f

scattered through angles grater than θ :

f = πb2 nt

On the base of that nuclear model Rutherford



On the base of that nuclear model Rutherford

derived an expression for the number of α particles

ΔN that would be scattered at any angle θ :

All this predictions was verified in Geiger

experiments, who observe several hundredsthousands α particles.



(a)Geiger data for α scattering from thin Au and Ag foils.

The graph is in log-log plot to cover over several orders ofmagnitudes.

(b)Geiger also measured the dependence of ΔN on t fordifferent elements, that was also in good agreement with

Rutherford formula.



Data from Rutherford’s group showing observed α

scattering at a large fixed angle versus values of r d

computed from for various kinetic energies.

Th Si f th N l



The Size of the Nucleus

This equation can be used to estimate the

size of the nucleus

For the case of 7.7-MeV α particles the

distance of closest approach for a head-on collision is

The Classical Atomic Model



Consider an atom as a planetary system.

The Newton’s 2nd Law force of attraction

on the electron by the nucleus is:

where v is the tangential velocity of the electron:

The total energy is then:

This is negative, so

the system is bound,

which is good.

The Planetary Model is Doomed



The Planetary Model is Doomed

From classical E&M theory, an accelerated electric

charge radiates energy (electromagnetic radiation),which means the total energy must decrease. So the

radius r must decrease!!

Electron

crashes

into the

nucleus!?



According to classical physics, a charge e moving with anacceleration a radiates at a rate

(a) Show that an electron in a classical hydrogen atom spiralsinto the nucleus at a rate

(b) Find the time interval over

which the electron will reach r = 0,starting from r 0= 2.00 × 10 –10 m.

The Bohr’s Postulates



The Bohr s Postulates

Bohr overcome the difficulty of the collapsing atom by

postulating that only certain orbits, called stationary

states, are allowed, and that in these orbits the electrondoes not radiate. An atom radiates only when the electron

makes a transition from one allowed orbit (stationary state)

to another:

▪ 1. The electron in the hydrogen atom can move only in

certain nonradiating, circular orbits called stationary

states.

▪ 2. The photon frequency from energy conservation is

given by

where E i and E

f are the energy of initial and final state, h is the

Plank’s constant.



Such a model is mechanically stable , because the

Coulomb potential

provides the centripetal force

The total energy for a such system can be written as the

sum of kinetic and potential energy:





Combining the second postulate with the

equation for the energy we obtain:

where r 1 and r

2 are the radii of the initial and

final orbits.





To obtain the frequencies implied by the

perimental Rydberg-Ritz formula,

it is evident that the radii of the stable orbits must be

proportional to the squares of integers.





Bohr found that he could obtain this condition if he

postulates the angular momentum of the electron in astable orbit equals an integer times ħ=h/2π . Since the

angular momentum of a circular orbit is just mvr , the

third Bohr’s postulate is:

3. n=1,2,3……….

where ħ=h/2π=1.055 x 10 -34J·s=6.582x10 -16 eV·s





The obtained equation mvr = nh/2π=nħ relates the

speed of electron v to the radius r . Since we had

or

We can write

or





Solving for r , we obtain

where a0 is called the first Bohr’s radius

Bohr’s Postulates



Substituting the expression for r in equation for

frequency:

If we will compare this expression with Z=1 for f=c/λ

with the empirical Rydberg-Ritz formula:

we will obtain for the Rydberg constant

Bohr’s Postulates



Using the known values of m, e, and ħ, Bohr

calculated R and found his results to agree with the

spectroscopy data.The total mechanical energy of the electron in the

hydrogen atom is related to the radius of the circular

orbit

Energy levels



Energy levels

If we will substitute the quantized value of r

as given by

we obtain

Energy levels



or

where

The energies E n with Z=1 are the quantized allowed

energies for the hydrogen atom.



Energy level diagram for hydrogen showing the seven

lowest stationary states. The energies of infinite number of

levels are given by E n

= (-13.6/n2 )eV , where n is an integer.

A hydrogen atom is in its first excited state (n = 2). Using the



y g ( ) g

Bohr theory of the atom, calculate (a) the radius of the orbit,

(b) the linear momentum of the electron, (c) the angular

momentum of the electron, (d) the kinetic energy of theelectron, (e) the potential energy of the system, and (f) the

total energy of the system.

Energy levels



gy

Transitions between this allowed energies

result in the emission or absorption of aphoton whose frequency is given by

and whose wavelength is

Energy levels



gy

Therefore is convenient to have the value

of hc in electronvolt nanometers! hc = 1240 eV∙nm

Since the energies are quantized, the

frequencies and the wavelengths of theradiation emitted by the hydrogen atom are

quantized in agreement with the observed

line spectrum.



(a) In the classical orbital model , the electron orbits

about the nucleus and spirals into the center because ofthe energy radiated.(b) In the Bohr model, the electron orbits without

radiating until it jumps to another allowed radius of lowerenergy, at which time radiation is emitted.



Energy level diagram for hydrogen showing the seven loweststationary states and the four lowest energy transitions for theLyman, Balmer, and Pashen series. The energies of infinitenumber of levels are given by E

n = (-13.6/n2 )eV , where n is an

integer.

λ21

=hc / (E1-E

2)



The spectral lines corresponding to the

transitions shown for the three series.



λ21

=hc / (E1-E

2)

Compute the wavelength of the Hβ

spectral line of the



Compute the wavelength of the Hβ

spectral line of the

Balmer series predicted by Bohr model.

A hydrogen atom at rest in the laboratory emits a photon of



yd oge ato at est t e abo ato y e ts a p oto o

the Lyman α radiation. (a) Compute the recoil kinetic energy

of the atom. (b) What fraction of the excitation energy of the

n = 2 state is carried by the recoiling atom? (Hint: Useconservation of momentum.)

In a hot star, because of the high temperature, an atom can



absorb sufficient energy to remove several electrons from the

atom. Consider such a multiply ionized atom with a single

remaining electron. The ion produces a series of spectrallines as described by the Bohr model. The series

corresponds to electronic transitions that terminate in the

same final state. The longest and shortest wavelengths of the

series are 63.3 nm and 22.8 nm, respectively. (a) What is the

ion? (b) Find the wavelengths of the next three spectral linesnearest to the line of longest wavelength.

A stylized picture of Bohrf 1 2 3



circular orbits for n=1,2,3,4.The radii r

n≈n2 .

In high Z -elements(elements with Z ≥12 ),electrons are distributed overall the orbits shown. If anelectron in the n=1 orbit is

knocked from the atom (e.g.,by being hit by a fast electronaccelerated by the voltageacross an x-ray tube) thevacancy that produced is filed

by an electron of higher energy(i.e., n=2 or higher).

The difference in energy between the two orbits is emitted as aphoton, whose wavelength will be in the x-ray region of thespectrum, if Z is large enough.



Characteristic x-ray spectra. (a) Part the spectra of neodymium

(Z=60) and samarium (Z=62).The two pairs of bright lines arethe Kα and K β lines. (b) Part of the spectrum of the artificiallyproduced element promethium (Z=61), its Kα and K β lines fallbetween those of Nd and Sm. (c) Part of the spectrum of allthree elements.

(a)

(b)

(c)

The Franck-Hertz Experiment



While investigating the inelastic scattering of

electrons, J.Frank and G.Hertz in 1914 performed

an important experiment that confirmed by direct

measurement Bohr’s hypothesis of energy

quantization in atoms.

The experiment involved measuring the plate

current as a function of V 0 .




Schematic diagram of the

Franck-Hertz experiment.

Electrons ejected from theheated cathode C at zero

potential are drawn to the

positive grid G . Those

passing through the holes

in the grid can reach theplate P and contribute in

the current I , if they have

sufficient kinetic energy to

overcome the small backpotential ΔV . The tube

contains a low pressure

gas of the element being

studied.




p

As V 0 increased from 0, the currentincreases until a critical value (about 4.9 V for

mercury) is reached. At this point the current

suddenly decreases. As V 0 is increased further,

the current rises again. They found that when the electron’s

kinetic energy was 4.9 eV or greater, the vapor

of mercury emitted ultraviolet light of wavelength

0.25 μm.

Current versus acceleration voltage in the Franck-Hertz

experiment



The currentdecreases becausemany electrons lose

energy due toinelastic collisionswith mercury atomsin the tube andtherefore cannot

overcome the smallback potential.

C u r r e n t , ( m A m p )

experiment.



The regular spacing ofthe peaks in this curve

indicates that only acertain quantity of energy,4.9 eV, can be lost to themercury atoms. Thisinterpretation is confirmed

by the observation ofradiation of photon energy4.9 eV, emitted bymercury atoms.

C u r r e n t , ( m A m p )

Suppose mercury atoms have an energy level

4 9 V b h l l l A



4.9 eV above the lowest energy level. An atom can

be raised to this level by collision with an electron; it

later decays back to the lowest energy level byemitting a photon. The wavelength of the photon

should be

This is equal to the measured wavelength,

confirming the existence of this energy level of themercury atom.

Similar experiments with other atoms yield the

same kind of evidence for atomic energy levels.

Lets consider an experimental tube filled by



p y

hydrogen atoms instead of mercury. Electrons

accelerated by V 0

that collide with hydrogen

electrons cannot transfer the energy to letter

electrons unless they have acquired kinetic

energy

eV0=E

2 – E

1=10.2eV



If the incoming electron does not have

sufficient energy to transfer ΔE = E 2 - E 1 to thehydrogen electron in the n=1 orbit (ground state),

than the scattering will be elastic.



If the incoming electron does have at least ΔE

kinetic energy, then an inelastic collision can occur inwhich ΔE is transferred to the n=1 electron, moving itto the n=2 orbit. The excited electron will typicallyreturn to the ground state very quickly, emitting a

photon of energy ΔE .



Energy loss spectrum measurement. A well-definedelectron beam impinges upon the sample. Electronsinelastically scattered at a convenient angle enter the slit of the

magnetic spectrometer, whose B field is directed out of thepage, and turn through radii R determined by their energy (E

inc

– E 1 ) via equation



An energy-loss spectrum for a thin Al film.

Reduced mass correction



The assumption by Bohr that the nucleus is fixed is

equivalent the assumption that it has an infinity mass.

If instead we will assume that proton and electron both

revolve in circular orbits about their common center of mass we

will receive even better agreement for the values of the

Rydberg constant R and ionization energy for the hydrogen.

We can take in account the motion of the nucleus (the

proton) very simply by using in Bohr’s equation not the electron

rest mass m but a quantity called the reduce mass μ of the

system. For a system composed from two masses m1

and m2

the reduced mass is defined as:




If the nucleus has the mass M its kinetic energy

will be ½Mv 2 = p2 /2M , where p = Mv is themomentum.

If we assume that the total momentum of the

atom is zero, from the conservation of momentumwe will have that momentum of electron and

momentum of nucleus are equal on the magnitude.




The total kinetic energy is then:

The Rydberg constant equation than changed to:

The factor μ was called mass correction factor .

As the Earth moves around the Sun, its orbits are quantized. (a) Follow the

steps of Bohr’s analysis of the hydrogen atom to show that the allowed radii



steps of Bohr s analysis of the hydrogen atom to show that the allowed radii

of the Earth’s orbit are given by

where MS is the mass of the Sun, M

E is the mass of the Earth, and n is an

integer quantum number. (b) Calculate the numerical value of n. (c) Find the

distance between the orbit for quantum number n and the next orbit out fromthe Sun corresponding to the quantum number n + 1

nuclear atom

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