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Structure of the Atom

CHAPTER 4Structure of the Atom

- The Atomic Models of Thomson and Rutherford
- 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

Niels Bohr (1885-1962)

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

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 Avogadros hypothesis that all gases at

given temperature contain the same number of

molecules per unit volume. - 1900 Kinetic theory of gases.
- Consequence Great three quantization

discoveries of XX century (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

Henri Becquerel - 1897 Discovery of electron by J.J.Thomson
- 1900 Derivation of black-body radiation formula

by Max Plank. - 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,.

The Structure of Atoms

- There are 112 chemical elements that have been

discovered, and there are a couple of additional

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 Georgy

Flerov, the founder of the Joint Institute for

Nuclear Research in Dubna, Russia, where the

element was discovered.

Georgi Flerov (1913-1990)

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 five neighbouring isotopes with mass

numbers 285289. 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 that

contains 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 Z1 the

next lightest atom, helium (He), has Z2 the

third lightest, lithium (Li), has Z3 and so

forth.

The Nuclear Atoms

- Nearly all the mass of the atom is concentrated

in a tiny nucleus which contains the protons and

neutrons. - Typically, the nuclear radius is approximately

from 1 fm to 10 fm (1fm 10-15m). The distance

between the nucleus and the electrons is

approximately 0.1 nm100,000fm. This distance

determines the size of the atom.

Nuclear Structure

An atom consists of an extremely small,

positively charged nucleus surrounded by a cloud

of negatively charged electrons. Although

typically the nucleus is less than one

ten-thousandth the size of the atom, the nucleus

contains more than 99.9 of the mass of the atom!

The number of protons in the nucleus, Z, is

called the atomic number. This determines

what chemical 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.

Structure of the Atom

- Evidence in 1900 indicated that the atom was not

a fundamental unit

- There seemed to be too many kinds of atoms, each

belonging to a distinct chemical element (way

more than earth, air, water, and fire!). - Atoms and electromagnetic phenomena were

intimately related (magnetic materials

insulators vs. conductors different emission

spectra). - Elements combine with some elements but not with

others, a characteristic that hinted at an

internal atomic structure (valence). - 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 by discussing

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 data has been collected on the emission

of light by atoms of individual elements in a

flame or in a gas exited by electrical discharge.

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 all the 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, ?2gt?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 of the 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 a line 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 by atoms of

individual elements in flame or in gas exited by

electrical charge, appears as a set of discrete

lines, each of a particular color or wavelength.

- The positions and intensities of the lines are

a characteristic of the element. The wavelength

of these lines could be determined with great

precision.

- Emission line spectrum of hydrogen in the

visible and near ultraviolet. The lines appear

dark because the spectrum was photographed. The

names of the first five lines are shown. As is

the point beyond which no lines appear, H8 called

the limits of the series.

Atomic Spectra

- In 1885 a Swiss schoolteacher, Johann Balmer,

found that the wavelengths of the lines in the

visible spectrum of hydrogen can be represented

by formula - Balmer suggested that this might be a special

case 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-Ritz formula, gives

the reciprocal wavelengths as - where m and n are integers with ngtm, and R is

the Rydberg constant.

Atomic Spectra

- The Rydberg constant is the same for all

spectral series. - For hydrogen the RH 1.096776 x 107m-1.
- For very heavy elements R approaches the value

R8 1.097373 x 107m-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 m2 and n3,4,5,
- Other series of hydrogen spectral lines were

found for m1 (by Lyman) and m3 (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 the spectrum was photographed. The

names of the first five lines are shown, as is

the point beyond which no lines appear, H8 called

the limits of the series.

The Limits of Series

- Find the predicted by Rydberg-Ritz formula

for Lyman, Balmer, and Paschen series.

- A portion of the emission spectrum of sodium.

The two very close bright lines at 589 nm are the

D1 and D2 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 at those points. Note

that frequency increases toward the right ,

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 much

lighter than the atom, and that it was

electrically neutral. - The most popular model was that of J.J.Thomson,

already quite successful in explaining chemical

reactions.

Knowledge of atoms in 1900

Electrons (discovered in 1897) 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 charge neutrality.

Thomsons Atomic Model

- Thomsons plum-pudding model of the atom had

the positive charges spread uniformly throughout

a sphere the size of the atom, with electrons

embedded in the uniform background.

- In Thomsons view, when the atom was heated, the

electrons could vibrate about their equilibrium

positions, thus producing electromagnetic

radiation. - Unfortunately, Thomson couldnt explain spectra

with this model.

- 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 experiments

conducted by Ernest Rutherford and his student

H.W.Geiger.

Experiments of Geiger and Marsden

- Rutherford, Geiger, and Marsden conceived a new

technique for investigating the structure of

matter by scattering a particles from atoms.

- Rutherford was investigating radioactivity

and had shown that the radiations from uranium

consist of at least two types, which he labeled a

and ß. - He showed, by an experiment similar to that of

Thompson, that q /m for the a - particles was

half that of the proton. - Suspecting that the a 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 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

- The 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

Gamma Decay

- Gamma 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 a gamma 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 a

and ß. - He showed, by an experiment similar to that of

Thompson, that q /m for the a - particles was

half that of the proton. - Suspecting that the a 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 a - particles is defined

by the small hole D in the shield surrounding the

radioactive source R of 214Bi . The a beam

strikes an ultra thin gold foil F, and a

particles are individually scattered through

various angels ?. The experiment consisted of

counting the number of scintillations on the

screen S as a function of ?.

- A diagram of the original apparatus as it appear

in Geigers paper describing the results.

- An a-particle by such an atom (Thompson model)

would have a scattering angle ? much smaller than

10. In the Rutherfords scattering experiment

most of the a-particles were either undeflected,

or deflected through very small angles of the

order 10, however, a few a-particles were

deflected through angles of 900 and more.

- An a-particle by such an atom (Thompson model)

would have a scattering angle ? much smaller than

10. In the Rutherfords scattering experiment a

few a-particles were deflected through angles of

900 and more.

Experiments of Geiger and Marsden

- Geiger showed that many a particles were

scattered from thin gold-leaf targets at backward

angles greater than 90.

Electrons cant back-scatter a particles.

- Calculate the maximum scattering angle -

corresponding to the maximum momentum change.

It can be shown that the maximum momentum

transfer to the a particle is

Determine qmax by letting Dpmax be perpendicular

to the direction of motion

too small!

Try multiple scattering from electrons

- If an a particle is scattered by N electrons

N the number of atoms across the thin gold

layer, t 6 10-7 m

n

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

N t / d

still too small!

- 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

scattering deflection angle could be observed. - Such model could not possibly account for the

large angles scattering. The unexpected large

angles a-particles scattering was described by

Rutherford with these words - It was quite incredible event that ever

happened to me in my life. It was as incredible

as if you fired a 15-inch shell at a piece of

tissue paper and it came back and hit you.

Rutherfords Atomic Model

even if the a particle is

scattered from all 79 electrons in each atom of

gold. Experimental results were not consistent

with Thomsons atomic model. Rutherford proposed

that an atom has a positively charged core

(nucleus!) surrounded by the negative

electrons. Geiger and Marsden confirmed the idea

in 1913.

Ernest Rutherford (1871-1937)

- Rutherford concluded that the large angle

scattering could result only from a single

encounter of the a 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

for the scattered a particles. - His predictions on the dependence of scattering

probability 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 a particle experiences a

repulsive force kqaQ/r2. The a 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 ?

with OA.

Rutherford Scattering

Theres a relationship between the impact

parameter b and the scattering angle q.

When b is small, r is small. the Coulomb force

is large. ? can be large and the particle can be

repelled backward.

where

Rutherford Scattering

- Any particle inside the circle of area p b02

will be similarly (or more) scattered.

The cross section s p 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 the sphere the force is

proportional to Q/r2, inside the sphere the force

is proportional to qI/r2 Qr/R3, where qI

Q(r/R)3 is the charge within a sphere of radius

r. The maximum force occurs at r R

- Two a particles with equal kinetic energies

approach the positive charge Q Ze with impact

parameters b1 and b2, where b1ltb2. According to

equation for impact parameter in this case ?1 gt

?2.

- The path of a particle can be shown to be a

hyperbola, and the scattering angle ? can be

relate to the impact parameter b from the laws of

classical mechanics. - The quantity pb2, which has the dimension of

the area , is called the cross section s for

scattering.

- The cross section s is thus defined as the

number of particles scattered per nucleus per

unit time divided by the incident intensity.

The total number of nuclei of foil atoms in the

area covered by 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 is obtained by multiplying pb2I0 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 total number scattered

per second through angles grater than ? is thus

pb2I0ntA. If divide this by the number of a

particles incident per second I0A we get the

fraction of a particles f scattered through

angles grater than ? - f pb2nt

On the base of that nuclear model

Rutherford derived an expression for the number

of a particles ?N that would be scattered at any

angle ? All this predictions was

verified in Geiger experiments, who observe

several hundreds thousands a particles.

- (a)Geiger data for a scattering from thin Au and

Ag foils. The graph is in log-log plot to cover

over several orders of magnitudes. - (b)Geiger also measured the dependence of ?N on

t for different elements, that was also in good

agreement with Rutherford

formula.

- Data from Rutherfords group showing observed a

scattering at a large fixed angle versus values

of rd - computed from for various

kinetic energies.

The Size of the Nucleus

- This equation can be used to estimate the size

of the nucleus - For the case of 7.7-MeV a particles the

distance of closest approach for a head-on

collision is

The Classical Atomic Model

- Consider an atom as a planetary system.
- The Newtons 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

- From classical EM 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 an acceleration a radiates at a rate - Show that an electron in a classical hydrogen

atom spirals into the nucleus at a rate

(b) Find the time interval over which the

electron will reach r 0, starting from r0

2.00 1010 m.

The Bohrs 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 electron does 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 Ei and Ef are the energy of initial and

final state, h is the Planks 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

The Bohrs Postulates

- Combining the second postulate with the

equation for the energy we obtain

where r1 and r2 are the radii of the initial

and final orbits.

The Bohrs Postulates

To obtain the frequencies implied by the

experimental Rydberg-Ritz formula,

it is evident that the radii of the stable orbits

must be proportional to the squares of integers.

The Bohrs Postulates

- Bohr found that he could obtain this condition

if he postulates the angular momentum of the

electron in a stable orbit equals an integer

times hh/2p. Since the angular momentum of a

circular orbit is just mvr, the third Bohrs

postulate is - 3.

n1,2,3. - where hh/2p1.055 x 10-34Js6.582x10 -16eVs

The Bohrs Postulates

- The obtained equation mvr nh/2pnh relates the

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

or

We can write

or

The Bohrs Postulates

- Solving for r, we obtain

where a0 is called the first Bohrs radius

Bohrs Postulates

- Substituting the expression for r in equation for

frequency - If we will compare this expression with Z1 for

fc/? with the empirical Rydberg-Ritz formula - we will obtain for the Rydberg constant

Bohrs Postulates

- Using the known values of m, e, and h, 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

- If we will substitute the quantized value of r

as given by - we obtain

Energy levels

- or
- where
- The energies En with Z1 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 En

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

- A hydrogen atom is in its first excited state (n

2). Using the 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 the electron, (e) the potential energy of the

system, and (f) the total energy of the system.

Energy levels

- Transitions between this allowed energies

result in the emission or absorption of a photon

whose frequency is given by - and whose wavelength is

Energy levels

- Therefore is convenient to have the value of hc

in electronvolt nanometers! - hc 1240 eVnm
- Since the energies are quantized, the

frequencies and the wavelengths of the radiation

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 of the energy radiated. - (b) In the Bohr model, the electron orbits

without radiating until it jumps to another

allowed radius of lower energy, at which time

radiation is emitted.

?21hc / (E1-E2)

- Energy level diagram for hydrogen showing the

seven lowest stationary states and the four

lowest energy transitions for the Lyman, Balmer,

and Pashen series. The energies of infinite

number of levels are given by En (-13.6/n2)eV,

where n is an integer.

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- The spectral lines corresponding to the

transitions shown for the three series.

?21hc / (E1-E2)

- Compute the wavelength of the Hß spectral line

of the Balmer series predicted by Bohr model.

- A hydrogen atom at rest in the laboratory emits

a photon of the Lyman a 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

Use conservation 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 spectral

lines 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 lines nearest to the line of longest

wavelength.

- A stylized picture of Bohr circular orbits for

n1,2,3,4. The radii rnn2. - In high Z-elements (elements with Z 12),

electrons are distributed over all the orbits

shown. If an electron in the n1 orbit is knocked

from the atom (e.g., by being hit by a fast

electron accelerated by the voltage across an

x-ray tube) the vacancy that produced is filed by

an electron of higher energy (i.e., n2 or

higher).

The difference in energy between the two orbits

is emitted as a photon, whose wavelength will be

in the x-ray region of the spectrum, if Z is

large enough.

(a)

(b)

(c)

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

spectra of neodymium (Z60) and samarium

(Z62).The two pairs of bright lines are the Ka

and Kß lines. (b) Part of the spectrum of the

artificially produced element promethium (Z61),

its Ka and Kß lines fall between those of Nd and

Sm. (c) Part of the spectrum of all three

elements.

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 Bohrs hypothesis

of energy quantization in atoms. - The experiment involved measuring the plate

current as a function of V0.

The Franck-Hertz Experiment

Schematic diagram of the Franck-Hertz experiment.

Electrons ejected from the heated cathode C at

zero potential are drawn to the positive grid G.

Those passing through the holes in the grid can

reach the plate P and contribute in the current

I, if they have sufficient kinetic energy to

overcome the small back potential ?V. The tube

contains a low pressure gas of the element being

studied.

The Franck-Hertz Experiment

- As V0 increased from 0, the current increases

until a critical value (about 4.9 V for mercury)

is reached. At this point the current suddenly

decreases. As V0 is increased further, the

current rises again. - They found that when the electrons 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 current decreases because many electrons

lose energy due to inelastic collisions with

mercury atoms in the tube and therefore cannot

overcome the small back potential.

Current, (mAmp)

- The regular spacing of the peaks in this curve

indicates that only a certain quantity of energy,

4.9 eV, can be lost to the mercury atoms. This

interpretation is confirmed by the observation of

radiation of photon energy 4.9 eV, emitted by

mercury atoms.

Current, (mAmp)

- Suppose mercury atoms have an energy level 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 by emitting a photon. The wavelength

of the photon should be

This is equal to the measured wavelength,

confirming the existence of this energy level of

the mercury atom.

Similar experiments with other atoms yield the

same kind of evidence for atomic energy levels.

- Lets consider an experimental tube filled by

hydrogen atoms instead of mercury. Electrons

accelerated by V0 that collide with hydrogen

electrons cannot transfer the energy to letter

electrons unless they have acquired kinetic

energy - eV0E2 E110.2eV

- If the incoming electron does not have

sufficient energy to transfer ?E E2 - E1 to the

hydrogen electron in the n1 orbit (ground

state), than the scattering will be elastic.

- If the incoming electron does have at least ?E

kinetic energy, then an inelastic collision can

occur in which ?E is transferred to the n1

electron, moving it to the n2 orbit. The excited

electron will typically return to the ground

state very quickly, emitting a photon of energy

?E.

- Energy loss spectrum measurement. A

well-defined electron beam impinges upon the

sample. Electrons inelastically scattered at a

convenient angle enter the slit of the magnetic

spectrometer, whose B field is directed out of

the page, and turn through radii R determined by

their energy (Einc E1) 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

Bohrs 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

Reduced mass correction

- If the nucleus has the mass M its kinetic

energy will be ½Mv2 p2/2M, where p Mv is the

momentum. - If we assume that the total momentum of the

atom is zero, from the conservation of momentum

we will have that momentum of electron and

momentum of nucleus are equal on the magnitude.

Reduced mass correction

- 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 Bohrs

analysis of the hydrogen atom to show that the

allowed radii of the Earths orbit are given by - where MS is the mass of the Sun, ME 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 from the Sun

corresponding to the quantum number n 1