39 Part 2 Office2003

8/2/2019 39 Part 2 Office2003

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Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Chapter 39. Quantization

(Part II)The image shows the electron

density in the vicinity of a

circle of 60 iron atoms. The

circular ripple-like rings in

the center of the corral arepart of an electron standing

wave, rather like the standing

wave on the head of a

vibrating drum.Chapter Goal: To

understand the quantization

of energy for light and matter.

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Topics:• Bohr’s Model of Atomic Quantization

• The Bohr Hydrogen Atom

•

The Hydrogen Spectrum

Chapter 39. Quantization

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Bohr’s Model of Atomic Quantization

The Rutherford model (positively charged nucleus

surrounded by electron cloud) could not explain:

1) The stability of the atom: The electrons on orbits

should radiate and lose energy in

2) The observed atomic emission spectra

s1010

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The Bohr postulates

1. An atom consists of negative electrons orbiting a very

small positive nucleus.

2. Atoms can exist only in certain stationary states. These

states can be numbered n=1, 2, 3, 4, . . . , where n is the

quantum number.

3. Each stationary state has an energy E n. The stationary

states of an atom are numbered in order of increasing

energy: E 1 < E 2 < E 3 < …

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The Bohr postulates

4. The lowest energy state of the atom E 1 is stable and can

persist indefinitely. It is called the ground state of theatom. Other stationary states with energies E 2, E 3, E 4,. . .

are called excited states of the atom.

5. An atom can “jump” from one stationary state to another by emitting or absorbing a photon of frequency

where h is Planck’s constant and and Δ E atom = | E f – E i|.

E f and E i are the energies of the initial and final states. Such

a jump is called a transition or, sometimes, a quantum

jump.

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6. An atom can move from a lower energy state to a higherenergy state by absorbing energy Δ E atom = E f – E i in an

inelastic collision with an electron or another atom.

The Bohr postulates

This process, called collisional excitation, is shown.

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EXAMPLE 39.10 Emission and absorption

eVsh15

10136.4

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The Stationary States of the Hydrogen Atom

Assume a single electron orbiting (with radius

r and speed v) a proton (proton remains

stationary).

Circular orbit- centripetal force is Coulomb

force:

chargeelectron-eelectron,anof mass-m

4

4

0

22

2

2

0

2

mr

ev

r

va

mr

e

m

F a cent

elecelec

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Bohr quantization - a de Broglie wave forthe electron set up a standing wave around

the circumference (just like the particle in the

box).An integer number of wavelength to fit around

circumference of the electron’s orbit

The mathematical condition is found byreplacing the round-trip distance 2L in a box

with the round-trip distance on a circle.r 21,2,3..n 2 nr

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De Broglie wavelength

1,2,3..n 2 nr

mv

h

p

h

eVs Js

mr n

mr hnv

mv

hnr

n

1634 1058.61005.1

2

h

1,2..n 2

2

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1,2..n 4

4

2

2

02

22

22

0

22

me

nr

r m

n

mr

ev

n

ma B11

1029.5radiusBohr

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...2,1 1 1

nn

v

manmr

nv

Bn

n

smmav B / 1019.2 / 6

1

)24

1(1randfor vformulausing

42

1UTE:Since

2

0

2nn

0

22

B

n

n

n

ae

n E

r

emv

...3,2,1 6.13

22

1 n

n

eV

n

E E n

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SummaryThe radius of the electron’s orbit in Bohr’s hydrogen atom

is

where a B is the Bohr radius, defined as

The possible electron speeds and energies are

smmav B / 1019.2 /

6

1

eV a

e

E B 6.1324

1 2

01

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The first four

stationary states,or allowed orbits,

of the Bohr

hydrogen atom

drawn to scale.

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Binding Energy and Ionization Energy

Why the energies of the stationary states are

negative?

Because the potential energy between two

charged particles is zero when they are

infinitely far apart. We have to add energy into

electron-proton bounded system in order to

pull them apart.

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Binding Energy and Ionization Energy

n E - the binding energy of the electron in

stationary state n.

It is an energy that we should add to the electronto free it from the proton and reach the zero

energy of two particles infinitely apart.

The binding energy of the ground state is

called ionization energy of an atom.1

E

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Quantization of Angular Momentum

Angular momentum – L=mvr -is conserved in

orbital motion.

The condition that a de Broglie wave for the

electron set up a standing wave around the

circumference:

nh

nmvr

mv

hnnr

2

2 ...3,2,1 nn L

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The Hydrogen Spectrum

According to the fifth assumption of Bohr’s model of

atomic quantization, the frequency of the photon emitted in

an n → m transition is

The corresponding wavelengths in the hydrogen spectrum

are then

BALMER

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Hydrogen-Like ions

91....,,........., U Li He All have 1 electron orbiting a Z

nucleus

r Zer 02

02 4 / 4 / e-

EnergypotentialCoulomb

eV n

Z E

nv Z v

Z

anr

n

n

Bn

2

2

1

2

6.13

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Success and Failure

Predict the Balmer formula

Bohr Hydrogen atom, unlike Rutherford’s, was

stable.

Completely unsuccessful at explaining the

spectra of any other neutral atom. Here, Bohr’s

model of the atomic quantization remains valid

but the procedure of fitting standing waves to a

circle is just too simple to find the stationary

states of complex atoms.

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Example-1 electron atom

Which is the ground state?

Which are excited states?

What is the ionizationenergy of this atom?

Can an atom in state n = 1

emit a photon? If so, what is

the wavelength? If not, whynot?

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Example

What wavelengths are

seen in the emission

spectrum of this atom?

What wavelengths are

seen in the absorption

spectrum of this atom? J E

Jssm

eV E

ch

f

c

atomatom

19

348

106.1

106.6 / 103

)(

nm E

atom

1240

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Chapter 39. Summary Slides

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General Principles

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Important Concepts

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Applications

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Important Concepts

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Applications

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Chapter 39. Part 2

Reading Quizzes

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A photon with a wavelength of 414 nm has energy

Ephoton = 3.0 eV. Do you expect to see a spectral line with

= 414 nm in the emission spectrum of the atomrepresented by this energy-level diagram? If so, what

transition or transitions will emit it?

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A photon with a wavelength of 414 nm has energy

Ephoton = 3.0 eV. Do you expect to see a spectral line with

= 414 nm in the emission spectrum of the atomrepresented by this energy-level diagram? If so, what

transition or transitions will emit it?

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What is the quantum number of this

hydrogen atom?

A. n = 5B. n = 4

C. n = 3

D. n = 2

E. n = 1

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A. n = 5B. n = 4

C. n = 3

D. n = 2

E. n = 1

What is the quantum number of this

hydrogen atom?

nh

nmvr L

mv

hnnr

n

2

2

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39 Part 2 Office2003