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Chapter 27
Quantum Physics
Need for Quantum Physics
Problems remained from classical mechanics
that relativity didn’t explain
 Blackbody Radiation


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Photoelectric Effect


The electromagnetic radiation emitted by a heated
object
Emission of electrons by an illuminated metal
Spectral Lines

Emission of sharp spectral lines by gas atoms in
an electric discharge tube
Development of Quantum
Physics

1900 to 1930

Development of ideas of quantum mechanics




Also called wave mechanics
Highly successful in explaining the behavior of atoms,
molecules, and nuclei
Quantum Mechanics reduces to classical
mechanics when applied to macroscopic systems
Involved a large number of physicists


Planck introduced basic ideas
Mathematical developments and interpretations
involved such people as Einstein, Bohr, Schrödinger, de
Broglie, Heisenberg, Born and Dirac
Blackbody Radiation

An object at any temperature is known
to emit electromagnetic radiation
Sometimes called thermal radiation
 Stefan’s Law describes the total power
radiated
 The spectrum of the radiation depends on
the temperature and properties of the
object

Blackbody Radiation Graph


Experimental data for
distribution of energy in
blackbody radiation
As the temperature
increases, the total
amount of energy
increases


Shown by the area under
the curve
As the temperature
increases, the peak of
the distribution shifts to
shorter wavelengths
Wien’s Displacement Law

The wavelength of the peak of the
blackbody distribution was found to
follow Wein’s Displacement Law

λmax T = 0.2898 x 10-2 m • K
λmax is the wavelength at the curve’s peak
 T is the absolute temperature of the object
emitting the radiation

The Ultraviolet Catastrophe


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

Classical theory did not
match the experimental
data
At long wavelengths, the
match is good
At short wavelengths,
classical theory predicted
infinite energy
At short wavelengths,
experiment showed no
energy
This contradiction is called
the ultraviolet catastrophe
Planck’s Resolution

Planck hypothesized that the blackbody
radiation was produced by resonators


Resonators were submicroscopic charged
oscillators
The resonators could only have discrete
energies

En = n h ƒ




n is called the quantum number
ƒ is the frequency of vibration
h is Planck’s constant, 6.626 x 10-34 J s
Key point is quantized energy states
Photoelectric Effect

When light is incident on certain metallic
surfaces, electrons are emitted from the
surface


This is called the photoelectric effect
The emitted electrons are called photoelectrons
The effect was first discovered by Hertz
 The successful explanation of the effect was
given by Einstein in 1905


Received Nobel Prize in 1921 for paper on
electromagnetic radiation, of which the
photoelectric effect was a part
Photoelectric Effect Schematic



When light strikes E,
photoelectrons are
emitted
Electrons collected at C
and passing through the
ammeter are a current
in the circuit
C is maintained at a
positive potential by the
power supply
Photoelectric Current/Voltage
Graph


The current increases
with intensity, but
reaches a saturation
level for large ΔV’s
No current flows for
voltages less than or
equal to –ΔVs, the
stopping potential

The stopping potential is
independent of the
radiation intensity
Features Not Explained by
Classical Physics/Wave Theory
No electrons are emitted if the incident
light frequency is below some cutoff
frequency that is characteristic of the
material being illuminated
 The maximum kinetic energy of the
photoelectrons is independent of the
light intensity

More Features Not Explained
The maximum kinetic energy of the
photoelectrons increases with
increasing light frequency
 Electrons are emitted from the surface
almost instantaneously, even at low
intensities

Einstein’s Explanation

A tiny packet of light energy, called a photon, would
be emitted when a quantized oscillator jumped from
one energy level to the next lower one

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Extended Planck’s idea of quantization to
electromagnetic radiation
The photon’s energy would be E = hƒ
Each photon can give all its energy to an electron in
the metal
The maximum kinetic energy of the liberated
photoelectron is KE = hƒ – Φ
Φ is called the work function of the metal
Explanation of Classical
“Problems”

The effect is not observed below a
certain cutoff frequency since the
photon energy must be greater than or
equal to the work function


Without this, electrons are not emitted,
regardless of the intensity of the light
The maximum KE depends only on the
frequency and the work function, not
on the intensity
More Explanations
The maximum KE increases with
increasing frequency
 The effect is instantaneous since there
is a one-to-one interaction between the
photon and the electron

Verification of Einstein’s
Theory
Experimental
observations of a
linear relationship
between KE and
frequency confirm
Einstein’s theory
 The x-intercept is
the cutoff frequency

Photocells
Photocells are an application of the
photoelectric effect
 When light of sufficiently high
frequency falls on the cell, a current is
produced
 Examples


Streetlights, garage door openers,
elevators
X-Rays

Electromagnetic radiation with short
wavelengths
Wavelengths less than for ultraviolet
 Wavelengths are typically about 0.1 nm
 X-rays have the ability to penetrate most
materials with relative ease


Discovered and named by Roentgen in
1895
The Compton Effect
Compton directed a beam of x-rays toward a
block of graphite
 He found that the scattered x-rays had a
slightly longer wavelength that the incident xrays


This means they also had less energy
The amount of energy reduction depended on
the angle at which the x-rays were scattered
 The change in wavelength is called the

Compton shift
Compton Scattering



Compton assumed the
photons acted like
other particles in
collisions
Energy and
momentum were
conserved
The shift in
wavelength is
h
     o 
(1  cos )
m ec
Compton Scattering, final

The quantity h/mec is called the Compton
wavelength


Compton wavelength = 0.00243 nm
Very small compared to visible light
The Compton shift depends on the scattering
angle and not on the wavelength
 Experiments confirm the results of Compton
scattering and strongly support the photon
concept

QUICK QUIZ 27.1
An x-ray photon is scattered by an electron.
The frequency of the scattered photon
relative to that of the incident photon (a)
increases, (b) decreases, or (c) remains the
same.
QUICK QUIZ 27.1 ANSWER
(b). Some energy is transferred to the
electron in the scattering process. Therefore,
the scattered photon must have less energy
(and hence, lower frequency) than the
incident photon.
QUICK QUIZ 27.2
A photon of energy E0 strikes a free
electron, with the scattered photon of energy
E moving in the direction opposite that of
the incident photon. In this Compton effect
interaction, the resulting kinetic energy of
the electron is (a) E0 , (b) E , (c) E0  E , (d)
E0 + E , (e) none of the above.
QUICK QUIZ 27.2 ANSWER
(c). Conservation of energy requires the
kinetic energy given to the electron be equal
to the difference between the energy of the
incident photon and that of the scattered
photon.
QUICK QUIZ 27.3
A photon of energy E0 strikes a free electron
with the scattered photon of energy E
moving in the direction opposite that of the
incident photon. In this Compton effect
interaction, the resulting momentum of the
electron is (a) E0/c
(b) < E0/c
(c) > E0/c
(d) (E0  E)/c
(e) (E  Eo)/c
QUICK QUIZ 27.3 ANSWER
(c). Conservation of momentum requires
the momentum of the incident photon
equal the vector sum of the momenta of
the electron and the scattered photon.
Since the scattered photon moves in the
direction opposite that of the electron, the
magnitude of the electron’s momentum
must exceed that of the incident photon.
Photons and Electromagnetic
Waves

Light has a dual nature. It exhibits both
wave and particle characteristics


The photoelectric effect and Compton
scattering offer evidence for the particle
nature of light


Applies to all electromagnetic radiation
When light and matter interact, light behaves as if
it were composed of particles
Interference and diffraction offer evidence of
the wave nature of light
Wave Properties of Particles

In 1924, Louis de Broglie postulated
that because photons have wave and
particle characteristics, perhaps all
forms of matter have both properties

Furthermore, the frequency and
wavelength of matter waves can be
determined
de Broglie Wavelength and
Frequency

The de Broglie wavelength of a particle
is
h

mv

The frequency of matter waves is
E
ƒ
h
The Davisson-Germer
Experiment

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They scattered low-energy electrons from a
nickel target
They followed this with extensive diffraction
measurements from various materials
The wavelength of the electrons calculated
from the diffraction data agreed with the
expected de Broglie wavelength
This confirmed the wave nature of electrons
Other experimenters have confirmed the
wave nature of other particles
QUICK QUIZ 27.4
A non-relativistic electron and a nonrelativistic proton are moving and have the
same de Broglie wavelength. Which of the
following are also the same for the two
particles: (a) speed, (b) kinetic energy, (c)
momentum, (d) frequency?
QUICK QUIZ 27.4 ANSWER
(c). Two particles with the same de Broglie wavelength
will have the same momentum p = mv. If the electron
and proton have the same momentum, they cannot
have the same speed because of the difference in their
masses. For the same reason, remembering that KE =
p2/2m, they cannot have the same kinetic energy.
Because the kinetic energy is the only type of energy
an isolated particle can have, and we have argued that
the particles have different energies, Equation 27.15
tells us that the particles do not have the same
frequency.
QUICK QUIZ 27.5
We have seen two wavelengths assigned to
the electron, the Compton wavelength and
the de Broglie wavelength. Which is an
actual physical wavelength associated with
the electron: (a) the Compton wavelength,
(b) the de Broglie wavelength, (c) both
wavelengths, (d) neither wavelength?
QUICK QUIZ 27.5 ANSWER
(b). The Compton wavelength, λC = h/mec,
is a combination of constants and has no
relation to the motion of the electron. The
de Broglie wavelength, λ = h/mev, is
associated with the motion of the electron
through its momentum.
The Electron Microscope



The electron microscope
depends on the wave
characteristics of electrons
Microscopes can only
resolve details that are
slightly smaller than the
wavelength of the radiation
used to illuminate the object
The electrons can be
accelerated to high energies
and have small wavelengths
The Uncertainty Principle

When measurements are made, the
experimenter is always faced with
experimental uncertainties in the
measurements
Classical mechanics offers no fundamental
barrier to ultimate refinements in
measurements
 Classical mechanics would allow for
measurements with arbitrarily small
uncertainties

Importance of Hydrogen Atom
Hydrogen is the simplest atom
 The quantum numbers used to
characterize the allowed states of
hydrogen can also be used to describe
(approximately) the allowed states of
more complex atoms


This enables us to understand the periodic
table
More Reasons the Hydrogen
Atom is so Important

The hydrogen atom is an ideal system for
performing precise comparisons of theory and
experiment


Also for improving our understanding of atomic
structure
Much of what we know about the hydrogen
atom can be extended to other singleelectron ions

For example, He+ and Li2+
Early Models of the Atom

J.J. Thomson’s model of
the atom



A volume of positive
charge
Electrons embedded
throughout the volume
A change from
Newton’s model of the
atom as a tiny, hard,
indestructible sphere
Early Models of the Atom, 2

Rutherford




Planetary model
Based on results of
thin foil experiments
Positive charge is
concentrated in the
center of the atom,
called the nucleus
Electrons orbit the
nucleus like planets
orbit the sun
Difficulties with the Rutherford
Model

Atoms emit certain discrete characteristic
frequencies of electromagnetic radiation


The Rutherford model is unable to explain this
phenomena
Rutherford’s electrons are undergoing a
centripetal acceleration and so should radiate
electromagnetic waves of the same frequency


The radius should steadily decrease as this
radiation is given off
The electron should eventually spiral into the
nucleus

It doesn’t
Emission Spectra
A gas at low pressure has a voltage applied
to it
 A gas emits light characteristic of the gas
 When the emitted light is analyzed with a
spectrometer, a series of discrete bright lines
is observed



Each line has a different wavelength and color
This series of lines is called an emission spectrum
Examples of Spectra
Emission Spectrum of
Hydrogen – Equation

The wavelengths of hydrogen’s spectral lines
can be found from
1
1
 1
 RH  2  2 

2 n 

RH is the Rydberg constant



RH = 1.0973732 x 107 m-1
n is an integer, n = 1, 2, 3, …
The spectral lines correspond to different values of
n
Spectral Lines of Hydrogen
The Balmer Series
has lines whose
wavelengths are
given by the
preceding equation
 Examples of spectral
lines



n = 3, λ = 656.3 nm
n = 4, λ = 486.1 nm
Absorption Spectra
An element can also absorb light at specific
wavelengths
 An absorption spectrum can be obtained by
passing a continuous radiation spectrum
through a vapor of the gas
 The absorption spectrum consists of a series
of dark lines superimposed on the otherwise
continuous spectrum


The dark lines of the absorption spectrum coincide
with the bright lines of the emission spectrum
Applications of Absorption
Spectrum

The continuous spectrum emitted by
the Sun passes through the cooler
gases of the Sun’s atmosphere
The various absorption lines can be used to
identify elements in the solar atmosphere
 Led to the discovery of helium

The Bohr Theory of Hydrogen
In 1913 Bohr provided an explanation
of atomic spectra that includes some
features of the currently accepted
theory
 His model includes both classical and
non-classical ideas
 His model included an attempt to
explain why the atom was stable

Bohr’s Assumptions for
Hydrogen

The electron moves
in circular orbits
around the proton
under the influence
of the Coulomb
force of attraction

The Coulomb force
produces the
centripetal
acceleration
Bohr’s Assumptions, cont

Only certain electron orbits are stable



These are the orbits in which the atom does not
emit energy in the form of electromagnetic
radiation
Therefore, the energy of the atom remains
constant and classical mechanics can be used to
describe the electron’s motion
Radiation is emitted by the atom when the
electron “jumps” from a more energetic initial
state to a lower state

The “jump” cannot be treated classically
Bohr’s Assumptions, final

The electron’s “jump,” continued
The frequency emitted in the “jump” is
related to the change in the atom’s energy
 It is generally not the same as the

frequency of the electron’s orbital motion

The size of the allowed electron orbits
is determined by a condition imposed
on the electron’s orbital angular
momentum
Mathematics of Bohr’s
Assumptions and Results

Electron’s orbital angular momentum


me v r = n ħ where n = 1, 2, 3, …
The total energy of the atom
2
1
e
2
 E  KE  PE  me v  k e
2
r

The energy can also be expressed as

k ee2
E
2r
Bohr Radius

The radii of the Bohr orbits are
quantized
n2  2
rn 
n  1, 2, 3, 
2
m ek e e
 This shows that the electron can only exist
in certain allowed orbits determined by the
integer n
When n = 1, the orbit has the smallest radius,
called the Bohr radius, ao
 ao = 0.0529 nm

Radii and Energy of Orbits

A general expression
for the radius of any
orbit in a hydrogen
atom is


rn = n2 ao
The energy of any
orbit is

En = - 13.6 eV/ n2
Specific Energy Levels

The lowest energy state is called the ground
state



The next energy level has an energy of –3.40
eV


This corresponds to n = 1
Energy is –13.6 eV
The energies can be compiled in an energy level
diagram
The ionization energy is the energy needed to
completely remove the electron from the
atom

The ionization energy for hydrogen is 13.6 eV
Energy Level Diagram
The value of RH
from Bohr’s analysis
is in excellent
agreement with the
experimental value
 A more generalized
equation can be
used to find the
wavelengths of any
spectral lines

Generalized Equation
 1 1
1
 RH  2  2 

 nf ni 



For the Balmer series, nf = 2
For the Lyman series, nf = 1
Whenever an transition occurs between a
state, ni to another state, nf (where ni > nf), a
photon is emitted

The photon has a frequency f = (Ei – Ef)/h and
wavelength λ
Bohr’s Correspondence
Principle

Bohr’s Correspondence Principle states
that quantum mechanics is in
agreement with classical physics when
the energy differences between
quantized levels are very small

Similar to having Newtonian Mechanics be
a special case of relativistic mechanics
when v << c
Successes of the Bohr Theory

Explained several features of the hydrogen
spectrum






Accounts for Balmer and other series
Predicts a value for RH that agrees with the
experimental value
Gives an expression for the radius of the atom
Predicts energy levels of hydrogen
Gives a model of what the atom looks like and how it
behaves
Can be extended to “hydrogen-like” atoms


Those with one electron
Ze2 needs to be substituted for e2 in equations

Z is the atomic number of the element
Modifications of the Bohr
Theory – Elliptical Orbits

Sommerfeld extended the results to
include elliptical orbits
Retained the principle quantum number, n
 Added the orbital quantum number, ℓ


ℓ ranges from 0 to n-1 in integer steps
All states with the same principle quantum
number are said to form a shell
 The states with given values of n and ℓ are
said to form a subshell

Modifications of the Bohr
Theory – Zeeman Effect

Another modification was needed to account
for the Zeeman effect



The Zeeman effect is the splitting of spectral lines
in a strong magnetic field
This indicates that the energy of an electron is
slightly modified when the atom is immersed in a
magnetic field
A new quantum number, m ℓ, called the orbital
magnetic quantum number, had to be introduced

m ℓ can vary from - ℓ to + ℓ in integer steps
Modifications of the Bohr
Theory – Fine Structure

High resolution spectrometers show
that spectral lines are, in fact, two very
closely spaced lines, even in the
absence of a magnetic field
This splitting is called fine structure
 Another quantum number, ms, called the
spin magnetic quantum number, was
introduced to explain the fine structure

de Broglie Waves
One of Bohr’s postulates was the
angular momentum of the electron is
quantized, but there was no explanation
why the restriction occurred
 de Broglie assumed that the electron
orbit would be stable only if it contained
an integral number of electron
wavelengths

de Broglie Waves in the
Hydrogen Atom


In this example, three
complete wavelengths
are contained in the
circumference of the
orbit
In general, the
circumference must
equal some integer
number of wavelengths

2  r = n λ n = 1, 2, …
de Broglie Waves in the
Hydrogen Atom, cont

The expression for the de Broglie wavelength
can be included in the circumference
calculation


me v r = n 
This is the same quantization of angular
momentum that Bohr imposed in his original
theory
This was the first convincing argument that
the wave nature of matter was at the heart of
the behavior of atomic systems
 Schrödinger’s wave equation was
subsequently applied to atomic systems

QUICK QUIZ 28.1
In an analysis relating Bohr's theory to the de Broglie
wavelength of electrons, when an electron moves
from the n = 1 level to the n = 3 level, the
circumference of its orbit becomes 9 times greater.
This occurs because (a) there are 3 times as many
wavelengths in the new orbit, (b) there are 3 times as
many wavelengths and each wavelength is 3 times as
long, (c) the wavelength of the electron becomes 9
times as long, or (d) the electron is moving 9 times as
fast.
QUICK QUIZ 28.1 ANSWER
(b). The circumference of the orbit is n times the
de Broglie wavelength (2πr = nλ), so there are
three times as many wavelengths in the n = 3
level as in the n = 1 level. Also, by combining
Equations 28.4, 28.6 and the defining equation
for the de Broglie wavelength (λ = h/mv), one
can show that the wavelength in the n = 3 level
is three times as long.
Quantum Mechanics and the
Hydrogen Atom
One of the first great achievements of
quantum mechanics was the solution of the
wave equation for the hydrogen atom
 The significance of quantum mechanics is
that the quantum numbers and the
restrictions placed on their values arise
directly from the mathematics and not from
any assumptions made to make the theory
agree with experiments

Quantum Number Summary
The values of n can range from 1 to  in
integer steps
 The values of ℓ can range from 0 to n-1
in integer steps
 The values of m ℓ can range from -ℓ to ℓ
in integer steps


Also see Table 28.2
QUICK QUIZ 28.2
How many possible orbital states are there for
(a) the n = 3 level of hydrogen? (b) the n = 4 level?
QUICK QUIZ 28.2 ANSWER
The quantum numbers associated with orbital states
are n, , and m . For a specified value of n, the allowed
values of  range from 0 to n – 1. For each value of ,
there are (2  + 1) possible values of m.
(a) If n = 3, then  = 0, 1, or 2. The number of possible
orbital states is then
[2(0) + 1] + [2(1) + 1] + [2(2) + 1] = 1 + 3 + 5 = 9.
(b) If n = 4, one additional value of  is allowed ( = 3)
so the number of possible orbital states is now
9 + [2(3) + 1] = 9 + 7 = 16
QUICK QUIZ 28.3
When the principal quantum number is n =
5, how many different values of (a)  and
(b) m are possible?
QUICK QUIZ 28.3 ANSWER
(a) For n = 5, there are 5 allowed values of ,
namely  = 0, 1, 2, 3, and 4.
(b) Since m ranges from – to + in integer
steps, the largest allowed value of  ( = 4 in
this case) permits the greatest range of values
for m. For n = 5, there are 9 possible values
for m: -4, -3, -2, –1, 0, +1, +2, +3, and +4.