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General Physics (PHY 2140)
Lecture 16
 Modern Physics
Atomic Physics
Early models of the atom
Atomic spectra
Bohr’s theory of hydrogen
De Broglie wavelength in the atom
Quantum mechanics and Spin
Chapter 28
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http://www.physics.wayne.edu/~alan/2140Website/Main.htm
1
Lightning Review
Last lecture:
1. Quantum physics
 Wave function
 Uncertainty relations
h
2
h
E t 
2
xp 
Review Problem:
If matter has a wave structure, why is this not observable in our daily
experiences?
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(baseball)  10-34 m
2
27.9 The Uncertainty Principle
When measurements are made, the experimenter is
always faced with experimental uncertainties in the
measurements


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Classical mechanics offers no fundamental barrier to
ultimate refinements in measurements
Classical mechanics would allow for measurements with
arbitrarily small uncertainties
3
The Uncertainty Principle
Quantum mechanics predicts that a barrier to measurements
with ultimately small uncertainties does exist
In 1927 Heisenberg introduced the uncertainty principle

If a measurement of position of a particle is made with precision Δx
and a simultaneous measurement of linear momentum is made with
precision Δp, then the product of the two uncertainties can never be
smaller than h/4
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4
The Uncertainty Principle
Mathematically,
h
xp x 
4
It is physically impossible to measure simultaneously the
exact position and the exact linear momentum of a
particle
Another form of the principle deals with energy and time:
h
Et 
4
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Thought Experiment – the Uncertainty
Principle
A thought experiment for viewing an electron with a powerful
microscope
In order to see the electron, at least one photon must bounce off it
During this interaction, momentum is transferred from the photon to
the electron
Therefore, the light that allows you to accurately locate the electron
changes the momentum of the electron
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Problem: macroscopic uncertainty
A 50.0-g ball moves at 30.0 m/s. If its speed is measured to an
accuracy of 0.10%, what is the minimum uncertainty in its
position?
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A 50.0-g ball moves at 30.0 m/s. If its speed is measured to an accuracy of 0.10%,
what is the minimum uncertainty in its position?
Given:
v = 30 m/s
v/v = 0.10%
m = 50.0 g
Notice that the ball is non-relativistic. Thus, p = mv,
and uncertainty in measuring momentum is
dv 
p  m  v   m   v 
 v 



 50.0 102 kg 1.0 103  30 m s  1.5 10 2 kg  m s
Thus, uncertainty relation implies
Find:
dx = ?
h
6.63 1024 J  s
32
x 


3.5

10
m
3
4  p  4 1.5 10 kg  m s


Too small to measure!!
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Problem: Macroscopic measurement
A 0.50-kg block rests on the icy surface of a frozen pond, which we
can assume to be frictionless. If the location of the block is measured
to a precision of 0.50 cm, what speed must the block acquire because
of the measurement process?
Recall:
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h
 x p 
4
x
and
p  mv
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Atomic physics:
Chapter 28
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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
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 single-electron ions

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For example, He+ and Li2+
11
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
“watermelon” model
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Experimental tests
Expect:
1.
2.
Mostly small
angle scattering
No backward
scattering events
Results:
1.
2.
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Mostly small
scattering events
Several
backward
scatterings!!!
13
Early Models of the Atom
Rutherford’s model




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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
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Problem: Rutherford’s model
The “size” of the atom in Rutherford’s model is about 1.0 × 10–10 m.
(a) Determine the attractive electrical force between an electron
and a proton separated by this distance.
(b) Determine (in eV) the electrical potential energy of the atom.
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The “size” of the atom in Rutherford’s model is about 1.0 × 10–10 m. (a) Determine the
attractive electrical force between an electron and a proton separated by this
distance. (b) Determine (in eV) the electrical potential energy of the atom.
Electron and proton interact via the Coulomb force
Given:
r = 1.0 ×
10–10
m
F  ke
q1q2
r2
8.99 10


9
N m C
2
1.0 10
2
1.60 10
m
10
19
C

2
2
 2.3 108 N
Find:
(a) F = ?
(b) PE = ?
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Potential energy is
q1q2
1eV
18 
PE  ke
 2.3 10 J 
19
r
 1.6 10 J

  14 eV

16
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, thus leading to electron
“falling on a nucleus” in about 10-12 seconds!!!


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The radius should steadily decrease as this radiation is given off
The electron should eventually spiral into the nucleus
It doesn’t
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28.2 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
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Emission Spectrum of Hydrogen
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
A.k.a. Balmer series
Examples of spectral lines


n = 3, λ = 656.3 nm
n = 4, λ = 486.1 nm
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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
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Applications of Absorption Spectrum
The continuous spectrum emitted by the Sun passes
through the cooler gases of the Sun’s atmosphere


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The various absorption lines can be used to identify elements in
the solar atmosphere
Led to the discovery of helium (from helios)
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28.3 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
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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
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

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Ei  E f  hf
The “jump” cannot be treated
classically
23
Bohr’s Assumptions
More on the electron’s “jump”:


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
Ei  E f  hf
The size of the allowed electron orbits is determined by a
condition imposed on the electron’s orbital angular
momentum
 h
me vr  n 
 2
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
 , n  1, 2,3,...

24
Results
The total energy of the atom

2
1
e
E  KE  PE  me v 2  ke
2
r
Newton’s law
e2
v2
F  me a or ke 2  me
r
r
This can be used to rewrite kinetic energy as
mv 2
e2
KE 
 ke
2
2r
Thus, the energy can also be expressed as
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k ee2
E
2r
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Bohr Radius
The radii of the Bohr orbits are quantized (  h
n2  2
rn 
m ek e e 2
2
)
n  1, 2, 3, 
(from quantized angular momentum: mevr = nħ and Newtons law)

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
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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
The lowest energy state is called the
ground state


This corresponds to n = 1
Energy is –13.6 eV
The next energy level has an energy of
–3.40 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 (n = ∞, E = 0)

The ionization energy for hydrogen is
13.6 eV
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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
 1 1
1
 RH  2  2 

 nf ni 


For the Balmer series, nf = 2
For the Lyman series, nf = 1
Whenever a transition occurs between a
state, ni and another state, nf (where ni > nf),
a photon is emitted

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The photon has a frequency f = (Ei – Ef)/h
and wavelength λ
28
Problem: Transitions in the Bohr’s model
A photon is emitted as a hydrogen atom undergoes a transition from
the n = 6 state to the n = 2 state. Calculate the energy and the
wavelength of the emitted photon.
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A photon is emitted as a hydrogen atom undergoes a transition from the n = 6 state to
the n = 2 state. Calculate the energy and the wavelength of the emitted photon.
From
Given:
ni = 6
nf = 2
 1
1 
 RH  2  2  , or
 n f ni 



1
1

RH
 ni2 n 2f
 2
2
n

n
f
 i



With ni = 6 and nf = 2 we have:

1
 (36)(4) 
7

4.10

10
m  410nm

7
1 
1.09737 10 m  36  4 
Find:
a  = ?
(b) Eg = ?
The Photon energy is:
E
hc

(6.63 1034 J s)(3.00 108 m/s)
19


4.85

10
J  3.03 eV
9
410 10 m
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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

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Similar to having Newtonian Mechanics be a special case of
relativistic mechanics when v << c
31
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
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Recall Bohr’s Assumptions
Only certain electron orbits are stable. Radiation is
emitted by the atom when the electron “jumps” from a
more energetic initial state to a lower state
Ei  E f  hf
The size of the allowed electron orbits is determined by a
condition imposed on the electron’s orbital angular
momentum
mevr  n , n  1, 2,3,...
Why is that?
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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


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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
34
Modifications of the Bohr Theory – Zeeman
Effect and fine structure
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
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
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28.5 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
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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 ,   1, 2,3,...
but
h

me v
, so
mevr  n , n  1, 2,3,...
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This was the first convincing argument that the wave nature of
matter was at the heart of the behavior of atomic systems
37
QUICK QUIZ 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.
(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.
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28.6 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
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Problem: wavelength of the electron
Determine the wavelength of an electron in
the third excited orbit of the hydrogen
atom, with n = 4.
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Determine the wavelength of an electron in the third excited orbit of the hydrogen
atom, with n = 4.
Given:
Recall that de Broglie’s wavelength of electron
depends on its momentum,  = h/(mev). Let us find it,
me vrn  n ,
n=4
Recall that
so me v 
n
rn
rn  n2 a0 , so mev 
h
 2 a0  n
Find:
e = ?
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Thus,

h
  2 a0  n  8  0.0529nm   1.33nm
me v
41
Quantum Number Summary
The values of n can increase from 1 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
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QUICK QUIZ 2
How many possible orbital states are there for
(a) the n = 3 level of hydrogen? (b) the n = 4 level?
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
(each is n2)
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Spin Magnetic Quantum Number
It is convenient to think of the
electron as spinning on its axis

The electron is not
physically spinning
There are two directions for
the spin

Spin up, ms = ½

Spin down, ms = -½
There is a slight energy
difference between the two
spins and this accounts for the
Zeeman effect
Sodium D lines are an
example of this splitting.
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Electron Clouds
The graph shows the solution
to the wave equation for
hydrogen in the ground state

The curve peaks at the
Bohr radius

The electron is not
confined to a particular
orbital distance from the
nucleus
The probability of finding the
electron at the Bohr radius is a
maximum
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Electron Clouds
The wave function for
hydrogen in the ground state is
symmetric

The electron can be found
in a spherical region
surrounding the nucleus
The result is interpreted by
viewing the electron as a cloud
surrounding the nucleus

The densest regions of the
cloud represent the highest
probability for finding the
electron
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