Transcript Chapter 27

Chapter 41
Quantum Mechanics
de Broglie’s Idea


Particles had wave characteristics just like waves had particle
characteristics (e.g. E&M wave  photon)
The de Broglie wavelength of a particle of mass, m, is
h

mv

p
h

The frequency of matter waves is
E
ƒ
h
Uncertainty Principle and wave
character
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Mathematically,
h
xp x 
4
For a wave with λ one has Δx~λ and by
deBroglie Δp~h/λ
Thus ΔxΔp~h
Quantum mechanical dual
nature (Light)
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All things have a dual particle/wave
nature. Which property is
important/noticeable depends on the
experiment
The light two slit experiment reveals
wave nature of light
The Compton experiment, photoelectric
effect reveals the particle nature of light
Quantum mechanical dual
nature (particles)
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De Broglie’s hypothesis suggested a
wave character for particles.
This wave nature revealed itself in the
Davisson-Germer experiment.
Also in the particle two-slit experiment
The particle two slit
experiment I.
The particle two slit
experiment II.
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Which slit does a particle go through?
The answer is both!  Probability wave
The Wave Function ψ
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What are these matter waves of?
They are probability waves described by
a complex valued function of x and t i.e.
Ψ(x,t)
The probability to find the particle in
some dx is given
P( x)dx   dx
2
Properties of the Wave
Function

Normalization


 dx  1
2


Finite probability
b

 dx  Pba
2
a

Expectation value (average)
b
 x dx  x
2
a
b

a
f ( x) dx  f ( x)
2
Particle in 1D Hard Box I
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Classically the particle has equal
probability to be anywhere in the
box
But if xi and pi are known then
one knows exactly where the
ball is at any later time
Particle in a 1D Hard Wall Box II
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Quantum mechanically L=nλ/2 or λ=2L/n (n=1,2,3…)
The wave function (guess) ψ(x)=Asin(kx) with k=2π/λ=nπ /L
Giving ψ(x)=Asin(n πx/L). A=√(2/L) by normalization
Particle in a 1D Hard Wall Box
III
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The energy spectrum is found as
follows
De Broglie  p=h/λ=hn/2L (n=1,
2, 3, …)
Energy En = p2 /2m = h2 n2 / (8
m L2 )
n=0 not allowed. Lowest energy is
n=1 the zero point energy. This is
different from classical mechanics
where E=0 is possible. This is
connected with the Heisenberg
uncertainty principle.
The Schrödinger Equation for ψ I
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In 1926 Schrödinger proposed
a wave equation that
describes the manner in which
de Broglie matter waves
change in space and time
Schrödinger’s wave equation is
a key element in quantum
mechanics
Schrödinger’s wave equation is
generally solved for the wave
function, Ψ
The Schrödinger Equation for ψ II
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The 1D Schrödinger Equation for a particle
with energy E, moving in a potential U(x),
having mass, m, is
d  ( x)
2m
  2 E  U ( x)  ( x)
2
dx

2
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The full (complex) wave function is
( x, t )   ( x)e
iEt / 
Particle in 1D box (again)
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In terms of Schrödinger’s Equation the 1D box
problem becomes
2
d
2m E


  k 2
2
2
dx

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Where
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Solutions
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Boundary Conditions give
kL 
k
2m E

 ( x)  A sin(kx) 
2m E
L  n

2
sin(kx)
L
 h2n2 

E n  
2 
 8m L 
The Schrödinger Cat Problem
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It’s OK for electrons and other small things but
what about marcoscopic things like a cat?
Emission Spectra
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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
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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 
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RH is the Rydberg constant
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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
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The Balmer Series
has lines whose
wavelengths are
given by the
preceding equation
Examples of spectral
lines
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n = 3, λ = 656.3 nm
n = 4, λ = 486.1 nm
The Bohr Theory of Hydrogen
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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
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The electron moves
in circular orbits
around the proton
under the influence
of the Coulomb
force of attraction
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The Coulomb force
produces the
centripetal
acceleration
Bohr’s Assumptions, cont
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Only certain electron orbits are stable
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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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The “jump” cannot be treated classically
Bohr’s Assumptions, final
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The electron’s “jump,” continued
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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
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Electron’s orbital angular momentum
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me v r = n ħ where n = 1, 2, 3, …
The total energy of the atom
2
1
e
E KE  PE  me v 2  k e
2
r
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The energy can also be expressed as
k ee2
 
E
2r
Bohr Radius
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The radii of the Bohr orbits are
quantized
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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
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When n = 1, the orbit has the smallest radius,
called the Bohr radius, ao
ao = 0.0529 nm
Radii and Energy of Orbits
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A general expression
for the radius of any
orbit in a hydrogen
atom is
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rn = n2 ao
The energy of any
orbit is
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En = - 13.6 eV/ n2
Specific Energy Levels
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The lowest energy state is called the ground
state
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The next energy level has an energy of –3.40
eV
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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
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The ionization energy for hydrogen is 13.6 eV
Energy Level Diagram
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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 
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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
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The photon has a frequency f = (Ei – Ef)/h and
wavelength λ
Successes of the Bohr Theory
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Explained several features of the hydrogen
spectrum
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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
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Those with one electron
Ze2 needs to be substituted for e2 in equations
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Z is the atomic number of the element
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
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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