Transcript Lecture 18: Intro. to Quantum Mechanics

Lecture 16: Intro. to Quantum
Mechanics
• Reading: Zumdahl 12.5, 12.6
• Outline
– Basic concepts.
– A model system: particle in a box.
– Other confining potentials.
Quantum Concepts
• The Bohr model was capable of describing the
discrete or “quantized” emission spectrum of H.
• But the failure of the model for multielectron
systems combined with other issues (the
ultraviolet catastrophe, workfunctions of metals,
etc.) suggested that a new description of atomic
matter was needed.
Quantum Concepts
• This new description was known as wave
mechanics or quantum mechanics.
• Recall, photons and electrons readily demonstrate
wave-particle duality.
• The idea behind wave mechanics was that the
existence of the electron in fixed energy levels
could be though of as a “standing wave”.
Exercise
• What is the wavelength of an electron (mass 9.11
x 10-31 kg) traveling at a speed of 1.0 x107 m/s?
l = h / p = h/mv
l = 6.626x10-34 Js /(9.11x10-31kg)(1x107m/s)
l = 6.626x10-34 Kgm2/s /(9.11x10-31kg)(1.x107m/s)
l = 7.3x10-11 m
Exercise
• What is the wavelength of a baseball (mass 0.1 kg)
traveling at a speed of 35 m/s?
l = h / p = h/mv
l = 6.626x10-34 Js /(0.1kg)(35m/s)
l = 6.626x10-34 Kgm2/s /(0.1kg)(35m/s)
l = 1.9x10-34 m
Uncertainty Principle
• Another limitation of the Bohr model was that
it assumed we could know both the position
and momentum of an electron exactly.
• Werner Heisenberg development of quantum
mechanics leads him to the observation that
there is a fundamental limit to how well one
can know both the position and momentum of a
particle.
h
x  p 
4
Uncertainty in position
Uncertainty in momentum
Example
• Example:
What is the uncertainty in velocity for an electron in a
1Å radius orbital in which the positional uncertainty is
1% of the radius.
x = (1 Å)(0.01) = 1 x 10-12 m
34
6.626x10
J.s

h
23
p 


5.27x10
kg.m /s
12
4 x
4 1x10 m
p 5.27x1023 kg.m /s
7m
v 


5.7x10
s
m
9.11x1031 kg
huge
Example
• Example (you’re quantum as well):
What is the uncertainty in position for a 80 kg student
walking across campus at 1.3 m/s with an uncertainty
in velocity of 1%.
p = m v = (80kg)(0.013 m/s) = 1.04 kg.m/s
34
6.626x10
J.s

h
x 

 5.07x1035 m
4p 4 1.04kg.m /s
Very small……we know where you are.

De Broglie’s wavelength
• He provided a relationship between the electron properties
and their ‘wavelength’ which experimentally demonstrated
by diffraction experiments
l = h / p = h/mv
Quantum Concepts (cont.)
• What is a standing wave?
• A standing wave is a motion in
which translation of the wave does
not occur.
• In the guitar string analogy
(illustrated), note that standing
waves involve nodes in which no
motion of the string occurs.
• Note also that integer and halfinteger values of the wavelength
correspond to standing waves.
Quantum Concepts (cont.)
• Louis de Broglie suggests that for the e- orbits envisioned
by Bohr, only certain orbits are allowed since they satisfy
the standing wave condition.
not allowed
Schrodinger Equation
• Erwin Schrodinger develops a mathematical
formalism that incorporates the wave nature of
matter:
Hˆ   E
The Hamiltonian:

The Wavefunction:
 
Kinetic Energy
pˆ
ˆ
H   (PE)
2
x

2m
E = energy
d2/dx2
Wavefunction
• What is a wavefunction?

= a probability amplitude
• Consider a wave:
y  Ae

Intensity =
2

y  Ae
i2  t  
i2  t  
Ae
i2  t  
• Probability
of finding a particle in space:

Probability =

*

• With the wavefunction, we can describe
spatial distributions.
 A
2
Potential Energy and Quantization
• Consider a particle free to move in 1 dimension:
p
x
“The Free Particle”
Potential E = 0
• The Schrodinger Eq. becomes:
2
2
 pˆ 2

ˆ
p
p
1 2
ˆ
H    PE  ( p) 
 ( p) 
 ( p)  mv  ( p)  E ( p)
2m
2m
2
2m

0
• Energy ranges from 0 to infinity….not quantized.
Potentials and Quantization (cont.)
• What if the position of the particle is constrained
by a potential:
“Particle in a Box”
inf.
Potential E
0
0
x
L
= 0 for 0 ≤ x ≤ L
=  all other x
• Now, position of particle is limited to the dimension
of the box.
Potentials and Quantization (cont.)
• What do the wavefunctions look like?
2 nx 
 x  
sin 

L  L 
n = 1, 2, ….

Like a standing wave

*
Potentials and Quantization (cont.)
• What does the energy look like?
2
2
n h
E
8mL2
n = 1, 2, …

E

*
Energy is quantized
Potentials and Quantization (cont.)
• Consider the following dye molecule, the length of which
can be considered the length of the “box” an electron is
limited to:
+
N
L=8Å
N
What wavelength of light corresponds to E from n=1 to
n=2?
2
h2
h
2
2
2
19
E 
n

n

2
1

2.8x10
J

initial
2  final
2 
8mL
8m(8Å)
l  700nm
(should be 680 nm)
Potentials and Quantization (cont.)
• One effect of a “constraining potential” is that the
energy of the system becomes quantized.
• Back to the hydrogen atom:
er
0
r
e 2
V (r) 
r
P+
constraining potential

Potentials and Quantization (cont.)
• Also in the case of the hydrogen atom, energy
becomes quantized due to the presence of a
constraining potential.
0
r
0
e
V (r) 
r
2
Schrodinger
Equation
Recovers the “Bohr” behavior
