Download Lecture XVI

Application of quantum in chemistry
The Particle in A Box
The ‘Classical’ Case
The ‘Quantum’ Case
The absolutely small particle in the nanometer size box is a quantum particle,
and it must obey the Uncertainty Principle, that is, ΔxΔp= h/4π. If V=0 and x=
L/2, we know both x and p. The result would be ΔxΔp 0, the same as the classical
racquetball.
This is impossible for a quantum system. Therefore, V cannot be zero. The
particle cannot be standing still at a specific point. If V cannot be zero, then Ek
can never be zero. The Uncertainty Principle tells us that the lowest energy that a
quantum racquetball can have cannot be zero. Our quantum racquetball can
never stand still.
Energies of a Quantum Particle in a Box
Wave functions must be zero at the walls
Nodes are the points where the wave function crosses zero
Energies are quantized
A Discreet set of energy levels
Why are Cherries Red and Blueberries blue ?
The Colour of Fruit
This energy corresponds to
Deep Red Colour
If L=0.7 nm, =540 nm
If L=0.6 nm, =397 nm
Green Colour
Blue Colour
Particle in a box
Step 1: Define the potential energy
Step 2: Solve the Schrodinger equation
Step 3: Define the wave function
Step 4: Determine the allowed energies
Step 5: Interpret its meaning
Particle in 1-dimensional box
• Infinite walls
Time Independent Schrödinger Equation
  2 d 2 ( x)
 V ( x)  E
2
2m
dx
Region II
Region I
Region III
PE
KE
TE
Applying boundary conditions:
V(x)=0
V(x)=∞
V(x)=∞
Region I and III:
0
L
x
  2 d 2 ( x)
  *  E
2
2m
dx
 
2
0
d 2 ( x )
2mE

 ( x)
2
2
dx
Second derivative of a function equals a
negative constant times the same function.
Functions with this property
sin and cos.
d 2 sin(ax )
 a 2 sin(ax )
2
dx
d 2 cos(ax )
 a 2 cos(ax )
dx
Copyright – Michael D. Fayer, 2007
b) x=L ψ=0
0  Asin kL
Region II:
  2 d 2 ( x)
 E
2
2m
dx
d 2 ( x)
2m
 

E
2
2
dx

This is similar to the general differential equation:
d 2 ( x)

 k 2
2
dx
  A sin kx  B cos kx
Applying boundary conditions:
a) x=0 ψ=0
0  A sin 0k  B cos 0k
0  0  B *1
 B  0
ButA  0
kL  n
Thus, wave function:
 II  A sin
nx
L
But what is ‘A’?
Normalizing wave function:
Calculating Energy Levels:
L
2
(
A
sin
kx
)
dx  1

0
k2 
2mE
2
L
 x sin 2kx 
A  
 1
2
4
k

0
2
n 

sin
2
L
2L
L
A  
 1
n
2


4
L


2 L 
A   1
2
A 
k 2h2
E
2m4 2
( 
n 2 2
h2
E
L2 2m4 2
2
L
Thus normalized wave function is:
 II 
k 2 2
E
2m
2
nx
sin
L
L
Thus Energy is:
n2h2
E
8mL2
h
)
2
Particle in a 1-Dimensional Box
2
nx
sin
L
L
 II 
 II
2
2
nx 
  sin

L
L 
2
1) Difference b/w adjacent
energy levels:
2) Non-zero zero point energy
+
+
+
+
+
E

*
3) Probability density is structured
with regions of space demon-strating enhanced probability.
At very high n values, spectrum
becomes continousconvergence with CM
(Bohr’s correspondance
principle)
Particle in a 3-D box
Question: An electron is in 1D box of 1nm length. What is the probability of
locating the electron between x=0 and x=0.2nm in its lowest energy state?
Question: An electron is in 1D box of 1nm length. What is the probability of
locating the electron between x=0 and x=0.2nm in its lowest energy state?
Solution:
Question: What are the most likely
locations of a particle in a box of length L
in the state n=3
Example: What are the most likely locations of a particle in a box of length L in
the state n=3
Expectation value of position and its
uncertainty
Expectation values
Position
Uncertainity
Expectation value of Momentum
And square of momentum
Momentum
Estimating pigment length
Assumptions:
Wavelength of transition for Anthracene
Particle in a Box
Simple model of molecular energy levels.
Anthracene
L

L6 A
 electrons – consider “free”
in box of length L.
Ignore all coulomb interactions.
Pigments and Quantum mechanics
High degree of conjugation!!
• Electrons have wave properties and they don’t jump off the pigments when they reach its ends.
• These electrons resonances determine which frequencies of light and thus which colors, are absorbed
& emitted from pigments
Electron resonances in a cyclic conjugated molecule
A crude quantum model for such molecules assumes that electrons move freely in a ring.
Resonance condition:
R: radius of molecule, λ wavelength of electron
Energy is once again quantized. It depends on variable n which posseses
discrete values only