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 nmIf L=0.6 nm, =397 nm

Green ColourBlue 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

V(x)=0 V(x)=∞V(x)=∞

0 L x

Region I Region II Region III KE PE TE

ExV

dx

xd

m

)(

)(

2 2

22

Time Independent Schrödinger Equation

Applying boundary conditions:

E

dx

xd

m

*

)(

2 2

22

Region I and III:

02

2

2 2

( ) 2( )

d x mEx

dx

Second derivative of a function equals anegative constant times the same function.

22

2

sin( )sin( )

d axa ax

dx

22cos( )cos( )

d axa ax

dx

Functions with this property sin and cos.

Copyright – Michael D. Fayer, 2007

E

dx

xd

m

2

22 )(

2

Region II:

E

m

dx

xd22

2 2)(

22

2 )(k

dx

xd

kxBkxA cossin

Applying boundary conditions:

kBkA 0cos0sin0

0

1*00

B

B

b) x=L ψ=0

0ButA

kLAsin0

nkL

This is similar to the general differential equation:

a) x=0 ψ=0

L

xnAII

sin

Thus, wave function:

But what is ‘A’?

Normalizing wave function:

1)sin(0

2 L

dxkxA

14

2sin

20

2

L

k

kxxA

14

2sin

2

2

Ln

LLn

LA

12

2

L

A

LA

2

Thus normalized wave function is:

L

xn

LII

sin2

22 2

mE

k

m

kE

2

22

2

22

42 m

hkE )

2(

h

2

2

2

22

42

m

h

L

nE

2

22

8mL

hnE

Calculating Energy Levels:

Thus Energy is:

Particle in a 1-Dimensional Box

L

xn

LII

sin2

2

2sin

2

L

xn

LII

+

+ +

+

+

-

E

1) Difference b/w adjacent energy levels:

2) Non-zero zero point energy

3) Probability density is structured with regions of space demon- -strating enhanced probability.

At very high n values, spectrum becomes continous-

convergence 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 6 AL

electrons – consider “free”in box of length L.

Ignore all coulomb interactions.

Pigments and Quantum mechanics

• 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

High degree of conjugation!!

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