Application of Quantum Theory 1- Particle in 1-D box.

The particle is free to move inside the box

Consider the ethylene molecule as an example of the free particle move

in 1-D box, in which the two electrons, only, are free to move, while the

electrons are frozen in bonds with atoms. The length of the box is twice

the C=C bond length.

1- Schrodinger Equation is

( )

2- Using V=0

3- The solution of the 2nd

order ordinary differential equation is

( ) ( )

x = 0 x = L

V = 0 H

H H

H

L = 2 * 1 . 4 = 2 . 8 A

V = 0

4- For particle in box =0 at x =0 and =0 at x =L (boundary condition)

( ) ( )

( )

5- The last equation are mitts only if the argument of sin function is an

integer multiplier of π (0, 180, 360… )

(

)

6- Note that

7- The energies, wave functions and the probability densities are

ψ ψ2

Energy Level

Diagram

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

E

𝐸𝑛 𝑛

𝑚𝐿

8- Comments

a. The energy increase as the quantum number increase

b. The energy separation between energy levels increase as the

quantum number increase

c. The energy and the energy separation increases as the size of the

box decreases

d. ψ2 < ψ when ψ have small values while ψ

2 = ψ when ψ has the

maximum value. (why)

e. Node = the point where wave function passes through zero, or the

position where probability of finding particle = 0 (No. of nodes =

n-1)

f. The probability of finding the particle between two points x1 and

x2 are different when n have small values, while as n increases (n

>100) these 2 probabilities become the same. Generally the

probability density become uniform as n increase, that is to say

quantum mechanics results and classical mechanics results tend to

agree in the limit of the large quantum numbers

9- The particle in a box model can be applied to electrons moving freely

(π electrons) in a molecule

a. For example butadiene has an absorption band at 217 nm for the

1st π→ π* transition. As a simple approximation, consider

butadiene as being a 1-D box of length 4*1.4 = 5.6 Å. and

consider the 4 π electrons to occupy the levels calculated using

the particle in box model

b. The calculated excitation energy is

HOMO : Highest Occupied Molecular Orbital

LUMO : Lowest Unoccupied Molecular Orbital

H2C CH

CH

CH2

E1

E2

HOMO

LUMO

E3

E=ELUMO – EHOMO

( )

h (Plank Constant)= 6.626 x 10-34

JS, m (mass of electron)=9 x 10-31

Kg,

L (length of bonds = no. of carbons *1.4 Å , 1 Å =1 x10-10

m

c (speed of light)=3 x 108 m/S , 1 m = 10

9 nm

What are the causes of difference between Experimental value (217 nm) and

calculated value (204 nm)

That is due to the approximations done

1. Box length C=C equal C-C =1.4 Å

2. Assume it linear

Calculate (nm) for the 1st π→ π* transition for the following compounds

Compound Ethylene Butadiene Hexatriene Octatetraene -carotene

No. of C's 2 4 6 8 22

Exp. 162 217 274 304 425

Calculated