Download Application of Quantum Theory 1- Particle in 1

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.
V=0

V=0
H
H
H

H
x=0
1- Schrodinger Equation is
L = 2*1.4 =2.8 A
(
)
2- Using V=0
3- The solution of the 2nd order ordinary differential equation is
( )
( )
x=L
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
Energy Level
Diagram
15
𝐸𝑛
𝑛
𝑚𝐿
E
ψ
ψ2
10
5
0
0
0.1 0.2 0.3
0.4 0.5 0.6
0.7 0.8 0.9
1
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 ψ2 = 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 x 1 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
H2C C C CH2
the particle in box model
LUMO
H H
b. The calculated excitation energy is
E3
HOMO
HOMO : Highest Occupied Molecular Orbital
LUMO : Lowest Unoccupied Molecular Orbital
E2
E1
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 = 109 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