QUANTUM OR WAVE MECHANICS
Quantum chemistry mathematically describes the fundamental behaviour of matter at
the molecular scale.
Normal chemical reactions are actually the flow of electrons between atoms or
molecules
Na + Cl  NaCl  Na+ + Cl
e
Knowledge of the energy and position of electrons helps us understanding chemical
reactions and also the structures of different molecules.
The particle property of electrons:
2
22 e' 4 Z 2
 e' 4 Z 2
 e 4Z 2
18 Z
En  


  2.17868 x 10
n 2h 2
2n 2 2
8h 202
n2
En
er
gy
n=7
E7
n=5
E5
n=5
E5
n=4
E4
n=3
E3 = -1.4 eV
n=2
E2 = -3.4 eV
n=1
E1 = -13.6 eV
1. Calculate the groundstate energy of the hydrogen atom and convert the
result to electron volts. 1 eV = 1.602177 x 10-19 J. (Answer: -2,17868 x 10-18 J
= -13.568 eV. This is the minimum energy needed to ionise a ground-state
hydrogen atom)
2. Calculate the energy required to excite the hydrogen electron from an energy
level n = 1 to n = 6 (Answer: 2.118 x 10-18 J)
3. Calculate the wavelength of light that must be absorbed by a hygrogen atom
in its ground state (n = 1) to reach the first excited state (n = 2). (Answer:
1.217 x 10-7 m)
Electron waves:
 
 = “nu” = frequency in s-1 or Hz (hertz)
c
c = velocity of electromagnetic waves = 3 × 108 ms-1
 = wavelength

Derivation of de Broglie equation:
E  h 
h = 6,62608 x 10-34 J s
Planck’s constant, c = 3 × 108 m s-1
hc

E  h 
E = mc2
 
h

mc
h
mv
hc


 mc 2 or λ =
c


h
mc
h
p
Compare the wavelength for an electron (mass 9.11 x 10-31 kg) travelling at a speed of
1.0 x 107 m s-1 with that for a ball (mass 0.10 kg) travelling at 35 m s-1. (Answer: e =
7.3 x 10-11 m; b = 1.9 x 10-34 m)
Heisenberg’s uncertainty (indeterminacy) principle:
Our concepts of mass, position, velocity, momentum and energy is derived from
experience with the mechanical behaviour of relatively large macroscopic bodies
and we can describe the behaviour of such bodies to our satisfaction in terms of
these concepts. When used in the description of the very small particles
involved in atomic systems, however, the quantities defined by these concepts
cannot be measured with equally satisfactory certainty. Any observation or
measurement we make necessitates some interaction with the system being
studied. For example, when the position of an object is observed, there is no
avoiding the impingement of light upon it or it being touched by some other
object. The measurement itself therefore influences the system under
observation and introduces an uncertainty in the validity of the measurement.
(x)(p) 
h
4
x
 
p  p
x p   h

assume 

x
p
h

x
x
Planck' s constant

6.624 x 10 34 J s

Wavefunctions in describing p en x of an electron:
Since an electron has wave properties, it’s behavior is described as a wave
function, or (x, y, z), the latter meaning that  is a function of co-ordinates x,
y and z. The wave function can take on positive, negative, or imaginary values.
The probability of finding an electron in any volume element in space is
proportional to the square of the absolute value of the wave function,
integrated over that volume of space.
An example of a wave function describing the behavior of electrons is
2x
ψ = A sin
. The probability of finding an electron with wavelength λ

(momentum p = h/) at position x is proportional to A2: probability  A2.
The general wave-equation:
 (x ) = A sin
2x

general wave equation.
and
 (t )  A sin 2   t are used to derive the
+y
-y
Y(t)
A
t
y = A sin 2 t =  ( t ) with A = amplitude = maximum value of y
and  = frequency = number of cycles per second measured in s-1 = Hz
Y(x)
x

x
c
The velocity c =
in the wave property  

t
x
x
v=
=  , therefore t =
t

x = A sin 2  x
 (x) = A sin 2


can be displaced with
 ( x, t ) =  ( x) .  (t ) = A sin
 (x ) =

x
 
 x2
2
A sin
2x

and
 A sin 2  t

 (t )  A sin 2   t
 2 
2 x 

A
cos

 (t )
 
 

 4 
2 x 
A
sin

 t  
2
 
 
2

=  x 2 A cos 2t 
t
2
t 2
2 x

 4

2
2
 ( x) (t )
d (sin x)
dx
d cos x
dx
d sin kx
dx
d cos kx
dx
  x   4 2 2  A sin 2 t    ( x)  4 2 2  (t )
 cos x
  sin x
 k cos kx
  k sin kx
 2
x2
 2
t 2
1

 
2
  
2
met
2 

2
r
2

2

2
x2
1
c2
2
1

2
c
t 2





,
2
2
2
x
y
z

1
c2
2

t 2
  del.
2 
2 
2  1
2 


 2 
2
2
2
x
y
z
c
t 2
with
  ( x , y , z, t )
Derivation of the time-independent Schrödinger equation in one dimension:
 ( x, t )   ( x)  (t )   ( x)  A sin 2 t

x

 ( x)
x

A sin 2  t
2
x2
 2 ( x)
 A sin 2 t 
2
x


t
  ( x) 
2
2
t 2
  ( x) 
 4 
 2 x 
  t 
2
x
A cos 2 t 
2
  4 2 2 ( x) (t )
2
 A sin 2 t

 2
 x2

1
c2
 2

t2
 2 ( x)
 1
2 2




t

 2    4 
2
x
c 

 2 ( x)
 x2
 2 ( x)
x2


4 2 2
1c 2
4 2

2
  ( x)  0
  ( x)  0

   x   t 


c


  

h

p 
 p 
h
p
 
OR
1

2
2mE  PE 
E = KE + PE
1
mv 2  PE
2
2 2
= m v
 PE
2m
2
= p
 PE
2m
=
1
2
 h2m E  PE 

1
2
 2m E  PE 


h2


 2 ( x)
x2

2
8 2 m

h2
8 2 m

h2
 h2
  2
2
8 m
E  PE  ( x)  0
E  PE 
 U
 0
 E
The Hamilton operator H:
H
H
d

dx
d
dx

d
dx

E

E
 h2
2


8 2 m

 U
 e nx
 ne nx  n
d
d

E

 n
If   e is a solution of
, it follows from
dx
dx
that E = n, in other words Eigenvalue = n for the eigenfunction   e nx
nx
Relationship between the total energy and amplitude:
Consider the ball hanging from a spring. Hook's law states F(x) = -Kx with K =
constant of spring, x = displacement, F = force exerted on or by the spring.
If the spring is stretched and the ball moves, the following hold:
At x = 0 the PE = 0 and KE = max = 1 mv 2 . At x = A the PE = max = -Kxmax and
2
KE = 0. Therefore the total energy E of the spring: E  KE  PE
 0  PE maks
but PE (x) 
 F x  dx
x
0
Therefore E  PE maks


 Kx dx  
x
0
1

  Kx 2 
2
 x mak s  A
if (x = A)
1
Kx 2
2
1

  KA 2 
2

or E  A2
Solution of the time-independent Schrödinger equation of a particle in
a one- dimensional box:
d 2  x 
d x2

8 2 m
h2
 n x   A sin
 0
n x

E   x 

0
0 
x  
x  0, x  
n  1, 2, 3...
 n  x 
n x
 n 
  A
 cos
x

  
18
 2 n  x 
n x
 n 


A
sin



x2
  
16
2
E4 = 16h2/8ml2
14
 n 
 
  n x 
  
2
Energy levels
  n x 
 n 

   n x   0
2
x
  
2
2
12
10
E3 = 9h2/8ml2
8
6
En
n2 h2

8m  2
4
E2 = 4h2/8ml2
2
E1 = h2/8ml2
0
18
18
16
16
14
14
12
10
2
2
E3 = 9h /8ml
8
6
4
10
E3 = 9h2/8ml2
8
6
E2 = 4h2/8ml2
4
2
E1 = h2/8ml2
0
Energy levels
Energy levels
12
E2 = 4h2/8ml2
2
E1 = h2/8ml2
0
18
16
18
E4 = 16h2/8ml2
16
14
14
12
10
E3 = 9h2/8ml2
8
6
10
E3 = 9h2/8ml2
8
6
E2 = 4h2/8ml2
4
2
E1 = h2/8ml2
0
Energy levels
Energy levels
12
4
E4 = 16h2/8ml2
E2 = 4h2/8ml2
2
E1 = h2/8ml2
0
[Y4(x)]2
18
18
Y4(x)
E4 = 16h2/8ml2
16
14
12
Y3(x)
10
E3 = 9h2/8ml2
8
6
Y2(x)
E2 = 4h2/8ml2
4
Energy levels in h2/8ml2
14
Energy levels in h2/8ml2
E4 = 16h2/8ml2
16
12
[Y3(x)]2
10
E3 = 9h2/8ml2
8
[Y2(x)]2
6
E2 = 4h2/8ml2
4
[Y1(x)]2
Y1(x)
2
2
E1 = h2/8ml2
E1 = h2/8ml2
0
0
0
l /2
l
0
l /2
l
Electrons confined to a one-dimensional box
[Y3(x)]2
(2/a) 1/2
x = l /2
x= l
2
Y3(x) or Y3(x) for energylevel E3
(2/a)
Y3(x) = (2/a)1/2sin3x/l
c
The three dimensional box with dimensions a x b x c:
b
a
PE = 0
PE = 
0<x<a
x  0, x  a
0<y<b
y  0, x  b
0<z<c
z  0, x  c
 2
2
2 
8 2 m
 2  2  2  ( x, y, z )  2 ( E ) x, y, z   0
y
z 
h
 x
 n x 
x
n x
 A sin x
a
 n  y   B sin
y
 n z 
z
ny  y
b
n z
 C sin z
c
h 2 n x2

8m a 2
with E x
h 2 n y2
with E y

with E z
h 2 n z2

8m c 2
8 mb 2
ny  y
n x
n z
 x, y, z  =  x  y  z  = ABC sin x
sin
sin z
Etot = E x  E y  E z
2
= h  n x
8m  a 2
2

a
n
2
y
b
2

n 
c 
2
z
2
b
c
with



0
 n2 x  dx  1 and / en n x  is normalised
nx
dx  1
0

A 2 
2nx 
1

cos

 dx  1

0
2 
 

A 2 sin 2

1
2

A
A
2
n

2nx 



x0   
 sin

2
2




 0

2
 1
1

A 2  A 2  2n 
0
 
 sin 2n  sin 0   1
2
 2   

A 2
 0  00  1
2
2
A2 



 2
A   

or
1
2
1
1  cos 2 x 
2
 cos x dx  sin x
sin 2 x 
sin n
2

 n  x     sin
 0
 cos kx dx 
2
sin
 kx 
1
sin kx
k
1
 2 1  cos 2kxdx


1
2
1 
1

x

sin
2
kx


2 
2k
 0
nx

Three dimensional box:
 n  x, y , z    n  x   n  y   n  z 
x
y
 ABC sin

8
abc
sin
z
nxx
a
nxx
a
sin
sin
n yy
b
n yy
b
sin
sin
nzz
c
with
A
2
; B
a
2
; C
b
2
c
nzz
c
2
h 2  nx2 n y nz2 
with E 


8m  a 2 b 2 c 2 
Degeneration:
b
0
 2
x2
a

 2
 y2
8 2 m

h2
E 
 0
with  =  x, y    x .   y 
=
and Etot =
n xx
2

sin
a
b
4 sin n xx sin n y y
a
b
ab
2
a
sin
9
b
8
h 2  n x2 n 
Ex  E y 

8m  a 2 b 2 
2
y
(x, y, z) = (x)(y)(z)
=
n yy
8 sin n xx sin n y y sin n zz
a
c
b
abc
6
5
Etot
E12=E21=5h2/8ml2
4
3
2
1
2
h 2  n x2 n y n z2 
 Ex  E y  Ez 


8m  a 2 b 2 c 2 
E 22= 8h2/8ml2
7
Energy levels
=
10
0
E11=2 h2/8ml2
18
16
18
16h2/8ml2
16
14
14
12
10
2
2
9h /8ml
8
6
6h2/8ml2
4
Energy levels
Energy levels
12
E111
10
0
E121
E122
E212
E221
E122
8
6
4
3 h2/8ml2
2
E211
2
0
E222
Solutions of the Schrödinger equation for the hydrogen atom:
Y(r, , ) = R(r) () ()
n
l
ml
the principal quantum number
the angular momentum quantum number
the magnetic quantum number
Electromagnetic radiation:
Absorption of radiation by molecules:
Emolecule = Eelectronic + Evibrational + Erotational + Etranslational + other ways
E

hc

or
 
hc
E
En

h 2n 2
8ma2
18
18
16
16
14
14
12
12
10
E3 = 9h2/8ma2
8
6
4
10
E3 = 9h2/8ma2
8
6
E2 = 4h2/8ma2
2
4
E2 = 4h2/8ma2
2
2
E1 = h /8ma
0
Energy levels
Energy levels
Butadiene, CH2 = CH - CH = CH2
2
E1 = h2/8ma2
0
E  E excited  E groundstate
2 2
  h ng

  8ma 2
 
 h 2 C 2C 2120
 ne  n g
 
2  120º
C
8
ma


 h 2 ne2
 
2
 8ma

C C




120º
H2C
0
120
C C
0.5 x (1.54Å) 1
H2C CH CH CH2
C
0
0.5 x (1.54Å) 1.35Å 1.54Å 1.35Å
0.5 x (1.54Å)
a = (C - C bond length distance) + 2(C = C bond length distance) + 2(0.5 x C-C
bond length distance for C orbital at end of molecule)
a = 1.54 Å + (2 x 1.35 Å) + (2 x 0.5 x 1.54 Å) = 5.78 Å = 5.78 x 10-10 m
 
8cma 2
h( ne  n g )
2
2
8  3 10 8  9.11  10 31  (5.78 10 10 ) 2

6.6226  10 34 [(3) 2  (2) 2 ]
 2.20 x 10 7 m or 220nm