Physical Chemistry 2nd Edition

Transcript Physical Chemistry 2nd Edition

Chapter 18

A Quantum Mechanical Model for the Vibration and Rotation of Molecules Physical Chemistry 2 nd Edition Thomas Engel, Philip Reid

Objectives

• • • Solving Schrödinger Equation Introducing Angular Momentum Introducing Spherical Harmonic Functions

Outline

Solving the Schrödinger Equation for the Quantum Mechanical Harmonic Oscillator Solving the Schrödinger Equation for Rotation in Two Dimensions Solving the Schrödinger Equation for Rotation in Three Dimensions The Quantization of Angular Momentum The Spherical Harmonic Functions

18.1 Solving the Schrödinger Equation for the Quantum Mechanical Harmonic Oscillator • • • Translational motion in various potentials is described in the context of wave-particle duality.

In applying quantum mechanics to molecules, there are 2 motions for molecules to undergo: vibration and rotation. For vibration, the harmonic potential is

 1 2

2 where k = force constant

18.1 Solving the Schrödinger Equation for the Quantum Mechanical Harmonic Oscillator • The normalized wave functions are  

A n H n

 1 2

x e

 

2 / 2 , and

 0 , 1 , 2 ,...

•

18.1 The Classical Harmonic Oscillator Chapter 18: A Quantum Mechanical Model for the Vibration and Rotation of Molecules Physical Chemistry 2 nd Edition © 2010 Pearson Education South Asia Pte Ltd

Example 18.1

Show that the function satisfies the Schrödinger equation for the quantum harmonic oscillator. What conditions does this place on ? What is E ?

Solution

We have  

2 2 

d d

2 2 

n dx

2 (

)

 



   

n x e

 

  

E n



 

2 2 

2 2  

   2 

 

2 

 

2   1

2 2   2  

 

4  2

 

2   1 2

 

Solution

The function is an eigenfunction of the total energy operator only if the last two terms cancel:

total e

 

2 

2  

 

2 if  2   1 4



2 Finally,



2   

2  1

 4

2 



18.1 Solving the Schrödinger Equation for the Quantum Mechanical Harmonic Oscillator • Hermite polynomials states that  0  1  2  3     1 / 4

 1 /    4   3   1 / 4

 

2 

2   4   1 / 4  2 

2  1 

 

2     3 9    1 / 4  2 

3  3



   

18.1 Solving the Schrödinger Equation for the Quantum Mechanical Harmonic Oscillator • • • The amplitude of the wave functions approaches zero for large x values only when

E n



h k



1 2 

 1 2   with n  1,2,3,...

The frequency of oscillation is given by

 1 2 



18.2 Energy Levels and Eigenfunctions for the Harmonic Oscillator Chapter 18: A Quantum Mechanical Model for the Vibration and Rotation of Molecules Physical Chemistry 2 nd Edition © 2010 Pearson Education South Asia Pte Ltd

Example 18.2

a. Is an eigenfunction of the kinetic energy operator? Is it an eigenfunction of the potential energy operator?

b. What are the average values of the kinetic and potential energies for a quantum mechanical oscillator in this state?

Solution

 a. Neither the potential energy operator nor the kinetic energy operator commutes with the total energy operator. Therefore, because   4  3 /   1 4

   

2 is an eigenfunction of the total energy operator, it is not an eigenfunction of the potential or kinetic energy operators.

Solution

b. The fourth postulate states how the average value of an observable can be calculated. Because then

potential

(

)

E potential

  

 1 * (

) and (

)

V kinetic

(

) (

)  1 (

)

 

2 2 

2        4   3   1 / 4

   

2 1 2

2   4   3   1 / 4

 ( 1 / 2 ) 

 1 2

  4   3   1 / 2    

 



  4   3   1 2    

 

Solution

The limits can be changed as indicated in the last integral because the integrand is an even function of x . To obtain the solution, the following standard integral is used:  0 

n e

 

 1  3  2 5

 1



 1  1  

The calculated values for the average potential and kinetic energy are

E potential

 1 2

  4   3   1 / 2   

  3 4  2  3

4   3 4

h k



Solution

Thus

E kinetic

   1 *     

2 

2    1  

       4   3   1 / 4

1 / 

2   

2 2 

2     4   3   1 / 4

   

 

2 2    4   3   1 / 2       2

4  3 

2 

 

 

2 2    4   3   1 / 2  0    2

4  3 

2 

 

 

2 2    4   3   1 / 2  2      3 4  2     3      3 4

2    3 4

h k

   1 2    

Solution

In general, we find that for the n th state,

E kinetic



E potential



 1 2

18.1 Solving the Schrödinger Equation for the Quantum Mechanical Harmonic Oscillator • 18.3 Probability of Finding the Oscillator

18.2 Solving the Schrödinger Equation for Rotation in Two Dimensions • Consider rotation, the total energy operator can be written as a sum of individual operators for the molecule: ˆ

total



trans

 



vib

  int

ernal

 

rot

 

, 

 • Also the system wave function is a product of the eigenfunctions for the three degrees of freedom: 

total

 

trans

  

cm vib

 int

ernal

 

rot



, 



Example 18.3

The bond length for H molecular axis?

19 F is 91.68 × 10 -12 m. Where does the axis of rotation intersect the

Solution

If x H x H and

x H

=87.10

 x

Substituting we find that are the distances from the axis of rotation to the H and F atoms, respectively, we can × 10 for HI or HCl.

m x F 91 .

68 F = -12  12 x H m H =x F m F . 19.00 amu and m H =4.58×10-12 m and = 1.008 amu, m. The axis of rotation is very close to the F atom. This is even more pronounced

18.2 Solving the Schrödinger Equation for Rotation in Two Dimensions • • angle Ф.

     

 



im l

 and   

 



im l

 The solutions above correspond to clockwise and counterclockwise rotation.

Example 18.4

Solution

 2π. The following result is obtained: 2  0   *

m l

   

m l d

   2 2  0 



im l



e im l



  1    2 2  0 

  1

   1 2 

18.2 Solving the Schrödinger Equation for Rotation in Two Dimensions • • The energy-level spectrum is discrete and is given by

E m l



2 2

m l

2 

0 2 

m l

2 2

for

m l

 0 ,  1 ,  2 ,  3 ..

where m l = quantum number We say that the energy levels with are twofold degenerate .

18.2 Solving the Schrödinger Equation for Rotation in Two Dimensions • • • For rotation in the x-y plane, the angular momentum vector lies on the z axis. The angular momentum operator in these



Applying this operator to an eigenfunction,



   

2 

de im l



 

m l h

2 

e im l

 

m l h

 

18.3 Solving the Schrödinger Equation for Rotation in Three Dimensions • For molecule rotating in two dimensions:   



e im l

 and   



im l

 , for

m l

 0 , 1 , 2 , 3 ,...

• To make sure Y(θ,Ф) are single-valued functions and amplitude remains finite, the following conditions must be met. 

m l

 

  

,  , for  , 

l l

  0 , 1 , 2 , 3 ,...

and 2  ,..., 0 ,..., 

 2    ,

18.3 Solving the Schrödinger Equation for Rotation in Three Dimensions • • Both form l and m l must be integers and the spherical harmonic functions are written in the



Y l m l

 

l m l

   

m l

The quantum number l total energy observable, is associated with the

E l



2 2

I l

  , for

 0 , 1 , 2 , 3 ,...

18.4 The Quantization of Angular Momentum • • The spherical harmonic functions, are eigenfunctions of the total energy operator for a molecule that rotates freely in three dimensions.

The eigenvalue equation for the operator can be written as

ˆ 2

Y l m l





Y l m l

18.4 The Quantization of Angular Momentum

• The operators have the following

x y l

form in Cartesian coordinates:

l l

ˆ ˆ

y x z

  

   

  

z y

 



 

   

 

    

  

 



 

  

18.4 The Quantization of Angular Momentum

• The operators have the following form in spherical coordinates:

 

   sin      cot  cos      

 

   cos      cot  sin      

 

      

18.4 The Quantization of Angular Momentum

• For the operators in Cartesian coordinates, the commutators relating the operators are given by



ih l

z l

   

 

x z

 

ih ih l

x l

• Thus the spherical harmonics is as



Y l m l

for

m l

         

 0 ,  1 ,  2 ,  3 ,..., 

     1 2 



im l

      

m l h

     ,

18.5 The Spherical Harmonic Functions • For spherical harmonic functions, these are the first few values of l and m l :

0 0 ,  4  1   1 / 2

1 0

1  1

2 0

2  1

2  1 ,  3 4  1 / 2 cos  ,   5 16  1 / 2 sin 



 , ,  ,     5 16  1 / 2  3 cos 2   1  5 8  1 / 2 sin  cos 

15 32   1 / 2 sin 2 



 



18.5 The Spherical Harmonic Functions

• The functions which form an orthonormal set are given in the following equations:

p x



p y

 1 2 

1 1 

1  1   1 2



1 1 

1  1  

p z



1 0  3 4  cos  3 4  sin  cos  3 4  sin  sin 

d z

2 

1 0 

d xz



d yz

 5 16   3 cos 2   1  1 2 

2 1 

2  1   1 2



2 1 

2  1   15 4  sin  cos  cos  15 4  sin  cos  sin 

d x

2 

d xy

  1 2



2 2 

2  2   1 2



2 2 

2  2   15 16  15

6  sin 2  cos 2  sin 2  sin 2 

18.5 The Spherical Harmonic Functions

• 3D perspective plots of the p and d linear combinations of the spherical harmonics.

Physical Chemistry 2nd Edition

Transcript Physical Chemistry 2nd Edition

Chapter 18

Objectives

Outline

Example 18.1

Solution

Solution

Example 18.2

Solution

Solution

Solution

Solution

Solution

  

 



Example 18.3

Solution

Example 18.4

Solution





   

18.4 The Quantization of Angular Momentum

18.4 The Quantization of Angular Momentum

18.4 The Quantization of Angular Momentum

18.5 The Spherical Harmonic Functions

18.5 The Spherical Harmonic Functions

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