Download Chapter 3 - Illinois State Chemistry

13
Chapter 3. Introduction to Molecular Vibrations
3.1 Overview
In this class, we will apply the techniques of Fortran programming to investigate specific
applications in chemistry. Our first application will be to simulate the vibrational motion of a
chemical bond using classical mechanics. Strictly speaking, quantum mechanics should be applied
to the vibrational motion; however, the atoms are heavy enough that even though the vibrational
energies are quantized, the dynamical motion is described reasonably well by classical mechanics.
3.2 Review of Classical Mechanics
Newton's equations for a single object in one dimension have the form
F = ma.
(3.1)
The variable F is the force on the system, m is the mass, and a is the acceleration. The acceleration
a is related to the position by the equation
€
a = x˙˙ .
(3.2)
The double dot in the equation above indicates a second derivative of the position with respect to
time,
€
x˙˙ =
d 2 x(t)
dt 2
.
(3.3)
For many systems, the force F can be defined as
€
F = −
dV ( x )
,
dx
(3.4)
where V is the potential energy of the system.
Newton's equations describe the motion of the system according to classical mechanics.
Solving this set of equations€yields the position as a function of time, x ( t ) . In addition, the velocity
as a function of time, v ( t ) , may also be obtained by using the definition
v ( t ) = x˙ ( t ) .
€
€
3.3 Models of Molecular Vibrations – The Harmonic Oscillator
Consider a chemical bond in a molecule. If we plot the energy E of the molecule as a
function of the bond distance€R, we get a curve similar to the one shown in Figure 3.1.
(3.5)
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E
R
Figure 3.1. A plot of energy versus bond distance for a typical chemical bond.
For energies close to the equilibrium separation (or minimum), the curve appears to be quadratic. A
system that has a potential energy that is quadratic is called a harmonic oscillator. For a harmonic
oscillator, the potential energy is given by
V ( x) =
1 k x2,
2
(3.6)
where k is the force constant and x is the bond displacement,
€
x = R − Req .
(3.7)
In Eq. (3.7), R is the instantaneous bond length and Req is the equilibrium bond length. The force
constant k is related to the stiffness of the bond: k is larger for stiffer bonds (such as triple bonds)
€
and k is smaller for weaker bonds (such as single bonds).
3.4 Classical Mechanics of Harmonic €
Vibrational Motion
To solve Newton’s equations for the harmonic oscillator, we must first obtain the force.
The force F for the harmonic oscillator is
dV ( x )
dx
= − k x.
F = −
(3.8)
Substituting into Eq. (3.1), we have the equation
€
or
F = ma,
− k x = m x˙˙ .
(3.9)
For a system as simple as the harmonic oscillator, Newton’s equation may be solved
exactly. This is useful because it allows us to test algorithms and accuracy when the system is
€
programmed. The exact analytical
solutions of Newton’s equation for a harmonic oscillator with an
initial position of zero are
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x(t) = Asin ω t
(3.10)
v(t) = Aω cosω t .
In this set of equations, the parameter A equals
€
A =
2E
,
k
(3.11)
where E is the total energy and k is the force constant. The parameter ω is called the angular
velocity and is defined by the relation
€
ω =
k
,
m
(3.12)
where m is the mass. The angular velocity ω is related to the harmonic frequency of oscillation
νo,
€
νo =
€
ω
.
2π
(3.13)
In a system like the harmonic oscillator, the total energy is conserved throughout the
trajectory. The total energy E€at any time is given by
E =
p2
+ V (x).
2m
(3.14)
In classical mechanics, the momentum p is related to the velocity v via the equation
€
p = mv .
(3.15)
Energy conservation is another useful quantity that may be employed as a check of the accuracy of
€
the Fortran program.
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3.5 Beyond the Harmonic Approximation – The Morse Oscillator
Because chemical bonds can dissociate, the harmonic oscillator potential is not an accurate
representation of vibrational motion in molecules at high energies. A better model is given by the
Morse potential,
[
V ( x ) = D 1 − e−a x
]
2
.
(3.16)
The parameters D and a control the dissociation energy and the harmonic frequency of the model.
A plot of a typical Morse potential is shown in Figure 3.2.
€
0.6
0.5
V(x)
0.4
0.3
0.2
0.1
0
-0.5
0
0.5
1
1.5
2
x
Figure 3.2. The Morse oscillator potential.
At low energies, the Morse oscillator reduces to a harmonic potential. This result may be shown by
making a Taylor series expansion of the Morse oscillator,
V ( x ) = V (0) + V ′(0) ⋅ x +
V ′′(0) 2
V ′′′(0) 3
⋅x +
⋅x + K
2!
3!
(3.17)
From the equation for the Morse oscillator, it can be shown that the potential and its first derivative
both equal zero at x=0. Evaluating the second and third derivatives and substituting yields
€
1
V ( x) =
2a 2 D x 2 − a 3 D x 3 + K
(3.18)
2
(
)
( )
The leading term in the Taylor series expansion of the Morse oscillator is quadratic with a force
constant equal to 2a2 D. The harmonic frequency ν o of the Morse oscillator is then
€
1
νo =
2π
€
€
2a 2 D
.
m
(3.19)
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Chapter 3 Review Problems
1. By explicit substitution into Eq. (3.9), verify that one solution to Newton's equation for the
harmonic oscillator is given by
x ( t ) = A sin(ω t ) ,
where A is a constant and ω is the angular frequency. This solution assumes that the initial
position is 0; i.e., x(0) = 0. You should be able to demonstrate how the angular frequency ω is
€ k and the mass m; i.e., verify the relation given in Eq. (3.12).
related to the force constant
€ of momentum, p = mv , and the analytic solution for the position, x ( t ) ,
2. Using the definition
t ) , for a
determine the exact analytic solution for the momentum as a function of time, p(€
harmonic oscillator.
€
€
3. Substitute the exact solutions for position and velocity of the harmonic oscillator, Equation
(3.10), into the expression for the energy given in Equation (3.14) in€order to show that the
energy is constant as a function of time.
4. Obtain Newton’s equation for a cubic oscillator for which the potential energy is given by
V ( x) =
1α
3
x3 .
Note: you just need to obtain the equation. It is not necessary to determine a solution.
€ oscillator potential, Equation (3.15), show that the minimum
5. By differentiation of the Morse
occurs at x=0.
6. Obtain Newton’s equation of motion for a Morse oscillator potential.