Vibrational Spectroscopy Outlines Vibration of diatomic molecules Permanent vs Dynamics dipole moment Force and Harmonic potential energy Classical harmonic oscillator Exercise: calculate the force constant Anharmonicity Quantum harmonic oscillator
Comparison of the bond strength of the following molecules. 1) HF HCl HBr HI 2) H2 HD D2 3) F2 Cl2 Br2 I2 N2 O2
Permanent vs Dynamics dipole moment HCl Diatomic molecules Permanent vs Dynamics dipole moment HCl flexible Permanent Dynamics absorbing energy in the microwave region Absorb energy in the infrared region the spring allows the two masses to oscillate about their equilibrium distance. If the electric field and oscillation of the dipole moment have the same frequency, the molecule can absorb energy from the field. rigid What’s about homonuclear diatomic molecules?
Working model & harmonic potential energy Two atoms are connected by a spring. (The strength of the bond is related to the stiffness of the spring.) stretch (Equilibrium) bond length compress the spring allows the two masses to oscillate about their equilibrium distance. If the electric field and oscillation of the dipole moment have the same frequency, the molecule can absorb energy from the field. In spectroscopy, the vibrational frequency is related to the (potential) energy.
Force and Harmonic Potential Energy Force acts on the system (Hook’s law): k : the force constant - The negative sign shows that the force and the displacement are in opposite directions. The potential energy (V) of the vibration of the diatomic molecule: Simple harmonic potential
Electric dipole moment oscillates with vibration Electric field (external force to stretch or compress the molecule) Electric field interacts with a dipole moment within a molecule. oscillation of the dipole moment the spring allows the two masses to oscillate about their equilibrium distance. If the electric field and oscillation of the dipole moment have the same frequency, the molecule can absorb energy from the field.
Classical (Physics) Harmonic Oscillator Consider two masses m1 and m2 that are connected by a coiled spring. the deviation of the spacing between the masses from its rest position is denoted by , the motion of a system composed of 2 bodies was dissociated in the motion for the center of mass and the relative motion of one body with respect to the other one. Positive and negative values of x correspond to stretching and compression of the spring, respectively.
It is more convenient to view the vibration motion in the center of mass coordinates (because we consider only the relative motion). Introducing “The Center of Mass”, xcm The two mass is reduced into a single mass And the reduced mass,
Ex. Find the center of mass of 1H35Cl if the H-Cl bond length is 0 Ex. Find the center of mass of 1H35Cl if the H-Cl bond length is 0.129 nm. (atomic weight of H and Cl are 1.008 amu and 35.453 amu). H Cl z y x H Cl xcm=0.125
Solving the Newton’s second law of motion (to find x) Try the general solution Because therefore The amplitude of oscillation
Particle oscillates between +a and - a with frequency if if The general solution to the differential equation is c1 and c2 are arbitrary coefficients
Consider the Euler formula: The equation can be further simplified to
b1 and b2 are real because the amplitude of oscillation is real. If we set the boundary condition x(0) = 0 and v(0) = v0 then
Therefore, if x(0) = 0 and v(0) = v0 The specific solution takes the form: Thus, x(t) is a time-dependence function of sine wave . Note that if the boundary condition x(0) 0 and v(0) = v0 Then b1 and b2 are nonzero (start at different phase because of the mixed cosine and sine functions).
Last lecture : Important messages Intro. spectroscopy: EM : wave vs particle Spectroscopic techniques, range, type of transitions Boltzmann population Vibration of diatomic molecules Permanent vs dynamics dipole moments Harmonic oscillator: Classical physics (Newton’s law) treatment: k if x(0) = 0 and v(0) = v0
Frequency () vs Angular frequency ()
Where is the phase shift. Therefore, the general solution for harmonic oscillation motion is or Or equivalent to Where is the phase shift.
IMPORTANCE Force constants can be calculated from observed vibrational frequency. The strength of the bond is related to force constants.
Calculate the force constant, k, for this molecule. EXAMPLE A strong absorption of infrared radiation is observed for HCl at 2991 cm-1. Calculate the force constant, k, for this molecule. 1 N = 1 m kg s-2
The total energy of the oscillator: Assuming the total energy is conserved (no heat lost). KE and PE are transformed into each other. Therefore, At maximum stretching (x=A) : KE = 0 At maximum velocity (x=0) : PE = 0 The total energy is conserved:
Ex. When the displacement of the reduced mass of a diatomic molecule is A/2, what fraction of the mechanical energy is kinetic energy and what fraction of it is potential energy? For x=A/2, the fraction of its potential energy is the fraction of its kinetic energy is
Ex. At what displacement, as a fraction of A, is the energy half kinetic and half potential? Ex. Find the velocity v at a point where x = A/2 Use KE(max) for Etotal
For the classical treatment, we can assign any values of x (and v) to calculate the potential (and kinetic) energy. This is why the classical harmonic oscillator has a continuous energy spectrum.
EX. Calculate the force constant for HF and HBr if the absorption is observed at 3962 cm-1 for HF and at 2558 cm-1 for HBr. For HF For HBr
Ex. Based on the force constant of HF, HCl and HBr obtained previously, rearrange an order of these molecules according to the bond strength. Observed frequencies (cm-1) Force constant k (N/m) HF 3962 892.4 HCl 2991 520.8 HBr 2558 386.8 kHF > kHCl > kHBr Ordering the bond strength HF > HCl > HBr
EX. For an IR absorption of 1H35Cl at 2991 cm-1, if hydrogen is replaced by deuterium, by what factor this frequency would be shifted? Will the vibrational frequency of DCl be higher or lower? Assuming the force constant is unaffected by this substitution. The vibrational frequency for DCl is lower by.
Ex. the graphs below illustrate the vibrational potential energy of three diatomic gases. Identify which graphs belongs to HF, HCl and HBr using the force constant of HF, HCl and HBr obtained previously.
De = Dissociation energy, Do = bond energy Anharmonicity Morse potential De = Dissociation energy, Do = bond energy An simple analytical expression, called the Morse potential, represents the main features of the real potential energy for a molecule: - close to the minimum potential of depth De, the potential is close to be harmonic. - for large displacement, the potential represents the bond dissociation.
Useful mathematics When x xe; x-xe very small Use the expansion series:
Energy levels for anharmonic potential (quantum treatment) Negative sign: suggesting the lower energy than Eharmonic
The bond energy Do = De – E0
Ex. The vibration potential of HCl can be described by a Morse potential with De=7.41 X 10-19 J and = 8.967 X 1013 s-1 . De=7.41 X 10-19 J = 8.97 X 1013 s-1 h = 5.944 X 10-20 J (h)2/4De= 1.192 X 10-21 J
1) Calculate the force constants of diatomic gases Home Work (1 week) 1) Calculate the force constants of diatomic gases Observed frequencies (cm-1), Useful formula F2 892 Cl2 546 Br2 319 I2 213 N2 2331 O2 1556 H2 4160 HD 3632 D2 2989 2) Based on the force constant obtained in 1), rearrange an order of the molecules below according to the bond strength . a) H2 HD D2 b) F2 Cl2 Br2 I2 N2 O2
Calculate the reduced mass in amu and kg for the following molecules using given atomic mass in the table Atom Mass (amu) H 1.008 O 15.99 F 18.998 Br 79.904 N 14.007 gases (amu) (kg) HBr HF OH N2 O2
Calculate the vibrational state populations of diatomic gases in the first excited state (N1) relative to those in the ground state (N0) at 500K gases (cm-1) N0/N1 at 500K H2 4400 HF 4138 HBr 2649 N2 2358 C-O 2170
Calculate the force constants of diatomic gases from the spectrum Observed frequencies (cm-1) N2 2331 O2 1556
Quantum (Mechanical) Harmonic Oscillator A simple form of Schrödinger equation Operator Eigenvalue Eigenfunction This is an “Eigenvalue Equation” One Dimensional Schrödinger Equation
1D Schrödinger equation of a diatomic molecule For solving the Schrödinger Equation, we need to define the Eigen functions associated with the state. It is commonly known that the exponential function is a good one to work with. Let see a simple example
the Schrödinger equation Ex. Show that the function satisfies the Schrödinger equation for the quantum harmonic oscillator. What conditions does this place on ? What is E? Solution: and the Schrödinger equation Differential of production function
Thus from the product: Consider the solution for the Schrödinger eq. The last two terms must be cancelled in order to satisfy the Schrödinger eigenfunction Therefore, the energy eigenvalue:
To find β, we use the cancellation of the last two terms Therefore, the condition for β that makes the function satisfies the Schrödinger equation From the classical treatment : Substitute β, we get the energy :
They can be better described using Hermite polynomial wave functions : Unlike classical treatment, quantum harmonic oscillation of diatomic molecule exhibits several discrete energy states. They can be better described using Hermite polynomial wave functions : Vibrational energy states Eigenfunctions Normalization constant Herrmite Polynomials
The solution of quantum harmonic oscillation gives the energy: Where the vibrational quantum number (n) = 0, 1, 2, 3 … n vibrational state Evib (harmonic) Ground state 1/2 h 1 1st excited st. 3/2 h 2 2nd 5/2 h 3 3rd 7/2 h It should be noted that the constant Evib= h is good for small n. For larger n, the energy gap becomes very small. It is better to use anharmonic approximation.
Vibrational transitions h 1 2 3 4 n0 ni Transitions 01 Fundamental 02 First overtone 03 Second overtone 04 Third overtone - The fundamental transitions, n=1, are the most commonly occurring. - The overtone transitions n=2, 3, … are much weaker.
To plot the vibrational wave functions, we separate the functions into three terms, An , Hn and An and are not difficult to calculate The most complicate one is Herrmite Polynomials
Hermite polynomials The recurrence relation for Hermite polynomials For n > 2 The first few Hermite polynomials n = 0 n = 1 n = 2 n = 3 n = 4
For n = 0
For n = 1
For n = 2
The first four Hermite polynomial wave functions, (n= 0,1, 2, 3)
Plot of the first few eigenfunctions of the quantum harmonic oscillator (red) together with the potential energy The energy of a system cannot have zero energy. 1st exited vib. state ground vibrational state Zero point energy
The Probability (ψ2(x))
Example for the Hermite polynomial wave functions at n =1 for HCl What is the normalization constant at this state?
Example for the Hermite polynomial wave functions at n =1 for HCl 1) What is the normalization constant at this state?
Example for the Hermite polynomial wave functions at n =1 for HCl 1) What is the normalization constant at this state?
Example for the Hermite polynomial wave functions (n =0) for HCl The corresponding energy: Where n=0, 1, 2, 3, …,
EXAMPLE Answer the following questions. What are the units of A? What role does have in this equation Graph the kinetic and potential energies as a function of time Show that the sum of the kinetic and potential energies is independent of time.
What are the units of A? Because x(t) has the units of length and the sine function is dimensionless, A must have the units of length. The quantity sets the value of x at t = 0,because .
b) Graph the kinetic and potential energies as a function of time Show that the sum of the kinetic and potential energies is independent of time.
Solution For the J=0 → J=2 transition, The preceding calculations show that the J=0 → J=1 transition is allowed and that the J=0 → J=2 transition is forbidden. You can also show that is also zero unless MJ=0 .
Consider the relative motion via the center of mass coordinates. Angular Momentum (L) m Consider the relative motion via the center of mass coordinates.