Molecular Spectra By – P.V.Koshti

Here we need to determine (1) the eigenvalues of the Hamiltonian in order to know the frequencies of the possible transitions, (2) the selection rules to determine what transitions actually occur, (3) the transition dipole moments to determine the strengths of the transitions, (4) degeneracies, and (5) relative populations in order to determine the intensity of transitions. The simplest case beyond an atom is a diatomic molecule.

Spectroscopy is that part of science which uses emission and/or absorption of radiation to deduce atomic/molecular properties. Molecular spectroscopy, which is the study of the interaction of light with atoms and molecules, is one of the richest probes into molecular structure. Radiation from various regions of the electromagnetic spectrum yields different information about molecules. For example, microwave radiation is used to investigate the rotation of molecules and yields moments of inertia and therefore bond lengths. Infrared radiation is used to study the vibrations of molecules, which yields information about the stiffness or rigidity of chemical bonds. Visible and ultraviolet radiation is used to investigate electronic states of molecules, and yields information about ground and excited state vibrations, electronic energy levels, and bond strengths.

The features of the electromagnetic spectrum that are of interest to us are summarized below. Region Frequency/Hz Wavelength/m Wave number/cm- Energy/ J *molecule-1 Molecular Process Microwave 109-1011 3x10-1-3x10-3 0.033-3.3 6.6x10-25- 6.6x10-23 Rotation of polyatomic molecules Far Infrared 1011-1013 3x10-3-3x10-5 3.3-330 6.6x10-23- 6.6x10-21 Rotation of small Infrared 1013-1014 3x10-5-3x10-6 330-4000 6.6x10-21- 8.0x10-20 Vibration of flexible bonds Visible a 1015-1016 9 x 10-7-3x10-8 11,000-3.3x105 2.2x10-19- 6.6x10-18 Electronic Structure nd Ultraviolet

The electronic states of molecules are in analogy with electronic states of atoms. In addition molecules can vibrate and rotate. In the Born-Oppenheimer approximation these motions are initially treated separately Etot= Eel + Evib + Erot To more easily describe vibrational and rotational energies, a simpler model for the molecule will be used.

The rotation of a diatomic molecule can be described by the rigid rotor model. To imagine this model think of a spinning dumbbell. The dumbbell has two masses set at a fixed distance from one another and spins around its center of mass (COM). This model can be further simplified using the concept of reduced mass which allows the problem to be treated as a single body system.

Rigid Rotor Model A diatomic molecule consists of two masses bound together. The distance between the masses, or the bond length, (l) can be considered fixed because the level of vibration in the bond is small compared to the bond length. As the molecule rotates it does so around its COM with a frequency of rotation of νrot given in radians per second.

Note that there is no potential energy involved in free rotation. Here we consider only rotation restricted to a 2-D plane where the two masses (i.e., the nuclei) rotate about their center of mass. The rotational kinetic energy for diatomic molecule in terms of angular momentum Note that there is no potential energy involved in free rotation.

Reduced Mass The system can be simplified using the concept of reduced mass which allows it to be treated as one rotating body. 1.) The system can be entirely described by the fixed distance between the two masses instead of their individual radii of rotation. Relationships between the radii of rotation and bond length are derived from the COM given by:  M1R1 = M2R2 where l is the sum of the two radii of rotation: l=R1+R2. 2.) Through simple algebra both radii can be found in terms of their masses and bond length: R1=M2 l /(M1+M2) and R2=M1 l /(M1+M2). 3.) The kinetic energy of the system, T, is sum of the kinetic energy for each mass:  T= [M1v21+M2v22 /2] where v1=2πR1νrot and v2=2πR2νrot.

4.) Using the angular velocity, ω=2πνrot, the kinetic energy can now be written as:  T=ω [M1R1 2+M2R22 ]/2. 5.) With the moment of inertia, I=M1R21+M2R22, the kinetic energy can be further simplified: T=Iω2 /2. 6.) The moment of inertia can be rewritten by plugging in for R1 and R2:   I =[ M1M2 /(M1+M2)] l2, where (M1M2/M1+M2) is the reduced mass, μ. 7.) The moment of inertia and the system are now solely defined by a single mass, μ, and a single length, l: I=μl2.

Angular Momentum Another important concept when dealing with rotating systems is the the angular momentum defined by: L=Iω  The angular momentum can now be described in terms of the moment of inertia and kinetic energy: L2=2IT.

Setting up the Schrödinger Equation The wave functions for the rigid rotor model are found from solving the time-independent Schrödinger Equation:  Hψ =Eψ Where the Hamiltonian Operator is:  H=−(ℏ/2μ)∇2+V(r) where ∇2 is the Laplacian Operator and can be expressed in either Cartesian coordinates or polar coordinate. At this point it is important to incorporate two assumptions: 1) The distance between the two masses is fixed. This causes the terms in the Laplacian containing ∂/∂r to be zero. 2) The orientation of the masses is completely described by θ and ϕ and in the absence of electric or magnetic fields the energy is independent of orientation. This causes the potential energy portion of the Hamiltonian to be zero.  

The Hamiltonian solution to the rigid rotor is H=T since, H=T+V Where T is kinetic energy and V is potential energy. Potential energy, V, is 0 because the distance between particles does not change. In reality, however, V≠0 because even though the distance between particles does not change, the particles still vibrate. We will assume V = 0 to simplify our discussion. Since H=T, we can also say that:

This shows that the energy is quantized by expressing in terms of β: H = T with the Angular Momentum Operator being defined:  L=2IT The Schrödinger Equation can be solved using separation of variables. The result is Φ(ϕ)=√ (1/2π) eimϕ And β=J(J+1) This shows that the energy is quantized by expressing in terms of β: E=ℏ2β/2I On using the rotational constant, B=ℏ2/2I, the energy is further simplified: E=BJ(J+1)

Energy of Rotational Transitions When a molecule is irradiated with photons of light it may absorb the radiation and undergo an energy transition. The energy of the transition must be equivalent to the energy of the photon of light absorbed given by: E=hν. For a diatomic molecule the energy difference between rotational levels (J to J+1) is given by:  EJ+1−EJ=B(J+1)(J+2)−BJ(J=1)=2B(J+1) with J=0, 1, 2,... 

Because the difference of energy between rotational levels is in the microwave region (1-10 cm-1) rotational spectroscopy is commonly called microwave spectroscopy. In spectroscopy it is customary to represent energy in wave numbers (cm-1), in this notation B is written as  B=h/8πcI.

Figure predicts the rotational spectra of a diatomic molecule to have several peaks spaced by 2{B}. This contrasts vibrational spectra which have only one fundamental peak for each vibrational mode. From the rotational spectrum of a diatomic molecule the bond length can be determined. Because {1\{B} is a function of I and therefore a function of l (bond length), l is readily solved for:   l=√h/8π2cBμ Selection Rules only permit transitions between consecutive rotational levels: ΔJ=J±1, and require the molecule to contain a permanent dipole moment. Due to the dipole requirement, molecules such as HF and HCl have pure rotational spectra and molecules such as H2 and N2 are rotationally inactive.

Calculate the value of I and r of CO. B = 1.92118 cm -1 . Solution: I = h/(8π2 Bc) = 6.626 x 10-34 /(8 x 3.14152 x 1.92118 x 3 x 1010 ) = 1.45579 x 10-46 kg m 2 Since the value of B is in cm-1 , the velocity of light c is taken in cm/s. I = μr2 The atomic mass of C ≡ 12.0000 amu, O ≡ 15.9994 amu. 1 amu = 1.6604 x 10-27kg. The reduced mass of CO can be calculated to be 1.13836 x 10 -27 kg. Therefore r2= I/μ = 1.45579 x 10-46 /1.13826 x 10-27m2 Or r = 1.131 Ǻ

Solution: I = h/(8π2 Bc) = 6.626 x 10-34 /(8 x 3.14152 x 1.92118 x 3 x 1010 ) = 1.45579 x 10-46 kg m 2 Since the value of B is in cm-1 , the velocity of light c is taken in cm/s. I = μr2 The atomic mass of C ≡ 12.0000 amu, O ≡ 15.9994 amu. 1 amu = 1.6604 x 10-27kg. The reduced mass of CO can be calculated to be 1.13836 x 10 -27 kg. Therefore r2= I/μ = 1.45579 x 10-46 /1.13826 x 10-27m2 Or r = 1.131 Ǻ

Vibrations and Rotations of a diatomic You have noticed in your earlier studies that simple pendulums or stretched strings exhibit simple harmonic motion about their equilibrium positions. Molecules also exhibit oscillatory motions. A diatomic oscillates about its equilibrium geometry. The quantized vibration energies Eυ of a harmonic oscillator are E = υ = (u +½ ) h ν v = 0,1,2,............ The vibrational frequency ν is related to the force constant k through