Molecular Spectroscopy

Similar documents
Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Vibrations. Matti Hotokka Department of Physical Chemistry Åbo Akademi University

To get you thinking...

MANIPAL INSTITUTE OF TECHNOLOGY

Lecture 5: Equilibrium and Oscillations

( ) kt. Solution. From kinetic theory (visualized in Figure 1Q9-1), 1 2 rms = 2. = 1368 m/s

Sodium D-line doublet. Lectures 5-6: Magnetic dipole moments. Orbital magnetic dipole moments. Orbital magnetic dipole moments

d sinθ = mλ Interference and diffraction double slit or diffraction grating d sinθ = mλ d sinθ is the path difference x (small angle approximation)

Lecture 23: Lattice Models of Materials; Modeling Polymer Solutions

Chemistry 20 Lesson 11 Electronegativity, Polarity and Shapes

Solution to HW14 Fall-2002

Interference is when two (or more) sets of waves meet and combine to produce a new pattern.

Dispersion Ref Feynman Vol-I, Ch-31

Dr M. BROUARD. 5. Thermodynamic formulation of Transition State Theory Entropy of activation. Thermochemical kinetics. CHEMICAL REACTION RATES

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

LCAO APPROXIMATIONS OF ORGANIC Pi MO SYSTEMS The allyl system (cation, anion or radical).

Matter Content from State Frameworks and Other State Documents

Phys101 Final Code: 1 Term: 132 Wednesday, May 21, 2014 Page: 1

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

4 electron domains: 3 bonding and 1 non-bonding. 2 electron domains: 2 bonding and 0 non-bonding. 3 electron domains: 2 bonding and 1 non-bonding

General Chemistry II, Unit II: Study Guide (part 1)

Fall 2013 Physics 172 Recitation 3 Momentum and Springs

Chem 115 POGIL Worksheet - Week 12 Molecular Shapes

Kinetics of Particles. Chapter 3

1 The limitations of Hartree Fock approximation

Lecture 7: Damped and Driven Oscillations

Lecture 6: Phase Space and Damped Oscillations

Sections 15.1 to 15.12, 16.1 and 16.2 of the textbook (Robbins-Miller) cover the materials required for this topic.

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium

Lecture 18 Title : Fine Structure : multi-electron atoms

11. DUAL NATURE OF RADIATION AND MATTER

Phy 213: General Physics III 6/14/2007 Chapter 28 Worksheet 1

Kinematic transformation of mechanical behavior Neville Hogan

Physical Nature of the Covalent Bond Appendix H + H > H 2 ( ) ( )

Chapter 23 Electromagnetic Waves Lecture 14

lecture 5: Nucleophilic Substitution Reactions

Name: Period: Date: ATOMIC STRUCTURE NOTES ADVANCED CHEMISTRY

Module 4: General Formulation of Electric Circuit Theory

General Chemistry II, Unit I: Study Guide (part I)

**DO NOT ONLY RELY ON THIS STUDY GUIDE!!!**

15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O

Rigid Body Dynamics (continued)

Work, Energy, and Power

Hubble s Law PHYS 1301

TOPPER SAMPLE PAPER 2 Class XII- Physics

CLASS XI SET A PHYSICS

Principles of Organic Chemistry lecture 5, page 1

Supporting information

Chapter 9: Quantization of Light

Electric Current and Resistance

Q x = cos 1 30 = 53.1 South

Study Group Report: Plate-fin Heat Exchangers: AEA Technology

February 28, 2013 COMMENTS ON DIFFUSION, DIFFUSIVITY AND DERIVATION OF HYPERBOLIC EQUATIONS DESCRIBING THE DIFFUSION PHENOMENA

Unit 14 Thermochemistry Notes

Phys101 Second Major-061 Zero Version Coordinator: AbdelMonem Saturday, December 09, 2006 Page: 1

Plan o o. I(t) Divide problem into sub-problems Modify schematic and coordinate system (if needed) Write general equations

Name: Period: Date: BONDING NOTES ADVANCED CHEMISTRY

Bootstrap Method > # Purpose: understand how bootstrap method works > obs=c(11.96, 5.03, 67.40, 16.07, 31.50, 7.73, 11.10, 22.38) > n=length(obs) >

Yeu-Sheng Paul Shiue, Ph.D 薛宇盛 Professor and Chair Mechanical Engineering Department Christian Brothers University 650 East Parkway South Memphis, TN

SAFE HANDS & IIT-ian's PACE EDT-04 (JEE) Solutions

39th International Physics Olympiad - Hanoi - Vietnam Theoretical Problem No. 1 /Solution. Solution

A Mechanistic Approach to Bond Formation in H 2

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b

Exercise 3 Identification of parameters of the vibrating system with one degree of freedom

ECE 5318/6352 Antenna Engineering. Spring 2006 Dr. Stuart Long. Chapter 6. Part 7 Schelkunoff s Polynomial

Thermodynamics and Equilibrium

AP Chemistry Assessment 2

Session #22: Homework Solutions

More Tutorial at

Phys. 344 Ch 7 Lecture 8 Fri., April. 10 th,

FIELD QUALITY IN ACCELERATOR MAGNETS

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review

ELECTRON CYCLOTRON HEATING OF AN ANISOTROPIC PLASMA. December 4, PLP No. 322

Fundamental Concepts in Structural Plasticity

ECE 2100 Circuit Analysis

Differentiation Applications 1: Related Rates

Measurement of Radial Loss and Lifetime. of Microwave Plasma in the Octupo1e. J. C. Sprott PLP 165. Plasma Studies. University of Wisconsin DEC 1967

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY

Name: Period: Date: BONDING NOTES HONORS CHEMISTRY

A) 0.77 N B) 0.24 N C) 0.63 N D) 0.31 N E) 0.86 N. v = ω k = 80 = 32 m/s. Ans: (32) 2 = 0.77 N

Problem Set 6 & 7: Solutions

On Boussinesq's problem

Study Guide Physics Pre-Comp 2013

ENGI 4430 Parametric Vector Functions Page 2-01

ChE 471: LECTURE 4 Fall 2003

Thermodynamics Partial Outline of Topics

Name Honors Chemistry / /

Biplots in Practice MICHAEL GREENACRE. Professor of Statistics at the Pompeu Fabra University. Chapter 13 Offprint

The Solar Interior - The Standard Model. Topics to be covered: o Solar interior. Radiative Zone. Convective Zone

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command

Model of the Pulsing Atom

Q1. In figure 1, Q = 60 µc, q = 20 µc, a = 3.0 m, and b = 4.0 m. Calculate the total electric force on q due to the other 2 charges.

CHE 105 EXAMINATION III November 11, 2010

( ) + θ θ. ω rotation rate. θ g geographic latitude - - θ geocentric latitude - - Reference Earth Model - WGS84 (Copyright 2002, David T.

HW #2: 2.42, 2.44, 2.48, 2.50, 2.52, 2.58, 2.60, 2.62, 2.66, 2.68, 2.72, 2.82, 2.90, 2.96, 2.98

Introduction: A Generalized approach for computing the trajectories associated with the Newtonian N Body Problem

ELECTROSTATIC FIELDS IN MATERIAL MEDIA

[ ] [ ] [ ] [ ] [ ] [ J] dt x x hard to solve in general solve it numerically. If there is no convection. is in the absence of reaction n

NUMBERS, MATHEMATICS AND EQUATIONS

Transcription:

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Mlecular Spectrscpy Fr mlecules (e.g. diatmic), the ttal energy f mlecule is a cntributin f the fllwing: Translatin kinetic energy, E FOR THE CENTER OF MASS - ( ) - ( Excitatin energy, excit ) 3- ( Vibratinal energy, E vib ) fr the electrns in the E vib 4- ( Rtatinal energy, rt ) Ptential Energy Curve between Bnding Atms trans E fr the electrns in the mlecule 4 E excit ev 0 K. 3 0 ev t 0 ev E arund the center f mass f the mlecule 4 E rt 0 ev. Ntes abut the ptential energy curve The equilibrium bnd length is the distance where the electrn verlap balances the nuclear repulsin. At R =, the energy f the system is the energy f the atms themselves. - The bnding f the atms causes the system s energy t decrease. Once atms are pushed clser than equilibrium bnd length, nuclear repulsin causes ptential energy f the system t increase. Energy nuclear repulsin electrn verlap zer interactin at R = Atmic Distance [R] Equilibrium bnd length Mrse Ptential An excellent apprximatin f an actual ptential energy curve is the Mrse ptential. ax ( ) ( ) µω V x = De e where a = De and D e is the ptential well depth. V(x) harmnic (quadratic) x D 0 D e D anharmnic (Mrse) D 0 dissciatin energy D = ω zer pint energy e 0 Rtatinal_mtin.dc

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 THE ELECTROMAGNETIC SPECTRUM Acrnym Classificatin Frequency Range, f Wavelength THF Tremendusly High Freq. 300 3000 GHz mm 00 µm EHF Extremely High Freq. 30 300 GHz cm mm (Millimetric waves/micrwaves) SHF Super High Freq. 3 30 GHz 0 cm (Centimetric waves/micrwaves) UHF Ultra High Freq. (Radi frequency) 300 3000 MHz m 0 cm (Decimetric wave) VHF Very High Freq. 30 300 MHz 0 m (Metric waves) HF High Freq. 3 30 MHz 00 0 m (Decametric waves) MF Medium Freq. 300 3000 khz km 00 m (Hectmetric waves) LF Lw Freq. 30 300 khz 0 km km (Kilmetric waves) VLF Very Lw Freq. 3 30 khz 00 0 km (Myriametric waves) ULF Ultra Lw Freq. 300 3000 Hz 0 3 0 km SLF Super Lw Freq. 30 300 Hz 0 4 0 3 km ELF Extremely Lw Frequency 3 30 Hz 0 5 0 4 km TLF Tremendusly Lw Freq. 3 Hz 0 6 0 5 km Regin f the electrmagnetic spectrum regin Frequency Range, Wavelength f ( Hz) γ -ray 8 0 3 0 3 0? Energy changes invlve the rearrangement f nuclear cnfiguratin. x -ray 6 8 3 0 3 0? Energy changes invlving the inner electrns f an atm r mlecule. Visible and 4 6 3 0 3 0? T study electrnic spectrscpy. ultra-vilet Infra-red 4 3 0 3 0? T study the vibratins f mlecules and yield infrmatin cncerning the stiffness r rigidity f 0 Micrwaves 3 0 3 0 Radi frequency chemical bnd.? Investigatin the rtatin f mlecules and yields mment f inertia and bnd length, The energy change invlved is that arising frm the reversal f spin f a nucleus r electrn. 6 0 3 0 3 0 cm 0 m Rtatinal_mtin.dc

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Rtatinal_mtin.dc 3

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0. Classical descriptin. The Harmnic Oscillatr A particle f mass m is subject t a restring frce F x, which is prprtinal t its displacement frm the rigin (Hke s Law). dv(x) dx = F x = kx where k is the frce cnstant. If we take the zer f the ptential energy V t be at the rigin x = 0 and integrate, x x V (x) = dv = k xdx = 0 0 kx Nte that this ptential energy functin differs frm that in the particle-in-the-bx prblem in that the walls d nt rise steeply t infinity at sme particular pint in space (x = 0 and x = L), but instead apprach infinity much mre slwly. Frm Newtn s secnd law, d x F = ma = m = kx dt thus, d x = k x = ω x, ω = k dt m m This secnd-rder differential equatin is just like that fr the free particle, s slutins must be f the frm x() t = A sin[ ω t] + B cs[ ω t] where A and B are cnstants f integratin. If we assume that x = 0 at t = 0, then B = 0 and x() t = x 0 sin[ ω t] where x 0 = A is the maximum displacement amplitude. Since this can als be written as x(t) = x 0 sin(ωt) = x 0 sin(πνt), we see that the psitin f the particle scillates in a sinusidal manner with frequency ω k ν= = k = 4 π m ν. π π m The energy f the classical scillatr is E = T + V = mv + kx dx v( t) = = ω x 0 cs[ ωt] dt E = mω x 0 cs [ ωt] + k x 0 sin [ ωt] = kx 0, and is nt quantized. Rtatinal_mtin.dc 4

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Quantum mechanical descriptin. Fllwing ur prescriptin, we begin by writing dwn the classical energy expressin fr the scillatr px E = T + V = mv + kx = + kx m and then cnvert this t the quantum mechanical analg, the Hamiltnian peratr Ĥ, by replacing each f the dynamical variables (p x and x) by their peratr equivalents, px pˆ x = i, x xˆ = x x This yields Ĥ = kx m x + We then use this frm f Ĥ in the time-independent Schrödinger equatin Ĥ ψ = Eψ nt zer. New feature. yielding ψ(x) + kx ψ(x) = Eψ(x). m x Eigenvalues The energies (eigenvalues) f the ne-dimensinal harmnic has the frm: ( ) ( ) E = n + ω = n + hν, n = 0,,, n Here n is the vibratinal quantum number. Cnverting t the spectrscpic units, m -, we have, ( ) n E n E = = n + ν hc Here n is called term value.. ν is the vibratinal frequency f the scillatr in wavenumbers, s its units is m -. Zer-pint energy in m - = ν. The selectin rule fr the harmnic scillatr under ging vibratinal changes is n =±. Vibratinal energy changes will nly give rise t an bservable spectrum if the vibratin can interact with radiatin, i.e., if the vibratin invlves a change in the diple mment f the mlecule. Thus, vibratinal spectra will be bservable nly in heternuclear diatmic mlecules (like HF, HCl, HBr) since hmnuclear mlecules (like H, N and O ) have n diple mment. Example: Shw that the vibratinal absrptin spectrum f a diatmic mlecule in the harmnic scillatr apprximatin cnsists f just ne line whse frequency is given by Eq. ω k ν= =. π π m Slutin: Frm the Eq.: E = n ( n + ) ω = ( n + ) hν, n = 0,,, Accrding t the selectin rule, n =+ fr absrptin, the vibratinal energy change fr absrptin is (A) Rtatinal_mtin.dc 5

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 ( ) ( ) E = E E = n + + hν n + hν = hν = n+ n Thus, the spectrum cnsists f a single line whse frequency is E k ν = = bs h π m Using the last equatin, the bserved infrared frequency can yield the frce cnstant k, which is a direct measure f the stiffness f the bnd. E =ν It means that all the vibratinal lines btained frm harmnic scillatr are f the same frequency. The allwed vibratinal energy levels and transitins between them fr a diatmic mlecule underging simple harmnic mtin are shwn in Fig. 3.8. h π k m Pure vibratinal spectra are bserved nly in liquids. This is because interactins between neighbring mlecules prevent their rtatinal mtin. Example. in the near infra-red spectrum f HC mlecule there is single intense band at 885.9 cm -. Assuming that it is due t the transitin between vibratinal levels, shw that the frce cnstant k is 480 Nm -. (Given :M H =.68x 0-7 kg). Nw, Therefre, ω k ν= = π π µ where k is the frce cnstant and µ is the reduced mass f the mlecule. Rtatinal_mtin.dc 6

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Example. The frce cnstant f the bnd in CO mlecule is 87 Nm -. Calculate the frequency f vibratin f the mlecule and the spacing between its vibratinal energy level in ev. Given that reduced mass f CO =.4 x 0-6 kg, h = 6.6 x 0-34 Js and l ev.60 x 0-9 J. Slutin: The frequency f vibratin f the mlecule is Unlike the crrespnding classical result, we find that the quantum mechanical energy is quantized, in units f ω, where ω is the classical frequency ω = k/m. We als find that the lwest state, with n = 0, des nt have zer energy but instead has E = ω /, the s-called zer pint energy. We can summarize these results in the frm f an energy level diagram 3 ω ω E 3 = 7 ω/ E = 5 ω/ ω E = 3 ω/ ω E v = 0 0 = ω/ Rtatinal_mtin.dc 7

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Bltzmann Distributin As the numbers f mlecules (r atms) get larger, that is 0 3, the mst prbable distributin fr a given amunt f macrscpic energy amng micrscpic energy states is given by the Bltzmann distributin. ni N = e j Ei kt e E j kt n i number f mlecules in the i th energy state. N ttal number f mlecules. E i energy f the i th energy state. k Bltzmann s cnstant:.38 0-3 J/K Nte: R = kn A Let us examine hw the energy f 0 6 HCl -mlecules is distributed amng the different vibratinal states at T = 98 K, where ν=99cm. n 34 0 ( ) ( )( )( )( ) 0 = ( 5.67 0 J)( n + ) E = hcν n + = 6.66 0 J s.997 0 cm / s 99 cm n + ( )( ) 3 kt =.38 0 J / K 98 K = 4.8 0 J n E n En kt En e kt 0.80 0-0 J 6.8.09 0-3 8.45 0-0 J 0.46.3 0-9.404 0-9 J 34.0.55 0-5 3.966 0-9 J 47.74.85 0-4.58 0-9 J 6.38.0 0-7 6.8 3 n0 n0 e.09 0 6 6 = = = = n 6 6.8 0.46 34.0 47.74 6.38 3 0 = 0 = 0 N 0 e + e + e + e + e.09 0 0.46 9 n n e.3 0 6 6 6 = = = =. 0 n 6 6.8 0.46 34.0 47.74 6.38 3 = 0. 0 = N 0 e + e + e + e + e.09 0 The vibratinal energy is large cmpared t the thermal energy; therefre, apprximately 999,999 mlecules ut f a millin are in the grund vibratinal state at 98 K and mlecule is in the first excited state. Let us cnsider what happens t the energy distributin at a higher temperature such as 000 K? ( )( ) 3 0 kt =.38 0 J / K 000K =.76 0 J Rtatinal_mtin.dc 8

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 n E n kt n = 0 n 6 ν e En kt e En kt 0.0 868998 3.04 3843 5.08 49 3 7. 950 4 9.5 60 5.7 34 6 3. 4 Cnclusin: Mst diatmic mlecules are in the grund vibratinal state ( n = 0) at rm temperature. Rle f Degeneracy in the Bltzmann Distributin The Bltzmann distributin includes all degenerate states as equally prbable as well. Therefre the mre precise frmulatin f the Bltzmann distributin is ni N = Ei kt ge i E j kt ge j j, g i degeneracy f the i th energy state. Rtatinal_mtin.dc 9

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Rtatinal mtin f rigid linear mlecules Classical descriptin. A rigid rtr is a dumbbell-shaped bject cnsisting f tw masses, m and m, separated by a fixed distance r. m m r In general, the rtatinal energy f any three-dimensinal bject can be written as E rt = Ix ωx + Iy ωy + Iz ωz Here, ω x, etc., are the angular velcities f rtatin and I x, etc., are the mments f inertia, each referred t the principal axes f rtatin. Fr ur dumbbell, if we assume that m and m are pint masses, then ωx = ωy = ω, ωz = 0 I = I = I, I = 0 x y z x = + I mr mr If the rigin f the crdinate system is lcated at the center f mass (CM), use the relatins: mr = mr, r= r+ r, we can write: m m r = r, r = r m+ m m+ m mm mm I = r = µ r, µ = (reduced mass) m+ m m+ m Frm this we see that the rtatinal mtin f ur dumbbell can als be described as the rtatinal mtin f a mass µ, which is lcated at a distance R 0 frm the center f a spherical plar crdinate system. S Ert = I x ωx + I yωy = Iω = ( mr + mr) ω m m = m + m R0ω = µ R0 ω = I ω m+ m m+ m particle n a ring, n a sphere. Since L = Iω, where L is the angular mmentum f the bject, we can als write L L E rt = =. µ r I Rtatinal_mtin.dc 0

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Quantum mechanical descriptin. T describe the system by QM, we need the peratr equivalent f L. This is, in plar crdinates csθ Lˆ = + + θ sin θ θ sin θ φ S, Ĥ = Lˆ / µ r is the QM analg f the classical energy. ˆ LY = J J + Y, J = 0,, ( ) J, m J, m EJ = J( J + ) I With the degeneracy g J = J + When we cnsider bth rtatins and vibratins simultaneusly, we take advantage f the fact that these transitins ccur n different timescales. Typically, a mlecular vibratin takes n the rder 4 9 0 f 0 s. A mlecular rtatin is nrmally much slwer, taking n the rder f 0 s r 0 s. Hence, as a mlecule rtates ne revlutin, it vibrates many, many times. Since the vibratinal energies are large cmpared with the rtatinal energies, the apprpriate energy level diagram is: E J = 3 J = J = J = 0 v = Figure 4 ~ J = 3 In wave-numbers, the energy, ν ( J ), and the rtatinal cnstant, B ~, are related by the equatin: ν ( J) = B J( J + ) cm - h and B = cm - 8 π ci c Prf: EJ = J( J + ), but E = hν = h = hc ν I λ EJ h EJ = hc ν ν = = J( J + ) = J( J + ) = BJ ( J + ) hc hc I 8π ci B fr diatmic mlecules are in the rder cm -. Rtatinal spectra are always btained in absrptin. Hence fr a transitin frm sme initial state f quantum number J t the next higher state f quantum number J +, the wave number f the absrbed phtn is: ν = J J B ( J + )( J + ) B + J( J + ) = B ( J + ) J = J = J = 0 v = 0 Rtatinal_mtin.dc

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 The energy level diagram is Ntice that: - the levels are nt equally spaced, and that the lwest level has zer energy (unlike HO). By calculating B, we can calculatei, and then the distance R between the nuclei can be calculated. This gives the length f the chemical bnd between the atms.. - Rtatinal transitins ccur nly in thse mlecules which pssess a permanent electric diple mment. Fr this reasn nnplar diatmic mlecules such as H and symmetric plyatmic mlecules such as CO and CH 4 d nt exhibit rtatinal spectra. 3- The selectin rule fr rtatinal transitins is J =±. Example. The lines in the pure rtatinal spectrum f HC are spaced as 0.8 x 0 m. Calculate the mment f inertia and the internuclear distance. Mass f prtn =.67 x l0-7 kg ; mass f chlrine = 58.5 x 0-7 kg. Sl. The reduced mass is defined as Rtatinal_mtin.dc

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Example: In the CO mlecule the wavenumber difference between the successive absrptin lines in the pure rtatinal spectrum is384 m. Calculate the mment f inertia f the mlecule and the equilibrium bnd length f the mlecule. Masses f the C and O 6 atms are respectively.99 x 0-6 kg and.66 x 0-6 kg. Answer: Here ν = = 384 m π Ic Intensity f Rtatinal Transitins Degeneracy is imprtant when partitining the rtatinal energy f sample f mlecules. Remember that each rtatinal state J can have an M value that ranges frm J,, -J. Thus each rtatinal state J has J + M values; thus the degeneracy fr rtatinal energy levels is J +. The intensity f a transitin in the absrptin (micrwave) r Rtatinal Raman spectrum is prprtinal t the number f mlecules in the initial state (J ); i.e. Int. NJ Bltzmann Distributin: E J kt N g e N (J + ) e J J J hcbj ( J + ) kt EXAMPLE : Calculate E fr radiatin f wave number, ν =.00 cm. T what type f mlecular prcess will this radiatin crrespnd? Slutin: Recall that wave number is given by reciprcal wavelength r that Accrding table (), this value f energy crrespnds t rtatinal transitin. EXAMPLE : Shw that the rtatinal absrptin predicted by the rigid rtatr mdel cnsists f a series f equally spaced lines in the micrwave regin. Slutin: The energy levels are given by Eq. Rtatinal_mtin.dc 3

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 EJ = J( J + ). I Accrding t the selectin rule fr rtatinal transitins, J = fr absrptin. The energy change fr a rtatinal transitin is: E = EJ+ EJ = ( J + )( J + ) J( J + ) = ( J + ) I I I and E h ν = = ( J + ), J = 0,,, h 4π I Accrding table (), these lines ccur in the micrwave regin. h = 6.63x0-34 J s c = 3.00x0 8 m/s c = 3.00x0 0 cm/s NA = 6.0x0 3 ml - k =.38x0-3 J/K amu =.66x0-7 kg Example 3: The HCl bnd length is 0.7 nm. Calculate the spacing between lines in the rtatinal absrptin spectrum f HCl, in cm -. Answer: 7 m m ( amu )( 35 amu ).66x0 kg H Cl µ= = = 0.97 = 0.97 =.6 0 m + m amu + 35 amu amu H Cl ( )( ) I =µ r =.6x0 kg 0.7x0 m =.60x0 kg m 7 9 47 34 h 6.63x0 J s 8π 8 3.4.60 0 47 3.00 0 0 / 7 amu amu x kg B = = = 0.78 cm 0.8 cm Ic ( ) ( x kg m )( x cm s) As discussed abve, micrwave absrptin lines ccur at B, 4B, 6B,... Therefre, the spacing is B Spacing = B = 0.8 =.6 cm Rtatinal_mtin.dc 4

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 I 3 4 Example 4: calculate the rati f intensities I B = 0.8 cm Answer: in the abve example at 5 C, where Rtatinal_mtin.dc 5

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 The Harmnic Oscillatr d x = k x = ω x, ω = k dt m m ω k ν= = k = 4 π m ν. π π m Ĥ = + kx m x E n ( n ) ( n ) h n Summary = + ω = + ν, = 0,,, n =± n+ n ( ) ( ) E = E E = n + + hν n + hν = hν = E k ν = = bs h π m Bltzmann Distributin: ni = N Ei kt ge i E j kt ge j j E =ν, g i degeneracy f the i th 3 energy state, kt = (.38 0 J / K)( T K) Rtatinal mtin f rigid linear mlecules L L mm Ert = µ R0 ω = I ω, Ert = =, µ = (reduced mass) µ r I m + m h π EJ = J( J + ) I ν ( J) = B J( J + ) cm - h and B = cm - 8 π ci ν = J J B ( J + )( J + ) B + J( J + ) = B ( J + ) k m Rtatinal_mtin.dc 6

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Rtatinal Transitins Accmpany Vibratinal Transitins Within the rigid rtatr-harmnic scillatr apprximatin, the rtatinal and vibratinal energy f a diatmic mlecule is: E = ( n + ) hν + hcbj ( J + ) vib, rt n = 0,,, J = 0,,, Figure 0-. An energy diagram shwing the rtatinal levels assciated with each vibratinal state fr a diatmic mlecule. The lwer rtatinal levels are t clsely spaced t be shwn. where the rtatinal cnstant B =. Typical values f the spacing between rtatinal 4π Ic levels are arund 0-3 J mlecule - (cf. Table 0-) and f thse f vibratinal levels are arund 0 - J mlecule - (cf. Table 0-). This result is shwn schematically in Figure 0-. When a mlecule absrbs infrared radiatin, the vibratinal transitin is accmpanied by a rtatinal transitin. The selectin rules fr absrptin f infrared radiatin are: n =+ J =± The frequency assciated with the absrptin is: [ '( ' ) ( ) ] νbs = ν + cb J J + J J + Where J ' can be either ( J + ) r ( J ). If j ' = J +, then νbs ( J =+ ) = ν + cb ( J + ), J = 0,,, R - branch (A) If J ' = J ν ( J = ) = ν cbj, J =,,3, P - branch (B) bs In bth Eqs. (A) and (B), J is the initial rtatinal quantum number. Typically, ν 3 0 Hz 3 B 0 Hz, and s the spectrum predicted by Eqs. (A) and (B) typically cntains lines at 0 Hz ± integral multiples f 0 Hz. Ntice that there is n line at ν. The rtatinalvibratinal spectrum f HBr(g) is shwn in Figure 0-3. The gap centered arund 560 cm - and Rtatinal_mtin.dc 7

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 crrespnds t the missing line at ν. On each side f the gap is a series f lines whse spacing is abut 0 cm -. The series tward the high-frequency side is called the R branch and is due t rtatinal transitins with J =+. The series tward the lw frequencies is called the P branch and is due t rtatinal transitins with J =., r cb = 3.8 cm - (cf. Figure 0-3 fr HBr). Figure 0-3: The vibratin-rtatin spectrum f the 0 vibratinal transitin f HBr. The R- and P- branches are indicated in the figure. The vibratin-rtatin spectrum will cnsist f lines at: ν ± cjb, j=,,3, There will be n line at ν and the separatin f the lines in the P and R branches will be.5 x 0 s - r B = 3.8 cm - (See the abve figure). cb= Rtatinal_mtin.dc 8

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 If we cmpare the results f Example 0-6 t experimental data, r lk clsely at Figure 0-3, we see that there are several features in the vibratin-rtatin spectrum that we are nt able t explain. Fr examples: - The intensities, r heights, f the lines in the P and R branches shw a definite pattern and the spacing f the lines is nt equal. - The lines in the R branch are mre clsely spaced with increasing frequency and that the lines f the P branch becme further apart with decreasing frequency, i.e. the spacing between the lines is nt equal. Cmments: Using Bltzmann distributin ne can explain the fllwing: - mst diatmic mlecules are in the n = 0 vibratinal state at rm temperature. - the intensities f the lines in the P and R branches in a vibratin-rtatin spectrum. The Intensities f the Lines in the P and R Branches in a Vibratin-Rtatin Spectrum Are Explained By a Rtatinal Bltzmann Distributin If we assume that the intensities f the rtatinal lines in a vibratin- rtatin spectrum are prprtinal t the fractin f mlecules in the rtatinal level frm which the transitin ccurs, then we can use the Bltzmann distributin f rtatinal energies t explain the bserved intensities. We cannt use Eq. 0- directly because the rtatinal energy levels are (J + )-fld Rtatinal_mtin.dc 9

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Nn-Rigid Rtatr On the wavenumber scale, the frequency difference between tw successive lines in the pure rtatinal spectrum f a diatmic mlecule is given by: ν = π Ic The rtatinal spectrum can be recrded. The absrptin lines are equi-spaced. The separatin between the adjacent lines is identified as B. ν = = B B = π Ic 4π Ic By measuring ν, the rtatinal cnstant B can be calculated. Frm this, the mment f inertia f the mlecule I can be calculated. KnwingI, ne can calculate the reduced mass µ f the mlecule and R the bnd length. 35 H Cl. The The fllwing table lists sme f the bserved lines in the rtatinal spectrum differences listed in the third clumn clearly shw that the lines are nt exactly equally spaced as the rigid rtatr apprximatin predicts. Experimental investigatins have shwn that the successive lines in the far infra-red spectrum are nt evenly spaced, but the frequency separatin decrease slightly with increasing the value f J (Larger speed f rtatin). It shws that the bnd length R increases with J. Therefre, ur assumptin that the mlecule is a rigid rtatr is false. In fact, all bnds are elastic t sme extent. The mre quickly a diatmic mlecule rtates, the greater is the centrifugal frce tending t mve the atms apart. Nw we discuss the cnsequences f the change in bnd length with J. I- When a bnd is elastic, it will stretch and cmpress peridically with a certain functinal frequency dependent upn the masses f the atms and the elasticity (r frce cnstant k) f the bnd. This means that the mlecule may have vibratinal energy. If the vibratinal mtin is simple harmnic, the frce cnstant k is given by: k = 4π ω c µ () Rtatinal_mtin.dc 0

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Here,ω is the vibratin frequency (in m - ). µ is the reduced mass f the mlecule. The variatin f B with J is determined by the frce cnstant, i.e., the weaker the bnd, mre readily will it distrt under centrifugal frces. II- The secnd cnsequence f elasticity is that the quantities r and B vary during a vibratin. Energy Levels Cnsider a single particle f mass µ rtating abut a fixed pint with an angular velcity, ω. Let the particle be at a distance R frm the fixed pint when there is n rtatin. Let this length increase t R, when the particle rtates. Centrifugal frce during rtatin = µ R ω. Restring frce due t bnd stretching (Hk s law) = k ( R R ) The abve tw frces balance each ther at any instant f rtatin. kr µ Rω = k ( R R ) R = ( k µω ) This gives the distrted bnd length. Ttal energy f the rtating system = K. E. + P. E 4 4 µ R ω L L E = Iω + k ( R R ) = Iω + = +. k I I kr Where we used: L = Iω=µ R ω. The quantum restrictin that the angular mmentum I ω be quantized accrding t J( J + ) will cnvert this classical result t a quantum-mechanical result: L = J( J + ). The crrect allwed energies are, 4 E = J J( J ) J ( J ) I + + I kr + Using: Imprtant nte: Then ( k ) ( k ) 4 kr µω / µω µω L L R = = = + = + 4 8 µω R R R k k R µ kr µ k R E J J J J I R 4 J = ( + ) ( + ) +, = µ I kir E J E J = = BJ( J + ) DJ ( J + ) hc In the last Eq., the first term is f majr imprtance. The secnd is a minr term. The quantity D is called the centrifugal distrtin cnstant, D << B. H.W. Shw that 4 B k D =, ν =. ν π c µ H.W. Check the fllwing: Rtatinal_mtin.dc

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 Fig. 3.5 shws the lwering f rtatinal evels when passing frm the rigid t the nn-rigid diatmic mlecule. The difference between the energy levels increases with increasing value f J. The rtatinal absrptin spectrum is prduced due t mlecular transitin frm the state J t the state (J+ ). The selectin rule is J =+. Thus, we see that the spectrum f a nn-rigid diatmic mlecule is similar t that f the rigid mlecule except that each line is displaced slightly t lw frequency (Fig. 3.6). We nte frm Eq. (8) that the lines are n lnger exactly equidistant but their separatin decreases slightly with increasing J. The effect, hwever, is small wing t the smallness f D as cmpared t B. Rtatinal_mtin.dc

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 MOLECULAR SPECTRA Determinatin f Frce Cnstant frm Rtatinal Spectrum By making bservatins n a number f lines in the rtatin spectrum f a mlecule, and by curve fitting methd, the values f B and D are fund ut. Once D is knwn, the J values f lines in the bserved spectrum can be determined. Further, knwing B and D, the angular frequency f vibratin f the rtating mlecule can be calculated using the relatin Fr hydrgen fluride, the frce cnstant is 960 Nm - which indicates that H-F is a relatively strng bnd. Example: HC mlecule has a rtatinal cnstant B value f 059.3 m - and a centrifugal cnstant D f 5.3 x 0 - m -. Estimate the vibratinal frequency and frce cnstant f the mlecule. Rtatinal_mtin.dc 3

Prf. Dr. I. Nasser Atmic and mlecular physics -55 (T-) April 0, 0 EJ h EJ = J( J + ) ν = = BJ( J + ), B = I hc 8π ci k En = ( n + ) hν E = E n+ En = hν = νbs = m π The rtatinal and vibratinal energy f a diatmic mlecule is given by: k m vib, rt E = ( n + ) hν + hcbj( J + ) n J = 0,,, = 0,,, Rtatinal_mtin.dc 4