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CHAPTER 11 –Energy Levels

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CHAPTER 11 –Energy Levels In Chapter 10 we learned about energy distributions and ensembles. In chapter 11 we build on those ideas by using silimple modldels for a gas moll’lecule’s (or at’)tom’s) lddladder of energy lllevels. Salient points from Chapter 10 For independent subsystems (i.e., gas atoms or molecules), Q=qN if distinguishable (Eqn10.28); and Q=qN/N! if distinguishable (Eqn 10.30) Translational Energy Levels Use Particle‐in‐box Model (1‐dimensional) nh22 ε = n 8mL2 ∞ V = 0 V = 22 2 2 V=V=∞ V=∞ −nh/8 mLkT −nTθtr / qetr ==∑∑ e n=1 IIIIII For Ar in 1 cm box 0 L x A potential energy diagram for the one‐dimensional particle in a box is drawn here. if θ/T << 1 ∞ 2 ∞ 2 −ax 1 π −nTθtr / edx= ∫0 qedntr ≈ 2 a ∫0 1/2 ⎛⎞2π mkT qLtr = ⎜⎟2 ⎝⎠h What about a 3D box? For 3D box of sides of length a, b, and c, the energy can be written as So that qtr is a product of three one dimensional translational partition functions 3/2 3/2 ⎛⎞⎛⎞22ππmkT mkT V qqrxyz== q q q⎜⎟⎜⎟223 abc = V = tr ⎝⎠⎝⎠hhΛ For Argon T = 273 K, V = 224 x 10‐2 m3 for 1 mole, p = 1 atm 32 39 qxVxtr ==2.14 10 4.79 10 states/atom Even for the lightest molecules, H2, this approximation is good at room temperature 24 Λ = 0.714 Å for H2@T = 300 K so that qtr = 2.7 x 10 states per molecule What is the average energy of a monatomic ideal gas atom? Use the translational partition function to prove that the average energy of an ideal monatomic gas atom given by 1.5 k T. Use What about ideal gas molecules ? A potential energy surface is sketched for the ground electronic state of a diatomic molecule. Assume separability for different degrees of freedom ε j = ε j,trans + ε j,rot + ε j,vib + ε j,elec 0 n =1 -D o n =0 For their partition functions we find -De distance R o M q = exp - β (ε + ε + ε + ε ) = ∑ gi ()j,trans j,rot j,vib j,elec qtrans qrot qvib qelec j=1 The average energies become < ε > = < ε trans > + < ε rot > + < ε vib > + < ε elec > Can we find qelec? Each molecule (or atom) has its own unique ladder of electronic energy states. Thus, we have no simple model and we must evaluate the electronic partition function sum explicitly. gi −Δεε12//kT −Δ kT qggel =+01 + ge 2 +... degeneracies Fortunately, most electronic states are separated from the ground state by a large energy gap, so that we only need to Hg20 ==1, Hg 02 (considering L, S) evaluate the first term, or the NO g==2, O g 9 00 first few terms in the partition Og20 = 3 3 P0 0.03 eV function summation. OP33→ P 0.02 eV () 1 J quantum # 3 PeV2 0 For qvib we can find a closed for expression in the harmonic oscillator approximation Vibration model ⎛⎞1 ευV =+⎜⎟vhv, =0,1,2,... ⎝⎠2 1 kmm υμ==, 12, where k = force constant 2πμ mm12+ ∞ Start zero from ground vibrational level. where k = Boltzmann constant 11 O2: T = 298 K, qvib = 1.0005 qvib ==−hkTυ/ 11−−xe T = 1000 K, qvib = 1.11 Finding the average vibrational energy of a diatomic The average energy is g iven by 11 q == vib 11−−xe−hkTυ/ −θvib −θvib T T θvib k e h ν e < ε vib > = = ()1 - exp(-θvib /T) ()1 - exp(-θvib /T) O2: ‐1 T = 298 K, <εvib > = 0. 814 cm ‐1 T = 1000 K, <εvib > = 183 cm In the high temperature limit, we find that qvib Æ kT/hnu and <εvib> = kT. For polyatomics 3n-6(5) 3n-6(5) ⎛ 1 ⎞ 3n-6(5) ⎛ h ν i ⎞ ε vib = ε i and q = ⎜ ⎟ ∑ vib ∏ ⎜ ⎟ < ε vib > = ⎜ ⎟ i=1 i=1 1 - exp(- θ /T) ∑ ⎜ ⎟ ⎝ i ⎠ i=1 ⎝ () exp(θ i /T) - 1 ⎠ Populations in vibrational energy levels of diatomics Pm = f m = exp (−m θ vib / T) (1 − exp (−θ vib / T ) ) hν / cm‐1 θ / K vib 19F35Cl 1Σ+ 786 1131 1 + 19F79Br Σ 671 965 0 10 127 35 I Cl 1Σ+ 384 552 1 + 10-1 127I79Br Σ 269 387 10-2 f m 10-3 10-4 FCl ICl FBr IBr 10-5 10 -6 012345678910 m This figure shows the fractional population of vibrational levels for FCl (○), FBr (▲), ICl (□), and IBr (●). Can we find qrot for rotational energy levels ? angular momentum Rigid rotor model: A = quantum # I = moment of inertia for T >> θrot 2 σ = symmetry factor (accounts for ∞ −ε /kT TIkT8π qed=+(21AA) A == overcounting of states) rot ∫0 σθθ σ h2 rot AB− σ =1 H AA− σ = 2 OKq2 , 298 , rot = 72 HO σ = 2 C 2 CH σ =12 H H 4 CH σ =12 H 66 For nonlinear molecules 2 3/2 π IIIabc⎛⎞8π kT qrot = ⎜⎟2 σ ⎝⎠h Populations in rotational energy levels of diatomics ⎛ θ ⎞ (2J + 1) exp ⎜ - J (J + 1) rot ⎟ nJ ⎝ T ⎠ f = = 0.04 FCl J N qrot 0.03 FBr f ICl J 0.02 0.01 IBr 0.00 0 20406080100120140 J The population distribution of rotational energy levels is shown here for four interhalogen molecules. Let’s put the pieces together and build the total partition function qqqqq= tr rot vib el q 1 Consider a linear diatomic with el = 3/2 ⎛⎞281ππmkT⎛⎞2 IkT ⎛ ⎞ qV= ⎜⎟22/⎜⎟ ⎜−hkTυ ⎟ ⎝⎠hhe⎝⎠σ ⎝1− ⎠ Ideal gas (indistinguishable particles) ⎡⎤q N An ⎢⎥ ⎛⎞qeqN ⎛⎞ ⎣⎦N! F=− kTnQAA =− kTn =− NkTn A using Stirling’s ⎜⎟ ⎜⎟ =−NnqNnNNAA + ⎝⎠N! ⎝⎠N approximation =−NnqNnNAA() + Nne A () ⎛⎞eq = NnA ⎜⎟ ⎝⎠N Now we can find thermodynamic relations 1) Use to find the ideal gas equation of state or pV= NkT 2) Use to find the internal energy N 3(2) 3n-6(5) 3 1 ⎛ Nhν i ⎞ U = < E > = < ε total > = NkT + NkT + ⎜ ⎟ - N D0 ∑ ∑ ∑ θ i /T i=1 2 n=1 2 i=1 ⎝ e - 1⎠ 3) For the constant volume heat capacity we find ⎛ ∂ < E > ⎞ 3 3(2) 1 3n-6(5) CV = ⎜ ⎟ = N k + ∑ N k + ∑ ()N k ⎝ ∂ T ⎠V 2 n=1 2 i=1 In Class Problem A. The heat capacities of Ne and Xe are the same at T = 300 K! What is the value of their constant volume heat capacity? Why are their heat capacities the same? B. The heat cappyacity of a diatomic gas is always higher than that of a monatomic gg,as, if the total mass is the same and if the conditions are similar. Why is the heat capacity of a diatomic gas higher and what range of values may it take? C. The heat capacity of molecules is temperature dependent, even for an ideal gas of molecules. Explain why this is so! The constant pressure heat capacity ⎛ ∂ H ⎞ C P = ⎜ ⎟ = C V + R ⎝ ∂ T ⎠P 40 Heat capacity data for monatomic gases and diatomic oxygen are shown here. The data were 30 taken from I. Barin, Thermochemical Data of Pure 3.5R Substances. 20 bbbb bbb b bb b bb b b b b b b 2.5R 10 0 100 300 500 700 900 1100 1300 1500 1700 1900 2100 Temperat ure (K) We can calculate the Entropy ⎛ qN ⎞ < E > < E > ⎛ ∂ lnQ ⎞ < S > = k ln⎜ ⎟ + = k lnQ + = k lnQ + k T ⎜ ⎟ ⎜ ⎟ ⎝ N! ⎠ T T ⎝ ∂ T ⎠V 3 / 2 ⎛ 2 π m k T ⎞ 3 Use q = ⎜ ⎟ V ; < E > = Nk T ⎝ h2 ⎠ 2 3 / 2 to show that ⎛ (2 π m k T ) V e ⎞ 3 < S > = Nk ln⎜ ⎟ + Nk ⎜ 3 ⎟ ⎝ N h ⎠ 2 or 3 / 2 5 ⎛⎛ 2 π m k T ⎞ V ⎞ 3 3 < S > = Nk + Nk ln⎜⎜ ⎟ ⎟ = Nk lnM + Nk lnT + Nk lnV + K′ ⎜ 2 ⎟ 2 ⎝⎝ h ⎠ N ⎠ 2 2 In Class Problem Consider the absolute entropies for the gases shown on the right. Explain the trend in the entropies for the He 126.0 monatomic series of gases. Ne 146. 2 Ar 154.7 Kr 164.0 Xe 170.0 HHg 175. 0 We can calculate the chemical potential of an ideal gas ⎛⎞∂∂F ⎡⎤ ⎛⎞eq ⎛⎞ eq 1 μ ==−⎜⎟⎢⎥NkTAA n ⎜⎟ =−+ kT n ⎜⎟ NkT ⎝⎠∂∂NNTV, ⎣⎦ ⎝⎠ NTV, ⎝⎠ NN ⎛⎞Nq⎛⎞ =+=kTAA n⎜⎟ kT− kT n ⎜⎟ ⎝⎠eq⎝⎠ N If we write ⎛⎞qkT 0 pqkT0 = then we find that μ =−kTA n⎜⎟ with int 0 ⎝⎠p 0 pint = standard state pressure In an alternative form, we find 00 0 μμμ0int=−kTAA np ⇒ = + kT np μ = standard state chemical potential Equipartition theorem −ε xkT/ ∑ε ( xe) ( ) ε = x , ∑e−ε(xkT)/ x a degree of fdfreedom x ∞ −ε xkT/ ε ()xe() dx ∫−∞ ε = ∞ edx−ε()xkT/ ∫−∞ trans 2 2 Suppose ε ( x) = cx εαn n 1 εαrot AA+1 ε = kT e ( ) 2 vib ⎛⎞1 2 kT εαv =+⎜⎟vhv x = : ⎝⎠2 2c fluctuations ε vib = kT (both KE, PE) only holds at very high T Einstein model of solids CNk= 3 N atoms ⇒ v Dulong+ Petit equipartition actual at low T, Cv → 0 (as explained by Einstein) Einstein: solid = 3N oscillators −βhv −βhv −1 ⎛⎞e q =−(1 ehv) →ε = ⎜⎟−βhv ⎝⎠1− e 0 as T → 0 2 ∂ ε ⎛⎞hv e−hv/ kT CNv ==33 Nk⎜⎟ 2 ∂T ⎝⎠kT −hv/ kT ⎛⎞hv⎛⎞1/− hv kT (1− e ) 33Nk⎜⎟= Nk as T → 0 ⎜⎟⎜⎟2 ⎝⎠kT ⎝⎠11−+()hv / kT 3 actually CTv ~ at low T (explained by Debye model).
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    Abstract Laser cooling and slowing of a diatomic molecule John F. Barry 2013 Laser cooling and trapping are central to modern atomic physics. It has been roughly three decades since laser cooling techniques produced ultracold atoms, leading to rapid advances in a vast array of fields and a number of Nobel prizes. Prior to the work presented in this thesis, laser cooling had not yet been extended to molecules because of their complex internal structure. However, this complex- ity makes molecules potentially useful for a wide range of applications. The first direct laser cooling of a molecule and further results we present here provide a new route to ultracold temperatures for molecules. In particular, these methods bridge the gap between ultracold temperatures and the approximately 1 kelvin temperatures attainable with directly cooled molecules (e.g. with cryogenic buffer gas cooling or decelerated supersonic beams). Using the carefully chosen molecule strontium monofluoride (SrF), decays to unwanted vibrational states are suppressed. Driving a transition with rotational quantum number R=1 to an excited state with R0=0 eliminates decays to unwanted ro- tational states. The dark ground-state Zeeman sublevels present in this specific scheme are remixed via a static magnetic field. Using three lasers for this scheme, a given molecule should undergo an average of approximately 100; 000 photon absorption/emission cycles before being lost via un- wanted decays. This number of cycles should be sufficient to load a magneto-optical trap (MOT) of molecules. In this thesis, we demonstrate transverse cooling of an SrF beam, in both Doppler and a Sisyphus-type cooling regimes.
  • Chapter 10 100 Years of Progress in Gas-Phase Atmospheric Chemistry Research

    Chapter 10 100 Years of Progress in Gas-Phase Atmospheric Chemistry Research

    CHAPTER 10 WALLINGTON ET AL. 10.1 Chapter 10 100 Years of Progress in Gas-Phase Atmospheric Chemistry Research T. J. WALLINGTON Research and Advanced Engineering, Ford Motor Company, Dearborn, Michigan J. H. SEINFELD California Institute of Technology, Pasadena, California J. R. BARKER Climate and Space Sciences and Engineering, University of Michigan, Ann Arbor, Michigan ABSTRACT Remarkable progress has occurred over the last 100 years in our understanding of atmospheric chemical composition, stratospheric and tropospheric chemistry, urban air pollution, acid rain, and the formation of airborne particles from gas-phase chemistry. Much of this progress was associated with the developing un- derstanding of the formation and role of ozone and of the oxides of nitrogen, NO and NO2, in the stratosphere and troposphere. The chemistry of the stratosphere, emerging from the pioneering work of Chapman in 1931, was followed by the discovery of catalytic ozone cycles, ozone destruction by chlorofluorocarbons, and the polar ozone holes, work honored by the 1995 Nobel Prize in Chemistry awarded to Crutzen, Rowland, and Molina. Foundations for the modern understanding of tropospheric chemistry were laid in the 1950s and 1960s, stimulated by the eye-stinging smog in Los Angeles. The importance of the hydroxyl (OH) radical and its relationship to the oxides of nitrogen (NO and NO2) emerged. The chemical processes leading to acid rain were elucidated. The atmosphere contains an immense number of gas-phase organic compounds, a result of emissions from plants and animals, natural and anthropogenic combustion processes, emissions from oceans, and from the atmospheric oxidation of organics emitted into the atmosphere.