Lecture #11: Pictures of Spectra and Notation

Lecture #11: Pictures of Spectra and Notation Last Time Selection, propensity and intensity rules   αSb(θ,φ) M(R)if absolute rotational linestrengths S JΩi JifΩf  rotation permanent dipole P ∝ µ2 gi or S Sum rule vibration dM dQ ⎛ dM ⎞ P ∝⎜ ⎟ vg i ⎝ dQ ⎠ electronic Mif 〈vi | vf〉 P ∝ M 2 q vi vf g i typical form for universal angular factor → must sum to 1 or to total degeneracy 2 TODAY: Patterns in spectra Typical constants How to assign spectra - problems and tricks Notation PURE ROTATION µ-wave sources limit range of J sampled (more difficult to assign than expected) superposition of lines from vastly different energy regions, isotopomers population effects. How to tell up from down. ← Think about this! VIBRATION-ROTATION J range limited by T, not by radiation source. P and R branch structure - open, no heads, zero gap (easy to assign) PQR notation hot bands and isotopomers ELECTRONIC Band heads due to large |∆B| blue vs. red degraded → sign of ∆B (hard to assign because of overlapping lines) JHead vibrational Sequences vs. Progressions Qualitative Franck-Condon Principle Universal notation for all molecular spectroscopy. Upper level denoted by ′ Lower level denoted by ″ always stated first always stated second ∆Z ≡ Z′ – Z″ (for any quantity, Z) J+1 ∆J = J′ – J″ O –2 P –1 i.e. R(J) means J Q 0 R 1 S 2 Denote whether transition is absorption or emission by direction of arrow. Always state upper level first. PURE ROTATION SPECTRA – µ-wave (e.g. heteronuclear diatomic) requires permanent dipole moment c = 3 × 1010 cm/s] ν 10-300 GHz [ ν = ν/c λ 3cm – 1mm “cm-wave” and “mm-wave” regions 1 + for ∑ ∆J = ±1 only because Ω = 0. recall S JΩΩ i Jf centrifugal distortion Fv(J) = Bv J(J + 1) – DvJ2(J + 1)2 J←J–1 4B3 D ≈ ω2 ~ 10–6B J of higher level Fv(J) – Fv(J – 1) = 2JBv – 4J3Dv Typical B-value B: Cs2 B ( cm −1 ) = ≈ 1 cm–1 = 30 GHz 0.01 cm–1 I2 0.04 cm–1 H2 61 cm–1 1 1 −15 1.685763 ×10 R 2 (cm) µ (amu ) 12 µ= m1m 2 m1 + m 2 C = 12.000… Microwave Spectrum: “Klystron”, “Backward Wave Oscillator” (BWO) typical tuning range < 1/2 octave ± ~25% 8-12 GHz 12-18 18-24 24-40 not DC to 100 GHz. See only a small portion of pure rotation spectrum for any given Exptl. setup. 2 1 Level Diagram ω + 6B1 1 1 ω + 2B1 Spectrum typical Klystron tuning range 2B0 ω + 0B1 J = 0, v = 1 2αe 0 4B0 4αe 1←0 6B0 6αe 2←1 v=1 3←2 ν isotope heavy isotope larger µ smaller B 2 0 6B0 Bv = Be – αe(v + 1/2) 1 0 2B0 distinguish hot band from isotope by size of expected isotope fractional shift. αe ≈ 0.01 Be J = 0, v = 0 0B0 “hot bands” interspersed among cold band, even though ω  B Almost regular pattern of lines separated by 2B (–4J3D makes lines draw closer together at high J) intensity ΩΩ − S J′J′ ′ ∝ N vgJ < (N J ′′M − N J ′M ) ∝ N v=0 e 2J′′ +1 specific M value ωe v kT (2J′′ +1) νROT kT  200 cm–1 vibration at RT population Derive below NON-LECTURE J+1←J e −F( J ′′ )/kT −e −F( J ′ )/kT ∆M = 0 (Z-polarized) (2J″ + 1) from M-degeneracy ⎡ F( J′′) F( J′) ⎤ F( J′) − F( J′′) νROT = ⎢1− −1+ = = ⎣ kT kT ⎥⎦ kT kT How to assign? Can’t see entire pattern. ∆ν = 2BJ Can usually guess Re to ~10% so Be to ~20% given a line at 15GHz, probably J = 2 ← 1 next line would be at 3/2 15 = 22.5 GHz if it was 1 ← 0, next line would be at 2/1 15 = 30 GHz if it was 3 ← 2, next line would be at 4/3 15 = 20 GHz So assignment is based on a guess followed by at least one confirming measurement. Non-1∑+ states and polyatomic molecules: other kinds of transitions possible (always Mz,ii ∆Ω = 0) for linear polyatomics and symmetric tops For Ω ≠ 0, etc. (parity doubling) J=Ω+1 – + can have ∆J = 0 “Q branches” Asymmetric tops → very complicated J=Ω No electric dipole? e.g. O2 + – X 3 ∑−g ∆∑ = ∆Ω ≠ 0 fine structure transitions due to magnetic dipole. VIBRATION-ROTATION spectra — IR requires dMz,ii/dQ ≠ 0 for 1∑+ ∆J = ±1 only 100 - 5000 cm–1 ∆v = ±1 strongest G(v) = ωe(v + 1/2) – ωexe(v + 1/2)2 + … Fv(J) = BvJ(J + 1) – DvJ2(J + 1)2 Bv = Be – αe(v + 1/2) NOTE: Signs. Because almost always see ωexe, Dv, αe > 0 as defined above! typical ωe ≈ 1000 cm–1 Cs2 42cm–1, I2 215 cm–1, H2 4400 cm–1 J′ 3 G(v′) + Fv′(J′) v′ 2 1 0 R branch P branch J″ 3 G(v″) + Fv″(J″) v″ 2 1 0 1st line in R branch J′ = J″ + 1 J′ = 1 ← J″ = 0 2B R(0) R(J″) 1st line in P branch J′ = J″ – 1 J′ = 0 ← J″ = 1 2 1 0 P(1) P(J″) 1 2 3 zero gap “blue on left” “red on right” Lines spaced by ≈2B with “zero-gap” of 4B where Q(0) would be. always negative R branch P branch ∆J = +1 ∆J = –1 R(J″) = ∆G + ∆F = ∆G + ∆BJ″2 + (2B′ + ∆B)J″ + 2B′ P(J″) = ∆G + ∆F = ∆G + ∆BJ″2 – (2B′ – ∆B)J″ –∆vαe small negative quadratic term notation ⎛v> + v<⎞ ∆G = G(v>) – G(v<) ≡ ∆G⎜⎝ 2 ⎟⎠ ≈ 2B′ large linear term ∆G(1/2) = ωe – 2ωexe = ν lines in R branch gradually pull closer together, branch goes to blue lines in P branch gradually pull further apart, branch goes to red Double Humped Appearance of Vibration-Rotation band gradually pulling together gradually spreading seldom see “band heads” gap blue ←ν λ→ R P red I ~ (2J + 1)e–BJ(J + 1)/kT Finer details: Hot Bands: 1 ← 0 looks like 2 ← 1 except 1. 2 ← 1 is shifted 2ωexe to red 2. “constant” difference between rotational lines is smaller by 2αe (< ~1%) 3. identical quadratic term ∆B = αe 4. intensity down by e−ωe kT Isotopomer Bands Bv ∝ µ–1 ωe ∝ µ–1/2 intensity depends on isotopic abundance, NOT on T Overtone Bands 1. 2 ← 0 near 2ωe 2. linear term (2B′ ± ∆B) in rotational branches almost same as for ∆v = +1 band. 3. quadratic term 2× as large → can sometimes see bandheads for high overtone bands 4. intensity weaker than ∆v = 1 band, usually by 10 to 100× ROTATIONAL ASSIGNMENT trivial because of * presence of zero gap * no overlap of band with itself Electronic Spectra 1 ∑+ – 1∑+ for now — deal with non-1∑+ soon. No restriction on types of molecules. [Symmetry restrictions, g↔u, ∑+↔∑+, ∆S = 0, ∆Λ = ∆Ω = 0,±1] ν 10,000 cm–1 → 100,000 cm–1 VIS & UV & VUV (not X-ray) because not sharp lines Rotational Structure of one vibrational band of electronic transition: just like VIBRATION ROTATION band but more can happen. ∆J = ±1 only ↓ Band origin ∆Te + G′(v′) – G″(v″) R(J) = ν0 + ∆BJ2 + (2B′ + ∆B)J + 2B′ starts toward blue P(J) = ν0 + ∆BJ2 – (2B′ – ∆B)J starts toward red Fortrat parabola — ambiguity about assignment Q(J) = ν0 + ∆BJ2 + ∆BJ (not for 1∑+ – 1∑+) usually compact ∆B can be large and have either sign because B′ – B″ ≠ –αe Heads form! ∆B > 0 R′e < R′′e Head in P degrades to blue blue ∆B < 0 R′e > R′′e Head in R degrades to red red blue red By inspection, can tell sign of ∆B ⇒ bonding nature of excited state. At what J does head form? How far is head from band origin? ν= Extrenum dP(J) = 2 ∆ BJ − (B′ + B′′) dJ B′ + B′′ B J PHEAD = if ∆ B > 0, then J PHEAD ~ 2  ∆B ∆B >0 J RHEAD = − (3B′ + B′′) 2  ∆B if ∆ B < 0, then J RHEAD ~ − B ∆B <0 2 ⎧ ⎡ B′ + B′′ ⎤ ⎡ B′ + B′′ ⎤ P − (B′ + B′′) ⎢ ⎪ P ( J HEAD ) − ν0 = ∆ B⎢ ⎥ ⎣ 2∆B ⎦ ⎣ 2 ∆ B ⎥⎦ ⎪ ⎪ (B′ + B′′)2 ⎨ =− <0 seldom negligible 4∆B ⎪ ⎪ 2 ⎪R ( J RHEAD ) − ν0 = − (3B′ − B′′) + 2B′ > 0 ⎩ 4∆B ~− B2 ∆B ~+ B2 ∆B Can't use easily picked-out head as approximation for band origin! Hard to find origin because no zero gap. Covered by returning branch. || type transition ν0 Q ⊥ type transition ν0 Vibrational structure next time.