Molecular Spectra and Molecular Structure - Lec 4

Page 1 of 6 pages Lecture #4: How Do We Get Information About V(Q) from Molecular Constants?    →      (diatomic molecules)   erturbation Theory always gives some insight. Non-rigid anharmonic vibrating rotor is a beautiful illustration of how we use perturbation theory, even though it is a 1-D problem. Perturbation Theory E vJ = E 0vJ + E1vJ + E 2vJ E1vJ = vJ H′ vJ E =∑ 2 vJ v′ H′vJ, v′J 2 E vJ − E 0v′J Harmonic Oscillator Matrix Elements v-scaling selection rules What do you expect will happen to the Bv for a non-rigid harmonic oscillator? What do you expect ax3 term to do to ∆G(v + 1/2)? How will anharmonicity affect this? B Will the effect of ax3 on ∆G depend on the sign of a? Page 2 of 6 pages Review of Anharmonic Vibrating-Rotor Handout x = R – Re The harmonic oscillator displacement. 2 2 R pˆ k 2 J 3 + x + ax + H= 2 2µ 2 2 µR momentum conjugate to x Harmonic Oscillator H° 1st anharmonic correction non-rigid rotor (use rigid rotor basis) a < 0 gives skew in usual sense 1. Product basis set H-O Rigid Rotor v ΩJ M These allow us to evaluate matrix elements of xn and J 2. 2 1 First job is to manipulate R2 into convenient form for our H-O basis set. 1 1 1 ⎡ 2 x 3x 2 ⎤ 1 = = − + ⎢ ⎥ R 2 ( Re + x )2 Re2 ⎢ rigid Re Re2 ⎥ rotor non-rigid ⎥ ⎢ corrections ⎦ ⎣ 3. 2 J commutes with p̂ 2R and x̂ n so H is block diagonalized according to J (no ∆J ≠ 0 matrix elements) but not according to v. We consider each H J(x) separately. 4. We are going to use 1st and 2nd order nondegenerate perturbation theory to compute effect on EvJ of the ax3 and the 1/R2 terms. Page 3 of 6 pages A General procedure: write EvJ as empirical power series in (v + 1/2) and J(J + 1). EvJ ≡ y00 + ωe(v + 1/2) – ωexe(v + 1/2)2 + BvJ(J + 1) – DvJ2(J + 1)2 7 fit parameters, each is coefficient of a Bv ≡ Be – αe(v + 1/2) term with characteristic v,J dependence Dv ≡ De – βe(v + 1/2) Y00, ωe, ωexe, Be, αe, De, and βe all cause characteristic v,J dependence. B B B. Apply perturbation theory to 1/ 2 l = =⎡k ⎤ H ⎢µ⎥ ⎣ ⎦ ⎡ 2 x 3x 2 ⎤ (v + 1/ 2) v v + ax + Be J ( J + 1) ⎢1 − + 2⎥ ⎣ Re Re ⎦ 3 convenient notation ∆v = ±1 ∆v = ±3, ±1 ∆v = ±2, 0 and compute perturbation theory results, term by term, with above explicitly v,J dependent terms. x n →∝ v n / 2 (kµ) − n / 4 = n / 2  H–O matrix elements of ωe B αe ~ Be B2e ωe ⎡ ⎡ k ⎤1/ 2 3 = ⎛ ⎞⎤ l H = ⎢ = ⎢ ⎥ ( v + 1/ 2 ) + Be J ( J + 1) ⎜1 + 2 ( v + 1/ 2) ⎥v v 1/ 2 ⎟ ⎝ Re (kµ) ⎠⎦ ⎣ ⎣µ⎦ ⎛ 3 ⎡ 2 x 3x 2 ⎤ ⎞ + ⎜ ax + Be J ( J + 1) ⎢ − + 2 ⎥ ⎟ v′ v ⎣ Re Re ⎦ ⎠v′ ≠ v ⎝ zero’th and first order second order beware of cross terms for ∆v = ±1 xn ∝ vn/2(kµ)–n/4 n/2 ωe = (k/µ)1/2 Re = (2µBe)1/2 ax → av (kµ) 3 v′v 3/ 2 −3/ 4 3/ 2 a2 3 –3/ 2 ⎯⎯⎯⎯ → ( ) v kµ (k / µ)1/ 2 ωe 2nd order perturbation a 2 h2 3 v ~ k2 µ ∝ a 2 ω e4 3 possible contribution to [J(J + 1)]0(v + 1/2)3,2,1,0 Page 4 of 6 pages ⎛ −2x vv′ ⎞ Be J(J +1)v1/2 B2e h(kµ)−1/2 2 −1/4 1/2 2 2nd order (kµ) h ⎯perturbation ⎯⎯⎯ → 2 J (J +1) v Be J(J +1) ⎜ ⎟→ 1/2 R R R h(k / µ) ⎝ e ⎠ e e B2e R 2e k possible contribution to [J(J + 1)]2(v + 1/2)1,0 ∝ Be3 ω 2e Cross term between E (2 ) v −Be 2xJ(J + 1) and ax3 → Re ⎛ −2x vv′ Be J(J +1) ⎞ 3 2⎜ ⎟ (ax v′v ). Re ⎝ ⎠ Be J(J + 1)av2 (kµ)−1 h2 Be ah 2 ~ ~ v J(J +1) 1/2 3/2 1/2 Rek µ R e h(k / µ) Be3/ 2 a ∝ 3 ωe ⎛ 3x 2 ⎞ –1/2 ⎟ → BeJ(J + 1)v(kµ) ⎝ Re ⎠ possible contribution to [J(J+1)]1(v + 1/2)2,1,0 ∆v ≠ 0 terms from BeJ(J + 1) ⎜ 2nd order perturbation B B2e J 2 (J +1)2 v2 h R e4 k 3/2 µ1/2 possible contribution to [J(J+1)]2(v + 1/2)2,1,0 B Put all of this together. Very complicated algebra. Always start with highest power of (v + 1/2) and J(J + 1). 15 2 a 2 ωe xe = 2 4 k µ ωe4 Note that regardless of sign of a, the ax3 term leads to a negative coefficient in front of (v + 1/2)2 (because ax3 makes potential get wider as E increases). a ∝ ω 2e ω e x1/e 2 ωe x1/e 2 ω 2e ⎤ 6Be ⎡ ah2 αe = − Be ⎥ ⎢− ω e ⎣ kµR e ⎦ Two contributions to Bv = Be – αe (v + 1/2) Second one is from non-rigidity of rotor (no anharmonicity) and causes Bv to increase with v, not to decrease. Page 5 of 6 pages 1 h−o weights χ v more at short R than long, so as v increases, Bv 2 R increases because 〈R–2〉 increases. First one is from anharmonicity and does depend on sign of a. a>0 (unphysical) a < 0 (most potentials have this asymmetry) softer wall R R a < 0 means contribution to αe with sign such that αe > 0, thus Bv decreases with v. This is almost always the dominant term in αe ~ 4Be3 De = + 2 2 h ωe Be3/2 ωe x1/2 e . ωe “Kratzer” Relation comes exclusively from non-rigidity, not anharmonicity. Always leads to effective decrease in B as J increases. Centrifugal distortion would cause molecule to stretch as it rotates. −18Be4 βe = 3 3 h ωe Y00 = − other small but unimportant terms 2 2 7 ha 7 = − ωe x e 2 16 k µ 60 (v + 1/2)3 term disappeared in perturbation sum because of cancellation between ∆v and –∆v pairs of terms. Page 6 of 6 pages Morse Potential V(R) = D e [1 − e−a (R −R ) ] e ⎛ µ ⎞ 1/ 2 a = 2πω e ⎜ e ⎟ = 2π (2µω e x e ) ⎝ 2D ⎠ 1 ωe = vD + 1 / 2 2 ωe x e 1/ 2 2 of all Y 0 only Y00, Y10, and Y20 are ≠ 0. De = 4Be3 ω 2e 1/2 6 −Y11 = α e = ⎡⎣(ωe x e Be3 ) − B2e ⎤⎦ ωe −Y02 = D e = 1 ω 2e 4 ωe x e Kratzer Pekeris Herzberg p. 109, Jevons, p. 27-29, Gaydon, p. 36 1/ 2 B5e ⎡ ⎛ ω e x e ⎞ ⎤ Y03 = H e = 16 4 ⎢ 3 − ⎜ ⎟ ⎥ ωe ⎣ ⎝ Be ⎠ ⎦ 1/ 2 ⎛ ωe x e ⎞ ⎤ Be7 ⎡ 5 ωe x e − 9⎜ Y04 = L e = −64 6 ⎢13 + ⎟ ⎥ ωe ⎣ 3 Be ⎝ Be ⎠ ⎦ Birge 1/ 2 ⎛ ωe x e ⎞ Be4 ⎡19 13 ω e x e ⎤ + Y12 = −βe = −12 3 ⎢ − 9 ⎜ ⎥ ⎟ ωe ⎣ 2 6 Be ⎦ ⎝ Be ⎠ 1/ 2 3/2 ⎛ ωe x e ⎞ 6Be3 ⎡ 23 ⎛ ω e x e ⎞ 7 ⎛ ω e x e ⎞ ⎤ Y21 = γ e = 2 ⎢5 −10 ⎜ ⎟ + ⎜ ⎟− ⎜ ⎟ ⎥ 4 ⎝ Be ⎠ 6 ⎝ Be ⎠ ⎦ ωe ⎣ ⎝ Be ⎠ 1/2 3/ 2 ⎛ ωe x e ⎞ ⎛ ωe x e ⎞ 20Be4 ⎡ 139 ⎛ ωe x e ⎞ Y31 = ⎢7 + 21⎜ ⎟ + ⎜ ⎟ + 22.12 ⎜ ⎟ ωe3 ⎣ B B 6 ⎝ e ⎠ ⎝ e ⎠ ⎝ Be ⎠ 2 5 /2 ⎛ ωe x e ⎞ 1 ⎛ ωe x e ⎞ ⎤ +3.633⎜ ⎟ − ⎜ ⎟ ⎥ ⎝ Be ⎠ 2 ⎝ Be ⎠ ⎦ Y00 = 0 for Morse, but Dunham’s result is B − ωe x e α e ωe ⎛ α e ωe ⎞ 1 Y00 = e + +⎜ ⎟ 4 12Be ⎝ 12Be ⎠ 2 2 ⎡error in ⎤ ⎢ ⎥ ⎣Herzberg⎦