Rotational spectroscopy: The rotation of molecules z (IC) symmetric: ≠ ≠ IA=IB IC 0 x (IA) linear: IA=0; IB=IC asymmetric: ≠ ≠ ≠ IA IB IC 0 y (IB) Rotor in 3-D Nils Walter: Chem 260 ≠ spherical: IA=IB=IC 0 The moment of inertia for a diatomic (linear) rigid rotor r definition: r = r1 + r2 m1 m2 C(enter of mass) r1 r2 Moment of inertia: Balancing equation: m1r1 = m2r2 2 2 I = m1r1 + m2r2 m m ⇒ I = 1 2 r 2 = µr 2 µ = reduced mass + m1 m2 Nils Walter: Chem 260 Solution of the Schrödinger equation for the rigid diatomic rotor h2 (No potential, − V 2Ψ = EΨ µ only kinetic 2 energy) ⇒ EJ = hBJ(J+1); rot. quantum number J = 0,1,2,... h B = [Hz] 4πI ⇒ EJ+1 -EJ = 2hB(J+1) Nils Walter: Chem 260 Selection rules for the diatomic rotor: 1. Gross selection rule Light is a transversal electromagnetic wave a polar rotor appears to have an oscillating ⇒ a molecule must be polar to electric dipole Nils Walter: Chem 260 be able to interact with light Selection rules for the diatomic rotor: 2. Specific selection rule rotational quantum number J = 0,1,2,… describes the angular momentum of a molecule (just like electronic orbital quantum number l=0,1,2,...) and light behaves as a particle: photons have a spin of 1, i.e., an angular momentum of one unit and the total angular momentum upon absorption or emission of a photon has to be preserved ⇒⇒∆∆J = ± 1 Nils Walter: Chem 260 Okay, I know you are dying for it: Schrödinger also can explain the ∆J = ± 1 selection rule Transition dipole µ = Ψ µΨ dτ moment fi ∫ f i initial state final state ⇒ only if this integral is nonzero, the transition is allowed; if it is zero, the transition is forbidden Nils Walter: Chem 260 The angular momentum 1 E = Iω 2 P = Iω ⇒ P = 2EI 2 classical rotational classical angular momentum energy = + and EJ hBJ (J 1) m h J with B = π 4 I J=1 ⇒ P = J (J +1)h But P is a vector, i.e., has magnitude and direction Nils Walter: Chem 260 Degeneracy J=3 Energy only J=2 determined by J ⇓ all mJ = -J,…,+J m = mJ = J share the same energy ⇓ 2J+1 degeneracy ∆ ± Selection rule: mJ = 0, 1 Nils Walter: Chem 260 The allowed rotational transitions of a rigid linear rotor and their intensity ∆J = ±1 degenerate increasingly ∆ N − E 2J + 1 upper = e kT states Nlower Nils Walter: Chem 260 The non-rigid linear rotor rotor: Centrifugal forces elongate the bond and increase the moment of inertia ∆J = ±1 still holds! Nils Walter: Chem 260 The symmetric rotor = + + − 2 EJ ,K hBJ (J 1) h(A B)K h h A = B = π π 4 Ill 4 I⊥ J = 0,1,2,… K = J,J-1,…,-J Nils Walter: Chem 260.