Methods Sensitive to Free Radical Structure Resonance Raman Electron-Spin Resonance (ESR) or Electron Paramagnetic Resonance (EPR)

Motivation Absorption spectra of free radical and excited states are generally broad and featureless Conductivity is not species specific Conductivity is additive with respect to ionic content of the cell

Specific Vibrations? Now have vibrational spectroscopy in laser flash photolysis, usually in organic solvents Water is a good filter of infrared and masks vibrational features of free radicals Raman is weak, second-order effect What about Resonance Enhanced Raman?

Medium Emergent light +-+- EiEi Incident light s = 0 0 s = 0  mn s Scattered light Raman Rayleigh LIGHT SCATTERING P i = α ij E j P = Induced electric dipole moment E = Electric field of the electromagnetic radiation α ij = Elements of polarizability tensor G.N.R. Tripathi

I mn = Const. I 0 ( 0  mn ) 4  I (   ) mn I 2  (   ) mn = (1/h)  M me M en / ( em  0 + i  e ) e + non-resonant terms ee nn mm mn em 0 ENHANCEMENT OF RAMAN SCATTERING (via α ij ) Probability Amplitude G.N.R. Tripathi

RESONANCE RAMAN em >> 0 Normal Raman em - 0 ~ 0 Resonance Raman |(   ) mn  2 = Const. × (M me M en ) 2 /  2 Enhancement up to Pulse radiolysis time-resolved resonance Raman Identification, structure, reactivity and reaction mechanism of short-lived radicals and excited electronic states in condensed media Relevance: Theoretical chemistry, chemical dynamics, biochemistry,,paper and pulp-industry, etc. G.N.R. Tripathi

Raman shift (cm -1 ) [Ru(bpy) 2 dppz] 2+ bound to DNA NO DNA present 2,2’- bipyridyl dppz = dipyridophenazine

Two-slit experiment

Selection Rules for the Amplitudes of Transitions Electronic Transition Elements (Dipole allowed) Franck-Condon Factor For Resonance enhancement both must be non-zero

Relationship to Radiationless Transitions and Absorption dP(nm)/dt = (4 2 /h) |V mn | 2 FC (E m ) This is a probability. Quantum mechanics usually calculates amplitudes which are “roughly the square root” (being careful about complex numbers) Taking the square root, shows that the amplitudes for radiationless transitions are first-order in the interaction V Likewise, simple absorption and spontaneous emission are first-order processes with regard to an interaction V rad

Connection to Wavefunctions So we can use the path integral to see how one non-stationary state (f) at time t a propagates into another  at time t b In terms of the stationary states of the system

Expansion of part of exponential for small potentials Putting this back into the Amplitude K v (b,a) gives a perturbation expansion of the path integral

Interpretation of First Term Represents propagation of a free particle from (x a,t a ) to (x b,t b ) with no scattering by the potential V a b

Second Term

Interpretation of Second Term x t tbtb tctc tata a c b Particle moves from a to c as a free particle. At c it is scattered by V[x(s),s] = V c. After it moves as a free particle to b. The amplitude is then integrated over x c, namely over all paths.

Physical Meaning of 2 nd Term Represents propagation of a particle from (x a,t a ) to (x b,t b ) that may be scattered once by the potential at (x c,t c ) V a b c

Interpretation of Third Term Represents propagation of a particle from (x a,t a ) to (x b,t b ) that may be scattered twice by the potential, once at (x(s),s) and once at (x(s),s) V a b

Selection Rules (A-term) A-term: Condon approximation - the transition polarizability is controlled by the pure electronic transition moment and vibrational overlap integrals The A-term is non-zero if two conditions are fulfilled: (i) The transition dipole moments [   ] ge 0 and [   ] eg 0 are both non-zero. (ii) The products of the vibrational overlap integrals, i.e. Franck-Condon factors, are non-zero for at least some values of the excited state vibrational quantum number.

Consideration of Franck-Condon Factors = 0 orthogonal  0 Non-symmetric Or Symmetric Totally Symmetric Vibrational Mode Totally Symmetric Vibrational Mode  0

Why must these modes totally symmetric vibrations? H g (Q) = Q g + (k/2)Q 2 H e (Q) = Q g +  Q + (k/2)Q 2 All terms in the Hamiltonian must be totally symmetric, Therefore, the displacement  Q must also be totally symmetric

G.N.R. Tripathi