Chemistry 24b Lecture 23 Spring Quarter 2004 Instructor: Richard Roberts. (1) It induces a dipole moment in the atom or molecule.

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1 Chemistry 24b Lecture 23 Spring Quarter 2004 Instructor: Richard Roberts Absorption and Dispersion v E * of light waves has two effects on a molecule or atom. (1) It induces a dipole moment in the atom or molecule. v µ in = α v E * α molecular polarizability v E * v E * = 0 Figure 9-9 (2) If the molecule has a permanent dipole moment, v µ, it can attempt to align the molecular dipoles (counteracted by thermal motions). Debye showed that for molecules in a gas or in a dilute solution, v µ and α are related to the dielectric constant ε of the substance. ( ε 1) ( ε + 2) M ρ = 4π N A 3 α + µ 2 3k B T ε = dielectric constant At high temperatures, v E * cannot align v µ very effectively so α dominates. Also, when the frequency of light waves becomes very high, permanent dipoles of molecules cannot follow the oscillations of E v * and therefore do not become aligned, even partially, with E v *. So, ( ε hf 1) ( ε hf + 2) M ρ = 4π N A α 3 and it iturns out ε hf = n 2 or the square of the refractive index of the substance.

2 Chemistry 24b Lecture 23 2 Dielectric Absorption of a Hemoglobin Solution. Dispersion 2 compounds, say E * = E 0 * sin ω t Hb ν rot 10 7 Hz O H H H 2 O ν rot Hz Figure 9-10 At ω /2π < 10 7 Hz, both Hb and H 2 O will contribute to µ 2 and to ε of solution. As ω /2π is swept through 10 7 Hz, the solution will absorb power from rf circuit. Absorption is caused by the lagging of Hb dipoles behind the oscillation of the alternating electric field. As ω /2π passes through this region, the dipoles of the Hb can no longer come to alighment with the a.c. field. So, at frequencies above 10 7 Hz, µ 2 of Hb no longer contributes to ε of solution. This decline in dielectric constant is called a dispersion, and the absorption of energy associated with it is called a dielectric loss. Similarly, as ω /2π passes through Hz, similar dielectric absorption and dispersion occurs as the water dipoles can no longer come into alignment with E v *. ε Hz contribution from polarization of the electron distribution induced by electric field ν Figure 9-11

3 Chemistry 24b Lecture 23 3 Dipoles Can Absorb Power from rf Circuit when Their Oscillations Lag 90 Behind v the Polarizing E * Consider k e E * = E 0 * sin ω t Driving electric field: E * = E 0 * sin ω t At resonance, Vibrational frequency of e : ω 0 = k / m x = x 0 sin ω t + x 90 cos ω t where x = displacement of e from equilibrium position x 0 sin ω t ( ) is in phase with force E 0 * sin ω t x 90 cos ω t is 90 o out of phase with force if x 90 < 0, lags if x 90 > 0, leads The work done on e in a complete cycle of the electrostatic field is zero, if the oscillation of the e stays in phase with the driving field ( x = x 0 sin ω t). Work done by field on charge = Fdx cycle = qe * dx = q E cycle * 0 sin ω t cycle 2π = qe * 0 x 0 sin θ d sinθ θ = ω t 0 = qe 0 * x 0 sin 2 θ 2 2 π 0 = 0 dx ( 0 sin ω t) Work is done on e in a complete cycle of the electrostatic field, if the oscillation of the e lags 90 out of phase with the driving field ( x = x 90 cos ω t).

4 Chemistry 24b Lecture 23 4 Work done by field on charge = = q E * 0 sin ω t d x 90 cos ω t = qe * 0 x 90 sin ω t cycle 2 π = +qe * 0 x 90 sin 2 θ dθ 0 Fdx cycle ( ) d cos ω t = π qe 0 * x 90 > 0 e or charge absorbs power from rf circuit Note that when oscillation of e (response) leads the driving field by 90, Work done by field on charge = π qe 0* x 90 So e or charge does work on the rf circuit. x = x 90 cos ω t F, x condition for absorption F = qe 0 * sin ω t Figure 9-12

5 Chemistry 24b Lecture 23 5 Classical Mechanical Description of Absorption and Dispersion Detailed Treatment Problem k e m E * = E 0 * sin ω t (1) Electromagnetic wave: E 0 * sin ω t ω = 2 πν (2) Natural frequency of oscillator: ω 0 = k / m (3) Forces on e (a) Driving force of electromagnetic wave: e E 0 * sin ω t (b) Restoring force between negative electron and positive charge: kx (c) Damping force, which is proportional to the velocity of the moving electron and arises from frictional and radiative energy losses: η dx dt Newton s Equation of Motion Applied to e Force = m d2 x dt 2 e E 0 * sin ω t kx η dx dt = m d2 x dt 2 General solution to linear second-order differential equation: Differentiating, x = x 0 sin ω t + x 90 cos ω t ; let β = x 90 x 0 = x 0 sin ω t + β x 0 cos ω t dx dt = ω x 0 cos ω t ωβx 0 sin ω t d 2 x dt 2 = ω 2 x 0 sin ω t ω 2 β x 0 cos ω t

6 Chemistry 24b Lecture 23 6 Substituting into F = ma and collecting terms, * [ e E 0 kx0 + ηωβx 0 + ω 2 x 0 m]sin ω t + [ k β x 0 ηω x 0 + mω 2 β x 0 ]cos ω t = 0 True for all values of time. Therefore e E * 0 kx 0 + ηωβx 0 + mω 2 x 0 = 0 and k β x 0 ηω x 0 + mω 2 β x 0 = 0 Thus β = x 0 = ηω mω 2 and k * ee 0 mω 2 k + ηωβ Since k = mω 0 2 and Define ω = η 2m 2mω ω β = mω 2 2 = 2ω ω mω 0 ω 2 0 ω 2 x 0 = * e E 0 mω 2 mω mω ω β = e E * 0 m ω 2 0 ω 2 ( ω 2 0 ω 2 ) ω 2 ω 2 Note that when the periodic driving force is removed, one has a damped oscillator, in which case x = x 0 e ω t cos ω 02 η where ω = 2m Polarizabilities ( ω 2 ) 1 2 t α = µin E * = e x E * Static field E *

7 Chemistry 24b Lecture 23 7 For our problem here, µ in = e x 0 sinω t e β x 0 cosω t in phase out of phase component component (90 o ) Define α 0 = e x 0 E 0 * in phase polarizability α 90 = e β x 0 E 0 * out of phase polarizability (choose sign such that it lags) Then µ in = α 0 E 0 * sinω t α 90 E 0 * cosω t Substituting α 0 = e2 m α 90 = 2e2 m ( ω 2 0 ω 2 ) ( ω 2 0 ω 2 ) ω 2 ω 2 ω ω ( ω 2 0 ω 2 ) ω 2 ω 2 ω = ω 0 ω ω = ω 0 max e 2 α 90 = 2m 1 ωω 0 ω = ω 0 + ω α 0 ω, driving frequency Figure 9-13 It turns out

8 Chemistry 24b Lecture 23 8 Extinction coefficient εω ( ) = 4π N A 2303c ωα 90 I s n π N A ρ M α 0 θ = 90 o ( ) = I 0 16π 4 α 0 2 ν 4 r 2 c 4 n refractive index