L8 - Biophysical Chemistry Chemistry 24a Winter Term...

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Biophysical Chemistry Chemistry 24a Winter Term 2009-10 Instructor: Sunney I. Chan Lecture 8 January 27, 2010 Spontaneous Emission. Fluorescence Spectroscopy. Fluorescence spectroscopy Consider the following two-level system: ─────────── S b (singlet) B ab A ba = coefficient of spontaneous emission ─────────── S a (singlet) Δ E ab = h ν ab Light of radiation density I( ν ) induces transition from S a to S b (and from S b to S a ) at the rate of B ab (or B ba ) per molecule per sec (per unit energy density of radiation). B ba How does the system reach thermal equilibrium in the dark? 1. In the dark, there is only blackbody radiation. 2. At temperature T, the blackbody radiation is given by the Planck radiation law: I thermal radiation ( ν ) = (8 π h ν ³/c³) [exp(h ν /k B T) – 1] 1 3. There must be spontaneous emission from the upper state. Fluorescence Spectroscopy Emission of light from excited electronic state : In contrast to absorption, which occurs ~10 -15 sec , or femtoseconds, emission occurs ~ 10 -12 to 10 -9 sec , or picoseconds to nanoseconds.
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Coefficient of Spontaneous Emission Einstein derived A ba in terms of B ba using the principle of detailed balance: Suppose states b and a are at thermal equilibrium. Then, n b /n a = exp-( ε b ε a )/k B T If the system is at thermal equilibrium, it must be at thermal equilibrium with the blackbody radiation. That is, the blackbody radiation induces a b and b a transitions at such a rate that they compensate for spontaneous emission. That is, the emission and absorption of radiation density must be equal. Thus, we must have n a eq I blackbody ( ν ab )B ab = n b eq I blackbody ( ν ab )B ba + n b eq A ba or n a eq /n b eq = [B ba I blackbody ( ν ab )+ ±A ba ]/ B ab I blackbody ( ν ab ) = 1 + (A ba /B ab ) / I thermal radiation ( ν ab ) Now, I thermal radiation ( ν ) = (8 π h ν ³/c³) [exp(h ν /k B T) – 1] 1 (see data on following slide) Substituting, n a eq /n b eq = 1 + (A ba /B ab )[exp(h ν ab /k B T) -1](c³/8 π h ν ab ³) = exp(h ν ab /k B T) Therefore, A ba = B ab (8 π h ν ab ³/c³).
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Important conclusions : Note ν ab ³ dependence in A ba ! Spontaneous emission is important for transitions in the visible, uv, x-ray, and γ -ray regions, not in IR, microwave, and radiofrequency regions. Also, direct proportionality to B ab or B ba , so A ba = 0 if B ab = B ba = 0! Spontaneous emission does not occur if the induced transition is not allowed. Dipole strengths Spectroscopists have introduced the dipole strength D ab = (3 ħ ²/2 π )B ab so that A ba = B ab (8 π h ν ab ³/c³) = (32 π ³ ν ab ³/3c³ ħ )D ab In the absence of radiation, or any other perturbations or interactions, the rate of de- excitation of molecules initially in state S b will be dn b /dt = - A ba n b. Thus,
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This note was uploaded on 10/27/2010 for the course BI 110 taught by Professor Richards,j during the Winter '08 term at Caltech.

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L8 - Biophysical Chemistry Chemistry 24a Winter Term...

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