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Wickstrom PR 613 Fluorescence. Source Dispersing Sample Detector Computer Lens Dispersing Instrumentation.

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Presentation on theme: "Wickstrom PR 613 Fluorescence. Source Dispersing Sample Detector Computer Lens Dispersing Instrumentation."— Presentation transcript:

1 Wickstrom PR 613 Fluorescence

2 Source Dispersing Sample Detector Computer Lens Dispersing Instrumentation

3 Electronic Excitation S0S0 V1V1 V0V0 V2V2 S1S1 V1V1 V0V0 V2V2 Vibrational relaxation, 10 -11 to 10 -10, is stepwise,  V=1. Excess energy goes to the solvent and dissipated as heat. Fluorescence, 10 -9 to 10 -7 s.

4 S0S0 V1V1 V0V0 V2V2 S1S1 V1V1 V0V0 V2V2 Spectrum

5 S0S0 V1V1 V0V0 V2V2 S1S1 V1V1 V0V0 V2V2 Collisional deactivation (external conversion), 10 -9 to 10 -7 s. Excess energy given to the solvent. Competition with vibrational relaxation

6 S0S0 V1V1 V0V0 V2V2 S2S2 V1V1 V0V0 V2V2 Vibrational relaxation Internal conversion, 10 -12 s. V1V1 V0V0 V2V2 S1S1 Competition IC

7 S0S0 V1V1 V0V0 V2V2 S1S1 V1V1 V0V0 V2V2 Vibrational relaxation Intersystem crossing, singlet to triplet. V1V1 V0V0 V2V2 T1T1 Phosphorescence, 10 -4 to 10 4 s. Oxygen is a cause. Competition ISC

8 Summary: Fate of Excited State  Vibrational relaxation  Fluorescence  External Conversion (collisional deactivation)  Intersystem Crossing  Internal conversion  Phosphorescence  Dissociation  Pre-dissociation

9 Summary of Timescales  Photon Absorption: 10 -15 to 10 -14 s.  Vibrational Relaxation: 5 x 10 -12 s  Internal and external conversion, intersystem crossing: 10 -12 to 10 -7 s.  Fluorescence: 10 -9 to 10 -5 s.  Phosphorescence: 10 -4 to 10 s.

10 Intensity of fluorescence  Intensity is proportional to the ground-state population of the fluorophore, P 0, the rate of absorption, the fluorescence quantum efficiency, and the volume element of the sample illuminated.

11 Excitation  A fluorescent molecule can be irradiated with different wavelengths within its excitation spectrum and, accordingly, will emit light with a characteristic emission spectrum. Its amplitude is determined by the intensity of radiation and the excitation efficiency, which is a function of the excitation wavelength.

12 Factors Affecting Fluorescence  Molecular structure and the environment of the molecule affect fluorescence. The nature of the lowest-lying excited singlet (S 1 ) is critical in determining the luminescence behavior of a molecule because fluorescence and intersystem crossing usually occur from this state. For organic molecules, the transitions between S 0 and S 1 can involve  -  * or n-  * transitions. The most efficient fluorescence usually involves  -  * transitions because the transition probability is high (i.e., large , k A, and k f ). The rate of intersystem crossing is generally 1,000 times faster between states of different electronic origin (S 1 (n,  *)-T 1 ( ,  *) or S 1 ( ,  *)-T 1 (n,  *)).

13 Structural Effects

14 Efficiency of Fluorescence  Depends on the relative competition between radiative and non-radiative routes of deactivation. Can be represented as quantum efficiency or for fluorescence quantum efficiency or yield . This efficiency is related to the rate of absorption and the rate of deactivation of the first excited singlet state S 1. If it is assumed that all processes are first order with respect to number densities of S 0 and S 1 (nS 0 and nS 1 in molecules per cm 3 ) the rate of absorption is K A nS O and the rate of deactivation is (K f + K nr )nS 1, where K A, K F, and K nr are the first order rate constants in s -1 for absorption, fluorescence, and nonradiative deactivation, respectively. Therefore, the rate of change of the number density of the S 1 state is given by:

15 Fluorescence rates  If the analyte is contained in a sample volume V that is fully illuminated with photons of constant intensity, a steady-state concentration of S 1 is rapidly achieved (dnS 1 /dt = 0).  Then:

16 Environmental Factors  Temperature: Increasing the temperature will decrease fluorescence because the rate of dynamic quenching is increased.  Solvent: The viscosity, polarity, and hydrogen bonding characteristics of the solvent can significantly affect fluorescence. Fluorescence usually increases with an increase in solvent viscosity due to reduced rate of bimolecular collisions and rate of dynamic quenching. Solvent polarity and hydrogen bonding are critical because they affect the nature of the excited state. There is a rapid (10 -11 to 10 -12 s) reorientation of solvent molecules around the excited state species. Causes a shift in the excitation wavelength.  pH: pH of solutions in protic solvents can be critical for aromatic molecules with acidic or basic functional groups (phenols, amines). In some cases only the protonated or unprotonated form of the acid or base may be fluorescent. For example, many phenols are fluorescent only in the nonionized form.

17 Structural Factors  Substituents that delocalize the pi-electrons such as –NH 2, -OH, -F, -OCH 3, -NHCH 3, and –N(CH 3 ) 2 groups often enhance fluorescence because they tend to increase the transition probability between the lowest excited singlet state and the ground state.  Electron withdrawing groups such as –Cl, -Br, -I, -NHCOCH 3, -NO 2, or –COOH decrease or quench fluorescence completely.

18 Solvent affects  Fluorescence intensity and wavelength often vary with the solvent. Solvents capable of exhibiting strong van der Waals binding forces with the excited state species prolong the lifetime of a collisional encounter and favor deactivation.  On the other hand, the hydrophobic interior of a protein or DNA usually increases the fluorescence intensity (quantum yield) significantly.

19 Rules for Fluorescence  1.Not observed from saturated hydrocarbons as there are no  or n electrons. Weak fluorescence is sometimes observed in the vacuum UV due to  -  * transitions.  2.Is rarely observed from nonaromatic hydrocarbons that have some double bonds. Weak fluorescence in the UV is observed for some aliphatic carbonyl compounds having n-  * transitions.  3.Many aromatic compounds are intensely fluorescent because of low lying  -  * singlet states. The energy required for transition is often low enough to prevent bond rupture. Phosphorescence is less likely without atoms providing n electrons or substituent groups to promote ISC.  4. Phosphorescence is often favorable in aromatic molecules containing carbonyl groups. These groups introduce nonbonding electrons and increase ISC.

20 Rules for Fluorescence II  5.Substituents attached to aromatic rings can dramatically influence fluorescence. These groups often change the nature of the lowest-lying excited states. Electron donating groups such as -OH in general increase yeild relative to the parent group. Electron withdrawing groups such as -NO 2 decrease yeild by introducing low-lying n,  * states.  6.Effect of halide is specifically to increase ISC.  7.Addition of rings increase the fluorescence yield.  8.Fluorescence is favored for rigid molecules that are planar. They increase conjugation of pi electron that result in decrease interaction of solvent.  9.For molecules consisting of 2 or more aromatic rings separated by alkyl groups, the fluorescence properties for the compound are close to those of the separate rings.  10.Fluorescence from metals usually occurs only in organometallic complexes, basically rigid metal chelates. Often the ligand is non fluorescence. The complex exhibits fluorescence if the lowest-lying singlet state of the ligand is changed from n pi* to a  -  *. Sometimes fluorescence involves charge transfer with metal d electrons and ligand orbitals. In some rare earths, fluorescence is due to f-f transitions of metal electrons.

21 Concentration Determination  The fluorescent power and concentration is proportional to the number of molecules in the excited states which is proportional to the radiant power absorbed by the sample.  P F =  F P 0 where P F is the radiant power of fluorescence,  F is fluorescence efficiency or quantum yield.   F =photons emitted as fluorescence/photons absorbed.

22 Concentration Determination  Applying Beer’s law:  Expanding in power series:

23 Concentration Determination Concentration, M (very low 10 -6 to 10 -3 M) Fluorescence intensity  If  bc is small, only the first term in the series is significant with an error of 2.5%.  Inner filter effect

24 Concentration Determination  Use dilute solutions.  Sample should be degassed.  Choose the correct solvent.

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