CHAPTER 15- MOLECULAR LUMINESCENCE.ppt
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1、CHAPTER 15: MOLECULAR LUMINESCENCE,Chapter 15 - 2,LUMINESCENCE TECHNIQUES,Emission of light is used to determine certain properties,e e.g.structure and concentration, of the emitting species. Deactivation processes involved in converting a substance from excited state to ground state: the emission o
2、f heat, activation of a chemical reaction or emission of radiation of the same or a modified wavelength. Forms of photoluminescence (luminescence after absorption) are fluorescence (short lifetime) and phosphorescence (long lifetime). Approximately 10x more sensitive than absorption techniques:ppb d
3、etection limit Limited number of systems that photoluminesce. Luminescence observed for simple and complex systems and for all three phases.,Chapter 15 - 3,Theory,Atoms: e.g. dilute Na (g) the 3s 3p transition occurs by absorption at l = 5895 and 5890 A. with a lifetime 10-8 sec, the electron return
4、s to the ground state isotropically (isotropically) emitting hn with the wavelength of emission being the same as the wavelength of excitation. resonance fluorescence. Polyatomic Systems Resonance fluorescence observed Emission of radiation of longer l (called a Stokes shift) more common Most fluore
5、scent systems are complex organic compounds with 1 or more aromatic functional groups so that the commonly observed transitions are: p* p, p* n, s* s, s* s.,Chapter 15 - 4,EXCITED STATES,Multiplicity (number of lines observed when the molecule is placed in a magnetic field) is related number of unpa
6、ired spins in the molecule (S):. M = 2S + 1. Most molecules have an even number of electrons which means that all of their electrons in the ground state must be paired: Singlet state (M = 20 + 1);. all electrons in ground state paired. Doublet state : (M = 2 + 1 = 2)a free radical (substance that ha
7、s an odd number of electrons);. electrons can have 2 orientations in the magnetic field-opposed to the field and aligned. Triplet state: (M = 21 + 1 = 3), excited state in which excited electron spin is flipped so that the spins are parallel.Singlet state Triplet stateDiamagnetic ParamagneticProbabl
8、e Less probableLifetime 10-8-10-13sec Lifetime up to 1 s,Chapter 15 - 5,ENERGY LEVEL DIAGRAM,Let ground , the first excited, second excited etc. electronic states be: S0, S1, S2 and etcfor all of the possible singlet states. Triplet states would then be T1, T2 and so on . All electronic state has se
9、veral vibrational and rotational states.,Chapter 15 - 6,Decay Processes,Internal conversion Movement of electron from one electronic state to another without emission of a photon, e.g. S2 S1) lasts about 1012 sec. Predissociation internal conversion electron relaxes into a state where energy of that
10、 state is high enough to rupture the bond. Vibrational relaxation (1010-1011sec)- Energy loss associated with electron movement to lower vibrational state without photon emission. Intersystem crossing: Conversion from singlet state to a triplet state. e.g. S1 to T1 External conversion is a non-radia
11、tive process in which energy of an excited state is given to another molecule (e.g. solvent or other solute molecules). Related to the collisional frequency of excited species with other molecules in the solution. Cooling the solution minimizes this effect.,Chapter 15 - 7,QUANTUM YIELD,Only a fracti
12、on of the photon absorbed result in fluorescence. Fraction called Quantum yield (efficiency), F: Excite molecule in say the S1 state can undergo a transition back to the ground state: S1 S0 + hn. The emitted photon is the useful fluorescence line and takes about 10-6-10-10 sec to occur. Rate of all
13、processes which involved the absorption or emission of a photon can be written in terms of a first order rate equation:,Instrumental Analysis,Christian and OReilly, p. 251,Chapter 15 - 8,KINETICS OF ADSORPTION,The intensity of the light absorbed is .DP = Po - PT = rate of absorption where Po = photo
14、n flux to sample, PT = photon flux out of sample. At steady state rate of absorption equals rate of fluorescence or: DP = (kIC + kISC + kf + kQQ)S1 where kQ = rate constant for quenching process-a second order process since the Q is also important in determining the relative rate of this process. Le
15、t S1 = steady state concentration of S1 molecules. The rate of fluorescence Pf = FDP = F(kIC + kISC + kf + kQQ)S1 and Pf = kfS1. Combine and rearrange: F = large F means a large kf. The lifetime of the fluorescing state is given by The inverse relationship between the rate constant and the lifetime
16、tells us that a process having a large rate constant has a short lifetime and will have the largest fluorescence intensity.,Chapter 15 - 9,FLUORESCENCE INTENSITY VS CONCENTRATION,Before fluorescence occurs absorption must occur. The absorption process given by Beers Law: where e = k/2.303 = molar ab
17、sorptivity. We will use this in the development of the fundamental equation for fluorescence. Earlier we stated that Pf = rate of fluorescence = F(kIC + kISC + kf + kQQ)S1 = FDP = FPo - PT. Beers law can be written as: PT = Po10-ebC = Poe-2.303ebC. Substituting into the fluorescence equation gives:
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