Molecular Fluorescence Spectroscopy Jose Hodak Introduction This notes were prepared based mainly on Bernard Valeur molecular luorescence book. I would recommend this book to the students who need learn this subject. Joseph Lakowitz book is also very complete and useul. We are concerned with electronic transitions that lead to an excited state o a molecule. Further more, the electronic transition being considered here the promotion o an electron rom an orbital o the molecule in its ground state, to an unoccupied orbital by absorption o light in the UV and visible range. The state created by this transition is termed an excited state. Example: Let s consider the case ormaldehyde molecule: The sigma orbitals are present in all molecules, have cylindrical symmetry, and orm the backbone o the molecule. The pi orbital has two lobules and a nodal plane along the axis o the bond. Formaldehyde has sigma and pi orbitals. In addition, non-bonding electron pairs reside on sp hybrid orbitals localized on the oxygen: H s H C p s p O Formaldehyde There are several possible electronic excitations in ormaldehyde:
Energy s* p* LUMO n p s HOMO Ground State s s* n s* p p* n p* Excited states o ormaldehyde The promotion o a sigma electron to an antibonding (sigma*) orbital drastically reduces the bond strength, causing some o the molecules to dissociate beore they can relax to the ground state (photodissociation). This excited state is called sigma-sigma*. The pi orbital can be excited to an antibonding orbital (pi*). The molecule is still held together by the sigma orbital. This state is called pi-pi*. Another possible excitation is the promotion o a non bonding electron to either a sigma* (n-sigma* state) or to a pi* orbital (n-pi* state). Because the non-bonding electrons reside close to the oxygen atom in the ground state o ormaldehyde, the excitation spreads them over the molecule. For instance, the pi* state has electronic density in our lobules directed away rom the center o the molecule. Thus, the n-pi* transition has charge transer character because it transers electron density rom the oxygen to other parts o the molecule. The lowest energy requiring excitation is that rom the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). Spin and multiplicity: When an electron is promoted to a higher energy orbital its SPIN remains unchanged. The total molecular spin quantum number remains equal to zero upon a transition: 1 S = si = 0 with si =± 2 The multiplicity o the state is deined as M = 2S + 1. For states with S=0, M=1. These states are called SINGLETS. Most o the molecules that we are amiliar with are singlets in their ground state. Question: Give an example o a molecule that is not a singlet in the ground state.
There are some circumstances in which the spin o the electron changes when it undergoes excitation, or ater that process. Now we have S=1, and M = 2S + 1=3, and there are 3 possible states, or a TRIPLET state. Probability o transitions: Absorption o light. Lets estimate how eiciently a molecule can uptake a photon and become an excited state. A situation in which a source o light illuminates a slab o liquid solution containing absorbing molecules is shown below: Absorbing molecules source I0 I D dl Each molecule has an associated photon capture cross-section σ. The total eective absorption α o the layer will be the sum o the absorption cross-sections o the molecules present in the volume o the layer α = σ dn. I we illuminate an area A, the volume is dv = A dl and the number o molecules in this volume is dn = A dl C N Avogadro. The probability o absorption is σ Pabs = dn. This probability must be equal to the raction absorbed o photons out o A the total number i incident photons: I0 I di NAvogadroCAσ dl = σ dn = = N AvogadroCσ dl I I A
We now integrate this expression or a particular sample thickness b: I di b I0 = N AvogadroCσdl ln N AvogadroσCb I0 I = 0 I transorming the logarithm to base 10 we have I NAvogadroσ bc Absorbance = log = = εbc (1) I 2.303 0 The absorbance is a commonly measured quantity in chemistry. The ratio I/I0 is called the transmittance. Both quantities depend on the wavelength o the light. I0 The ratio T = is known as Transmittance, so that Absorbance = log( T ). I In class exercise: How big isε?. A typical bond length is on the order o 1 A o. Let s say a molecule is 2A in diameter. A zero order approximation to the absorption cross-section is the area occupied by one molecule. Thus, 2 7 2 16 2 σ 2πr = 2 π(0.1 10 cm) = 6.2 10 cm Now we can estimate ε, 2 2 N Avogadroσ 23 molecules 16 cm 1 8 cm ε = = 6.02 10 6.2 10 = 3.73 10 2.303 mol molecule 2.303 mol Now we should convert the units to those used the lab so we multiply and divide by litre, 2 2 2 8 cm 8 l cm 8 l cm ε = 3.73 10 = 3.73 10 = 3.73 10 1000 3 mol l mol cm mol l = 3.73 10 = 3.73 10 M cm mol cm 5 5 1 1 Comments: In these units epsilon receives the name molar absorption coeicient or molar absorptivity. This igure is about a actor o two bigger than the largest observed values 5 1 1 or ε. The blue dye Cy5 has a molar absorption o ε Cy 5 = 2.5 10 M cm, the dye Cy3 5 1 1 has a molar absorption coeicient o ε Cy 5 = 1.5 10 M cm. The agreement o our cheesy calculation is remarkably good!. However, this is only to get a coarse idea o orders o magnitude. This is by no means the way the absorption cross-sections are calculated at all!. The absorption cross-section is a sharp unction o the wavelength.
These coeicients are exceptionally large, and most molecules have much smaller absorption coeicients or reasons that will be discussed in the ollowing section. Probabilities and rates o transitions To simpliy things, we do not worry about all the possible transitions that can occur in a molecule, and consider a system containing only two energy levels, the ground state(l) and the excited state(u). Only one electron is being excited by absorption o a photon o requency ν, and this requires an energy E = hν. Einstein proposed that i P is the probability o being in the excited state, then the rate o change o that probability is dp w = dt For a system with N molecules the overall rate o change o the probability o being in the excited state is W = Nw. Einstein proposed that a photon can stimulate the excitation o the molecule, and that the rate W is proportional to the radiation density ρ( ν ), which is the amount o energy available at a given requency: W = Nw= NBρ( ν ). For the black body radiation, (which accurately describes the emission o an incandescent ilament) the radiation density is 3 8πν h 1 ρν ( ) = (2) 3 hν c kt B e 1 Here, h is Planck s constant, c is the speed o light and kb is Boltzmann s constant. Einstein also proposed that radiation will stimulate a molecule that is in the excited state to return to the ground state with emission o a photon. The rate or this stimulated emission has a rate W ' = N' w= N' B' ρ( ν ), where N is the number o molecules in the excited state. When the system is being illuminated, an equilibrium state will be achieved, when the rate o change o the probability o being in the excited state equals the rate at which the excited state is de-populated by stimulated emission: NBρ( ν) = N ' B ' ρ( ν) but this implies that the ratio o the populations in the excited and ground state does not obey a Boltzmann distribution, which is known to be true: hν N' B kt B = e Inconsistent with experiments!. The ratio o the populations is N B' known to depend on T, while Einstein s prediction does not depend on T. To solve this discrepancy Einstein proposed that there must be another process that removes population rom the excited state. This additional process is independent o the radiation density and it is calls SPONTANEOUS emission. Then the total rate o change o the probability o being in the excite state is: w' = B' ρ( ν) + A and W ' = N'( B' ρ( ν) + A) At thermal equilibrium, the rate o populating the excited state must equal the rate o removing population rom it:
NBρ( ν) = N '( B ' ρ( ν) + A) NBρ( ν) = N ' B ' ρ( ν) + N ' A NBρ( ν) N ' B ' ρ( ν) = N ' A N' A A N' ρν ( )( NB N ' B ') = N ' A ρν ( ) = = NB N ' B ' B B ' ( N N ' ) B A N' A 1 ρν ( ) = = BN ' N B' N B' ( ) B ( ) N' B N' B now we replace the ration N/N by a Boltzmann distribution which is known to be true, ρν ( ) = A B hν 1 kt B ' B ( e ) B In order or this radiation density to be equal to that o the black body radiation, we must 3 8πν h have B = B' and A= B. Thus A, B and B are all related. 3 c The most important consequence o this result is the dependence o the rate o spontaneous emission on the requency to the third power. This means that the larger the energy involved in a excitation, the larger the rate at which this state will SPONTANEOUSLY return to the ground state. For this reason excited states prepared by excitation in the UV and visible live shorter that states that require an inrared or microwave photon to be excited. Question: What consequences does this have i you are trying to build a laser? A, and B should only depend on the wave unctions describing the electron under consideration. It can be shown by using perturbation theory that 2 2 2 U μ L τ U μ L where μ qr B Ψ Ψ d = M = Ψ Ψ = is the dipole moment. This integral is called transition dipole moment, and it determines how likely a transition is. Thus, this is what determines the magnitude o the molar absorption coeicient ε. To estimate how likely a transition is we will assume that the wave unctions are real unctions. Even with this assumption M is diicult to calculate because we do not know the exact orm o the wave unctions o any molecule. Approximations must be made to reduce this problem. Born-Oppenheimer approximation
Because the nuclei move much slowly that the electrons, we can consider that they move in a potential energy surace o the nearly static nuclei. Then, we can solve Schrödinger equation or a large number o nuclear conigurations. The potential energy o the electrons at each o the nuclear positions could now be plotted as a unction o the nuclear coordinates. This assumption is equivalent to actorizing the total wave unction as a product o the nuclear and electronic wave unctions: Ψ tot = θψ = (nuclear wave unction) (electronic wave unction) The nuclear wave unction describes the vibrational motion o the molecule. M = Ψ μ Ψ = θ ψ μ θ ψ = θ θ dτ ψ μ ψ = θ θ dτ ψ μψ dτ U L U U L L U L nuclei U L U L nuclei U L e The last equality is obtained by realizing that the dipole moment operator only operates on the electrons. The irst integral is over the nuclear coordinates, and the second one is over the electronic coordinates. The last expression is still too diicult to solve. We assume that ψ is can be represented as a product o one electron wave unctions (orbitals), which are in turn linear combinations o atomic orbitals. Furthermore, we assume that these atomic orbitals are the same in the ground(l) and excited state (U). Additionally, we assume that only one electron is being promoted during the transition, so φu and φl are the atomic orbitals in the upper (excited) and lower (ground) states. With these new assumptions M can be written as: M θ θ dτ φ μφ dτ = U L nuclei U L e Finally, we actorize the electron wave unctions into spatial and spin wave unctions φ = ϕ S. Since the dipole moment only operates on the spatial coordinates, the spin wave unctions can be separated and the inal orm or M is M θ θ dτ S S dτ ϕ μϕ dτ = U L nuclei U L spin U L e Te irst actor is the overlap between the vibrartional wave unctions in the ground and excited state. It is called Franck-Condon overlap The second actor is the spin overlap integral, which depends on the spin o the ground and excited states. The third actor is called the electronic transition moment, and its value depends on the symmetries o the ground and excited wave unctions. This separation o the wave unction in actors is equivalent to separating the complete Schrödinger equation HΨ= EΨinto 3 equations: Hϕ = Eϕ Hθ = Eθ HS = ES
the total energy is the sum o an electronic, a vibrational and spin energies E = Ee + Ev + ES. We used real unctions. The real wave unctions are complex, but the reasoning is the same in the case o complex unctions. The transition dipole moment M is an integral o a product o 3 integrals. I any one o them is zero, then M is zero, and the transition is orbidden (and it is not observed). For situations in which some o the actors are small, the transition will occur with low probability, and the value o ε will be small. Problem: Analyze qualitatively the eect o the vibrational overlap between the ground and excited states on the magnitude o ε. Hint: You have to consider the behavior o the integral θθ U Ldτnuclei Assume that the vibrational waveuncitons o the molecule are those or the harmonic 1 oscillator, θ = θ ( v) with energies Ev () = hν v+. Start with a sketch o the wave 2 unctions or the harmonic oscillator or the ground state, and or the excited state. Consider that in the ground state most molecules are in the lowest vibrational level.
(Figure scanned From Molecular Fluorescence: Principles and Applications. Bernard Valeur 2001 Wiley-VCH Verlag GmbH.) Spin The spin wave unction can only have to values, which are called α and β. For the transition between two states with the same spin, we have two cases singlet singlet and triplet triplet the integral SUSLdτ spinis either ααdτ spin = 1 or ββdτ spin = 1 because the spin wave unctions are normalized. Thus singlet singlet and triplet triplet transitions are ully allowed and can be readily observed. The case o singlet triplet and triplet singlet leads to αβdτspin = βαdτ spin = 0 because the α and β spin wave unctions are orthogonal. Thus singlet triplet and triplet singlet transitions are orbidden, and should not occur. However
Since the spin carries a magnetic moment, and the orbiting electron behaves as a current around the nucleus, these two are coupled. It is possible that some loss o spin moment is lost, and dome orbital moment is gained by the electron. Under this circumstance, a state is not purely a triplet but it might have a little o singlet character. Thus a transition that should be orbidden, occurs, but with low probability. Thus ε or an excitation that involves change in spin is usually very weak. For the electronic transition moment consider that the integral o the wave unctions is over the entire domain. Since = qris an odd unction o the coordinates, the triple product will have to be totally symmetric in order or M not to be zero. To analyze this actor, symmetry must be invoked. More over, the dipole moment is split into the x, y and z components,, and the integral is then actorized. This actor also encodes the act that the emission o the molecule is polarized along some direction with respect to the molecular rame o reerence. Radiative Lietimes: μ I we start with a population o molecules, all in the excited state, they will decay, with emission o energy in the orm o electromagnetic radiation o the proper requency, EUL = hν. The population o excited states will decay in time in a similar way as the decay o radioactive atoms (which are excited nuclear states). To see this we write dn '( t) dn '( t) = AULN'( t) = AULdt dt N '( t) N'( t) N'( t) ln N' = A ( 0) ln N'( t 0) UL t = AULt = AULt = N '(0) or: UL N'( t) = N'(0) e A t (3) This is an exponential decay wit a rate equal to the Einstein coeicient or spontaneous 1 emission. AUL has units o s (reciprocal seconds), so it is a requency. The reciprocal o 1 AUL is τ 0 = is a time in seconds per transition, and it represents the average time that a A UL molecule will spend in the excited state. It is called radiative lietime or Natural lietime. What else can an excited state do with the extra energy? Jablonski Diagrams: A convenient way o representing phenomena involving excited states is an energy diagram with arrows connecting dierent states. These are called Jablonski diagrams.
(Figure scanned From Molecular Fluorescence: Principles and Applications. Bernard Valeur 2001 Wiley-VCH Verlag GmbH.) So is the SINGLET ground state, S1 s the irst excites SINGLET state, and so orth. T1 is the lowest TRIPLET state. For each electronic level, there is a maniold o vibrational levels. In turn (not drawn), or each vibrational level there is a maniold o rotational energy levels. Question: The energy o the TRIPLET state is slightly lower than that or the S1 level. Can you give a qualitative explanation or this act? A straight vertically going arrow is an excitation (stimulated absorption), and it accompanies the absorption o one photon. Straight vertical arrow pointing down represents a spontaneous or stimulated decay, and it is accompanied by the emission o one photon. Wavy lines are processes in which no photon is emitted or absorbed.
Most molecules will be in the lowest vibrational level o their ground electronic state. This is why all the absorption processes start at v=0 (lowest vibrational level o the ground electronic state). The promoted electron arrives at some high vibrational level o the S1 (no spin change) (irst excited state). From there, a ast vibrational relaxation occurs, until the molecule reaches the v=0 vibrational level o the S1 electronic state. The SPONTANEOUS emission that arises rom the S(v=0) 1 S(v) 1 is called MOLECULAR FLUORESCENCE or simply FLUORESCENCE. Notice that the emission returns the molecule to the ground electronic state, but not to the lowest vibrational level. Further vibrational relaxation occurs, which brings the molecule to its lowest energy. Because the energy o the luorescence photon emitted is lower than the energy o the absorbed photon, the emission is said to be red shited with respect to the absorption spectrum. This shit is called Stokes shit On excitation, dierent transitions can occur, and they have dierent energies depending o the vibrational level they terminate on. This gives rise to broad absorption spectrum. A similar process occurs during luorescence, where all emissions start rom the same vibration in the excited state, but end in dierent vibrations o the ground state. Thus the luorescence spectrum is broad. The absorption and emission spectra are the envelope o all the possible (allowed) electronic-vibrational transitions. On the bottom part o the Jablonski diagram we have a representation o the absorption and luorescence spectra. Notice the red shit o the luorescence spectrum with respect to the absorption spectrum. Once the singlet excited state is prepared, it can undergo a number o dierent processes. Vibrational Relaxation: First, it may non-radiatively decay by vibrational relaxation: This occurs when a low vibrational level o the irst excited singlet state has nearly the same energy as that o a high vibrational level on the ground electronic state (So). This means that the molecule transitions to the ground electronic state, but all its electronic energy is now present as a vibration with a high quantum umber. This is a very unstable, since any molecule that collides with this vibrationally excited molecule can acquire momentum, and thus the excited molecule quickly looses all the excitation energy, without emission o a photon. This process may compete with luorescence and it in act the reason why most molecules do not luoresce. Thus only very rigid molecules or which vibrations take a lot o energy are more likely to be luorescent. Intersystem Crossing (ISC): I it occurs that there is a vibration on the triplet state which has the same energy as that o the vibrational level o the singlet state in which the excited molecule resides on, then the molecule can transition (at constant energy) rom the singlet state, to the triplet state, which involves changing the spin o the electron. This is called intersystem crossing. Once in the triplet state, the molecule will irst relax to the bottom o the vibrational ladder and rom the v=0 vibration o the triplet state it can decay to the ground state by emission o a photon. This process is called PHOSPHORESCENCE. Notice that the spin o the electron has changed twice or each emitted phosphorescence photon. Because this spin change is orbidden, it has a low probability. Thus, the molecule needs to attempt the process many times beore it can happen. Thus Phosphorescence is a much slower
process, and the molecule may be trapped in the triplet state or a time between microseconds to second. Once in the triplet state, the molecule can also undergo ISC back to the irst excited state, and then luorescence may occur. Because the luorescence photon has been produced ater some time in the triplet state, this is called DELAYED FLUORESCENCE. In addition, ISC to a high vibrational level o the ground state (So) can occur, leading to vibrational relaxation. Photochemistry A molecule in the excited state can undergo a chemical transormation. Because luorescence and vibrational relaxation are very ast, the only way the reaction can happen is that it occurs beore the molecule can relax, and this means that most likely some rearrangement that does not require the interaction with another molecule will take place. Bringing in another molecule will be a diusion controlled process, and it will be much more unlikely. Triplet states have a larger chance to react, and this indeed occurs. One typical reaction o triplets is the reaction with oxygen. This molecule is also a triplet, and it can react chemically with the triplet oxidizing the molecule beore phosphorescence. The triplet and the molecule- are lost, and a dierent molecule is ormed. Oxygen may also take the energy o the triplet without destroying the molecule. This deactivation produces oxygen in a vibrationally excited state, which releases the energy to the surrounding molecules by collisions. This process is called TRIPLET QUENCHING (by oxygen). In general when phosphorescence is to be studied, rigorous exclusion o oxygen must be done to prevent this. All the photophysical processes are summarized in the ollowing table: Avrev. Name Timescale /s Timescale typical units A Absorption 15 10 s Femtoseconds VR Vibrational Relaxation 12 10 10 10 s Picoseconds F Fluorescence 10 7 10 10 s ps-ns ISC Intersystem crossing 10 8 10 10 s ps-ns IC Internal Conversion 11 9 10 10 s ps-ns 6 P Phosphorescene 10 1s Microsecondsseconds Quantum Yield o luorescence The number o excited states that return to the ground state by emission o a luorescence photon divided by the total number o excited states is a measurement o how EFFICIENT the molecule was at producing luorescence. Since it takes one absorbed photon to create an excited state, the quantum yield is deined as Φ = nluorescence photons Iluorescence n I (4) absorbed photons Absorbed
The quantum yield relects the competition o luorescence with other processes that remove excited states without producing photons. It is more useul to have a measurement that is based on the rates o these processes. Lets consider that the excited state can relax by luorescence with a rate k or non radiatively, with a rate knr We can write a dierential equation or the change in the number o molecules that are in the excited state at a given time as dn '( t) dn '( t) = kn'( t) kn'( t) = ( k + knr) N'( t) = ( k + knr) dt dt N '( t) N'( t) N'( t) ln N' = ( k )( 0) ( ) ln ( ) '( 0) + k N t t = k + knr t = k + knr t = N '(0) or: ( k knr) t N'( t) = N'(0) e + (5) This s similar to our discussion o the radiative lietime except that now the excited states decay aster. But it is still a single exponential decay. The reciprocal o ( k + knr) has units o times, and it is called FLUORESCENCE LIFETIME 1 τ = (6) k + k NR The rate o emission o photons by luorescence will be proportional to the number o excited states present at a given time. Because the intensity o luorescence is the number o luorescence photons emitted per unit time, it is equal to the rate o luorescence, k multiplied by the number o excited states present N. Then the quantum yield can be written as: nluorescence photons Iluorescence k N' k τ Φ = = = = n I ( k + k ) N' k + k τ absorbed photons Absorbed NR NR 0 (7) Thereore there are several ways to measure the quantum yield o luorescence. One is by measuring intensities, and another is by measuring the rate o decay o the molecule and comparing this to k which can be calculated or even estimated rom other experiments. Time resolved measurements: Fluorescence occurs ast, but it still takes rom hundreds o ps to several ns to occur. Thus, we can use a very short pulse o light to excite the molecules, in less than say 50 ps. Then we monitor the luorescence in real time and plot the intensity as a unction o time. This is called TIME RESOLVED FLUORESCENCE and it is a very valuable tool. Ater the excitation pulse, equation 5 applies. Then we can measure the sum o the radiative and non radiative rates. Special equipment is needed to detect ast changes in the luorescence intensity (time correlated photon counting) Steady state measurements:
Normally luorescence is measured in a simple instrument in which light rom a lamp passes through a monochromator to select a narrow excitation wavelength range. This is then sent to the sample, and luorescence which radiated in all directions is collected at 90 degree rom the illumination direction. This way, the luorescence light which is much weaker than the excitation beam can be measured. The luorescence is passed through a second monochromator and then it reaches a detector. Usually a photomultiplier tube is used as a detector. To collect a luorescence spectrum, the second monochromator is scanned and the emission intensity is recorded as a unction o the wavelength o the second monochromator. I the excitation intensity is constant, then there will be a constant number o molecules per unit time that are being excited. At the same time, they are decaying rom the excited state at a rate k multiplied by the number o excited states present N (t). We have or the absorption process (excitation) I abs A+ hν A* and the rate o excitation equals the absorbed intensity I abs. On the other hand, or the emission we have k * A A+ hν and the rate o change o the number o excited states N is dn '( t) = ktotaln'( t) dt where k total = k is the sum o all the rates o processes that remove population rom i i the excited state (radiative and non radiative). A steady state is achieved when the rate o decay and excitation are equal, dn '( t) = Iabs ktotal N'( t) =0 dt rom this we get I I εbc I ( t) k N'( t) N'= abs abs 0 = total (8) ktotal ktotal Since the intensity o luorescence is proportional to N with proportionality constant k, it ollows that the intensity o luorescence is k I0εbC I = kn' = (9) k total
We can also combine (9) with the deinition o quantum yield to get rid o the rate k, k I0εbC I = =Φ I0εbC (10) k total This result shows that the luorescence intensity is proportional to the excitation intensity and to the concentration C o ground states. This assumes that the raction o= excited states is kept small, otherwise there will be no molecules to excite!. Also (8) is an approximation o the absorbed intensity that implies that the absorbance is small (typically smaller than 0.05). A series expansion o the exponential was done to get eqn (8) rom (1). The linear dependence o luorescence intensity with concentration grants the grounds or analytical chemistry applications.