2 Learning outcomes from lecture 11 Be able to use S, D and T to label the spin multiplicity of an electronic stateBe able to describe how energy is lost after absorption by radiative transitions and non-radiative transitionsBe able to explain the “mirror symmetry” and Stokes shift of absorption and fluorescence spectra explained using a Jablonski diagramAssumed knowledgeElectronic states are labelled using their spin multiplicity with singlets having all electron spins paired and triplets having two unpaired electrons. After absorption, energy is lost by radiative transitions and non-radiative transitions. Fluorescence spectra are red-shifted compared to absorption spectra but commonly have “mirror symmetry”
3 (light absorber in eye) last lecture…S1T1OCH3trans-retinal(light absorber in eye)S0
4 Absorption spectrum of a dye 200S3150S2S3Absorbance100S2S150S1400450500550600S0Wavelength (nm)
6 Showing singlet and triplet absorption Triplet statesS3S2T2S1T1S0Absorption to triplet states from single states is formally forbidden, thus very weak.
7 Fluorescence spectrometer Key features:Two monochromators are needed- one to select a single wavelength to excite the molecule- the other to resolve the emitted wavelengthsExcitation and detection occur at 90 to each other- this minimises the amount of incident light gets into the detector.
8 Fluorescence spectrometer Two types of spectra:Fluorescence spectrum- select excitation l with mono #1- scan mono #2 to measure fluorescence spectrumExcitation spectrum- later
9 Real data… Absorption Fluorescence 500550600650700Wavelength (nm)AbsorptionFluorescence- Fluorescence is always to longer wavelength - Stokes shift = (absorbance max) – (fluorescence max) = 50 nm here - Mirror symmetry
10 Fluorescence spectrum f(lexc) NRDNote: the non-radiative decay (NRD) only occurs in the condensed phase, where the fluorophore can transfer vibration energy to the solvent.
11 A bigger Stokes shift will produce more dissipation of heat AbsorptionThe shift between lmax(abs.) and lmax(fluor) is called the STOKES SHIFTA bigger Stokes shift will produce more dissipation of heat
13 Franck-Condon Principle (in reverse) Note: If vibrational frequencies in the ground and exited state are similar, then the spectra look the same, but reversed -> the so-called “mirror symmetry”
14 Absorption spectrum of another dye Excite dye atdifferent wavelengths40045050055060050100150200AbsorbanceWavelength (nm)l = 550 nml = 440 nml = 400 nmWhat will emission (fluorescence) spectra look like?
15 The emission is the same! lex= 400 nmlex = 440 nmlex = 550 nm***400450500550600650700750800Wavelength (nm)
16 Kasha’s Law: “Emission always occurs from the lowest excited electronic state”InternalConversion(IC)S2NRDS1EmissionS0“Internal conversion (IC)” is the spontaneous relaxation of an electron to a lower energy state, accompanied by a simultaneous increase of vibrational energy of the molecule. IC is a non-radiative process.
17 Born-Oppenheimer Breakdown Terms in the Hamiltonian operator which are ignored when making the BO approximation promote transitions between states of the same energy in the molecule.Small energy gap= Good Franck-Condon factorLarge energy gap= Bad Franck-Condon factorA molecule can make a non-radiative transition to an isoenergetic state. If this state can then lose vibrational energy to the solvent, it is irreversible.
18 Fluorescence spectrometer Two types of spectra:Fluorescence spectrum- select excitation l with mono #1- scan mono #2 to measure fluorescence spectrumExcitation spectrum- select fluorescence l with mono #2- scan excitation l using mono #1.Under what conditions will the excitation spectrum resemble the absorption spectrum?
19 Excitation Spectroscopy Absorption is a DIRECT technique for measuring an electronic transition – the direct loss of transmitted light is measured.There are other techniques to infer the absorption of light:- fluorescence excitation- phosphorescence excitation- resonant ionisation- photofragment excitation
20 Explanation of fluorescence excitation Because the fluorescence is the same no matter where the molecule is excited, then any (or all) fluorescence transitions can be monitored.N (fluorescence photons) N (absorbed photons)N (fluorescence photons) = ff x N (absorbed photons)The proportionality constant, ff is called the“fluorescence quantum yield”ff can vary between 0 and 1If ff is constant with l, then fluorescence excitation spectrum has the same shape (intensity vs l) as the absorption spectrum
21 Comparison of Absorption and Excitation Spectra Note the extended p-chromophore, which is responsible for the absorption.Multiple electronic statesRhodamine 6G
22 When absorption excitation… Especially note this difference??Pyrenesulfonic acid
23 More on Internal Conversion ICS2Smaller energy gapICS1Larger energy gapIC is usually very fast between excited states and slower between S1 and the ground state.S0Internal conversion is the relaxation of the electron to a lower level, but, accompanied by no radiation, the equivalent amount of energy is converted to vibrational energy. IC is a radiationless process
24 Intersystem Crossing (ISC) and Phosphorescence NRD & ICS1ISCT1PhosphorescenceS0“Intersystem crossing” is the flipping of an electron spin so that the molecule changes from singlet to triplet state (or vice versa). ISC is a non-radiative process and is typically 106 times slower than IC, other things being equal.
25 More on phosphorescence T1T2Triplet statesS2S1FluorescencePhosphorescenceS0The lowest triplet is nearly always below the first excited singlet state. Therefore phosphorescence is nearly always “red shifted” (i.e. at lower energy) than fluorescence. It is formally forbidden and about 106 times slower than fluorescence.
26 Everything together… ISC IC S2 T2 ISC IC S1 ISC T1 Phosphorescence S0 AbsorptionS2T2ISCICS1FluorescenceISCT1PhosphorescenceS0
27 Phosphorescence in research AbsorptionspectrumphosphorescenceS1←S0S2←S0This porphyrin molecule exhibits a huge p-system and absorbs across the visible region. The palladium metal centre promotes intersystem crossing. The molecule was synthesized in the Crossley group, and used by Schmidt’s group in solar research.
28 Ultrafast fluorescence Using ultrafast lasers, we can observe the porphyrin in the act of fluorescing. Here, after excitation at 300 nm, fluorescence at 680 nm builds up within 3 ps but gone after only 20 ps after which the molecules which are left are all in the T1 state and then phosphoresce on the 20 ms timescale.
29 How fast?One picosecond is one millionth of one millionth of a second. There are as many picoseconds in a second as there have been seconds since our species invented shoes, years ago. The bisporphyrin below has a very low S1 state. Ultrafast transient absorption experiments show that it does not undergo ISC, but rather IC is complete in 50ps. This experiment is 2 weeks old.S1←S0
30 Learning outcomesBe able to explain Kasha’s law by describing internal conversionBe able to define fluorescence quantum yieldBe able to describe intersystem crossing and how it leads to phosphoresenceBe able to explain why the phosphorescence occurs at lower energy (“red-shifted”) and is slower than fluorescence
31 Next lecture Week 13 homework Course wrap upWeek 13 homeworkElectronic spectroscopy worksheet in the tutorialsComplete the practice problems at the end of the lecturesNote: ALL of the relevant past exam problems have been used as practice problems. Other questions on past papers include parts which are no longer part of the course.
32 Practice QuestionsThe figure opposite shows the absorption, fluorescence and phosphorescence spectra of a common organic dye. Why is the phosphorescence spectrum significantly red shifted compared to the fluorescence spectrum?The spectra below show the fluorescence excitation (blue) and fluorescence emission spectrum (red) of two large molecules. Explain, the following features of the spectra, using a Jablonski diagram to illustrate your answer.The Stokes shift is quite different for molecules A and B. Explain how this difference arises, and give an example of what molecular property might give rise to a large Stokes shift.Molecule B in particular is a nice example of “mirror symmetry” between excitation and emission spectra. How does this mirror symmetry arise?
33 Practice QuestionsThe two spectra below show the fluorescence excitation and absorption spectra of pyrenesulfonic acid.In addition to the identified electronic origin transitions, there are other peaks in the absorption spectrum, as indicated by “a)” in the figure. Using a Jablonski diagram, explain how these other peaks arise.In the absorption spectrum, for the S1 and S2 transitions, the origin band is stronger than the two satellite bands marked by “a)”. In the S3 transition, the origin band is weaker than the satellite, marked by “b”. Explain, using the Franck-Condon principle, how this arises.Using a Jablonski diagram, describe one such process that can give rise to the observed difference in the relative intensities of the fluorescence excitation and absorption spectra at λ < 245 nm