Chemistry 2

Lecture 12 Molecular Photoppyhysics Assumed knowledge Electronic states are labelled using their 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. spectra are red‐shifted compared to absorption spectra but commonly have “mirror symmetry”

Learning outcomes • Be able to explain Kasha’s law by describing internal conversion • Be able to define fluorescence quantum yield • Be able to describe intersystem crossing and how it leads to phosphoresence • Be able to explain why the occurs at lower energy (“red‐shifted”) and is slower than fluorescence last lecture…

S1

T1

CH3 CH 3 CH3 O S0 CH3 trans-retinal CH3 (light absorber in eye) Absorption spectrum of a dye

200 S3

150

S2 nce S3 aa 100 S2 S1 bsorb AA

50 S1

0 S 400 450 500 550 600 0 Wavelength (nm) Absorption to several electronic states A = εcl

Benzene Showing singlet and triplet absorption

S 3 Triplet st

S2

T2

S1

T1

S0

Absorption to triplet states from single states is formally forbidden, thus very weak. Fluorescence spectrometer

Key features: Two monochromators are needed ‐ one to select a single wavelength to excite the molecule ‐ the other to resolve the emitted wavelengths Excitation and detection occur at 90° to each other ‐ this minimises the amount of incident light gets into the detector. Fluorescence spectrometer

Two types of spectra: Fluorescence spectrum ‐ select excitation λ with mono #1 ‐ scan mono #2 to measure fluorescence spectrum Excitation spectrum ‐ later Real data…

Absorption Fluorescence

500 550 600 650 700

WlWavelength th() (nm) ‐ Fluorescence is always to longer wavelength ‐ Stokes shift = (absorbance max) – (fluorescence max) = 50 nm here ‐ Mirror symmetry Fluorescence spectrum

≠ f(λexc)

NRD

Note: the non-radiative decay (NRD) only occurs in the condensed phase, where the can transfer vibration energy to the solvent. Stokes shift

Absorption

The shift between λmax(()abs.) and λmax(()fluor) is called the STOKES SHIFT A bigger Stokes shift will produce more dissipation of heat Franck‐Condon Principle (in reverse) y gg Ener

012345 R Franck‐Condon Principle (in reverse) y gg Ener

Note: If vibrational ffqrequencies in the ground and exited state are similar, then the spectra look the same, but

012345reversed -> the so-called R “mirror symmetry” Absorption spectrum of another dye

Excite dye at 200 different wavel ength s λ = 400 nm

150 λ = 440 nm rbance

oo 100

Abs λ = 550 nm

50

0 400 450 500 550 600 Wavelength (nm)

What will emission (fluorescence) spectra look like? The emission is the same!

λex= 400 nm

λex = 440 nm λex = 550 nm

* * *

400 450 500 550 600 650 700 750 800 Wavelength (nm) Kasha’s Law: “Emission always occurs from th thle lowes t excit itdled elec troni c s tt”tate”

Internal S2 Conversion (IC)

NRD

S1 ission Em

S0

“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. 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 factor Large energy gap = Bad Franck-Condon factor

A 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. Fluorescence spectrometer

Two types of spectra: Fluorescence spectrum ‐ select excitation λ with mono #1 ‐ scan mono #2 to measure fluorescence spectrum Excitation spectrum - select fluorescence λ with mono #2 Under what conditions will the excitation - scan excitation λ using mono #1. spectrum resemble the absorption spectrum? Excitation Spppyectroscopy

• 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 lig ht: ‐ fluorescence excitation ‐ phhhosphorescence excitati on ‐ resonant ionisation ‐ phfhotofragment exciiitation 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) = φf x N (absorbed photons)

• The proportionality constant, φf is called the “fluorescence quantum yield”

• φf can vary between 0 and 1

• If φf is constant with λ, then fluorescence excitation spectrum has the same shape (intensity vs λ) as the absorption spectrum Comparison of Absorption and Excitation Spectra

Note the extended π‐chromophore, which is responsible for the absorption. Multiple electronic states

1500 Excitation Abs orption

Rhodamine 6G

0 200 250 300 350 400 450 Wavelength (nm) When absorption ≠ excitation…

Especially note this difference??

2000

Pyrenesulfonic acid

0 200 250 300 350 Wavelength (nm) More on Internal Conversion

IC

S2

Smaller energy gap IC

S1

Larger energy gap

IC is usually very fast between excited states and slower between

S1 and the ground state. S0

Internal 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 Intersystem Crossing (ISC) and Phosphorescence

ISC

S2 T2 NRD & IC S1

ISC

T1 Phosph orescence

S0

“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. More on phosphorescence Triplet states

S2 T2

S1

T1 scence ee Phosph orescence Fluor

S0 The 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. Everything together…

ISC IC

S2 T2 ISC IC

S1 ISC rption cence T oo ss 1

Abs Phosphorescence Fluore

S0 Phosphorescence in research

Absorpti on phosphorescence spectrum

S1←S0 S2←S0

This porphyrin molecule exhibits a huge π-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. 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 μs timescale. 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, 40000 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 Summary • Emission always occurs from the lowest excited electronic state (Kasha’s law) • Internal conversion (()IC) is the non‐radiati ve change process by which an electron relaxes to a lower energy state with the energy converted to vibrational energy

• Fluorescence quantum yield, φf, is the proportionality constant linking the number of photons emitted by fluorescence to the number absorbed

• If φf is constant with the wavelength of the excitation, the fluorescence spectrum has the same shape as the absorption spectrum • Intersystem crossing is the non‐radiative change process by which a molecule changes spin • Phosphorescence normally occurs from the lowest triplet state and is nearly always “red shifted” and slower than fluorescence. Next lecture

• Course wrap up

Week 13 homework

• Electronic spectroscopy worksheet in the tutorials • Complete the practice problems at the end of the lectures • NtNote: ALL of the reltlevant past exam problems have been used as practice problems. Other questions on past papers include parts which are no longer part of the course. Practice Questions 1. The figure opposite shows the absorption, fluorescence and phosphorescence spectra of a common organic dye. Why is the ppphosphorescence spectrum significantly red shifted compared to the fluorescence spectrum?

2. 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 to illustrate your answer.

(a) The Stokes shift is quite different for molecule A and B. Explain how this dffdifference arises, and give an example of what molecular property might give rise to a large Stokes shift. ()(b) Molecule B in particular is a nice example of “mirror symmetry” between excitation and emission spectra. How does this mirror symmetry arise? Practice Questions 3. The two spectra below show the fluorescence excitation and absorption spectra of pyrenesulfonic acid. (a) 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.

(b) 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. (c) 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