Chemistry 2

Lecture 12 Molecular Photophysics

Electronic 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”

Assumed knowledge

Learning outcomes from lecture 11• Be able to use S, D and T to label the spin multiplicity of an

electronic state• Be able to describe how energy is lost after absorption by

radiative transitions and non-radiative transitions• Be able to explain the “mirror symmetry” and Stokes shift of

absorption and fluorescence spectra explained using a Jablonski diagram

last lecture…

S0

S1

T1

O

CH3

CH3CH3

CH3 CH3

trans-retinal(light absorber in eye)

Absorption spectrum of a dye

S2

S1

S0400 450 500 550 600

0

50

100

150

200

S2

Ab

so

rban

ce

Wavelength (nm)

S3

S1

S3

Absorption to several electronic states

Benzene

clA

T1

T2

Triplet states

S2

S1

S0

Showing singlet and triplet absorption

S3

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 wavelengthsExcitation 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 l with mono #1 - scan mono #2 to measure fluorescence spectrumExcitation spectrum - later

500 550 600 650 700

Wavelength (nm)

Absorption Fluorescence

Real data…

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

Fluorescence spectrum f(lexc)

NRD

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

Stokes shift

Absorption

The shift between lmax(abs.) and lmax(fluor) is called the STOKES SHIFTA bigger Stokes shift will produce more dissipation of heat

Franck-Condon Principle (in reverse)

0 1 2 3 4 5

En

erg

y

R

0 1 2 3 4 5

En

erg

y

R0 1 2 3 4 5

En

erg

y

R

Franck-Condon Principle (in reverse)

0 1 2 3 4 5

En

erg

y

R

0 1 2 3 4 5

En

erg

y

R0 1 2 3 4 5

En

erg

y

R

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”

Absorption spectrum of another dye

Excite dye at different wavelengths

400 450 500 550 600

0

50

100

150

200

Ab

sorb

ance

Wavelength (nm)

l = 550 nm

l = 440 nm

l = 400 nm

What will emission (fluorescence) spectra look like?

*

**

lex= 400 nm

400 450 500 550 600 650 700 750 800

Wavelength (nm)

The emission is the same!

lex = 440 nm lex = 550 nm

Em

issi

on

NRD

Kasha’s Law: “Emission always occursfrom the lowest excited electronic state”

S2

S1

S0

InternalConversion

(IC)

“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 BreakdownTerms 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 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?

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

• 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 1

• If ff is constant with l, then fluorescence excitation spectrum has the same shape (intensity vs l) as the absorption spectrum

Explanation of fluorescence excitation

Comparison of Absorption and Excitation Spectra

0

1500

200 250 300 350 400 450

Wavelength (nm)

Excitation

Absorption

Rhodamine 6G

Note the extended p-chromophore, which is responsible for the absorption. Multiple electronic states

0

2000

200 250 300 350

Wavelength (nm)

When absorption excitation…

Pyrenesulfonic acid

Especially note this difference??

More on Internal Conversion

S2

S1

S0

IC

IC

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

Smaller energy gap

Larger energy gap

IC is usually very fast between excited states and slower between S1 and the ground state.

T1

T2

S2

S1

S0

Intersystem Crossing (ISC) and Phosphorescence

Phosphorescence

“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.

NRD & IC

ISC

ISC

T1

T2

Triplet states

S2

S1

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.

Phosphorescence

Flu

ore

scen

ce

More on phosphorescence

S2

S1

S0

IC

IC

T1

T2

Ab

sorp

tion

Flu

ore

scen

ce

ISC

ISC

Everything together…

ISC

Phosphorescence

Phosphorescence in research

Absorptionspectrum

phosphorescence

S1←S0S2←S0

This 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.

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.

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

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 phosphorescence occurs at lower

energy (“red-shifted”) and is 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

• Note: 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.

Practice Questions

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 Jablonski diagram to illustrate your answer.

(a) 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.

(b) Molecule B in particular is a nice example of “mirror symmetry” between excitation and emission spectra. How does this mirror symmetry arise?

1. The 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?

Practice Questions3. 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