Chemistry 2
Lecture 12
Molecular Photophysics
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
Assumed knowledge
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”
last lecture…
S1
T1
CH3
CH3
CH3
O
trans-retinal
CH3
CH3
(light absorber in eye)
S0
Absorption spectrum of a dye
200
S2
S1
Absorbance
S3
150
S3
100
S2
50
S1
0
S0
400
450
500
550
Wavelength (nm)
600
Absorption to several electronic states
A   cl
Benzene
Showing singlet and triplet absorption
S3
Triplet
states
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 l with mono #1
- scan mono #2 to measure fluorescence spectrum
Excitation spectrum
- later
Real data…
Absorption
500
550
Fluorescence
600
650
700
Wavelength (nm)
- 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 SHIFT
A bigger Stokes shift will produce more dissipation of heat
E nergy
Energy
Energy
Franck-Condon Principle (in reverse)
0
1
2
3
4
5
0
1
2
3
4
5
R
R
E nergy
Energy
Energy
Franck-Condon Principle (in reverse)
0
1
2
3
4
5
0
1
2
3
4
5
R
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
Absorbance
200
l = 400 nm
Excite dye at
different wavelengths
150
l = 440 nm
100
l = 550 nm
50
0
400
450
500
550
Wavelength (nm)
What will emission (fluorescence) spectra look like?
600
The emission is the same!
lex= 400 nm
lex = 440 nm
lex = 550 nm
*
*
*
400
450
500
550
600
650
700
Wavelength (nm)
750
800
Kasha’s Law: “Emission always occurs
from the lowest excited electronic state”
Internal
Conversion
(IC)
S2
NRD
Emission
S1
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 l with mono #1
- scan mono #2 to measure fluorescence spectrum
Excitation 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
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 1
• If ff is constant with l, then fluorescence excitation spectrum
has the same shape (intensity vs l) as the absorption spectrum
Comparison of Absorption and
Excitation Spectra
Note the extended p-chromophore, which is
responsible for the absorption.
Multiple electronic states
1500
Excitation
Absorption
Rhodamine 6G
0
200
250
300
350
Wavelength (nm)
400
450
When absorption  excitation…
Especially note this
difference??
2000
Pyrenesulfonic acid
0
200
250
300
Wavelength (nm)
350
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
Phosphorescence
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
Fluorescence
S1
T2
T1
Phosphorescence
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
ISC
Absorption
Fluorescence
IC
S0
T2
S1
ISC
T1
Phosphorescence
Phosphorescence in research
Absorption
spectrum
S2←S0
phosphorescence
S1←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
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?
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?
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 FranckCondon 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