Physical Chemistry VIII_ Photochemistry

Physical Chemistry VIII
Course Code 5449 Prerequisite 5342 Credit Hours 2
Lectured by: Dr. Fateh Eltaboni
Assistant Professor of Physical Chemistry at the University of Benghazi
University of Benghazi
Faculty of Science
Chemistry Department
5449 Lecture Notes (Dr Fateh Eltaboni) 1

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Photochemistry
Photochemical
principles; Rates of
intramolecular
processes and
Intermolecular
processes
Energy transfer
Photochemical
reactions and
their quantum
yields
Photochemistry
in nature
Special topics in
photochemistry.

What is
Photochemistry?
Its reaction of matter with light!
Mr. Photo

CHARACTERISTICS OF LIGHT
• Light = band of waves of different wavelengths
Increase in wavelengths (nm)
Increase in frequency (energy)

(Light) Electromagnetic spectrum important for photochemistry:

• The dual nature of light: It means that, in some experiments, light behaves
as a wave, In other experiments, light behaves as a particle.
• Wavelength (): Distance between successive crests or troughs
• Frequency (): Number of crests or troughs that pass a point per second
• Speed of light (c) c =  
• Solar energy (Es): (Es = h) It’s the energy come from the sun or other
radiant source.

PARTICLE CHARACTERISTICS OF LIGHT:
• Light = flux of discrete units (i.e quanta) called photons.
Fireworks:

ABSORBANCE, COLOUR AND TRANSPARENCY OF MATERIALS:
Transparent materials absorb no light in the visible wavelength range, and
hence have no energy states separated by between 1.65 and 3.10eV.
Examples:
SiO2 (glass, quartz);
Water soluble polymers like:
Poly(acrylic acid) (PAA)C(Diamond);

ABSORBANCE, COLOUR AND TRANSPARENCY OF MATERIALS:
Opaque materials: are absorb light in all the visible wavelength range,
and hence have many energy states separated by between 1.65 and
3.10eV.
Examples:
• C(graphite) also absorbs light at all wavelengths, but is black and not
reflective like a metal.
• Some metals (Cu, Fe…)
are coloured as well as reflective.

• Dyes and many other materials may
be coloured because they absorb
some of the wavelengths in the
visible range.
UV Absorbance – Sunscreens:
• Absorbance in the ultraviolet (UV) range in
important for transparent materials. UV is
higher energy light, and UV absorbance
can lead to photochemical reactions and
the formation of highly reactive free
radicals. As with ionizing radiation these
can lead to cell damage and (skin) cancer.

Mechanism of UV-degradation
of polymer chain In presence
of O2:
UV-light

• Sunscreens contain a mixture of compounds, like polymers, that
absorb UV light from the solar spectrum:

• The "Z" in Structure I is selected from a UV-chromophore:

Relation between wavelength and conjugation
Absorbance(a.u)

o Ordinary reaction (thermal or dark reaction):
o Its occurs by absorption of heat energy from outside in absence of light.
1. The reacting molecules are energised
2. Molecular collision is occurred
3. Product is formed
Steps of Ordinary reactions

o Photochemical reaction:
o Its a reaction which takes place by absorption of the visible and ultraviolet
radiations (200-800 nm)
o Photochemical reaction
leads to the heating of
the atmosphere during
the daytime by
absorption of ultraviolet
radiation.
• Mechanism of photochemical reactions occurring during atmospheric:
Photochemical change occurs only by absorption of photons.

o Photochemical reaction includes the absorption of visible radiation
during photosynthesis.
• Without photochemical processes, the Earth would be simply a warm,
sterile, rock.
Schematic of photosynthesis in plants.
The carbohydrates produced are stored
in or used by the plant.
Overall equation for the type of
photosynthesis that occurs in plants

o Photochemistry:
o Its the branch of chemistry which deals with the study of photochemical
reactions.
o Demonstration of a Photochemical reaction

o Difference between photochemical and thermochemical reactions

ENERGY TRANSITIONS:
• Molecules absorb radiation by increasing internal energy:
Internal energy  electronic, vibrational, & rotational states
• Energy requirements:
(1). Electronic transitions
 UV-vis ( 400-700 nm)
(2). Vibrational transitions
 Near-IR (700-2000 nm)
(3). Rotational transitions
 Far-IR-Mic (> 2000 nm)

Typical electronic transitions: Typical Molecular vibrations
Typical Molecular rotations

What Happens When Radiation Hits a Molecule?

Absorption and Emission:
• An electronic transition occurs when an electron changes from state to
another.
• The orbital it is leaving must be partly filled (it contains at least one
electron), and the orbital it is entering must be partly unfilled
(it contains less than two electrons).

Absorption or Absorbance: An electron absorbs the energy of a
photon, and jumps from a lower into a higher energy orbital.
Emission: An electron jumps from a higher into a lower orbital, and
releases (emits) a photon of energy equal to the difference.

o Light absorption
o When light is passed through a medium, a part of it is absorbed. This
absorbed portion of light which causes photochemical reactions.
o Let a beam of monochromatic light pass through a thickness dx of the
medium. The intensity of radiation reduces from I to I-dI.
o The intensity of radiation can be defined as the
number of photons that pass across a unit area
in unit time.
o Let us denote the number of incident photons
by N and the number absorbed in thickness dx
by dN.
o The fraction of photons absorbed is then dN/N
which is proportional to thickness dx. That is,

α
o where b is proportionality constant called absorption coefficient.
o Let us set I = I0 at x = 0 and integrate. This gives
o Lambert first derived equation (1) and it is known as Lambert Law. Beer
extended this relation to solutions of compounds in transparent solvents.
The equation (1) then takes the form (2).
Lambert-Beer Law. This law forms the basis
of spectrophotometric methods of chemical
analysis.
o where C = molar concentration; ∈ is a constant characteristic of the solute called the
molar absorption coefficient.

b = x
Beer-Lambert Law :
The attenuation of a
beam by absorbance
is typically
represented in two
ways:
1. Fraction or Percent
Transmission (%T)
2. Absorbance (A)

ABSORBANCE SPECTROMETRY
• Determines concentration of a substance in solution
• Measures light absorbed by solution at a specific wavelength
• Visible region: low energy electronic transition due to:
a. Compounds containing transition metals
b. Large aromatic structures & conjugated double bond systems
• UV region (200-400 nm):
a. Small conjugated ring systems
b. Carbonyl compounds

o Determination of absorbed intensity
o A photochemical reaction occurs by the absorption of photons of light by the
molecules.
o Therefore, it is essential to determine the absorbed intensity of light for a
study of the rate of reaction.
Schematic diagram of the spectrophotometer used for measurement of light intensity
A sample UV-Vis spectrum.

1. Light beam from a suitable source (tungsten filament or mercury
vapour lamp) is rendered parallel by the lens.
2. The beam then passes through a ‘filter’ or monochrometer which yields
light of one wavelength only. The monochromatic light enters the reaction
cell made of quartz.
3. The part of light that is not absorbed strikes the detector. Thus the
intensity of light is measured first with the empty cell and then the cell
filled with the reaction sample.
4. The first reading gives the incident intensity, I0, and the second gives
the transmitted intensity, I. The difference, I0 – I = Ia, is the absorbed
intensity.
o Steps of absorbance measurements

o The detector generally used for the measurement of intensity of
transmitted light is :
(a) a thermopile (b) photoelectric cell (c) a chemical actinometer.

o LAWSOF PHOTOCHEMISTRY
1) Grothus–Draper Law (qualitative aspect)
It is only the absorbed light radiations that are effective in producing a
chemical reaction.
2) Stark-Einstein Law of Photochemical Equivalence
Stark and Einstein (1905) studied the quantitative aspect of photochemical
reactions by application of Quantum theory of light.
Illustration of Law of Photochemical
equivalence; absorption of one
photon decomposes one molecule.

o In a photochemical reaction, each molecule of the reacting
substance absorbs a single photon of radiation causing the
reaction and is activated to form the products.
o In practice, we use molar quantities. That is, one mole of A absorbs one
mole of photons or one einstein of energy, E. The value of E can be
calculated by using the expression given below:

o Primary and Secondary reactions:
The overall photochemical reaction may consist of :
(a) a primary reaction: proceeds by absorption of radiation.
(b) secondary reaction: is a thermal reaction which occurs subsequent to
the primary reaction.
For example, the decomposition of HBr occurs as follows :
o Evidently, the primary reaction only obeys the law of photochemical equivalence strictly. The
secondary reactions have no concern with the law.

o Quantum yield (or Quantum efficiency) (ᶲ)
o For a reaction that obeys strictly the Einstein law, one molecule
decomposes per photon, the quantum yield φ = 1.
o When two or more molecules are decomposed per photon, φ > 1 and the
reaction has a high quantum yield.
o If the number of molecules decomposed is less than one per photon, φ <
1 the reaction has a low quantum yield.

o Causes of high quantum yield
The chief reasons for high quantum yield are :
(1) Reactions subsequent to the Primary reaction:
(2) A reaction chain forms many molecules per photon:
✓ A reaction chain steps:
1) Initiation……………….>
2) Self-propagating……>
3) Termination………….>
The number of HCl molecules formed
for a photon of light is very high. The
quantum yield of the
reaction varies from 104 to 106

o Causes of low quantum yield
(1) Deactivation of reacting molecules:

(2) Occurrence of reverse of primary reaction:
Dimerization of Anthracene:
o The quantum yield should be 2 but it is actually found to be 0.5.
The low quantum yield is explained as the reaction is
accompanied by fluorescence which deactivates the excited
anthracene molecules. Furthermore, the above reaction is
reversible.

(3) Recombination of dissociated fragments:
Combination of H2 and Br2.
2
o The reaction (2) is extremely slow. The reactions (3), (4) and (5), depend directly or
indirectly on (2) and so are very slow. Therefore most of the Br atoms produced in the
primary process recombine to give back Br2 molecules.
o Thus the HBr molecules obtained per quantum is extremely small. The quantum yield of
the reaction is found to be 0.01 at ordinary temperature.

CALCULATION OF QUANTUM YIELD
By definition, the quantum yield, , of a photochemical reaction is expressed as :
Thus we can calculate quantum yield from :
(a) The amount of the reactant decomposed in a given time and
(b) The amount of radiation energy absorbed in the same time
o The radiation energy is absorbed by a chemical system as photons.
Therefore we should know the energy associated with a photon or a mole
of photons.

Theenergyof photons; einstein
o We know that the energy of a photon (or quantum), ∈ , is given by the
equation:
h = Planck’s constant (6.624 × 10– 27 erg-sec)
v = frequency of radiation
λ = wavelength of radiation
c = velocity of light (3 × 1010 cm sec– 1) o If λ is given in cm, the energy is expressed in ergs.
The energy, E, of an Avogadro number (N) of photons is referred to as one einstein:
Substituting the values of N (= 6.02 × 1023), h and c, in (2), we have:

If λ is expressed in Å units (1Å = 10– 8 cm),
Since 1 cal = 4.184 × 107 erg, energy in calories would be:
o It is evident from (3) that the numerical value of einstein varies inversely as the
wavelength of radiation.
o The higher the wavelength, the smaller will be the energy per einstein.
Electron-volt (eV) is another
commonly used energy unit,
1 eV = 1.6 x 10-19 J.

SOLVED PROBLEM 1. Calculate the energy associated with (a) one photon;
(b) one einstein of radiation of wavelength 8000 Å. h = 6.62 × 10–27 erg-sec; c =
3 × 1010 cm sec–1.
SOLUTION
(a) Energy of a photon
(b) Energy per einstein

SOLVED PROBLEM 2. When a substance A was exposed to light, 0.002 mole of it
reacted in 20 minutes and 4 seconds. In the same time A absorbed 2.0 × 106 photons
of light per second. Calculate the quantum yield of the reaction. (Avogadro number
N = 6.02 × 1023)
SOLUTION
Number of molecules of A reacting = 0.002 × N = 0.002 × 6.02 × 1023
Number of photons absorbed per second = 2.0 × 106
Number of photons absorbed in 20 minutes and 4 seconds = 2.0 × 106 × 1204
Quantum yield

SOLVED PROBLEM 3. When irradiated with light of 5000 Å wavelength, 1 × 10–
4 mole of a substance is decomposed. How many photons are absorbed during the
reaction if its quantum efficiency is 10.00. (Avogadro number N = 6.02 × 1023)
SOLUTION
Quantum efficiency of the reaction = 10.00
No. of moles decomposed = 1 × 10–4
No. of molecules decomposed = 1 × 10–4 × 6.02 × 1023 = 6.02 x 1019
we know that,

SOLVED PROBLEM 4. When propionaldehyde is irradiated with light of λ =
3020 Å, it is decomposed to form carbon monoxide.
CH3CH2CHO + hv ⎯⎯→ CH3CH3 + CO
The quantum yield for the reaction is 0.54. The light energy absorbed is 15000 erg
mol in a given time. Find the amount of carbon monoxide formed in moles in the
same time.
SOLUTION

PHOTOSENSITIZED REACTIONS
o In many photochemical reactions the reactant molecule does not absorb the
radiation required for the reaction. Hence the reaction is not possible.
o In such cases the reaction may still occur if a foreign species such as mercury
vapour is present.
o A species which can both absorb and transfer radiant energy for activation of the
reactant molecule, is called a photosensitizer. The reaction so caused is called a
photosensitized reaction.
Hg + hv ⎯⎯→ Hg*
Hg* + A ⎯⎯→ A* + Hg
Reaction between H2 and CO:
Hg + hv ⎯⎯→ Hg* Primary absorption
Hg* + H2 ⎯⎯→ 2H * + Hg Energy transfer
H + CO ⎯⎯→ HCO
Some glyoxal, CHO-CHO, is also formed by dimerization of formyl radicals, HCO.

EMISSION:
• An electron jumps from a
higher into a lower orbital,
and releases (emits) a photon
of energy equal to the
difference (∆E = hv).
• Luminescence (emission of
light) works on the basis that
if a molecule is excited above
its ground state, then it must
de-excite somehow and
release excess energy as
either non-radiative or
radiative energy.

 A good starting point for a discussion of
luminescence (fluorescence or
phosphorescence) principle is a simplified
Jablonski diagram.
• The Jablonski diagram is employed to
represent the energy levels of a molecule. As
depicted in JD, S0, S1 and S2 represent ground,
first and second singlet electronic states,
respectively, whereas T1 and T2 describe first
and second triplet electronic states,
respectively.
• Each electronic energy level of a molecule also
has numerous vibrational (v) and rotational (n)
energy sublevels.
Jablonski diagram (JD)

(JD)
10-15 s
10-11 s
10-9 to 10-7 s
10-5 to 10-s
Absorbanceenergy

So
S1
VR
VR
F
A
T1
VR
P
DE
DE
Relationship between fluorescence, phosphorescence and absorption spectra
-spectrum: intensity
of luminescence vs
wavelength
Similarly, a molecule
can relax back to
a vibrationally excited
level of So from T1.
- have vibrational fine
structure in absorption
and in emission
molecule relaxes back
from the S1 state to a
vibrationally excited
level of the ground
state So.

• Mirror-image rule and Franck-
Condon factors. The absorption
and emission spectra are for
anthracene. The numbers 0, 1,
and 2 refer to vibrational energy
levels.
• Phosphorescence and ISC are
spin-forbidden transitions. In
contrast, fluorescence and IC
are spin-allowed transitions.
“Stokes” shift
Absorption vs Emission

PHOTOPHYSICAL PROCESSES:
o If the absorbed radiation is not used to cause a chemical change, it is re-emitted
as light of longer wavelength. The three such photophysical processes which can
occur are :
(1) Fluorescence (2) Phosphorescence (3) Chemiluminescence
o Fluorescence
✓ Process at which certain molecules (or atoms) when exposed to light radiation of
short wavelength (high frequency), emit light of longer wavelength.
✓ Substance that exhibits fluorescence is called fluorophore.
✓ Florescence stops as soon as the incident radiation is cut off.
Examples:
(a) a solution of quinine sulphate on exposure to visible light, exhibits blue
fluorescence.
(b) a solution of chlorophyll in ether shows blood red fluorescence.

Molecular formulas of common dyes of fluorescent probes (fluorophores)

o Fluorescence depends on:
1. pH
2. Solvent
3. Temperature
4. Salts
5. Concentration

Demonstration of a Solvent Sensitive Fluorophore
Corrected fluorescence emission spectra of DOS in
cyclohexane(CH), toluene (T), ethylacetate(EA) and
butanol(Bu). The dashed line shows the emission of
DOS from DPPC vesicles
Solvent Polarity

Fluorescent minerals, shown under ultraviolet light:

❖ The colour of fluorescence depends on the wavelength of light emitted.

Phosphorescence
❑ When a substance absorbs radiation of high frequency and emits light even after the
incident radiation is cut off, the process is called phosphorescence.
❑ The substance which shows phosphorescence is called phosphorescent substance.
❑ Phosphorescence is chiefly caused by ultraviolet and visible light. It is generally shown
by solids.
Examples:
(a) Sulphates of calcium, barium and strontium exhibit phosphorescence.
(b) Fluorescein in boric acid shows phosphorescence in the blue region at 570nm
wavelength.
o Phosphorescence could be designated as delayed fluorescence.
A + hv ⎯⎯→ A* ⎯slow⎯→ hv′

Phosphorescent powder under:
visible light, ultraviolet light, darkness

How much luminescence is observed depends on efficiency of the various
processes shown in the Jablonski Diagram.
- which compete with one another to deactivate the excited state
Consider fluorescence:
-amount of fluorescence observed is dependent on relative rates of
the following unimolecular processes:
process rate
hu1M 1M * absorption Ia
k F
1M * 1M + hu fluorescence k F [ 1M *]
k IC1M * 1M internal conversion k IC [ 1M *]
k ISC1M * 3 M * intersystem crossing k ISC [ 1M *]
Consider a molecule, M
(* denotes electronically excited)

Quantum Yield of Fluorescence
FO = number of photons emitted as fluorescence
total number of photons absorbed
rate
follow law of energy conversion
(i.e. what goes up must come down!)
rate of
absorption
rate of all deactivation processes
(i.e. total rate of deactivation)
FO rate of fluorescence
=
=
k F [ 1M *]
k F [ 1M *] + k IC [ 1M *] + k ISC [ 1M *]
(PHI)
rate

=
k F [ 1M *]
(k F + k IC + k ISC ) [ 1M *]
=
k F
k F + k IC + k ISC
Fo
since k n r = k IC + k ISC
rate of non-radiative deactivation
then
Fo
=
k F
k F + k nr

Now, since k = rate of all deactivation processes for the excited state
then
to =
1
k F + k nr
1
k
=
What does a lifetime mean?
This is a Relaxation Process
1 M * population relaxes back to the ground state lifetime to relaxation time
78.
Physically it corresponds (due to the exponential nature of the decay)
to the time taken for [ 1M *] to decay from [ 1M *]o
to 1/e [ 1M *]o
=
Fo
k F
k F
k F + k nr
x =
1
k F
Example: In water, the fluorescence quantum yield and observed fluorescence lifetime of
tryptophan are φo = 0.20 and τ0 = 2.6 ns, respectively. The fluorescence rate constant kf is?
Solution: to =
Fo
k F
So, k F =
Fo
to
=
0.20
2.6 x 10-9 s
= 7.7 x 107 s-1

First, the sample is excited with a short light pulse from a laser
using a wavelength at which S absorbs strongly. Then, the
exponential decay of the fluorescence intensity after the pulse is
monitored. From eqns 21.80 and 21.81, it follows that
Example: In water, the fluorescence quantum yield and observed
fluorescence lifetime of tryptophan are φf = 0.20 and τ0 = 2.6 ns,
respectively. The fluorescence rate constant kf is?

Fluorescence Lifetime (Lifetime of the Excited State)
-if use light of intensity Io
create an excited state population [ 1 M *]
(M absorbs the light to create 1 M *)
[ 1 M *] will remain constant (i.e. in a steady state) until light is removed
At the instant light is removed concentration of excited
states is [ 1 M *] o
t=0 time (t)t
[1M *] t
[1M *] o
Io
[1M *]
The excited state population [ 1M * ]o will
decay exponentially
- after some time, t, it’s [ 1M * ] t

The observed fluorescence lifetime can be measured by
using a pulsed laser technique:

This is a 1st order rate process & obeys 1st order kinetics.
Shape of decay can be described by:
[ 1M * ] t = [ 1M * ] o exp ( - k t )
[ 1M * ] t = [ 1M * ] o exp ( - t / t )
concentration of
excited states at
some time, t
initial concentration
of excited state
rate constant for process
time
Alternatively a physicist would use
where, t =
1
k
= “lifetime” of 1 M *
77.

Bimolecular Processes (Quenching)
-can expand kinetic scheme to include bimolecular processes.
Certain molecules, through collisions with M * can remove excitation energy
- called quenchers since they reduce the excitation state population
(in addition to that from unimolecular processes).
Examples of quenchers:
O2 ; CCl4 ; CH3NO2 ; Tl+
Mechanism can be expanded to include steps of the type:
1 M * + Q M + Q*
process
bimolecular
quenching
rate
k q [ 1 M *] [Q]
k q
bimolecular quenching constant
usually non fluorescent
• The shortening of the lifetime of the excited state by the presence of another species is
called quenching

F
rate of fluorescence
=
=
k F [ 1M *]
k F [ 1M *] + k IC [ 1M *] + k ISC [ 1M *] + k q [ 1M *] [Q]
=
k F [ 1M *]
(k F + k IC + k ISC + k q [Q]) [ 1M *]
Quantum Yield in the Presence of Quencher (F )
F =
k F
(k F + k IC + k ISC + k q [Q])

Derivation of an expression which relates quantum yields in presence ( F )
& absence ( F o ) of quencher
k F
F o
F
=
k F
k F + k IC + k ISC
k F + k IC + k ISC
=
k F
(k F + k IC + k ISC + k q [Q])k F
k F + k IC + k ISC
X
k F + k IC + k ISC + k q [Q])
(k F + k IC + k ISC )
=
(k F + k IC + k ISC)
+

k q [Q])
(k F + k IC + k ISC)
1 +
F o
F
=
Such dynamic quenching can be described (simply) by Stern-Volmer Kinetics.
since t F
o
=
1
k F + k nr
1
k
=
lifetime in the absence of
quencher
quantum yield in
the absence of quencher
quantum yield in
the presence of quencher
t F
o is the lifetime in
the absence of quencher
quenching constant
given concentration
of quencher ([Q])
Stern-Volmer Equation
F o
F
= 1 + k q t F
o [Q]

FF
Fo
F
=
IF
Io
F
=
t F
t o
F
= 1 + kq tF
o [ Q ]
Obviously, a plot of
FF
Fo
F
IF
Io
F
or or
t F
t o
F
vs. [Q] is linear
with a slope = kq tF
o and intercept = 1
If we know tF
o can determine kq (the Bimolecular Quenching Constant)
- from the slope.
Since, I F  FF
(I F is the intensity of fluorescence and is what you measure “directly” via fluorescence
spectroscopy)
( y = c + m x )
In addition, for dynamic quenching:
FF
Fo
F
=
IF
Io
F
=
t F
t o
F
( t f data can be determined by a fluorescence lifetime spectrometer)

Example
(i) Show (graphically) that the quenching obeys Stern-Volmer kinetics.
84.
FF
Fo
F
=
IF
Io
F
=
t F
t o
F
= 1 + kq tF
o [ Q ]
(naphthalene)
When carbon tetrachloride is added to a dilute (10-5 M) solution of naphthalene in toluene, the
fluorescence, observed from the naphthalene, is quenched. The following table shows the effect
of varying amounts of carbon tetrachloride upon t F observed at 340 nm at 298 K.
[CCl4] / M t F / ns
0 18.0
0.0018 14.4
0.0049 10.8
0.0074 9.0
0.0110 7.2
t o
F
t F
vs. [Q] is linear

slope = 136 M-1
[CCl4]
/ M t F / ns to
F / t F
0 18.0 1.00
0.0018 14.4 1.25
0.0049 10.8 1.67
0.0074 9.0 2.00
0.0110 7.2 2.50
. : .
[CCl 4] / M
0.004 0.008 0.0120
1
2
3
to
F / t F
(ii) Estimate the value for k q, the bimolecular quenching rate constant, at 298K.
slope = k q to
F
k q = 136 M-1 / 18 x 10 –9 s
k q = 7.6 x 109 M-1 s-1

Since to
F of acetone is ~ 2 x10 -9 s
then k q = 7 M -1 / 2 x 10 -9 s
k q = 3.5 x 10 9 M-1 s-1 rate constant for the reaction
Since k q [ 1 M *] [Q] = rate of reaction
Can measure reaction rate
88.
0 0.1 0.2 0.3
1.0
2.0
Fo
F
FF
[dicyanoethylene] / M
slope = kq to
F = 7 M -1

e.g. photoaddition reaction
Stern-Volmer experiments can be used to study photochemical reactions:
fluorescent
-acetone fluorescence quenched by dicyanoethylene
C=O *
CN
CN
O
NC CN
+
k q
acetone dicyanoethylene dicyanooxetane
non-fluorescent
-can detect fluorescence from acetone to monitor progress of reaction

Energy Transfer
Energy exchange occurs through short distances ~ 2nm (collisions).
However, energy can be transferred over longer distances than collision diameters.
In general:
D* + A D + A*
-energy transfer can occur up to distances of 10 nm
FÖrster Transfer
-depends on overlap of the spectral bands of emission from donor
& absorption by the acceptor.
donor acceptor
M* + Q M + Q*
Fluorescence quenching - example of energy transfer.

• The excitation frequency of the DONOR must not excite the ACCEPTOR.!
• The emission spectrum of the DONOR must overlap the excitation spectrum of the ACCEPTOR.!
• The emission spectrum of the DONOR cannot significantly overlap with emission of ACCEPTOR. !
• DONOR and ACCEPTOR molecules must be in close proximity (typically 10–100 Å). !
Fluorescence Resonance Energy Transfer (FRET)

For example, 2 solutions in benzene:
(a) 10-3M in anthracene (b) 10-3M in perylene
-excitation of either solution within the chromphore’s excitation band
produces fluorescence from the fluorophore.
e.g. excitation of anthracene solution at 340 nm gives relatively
intense fluorescence between ~ 370-470 nm.
90.
(a) anthracene

Similarly, excitation of the perylene containing solution at 410 nm “see”
fluorescence from perylene (430-550 nm)
(b) perylene

Now, a solution containing both 10-3M anthracene and 10-3M perylene:
(i) excitation at 410 nm (where only perylene absorbs)
gives perylene fluorescence (unperturbed by presence of anthracene).
(ii) excitation at 340 nm (hardly absorbed at all by perylene) produces fluorescence
form both anthracene and perylene.
However, anthracene fluorescence quenched in mixture.
(c)
0
4 0
8 0
1 2 0
1 6 0
2 0 0
400 500
Wavelength / nm
Intensity
perylene
anthracene

FÖrster Transfer-form of non-radiative transfer
- depends upon spectral overlap of donor (anthracene) & absorption band of
acceptor (perylene)
Involves dipolar interactions:
-effectively electronic oscillations in D * induce oscillations in electron to be
promoted in A
(like resonance interactions between 2 tuning forks).
(iii) large eA for A
-also known as fluorescence resonance energy transfer (FRET).
Consequently, D * returns to D and A is promoted to A *
D * + A D + A *
This form of energy transfer is enhanced by:
(i) good overlap of fluorescence spectrum of D * & absorbance spectrum of A
(ii) large FF for D * (in absence of A)

Energy transfer Efficiency
E T =
Ro
6
Ro
6 + R 6
distance between donor & acceptor
- ET increases with decreasing distance (R)
Donor & acceptor must be within 10 nm (critical transfer distance)
-sometimes referred to as Spectroscopic Ruler Technique
-used extensively to measure distances in biological systems & across
science in general.
Fo
F
FF
E T = 1 -
for D in absence of acceptor
transfer
efficiency
for D in presence of acceptor
spectroscopic characteristic
of each donor/acceptor pair

Applications of FRET in Chemistry & Biology
Example: rhodopsin -protein (primary absorber of light contained in the eye)
-allows vision to take place
rhodopsin- consists of an opsin protein
molecule attached to 11- cis retinal
D A
-amino acid on the surface of rhodopsin was
labelled covalently with the energy
donor I AEDANS
Iodoacetyl amino ethyl amino naphthalene sulphonic acid (I AEDANS)

FF of label decreased from 0.75 to 0.68 due to quenching by the cis retinal acceptor
Ro is 5.4 nm for this pair from E T =
Ro
6
Ro
6 + R6
Can calculate the distance (R) between the surface of the rhodopsin protein
& cis retinal as 7.9 nm
AEDANS (D)
11- cis retinal (A)
opsin protein
rhodopsin
can calculate R
7.9 nm
ET = 1 –
0.75
0.68
= 0.093

• Example: Consider a study of the protein rhodopsin. When an amino acid
on the surface of rhodopsin was labelled covalently with the energy donor
1.5-I AEDANS (3), the fluorescence quantum yield of the label decreased
from 0.75 to 0.68 due to quenching by the visual pigment 11-cis-retinal (4).
Calculate the distance between the surface of the protein and 11-cis-
retinal (Ro = 5.4 nm for the 1.5-I AEDANS/11-cis-retinal pair )

Chemiluminescence:
o The emission of light as a result of chemical action is called chemiluminescence. The
reaction is referred to as a chemiluminescent reaction.
o Chemiluminescent reaction is the reverse of a photochemical reaction which proceeds
by absorption of light.
o The light emitted in a chemiluminescent reaction is also called ‘cold light’ because it is
produced at ordinary temperature.
o In a chemiluminescent reaction, the energy released in the reaction makes the product
molecule electronically excited. The excited molecule then gives up its excess energy
as visible light while reverting to ground state.

Chemiluminescence of fireflies and luminol
Examples:
(a) The glow of fireflies due to the aerial oxidation of luciferin (a protein) in the presence of
enzyme luciferase.
(b) The oxidation of 5-aminophthalic cyclic hydrazide (luminol) by hydrogen peroxide in
alkaline solution, producing bright green light.

Notes
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Physical Chemistry VIII_ Photochemistry

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