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PHOTOCHEMISTRY
SURIYA MARIMUTHU
Ph.D. Scholar,
C/O Prof. Chia-Hsiang Chen,
KMU-Taiwan
PART-1
 Photochemistry is concerned with the absorption, excitation and emission of photons by
atoms, atomic ions, molecules and molecular ions etc.
 It deals with the study of interaction of radiation with matter resulting into physical
changes or into a chemical reaction.
 Photochemistry start with the absorption of light radiation by atom or molecules which
brings about the excitation of atoms or molecules followed by physical or chemical
changes.
 According to the quantum theory, both matter and light are quantised, and only certain
specific energies of light are absorbed by specific organic molecule for its excitation.
The absorption or emission of light occurs by the transfer of energy as photons.
INTRODUCTION
PHOTOCHEMICAL ENERGY
Photochemical energy of electromagnetic radiation is given as:
E = hν ----------- 1
ν = c/λ ----------- 2
Equation 2 substitute in equation 1
E = hc/ λ
Where
ν = frequency of electromagnetic radiation.
λ = wavelength of electromagnetic radiation.
C = velocity of light
h = Planck constant
The two main processes
1. Photo physical process
2. Photochemical process
1. Photo physical process
In this process, the absorption of light does not result into any chemical reaction.
2. Photochemical process
In this process, the light that is absorbed by a system results into chemical change.
LAW GOVERNING ABSORPTION OF LIGHT
Lambert’s law
When a monochromatic light is passed through a pure homogeneous
medium, the decrease in the intensity of light with thickness of the absorbing medium at
any point is proportional to the intensity of the incident light.
Mathematically:
dI/dx α I
or dI/dx = kI
Beer’s law
When a monochromatic light is passed through a solution, the decrease
in the intensity of light with thickness of the solution is directly proportional to the
intensity of the incident light and the concentration of the solution.
Mathematically :
dI/dx α I × c
or dI/dx = Icɛ
Combined Lambert-Beer’s Law is given as:
log I0 / I = ɛcl = A
Where
I0 = Intensity of the incident light.
I = Intensity of the transmitted light.
c = Concentration of the solution in moles/litre.
l = Path length of the sample usually 1cm.
ɛ = Molar absorptivity or molar extinction coefficient.
A = Absorbance or optical density.
MOLAR ABSORPTIVITY
The molar absorptivity (formerly called the extinction coefficient) of a compound
constant that is characteristic of the compound at a particular wavelength.
ε = A/cl
So the units of ε usually given are:cm-1 x (mol-1 ) -1 = l mol-1 cm-1 .
Values above 104 are termed high intensity absorption while values below 103 are
called low intensity absorption.
LAWS OF PHOTOCHEMISTRY
There are two laws of photochemistry which are
1. Grotthuss-Draper law
2. Stark–Einstein law.
Grotthuss –Draper Law
When light falls on a cell containing a reaction mixture, some
light is absorbed and the remaining light is transmitted. Obviously, it is the absorbed
component of light that is capable of producing the reaction. The transmitted light is
ineffective chemically.
This is known as Grotthuss-Draper law and may be stated as follows.
“It is only the absorbed light radiations that are effective in
producing a chemical reaction. However, it does not mean that the absorption of
radiation must necessarily be followed by a chemical reaction. When the conditions
are not favourable for the molecules to react, the light energy remains unused. It
may be re-emitted as heat or light.”
Stark-Einstein Law of Photochemical Equivalence
According to this law each quantum of light absorbed by a
molecule, activate only one molecule in the primary step of photochemical process or
briefly one molecule one quantum.
One photon (hv)
AB AB# C
Reactant molecule Excited molecule Product
Figure 1. Illustrating law of photochemical equivalence
The law of photochemical equivalence is illustrated in (Figure 1) where a molecule
‘AB’ absorbs a photon of radiation and gets activated. The activated molecule (AB#)
then decomposes to yield C.
QUANTUM YIELD
Quantum yield is defined as “the number of molecules of the substance
undergoing photochemical change per quantum of radiation absorbed.
Thus Number of molecules reacting in a given time
Cause of high quantum yield
 Absorption of radiations in the first step involves production of atoms or
free radicals, which initiate a series of chain reactions.
 If the reactions are exothermic, the heat evolved may activate other
molecules without absorbing the additional quanta of radiation.
 Formation of intermediate products will act as a catalyst.
 The active molecules, produced after absorption of radiation, may collide
with other molecules and activate them which in turn activate other
reacting molecules.
Examples:
Decomposition of HI:
In the primary reaction, one HI molecule absorbs a photon and
dissociated to produce one H and one I. This is followed by the second reaction as shown
below
HI + hγ → H*+ I* Primary reaction
H * + HI → H2 + I* Secondary reaction
I *+ I* → I2
Overall reaction: 2HI + hγ → H2 + I2
The overall reaction shows that the two HI are decomposed for one photon (hγ).
Thus, the quantum yield (ϕ) = 2
Formation of HCl:
In the primary step, one Cl2molecule absorbs a photon anddiscussed into
two Cl atoms. This is followed by the secondary reaction as shown below
Cl2 + hγ → 2Cl* Primary reaction
Cl* + H2 → HCl + H*
H * + Cl2 → HCl + Cl * Secondary reaction
The Cl atom consumed in step 2 is regenerated in step 3, this will propagate the chain
reaction. The chain reaction gets terminated when the Cl atoms recombine at the walls of
the vessel, where they lose their excess energy. 2Cl*→ Cl2 .
Thus the quantum yield varies from 104 to 106 .
 Excited molecules may get deactivated before they form products.
 Excited molecules may lose their energy by collisions with non-excited molecules.
 Molecules may not receive sufficient energy to enable them to react.
 The primary photochemical reaction may be reversed.
 Recombination of dissociated fragments will give low quantum yield.
Cause of low quantum yield
Example:
Dimerization of anthracene to dianthracene
2C14H10 + hγ→C28H20
The quantum yield = 2, but actually it is found to be = 0.5.
The reason is the above reaction is reversible.
Conditions for high and low quantum yield
 All the reactant molecules should be initially in the same energy state and hence
equally reactive.
 The reactivity of the molecules should be temperature independent.
Jablonski diagram
 When the molecule absorbs UV light radiations it get promotes to excited singlet state
S1, S2 S3. . In the above state molecules are collides and after 10-13 to 10-11second it
return back to the S1, and release energy. Such process is called energy cascade.
 In a similar manner the initial excitation and the decay from higher singlet states
initially populate many of the vibration levels of S1, but these also cascade, down to
the lowest vibrational level of S1. This cascade is known as vibrational cascade.
 The life time of singlet excited state S1 is long hence in this state has done many
physical and chemical processes. Molecules returns to its ground state, S0 from
excited singlet S1,/ S2 state by release energy as heat, but this is generally quite slow
because the amount of energy is large between S0 and S1. This process is called
internal conversion.
 When molecules return to its ground state S0 from excited state S1,/ S2 by giving off
energy in the light form within 10-9 seconds. This process is known as Fluorescence.
 Most molecules in the S1 state may drop to triplet state (T1) (S1→T1). This is
energetically slow process. However, if the singlet state S1 is long lived, the S1 → T1
conversion occurs by a process called intersystem crossing.
 When molecule in the T1 state may return to the So state by giving up heat or light this
is called Phosphorescence.
 When fluorescence and phosphorescence occur in same molecule, phosphorescence is
found at lower frequencies than fluorescence. This is because of the higher difference in
energy between S1 and S0 than between T1and So and is longer-lived.
Thank You !
Suriya Marimuthu
2023.12.18

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PHOTOCHEMISTRY BASIC PRINCIPLE AND JABLONSKI DIAGRAM

  • 1. PHOTOCHEMISTRY SURIYA MARIMUTHU Ph.D. Scholar, C/O Prof. Chia-Hsiang Chen, KMU-Taiwan PART-1
  • 2.  Photochemistry is concerned with the absorption, excitation and emission of photons by atoms, atomic ions, molecules and molecular ions etc.  It deals with the study of interaction of radiation with matter resulting into physical changes or into a chemical reaction.  Photochemistry start with the absorption of light radiation by atom or molecules which brings about the excitation of atoms or molecules followed by physical or chemical changes.  According to the quantum theory, both matter and light are quantised, and only certain specific energies of light are absorbed by specific organic molecule for its excitation. The absorption or emission of light occurs by the transfer of energy as photons. INTRODUCTION
  • 3. PHOTOCHEMICAL ENERGY Photochemical energy of electromagnetic radiation is given as: E = hν ----------- 1 ν = c/λ ----------- 2 Equation 2 substitute in equation 1 E = hc/ λ Where ν = frequency of electromagnetic radiation. λ = wavelength of electromagnetic radiation. C = velocity of light h = Planck constant
  • 4. The two main processes 1. Photo physical process 2. Photochemical process 1. Photo physical process In this process, the absorption of light does not result into any chemical reaction. 2. Photochemical process In this process, the light that is absorbed by a system results into chemical change.
  • 5. LAW GOVERNING ABSORPTION OF LIGHT Lambert’s law When a monochromatic light is passed through a pure homogeneous medium, the decrease in the intensity of light with thickness of the absorbing medium at any point is proportional to the intensity of the incident light. Mathematically: dI/dx α I or dI/dx = kI
  • 6. Beer’s law When a monochromatic light is passed through a solution, the decrease in the intensity of light with thickness of the solution is directly proportional to the intensity of the incident light and the concentration of the solution. Mathematically : dI/dx α I × c or dI/dx = Icɛ
  • 7. Combined Lambert-Beer’s Law is given as: log I0 / I = ɛcl = A Where I0 = Intensity of the incident light. I = Intensity of the transmitted light. c = Concentration of the solution in moles/litre. l = Path length of the sample usually 1cm. ɛ = Molar absorptivity or molar extinction coefficient. A = Absorbance or optical density.
  • 8. MOLAR ABSORPTIVITY The molar absorptivity (formerly called the extinction coefficient) of a compound constant that is characteristic of the compound at a particular wavelength. ε = A/cl So the units of ε usually given are:cm-1 x (mol-1 ) -1 = l mol-1 cm-1 . Values above 104 are termed high intensity absorption while values below 103 are called low intensity absorption.
  • 9. LAWS OF PHOTOCHEMISTRY There are two laws of photochemistry which are 1. Grotthuss-Draper law 2. Stark–Einstein law. Grotthuss –Draper Law When light falls on a cell containing a reaction mixture, some light is absorbed and the remaining light is transmitted. Obviously, it is the absorbed component of light that is capable of producing the reaction. The transmitted light is ineffective chemically.
  • 10. This is known as Grotthuss-Draper law and may be stated as follows. “It is only the absorbed light radiations that are effective in producing a chemical reaction. However, it does not mean that the absorption of radiation must necessarily be followed by a chemical reaction. When the conditions are not favourable for the molecules to react, the light energy remains unused. It may be re-emitted as heat or light.” Stark-Einstein Law of Photochemical Equivalence According to this law each quantum of light absorbed by a molecule, activate only one molecule in the primary step of photochemical process or briefly one molecule one quantum.
  • 11. One photon (hv) AB AB# C Reactant molecule Excited molecule Product Figure 1. Illustrating law of photochemical equivalence The law of photochemical equivalence is illustrated in (Figure 1) where a molecule ‘AB’ absorbs a photon of radiation and gets activated. The activated molecule (AB#) then decomposes to yield C.
  • 12. QUANTUM YIELD Quantum yield is defined as “the number of molecules of the substance undergoing photochemical change per quantum of radiation absorbed. Thus Number of molecules reacting in a given time
  • 13. Cause of high quantum yield  Absorption of radiations in the first step involves production of atoms or free radicals, which initiate a series of chain reactions.  If the reactions are exothermic, the heat evolved may activate other molecules without absorbing the additional quanta of radiation.  Formation of intermediate products will act as a catalyst.  The active molecules, produced after absorption of radiation, may collide with other molecules and activate them which in turn activate other reacting molecules.
  • 14. Examples: Decomposition of HI: In the primary reaction, one HI molecule absorbs a photon and dissociated to produce one H and one I. This is followed by the second reaction as shown below HI + hγ → H*+ I* Primary reaction H * + HI → H2 + I* Secondary reaction I *+ I* → I2 Overall reaction: 2HI + hγ → H2 + I2 The overall reaction shows that the two HI are decomposed for one photon (hγ). Thus, the quantum yield (ϕ) = 2
  • 15. Formation of HCl: In the primary step, one Cl2molecule absorbs a photon anddiscussed into two Cl atoms. This is followed by the secondary reaction as shown below Cl2 + hγ → 2Cl* Primary reaction Cl* + H2 → HCl + H* H * + Cl2 → HCl + Cl * Secondary reaction The Cl atom consumed in step 2 is regenerated in step 3, this will propagate the chain reaction. The chain reaction gets terminated when the Cl atoms recombine at the walls of the vessel, where they lose their excess energy. 2Cl*→ Cl2 . Thus the quantum yield varies from 104 to 106 .
  • 16.  Excited molecules may get deactivated before they form products.  Excited molecules may lose their energy by collisions with non-excited molecules.  Molecules may not receive sufficient energy to enable them to react.  The primary photochemical reaction may be reversed.  Recombination of dissociated fragments will give low quantum yield. Cause of low quantum yield
  • 17. Example: Dimerization of anthracene to dianthracene 2C14H10 + hγ→C28H20 The quantum yield = 2, but actually it is found to be = 0.5. The reason is the above reaction is reversible. Conditions for high and low quantum yield  All the reactant molecules should be initially in the same energy state and hence equally reactive.  The reactivity of the molecules should be temperature independent.
  • 19.  When the molecule absorbs UV light radiations it get promotes to excited singlet state S1, S2 S3. . In the above state molecules are collides and after 10-13 to 10-11second it return back to the S1, and release energy. Such process is called energy cascade.  In a similar manner the initial excitation and the decay from higher singlet states initially populate many of the vibration levels of S1, but these also cascade, down to the lowest vibrational level of S1. This cascade is known as vibrational cascade.  The life time of singlet excited state S1 is long hence in this state has done many physical and chemical processes. Molecules returns to its ground state, S0 from excited singlet S1,/ S2 state by release energy as heat, but this is generally quite slow because the amount of energy is large between S0 and S1. This process is called internal conversion.
  • 20.  When molecules return to its ground state S0 from excited state S1,/ S2 by giving off energy in the light form within 10-9 seconds. This process is known as Fluorescence.  Most molecules in the S1 state may drop to triplet state (T1) (S1→T1). This is energetically slow process. However, if the singlet state S1 is long lived, the S1 → T1 conversion occurs by a process called intersystem crossing.  When molecule in the T1 state may return to the So state by giving up heat or light this is called Phosphorescence.  When fluorescence and phosphorescence occur in same molecule, phosphorescence is found at lower frequencies than fluorescence. This is because of the higher difference in energy between S1 and S0 than between T1and So and is longer-lived.
  • 21. Thank You ! Suriya Marimuthu 2023.12.18