Excited states and Photochemistry

Question 1

What is photochemistry and where does it take place?

Answer 1

Chemistry initiated by light

Takes place in the excikted state created by absorbtion of a photon with energy hv

Question 2

When does an elevctronic transition take place?

Answer 2

• When an electron moves from one MO to another:

the electron density is redistributed

the bonding and structure chage

photochemical reactions can be quite different from thermal reactions of the same molecule

• Typically indiced by UV light

wavelengths in the range of λ = 200-800 nm

absorbers called chromophores

Question 3

Equations for:

energy of one photon

energy of a mol of photons

wavenumber converting m - cm and m^-1 - cm^-1

Realtionship between energy, frequency, wavenumber and wavelength

Answer 3

Question 4

Units and symbols of frequency, wavenumber, wavelength, energy (of one photon and a mol of photons), speed of light, planck’s constant, avogadro’s constant

Answer 4

Question 5

What is the beer lambert law and what are the units of the components

Answer 5

Question 6

Typical long-wavelength absorption properties of saturated and unsaturated systems and some typical transitions

Answer 6

saturated systems: usually absorb in the vacuum UV

unsaturated systems: increasing conjugation shifts the absorbtion to longer λ

σ—–>σ*

π—–>π*

n—–>π*

Transitions invloving σ* prbitals usually give higher bands (λ<200 nm)

Question 7

Describe the states a molecule can be in

Answer 7

Ground state:

the lowest energy electronic state

electrons generally occupy bonding MOs and in some cases non bonding orbitals

Excited state:

Photon absorbtion induces an electronic transtion where an electron is excited to a higher energy level usually an antibonding MO

Question 8

Explain and draw the different types of transition that can occur between MOs

Answer 8

When electrons occupy the same MOs but with different spins different excited states can be achieved.

Question 9

What is multiplicity

Answer 9

Multiplicuty is used to describe the electron spin properties of a state

Defined as 2S+1

When an electron retains its spin on transtion multiplicity remains constant

When an excited state has a multiplicity of 1 it is called a singlet transition, when a transition has a multiplicity of 3 it is called a triplet transition

Question 10

Give the labels for singlet and triplet transitions and the ground and excited state

How many triplet states are there?

Answer 10

Singlet = S

Triplet = T

Ground = 0

Excited = 1,2,3…

Each excited singlet state has an equivalent triplet state at lower energy (hunds rule states that subshells will preferentially fill singly beofre doubly).

There is no T₀ state (Pauli exclusion principle says that 1 orbital cannot be doubly occupied with 2 electrons of the same spin).

Question 11

Describe the rotational and vibrational degrees of freedom of atoms and molecules?

Answer 11

Atoms: no rotational or vibrational degrees of freedom

electronic energy levels are clearly defined

UV-visible spectra has sharp lines

Molecules: has rotational and vibrational degrees of freedom

eavh electronic state has many energy levels

UV-visible spectra has many lines or braod bands

Question 12

Describe the Born-Oppenheimer approximation

Answer 12

states that the elctronic, vibrational and rotational properties are considered to be independent of each other

Question 13

Where do electronic and vibrational transitions usually occur from and to?

Answer 13

Electronic trnsitions occur from the ground state (S₀) to the first excited state (S₁) when the longest wavelenght absorption arises

From v’’ = 0 in the S₀ state (Boltzman states this is the most populated)

To: several v’ levels in the s₁ state

Question 14

Describe typical rotational energy transitions

Answer 14

Transition genrally occur from several J’’ rotational levels in the v’’ = 0 level (as Boltzman states that many J levels are populated) to several J’ leve;s in each v’ level in S₁

Question 15

Describe what typical spectra can look like

Answer 15

Rotational fine structure is seen from gases at high spectroscopic resolution

Vibrational structure seen from gases and liquids/solids in some cases but peak can be broad - particularly in polar solvents due to strong interations

Question 16

What determines the required energy of a photon for a transition and the probability of the transition?

Answer 16

The energy of the photon (wavelength of light) is determined by the energy gap between the ground and excited states

The probability (stregth of absorption coeffiecient) is determined by selection rules

Question 17

What is the overall selection rule for an electronuc transition?

Answer 17

The elctric field of the light must cause a displacement if charge giving a transition dipole moment (TM)

Question 18

How do you apply the born oppenheimer approximation to the overall selection rule?

Answer 18

separate thr electronic anf vibrational wavefunction:

ψ = ψ_e . ψ_v

Separate ψ_e into electron spin (S) and orbital wavefuntions:

ψ = S . φ . ψ_v

Substitute into the original TM equation noting that the TM operator (μ)only acts on the orbital wavefunction

Tis gives three separate selection rukles: spin (S), orbital (φ) and vibrattion (ψ_v)

IF any of the integrals are 0 the TM is 0 and the transition if disallowed

Question 19

What is the spin selection rule?

Answer 19

An electron has two possible spin states:

spin up (α) and spin down (β)

When the electron retains its spin on transition S’* and S’’ are both S_α​ or S_β - The integral is normalised to and therefore = 1

If the spin changes on transition S’* = S’*_α and S’’ = S’‘_β or vice versa and the integral is orthogonal and therefore = 0

Rule: the electron must retain its spin on transition for the transition to be allowed (triplet excited state cannot be created efficiently by absorption.

Question 20

Explain the orbital selection rule

Answer 20

The rbiotal selection rule has two parts:

symmetry and spatial

Symmetry rule: The orbital part of the transition can be split into three parts:

TM_orbital = TM_x + TM_y + TM_z

All three components must be 0 for the transition to be forbidden

Spatial: If the initial and final orbitals occupy a region of space that is similar the transition is allowed

Question 21

Describe how to determine which transiotion are symmetry allowed

Answer 21

Deduce the point group

Deduce the symmetries (irreducibles) of the initial and final states (Γ_i and Γ_f)

Deduce the symmetries of the x, y and z components of the dipole moment operator (read from character table) (Γ_x, Γ_y, Γ_z)

Deduce three direct products

Γ_f x Γ_x x Γ_i - repeat for y and z

If any direct products gives totally symmetric represention (A1/Ag) the transition is symmetry allowed

Question 22

Which transitions are allowed and forbidden under the spatial selection rule and what can affect the size of μ

Answer 22

π —–> π* have good spatial overlap and therefore are spatially allowed

n —-> π* occupy different regions of space and therefore are spatially forbidden

If there is a l;arge displacement of charge μ will be large

Question 23

For a harmonic oscillator give the most probable geometries for v = 0 and v >> 0and give the most probable transitions

Answer 23

v = 0: most probably geometry (most probable point of finding particle) is at equilibrium geometry

When v >> 0 the most probably geometry is at the turning points

electronic transitions occur mainly from v’’ = 0 in the S₀ state to v’ levels that give large vibrational overlap

Question 24

What is the Franck-Codon Principal?

Answer 24

electrons move faster than nuclei because of their relative masses

There is no change in the position of atoms during a transition

Any change in geometry occurs after the transition.

Question 25

Why is r_e (S₁) often > r_e(S₀)

Answer 25

an electron is promoted to an sntibonding orbital hence the excited state often has weaker, longer bond

Question 26

What is the probability of a transition given by?

Answer 26

Oscillator strenght (f)

f is proprtional to the dipole moment (TM)

f is proportional to the integral of the absorbtion coefficient

oscillator strength for an electronic component:

f = f_spin . f_symmetry . f_spatial

Question 27

What values of f are expected for allowed and disallowed transitions

Answer 27

if fully allowed: f_{x = 1}

If disallowed:

f_spin = 10^-8 - 10^-5

f_symmetry = 10^-3 - 10^-1

f_spatial = 10^-3 - 10^-1

Allowed means high probability

Disallowed means lower probability

Question 28

What is the breakdown of selection rules and explain the breakdown of the spin selection rule

Answer 28

The breakdown of selection rules is used to describe scenarios where the Born-Oppenheimer approximation does not hold well

The spin and orbital motions of electrons are not independant of each other

Spin orbit coupling magnitude is proportional to z⁴ (atomic number)

A spin flip transition is strongly forbidden but weakly allowed in the presence of heavy atoms

Internal - within a molecule

External - within other molecules (solvent)

Question 29

Describe the breakdown of the symmetry selection rule

Answer 29

The vibrational and electronic motions are vibronic coupling

An electronic transition that is forbidden can be weakly allowed if the point group changes (symmetry lowers)due to an asymmetric vibration distorting the molecule

Question 30

What is photo physics and what are the two processes of returning to ground state

Answer 30

Photophysics - the non radioactive decay of excited states

Excited states have a large excess of energy and are short lived

Losing ecess energy and returning to the ground state can accurr by two methods:

radiative

non radiative

Question 31

Describe vibrational relaxation

Answer 31

Excited electronic states can be formed in a vibrationally excited state when excess energy is given to suurpunding molecules (sovent) in collisions - non radiative process

Rate constant (in liquids) = 10¹³ collisions s^-1

Time scale = reciprocal of rate constant = 10^-13 s

This usually occurs amoung the v’ levels as v’ =- 0 is the most populated

Question 32

Describe luminesence and Fluorescence

Answer 32

Luminesence - a radiatuve transition (emission)

Fluoroesence - a radiative transition between states if the same multiplicity (S₁ —-> S₀)

Question 33

What are the selection rules for decay processes?

Answer 33

The selection rules are the same as electronic absorption: spin, orbital and vibrational overlap integral

By definition fluoresence is spin allowed - typically the rate constant is high (k_f = 10⁶ - 10⁹ s^-1)

Genrrally is the decay process is allowed the rate constant is high

Question 34

What does a gerneal absorbtion emission spetcrum look like

Answer 34

A vibrational progression from S₁ (v’ = 0) to several S₀ (v’’) levels

The probabilities are determines by the vibrational overlap integrals

The spectra are often mirror images

Question 35

What are some general notes of spectra?

Answer 35

The spacing between individual peaks gives the spacing of vibrational energy levels in the final state

Absorbtion —-> v’ levels in S₁ Emission —-> v’’ levels in S₀

The spacing in absorption and emission spectra are slightly different because the vibrational energies are different in the S₁ and S₀ states - vibrational energies typically smaller in the S₁ state because bonds are weaker

In gases the collision rate is lower in the S₁ state so vinrational relaxation is slower - unrelaxed resonance fluoresence can be observed in higher v’ levels in gases

Question 36

What is Stokes shift?

Answer 36

In solution the 0-0 band is typically shifted to a lightly lower energy in emission than absorption

This is because the solvent rearranges to stabalise the excited state after absorption but before emission

The extent of stabalisation varies with molecule and soolvent - polar systems generally give large shifts

Question 37

What is internal conversion?

Answer 37

A non radiative isoenergetic transition between states of the same multiplicity - elevtron moved to s lower MO with dE = 0 and dS = 0

State formed is vibrationally excited

vibrational relaxation follows rapidly

Question 38

What are the selection rules for non radiative transitions and what is vibrational overlap integral dependant on?

Answer 38

Similar rules to radiative transiotion: spin, orbital, vibrational overlap integral

By definition internal conversion is spin allowed

Vibrational overlap integral is based on energy and equilibirum geometry

Small energy gap = large overlap integral, very large rate constant

Large energy gap = small overlap integral, large rate constant

Question 39

What is Kasha’s rule

Answer 39

Even if higher excited states are achieved internal conversion and vibrational relaxation occur very rapidly

fluoresence amd other processes usually occur exclusively from S₁

In S₁ state internal conversion and fluroesence are in competition

Question 40

What is intersystem crossing

Answer 40

a non radiative, isoenergeitc transition between states of differing multiplicity - the molecule chnges state but may also change MO

vibrational relaxation follows rapidly

By definition is spin disallowed

For overlap ntegral Rate constants K_isc<<k>ic</k>

Small energy gap = large overlap integral, large rate constant

LArge energy gap = small overlap integral, very small rate constant

Question 41

What do the relative rates of the different process depend on?

Answer 41

Depend of the rate constants

For many molecules intersystem conversion of an efficient way to form T₁ states

T₁ states arec relatively long loved

Question 42

What is phosphoresence?

Answer 42

a radiative transition between states of different multiplicity (one step)

By definition spin disallowed - typically low rate constant

profile usually similar to fluoroesence spectra - vibrational overlap integrals are similar

Longer wavelength than in fluoroesence because the T₁ state is lower energy tha fluoroesence

Question 43

What are Jablonski diagrams?

Answer 43

Simple schematic diagrams taht summarise photophysical processes

Y axis = energy - usually qualitative but actual energies sometimes used

-v levels often shown butsome diagrams omit them

No parameter as x axis - states separated into stacks for ease

Question 44

What is chemiluminescence?

Answer 44

Luminescence from an excited state formed asthe product of a chemical reaction

REleasing enough energy for this requires a strongly exothermic reaction

Question 45

What is a quantum yield and what can it be applied to?

Answer 45

Quantum yield Φ: usually varies between 0 and 1

Φ = Number of molecules undergoing a process / number of photons absorbed

= rate of process / rate of absorption

Question 46

What is Φf in terms of rate constants?

Answer 46

In the S₁ state ; Φf = rate of S₁ emission / Rate of S₀ absorption

If S₁ only decays by photophysics:

Φ_f + Φ_ic + Φ_ics = 1 and k_f + k_ic + k_ics = k_tot

Rate of absorption = R_abs, Rate of S₁ emission = k_f[S₁] and Rate of S₁ decay = (k_f + k_ic + k_ics) [S₁]

Under steady state conditions Rate of S₀ absorption = Rate of S₁ decay and R_abs = (k_f + k_ic + k_ics )[S₁)

Φ_f = k_f [S₁] / R_abs

Therefore: Φ_f = k_f [S₁] / (k_f + k_ic + k_isc) [S₁]

= k_f / (k_f + k_ic + k_isc)

Question 47

What are quantum yields determined by and what steps need to be considered in a particular transition?

Answer 47

Determined by how the rate constant for a specific prpcess competes with the sum of the rate constants for all the decay processes

For a process from the T₁ state the quantum yield of T₁ formation (Φ_T) and the fraction of molecules in the T₁ state that decay by the process called a quantum efficiency θ_process

θ = No. molecules undergoing process / No. molecules undergoing all processes

= Rate of process / rate of all processes

Question 48

What is the quantum yield of phosphorescence?

Answer 48

Φ_p = Rate of T₁ emission / Rate of S₀ absorption

Considering two steps: Φ_p = ΦT . θ_p

T₁ is formed by isc so: Φ_T = Φ_isc =

k_isc / (k_f + k_ic + k_isc)

If T₁ only decays by photophysics the θ_p = k_p / k_p + k_isc

Φ_p = k_isc / (k_f + k_ic + k_isc) . k_p / k_p + k’_isc

Question 49

How do you find the lifetime of a decay process

Answer 49

An excited state A* can decau by radiatuve processes, non radiative processes or other possible processes B and C

Rateof decay = ( k_r + k_nr + k_B + k_C) [A*] = ΣK[A*]

Integrating gives: [A*]_t = [A*]₀e^-Σkt

Lifetime τ = 1 / Σk

Lifetime if the time when [A*] falls to 1/e of its initial value

it enables conversion of rate constants to timescales

Question 50

What are the types of lifetime?

Answer 50

Two types of lifetime:

• observed lifetime (actual) τ_obs

Can bemeasure in experiment τobs = 1/Σk = 1/kobs

• The radiative lifetime τ⁰

would be measured in experiment if decay was emission only - rarely measured as pther processes compete

Fluorescence and phosphorescence radiative lifetimes:

τ_f⁰ = 1/k_f τ_p⁰ = 1/k_p

Question 51

Why are T₁ states normally lopnger loived and what are typical observed lifetimes?

Answer 51

T₁ states are usualluy longer lived because their photophysical decay routes are spin disallowed (low rate constant)

Typical observed lifetimes:

S₁ τ_obs = pico seconds - nanoseconds

T₁ τ_obs = micro seconds - seconds

Photophysical rate constants depend of the olecule as there may be competing non-photophysical processes

Question 52

How does temperature affect rate constants?

Answer 52

generally has small effect on photophysical rate constants:

k_f, k_ic, k_isc, k_p and k’_isc

Can have a large effect on rate constants for other processes

when T decreases often k_other decreases and Φ_{photophysical} increases

Question 53

Why can T₁ states in the fluid solution decay faster than determined by k_p and k_isc

Answer 53

Non reactived decay can be caused by physical quenching

Question 54

How do rate constants show that T₁ lifetimes are shorter?

Answer 54

A typical quenchng rate constant = k_q = 10⁹ mol^-1dm³s^-1

even at low concentration [Q] = 10^-9 mol dm^-3 gives k_q[Q] = 1s^-1

This can out compete typical T₁ state photophysical rate constant of k_p = 0.05 s^-1

Hence T₁state lifetimes are shorter and photophysical quantum yields are smaller than determined by k_p and k’_isc

Question 55

How is phosphorescence studied?

Answer 55

Generally physical quenching makes phosphorescence too weak to study in fluid solution

Instead it is studies in a rigid matrix when even if quens=chers are present they cannot diffuse

Typically studied at 77 K using a solvent that becomes a rigid c

glass (EtOH)

Question 56

How do heavy atoms affect rate cinstants?

Answer 56

They incerase the rate cnstants for all spin flip processes due to spin-orbit couplinhg

k_isc, k_p, k’_isc

Question 57

What are steady state experimental techniques?

Answer 57

UV-visible absorption - gives spectrukm and aborption coefficients

UV-visible emission - gives spectrum and quantum yields

Question 58

HWat are time-resolved techniques?

Answer 58

Excite with a light pulse and measure the time dependence of

wither emission intensity or absorbanvce

both are proportional to [A*]

Can give observed t

lifetime and rate constants or absorption spectra of excited states

Various designs with pulsed lasers and lamps: milli, nano … femtosecond time resolution possible

Question 59

What is photochemistry?

Answer 59

The reactive decay of ezcited states

reactive processes compete with photophysical porcesses

Question 60

Hiw can the reactivity of excited states differ from ground state

Answer 60

Electronic configuration: bonding, structure, redox ability

Excess energy: ability to overcome activation barriers

Lifetime: S₁ states short lived

T₁ states longer lived

Question 61

What are some possibilities of reactions that can happen to an excited state

Answer 61

Reaction: subsitution, addition:

AB* + C —-> AC + B or ABC

Electron transfer:

AB* + C —-> AB.+ + C.- or AB^{<strong>.</strong>-} + C^{<strong>.</strong>+}

Energy transfer:

AB* + C —–> AB + C*

Isomerisation: AB* —-> AB’

Question 62

What is unimolecular photochemistry and describe photodissociation.

Answer 62

Unimolecular photochemistry: an excited state elementary reaction involving one molecule as reactant

Photodissociation: on excitatiom the molecule dissociates

Above the dissociation limit the absorption band loses vibrational structure and becomes continuous

The fragments carry a continuous range of translational energies

Question 63

What is photopredissociation?

Answer 63

a bound excited state is created initially but interval conversion to a dissociative state rapidly

due to the crossing of the potential energy curve from a dissciative state

the absorption band loses vibrational structure in the region of the crossing

Question 64

How can sample conditions affect dissociation

Answer 64

In the gas phase the fragments mopve apart freely

In solution the solvent cage can keep fragments together long enough for geminate recombination to occur lowering the fragment yield

Question 65

Examples of photodissociation

Answer 65

Question 66

What is photoisomerisation?

Answer 66

On excitation the molecule undergoes a chang ein structure

In the ground state the barrier to isomerisation can be high

In the excited state the excess energy may overcome the barrier

A change in bonding may promote the change in structure

Question 67

How does photoismoerisation affect alkenes?

Answer 67

In alkenes a π —> π* transition weakens the C=C bond so the most stable structure of the excited state is a twisted form

In general isomerisation of a trans or cis isomer gives a mixture of both.

Excitation of a sample of either isomer can give a mixture of bpth called a photostationary state

The composition depends on the relative absorbance of the isomers

If only one absorbs the conversion to the other can be 100%

Question 68

What is photochromism?

Answer 68

A light induced reversible change in colour

This is often seen on switching between two isomers such as between cis and trans or ring opening

Question 69

What is bimolecular photochemistry

Answer 69

An excited state elementary reaction involving two molecules

usually

A* + B ——–> products

Question 70

What are excimers?

Answer 70

Systems hat form dimers in the excited state but usually not in the ground state

Excimer: an excited state dimer

Typically conjugated ring giving pi-stacking structures in the S₁ state

Atb a low concentration of M the monomer fluorescence with vibrational structure

At high structure of M excimer fluorescemnce at longer wavelengthb without vibrational structure

Question 71

What are Exiplexes?

Answer 71

Systems that form complexes in the excited state but usually not in the ground state

Exciplex: an excited state complex (MQ*)

At low concentrayion of Q monomer fluorescence with vibrational structure

At high concentration of Q exciplex fluorescence at longer wavelength without vibrational structure

Question 72

Similarities and differences between eximers and exciplexes

Answer 72

Sim: fluorescence wavelength and vibrational structure characteristics

Potential energy curve characteristics

concentration and time dependence

Differences:

exciplexes are polar, charge transfer comlexes

significant degree of elevtron transfer between the two species

Question 73

What is quenching?

Answer 73

The deactivation of an excited state

can be reactive or non reactive

the emission intensity (quantum yield ) is lowered

Question 74

Scheme for biomolecular quenching

Answer 74

Question 75

How do you fidn the Stern-Volmer equation

Answer 75

In the presemce of quencher Φ_f^q = kf / (kf + knr + k_q[Q])

In the absence of quencher: Φ_f⁰ = k_f / (k_f + k_nr)

Ratio = Φ_f⁰/Φ_f^q = (k_f + k_nr + k_q[Q]) / (k_f + k_nr)

Stern Volmer; Φ_f⁰/Φ_f^q = 1 + (k_q[Q]) / (k_f + k_nr)

Can also be written: 1 + τ_obs⁰k_q[Q]

τ_obs⁰ is the observed lifetime in the absence of quencher

Question 76

What is the Stern-Volmer Equantion used for

Answer 76

determines quenching rate constants k_q

• measure Φ_f at [Q] = 0 amd several [Q] concentrations
• plot Φ_f⁰/Φ_f^q versus [Q]
• determine the gradient = τ⁰_obsk_q
• using known value of τ⁰_obsk_q cslculate k_q

Question 77

What is the diffusion controlled limit?

Answer 77

For many quenchers in non viscous solvents k_q = 10⁹-10¹⁰ sm³mol^-1s^-1 approachung the diffusion controlled limtit

the r.d.s is the reactants diffusing to meet

Can be estimated by the Debye equation

General trend K_diff decreases as T decreases or as η increases

Question 78

What are the thermodynamic and kinetics of quenching by electron transfer?

Answer 78

Between M* and Q and can be oxidative or reductive

Theremodynamics: ΔG = ΔG(M,Q –> M^{<strong>.</strong>+},Q^{<strong>.</strong>-}) - E₀₀ = ox

ΔG = ΔG(M,Q –> M^{<strong>.</strong>-}, Q^{<strong>.</strong>+}) - E₀₀ = red

E₀₀ = electronic excitation energy

The reaction is more favourable in the excited state due to E₀₀

Kinetics: in general as ΔG = -ve, k_q = k_diff

The back electron transfer can also be very fast reforming the reactants

Question 79

When is the yield of electron transfer products the highest?

Answer 79

The ion yield is highest in polar solvents which can stabalise the products

Question 80

What is an energy transfer?

Answer 80

The transfer of electronic excitation energy from an excited state donor molecule to an acceptor

D* + A ——> D + A*

The donor is deactivated to the griund state and the acceptor is promoted to anexcited state

Any emission or photochemistry of D* is replaced by that of A*

Processes occurring from A* are called sensitised

Question 81

Describe the radiative mechanism of energy transfer

Answer 81

The donor emits a photon, the acceptor absorbs it

D* —-> D + hν emission

A + hν —–> A* absorption

For efficient energy transfer the emission of D* must iverlap with the absorption of A

The dominant energy transfer mechanism for molecules at large separation: solutions at low conc

Could occur over large distances - sun - earth

Question 82

Describe long range coulombic mechanism of energy transfer

Answer 82

Energy transfer by dipole dipole interaction between D* and A without photon emission (non radiative) or contact

For efficient energy transfer the A/A* and D/D* energy gaps must match - energy release matches energy acceptance, emission overlaps with absorption

D* - A distance typically 2-10 nm (compared with molecular size this is a large distance)

Question 83

Characteristics of long range coulombic energy transfer

Answer 83

Rate constant larger than diffusionm controlled limit because collision is not required: k_E.T>k_diff

experimental rate constants and distances can be compared with those derived from theory

generally observed between singlet states

Question 84

Describe the shirt range electron exchange mechanism of energy transfer

Answer 84

Energy transfer by an exchange of electrons between D* + A without photon emission at short distances < 2 nm

For efficient energy transfer

• the A/A* and D/D* energy gaps must match - energy released matches energy absorbed
• D* - A distance typically = 1 nm - similar to molecular size

Question 85

Characteristics of shirt range electron exchange

Answer 85

rate cinstant typically approaches the diffusion controlled limit if the energy of A* lies below D*

k_E.T = k_diff if E(A*) ≤ E(D*)

A wider range of possible transfers due to the detailed spin selection rules

Question 86

What is photosensitisation?

Answer 86

Energy transfer to create T₁ states that cannot be cerated readily

• cannot create directly by absorption (spin selection rule)
• quantum yield for intersystem crossing may be low

Ceate by triplet triplet energy transfer

D* (T₁) + A (S₀) —-> D(S₀) + A* (T₁)

Requires a donoro with high Φ_isc to create T₁ and a T₁ state energy that is higher than that of the acceptor T₁

Question 87

What is oxygen quenching?

Answer 87

Quenching by oxygen is quite common so air is often excluded from ophotochemical studies