Chlorophyll for Efficient Solar Energy Capture

Question 1

What are the four phases of photosynthesis light reactions?

Answer 1

1) antennas and light collection
2) photochemistry (electronic energy converted to chemical energy via photons)
3) electron transfers stabilise chemical energy
4) storage of energy in new chemical compounds

Question 2

Where is chlorophyll found?

Answer 2

-in high concentration in proteins in photosynthetic membranes which are folded to increase surface area

Question 3

Atomic Bohr Model / Quantum Theory

Answer 3

• single electron states at different energy levels

- only photons with exact energy of transition (e.g. E2-E1) can be absorbed or emitted

Question 4

Molecular Bohr Model / Quantum Theory

Answer 4

• electrons can interact with multiple nuclei which can rotate and vibrate
• a single electron state is split into many sub-states so transitions are possible for a range of energies
• the distribution of electrons across sub-states is governed by thermal equilibrium

Question 5

Cholorophyll Electron Transitions

Absorption

Answer 5

• at room temperature, most electrons are near the bottom of the ground state
• most likely terminus after absorption is the centre of the excited state since the density of subtrates is greatest there
• this leads to broadening of peaks in the chlorophyll specta
• absorption is ultra-fast (fs)

Question 6

Cholorophyll Electron Transitions

Fluoresence

Answer 6

• higher vibrational states are unstable and decay to the lowest vibrational state in picoseconds (energy loss as heat)
• this means that the most likely starting point for emission is the bottom of the exctied state
• the higher electronic state in the lowest vibrational state is relatively stable, decays in ns
• the most likely teminus for emission is the centre of the ground state

Question 7

Cholorophyll Electron Transitions

Stokes Shift

Answer 7

• since ΔEa > ΔEf, the peak of absorption is at higher frequency than the emission peak
• the differenece in the location of the peaks is called Stokes Shift

Question 8

Singlet Ground State, S0

Answer 8

• both electrons in ground state
• one spin up, one spin down
• spins are paired, anti-parallel

Question 9

Singlet Excited State, S1

Answer 9

• one electron in ground state
• one electron in excited state
• one spin up, one spin down
• spins are paired, anti-parallel

Question 10

Triplet State, T1

Answer 10

• with low probability (very slow) the S1 excited state can become a triplet state, T1
• one electron in ground state, one electron in excited state
• electron in excited state flips spins so that has the same spin as the ground state electron
• this is allowed under the Pauli exclusion principle since the electrons have different energies
• spins unpaired, parallel

Question 11

Photoluminesence

Definition

Answer 11

• emission of a photon
• occurs by two mechanisms:
• -fluorescence
• -phosphoresence

Question 12

Jablonski Diagram

Fluorescence

Answer 12

• photon emission from singlet excited state
• radiative
• timescale: 10^(-10) - 10^(-7)s

Question 13

Jablonski Diagram

Phosphoresence

Answer 13

• photon emission form triplet excited state
• radiative
• timescale: 10^(-6)-10s

Question 14

Photosynthesis and Triplet States

Answer 14

• triplet states can be highly reactive
• this can be damaging to biological systems since they’re so long lived
• they must be dealt with (or prevented) during photosynthesis

Question 15

Absorption Spectroscopy

Experimental Set-up

Answer 15

• light source directed towards monochromator
• monochromatic light emerges at incident intensity, Io
• incident light passes through sample substance
• dectector measures intensity of transmitted light, I

Question 16

Absorption Spectroscopy

Absorbance Equation

Answer 16

A = log_10_(Io/I) = εCx
-where A is absorbance, C is the concentration of sample molcules, x is the thickness of the sample and ε is the molar absorption coefficient

Question 17

Absorption Spectroscopy

Estimating Absorbing Materials Concentration

Answer 17

-since A∝C, A can provide a simple estimate for concentration

Question 18

Absorption Spectroscopy

Calculating the Molar Absorption Coefficient

Answer 18

• generally, ε increases with the number of pi binds
• for known concentration range, can plot A against C
• will be a linear relationship, A(λ) ∝ ε(λ) for constant concentration
• calculate ε form gradient

Question 19

Absorption Spectroscopy

Absorption Spectrum

Answer 19

• plot of A against wavelength has important characteristics:
• -number, position and shape of peaks
• -wavelength of maximal absorption
• -width of absorption peaks
• the strength of a materials’ absorbancy is quanitified as a function of wavelength (or frequency)

Question 20

Estimating Effective Absorption Strength

Answer 20

-integrate ε(λ) over the wavelengths of absorption

Question 21

Chlorophylls Absorption

Wavelengths

Answer 21

• absorption over relatively large wavelength range, 400-900nm, requires extensive pi-system of conjugated bonds meaning a fairly large molecule
• when energy is absorbed it is distributed over the pi-system
• X and Y transitions for S0->S1 & S0->S2 means up to 4 absorption bands for one molecule

Question 22

Chlorophylls Absorption

Energy Transfer Between Chlorophylls

Answer 22

• the lowest excited singlet state is sufficiently long-lived (τ~5ns) to allow energy transfer to neighbouring chlorophyll molecule
• theory and experiment show this τ allows energy propagation between assemblies of many chlorophylls over relatively large distances (10-100nm)

Question 23

How stable is the excited state of a particular fluorophore?

Answer 23

-to answer this consider how long an electron stays in the excited state before decaying
-the lifetime of the excited state is equal to the reciprocal of the combined decay rates, typically 10ps-10ns
τ = 1 / (kf + kπr)
-where kf is the rate of radiative decay and kπr is the rate of all non-radiative dissipation

Question 24

How bright is a particular fluorophore?

Answer 24

-consider reemission of photons vs. non-radiative dissipation
-fluorescence quantum yield is the rate of fluorescence over all other processes
-can be close to unity or <5%
φf = kf / (kf + kπr)
-where kf is the rate of radiative decay and kπr is the rate of all non-radiative dissipation

Question 25

What does amount of fluoresence depend on?

Answer 25

- fluoresence depends on: - -absorption i.e. concentration, ε - -efficiency (or rate) of the competing decay processes

Question 26

Intensity of Fluoresence

Answer 26

If = φf * Ia = kf / (kf + kπr) * Ia -where Ia is the intensity of absorbed light, φf is the fluoresence quantum yield, kf is the rate of radiative decay and kπr is the rate of all non-radiative dissipation

Question 27

Relative Fluoresence Intensity

Answer 27

If ~ Fr(λ) ∝ φf(λ) If ~ Fr(λ) ∝ A(λ) -most fluoresence spectrometers measure 'relative' fluoresence, Fr(λ), because it can be challenging to detect every photon in/out

Question 28

Steady State Fluoresence Spectroscopy

Answer 28

- fluoresence measured at 90' to light source so incident light isn't picked up - typically use a collection time of 10-60sec for reasonable signal to noise ratio - fluoresence can be measured as a function of excitation wavelength with fixed emission position or function of emission wavelength with fixed excitation position - if we excite a particular wavelength (e.g. where absorption is known to be high) then record fluoresence intensity at a range of emission wavelengths (integrate the signal over a defined time)

Question 29

Measuring Fluorescence Quantum Yield

Answer 29

-with a special accessory added to the fluoresence spectrometer, absolute fluoresence quantum yield can be measured -this is done by collecting every excitation vs emitted photon: φf = kf / (kf + kπr) = no. of photons emitted / no. of photons absorbed

Question 30

Time-Resolved Fluoresence Spectroscopy

Answer 30

- with an alternative instrument configuration we can measure fluoresence intensity as a function of time - to do this we need to measure at a ps-ns timescale - the measurement system needs a fast and well-defined light source which excites the sample coherently so the chromophore enters the excited state at a defined t=0 - use a pulsed laser with 50-100ps duration pulse - uses timing electronics with a cumulative histogram graphical display - need a single-photon-counting detector with high detection quantum efficiency which can be quickly rest ready for another photon, dead time

Question 31

Fluoresence Decay Histogram

Answer 31

- using time-resolved spectroscopy repeat the time measurement many times and count how many photons have arrived after what time - sort the photons into a histogram accordint to their arrival times - exponential fit of fluoresence decay to obtain fluorescence lifetime i.e. number of molecules in the excited state decreases at a rate proportional to the current number of electrons in the excited state

Question 32

Fluorescence Decay Curve | Equation

Answer 32

``` N(t) = No * exp(-kt) -where k is the rate constant -OR N(t) = No * exp(-t/τ) -where τ=1/k is the mean lifetime, the mean time at which the population of the assembly is reduced to 1/e~37% of the initial value ```

Question 33

Fluorescence Decay Curve | Multi-exponential Fitting

Answer 33

F(t) = Σ Ai * exp(-t/τi) -sum over i=1 to i=n -e.g. a bi-exponential fit, if there are two decay processes F(t) = A1exp(-t/τ1) + A2exp(-t/τ2) + z -where τ1 is the lifetime of component 1, A1 is the amplitude of component 1 and equivalently for τ2 and A2 -and z is the fitting constant (detector background noise?)

Question 34

Fluorescence Excited State Lifetime and Natural Excited State Lifetime

Answer 34

-the natural excited state lifetime is the excited state lifetime in the absence of non-radiative dissipation (knr=0), not possible to measure directly τ0 = 1/kf -the fluoresence lifetime of the excited state is the resiprocal of the combined decay rate of the excited state: τf = 1 / (kf + knr) = τo*φf

Question 35

What does relative fluorescence intensity depend on?

Answer 35

- emission wavelength obsrved - excitation wavelength - samples fluorescence QY - sample's concentration - the detector's quantum efficiency - age of excitation lamp

Question 36

Why is fluorescence relavant to studying photosynthesis?

Answer 36

- it helps us to understand the energy capture processes going on as part of 'light harvesting': - -energy absorbed / energy of excited state - -effectiveness of process (non-radiative decay ~wastage) - -to quanitfy energy transfer efficiency, indirectly from fluoresence

Question 37

Jablonski Diagram | Intersystem Crossing

Answer 37

- transition between vibrational levels belonging to electronic states of differnent spin multiplicity, e.g. S1 to T1 - non-radiative - timescale: 10^(-10) - 10^(-6)s

Question 38

Jablonski Diagram | Internal Conversion

Answer 38

- an electron in a higher energy singlet state transfers to a lower lying singlet state, it is immediately followed vibrational relaxation to the lowest vibrational level of the new electronic state - non-radiative - timescale: 10^(-11) - 10^(-9)s

Question 39

Jablonski Diagram | Absoprtion

Answer 39

- transition from lower to higher electronic state, energy of photon converted to internal energy - radiative - timescale: 10^(-15)s

Question 40

Jablonski Diagram | Vibrational Relaxation

Answer 40

- transition to lower vibrational level within the same electronic state - non-radiative - timescale: 10^(-12) - 10^(-10)s

Question 41

Jablonski Diagram | Fluorescence

Answer 41

- transition from the lowest vibrational level of one electronic state to a lower electronic state of the same spin multiplicity - radiative - timescale: 10^(-10) - 10^(-7)s