Chapter 5 Workshop Questions for Week 5

5.1 Short conceptual question - O2 quenching

  1. Given molecular oxygen has an emission with λmax around 1280 nm why is it such a good quencher of excited states?

Singlet oxygen emisison occurs principally from the \(^1 \Delta _g\) state but it has more than one transition available to form an excited state, and a large amount of excitation occurs to initially form the \(1 \Sigma _g^+\) state (722 nm), this state has a much higher energy transition. However oxygen also has absorption transitions (you don’t need to know these but just showing to really prove my point) with λmax 477 nm, 532 nm, 630 nm, 762 nm, 1063 nm and 1268 nm and so there is usually a good overlapintegral for emission from any state.

  1. Why is the efficiency of quenching lower for excited singlet states?

Singlet states have a much shorter lifetimes and so they are less likely to encounter a quenching agent before they fluoresce.

5.2 Short mathematical question - Förster resonance energy transfer

A donor, D, with an unquenched lifetime of 5.0 ns, was found to have a steady state emission intensity of 20.5 in free solution and 4.1 when in the presence of a quencher, Q. Assuming the Förster distance is 50 Å determine :

  1. the transfer efficiency, E.

\[\begin{equation} E = 1 - \frac{\tau_{DA}}{\tau_D} \tag{5.1} \end{equation}\]

This relationship also works for qunatum yield and consequently intensity (you can derive this from the definitions of lfietime and quantum yield, equations (2.1) and (2.2) for quenched and unquenched systems)

So \(E = 1 - \frac{4.1}{20.5} = 0.80\)

  1. the lifetime of D in the presence of the quencher Q

Again from the relationship between lifetime and quantum yield (this should be ‘derived’ from equations (2.1) and (2.2) if ever needed) then:

\[\begin{equation*} \frac{\phi _{quenched}}{\phi_{unquenched}}=\frac{\tau _{quenched}}{\tau_{unquenched}} \end{equation*}\]

therefore this can be rearranced to be get \(\tau_{DA}\)=1.0 ns

  1. the equilibrium separation of D & Q

\[\begin{equation} E = \frac{R_0^6}{R_0^6+r^6} \tag{5.2} \end{equation}\]

where \(R_0\) is the forster distance

rearranging gives \(r = \sqrt[6]{\frac{R_0^6-ER_0^6}{E}}\) and \(r = \sqrt[6]{\frac{1-E}{E}}R_0\) so r = 40 Å

  1. the rate constant for energy transfer

\[\begin{equation} k_{ET}=\frac{1}{\tau_D}\frac{R_0^6}{r^6} \tag{5.3} \end{equation}\]

\[\begin{equation*} k_{ET}=\frac{1}{5.0 \textrm{ ns}}\frac{50^6}{38^6} = 7.6 \times 10^8 \textrm{ s}^{-1} \end{equation*}\]

5.3 Short conceptual quesiton - Förster distance

Determine the quantum yield of emission of the donor in a FRET pair system where donor and acceptor are separated by the Förster distance, the lifetime of the unquenched system is 8.4 ns. The unquenched quantum yield is 0.58*

Efficiency is another term for quantum yield of energy transfer, by definition the forster distance is the point where this is 0.5 and so we can rearrange equations (2.1) and (2.2) to find that this effectively halves all other quantum yields and so \(\phi_{f, quenched}\)=0.29

5.4 Short conceptual quesiton - Donor-accceptor system

In a single molecule study of a donor bound to an acceptor via a flexible, short aliphatic chain it was found that lifetime of the excited state varied (time beween single photon detection of emission after an excitation pulse). Suggest why this is the case.

The two main things that affect lifetime for energy transfer quenching are distance and for forster energy transfer dipole dipole interaction which varies between 0 and a maximum set by the distance. With an aliphatic chain it could be that the distance is fixed by rigid rings and only dipole interactions are changing, or it could be that both factors are able to change depending on the system.

5.5 Short conceptual quesiton - Quenching of ruthenium

Ruthenium tris bipyridine [Ru(bpy)3]2+ has a λ~max, abs~ = 470 nm and λ~max, em~ = 465 nm. The natural lifetime is 13.6 µs.

Suggest why molecular oxygen is an efficient quencher of the emmission of [Ru(bpy)3]2+.

With that lifetime [Ru(bpy)3]2+ has a spin forbidden transition, it therefore has a very long lived excited state and so is much more probable to be quenched in this lifetime.

5.6 Short conceptual question - Energy transfer and spectra

How would you expect the absorption and emission spectra observed in an experiment to change when an emmisive acceptor is added?

The absorption spectrum shouldn’t change (if it forms a static complex it is likely there will be a shift in λmax) , it will just be a linear combination of the absorptions of donor and acceptor. If the donor is excited (and the acceptor not) then the emission for a fixed concentration of donor would decrease and an emission for the acceptor (which doesn’t absorb at the excitation wavelength) would be appeared.

If you keep the emisison wavelength the same and scan the intensity of this band with changing excitation wavelength how will this (excitation) spectrum vary from the absorbance and how can it be used to confirm energy transfer?

Scanning the excitation wavelength and measuring the emission intensity should give a spectrum similar to the absorbance spectra as the more something absorbs teh more excited states are generated and the more it emits.

If we scan at a wavelength where the acceptor emits then if there is energy transfer the ‘excitation spectrum’ would show both donor and acceptor ‘absorbance bands’ if there is energy transfer this clearly seen by apeparance of both bands.

5.7 Short conceptual question - Effect of separation

‘cyanine-3’ and ‘cyanine-5’ are cyanine dyes which are frequently used as markers in experiments with DNA.

Cyanine-3 has a λmax,abs = 554 nm and λmax,em = 568 nm

Cyanine-5 has a λmax,abs = 649 nm and λmax,em = 666 nm

In an experiment they are tethered to either end of a DNA oligomer of 4, 5, 6, 7, 8, 9, 10 11 & 12 base pairs. However the pattern expected of increasing lifetime as the separation increased was not observed. Suggest why this may be occuring.

This is because DNA is a helix and has a turn, and so the separation effects are also blended with dipole-dipole effects showing a pattern more complicated than just distance. The helix completes a full turn in just over 10 bases and so you would expect to see similarities to just the distance every 10 turns. This is more detail than I would expect but thought it told the picture nicely