Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 2A / Section B / 22

For the absorption spectrum, pass a scanned monochromatic beam through the solution and a matched solvent reference. Record the incident and transmitted intensities and calculate the absorbance $A(\lambda )={\log }_{10}[I_0(\lambda )/I(\lambda )]$, using an appropriate blank and concentration to avoid saturation.

For the fluorescence emission spectrum, excite at a fixed wavelength within an absorption band, collect emission away from the incident beam, commonly at $90^{\circ }$, and scan an emission monochromator. Correct the detected signal for wavelength-dependent detector sensitivity, subtract solvent/background light and exclude scattered excitation light. The two plotted vertical scales have been independently normalized; their equal peak heights do not imply equal absorption and emission rates or unit fluorescence quantum yield.

Vibrational and solvent relaxation remove energy before emission, producing a Stokes shift. Absorption initially excites a vertical transition from the populated ground-state vibrational levels to excited-state vibrational levels. Vibrational relaxation and solvent rearrangement then bring the molecule near the relaxed lowest excited singlet level before fluorescence. Emission ends in vibrationally excited levels of the electronic ground state, so the emitted photon generally has less energy and a longer wavelength. This explanation does not require a change of electronic multiplicity.

The approximate mirror relationship follows from the Franck-Condon principle when the ground and excited electronic states have similar vibrational frequencies and comparable potential shapes displaced along the active coordinates. Absorption samples $S_0,v\simeq 0\to S_1,v^{\prime }$ overlaps, while relaxed emission samples $S_1,v^{\prime }\simeq 0\to S_0,v$ overlaps. For displaced harmonic potentials, the corresponding overlap intensities form similar progressions on opposite sides of the transition between the lowest vibrational levels of the two electronic states.

The symmetry is approximate and is most naturally compared on an energy or wavenumber axis. A wavelength axis is nonlinear in photon energy, and solvent response, frequency changes, differing transition strengths and instrumental corrections can spoil an exact mirror image.

For a fluorescence excitation spectrum, hold the emission detector at a fixed wavelength within the fluorescence band and scan the excitation wavelength. Correct for the incident photon flux and instrumental response at each excitation wavelength.

For one emitting molecular species with an approximately excitation-independent fluorescence yield, it should follow the absorption spectrum. In a dilute sample the detected emission is proportional to the photons absorbed, $I_{\mathrm{exc}}(1-10^{-A}){\varphi }_F\simeq I_{\mathrm{exc}}(\ln 10)A{\varphi }_F$. Without the dilute approximation it follows absorptance rather than absorbance exactly. Impurities, photochemistry or multiple emitting states could invalidate that correspondence.

The phosphorescence quantum yield is the number of phosphorescence photons emitted per photon absorbed. In rate form,It is dimensionless. The denominator counts absorbed photons, not incident photons or the instantaneous triplet population.

Almost every absorbed photon generates a triplet, within the stated kinetic scheme. The branching fraction from the excited singlet isso $k_{\mathrm{ISC}}[S_1]\simeq I_{\mathrm{abs}}$ under steady illumination. This statement concerns the triplet formation yield; it does not mean that the triplet population or the phosphorescence yield is also one. The triplet can subsequently deactivate or be quenched.

The steady-state approximation for the excited singlet givesand that for the triplet givesCombining these with the quantum yield definition gives the more general expressionUnder the supplied limit $k_{\mathrm{ISC}}\gg k_F$, this becomesThe approximation is the near-unit triplet formation yield, not an assumption that phosphorescence is the only triplet loss process.

Without quencher, ${\varphi }_{P,0}=k_P/(k_P+k_D)$. Taking the ratio gives the Stern-Volmer equationThus the plot has unit intercept and slope $k_Q{\tau }_0$. A triplet formation yield less than one would cancel from this ratio if it remained independent of quencher concentration. The result assumes dynamic triplet quenching and no additional quencher-induced absorption or singlet loss.^[ref]

$k_Q=9.9\times 10^4\mathrm{d}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. The unquenched triplet lifetime isSince the slope is $k_Q{\tau }_0$, the bimolecular rate constant isFor comparison, the unquenched phosphorescence yield in the unit-triplet-yield approximation is $180/198=0.909$.

The measured decay is approximately consistent with the predicted $198{\mathrm{s}}^{-1}$ loss rate. Following rapid singlet-to-triplet population transfer, the unquenched phosphorescence intensity obeysA straight-line fit of $\ln I_P$ to the four times, expressed in seconds, givesThis differs from $198{\mathrm{s}}^{-1}$ by about $1.7\%$. For example, using the first point and the predicted rate gives subsequent intensities about $67.3,55.2,45.3$, close to the supplied $66.3,56.2,45.3$. The data support the expected millisecond exponential decay to this precision. With no measurement uncertainties supplied, a formal statistical acceptance or rejection is not justified.