Chapter 3 Emission of Light
3.0.1 Learning Objectives
At the end of this block you should be able to:
Define singlet and triplet states in terms of molecular orbitals
Explain the mirror symmetry observed in absorption/fluorescence spectra.
Construct and explain the Jablonski diagram.
Define the rates of the processes in the Jablonski scheme and give typical values for the rate constants.
Define singlet and triplet lifetimes and describe how they are measured.
Analyse decay curves to obtain lifetimes of excited states.
Define and derive the quantum yield of fluorescence and phosphorescence.
Discuss the heavy atom effect.
Only the light absorbed by a molecule can produce photochemical change in a molecule. If a species absorbs radiation then one particle is excited for each quantum of radiation absorbed. This concept of light having to be absorbed before anything else can happen is an obvious one, but is an important concept in itself, it goes on to define the concept of quantum yield of a process, effectively the proportion of excited states which ‘decay’ to any given pathway.
3.1 Fluorescence & Phosphorescence
Just as there are ‘spin forbidden’ and ‘spin allowed’ processes for absorption, when the molecule is in an excited state there are a allowed and forbidden path ways of deactivation of the excited state.
If we look at a Jablonski diagram, figure 19, which you should already be familiar with, it shows the range of photophysical deactivation pathways of an excited state.
A Jablonski diagram (figure 3.1) usually uses a short hand to refer to the excited states, where a ground state, if it is a singlet, is referred to as S0, and each subsequent excited state is referred to as S1, S2 etc.
Figure 3.1: Jablonski diagram showing: A, the absorption of a photon, and the photophysical deactivation pathways, IC, internal conversion, ISC, intersystem crossing, F, fluorescence & P, phosphorescence.
Triplet states are referred to as T1, T2… if the ground state is not a triplet then there is no state labelled T0. All excited states are referred to by integer vales of 1 or more.
A note on Jablonski diagrams: transitions which are either allowed or non-radiative and therefore not governed by selection rules appear as either horizontal or vertical transitions, forbidden transitions (such as phosphorescence) are indicated by transitions which occur at an angle. Radiative transitions (those that occur with either the absorption or emission of a photon) are shown as solid lines with the incident or excident photon shown as a wave like arrow. Non radiative transitions are shown as dashed (or alternatively wiggly) lines to clearly differentiate them from radiative transitions. Transitions with a change of spin are isoenergetic (shown as horizontal lines) before decaying via internal conversion (vertical lines). A Jablonski diagram is a sketch of the energy levels and should not include details of the potential wells of either the ground or excited state.
The spin selection rule which applies for absorption transitions \(\Delta S = 0\), so formally emission transitions from the S1 to S0 are spin allowed, allowed emission transitions are referred to a fluorescence.
Emission transitions that involve a change in spin of the electron, such as from T1 to S0, are spin forbidden, and referred to as phosphorescence. The excited state can also decay via non-radiative transitions and other routes.
Table 3.1 the deactivation pathways from excited states:
| ‘Allowed transitions’ | ||
| Singlet-singlet absorption Singlet-singlet emission |
fluorescence |
\(S_0 + h \nu \longrightarrow S_1\) \(S_1 \longrightarrow S_0 + h \nu '\) |
| ‘Forbidden transitions’ | ||
| Singlet-triplet absorption Triplet-singlet emission |
phosphorescence |
\(S_0 + h \nu \longrightarrow T_1\) \(T_1 \longrightarrow S_0 + h \nu ''\) |
| ‘Other transitions’ | ||
| Internal conversion |
(vibrational relaxation) (vibrational relaxation) |
\(S_1 \longrightarrow S_0 + heat\) \(S_1 \longrightarrow T_1 + heat\) \(T_1 \longrightarrow S_0 + heat\) |
| Other pathways | ||
| Quenching of excited state Chemistry from excited state |
\(S_1 + Q \longrightarrow S_0 + Q +heat\) \(S_1 + Q \longrightarrow S_0 + Q^\ast +heat\) \(T_1 + Q \longrightarrow S_0 + Q +heat\) \(T_1 + Q \longrightarrow S_0 + Q^\ast +heat\) \(S_1 \longrightarrow\) new/changed molecule |
The spin selection rule has important consequences on the lifetime of singlet and triplet states. The spin allowed process of fluorescence usually has a short lifetime (less than 10 ns), whereas the spin forbidden process of phosphorescence has a significantly longer lifetime, which can be as long as several seconds.
Vibrational relaxation occurs in the ps timescale, consequently the excited state decays vibrationally to the S1, v’ = 0 state and emission essentially always occurs from the ground vibrational electronically excited state. This is an important concept in photophysics and is called Kasha’s rule.
3.1.1 Kasha’s Rule
An excited state always emits from the lowest energy level of that spin multiplicity state.
The Frank-Condon principle applies equally to emission processes and the excited state will decay to a number of available vibrationally excited states of the ground electronic state. After relaxation to the \(v’ = 0\) level emission occurs to each of the available vibrational states of the ground state. The relative intensity of each band depends upon the coupling or overlap integral between the ground and excited states, as shown in figure 3.2.
Figure 3.2: The absorption and emission spectra of anthracene in ethanol solution. Both the absorption and emission spectra show vibrational fine structure, due to transitions to discrete vibrational energy levels in the excited and ground states respectively.
3.2 Stoke’s shift
The Franck-Condon factors in absorption and emission is clearly evident by the symmetry shown in the emission of most organic chromophores. The Stoke’s shift, the difference between the \(v=0\) to \(v’=0\) absorption peak and the \(v’=0\) to \(v=0\) emission peak, can simply be explained by the offset of the ground and excited state potential wells (figure 3.3). Absorption happens from the centre of the \(v=0\) ground state and emission from the centre of the \(v’=0\) excited state, however this does not explain why the two curves are offset from one another.
, accessed July 2020]](images/stokesspectrum.png)
Figure 3.3: The absorption and emission spectra of rhodamine 6G showing the Stokes shift of the two spectra.[Data from OMLC, accessed July 2020]
The axis title of ’reaction coordinate’ is used ubiquitously, but is poorly understood. The simple idea that a bond becomes longer (as would be the case for a diatomic molecule) does not apply as we are considering the whole molecule.
The Stokes shift can be far more easily accounted for if we consider what is going on in the molecule. Initially the molecule exists in a state optimised for for that state, the shape and structure of the dye molecule and the arrangement of solvent around it are all minimised to give the lowest energy configuration. We shall consider the most simple case of the transitions between the ground vibrational states of the HOMO & LUMO.
Absorption of a photon is a very fast process, in the order of femtoseconds, and is significantly faster than molecular vibrations or structural reorganisation. Consequently the molecule becomes excited into a state which has exactly the same structural configuration and salvation shell as the ground state, however this is now no longer the most optimal state, figure 3.4.
Figure 3.4: Sketch of the photophysical processes involved in the Stokes shift of emission. The ground state structure of the dye and solvent sphere (lower left) is maintained upon almost instantaneous absorption of a photon, however this then relaxes to an optimised configuration of dye and solvent for the excited state (top right), however this destabilises the ground state.
Since the excited state is relatively stable (ns) when compared with the time taken for vibrations and molecular restructuring (ps) the excited state dye molecule and the solvent will reorganise to minimise the energy of the excited state, however in doing so this destabilises the ground state of the molecule. The actual process of emission is again very fast (ps) and so there is no opportunity for the structure of the chromophore or solvation sphere to reorganise. Upon deactivation this state, optimised for the excited state reorganises, lowering the energy of the ground state and destabilising the excited state.
As a consequence of this stabilising action of the ground and excited states the energy gap of absorption is larger than the energy gap of emission, or the emission of a photon is redshifted from the absorption. This explains the origin of Stokes shift, and explains the offset the Morse curves of the ground and excited states which have previously been used to explain the origin of the Stokes shift.
3.3 Emission from S2
Kasha’s rule is almost always true, but there are known exceptions to this rule, one of these examples is azulene. Azulene has a broad absorption band between 500-600 nm, however emission from this molecule occurs at below 400 nm. If we look at figure 3.5 which shows the absorption and emission spectra of azulene it can be seen that the emission spectra is a mirror image of S0-S2 (excitation into the second excited electronic state) absorption band, indicating the emission is also from the S2 to S0 states.
Figure 3.5: The absorption (solid line) and emission (dashed line) spectrum of azulene in ethanol. Note the emission appears in the near UV part of the spectrum.
The molar extinction coefficient is also significantly higher for this S0-S2 absorption indicating a much greater overlap integral of these two states.
What causes the unusual photophysical example? In short there is a very large energy gap between the S2 and S1 excited states this has the effect of greatly reducing the rate of internal conversion. There is a very weak emission from the S1 to S0, in this case the energy gap is relatively small and there is an efficient deactivation of the excited state by internal conversion.
3.4 Rates of Photophysical Processes & Quantum Yield
Fluorescence (and phosphorescence) are first order processes, just like radioactive decay. The probability of a molecule decaying from an excited state depends only the concentration of that excited state, \([A^\ast]\).
Rate of decay of \([A^\ast]\):
\[\begin{equation} \frac{\textrm{d}[A^\ast]}{\textrm{d}t}=-k_f [A^\ast] \tag{3.1} \end{equation}\]
It is usual to refer to the ‘lifetime’ of photophysical processes, and it is a term already used at the start of this chapter, the lifetime of a process, usually given the symbol τ, is the reciprocal of the rate constant for that process, so τf = 1/kf. Often the natural lifetime of a dye is used, this assumes that there are no other deactivation pathways from the excited state and all of the energy is lost as fluorescence (or the fluorescence quantum yield is 1). The natural lifetime of fluorescence is given the symbol τf0, it is the maximum a lifetime can be and is solvent dependent.
If there is more than one possible deactivation pathway from the excited state is is often useful to be able to quantify these values. The quantum yield of a process describes the efficiency of each process, when a molecule absorbs a photon of light it generates an excited state; there are a number of different ways the molecule can decay from this excited state (see section 3.1, figure 3.1). The quantum yield for each process is essentially the ratio of reactions proceeding down each pathway and the number of photons absorbed, quantum yield has no units.
\[\begin{equation} \phi = \frac{\textrm{number of reactions proceeding down a given pathway}}{\textrm{number of photons absorbed}} \tag{3.2} \end{equation}\]
| Rate of Process | Quantum Yield | ||
|---|---|---|---|
| Absorption | \(S_0 + h \nu \longrightarrow S_1\) | \(I\) | |
| Fluorescence | \(S_1 \longrightarrow S_0 + h \nu'\) | \(k_f[S_1]\) | \(\phi _f\) |
| Internal conversion | \(S_1 \longrightarrow S_0 + \textrm{heat}\) | \(k_{ic}[S_1]\) | \(\phi _{ic}\) |
| Intersystem crossing singlet to triplet |
\(S_1 \longrightarrow T_1 + \textrm{heat}\) | \(k_{ST}[S_1]\) | \(\phi _{ST}\) |
| Phosphorescence | \(T_1 \longrightarrow S_0 + h \nu''\) | \(k_p[T_1]\) | \(\phi _p\) |
| Intersystem crossing internal conversion triplet to singlet |
\(T_1 \longrightarrow S_0 + \textrm{heat}\) | \(k_{TS}[T_1]\) | \(\phi _{TS}\) |
According to the Stark-Einstein law the total quantum yield of all processes is unity (one), however if secondary reactions occur (such as radical propagation) it may be more than 1. The quantum yield and rate of a process are intrinsically linked. It should be implicit that the faster the rate of a process (the bigger the rate constant) the bigger the quantum yield of that process; in fact the quantum yield of a process, such as fluorescence, is the rate of fluorescence divided by the sum of the rates of all of the deactivation processes (equation (3.3)).
\[\begin{equation} \phi_f = \frac{k_f^0}{k_f^0+k_{ic}+ k_{ST}+k_{\textrm{other}}} \tag{3.3} \end{equation}\]
(Examples of expressions for other processes may be found in Section 3.9 equations (3.8), (3.9) & (3.10).)
When there are more processes occurring then the singlet lifetime (sometimes called the fluorescence lifetime) will be smaller than the natural fluorescence lifetime, because now the denominator term is bigger (equation (3.4)).
\[\begin{equation} \tau_S = \frac{1}{k_f^0+k_{ic}+ k_{ST}+k_{\textrm{other}}} \tag{3.4} \end{equation}\]
If we assume that there is no quenching of the system (this is discussed later) and no chemistry the only decay pathways from the excited state are emission (fluorescence), internal conversion and intersystem crossing to the triplet state, and so consequently only these terms would appear in the equation for the quantum yield. Processes that occur from the triplet state are independent, and therefore none of these terms occur in quantum yield or lifetime equations from the singlet, fluorescent state.
For processes that occur from the triplet state, then obviously, the quantum yield will depend upon the total number of triplet states produced, which is given by the quantum yield of intersystem crossing. So the quantum yield of phosphorescence is given by:
\[\begin{equation} \phi_p = \frac{k_p^0}{k_p^0+k_{TS}+k_{\textrm{other}}} \tag{3.5} \end{equation}\]
Quantum yields of photophysical processes can be difficult to measure, in part because it is difficult to easily quantify the number of photons going into a sample, and because emission occurs at all angles from the sample. A technique called ‘actinometry’ is usually used where the emission from an unknown sample is compared with that of a well characterised sample.
A good actinometer should have a good overlap in both the absorption and emission spectra, a high molar absorption coefficient, it should be photochemically stable and have a high emission quantum yield, a selection of well characterised actinometers are included in figure 3.3. Measurements on the absorption at the excitation wavelength, and the full emission spectra are required for both the unknown and the actinometer, as well as details on the reactive index of the solvents used. For best results a monochromatic light source should be used (a laser) and the same excitation wavelength must be used for both samples; the absorbance of the solutions should also be low (<0.1).
Usually emission spectra are measured as intensity against wavelength, but when ‘integrating’ the emission of the unknown and actinometer it is important that they are integrated over a linear space (energy or frequency) which can make calculations more difficult (but perfectly doable in something like Excel). Equation (3.6) shows the relationship between the fluorescence quantum yield of an unknown, Φ, with that of a reference actinometer, ΦR. A is the absorbance at the excitation wavelength, and n the refractive index of the solvent. Here, because all terms appear as ratios there is no need to worry about units with anything.
\[\begin{equation} \phi=\phi_R \frac{\int{I}\textrm{d}\bar \nu}{\int{I_R}\textrm{d}\bar \nu}\frac{A_R}{A}\frac{n^2}{n^2_R} \tag{3.6} \end{equation}\]
The integrated emission intensity, \(\int{I}\textrm{d}\bar \nu\), of the unknown and reference is here shown as being integrated over wavenumber, but so long as it is integrated over linear space4 it does not matter if frequency or energy is used instead as any units from the terms cancel.
| Reference | Solvent | λex/nm | Φem | λem,max / nm |
|---|---|---|---|---|
| quinine sulfate | 0.1 M H2SO4 | 350 | 0.577 | 444 |
| fluorescein | 0.1 M H2SO4 | 496 | 0.95 | 521 |
| rhodamine 6G | ethanol | 488 | 0.94 | 530 |
| cresyl violet | methanol | 540 | 0.54 | 622 |
3.5 Fluorescence lifetime
Spontaneous emission (either fluorescence or phosphorescence) from an excited state is a first order process, and so shows similar kinetics to that which you have already studied, figure 3.6 & equation (3.7).
Figure 3.6: Emission from an excited state is a first order process, whereby an excited state decays spontaneously. The emission lifetime is the time required for the concentratio of an excited state to fall to 1/e of its initial value.
However, instead of using rate constants when talking about this decay (as seen in Section 3.4 photochemists tend to talk about emission lifetimes, \(\tau\).
The fluorescence lifetime is defined as the time required for the concentration of excited state to fall to 1/e (~0.3679) of the initial concentration.
So for an excited singlet state the fluorescence lifetime is related to the concentrations as follows:
\[\begin{equation} [S_1]_t=[S_1]_0e^-{\frac{t}{\tau}} \tag{3.7} \end{equation}\]
where \(\tau\) is the value given in equation (3.4).
3.6 The Energy Gap Law
The natural lifetime of fluorescence assumes that there is no other available pathway for deactivation of the excited state. However, the effects of internal conversion and intersystem crossing have already been seen by the presence of rate constants for internal conversion (IC) and intersystem crossing (ST) in the equations for quantum yield of fluorescence. The rate of non-radiative decay is given by the energy gap law (a formalised equation is not given here) which depends upon a number of factors but principally the energy gaps (ΔE) between electronic levels.
Non radiative decay become increasingly unfavourable as the energy gaps become bigger, and the rate of non-radiative decay decreases exponentially with increasing energy gaps. The energy gaps between higher electronic states (S2, S3, S4 etc.) are usually significantly smaller than the S1 - S0 energy gap leading to rapid internal conversion down to the lowest (singlet) excited state, which is the origin of Kasha’s rule.
| λabs / nm | λem / nm | ΔE / eV | τ/ nw | Φem | |
|---|---|---|---|---|---|
| [Os(phen)3]2+ | 650 | 720 | 0.186 | 260 | 0.016 |
| [Os(phen)2(dppene)]2+ | 455 | 609 | 0.69 | 1830 | 0.138 |
| [Os(phen)(dppene)2]2+ | 400 | 530 | 0.761 | 3600 | 0.518 |
The rate of non-radiative decay also depends upon the vibrations within a molecule (this should be logical since the energy is lost as heat), as well as the size of any structural changes required upon deactivation of the excited state. Since the frequency of vibrations is a factor in the rate of non-radiative decay there are clear isotope effects; particularly when hydrogens are replaced with deuteriums. C-H stretches occur with a frequency around 3000 cm−1, whereas this is much lower at around 2200 cm−1 for C-D stretches; because of this the efficiency of vibrational relaxation in the hydrogen sample is considerably greater and the lifetime of the excited state is consequently longer in the deuterated sample.
Now a quantitive value may be determined for the rate of internal conversion it is easy to quantify the rate of intersystem crossing.
For intersystem crossing the efficiency is again determined by the size of the energy gap between the singlet and triplet states.
El Sayed’s rule states:
the rate of intersystem crossing is relatively large if the radiationless transition involves a change in the orbital type.
Consequently intersystem crossing is considerably more efficient in molecules where there are hetero atoms as can be seen when comparing the rates of singlet triplet intersystem crossing between anthracene and 9-acetoanthracene, figure 3.4. However, the presence of a hetero atom does not guarantee a high rate of intersystem crossing, for systems where there is no change in orbital type there rate constants are in the order of 100 to 1000 times more slow than those with a change in orbital type.
| transition | transition type | kST / s−1 | |
|---|---|---|---|
| anthracene | S1 (π,π*) ⟶ T1 (π,π*) | forbidden | 1.4 × 108 |
| acetone | S1 (n,π*) ⟶ T1 (n,π*) | forbidden | 5 × 108 |
| benzil | S1 (n,π*) ⟶ T1 (n,π*) | forbidden | 5 × 108 |
| biacetyl | S1 (n,π*) ⟶ T1 (n,π*) | forbidden | 7 × 107 |
| 9-acetoanthracence | S1 (π,π*) ⟶ T1 (n,π*) | allowed | ~1010 |
| benzophenone | S1 (π,π*) ⟶ T1 (n,π*) | allowed | ~1011 |
3.7 Before Completing this Section
To support the material in this section it is suggested you read chapters 3 & 4 of Wardle ‘Principles and Application of Photochemistry’.
3.8 Self Assessment Questions
Explain why the absorption and emission spectra of a chromophore appear as mirror images of each other.
Why does the emission always occur at longer wavelengths than the absorption of a dye molecule.
What property of azulene allows for emission from the S2 state?
Why does the absorption and emission spectra of anthracene in the gas phase show much sharper lines than those seen in solution spectra, and in the gas phase even more detail in fine structure can be observed?
Write suitable statements relating the rates of internal conversion and intersystem crossing with the quantum yields of these processes.
The effects of halogen substitution in naphthalene on kp and kTS are shown below. Comment on the trends and calculate in each case the fraction of triplet states that are deactivated by emission of a photon.
| Compound | kp / s−1 | kST / s−1 |
|---|---|---|
| naphthalene | 0.05 | 0.39 |
| 1-fluoronaphthalene | 0.23 | 0.42 |
| 1-chloronaphthalene | 1.10 | 2.35 |
| 1-bromonaphthalene | 13.5 | 36.5 |
| 1-iodonaphthalene | 190 | 310 |
- 5,10-dihydroindeno[2,1-a]indene and trans-stilbene (figure 3.7) are similar in structure but have very different fluorescent quantum yields of 0.05 and 1.00 respectively, however for trans-stilbene this increases to 0.75 at 77 K. Suggest a reason for the difference in quantum yield of:
- the two molecules
- the two temperatures
![5,10-dihydroindeno[2,1-a]indene (left) and trans-stilbene (right)](images/stilbeneindene.png)
Figure 3.7: 5,10-dihydroindeno[2,1-a]indene (left) and trans-stilbene (right)
3.9 Self Assessment Question Brief Answers
- Absorption occurs almost entirely from the ground vibrational, \(v = 0\), state of the ground electronic state because at room temperature the proportion of the molecules in the \(v = 1\) state is very very low. Molecules are excited into a number of different vibrational levels in the first electronic excited state, the population in each depends upon the overlap integral of the vibrational levels of the ground and excited state.
According to Kasha’s rule, emission always happens from the lowest vibrational state, \(v’ = 0\), of the electronically excited state. Emission then happens to all the available vibrational states in the ground electronic state. Absorption transitions from \(v = 0\) to \(v’ = 0\) are lower in energy than transitions from \(v = 0\) to \(v’ = 2\), however emission transitions from \(v’ = 0\) to \(v = 0\) are higher in energy than those from \(v’ = 0\) to \(v = 2\). If were were to sketch these then the two spectra appear as mirror images of each other.
This is due to Stokes’ shift. Absorption occurs from an optimised ground state into an non optimised excited state, this excited state then has time to relax to an optimised configuration before emission of a photon, but by optimising the excited state it destabilises the ground state. Consequently the energy gap between the ground and excited state for absorption is bigger than for emission. Since energy and wavelength are inversely proportional the wavelength of emission is always longer than the wavelength of absorption.
There is a large energy gap between the \(S_2\) & \(S_1\) excited states which reduces the efficiency of internal conversion. Given no other deactivation pathway azulene will fluoresce from the \(S_2\) state.
In the vapour phase, anthracene molecules do not experience the strong and fluctuating interactions with solvent molecules that occur in a solution phase. The vibrational structure is therefore well-defined. The molecules can also rotate giving rise to rotational fine structure in the spectra.
Internal conversion \[\begin{equation} \phi_ic = \frac{k_ic}{k_f^0+k_{ic}+ k_{ST}+k_{\textrm{other}}} \tag{3.8} \end{equation}\]
Intersystem crossing (\(S_1\) - \(T_1\))
\[\begin{equation}
\phi_ST = \frac{k_ST}{k_f^0+k_{ic}+ k_{ST}+k_{\textrm{other}}}
\tag{3.9}
\end{equation}\]
Intersystem crossing (\(T_1\) - \(S_0\))
\[\begin{equation}
\phi_TS = \frac{k_TS}{k_f^0+k_{ic}+ k_{ST}+k_{\textrm{other}}}
\tag{3.10}
\end{equation}\]
As you work down the halides the amount of spin orbit coupling increases due to the heavy atom effect and the spin selection rules are softened. This leads to a large (nearly x1000) increase in the rate of triplet formation from intersystem crossing. The ratio kp / kST changes far less dramatically (0.11, 0.35, 0.32, 0.27, & 0.38 respectively), this has less to do with the heavy atom effect as it increases both the rate of phosphorescence and singlet to triplet intersystem crossing.
Essentially the chromophore of stilbene and indene are the same, however stilbene allows rotation of the benzene rings around a single bond allowing efficient internal conversion, whereas the additional structure in the indene does not allow rotational relaxation. With no easy route of internal conversion the quantum yield of fluorescence increases dramatically.
As the sample is cooled the rotation in stilbene is restricted and therefore decreases the rate of internal conversion. As the rate constant for internal conversion decreases then the quantum yield of emission increases as in equation (3.3).
The term linear space implies the steps are the same size, energy is the more fundamental unit, since wavelength is a reciprical function of energy wavelength space is non linear.↩︎