VP 2:

Fluorescence and Phosphorescence

Therese Challand

Abbildung 1: Aromatic hydrocarbon crystals

1

INHALTSVERZEICHNIS 2

Inhaltsverzeichnis

1 Theory 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Nature of Electronic States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Nature of Absorption Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Luminescence Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.1 Vibrational Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3.2 Fluorescence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3.3 Internal Conversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3.4 Phosphorescence and Intersystem Crossing . . . . . . . . . . . . . . . . . . . . . . 51.3.5 Delayed Fluorescence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Experiment 62.1 Experimental System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.1.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1.2 Experimental Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Conclusion 19

4 Sources and Free Software 19

1 Theory

1.1 Introduction

Luminescence is defined as the radiation emitted by a molecule or atom, after it had absorbed energyto go to an excited state. The main types of luminescence are fluorescence and phosphorescence.

The goal of the experiment is to investigate the properties of optical excitations(= excited states)in organic crystals, see figure 1, and understand the processes that lead to phosphorescence and delayedfluorescence.

1.2 Nature of Electronic States

We distinguish between electronic state and electronic orbital:

• An orbital is defined as the volume element in which there is a very high probability to findan e− (electron). The orbital is calculated from a one-e− wave function and it is assumed to beindependent of all other e−’s in the molecule.

• electronic states are concerned with the properties of all e−’s in all the orbitals. So the wavefunction of an electronic state is a combination of the wave functions of each of the e−’s in each ofthe orbitals of the molecule.

We distinguish also between excited electronic states and the transition state:

• A transition state corresponds to a vibrationally excited ground state, i.e. ground state in a strained(dt. angespannt) configuration.

• Excited electronic states may contain no excess vibrational energy, but are still much higher inenergy than the ground state. A molecule in an excited state is regarded as a completely newentity. This state will have a different e− distribution from the ground state, a different geometry,and will undergo chemical reactions different from those of the ground state.

Electronic states of organic molecules can be grouped in two categories:

• Singlet States: A singlet state is one in which all of the e−’s in the molecule have their spinspaired.

1 THEORY 3

• Triplet States: Triplet states are those in which one set of e− spin have become unpaired.

Triplet and singlet states differ significantly in their properties, e.g. their energies. A triplet state willalways lie lower in energy than its corresponding singlet state.

1.2.1 Nature of Absorption Processes

Figure 2 shows a partial energy diagram for a photoluminescence system. By convention, the singletstates are stacked in a column and the triplet states are stacked in a column to the right.

Abbildung 2: Partial energy diagram for a photoluminescence system, so called Jablonsky diagram

Each of the electronic states (ground or excited) has a number of vibrational levels superimposed on it.These levels arise because a molecule in a given electronic state absorbs small increments of energy. Thiscorresponds to changes in vibrational modes, but the molecule retains the same electronic configuration.

There exists a certain degree of overlap between the vibrational levels of excited states such as S2 andS1, and also an overlap between the vibrational levels of the triplet state T1 and its corresponding singletstate S1.

The energy of a γ, E = hcλ , required to produce a particular excited state, is the difference in energy

between that state and the ground state, compare figure 2. So there is a range of wavelength that leadsto a transition between any two electronic states. This shows that electronic absorption spectra occur asbroad bands, rather than as single lines.

Transitions from the ground state to the triplet state T1, singlet-triplet processes, are improbable, ≈10−6 of the probability of the corresponding singlet-singlet or triplet-triplet processes.

1 THEORY 4

1.3 Luminescence Processes

Luminescence emission occurs from excited states. A molecule absorbs a γ, followed by reemission of aγ to give luminescence. But there are also non radiative processes, which precede and/or compete withγ emission. The mean lifetimes of all processes are important for determining the luminescence behaviorof a molecule.

1.3.1 Vibrational Relaxation

Assume that all molecules are in the lowest vibrational level of the ground state at room temperature.The time required for a γ absorption, i.e. the time required to go from one electronic state to another, is10−15 s. This is short compared to the time required for all other electronic processes and also nuclearmotion. So right after excitation a molecule has the same geometry and environment as in the groundstate.

Now it can do two things. Which one of these processes is dominant depends on the environment ofthe molecule:

• Emit a γ from the vibrational level to which it was excited initially.

• Undergo changes in vibrational levels before emission of radiation.

For an isolated molecule of a gas, the only way to lose vibrational energy is to emit an infrared γ. Thisis less probable than undergo an electronic transition to return to the ground state. So at low pressuresyou probably see γ emission from higher vibrational levels of excited states in gas phase spectra.

In a solution, thermal relaxation of a vibrationally excited molecule is rapid, 10−13 to 10−11 s, throughtransfer of excess vibrational energy from the molecules to the solvent(= dt. Losungsmittel). Thermalrelaxation is so efficient that all excess vibrational energy of the excited state is lost.

So before the excited molecule in the solution can emit a γ, it will undergo vibrational relaxation.Therefore γ emission will always occur from the lowest vibrational level of an excited state.

1.3.2 Fluorescence

Fluorescence is when a molecule, which arrives at the lowest vibrational level of an excited singlet state,returns to the ground state by γ emission. Since the lifetime of an excited state is ≈ 10−9 - 10−7 s, thedecay time of fluorescence is of the same order of magnitude.

So the process of fluorescence consists of γ absorption by a molecule, which goes to an excited state.Then the molecule relaxes from higher vibrational levels of that state to its lowest vibrational level. Thenthe molecule emits a γ to a vibrationally excited level of the ground state and again relaxes to the lowestvibrational level of the ground state.

1.3.3 Internal Conversion

A molecule in an excited singlet state can return to the ground state without γ emission, converting allthe excitation energy into heat. This is a radiationless process and is called internal conversion,but which is inefficient and in our case of aromatic hydrocarbons a small fraction of the total excitationenergy.

A molecule can undergo internal conversion to the lowest vibrational level of its lowest excited singletstate in a time that is short, relative to γ emission,regardless of the singlet state to which it was excitedinitially.

Assume that a molecule is excited to a higher singlet state such as S2. The molecule will undergovibrational relaxation. What happens to the molecule when it reaches the zeroth vibrational level of S2

depends of the energy separation between the excited singlet states.In aromatic molecules, the separation between excited singlet states is smaller than the separation

between the lowest singlet state S1 and the ground state S0. So the lowest vibrational level of S2 willoverlap with higher vibrational levels of the S1 state. This is a high degree of coupling between thesevibrational levels of the two states, which provides an efficient path for crossing from the S2 state to theS1 state. The time needed to undergo internal conversion from S2 state to the lowest vibrational level ofthe S1 state is the same, ≈ 10−13 - 10−11 s, as for the conversion from an excited vibrational level of S1

state to its zeroth vibrational level.

1 THEORY 5

1.3.4 Phosphorescence and Intersystem Crossing

Population of triplet states by direct absorption from the ground state is not significant. The process ofpopulation of triplet states from the lowest excited singlet state, caused by intersystem crossing, ismore efficient. This process is a spin-dependent internal conversion process.

A singlet-triplet process is less probable than a singlet-singlet process. But surprisingly a singlet-tripletprocess such as intersystem crossing has a lifetime of 10−8 s, which is about the lifetime of an excitedsinglet state.

The mechanism for intersystem crossing involves vibrational coupling between the excited singlet stateand a triplet state.

Singlet-triplet processes are about 10−6 times less probable than singlet-singlet states. Radiationlessvibrational processes, such as internal conversion, occur in ≈ 10−13 s. The time required for a spin-forbidden vibrational process is ≈ 10−8 s, which is ≈ the lifetime of an excited singlet state. Intersystemcrossing competes with fluorescence emission from the zeroth vibrational level of an excited singlet statebut cannot compete with vibrational deactivation from higher vibrational level of a singlet state.

After intersystem crossing the molecule undergoes internal conversion process and falls to the zerothvibrational level of the triplet state. The energy difference between zeroth vibrational level of the tripletstate and the zeroth vibrational level of the lowest excited state is large compared to thermal energy. Sorepopulation of a singlet state from a triplet state is improbable.

Also radiationless transition can occur e.g. in the case of a molecule in a solvent. Two factors canlead to a radiationless transition between the lowest excited state and the ground state:

1. The energy difference between triplet state and the ground state is smaller than the one betweenthe lowest singlet state and the ground state. This leads to vibrational coupling between these twostates and so leads to internal conversion.

2. More important is that the lifetime of a triplet state is much longer, ≈ 10−4 to 10 s, than thatof an excited singlet state. So loss of excitation energy by collisional transfer is enforced. At roomtemperature this is the dominant pathway for the loss of triplet state excitation energy.

A molecule in a medium where collisional processes do not occur like it is the case of our experimentwhere we use crystals, a radiative transition between the lowest triplet state and the ground state isobserved. This emission is called phosphorescence and originates from the lowest triplet state. So ithas a decay time equivalent to the lifetime of the triplet state. It is characterized by an afterglow whichis not observed for fluorescence.

1.3.5 Delayed Fluorescence

Delayed fluorescence is a non-collision energy transfer process. It has the fluorescence characteristic emis-sion spectrum but a lifetime shorter than phosphorescence. The intensity of delayed fluorescence is pro-portional to the square of the intensity of the exciting radiation and the square of the phosphorescenceintensity. This indicates that delayed fluorescence is a biphotonic process involving two triplet states.

We also observe the so-called P-type delayed fluorescence in this experiment:

1. S0+h · ν −→ S∗

The molecule is in the ground state, then absorbs a γ and goes to the lowest excited singlet state,S∗.

2. S∗ −→ T

The molecule undergoes intersystem crossing to the triplet state T .

3. T + T −→ S∗ + S0

Long lifetime of interaction between the two triplet excitons and the molecule allows the quasiparticles to diffuse through the crystal and interact leading to the two singlet states.

4. S∗ −→ S0 + h · νf

The excited singlet state emits its normal fluorescence.

2 EXPERIMENT 6

2 Experiment

2.1 Experimental System

2.1.1 Kinematics

Phosphorescence and delayed fluorescence are both decays from the triplet state T1. So we are interestedin the time dependence of triplet exciton density n.

We start with the following Ansatz:dndt = a · i− β · n− γ · n2 + D ~∇2 · n,where the first term is the dependence from excitation intensity i, the second direct decay (phospho-

rescence), then annihilation (delayed fluorescence) and diffusion.

• Intensity of phosphorescence: Ip ∝ n

• Intensity of delayed fluorescence: If ∝ n2

We neglect αi, because we will not excite the crystal during observation. Also neglect the diffusionterm, because it can be considered as a small perturbation in first order. So

dn

dt= −β · n− γ · n2

=⇒ dn

n− dn

n + βγ

= −βd

Integration leads to:

n(t) = n0e−βt

[1

1 + γn0β (1− e−βt)

]︸ ︷︷ ︸

∗

,

where ∗ is a deviation from the exponential behavior and represents the influence of triplet annihilation.We measure intensities of phosphorescence and delayed fluorescence.The phosphorescence intensity is:

Ip = Aβn = n0

[Aβ

1 + γn0β (1− e−βt)

]· e−β·t, (1)

where A is the fraction of the first order decay, namely from the transition T1 → S0, which is radiative.And the intensity of delayed fluorescence is:

If =12(Bγ)n2 =

12n0(Bγ)e−2βt

[1

1 + γn0β (1− e−βt)

]2

(2)

The triplet lifetime is:

τT =1β

So the lifetime of phosphorescence and delayed fluorescence are:

τT = τp = 2τf ,

that means here the delayed fluorescence decays are twice faster than phosphorescence decays.

2.1.2 Experimental Setup

The side walls of the dark experimental chamber contain 2 wheels connected by a bar, see figure 3. In eachwheel there are two gaps. Using this setup the Hg lamp cannot cause any damage to the photomultiplierby shining at it. This setup also ensures that the results only arrive from phosphorescence and delayedfluorescence. The turning speed can be controlled, it should be slow enough to allow an observationinterval of ≥ 3 lifetimes in all cases.

The path of light from the right wheel is then focused on a crystal causing the desired excitations;the left wheel allows the γ’s to arrive to the photomultiplier, see figure 3.

2 EXPERIMENT 7

Abbildung 3: Experimental Setup

2.2 Measurement

I should measure the data for the first 100-500 ms for all three substances. This I should plot like in eq.(1) the phosphorescence intensity Iph and like in eq. (2) for the delayed fluorescence intensity Ifl.

2.3 Data Analysis

I import the data from the measurement using ’TDS 2002 Open Choice Desktop’ into Origin programand plot.

We know that τphosphorescence > τfluorescence. So we measure a superposition of Iphosphorescence andIfluorescence. Since these exponential functions do decay differently fast, we can treat them each after theother. So we divide the intensity we have from the measurement into Ifluorescence and Iphosphorescence.The point where to divide is arbitrary under the constraint that we want only to observe fluorescence orphosphorescence.

First I fit linear on a logarithmic scale the exponential decay in order to get an initial value for β inthe formulas (1) and (2), which corresponds to the slope B of the linear fit, see figures 5, 6, 15, 19 and10. These values I use as an initial guess for the real fit of the intensities. If I do not use this guess for βand all other parameters in eq. (1) and eq. (2), I will get very big errors for the parameters in the real fit.It is also possible to fall into another stable parameter configuration which fulfills the same measurementdata.

Then we fit both parts for fluorescence or phosphorescence in the graph like in the formula for theintensities. From this fit we get the real value for β. So we have a value for the phosphorescence decaytime τphosphorescence [s] and the fluorescence decay time τfluorescence [s] for each material.

For Anthracene see figures 4, 7, 8 and 9; for Naphthaline see figures 16, 17, 18, 20 and 21; and forPara-terphenyl see figures 11, 12, 13 and 14.

2 EXPERIMENT 8

Abbildung 4: Anthracene

Abbildung 5: Anthracene

2 EXPERIMENT 9

Abbildung 6: Anthracene

Abbildung 7: Anthracene

2 EXPERIMENT 10

Abbildung 8: Anthracene

Abbildung 9: Anthracene

2 EXPERIMENT 11

Abbildung 10: Para-terphenyl

Abbildung 11: Para-terphenyl

2 EXPERIMENT 12

Abbildung 12: Para-terphenyl

Abbildung 13: Para-terphenyl

2 EXPERIMENT 13

Abbildung 14: Para-terphenyl

Abbildung 15: Naphthaline

2 EXPERIMENT 14

Abbildung 16: Naphthaline

Abbildung 17: Naphthaline

2 EXPERIMENT 15

Abbildung 18: Naphthaline

Abbildung 19: Naphthaline

2 EXPERIMENT 16

Abbildung 20: Naphthaline

2 EXPERIMENT 17

Abbildung 21: Naphthaline

2 EXPERIMENT 18

So we calculate a mean β [s−1], a mean τphosphorescence [s] and a mean τfluorescence [s]:

material mean βfluorescence [s−1] standard deviation mean τfluorescence [s] standard deviationAnthracene 122,45 ± 9,04 ≡ 7,38 % 0,00819 ± 0,00600538 ≡ 7,34 %Para-terphenyl 10,80 ± 4,44 ≡ 41,1 % 0,11764 ± 0,08148 ≡ 69,62 %Naphthalene 97,43 ± 17,25 ≡ 17,71 % 0,01057 ± 0,00217 ≡ 20,53 %

material mean βphosphorescence [s−1] standard deviation mean τphosphorescence [s] standard deviationAnthracene 1,28 ± 0,23 ≡ 17,97 % 0,79 ± 0,14 ≡ 17,72 %Para-terphenyl 0,99 ± 0,05 ≡ 5,05 % 1,02 ± 0,05 ≡ 4,9 %Naphthalene 1,23 ± 0,5 ≡ 40,65 % 0,94 ± 0,52 ≡ 55,32 %

I chose the standard deviation to be a good indicator for the uncertainty of the result, because thestandard deviation tells how much the data values differ from each other since the data does differ muchindeed. This I also expressed in % to show the deviations from the result clearly.

2.4 Problems

When the oscilloscope is not correctly adjusted, you get pictures like in figure 22 and 23.

Abbildung 22: Bad measurement

Abbildung 23: Other bad measurement

3 CONCLUSION 19

In the theory part of my report I wrote that τphosphorescence = 2 · τfluorescence. This is actually nottrue in my fits of the data. The problem is to find the point where the fluorescence intensity goes overinto pure phosphorescence intensity. To solve this problem there is the following way: We suppose that inthe fit for the phosphorescence intensity we mostly observe the background noise. To get away the noisewe subtract from the data the value of the intensity before the window in the wheel opens, i.e. the valuebefore the peak in the graph. Then plot the intensity logarithmically and you will see that the intensitydiffers clearly in several magnitudes, which helps to find the point where fluorescence intensity goes overinto phosphorescence intensity under the constraint that τphosphorescence = 2 · τfluorescence.

3 Conclusion

By using a relatively simple experimental setup we can measure the decay time of phosphorescence andfluorescence. Please note the values for three crystals in the table above.

4 Sources and Free Software

• Handout of D. Keller; March 24, 2005