solution
0 2 4
3.2 3.6 4.0 4.4
E ne rgy (e V )
Vibronic number
0 2 4
3.2 3.6 4.0 4.4
E ne rgy (e V )
Vibronic number crystal
solution crystal solution
Figure 25: Analysis of the vibronic peaks of anthracene shown in Fig. 8.13 as considered in Exercise 8.10.
(8.10) The assignment of the vibronic peaks of the solution is given in Fig. 8.13.
The energies of the vibronic peaks are plotted in Fig. 25 and a good straight line is obtained. The linear fit according to eqn 8.3 gives the vibrational energy as 0.16 eV.
In the case of the crystal, we have first to identify the various peaks in the absorption spectrum. A strong vibronic progression with energies of 3.13, 3.30, 3.46 and 3.61 eV is observed in the data. These can be identified as the 0–0, 0–1, 0–2, and 0–3 transitions, and a linear fit according to eqn 8.3 gives the vibrational energy as 0.18 eV. (See Fig. 25.)
Two other features can also be identified in the absorption spectrum of the crystal at 3.18 and 3.35 eV. These have the same splitting as the other progression and therefore involve similar types of vibrations. The most probable cause is a splitting of the electronic states in the lower symme-try of the crystal, eg by the Davydov effect. (See Pope and Swenberg, Electronic processes in organic crystals and polymers, 2nd edn, Oxford University Press, 1999, Section I.D.5, pp 59–66.)
(8.11) When the ‘mirror symmetry’ rule works, we expect the emission spectrum to be a mirror image of the absorption spectrum about the 0–0 transition.
(See, for example, Figs 8.10 or 8.17.) We thus expect a broad vibronic
band extending downwards from the 0-0 transition at 3.13 eV. The width of the band will be about 1 eV. A series of vibronic peaks will occur with energies given by hν ≈ (3.13 − n~Ω), with ~Ω ≈ 0.18 eV. We would thus expect peaks at 3.13, 2.95, 2.77, 2.59 eV · · · .
(8.12) The S1 absorption band has a 0–0 transition at 1.9 eV and extends to
∼ 2.8 eV. The emission band would thus have a 0–0 transition at 1.9 eV and extend down to about 1.0 eV. The 0–1 vibronic peaks occurs at 2.1 eV in the absorption spectrum, which implies a vibrational energy of ∼ 0.2 eV, and hence a 0–1 transition in emission at around 1.7 eV.
(8.13) The difference between the absorption and photoconductivity edges is caused by excitonic effects. The absorption edge corresponds to the cre-ation of tightly-bound (Frenkel) excitons. Since excitons are neutral par-ticles, they do not contribute to the photoconductivity. The photoconduc-tivity edge therefore corresponds to the band edge where free electrons and holes are first created. The difference in the two edges gives the exciton binding energy, which works out to be 1.1 eV.
It is important to realize that this is a different situation to that encoun-tered for weakly-bound (Wannier) excitons. Weakly-bound excitons can be easily ionized to produce free electrons and holes, and hence produce a photocurrent. (See, for example, Figs 4.5 and 6.12.)
(8.14) The dominant vibrational frequency can be deduced by analysing the vi-bronic progression of either the absorption or emission spectra in Fig. 8.17 according to eqns 8.3 or 8.5 as appropriate. This gives ~Ω ∼ 0.17 eV, im-plying Ω ∼ 2.6 × 1014rad/s. With ω = 2.98 × 1015rad/s, we then find ωRaman= (ω − Ω) = 2.72 × 1015rad/s, which is equivalent to a wavelength of 693 nm.
Two points could be made here:
• The molecule will have other vibrational modes, and these will give additional Raman lines.
• Not all vibrational modes are Raman-active, (see Section 10.5.2) and it is not immediately obvious that the 0.17 eV mode responsible for the vibronic spectra will show up in the Raman spectrum. In fact, these modes are observed in the experimental Raman spectra, but it requires a careful analysis by group theory to demonstrate this theoretically.
(8.15) Optical excitation creates only singlets because the ground state is a singlet and optical transitions do not change the spin. With only singlet states excited, the recombination of the electrons and holes is optically allowed for all the carriers.
On the other hand, with electrical injection there is no control of the relative spin of the electrons and holes. The spins can be either parallel or anti-parallel, and this gives rise to four possible total spin wave functions, as indicated in Table 5. Three of these are triplets and only one is a singlet state. The relative number of triplet and singlet excitons created by electrical injection is therefore in the ratio 3:1, which implies that only
Wave function Sz S State
↑e↑h +1 1 triplet (↑e↓h+ ↓e↑h)/√
2 0 1 triplet (↑e↓h− ↓e↑h)/√
2 0 0 singlet
↓e↓h −1 1 triplet
Table 5: Possible arrangements of relative electron-hole spins as discussed in Exercise 8.15.
25% of the excitons are in singlet states. The remaining 75% are in triplet states with very low emission probabilities. Hence the emission is expected to be weaker than that for optical excitation by a factor of four.
The creation of triplets in electrically-driven organic LEDs is a serious issue that limits their efficiency. One way to enhance the efficiency is to increase the spin-orbit interaction to encourage inter-system crossing.
This is typically done by including a heavy metal atom in the molecule.
(See Exercise 8.8.)
(8.16) (i) As we have seen in Exercise 8.15, we expect that 75% of the excitons created will be in triplet states with very low emission probabilities. Hence the maximum quantum efficiency that we can expect corresponds to the number of singlet excitons that we create, namely 25%.
(ii) The number of electrons and holes flowing into a device carrying a current i is equal to i/e. The quantum efficiency is defined as the ratio of photons out to electrons in, and so the number of photons emitted will be equal to ηi/e. The power emitted is then equal to hν × ηi/e, and we have:
P = 2.25 eV × 25% × 10 mA/e = 5.6 mW .
(iii) The electrical power consumed by the device is equal to iV = 50 mW.
The power conversion efficiency is thus equal to 5.6/50 = 11 %. The efficiency of a real device would be much lower, mainly due to the difficulty of collecting the photons, which are emitted in all directions (ie over 4π solid angle). Only a small fraction of these would be collected by the optics. This latter point is exacerbated by the high refractive index of the molecular material, which tends to limit the effective collection efficiency even further. (See Exercise 5.13.)