---
title: "Electronic Spectroscopy and Photophysics"
book: "University Chemistry — Year 3"
subject: chemistry
language: en
chapter: 7
exercises: 12
source: https://one-course.com/books/chemistry/4/en/chapter/7-electronic-spectroscopy-and-photophysics
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 7 — Electronic Spectroscopy and Photophysics

Under the ultraviolet lamp of a dark bar, a glass of tonic water glows a pale, ghostly blue. The quinine it contains absorbs ultraviolet light, which the eye cannot see, and gives part of it back a few nanoseconds later as visible blue light. Absorption, a short life in an excited state, emission at a longer wavelength, the energy that does not come back as light: the fate of an excited molecule is a competition between processes with rate constants, and its spectrum is shaped by the vibrations of the molecule and by the symmetry rules of [Chapter 5](https://one-course.com/books/chemistry/4/en/chapter/5-group-theory-applied#ch-b3-group-theory-applied). This chapter studies electronic transitions and what happens after them, the subject called photophysics; [Chapter 14](https://one-course.com/books/chemistry/4/en/chapter/14-photochemistry#ch-b3-photochemistry) adds the cases in which the excited molecule reacts.

**You already know.**

The Year 1 volume measured absorbance with the Beer–Lambert law, $A =
\varepsilon\ell c$, defined the molar absorption coefficient, chromophores and conjugated systems, and related colour to the absorption maximum. The Year 2 volume named the frontier orbitals HOMO and LUMO. [Chapter 2](https://one-course.com/books/chemistry/4/en/chapter/2-many-electron-atoms-and-term-symbols#ch-b3-many-electron-atoms) defined singlet and [triplet states](https://one-course.com/books/chemistry/4/en/chapter/2-many-electron-atoms-and-term-symbols#def-b3-many-electron-atoms-singlet-triplet); [Chapter 5](https://one-course.com/books/chemistry/4/en/chapter/5-group-theory-applied#ch-b3-group-theory-applied) the vanishing-integral theorem; [Chapter 6](https://one-course.com/books/chemistry/4/en/chapter/6-rotational-and-vibrational-spectroscopy#ch-b3-rovibrational-spectroscopy) the vibrational levels of a bond.

![Left: tonic water glowing blue under an ultraviolet lamp. Right: fluorite in daylight and fluorescing under ultraviolet light; the mineral gave fluorescence its name. Photograph Masha Milshina, CC BY 4.0.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-electronic-spectroscopy/img-4270d06081fd.jpg)

![Left: tonic water glowing blue under an ultraviolet lamp. Right: fluorite in daylight and fluorescing under ultraviolet light; the mineral gave fluorescence its name. Photograph Masha Milshina, CC BY 4.0.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-electronic-spectroscopy/img-c553b2a9e4d5.jpg)

*Left: tonic water glowing blue under an ultraviolet lamp. Right: fluorite in daylight and fluorescing under ultraviolet light; the mineral gave [fluorescence](#def-b3-electronic-spectroscopy-luminescence) its name. Photograph Masha Milshina, CC BY 4.0.*

## 7.1 Electronic transitions and their selection rules

**Definition 7.1 (Transition dipole moment, oscillator strength).**

The *transition dipole moment* between states $\Psi_i$ and $\Psi_f$ is $\boldsymbol\mu_{fi} = \langle\Psi_f|\hat{\boldsymbol\mu}|\Psi_i
\rangle$, with $\hat{\boldsymbol\mu} = -e\sum_k\mathbf r_k$ (plus the nuclear charges, which cancel between orthogonal states). The *oscillator strength* $f$ is the dimensionless measure of the intensity of a band, about 1 for the strongest transitions.

**Proposition 7.2 (Intensity and oscillator strength).**

The integrated absorption of a band is proportional to $|\boldsymbol\mu_{fi}|^2$, and in practice

$$
f \approx 4.32\times10^{-9}\int\varepsilon(\tilde\nu)\,\dd\tilde\nu,
$$

with $\varepsilon$ in $\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{cm}^{-1}$ and $\tilde\nu$ in $\mathrm{cm}^{-1}$.

**Proof.** *Admitted at this level.* ∎

The proportionality comes from time-dependent perturbation theory and the numerical factor from the classical oscillator; both are treated in more advanced courses. What matters here is that $f$ is large only when $\boldsymbol\mu_{fi}$ is, and that symmetry can force it to zero.

**Theorem 7.3 (The spin selection rule).**

An electric-dipole transition between states of different total spin is forbidden: $\Delta S = 0$.

**Proof.** Each state is a product of a spatial and a spin function, $\Psi = \phi\,\chi$. The dipole [operator](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-operator) acts on positions only, so $\langle\phi_f\chi_f|\hat{\boldsymbol\mu}|
\phi_i\chi_i\rangle = \langle\phi_f|\hat{\boldsymbol\mu}|\phi_i\rangle\langle\chi_f|\chi_i\rangle$. Spin functions of different $S$ are [eigenfunctions](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-operator) of $\hat S^2$ with different [eigenvalues](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-operator), hence orthogonal ([Theorem 1.4](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#thm-b3-quantum-model-systems-hermitian-real)): the second factor vanishes. ∎

**Theorem 7.4 (The Laporte rule).**

In a molecule with a [centre of inversion](https://one-course.com/books/chemistry/4/en/chapter/4-symmetry-and-point-groups#def-b3-point-groups-inversion), an electric-dipole transition must change parity: $g \leftrightarrow u$ is allowed, $g \leftrightarrow g$ and $u
\leftrightarrow u$ are forbidden. This is the *Laporte rule*.

**Proof.** The components of $\hat{\boldsymbol\mu}$ change sign under inversion: they are $u$. The product $\Gamma_f\otimes\Gamma_\mu\otimes\Gamma_i$ is $g$ only if $\Gamma_f$ and $\Gamma_i$ have opposite parities; otherwise it is $u$ and cannot contain the [totally symmetric representation](https://one-course.com/books/chemistry/4/en/chapter/5-group-theory-applied#def-b3-group-theory-applied-reducible) ([Theorem 5.18](https://one-course.com/books/chemistry/4/en/chapter/5-group-theory-applied#thm-b3-group-theory-applied-vanishing-integral)). ∎

The two rules are strict for the idealised states; real molecules relax them — [spin–orbit coupling](https://one-course.com/books/chemistry/4/en/chapter/2-many-electron-atoms-and-term-symbols#def-b3-many-electron-atoms-spin-orbit) mixes singlet and triplet character, vibrations remove the centre of symmetry for an instant — and the “forbidden” transitions appear, weakly. The $d$–$d$ bands of octahedral complexes, Laporte forbidden and weak, owe their colour to this relaxation ([Chapter 18](https://one-course.com/books/chemistry/4/en/chapter/18-electronic-spectra-and-magnetism-of-complexes#ch-b3-complex-spectra-magnetism)).

**Definition 7.5 (Charge-transfer transition).**

A *charge-transfer transition* moves an electron from an orbital located mainly on one part of a system (a donor) to an orbital located mainly on another (an acceptor); its transition dipole is large, its band intense and broad, and its energy sensitive to the polarity of the solvent.

**Method 7.6 (Assigning an absorption band).**

1. $\varepsilon_{\max}$ above about $10^{4}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{cm}^{-1}$ : an allowed transition ( $\pi\to\pi^*$ of a conjugated system, or charge transfer).
2. $\varepsilon_{\max}$ of 10 to a few hundred: a forbidden one ( $n\to\pi^*$ , symmetry-forbidden; $d$ – $d$ , Laporte-forbidden).
3. $\varepsilon_{\max}$ below 1: spin-forbidden.
4. An $n\to\pi^*$ band moves to shorter wavelengths in protic solvents (the lone pair is stabilised by hydrogen bonds), a $\pi\to\pi^*$ band usually to longer ones.

## 7.2 The Franck–Condon principle

**Theorem 7.7 (Franck–Condon principle).**

Within the [Born–Oppenheimer approximation](https://one-course.com/books/chemistry/4/en/chapter/3-computational-chemistry-hartreefock-and-dft#def-b3-computational-chemistry-born-oppenheimer), and if the transition dipole depends little on the nuclear positions, the intensity of the [vibronic transition](#def-b3-electronic-spectroscopy-vibronic) from vibrational level $v''$ of the lower electronic state to level $v'$ of the upper one is proportional to $|\langle\chi_{v'}|\chi_{v''}\rangle|^2$, the square of the overlap of the two vibrational wavefunctions: this is the *Franck–Condon principle*. Electronic transitions are vertical: the nuclei do not move while the electrons jump.

**Proof.** With $\Psi = \psi_{\mathrm{el}}(\mathbf r;R)\chi_v(R)$, the transition moment is $\int\chi_{v'}^*\big[\int\psi'^*_{\mathrm{el}}\hat\mu\psi''_{\mathrm{el}}\dd\mathbf r\big]
\chi_{v''}\,\dd R$. The bracket is the electronic transition moment $\mu_{\mathrm{el}}(R)$; taken as constant (the Condon approximation), it comes out of the integral, leaving $\mu_{\mathrm{el}}\langle\chi_{v'}|\chi_{v''}\rangle$. ∎

**Definition 7.8 (Vibronic transition, progression, Franck–Condon factor).**

A *vibronic transition* changes the electronic and vibrational states together; the lines $v' \leftarrow 0$ for $v' = 0, 1, 2,
\dots$ form a *vibronic progression*; the *Franck–Condon factor* of a vibronic line is $|\langle\chi_{v'}|\chi_{v''}\rangle|^2$.

**Proposition 7.9 (Displaced oscillators).**

If both states are harmonic with the same frequency and the upper minimum is displaced by $\Delta$ (in units of $\sqrt{\hbar/m\omega}$), the [Franck–Condon factors](#def-b3-electronic-spectroscopy-vibronic) from $v'' = 0$ form a Poisson distribution, $|\langle v'|0\rangle|^2 = \eu^{-S}S^{v'}/v'!$, with $S = \Delta^2/2$; the strongest line is near $v' = S$.

**Proof.** *Admitted at this level.* ∎

The general proof uses the [ladder operators](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-ladder) of [Chapter 1](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#ch-b3-quantum-model-systems); this chapter’s figure data checks the formula against direct integration. A small displacement gives one strong 0–0 line; a large one a long progression whose strongest member is far from the 0–0 line.

![Left: a vertical transition from the lowest level of the ground state lands where the excited curve, displaced to longer bond length, is high above its minimum (drawn for S = 1.5). Right: Franck–Condon factors of the progression for S = 0.3 (black), 1.5 (blue) and 5 (red); each set adds up to 1.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-electronic-spectroscopy/fig-43c7df36a5f0.svg)

*Left: a vertical transition from the lowest level of the ground state lands where the excited curve, displaced to longer bond length, is high above its minimum (drawn for $S = 1.5$). Right: [Franck–Condon factors](#def-b3-electronic-spectroscopy-vibronic) of the progression for $S = 0.3$ (black), 1.5 (blue) and 5 (red); each set adds up to 1.*

**Example 7.10 (Iodine).**

Iodine vapour is violet because its B$\leftarrow$X band absorbs yellow-green light. The bond lengthens from $266.6\,\mathrm{pm}$ in the ground state to $302.5\,\mathrm{pm}$ in the B state. In the displaced-oscillator model with the B-state frequency ($125.7\,\mathrm{cm}^{-1}$), $\Delta R = 35.8\,\mathrm{pm}$ gives $S \approx 15$: the absorption is a long progression whose strongest lines end on high vibrational levels of the B state, and whose continuation runs into the dissociation continuum.

## 7.3 The fates of an excited state: the Jablonski diagram

**Definition 7.11 (Jablonski diagram).**

A *Jablonski diagram* shows the electronic states of a molecule (singlets $S_0, S_1, S_2, \dots$ in one column, triplets $T_1, T_2,
\dots$ in another), their vibrational levels, and the processes connecting them: radiative ones as straight arrows, radiationless ones as wavy arrows.

**Definition 7.12 (Radiationless processes).**

*Vibrational relaxation* brings an excited molecule to the lowest vibrational level of its electronic state by collisions, in picoseconds. *Internal conversion* is a radiationless transition between states of the same multiplicity ($S_2 \to S_1$, $S_1 \to S_0$); *intersystem crossing* one between states of different multiplicity ($S_1 \to T_1$, $T_1 \to S_0$).

**Definition 7.13 (Luminescence, fluorescence, phosphorescence, Stokes shift).**

*Luminescence* is emission of light by an excited molecule. *Fluorescence* is spin-allowed emission, usually $S_1 \to S_0$, fast (nanoseconds); *phosphorescence* is spin-forbidden emission, usually $T_1 \to S_0$, slow (milliseconds to seconds). The *Stokes shift* is the difference between the absorption and emission maxima, in energy.

![A Jablonski diagram. Straight arrows: absorption (blue), fluorescence (red), phosphorescence (orange); wavy arrows: vibrational relaxation and internal conversion (IC); dashed wavy arrows: intersystem crossing (ISC). Emission starts from the lowest vibrational level of S_1 or T_1 and may end on any level of S_0.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-electronic-spectroscopy/fig-aaa37de5fb38.svg)

*A [Jablonski diagram](#def-b3-electronic-spectroscopy-jablonski). Straight arrows: absorption (blue), [fluorescence](#def-b3-electronic-spectroscopy-luminescence) (red), [phosphorescence](#def-b3-electronic-spectroscopy-luminescence) (orange); wavy arrows: [vibrational relaxation](#def-b3-electronic-spectroscopy-radiationless) and [internal conversion](#def-b3-electronic-spectroscopy-radiationless) (IC); dashed wavy arrows: [intersystem crossing](#def-b3-electronic-spectroscopy-radiationless) (ISC). Emission starts from the lowest vibrational level of $S_1$ or $T_1$ and may end on any level of $S_0$.*

**Proposition 7.14 (Kasha’s rule).**

[Luminescence](#def-b3-electronic-spectroscopy-luminescence) occurs, with very few exceptions, from the lowest excited state of a given multiplicity ($S_1$ or $T_1$), whatever state was first excited: *Kasha’s rule*.

**Status.** Established experimentally; the reason is that [internal conversion](#def-b3-electronic-spectroscopy-radiationless) between upper excited states, which lie close together, is far faster than emission, while the gap $S_1$–$S_0$ is large. ∎

**Proposition 7.15 (Mirror image).**

If the excited and ground states have similar vibrational structure, the [fluorescence](#def-b3-electronic-spectroscopy-luminescence) band is the mirror image of the first absorption band about their common 0–0 line; the emission lies at longer wavelengths ([Stokes shift](#def-b3-electronic-spectroscopy-luminescence)).

**Proof.** Absorption goes from $v'' = 0$ to the levels $v'$ of $S_1$, at $E_{00} + v'h\nu'$; emission from $v' = 0$ to the levels $v''$ of $S_0$, at $E_{00} - v''h\nu''$. With equal frequencies and the same [Franck–Condon factors](#def-b3-electronic-spectroscopy-vibronic) (equal displacement), the two progressions are symmetric about $E_{00}$. ∎

![Absorption and fluorescence of a model molecule (S = 1, vibrational quantum 1400\, cm-1): mirror images about the 0–0 line; the maxima are separated by the Stokes shift.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-electronic-spectroscopy/fig-214715d875cf.svg)

*Absorption and [fluorescence](#def-b3-electronic-spectroscopy-luminescence) of a model molecule ($S = 1$, vibrational quantum $1400\,\mathrm{cm}^{-1}$): mirror images about the 0–0 line; the maxima are separated by the [Stokes shift](#def-b3-electronic-spectroscopy-luminescence).*

## 7.4 Kinetics of excited states

After excitation, the population of $S_1$ decays by every process open to it, each first order: radiative decay $k_r$, [internal conversion](#def-b3-electronic-spectroscopy-radiationless) $k_{\mathrm{IC}}$, [intersystem crossing](#def-b3-electronic-spectroscopy-radiationless) $k_{\mathrm{ISC}}$.

**Definition 7.16 (Excited-state lifetime, radiative lifetime).**

The *excited-state lifetime* is $\tau =
1/\sum k$, the time constant of the exponential decay of the excited population. The *radiative lifetime* $\tau_r =
1/k_r$ is the lifetime the state would have if emission were its only fate.

**Definition 7.17 (Quantum yield).**

The *quantum yield* of a process that follows the absorption of light is the number of events of that process divided by the number of photons absorbed. The *fluorescence quantum yield* $\Phi_F$ is the number of photons emitted as [fluorescence](#def-b3-electronic-spectroscopy-luminescence) per photon absorbed.

**Proposition 7.18 (Quantum yield and lifetime).**

$\Phi_F = k_r/(k_r + k_{\mathrm{IC}} + k_{\mathrm{ISC}}) = k_r\tau = \tau/\tau_r$; likewise $\Phi_{\mathrm{ISC}} = k_{\mathrm{ISC}}\tau$, and the yields of the competing processes add up to 1.

**Proof.** $\dd[S_1]/\dd t = -(k_r + k_{\mathrm{IC}} + k_{\mathrm{ISC}})[S_1]$, so $[S_1] = [S_1]_0\eu^{-t/\tau}$. The number of photons emitted is $\int_0^\infty k_r[S_1]\,\dd t = k_r\tau[S_1]_0$, out of $[S_1]_0$ excited molecules. The same integral with each $k$ gives each yield, and their sum is $\tau\sum k = 1$. ∎

**Definition 7.19 (Fluorescence quencher, dynamic and static quenching).**

A *fluorescence quencher* is a species that reduces [fluorescence](#def-b3-electronic-spectroscopy-luminescence). In *dynamic quenching* it deactivates the excited molecule on collision, adding a rate $k_q[Q]$ and shortening the lifetime; in *static quenching* it forms a non-fluorescent complex with the ground-state molecule, reducing the intensity without changing the lifetime of the molecules that still emit.

**Theorem 7.20 (Stern–Volmer equation).**

For [dynamic quenching](#def-b3-electronic-spectroscopy-quencher),

$$
\frac{I_0}{I} = \frac{\tau_0}{\tau} = 1 + k_q\tau_0[Q] = 1 + K_{SV}[Q],
$$

the *Stern–Volmer equation*, where $I_0$, $\tau_0$ are measured without quencher and $K_{SV} = k_q\tau_0$.

**Proof.** With quencher, $\tau = 1/(k_0 + k_q[Q])$ with $k_0 = 1/\tau_0$, so $\tau_0/\tau = 1 +
k_q\tau_0[Q]$. The intensity is proportional to $\Phi_F = k_r\tau$, so $I_0/I = \tau_0/\tau$. ∎

![Dynamic quenching of a model fluorophore (_0 = 10\, ns, k_q = 5 × 109\, L\, mol-1\, s-1). Left: fluorescence decays, straight lines on a log scale, steeper with more quencher. Right: the Stern–Volmer line, of slope K_SV = k_q _0 = 50\, L/ mol.](https://one-course.com/images/onecourse/chapters/chemistry-4/b3-electronic-spectroscopy/fig-8a93c3a1d588.svg)

*[Dynamic quenching](#def-b3-electronic-spectroscopy-quencher) of a model fluorophore ($\tau_0 = 10\,\mathrm{ns}$, $k_q =
5 \times 10^{9}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$). Left: [fluorescence](#def-b3-electronic-spectroscopy-luminescence) decays, straight lines on a log scale, steeper with more quencher. Right: the Stern–Volmer line, of slope $K_{SV} = k_q\tau_0 = 50\,\mathrm{L}/\mathrm{mol}$.*

**Method 7.21 (Relative fluorescence quantum yield).**

1. Choose a standard of known $\Phi_{\mathrm{std}}$ absorbing at the same excitation wavelength.
2. Prepare dilute solutions ( $A < 0.1$ , to avoid re-absorption) and record their absorbances $A$ and integrated emission spectra $F$ under identical conditions.
3. $\Phi = \Phi_{\mathrm{std}}\,\dfrac{F}{F_{\mathrm{std}}}\,\dfrac{A_{\mathrm{std}}}{A}\,  \dfrac{n^2}{n_{\mathrm{std}}^2}$ , the last factor correcting for the refractive indices of the solvents.

**Method 7.22 (Analysing quenching data).**

1. Plot $I_0/I$ against $[Q]$ ; a straight line through 1 gives $K_{SV}$ .
2. Measure lifetimes too: if $\tau_0/\tau$ follows the same line, the quenching is dynamic, and $k_q = K_{SV}/\tau_0$ ; if $\tau$ does not change, it is static.
3. Compare $k_q$ with the diffusion limit, about $10^{10}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$ in water ( [Chapter 12](https://one-course.com/books/chemistry/4/en/chapter/12-theories-of-reaction-rates#ch-b3-rate-theories) ): a $k_q$ close to it means nearly every encounter quenches.

**Example 7.23 (Quinine and anthracene).**

Quinine sulfate in $0.5\,\mathrm{M}$ sulfuric acid absorbs at $349\,\mathrm{nm}$ with $\varepsilon =
5700\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{cm}^{-1}$ and fluoresces with $\Phi_F = 0.546$, which makes it the classic standard for relative [quantum yields](#def-b3-electronic-spectroscopy-quantum-yield). Anthracene in cyclohexane absorbs at $356\,\mathrm{nm}$ and has $\Phi_F = 0.36$; most of the rest of its excited molecules cross to the triplet.

## 7.5 Applications

[Fluorescence](#def-b3-electronic-spectroscopy-luminescence) is detected against a dark background, which makes fluorimetry a hundred to a thousand times more sensitive than absorption spectroscopy: nanomolar concentrations of a fluorescent compound are measured routinely. Fluorescent probes report on their surroundings through their wavelength, yield or lifetime (polarity, pH, the binding of an ion); energy transfer between two fluorophores, efficient only over a few nanometres, measures distances in proteins. Phosphorescent complexes of heavy metals, in which [spin–orbit coupling](https://one-course.com/books/chemistry/4/en/chapter/2-many-electron-atoms-and-term-symbols#def-b3-many-electron-atoms-spin-orbit) makes [intersystem crossing](#def-b3-electronic-spectroscopy-radiationless) efficient, harvest triplets in the light-emitting diodes of display screens.

**In the lab — Time-correlated single-photon counting.**

To measure a nanosecond lifetime, the sample is excited by a pulsed laser diode at a megahertz repetition rate, and for each pulse the delay of the first [fluorescence](#def-b3-electronic-spectroscopy-luminescence) photon detected is recorded. After millions of pulses the histogram of delays is the decay curve, from which $\tau$ is fitted, with the instrument response measured on a scattering solution. The ultraviolet source is enclosed; the operator works with goggles that block its wavelength.

**History — Stokes, 1852.**

George Gabriel Stokes passed sunlight through a violet glass and a prism into a solution of quinine, and saw blue light emerge from the region lit by invisible ultraviolet rays: the emitted light always had a longer wavelength than the light absorbed. He called the phenomenon [fluorescence](#def-b3-electronic-spectroscopy-luminescence), after the mineral fluorite, which glows the same way.

## 7.6 Exercises

**Exercise 7.1 ★.**

Allowed or forbidden, and by which rule: (a) $S_0 \to T_1$ in benzene; (b) a $d$–$d$ transition of an octahedral complex; (c) the $\pi\to\pi^*$ transition of ethene ($B_{1u} \leftarrow A_g$ in $D_{2h}$); (d) the $n\to\pi^*$ transition of methanal ($A_2$)?

**Solution of Exercise 7.1.**

(a) Forbidden by the spin rule ($\Delta S = 1$). (b) Forbidden by the [Laporte rule](#thm-b3-electronic-spectroscopy-laporte) ($g \to g$); seen weakly through vibronic coupling. (c) Allowed: $B_{1u}$ is the representation of $z$ in $D_{2h}$. (d) Forbidden by symmetry: $A_2$ is none of $x$, $y$, $z$ in $C_{2v}$; the band is weak.

**Exercise 7.2 ★.**

A compound absorbs at $349\,\mathrm{nm}$ and fluoresces with a maximum at $450\,\mathrm{nm}$ (data of the exercise). Compute the [Stokes shift](#def-b3-electronic-spectroscopy-luminescence) in $\mathrm{cm}^{-1}$ and in eV.

**Solution of Exercise 7.2.**

$10^7/349 - 10^7/450 = 28653 - 22222 = 6431\,\mathrm{cm}^{-1}$, $0.80\,\mathrm{eV}$.

**Exercise 7.3 ★.**

Compute the absorbance at $349\,\mathrm{nm}$ of a $1.0 \times 10^{-5}\,\mathrm{mol}/\mathrm{L}$ quinine sulfate solution in a $1\,\mathrm{cm}$ cell, and the fraction of the light it absorbs.

**Solution of Exercise 7.3.**

$A = 5700 \times 1 \times 1.0\times10^{-5} = 0.057$; absorbed fraction $1 - 10^{-0.057} =
0.12$.

**Exercise 7.4 ★.**

A fluorophore has $\Phi_F = 0.60$ and $\tau = 4.0\,\mathrm{ns}$. Compute $k_r$, the [radiative lifetime](#def-b3-electronic-spectroscopy-lifetime) and the sum of the non-radiative rate constants.

**Solution of Exercise 7.4.**

$k_r = \Phi_F/\tau = 1.5 \times 10^{8}\,\mathrm{s}^{-1}$, $\tau_r = 6.7\,\mathrm{ns}$; $k_{nr} = (1 - \Phi_F)/\tau
= 1.0 \times 10^{8}\,\mathrm{s}^{-1}$.

**Exercise 7.5 ★★.**

A band is Gaussian with $\varepsilon_{\max} = 5700\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{cm}^{-1}$ and a full width at half maximum of $5000\,\mathrm{cm}^{-1}$. Estimate its [oscillator strength](#def-b3-electronic-spectroscopy-transition-dipole) (the integral of a Gaussian is $1.0645\,\varepsilon_{\max}\times$FWHM).

**Solution of Exercise 7.5.**

$\int\varepsilon\,\dd\tilde\nu = 1.0645 \times 5700 \times 5000 = 3.03 \times 10^{7}$, so $f \approx
4.32\times10^{-9} \times 3.03\times10^7 = 0.13$: an allowed transition of moderate strength.

**Exercise 7.6 ★★.**

For displaced oscillators, find the strongest line of the progression for $S = 0.3$, 1.5 and 5, and the ratio of its factor to that of the 0–0 line.

**Solution of Exercise 7.6.**

The Poisson factor $\eu^{-S}S^v/v!$ is maximal at $v = \lfloor S\rfloor$. $S = 0.3$: the 0–0 line itself. $S = 1.5$: $v' = 1$, 1.5 times the 0–0 line. $S = 5$: $v' = 4$ and 5 equally, 26 times the 0–0 line.

**Exercise 7.7 ★★.**

From $S_1$, $k_r = 1.0 \times 10^{8}\,\mathrm{s}^{-1}$, $k_{\mathrm{IC}} = 2.0 \times 10^{7}\,\mathrm{s}^{-1}$ and $k_{\mathrm{ISC}} =
3.0 \times 10^{8}\,\mathrm{s}^{-1}$. Compute $\tau$, $\Phi_F$ and $\Phi_{\mathrm{ISC}}$.

**Solution of Exercise 7.7.**

$\sum k = 4.2 \times 10^{8}\,\mathrm{s}^{-1}$: $\tau = 2.4\,\mathrm{ns}$, $\Phi_F = 0.24$, $\Phi_{\mathrm{ISC}} = 0.71$ (and $\Phi_{\mathrm{IC}} = 0.05$).

**Exercise 7.8 ★★.**

The [fluorescence](#def-b3-electronic-spectroscopy-luminescence) of anthracene in cyclohexane ($A = 0.040$) integrates to 0.46 times that of quinine sulfate in dilute sulfuric acid ($A = 0.050$, $\Phi =
0.546$). With refractive indices 1.4266 (cyclohexane) and 1.333 (water, for the dilute acid), compute $\Phi_F$ of anthracene.

**Solution of Exercise 7.8.**

$\Phi = 0.546 \times 0.46 \times (0.050/0.040) \times (1.4266/1.333)^2 = 0.36$.

**Exercise 7.9 ★★.**

The lifetime of a fluorophore is 8.0, 5.6 and $4.3\,\mathrm{ns}$ at quencher concentrations 0, 0.010 and $0.020\,\mathrm{mol}/\mathrm{L}$. Is the quenching dynamic? Find $K_{SV}$ and $k_q$.

**Solution of Exercise 7.9.**

$\tau_0/\tau = 1.429$ and 1.860: linear in $[Q]$, so the lifetime itself is shortened: [dynamic quenching](#def-b3-electronic-spectroscopy-quencher). $K_{SV} = 43$ L/mol (both points), $k_q = 43/8.0
\times 10^{-9} = 5.4 \times 10^{9}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$.

**Exercise 7.10 ★★★.**

Adding a quencher halves the [fluorescence](#def-b3-electronic-spectroscopy-luminescence) intensity while the measured lifetime stays at $6.0\,\mathrm{ns}$. What kind of quenching is it, and what does it imply about the ground state? Write the corresponding relation between $I_0/I$ and the association constant of the complex.

**Solution of Exercise 7.10.**

[Static quenching](#def-b3-electronic-spectroscopy-quencher): the molecules that still emit live as long as before; half of them are bound, in the ground state, in a non-fluorescent complex. With an association constant $K_a$, the free fraction is $1/(1 + K_a[Q])$ and $I_0/I = 1 +
K_a[Q]$, the same form as Stern–Volmer but with $\tau$ unchanged.

**Exercise 7.11 ★★★.**

Anthracene in cyclohexane has $\Phi_F = 0.36$ and, in the conditions of a measurement, $\tau = 5.0\,\mathrm{ns}$ (data of the exercise). Compute $\tau_r$, and the yield of the triplet if [internal conversion](#def-b3-electronic-spectroscopy-radiationless) is negligible. Why is the [radiative lifetime](#def-b3-electronic-spectroscopy-lifetime) longer than the measured one?

**Solution of Exercise 7.11.**

$\tau_r = \tau/\Phi_F = 14\,\mathrm{ns}$; $\Phi_T = 1 - 0.36 = 0.64$. The measured lifetime is shortened by the competing [intersystem crossing](#def-b3-electronic-spectroscopy-radiationless); $\tau_r$ is what emission alone would give.

**Exercise 7.12 ★★★.**

For two electrons, the singlet spin function is $\frac{1}{\sqrt2}[\alpha\beta - \beta\alpha]$ and the $M_S = 0$ triplet function $\frac{1}{\sqrt2}[\alpha\beta + \beta\alpha]$. Show that they are orthogonal, and conclude on the intensity of $T_1 \leftarrow S_0$ in the absence of [spin–orbit coupling](https://one-course.com/books/chemistry/4/en/chapter/2-many-electron-atoms-and-term-symbols#def-b3-many-electron-atoms-spin-orbit).

**Solution of Exercise 7.12.**

$\frac12\langle\alpha\beta - \beta\alpha|\alpha\beta + \beta\alpha\rangle = \frac12(1 + 0 - 0 - 1) = 0$, using $\langle\alpha|\alpha\rangle = \langle\beta|\beta\rangle = 1$, $\langle\alpha|\beta\rangle = 0$ for each electron. The dipole [operator](https://one-course.com/books/chemistry/4/en/chapter/1-quantum-mechanics-for-chemists-model-systems#def-b3-quantum-model-systems-operator) does not touch spin, so the transition moment contains this zero factor: $T_1 \leftarrow S_0$ has no intensity unless [spin–orbit coupling](https://one-course.com/books/chemistry/4/en/chapter/2-many-electron-atoms-and-term-symbols#def-b3-many-electron-atoms-spin-orbit) mixes the states.

## 7.7 Problem: Why Tonic Water Glows

**Problem 7.1.**

Weekend problem — absorption by quinine, the fate of its excited state, quenching by chloride ions, and whether every encounter quenches

Quinine sulfate in $0.5\,\mathrm{M}$ sulfuric acid: absorption maximum $349\,\mathrm{nm}$, $\varepsilon = 5700\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{cm}^{-1}$, $\Phi_F = 0.546$. Data of the problem: the [fluorescence](#def-b3-electronic-spectroscopy-luminescence) lifetime without quencher is $\tau_0 = 19\,\mathrm{ns}$; the emission maximum is near $450\,\mathrm{nm}$; adding sodium chloride gives the intensity ratios

| $[\ce{Cl-}]$ / mol L$^{-1}$ | 0 | 0.010 | 0.020 | 0.040 | 0.080 |
| --- | --- | --- | --- | --- | --- |
| $I_0/I$ | 1.000 | 1.594 | 2.188 | 3.376 | 5.752 |

Water at $25\,{}^{\circ}\mathrm{C}$ has viscosity $0.890\,\mathrm{mPa}\,\mathrm{s}$; $R =
8.314\,\mathrm{J}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1}$.

**Part I — Absorption.**

1. Compute the absorbance at $349\,\mathrm{nm}$ of a $1.0 \times 10^{-4}\,\mathrm{mol}/\mathrm{L}$ solution in a $1\,\mathrm{cm}$ cell.
2. What fraction of the incident light is absorbed?
3. Is the transition allowed? Use $\varepsilon$ .
4. Give the energy of the absorbed photon in eV.
5. Why does the eye see no absorption colour in tonic water?
6. Why is a solution for [fluorescence](#def-b3-electronic-spectroscopy-luminescence) work kept below $A = 0.1$ ?

**Part II — The excited state.**

7. Draw the [Jablonski diagram](#def-b3-electronic-spectroscopy-jablonski) of the processes from $S_1$ .
8. Compute the radiative rate constant $k_r$ and the [radiative lifetime](#def-b3-electronic-spectroscopy-lifetime) .
9. Compute the total non-radiative rate constant.
10. Compute the [Stokes shift](#def-b3-electronic-spectroscopy-luminescence) in $\mathrm{cm}^{-1}$ .
11. Why is the emission blue although the absorption is in the ultraviolet?
12. What fraction of the absorbed energy leaves as [fluorescence](#def-b3-electronic-spectroscopy-luminescence) , counting the energy of each photon?

**Part III — Quenching by chloride.**

13. Plot $I_0/I$ against $[\ce{Cl-}]$ and check that it is linear.
14. Determine $K_{SV}$ by least squares through the five points.
15. Assuming [dynamic quenching](#def-b3-electronic-spectroscopy-quencher) , compute $k_q$ .
16. What lifetime would you measure at $0.040\,\mathrm{mol}/\mathrm{L}$ ?
17. How would you prove experimentally that the quenching is dynamic?
18. Why is tonic water, which contains no chloride, a good place for quinine to glow?

**Part IV — Every encounter?**

19. The rate constant of encounters controlled by diffusion in a solvent of viscosity $\eta$ is $k_d \approx 8RT/3\eta$ . Compute it in $\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$ .
20. Compare $k_q$ and $k_d$ .
21. What fraction of the encounters lead to quenching?
22. Would the quenching be faster or slower in a more viscous solvent?
23. State the result: the ratio $k_q/k_d$ for the quenching of quinine by chloride.

**Solution of Problem 7.1.**

**1.** $A = 5700 \times 1 \times 1.0\times10^{-4} = 0.57$. **2.** $1 - 10^{-0.57} = 0.73$. **3.** $\varepsilon$ of a few thousand: an allowed $\pi\to\pi^*$ transition, of moderate strength. **4.** $1239.8/349 = 3.55\,\mathrm{eV}$. **5.** It absorbs in the ultraviolet only; the visible light passes. **6.** To keep the absorbed light proportional to the concentration and avoid re-absorption of the emitted light (inner-filter effects). **7.** $S_0 \to S_1$ (absorption, then [vibrational relaxation](#def-b3-electronic-spectroscopy-radiationless)); from $S_1$: [fluorescence](#def-b3-electronic-spectroscopy-luminescence), [internal conversion](#def-b3-electronic-spectroscopy-radiationless), [intersystem crossing](#def-b3-electronic-spectroscopy-radiationless) to $T_1$. **8.** $k_r = 0.546/19 \times 10^{-9} = 2.87 \times 10^{7}\,\mathrm{s}^{-1}$; $\tau_r = 35\,\mathrm{ns}$. **9.** $(1 - 0.546)/\tau_0 = 2.39 \times 10^{7}\,\mathrm{s}^{-1}$. **10.** $28653 - 22222 = 6431\,\mathrm{cm}^{-1}$. **11.** Emission starts from the relaxed $S_1$ and ends on vibrationally excited levels of $S_0$, after the molecule and solvent have relaxed: the photon has less energy, here enough to fall in the visible. **12.** $\Phi_F \times (349/450) = 0.42$: 42 % of the absorbed energy. **13.** The points rise by 0.594 per $0.010\,\mathrm{mol}/\mathrm{L}$: a straight line through 1. **14.** Least squares through the five points: slope $K_{SV} = 59.4\,\mathrm{L}/\mathrm{mol}$, intercept 1.000. **15.** $k_q = K_{SV}/\tau_0 = 3.1 \times 10^{9}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$. **16.** $\tau = \tau_0/(1 + 59.4 \times 0.040) = 5.6\,\mathrm{ns}$. **17.** Measure lifetimes: for [dynamic quenching](#def-b3-electronic-spectroscopy-quencher) $\tau_0/\tau$ falls on the same line as $I_0/I$. **18.** Tonic water contains no chloride to quench it, and is acidic enough to keep quinine protonated, its fluorescent form. **19.** $k_d = 8 \times 8.314 \times 298/(3 \times 0.890\times10^{-3}) =
7.4 \times 10^{6}\,\mathrm{m}^{3}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1} = 7.4 \times 10^{9}\,\mathrm{L}\,\mathrm{mol}^{-1}\,\mathrm{s}^{-1}$. **20.** $k_q$ is a little less than half of $k_d$. **21.** About 42 %. **22.** Slower: $k_d \propto 1/\eta$, and $k_q$ can only fall with it. **23.** **$k_q/k_d = 0.42$**: chloride quenches quinine at almost half the rate of diffusion-controlled encounters.