Lesson 6 · 35 min

Conjugation and colour: π electrons in a box

Three cyanine dyes that differ only by –CH=CH– units run from magenta through blue to green. Shut their π electrons in a one-dimensional box and you can calculate the wavelengths they absorb — then test the model against nine measured absorption maxima.

In this lesson

Write the energy levels and wavefunctions of a particle in a one-dimensional box, and explain why the levels fall and crowd together as the box grows
Count the π electrons on a conjugated chain and estimate a cyanine dye’s absorption wavelength with the free-electron model
Fit the model’s parameters to measured data by linearising and least squares, and say what the parameters mean physically
Explain why the free-electron model works for cyanine dyes but fails for polyenes

Dissolve a little of each of three cyanine dyes in ethanol: one solution is magenta, one deep blue, and the third an almost pale green. The molecules are nearly identical — two identical nitrogen-containing rings joined by a carbon chain — and differ only in the length of the chain. Every extra –CH=CH– unit moves the absorption peak about 100 nm towards longer wavelengths. Why should a slightly longer chain change the colour? This lesson answers with the simplest model in quantum mechanics — the particle in a box — and then tests the answer against nine measured absorption maxima.

Predict first

If the box is made twice as long, what happens to the energy levels of an electron inside it?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

They all rise to twice their value They stay the same They all fall to half their value They all fall to a quarter of their value and crowd closer together

A particle in a box

Imagine an electron shut in a one-dimensional segment of length $L$ with infinitely high walls at both ends. The electron is a wave, and a confined wave can only form standing waves: the box must hold a whole number of half-wavelengths, $L = n\lambda/2$ with $n = 1, 2, 3, \ldots$ Put the de Broglie relation $p = h/\lambda$ into the kinetic energy $E = p^2/2m$ and the allowed energies below follow. Three conclusions come out of that one expression:

the energy is quantised: only a set of values proportional to $n^2$ is allowed;
the lowest energy $E_1$ is not zero — a confined electron is never at rest, which is called the zero-point energy;
the longer the box, the lower and closer the levels: $E \propto 1/L^2$ .

The $n$ th standing wave has $n - 1$ nodes, points where the electron is never found. For an electron in a 1 nm box, $E_1 = 0.376$ eV.

E_n = \frac{n^2 h^2}{8 m L^2},\qquad \psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\frac{n\pi x}{L}\quad (0 \le x \le L)

Energy levels and wavefunctions of a particle in a one-dimensional box with infinitely high walls.

$n$: quantum number, 1, 2, 3, …
$h$: Planck constant
$m$: mass of the particle (here, an electron)
$L$: length of the box
$\psi_n$: wavefunction; |ψ|² is the probability density of finding the particle at x

A box for π electrons

A cyanine dye has identical nitrogen-containing rings at its two ends and a conjugated carbon chain between them: N–C=C–…–C=N⁺. The positive charge can sit on either nitrogen; the two resonance structures are equivalent, so the bonds along the chain are almost all the same length (about 1.4 Å). In 1948 Hans Kuhn proposed that these π electrons move freely along the chain, like particles in a one-dimensional box.

Count them: the chain between the two nitrogens has $p$ carbons (those of the bridge plus one ring carbon at each end); each carbon gives one π electron and the two nitrogens together give three, $N = p + 3$ in all. They fill the levels in pairs from $n = 1$ up to $n = N/2$ . The lowest-energy absorption lifts an electron from the highest occupied level (the HOMO, $n = N/2$ ) to the lowest empty one (the LUMO, $n = N/2 + 1$ ). How long is the box? The textbook choice lets it reach one bond beyond each nitrogen: $L = (p + 3)\ell$ with $\ell = 1.39$ Å. To test that choice, the expression below also carries an end correction, $\gamma$ .

\lambda = \frac{8 m c L^2}{h\,(N + 1)},\qquad L = (p + 3 + \gamma)\,\ell,\qquad N = p + 3

Absorption wavelength of the HOMO → LUMO jump: ΔE = E_{N/2+1} − E_{N/2} = (N + 1)h²/(8mL²).

$N$: number of π electrons on the chain
$p$: carbon atoms on the chain between the two nitrogens
$\ell$: length of box per bond
$\gamma$: end correction: extra bonds by which the box reaches beyond the chain ends
$c$: speed of light

Worked example

Estimating a thiacarbocyanine with the textbook box

A thiacarbocyanine has 3 carbons in its bridge. Estimate its absorption wavelength with the textbook box ( $\ell = 1.39$ Å, $\gamma = 0$ ) and compare with the measured 559 nm.

Test the free-electron model against nine measured absorption maxima

The electron box inside a dye

Thiacarbocyanine (3-carbon bridge): 5 chain carbons, 8 π electrons; box 11.1 Å long

Box length per bond, ℓ 1.390 Å Extra bonds at each end, γ 0.00

Model: 453 nm
Measured: 559 nm
Error: −106 nm

Root-mean-square error over this family’s three dyes: 82.5 nm

Linearise: plot the effective box length √(λ(N+1)/K)

Start with the textbook box: it gives the same wavelengths for all three families — up to 19 % short for the thiacyanines, yet almost right, by coincidence, for the oxacyanines. Then press the least-squares button: the three thiacyanines are matched to within 3 nm with ℓ ≈ 1.25 Å and γ ≈ 1.9. Switch family and fit again; tick “Linearise” to see why the fit is a straight line.

Rewrite the model as $\sqrt{\lambda(N + 1)/K} = \ell\,(p + 3) + \ell\gamma$ (with $K = 8mc/h$ ) and it is a straight line in $p$ : the slope is $\ell$ and the intercept $\ell\gamma$ . All three families give $\ell$ between 1.23 and 1.25 Å — the fit knows nothing about bond geometry, yet it returns a meaningful number: the chain is a zigzag with 120° angles, so each 1.39 Å bond advances the chain along its own direction by only $1.39 \times \cos 30^\circ = 1.20$ Å. The end correction differs by family: 1.3 bonds for the oxacyanines, 1.9 for the thia- and indocyanines — the π electrons reach further into the larger, more polarisable end groups. With two physically meaningful parameters, the model describes each family’s three absorption maxima to within 3 nm.

Concept check

Which change would move a cyanine dye’s absorption peak to shorter wavelength?

Add another –CH=CH– unit to the chain Remove a –CH=CH– unit from the chain Replace benzoxazole ends with benzothiazole ends Dissolve more dye in the solution

Try the calculation

With the fitted thiacyanine box ( $\ell = 1.249$ Å, $\gamma = 1.88$ ), predict the absorption maximum of the family’s next member, with a 9-carbon bridge ( $p = 11$ , $N = 14$ ).

Key takeaways

A particle in a box has levels Eₙ = n²h²/(8mL²): the energy is quantised, there is a zero-point energy, and a longer box gives lower, closer levels.
A cyanine’s N = p + 3 π electrons fill up to n = N/2; the lowest absorption is λ = 8mcL²/(h(N + 1)), so a longer chain absorbs longer wavelengths, about 100 nm per –CH=CH–.
The textbook box (1.39 Å per bond) knows nothing of the end groups and misses by up to 19 %; fitting gives ℓ ≈ 1.24 Å (the zigzag projection) and an end-group-dependent γ, cutting the error to within 3 nm.
The model relies on equal bonds; it fails for polyenes with alternating bond lengths, such as β-carotene.

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Data and references

H. Kuhn, “Free electron model for absorption spectra of organic dyes”, J. Chem. Phys. 16, 840 (1948); “A quantum-mechanical theory of light absorption of organic dyes and similar compounds”, J. Chem. Phys. 17, 1198 (1949).
J. Autschbach, “Why the particle-in-a-box model works well for cyanine dyes but not for conjugated polyenes”, J. Chem. Educ. 84, 1840 (2007).
PhotochemCAD spectra: H. Du, R. A. Fuh, J. Li, L. A. Corkan, J. S. Lindsey, Photochem. Photobiol. 68, 141 (1998); J. M. Dixon, M. Taniguchi, J. S. Lindsey, Photochem. Photobiol. 81, 212 (2005); compiled by S. Prahl at omlc.org. Original data: P. J. Sims et al., Biochemistry 13, 3315 (1974); C. J. Tredwell, C. M. Keary, Chem. Phys. 43, 307 (1979); A. Waggoner et al., Methods Cell Biol. 30, 449 (1989).
C. W. Garland, J. W. Nibler, D. P. Shoemaker, Experiments in Physical Chemistry, 8th ed., McGraw-Hill, 2009, “Absorption spectra of conjugated dyes”.