“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

The Particle in a Box by Hand — A dye molecule's color from one formula

Exercises

Introductory Quantum Mechanics › Unit 2 · Bound states in one dimension › The Particle in a Box by Hand › all problems

Worked example Problem 1 of 10

A dye molecule's color from one formula

Problem

Model the six

π

electrons of hexatriene as free particles in a 1D box of length

L = 0.70

nm (the length of the conjugated chain). Fill the levels per the Pauli principle, find the lowest-energy absorption, and compare the predicted wavelength with the measured 268 nm.

Solution

The box energies are $E_{n} = n^{2} h^{2} /8 m L^{2}$ — everything below is this one formula plus counting. First the energy scale of THIS box:

E_{1} = \frac{h ^{2}}{8 m _{e} L ^{2}} = \frac{( 6.626 \times 1 0 ^{- 34} ) ^{2}}{8 ( 9.109 \times 1 0 ^{- 31} ) ( 7.0 \times 1 0 ^{- 10} ) ^{2}} = 1.229 \times 1 0^{- 19} J = 0.767 eV

Predict before reading on. Six electrons, two per level (spin up + spin down). Before reading on: which level is the highest occupied, and which transition is the lowest-energy absorption?

Electrons fill $n = 1, 2, 3$ (two each). The lowest absorption promotes one electron from $n = 3$ to $n = 4$ :

Δ E = E_{4} - E_{3} = (16 - 9) E_{1} = 7 \times 0.767 = 5.37 eV

Predict before reading on. Before converting to wavelength, predict the direction of the model’s error: a real chain lets the electron leak slightly past the end carbons. Does that make the true box longer or shorter, and the predicted $Δ E$ too big or too small?

Convert with $λ = h c /Δ E$ :

λ = \frac{h c}{Δ E} = \frac{1240 eV \cdotp nm}{5.37 eV} = 231 nm (experiment: 268 nm)

13% short — because the box is effectively longer than the bare C–C skeleton (the electron leaks past the ends), which lowers $Δ E$ and lengthens $λ$ . A crude model with zero adjustable chemistry lands within 15% of a real UV spectrum, and predicts the right trend: every added conjugated bond stretches $L$ , shrinks $Δ E$ , and pushes absorption toward the red — exactly how dye chemists tune color.

Predict before reading on. Articulate in one sentence: what two ingredients did the entire calculation consist of?

Result

Δ E = 7 E_{1} = 5.37 eV, λ_{pred} = 231 nm (obs. 268 nm)