Daily answer writing · Thu, 29 Oct, 2026
Particle in a 1D box: energy levels and the λmax of 1,3-butadiene
Write the expression for the energy levels of a particle in a one-dimensional box of length $L$. Treating the 4 π electrons of 1,3-butadiene as particles in a box of length 578 pm, calculate the energy of the lowest-energy electronic transition and the wavelength of light absorbed. Compare with the observed $\lambda_{\max}$ of 217 nm.
($h = 6.626\times10^{-34}$ J s, $m_e = 9.109\times10^{-31}$ kg, $c = 2.998\times10^{8}$ m s−1)
Syllabus: Common to CSE Paper I, IFoS Paper I and Geo-Scientist Mains (physical): Schrödinger equation, particle in a one-dimensional box, application to conjugated systems.
Model answer
Energy levels
Inside the box $V = 0$, outside $V = \infty$. Solving $-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2} = E\psi$ with $\psi(0) = \psi(L) = 0$:
$$\psi_n = \sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}, \qquad E_n = \frac{n^2h^2}{8mL^2}, \quad n = 1, 2, 3, \dots$$
Lowest transition (n = 2 → 3)
$$\frac{h^2}{8mL^2} = \frac{(6.626\times10^{-34})^2}{8(9.109\times10^{-31})(578\times10^{-12})^2} = 1.80\times10^{-19}\ \text{J}$$
$$\Delta E = (3^2 - 2^2)\frac{h^2}{8mL^2} = 5 \times 1.80\times10^{-19} = \mathbf{9.02\times10^{-19}\ J}\ (5.63\ \text{eV})$$
Wavelength
$$\lambda = \frac{hc}{\Delta E} = \frac{(6.626\times10^{-34})(2.998\times10^{8})}{9.02\times10^{-19}} = 2.20\times10^{-7}\ \text{m} = \mathbf{220\ nm}$$
Comparison. Observed 217 nm; the error is about 1.5%. The model ignores electron–electron repulsion and the real (non-flat) potential, yet works because the π electrons really are delocalised along the chain. It also explains why $\lambda_{\max}$ increases with conjugation length ($\lambda \propto L^2/(N+1)$).
Conclusion. The free-electron model predicts λmax = 220 nm for butadiene, close to the observed 217 nm.
Where the marks are
- Boundary conditions, ψn and En = n²h²/8mL² (2 marks).
- Filling 4 electrons: HOMO n = 2, LUMO n = 3 (2 marks). A level diagram helps.
- ΔE = 5h²/8mL² = 9.02 × 10⁻¹⁹ J with units, L in metres (2 marks).
- λ = 220 nm (2 marks).
- Comparison with 217 nm and one limitation or the trend with chain length (2 marks).