An electron is confined to a one-dimensional infinite well of width L = 0.500 nm. Derive the allowed energies from standing-wave boundary conditions, then calculate E₁, E₂ and the photon wavelength for a 2→1 transition. Write the normalized ground-state wavefunction, state the number of nodes for n = 3, and explain why the electron does not have a single classical trajectory between walls. Use h = 6.626 × 10⁻³⁴ J s and m_e = 9.11 × 10⁻³¹ kg.

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E_n = n²h²/(8mL²). E₁ = 1.50 eV, E₂ = 6.02 eV; the 2→1 photon has energy 4.51 eV and wavelength about 275 nm. ψ₁ = √(2/L)sin(πx/L); n = 3 has two internal nodes. Zero wavefunction at x = 0 and L requires an integer number of half wavelengths: L = nλ/2, so momentum magnitude p = h/λ = nh/(2L). Non-relativistic kinetic energy is p²/(2m), yielding E_n = n²h²/(8mL²). For L = 0.500 nm, E₁ = 2.41 × 10⁻¹⁹ J = 1.50 eV and E₂ = 4E₁ = 6.02 eV. Transition energy is E₂−E₁ = 3E₁ = 4.51 eV. Photon wavelength is hc/ΔE ≈ 275 nm. Normalization gives ψ₁(x) = √(2/L)sin(πx/L) inside and zero outside. State n has n−1 internal nodes, so n = 3 has two. The wavefunction provides a probability amplitude; |ψ|² predicts position distributions while energy states are stationary. A classical path with definite position and momentum at every instant is incompatible with the spatially extended state and uncertainty relation.

Energy levels in an infinite quantum well

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