See the real quantum wavefunctions of hydrogen as glowing 3D probability clouds sampled from |ψ|², coloured by phase. Pick n, l and m, slice the cloud to reveal nodes, and calculate the energy, mean radius and emission wavelength of any electron transition.
The hydrogen atom is the only atom whose Schrödinger equation has an exact solution. Each stationary state factorises into a radial part and an angular part, ψₙₗₘ(r,θ,φ) = Rₙₗ(r) · Yₗₘ(θ,φ). The radial function uses a generalised Laguerre polynomial, Rₙₗ ∝ ρˡ e^(−ρ/2) L^(2l+1)ₙ₋ₗ₋₁(ρ) with ρ = 2r / (n a₀); the angular part uses associated Legendre polynomials. This tool shows the real orbitals chemists draw (pₓ, d_xy, d_z² …), which are sums of the complex ±m states.
Every dot is a random position drawn from the probability density |ψ|². The radius is sampled exactly from the radial distribution P(r) = r²R² (inverse CDF), and the direction by rejection sampling on |Y|², so dense regions really are where the electron is most likely to be found. Colour shows the sign of ψ — the phase that decides whether orbitals overlap constructively when bonds form.
Eₙ = −13.598 eV / n² (reduced-mass Rydberg). It depends only on n — all n² orbitals in a shell are degenerate.⟨r⟩ = a₀ [3n² − l(l+1)] / 2, with a₀ = 52.918 pm.1/λ = R_H (1/nₗ² − 1/nᵤ²), R_H = 1.09678×10⁷ m⁻¹. Hα (3→2) is 656.47 nm in vacuum (656.28 nm in air).Worked example: 2→1 (Lyman-α) gives 1/λ = 1.09678×10⁷ × 3/4 → λ = 121.57 nm, E = 10.20 eV — deep ultraviolet.