An electron held by a proton has no orbit. It has a standing-wave state ψ_nlm, labelled by three integers. This sim draws any state up to n = 7 as a cloud of dots or a flat slice, so you can see what each integer changes.

What's on screen

• Blue dots: places where ψ > 0.
• Pink dots: places where ψ < 0.
• Gaps: surfaces where ψ = 0 and no dots appear.
• X, Y, Z axes (grey).
• Orbital properties (sidebar): the state's name and eight readout tiles.

Press ⌖ on a row to see where that control sits in the sim. Drag the picture to rotate it and scroll to zoom.

Fixed: lengths are in a₀, the Bohr radius (0.0529 nm), and energies in eV. Nothing here moves in time; these are stationary states, and only the camera moves.

The electron feels only the 1/r pull of the proton. The Schrödinger equation for that pull has stationary solutions, each picked out by three integers.

• n = 1, 2, 3, … labels the radial pattern. l = 0 to n − 1 labels the angular shape and is written s, p, d, f, g for 0, 1, 2, 3, 4. m = −l to +l labels the orientation about the z axis.
• Dots. Dot density follows |ψ|², a probability per unit volume. The probability of a region is |ψ|² times its volume.
• Colour. ψ is real here: blue marks ψ > 0, pink ψ < 0. Only |ψ|² decides where dots land.
• Real orbitals. For m ≠ 0 the sim draws cos(|m|φ) for m > 0 and sin(|m|φ) for m < 0, so |2, 1, 1⟩ is p_x and |2, 1, −1⟩ is p_y.

In the 1s state the dots crowd at the proton and thin smoothly outward. No ring or shell is singled out. R₁₀ ∝ e^−r/a₀, so |ψ|² at r = 1 a₀ is e⁻² ≈ 0.14 of its value at the proton.

That does not make the proton the likeliest distance. The probability of finding the electron between r and r + dr is r²R² dr, because the shell at radius r has volume 4πr² dr. For 1s this peaks at r = a₀, the Bohr radius.

• Density |ψ|² (what the dots show) is largest at r = 0.
• Distance distribution r²R² (what a ruler would record) peaks at 1 a₀.

A node is a surface where ψ = 0, so no dots land there. ψ changes sign across it, so the colour flips from blue to pink. The two factors of ψ_nlm each supply nodes.

• Radial nodes are spheres where that polynomial is zero. There are n − l − 1 of them. 2s has one, at r = 2 a₀; 3p has one, at r = 6 a₀.
• Angular nodes are the zeros of the angular factor: l surfaces through the proton, planes or cones.
• Total = (n − l − 1) + l = n − 1. The sim's three node tiles show these counts.

Near the proton the radial factor behaves like ρ^l, the middle factor in section 05's formula. For l = 0 that is 1, so R and |ψ|² stay finite at r = 0 and dots reach the proton. For l ≥ 1, ρ^l is zero at r = 0, so |ψ|² = 0 there and ∝ r^2l close by.

• Size. At r = 0.1 a₀, R is 0.64 for 2s and 0.019 for 2p (units of a₀^−3/2).
• Why. Angular momentum adds a term ħ²l(l + 1)/2μr² to the radial equation. It repels the wave from r = 0, more strongly for larger l.
• Picture. A p orbital's two lobes meet only at a point, with a nodal plane through the proton.

• Energy. E_n = −13.6 eV / n². Negative means bound. The tile reads −13.600 eV at n = 1 and −3.400 eV at n = 2, and it ignores l and m. All 2s and 2p states sit on one level. This is a feature of a pure 1/r pull.
• |L| = √(l(l + 1)) ħ, never lħ: 1.414 ħ for l = 1.
• L_z. For m = 0 the tile reads 0 ħ. For m ≠ 0 the shape is an equal mix of +|m| and −|m|. L_z would come out +|m|ħ or −|m|ħ with equal chances, so the tile shows ⟨L_z⟩ = 0.
• ⟨r⟩ = (a₀/2)[3n² − l(l + 1)] grows roughly as n². That is why the sim's view box reaches ±(2.5n² + 4) a₀.

• Spectral lines. A jump between levels emits a photon with λ = hc/ΔE. n = 3 → 2 gives 1.89 eV, 656 nm, hydrogen's red line. n = 2 → 1 gives 10.2 eV, 122 nm, in the ultraviolet. The sim shows states, not jumps.
• Chemistry. The s, p and d shapes are the vocabulary of atomic orbitals. In atoms with more electrons the shapes stay similar, but the levels split by l.
• Rydberg atoms. For l = 0, ⟨r⟩ = 1.5n² a₀, so n = 100 gives 15 000 a₀, about 0.8 µm. The sim stops at n = 7.

Key points

• ψ_nlm = R_nl(r) · Y_l^m(θ, φ), with l = 0 to n − 1 and m = −l to +l.
• Dot density is |ψ|² per unit volume; colour is the sign of ψ.
• n − l − 1 spherical nodes plus l planes or cones. Only l = 0 reaches the proton.
• E_n = −13.6 eV / n², |L| = √(l(l + 1)) ħ, and ⟨r⟩ = (a₀/2)[3n² − l(l + 1)].

Common mix-ups

• "The electron circles the proton at the Bohr radius." a₀ is the likeliest distance for 1s, not an orbit.
• "Blue and pink are positive and negative charge." They are the sign of ψ. Only |ψ|² says where the electron is likely.
• "2p_x carries L_z = +ħ." It is an equal mix of m = +1 and −1: L_z is +ħ or −ħ with equal chances, averaging 0.

Things to try on your own

• Pick a state with n = 4. Work out its node counts from n and l, then check the tiles.
• Load 5g and compute ⟨r⟩ by hand. Compare with the tile.
• Compare 3d_z² with 3d_xy: count lobes and colour changes, then compare their energy tiles.