The hydrogen atom: energy levels and orbitals

Equipment

• 3D probability-cloud canvas (draggable to rotate)
• Orbital selector buttons “1s”, “2s”, “2p”
• On-screen node and symmetry indicators

Procedure

1. Compare 1s and 2s

   Select “1s”, then “2s”, and rotate the cloud. Both are spherically symmetric ($l=0$), but 2s carries one radial node — a spherical shell where $|\psi|^2$ vanishes — and extends farther out, consistent with the energy rising to $E_2=-3.4$ eV.

2. Switch to 2p

   Press “2p” and rotate: the cloud breaks into two lobes with a nodal plane, the signature of $l=1$. This dumbbell shape is what the three $m_l=-1,0,1$ orbitals share; the cloud is a probability density, not a classical orbit.

3. Compute a transition

   Using $E_n=-13.6\,\mathrm{eV}/n^2$, compute the photon emitted when an electron falls from $n=2$ to $n=1$: $h\nu=13.6(1/1^2-1/2^2)=10.2$ eV, the Lyman-α line. Compare with the energy gap you would get for a $3\to2$ (Balmer, red) transition.

Simulation

Experiment history

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