ISEGORIA / MATH ENCYCLOPEDIA
Atomic physics: orbitals and spectra
The shapes of the hydrogen orbitals, the spectral series they produce, and how a magnetic field splits the sodium D lines.
Before you begin: Quantum mechanics
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The shapes of the orbitals
Each stationary state of hydrogen is a radial function times an angular one. Choose n, l and m and look at a slice through the atom. The nodes, where ψ changes sign, come in two kinds: spheres from the radial factor and cones or planes from the angular factor. On the right, r²R² gives the probability of finding the electron at distance r; the dashed curves are the other l values with the same n.
Worked example. For \(3d_{z^2}\), \(Y\propto3\cos^2\theta-1\) vanishes where \(\cos^2\theta=1/3\), that is at \(\theta\approx54.7^\circ\): the two nodal cones. With \(n=3,\ l=2\) there is no radial node, and \(\langle r\rangle=(27-6)/2=10.5\,a_0\).
Watch out. The pictures use real orbitals (\(p_x\), \(d_{xy}\) and so on), which are combinations of \(+m\) and \(-m\); they are not eigenstates of \(L_z\) when \(m\neq0\). Colour intensity grows like \(|\psi|^{0.55}\) so that faint outer lobes stay visible.
Why does the 2s orbital have a node but 2p does not have a radial one?
Both have \(n-1=1\) node in total. For 2s (\(l=0\)) it must be radial, a sphere; for 2p (\(l=1\)) it is the angular nodal plane, which leaves none for the radial factor.
2. Spectral series
When the electron drops from level n to a lower level, it emits one photon carrying the energy difference. Transitions that end on the same level form a series: Lyman in the ultraviolet, Balmer partly in the visible, Paschen and beyond in the infrared. Choose a series and an upper level, and press play to watch the electron fall.
Worked example. Hα is the Balmer transition \(3\to2\): \(E=13.598\,(1/4-1/9)=1.889\ \text{eV}\), so \(\lambda=1239.84/1.889\approx656.5\ \text{nm}\) in vacuum, the red line that colours emission nebulae.
Watch out. These are Bohr levels with the reduced-mass Rydberg constant. Fine structure and the Lamb shift move the lines by about one part in \(10^5\), far too little to see here. Wavelengths are vacuum values; in air they are about 0.03% shorter.
Why do the lines of a series crowd together at one end?
Because \(1/n_u^2\to0\) as \(n_u\) grows, the photon energies approach the ionisation energy of the lower level from below. The limit wavelength is \(n_\ell^2/R_{\mathrm H}\).
3. A magnetic field splits the lines
The yellow sodium doublet comes from 3p → 3s. In a magnetic field each level splits into sublevels \(m_J\), shifted by \(g_J m_J \mu_B B\), and photons can only change \(m_J\) by 0 or ±1. Because the Landé factor \(g_J\) is different in the upper and lower levels, the D1 line splits into four components and D2 into six. Drag the field and see each component move in proportion to B.
Worked example. For \(^2P_{3/2}\): \(g_J=1+\frac{15/4+3/4-2}{15/2}=\tfrac43\); for \(^2S_{1/2}\): \(g_J=2\). The D2 component \(m_u=\tfrac32\to m_\ell=\tfrac12\) shifts by \(\tfrac43\cdot\tfrac32-2\cdot\tfrac12=1\) unit of \(\mu_B B\), about 16 pm at 1 T.
Watch out. The electron g factor is taken as exactly 2 and hyperfine structure is ignored. Line thickness shows the squared 3j symbol, the relative strength of each transition; the brightness actually seen also depends on the viewing direction and polarisation.
What would happen in a field of 100 T?
The Zeeman shift would exceed the 2.1 meV fine-structure splitting, so L and S decouple and the pattern goes over to the Paschen–Back effect, where the shifts are \((m_L+2m_S)\mu_B B\).