University Physics IV · The Hydrogen Atom · 10.6
Hydrogen Wavefunctions & Orbital Shapes
Once the radial and angular equations are solved, the orbital is just their product — and every feature of the picture is already sitting in it. This lesson reads the shape off ψₙₗₘ: how many nodes and of which kind, why a definite-m density has no lobes at all, and what the boundary drawn round a p orbital really is.
Build the model
Connect the measurement to the mechanism.
Separation has already done the hard work, so the orbital is a product: ψₙₗₘ(r, θ, φ) = Rₙₗ(r) Yₗm(θ, φ), with normalisability fixing n ≥ 1, l ≤ n − 1 and |m| ≤ l. What is left is to read the shape off that product. The radial factor goes as rl near the origin, carries an associated Laguerre polynomial of degree n − l − 1, and decays as e(−r/na₀), so it has exactly n − l − 1 spherical nodes and a size set by n; the angular factor supplies l further nodal surfaces, n − 1 in all.
Two consequences need saying out loud. A definite-m state hides its whole φ dependence in the phase e(imφ), so its density is axially symmetric — a sphere for s, a ring about z for m = ±1 — and the lobed pₓ and py pictures are real combinations of the degenerate m = ±1 pair: equally valid orbitals, but no longer eigenstates of Lz. And the drawing is not a path.
An orbital is a one-electron wavefunction, |ψ|² is a probability per unit volume, and the surface you see is a contour enclosing some agreed fraction, usually 90 per cent, of it. What the model costs: it is exact only for one electron in a 1/r field, and the feature most often left out of the picture — the sign of ψ — is the part that goes on to do the chemistry.
- Simple definition
- A hydrogen orbital ψₙₗₘ is a one-electron energy eigenfunction: a radial factor Rₙₗ(r) fixed by n and l, times the spherical harmonic Yₗm(θ, φ), whose modulus squared is the probability per unit volume of finding the electron there.
- Example
- The ground state is ψ₁₀₀ = (πa₀³)(−1/2) e(−r/a₀): no nodes, spherical, with |ψ|² = 2.15 × 10³⁰ m⁻³ at the nucleus, down to 1/e² of that at r = a₀ = 52.9 pm.
One label per separated equation. The energy depends on n alone, so a single level holds n² orbital states.
n = 1, 2, …; l = 0…n−1; m = −l…+l; ψ carries units m(−3/2)
Reads out three things at once: rl at the origin, n − l − 1 zeros in between, and a size growing as na₀.
a₀ = 52.9 pm; L is an associated Laguerre polynomial of degree n − l − 1
3dz²: no radial node, two cones at θ = 54.7° and 125.3°, two nodes in all — as n − 1 = 2 demands.
The angular surfaces are |m| planes through z plus l − |m| cones of fixed θ.
The 2s differs from the 1s by a sign change, not a shape change: ψ₂₀₀ is negative everywhere beyond 2a₀.
Both real and spherical; the 2s bracket vanishes at r = 2a₀ = 106 pm.
Every m eigenstate is a surface of revolution about z; m = +1 and m = −1 share one density and differ only in current.
Holds for any central potential, not just the Coulomb one.
Buys lobes that point at neighbours and a real, currentless wavefunction; costs a definite value of Lz.
Legitimate because the m = 0, ±1 states are degenerate in energy.
n, l and m each control a different feature
The three labels arrive from three different demands, and each governs a different part of the picture. Single-valuedness in φ gives the integer m, which enters only through the phase e(imφ); regularity at the poles gives the integer l ≥ |m|, which fixes the angular pattern; and demanding a normalisable radial solution gives n ≥ l + 1, which fixes the energy, −13.6 eV/n², and the overall size. So n sets the scale and the total node count n − 1; l divides those nodes between radial and angular and controls how hard the electron is held off the nucleus; m changes neither the energy nor the r and θ dependence of the density — it only winds the phase about the z-axis. That is why one level holds n² orbital states of identical energy: the sum of 2l + 1 over l = 0 to n − 1 is n².
The radial factor: rl at the origin, n − l − 1 zeros
Write Rₙₗ(r) = N (2r/na₀)l e(−r/na₀) L(2l+1)_(n−l−1)(2r/na₀) and three features follow with no further work. Near the origin only the rl survives, so ψ(0) is non-zero for s orbitals alone — which is why the Fermi contact term and the Darwin term reach l = 0 states and nothing else. Far out, the exponential of range na₀ sets the size: ⟨r⟩ = (a₀/2)[3n² − l(l+1)] makes 3s nine times the extent of 1s. In between, the Laguerre polynomial has degree n − l − 1 and exactly that many positive roots — 1s has none, 2s has one at r = 2a₀ = 106 pm, and 3s has two, at 1.90a₀ and 7.10a₀, the roots of 2ρ² − 18ρ + 27 with ρ = r/a₀. At each root the radial function passes through zero and changes sign, and the node is a whole sphere, not a point.
Count the nodes before drawing anything
n − l − 1 radial nodes, l angular nodal surfaces, n − 1 in total, and that total depends on n alone. The angular ones divide further: |m| planes containing the z-axis, and l − |m| cones of constant θ. Two checks. The 3dz² orbital (n = 3, l = 2, m = 0) has no radial node at all — the polynomial has degree zero — and two cones, at the roots of 3cos²θ − 1, that is θ = 54.7° and 125.3°; the doughnut round its waist is simply the region between those cones, where ψ has the opposite sign, not a separate orbital. A 4d orbital (n = 4, l = 2) adds one spherical node, at r = 12a₀ = 635 pm, to the same two cones: three nodes, matching n − 1 = 3. The count takes ten seconds and kills most wrong sketches.
A definite m has no lobes
Because |e(imφ)|² = 1, the density of any ψₙₗₘ is independent of φ: it is a surface of revolution about z. For m = 0 and l = 1 that gives the familiar pair of lobes along the axis, 2pz. For m = ±1 it gives a ring around z, and the m = +1 and m = −1 densities are identical — what distinguishes them is a circulating probability current, jφ = (mħ/mₑ r sinθ)|ψ|², and with it the orbital magnetic moment. The lobed pₓ and py of every chemistry text are not m states: they are the real combinations (ψ_(21−1) − ψ_(211))/√2 and i(ψ_(21−1) + ψ_(211))/√2. Both bases are legitimate, because degenerate states can be mixed without leaving the energy eigenspace, and the question decides which to use: a field along z picks the complex ones, since those are the Zeeman eigenstates, while a bond along x picks the real ones, which point where the neighbour is and carry no current.
The drawn surface is a contour, not an orbit
|ψ|² is a density spread through three dimensions, so any picture must choose something to draw, and the usual choice is an isosurface enclosing a fixed fraction of the probability — conventionally 90 per cent. For 1s, integrating 4πr²|ψ|² outwards gives P(<r) = 1 − e(−2r/a₀)(1 + 2r/a₀ + 2r²/a₀²); setting that to 0.90 gives r = 2.66a₀ = 141 pm. Nothing physical happens at that radius: just outside it the density is still 0.5 per cent of its value at the nucleus, and it never reaches zero. Draw the 99 per cent surface instead and the same orbital looks markedly bigger. So the boundary is a reporting convention, the shading inside it is |ψ|² or sometimes ψ, and none of it is a trajectory — nothing runs round the lobes, and asking how the electron crosses a node assumes a path the model never supplies.
The sign and the phase are physics
A node is where ψ changes sign, and that sign survives into every integral you will compute with the orbital. Parity first: ψₙₗₘ(−r) = (−1)l ψₙₗₘ(r), since inversion leaves Rₙₗ alone and Yₗm supplies the (−1)l — which is precisely what makes the dipole matrix element vanish unless l changes by one. Overlap second: a positive lobe brought against a positive lobe adds amplitude and lowers the energy, against a negative one it cancels, and that is bonding against antibonding, decided by a sign the density plot has already thrown away. The complex phase does its own work: e(imφ) carries the current and the moment mμB, so the real pₓ, an equal mixture of m = +1 and −1, has zero mean Lz and no current — yet a measurement of Lz on it still returns +ħ or −ħ, never 0.
Change one variable at a time
Make the relationship visible.
Leave β at 0 and the curve is flat: a definite-m orbital has no lobes anywhere. Push β to 45° and the density falls to zero at φ = 90° and 270° — a nodal plane neither parent state had. Then turn δ to swing the lobes from x towards y, and drop θ to watch the whole ring fade as sin²θ.
MODULATION sin 2β0.00
PEAK |ψ|² / f(r)²1.00
MINIMUM |ψ|² / f(r)²1.00
MINIMUM AT φ90 °
Live interpretationMODULATION sin 2β: 0.00. PEAK |ψ|² / f(r)²: 1.00. MINIMUM |ψ|² / f(r)²: 1.00. MINIMUM AT φ: 90 °
Catch the common trap
Explain before calculating.
A 3pₓ orbital is the usual real combination of the l = 1, m = ±1 hydrogen states. Which description of its nodes and of its angular momentum about z is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFor a 4d orbital, state n, l and the allowed m; count its radial and angular nodes and locate the radial one; and say what the radial factor does at the nucleus.
- The name fixes two labels: the 4 is n and the d is l = 2. That leaves m free over −2, −1, 0, +1, +2 — five degenerate orbitals, since 2l + 1 = 5.
- Radial nodes: n − l − 1 = 4 − 2 − 1 = 1. Here the Laguerre factor is 6 − ρ with ρ = 2r/na₀ = r/2a₀, so the single spherical node sits at ρ = 6, that is r = 12a₀ = 635 pm.
- Angular nodal surfaces: l = 2. For m = 0 they are |m| = 0 planes and l − |m| = 2 cones; for m = ±2 they are two planes through z and no cone.
- Total 1 + 2 = 3 = n − 1, as it must be: the total node count depends on n alone, never on how it splits.
- At the nucleus R ∝ rl = r², so ψ(0) = 0 and the density dies as r⁴ near the origin. Only l = 0 puts amplitude on the nucleus itself.
Answern = 4, l = 2, m = −2…+2 (five states); one spherical radial node at r = 12a₀ = 635 pm plus two angular surfaces, three nodes in all; ψ(0) = 0 because R ∝ r².
MediumThe 2s orbital is ψ₂₀₀ = (32πa₀³)(−1/2) (2 − r/a₀) e(−r/2a₀). Locate its node, then evaluate ψ at r = a₀ and at r = 4a₀ and compare both the values and the densities there.
- The exponential never vanishes, so the only zero comes from the bracket: 2 − r/a₀ = 0 at r = 2a₀ = 106 pm. That node is a whole sphere, and it is the one radial node n − l − 1 = 2 − 0 − 1 = 1 predicts.
- Prefactor: (32π)(−1/2) = 1/10.027 = 0.0997, in units of a₀(−3/2).
- At r = a₀: (2 − 1) e(−0.5) = 0.6065, so ψ = 0.0997 × 0.6065 = +0.0605 a₀(−3/2).
- At r = 4a₀: (2 − 4) e(−2) = −0.2707, so ψ = 0.0997 × (−0.2707) = −0.0270 a₀(−3/2). The sign has flipped across the node.
- Densities: (0.0605)² = 3.66 × 10⁻³ against (0.0270)² = 7.29 × 10⁻⁴ a₀(−3), a factor of 5.02. The outer point has a perfectly ordinary positive density; only ψ is negative there.
AnswerNode at r = 2a₀ = 106 pm; ψ(a₀) = +0.0605 a₀(−3/2) and ψ(4a₀) = −0.0270 a₀(−3/2). The sign change is real physics; the density is simply 5.0 times smaller at the outer point.
HardThe l = 1 angular factors with m = ±1 are Y₁(±1) = ∓√(3/8π) sinθ e(±iφ). (a) Show that the density of either one is axially symmetric. (b) Build the real combination whose density is lobed along x, and name the nodal surface it acquires. (c) In the equatorial plane, compare the density at φ = 30° with the peak density, first for that combination and then for the pure m = +1 state.
- (a) |Y₁(±1)|² = (3/8π) sin²θ |e(±iφ)|² = (3/8π) sin²θ. The phase drops out of the modulus, so nothing depends on φ: each state is a ring about z, and the two have identical densities, differing only in the direction of the circulating current.
- (b) Combine them so the two phases add instead of winding: (Y₁(−1) − Y₁(+1))/√2 = √(3/8π) sinθ (e(−iφ) + e(+iφ))/√2 = √(3/4π) sinθ cosφ.
- Since sinθ cosφ = x/r, this is 2pₓ: real, currentless, with density (3/4π) sin²θ cos²φ. That vanishes on the entire plane x = 0 — a nodal plane neither parent state possessed.
- (c) On the equator, θ = 90°, the pₓ density goes as cos²φ and peaks at φ = 0. At φ = 30°: cos²30° = (0.8660)² = 0.750, so it holds 75.0 per cent of the peak, down by a factor 1.33.
- For m = +1 the equatorial density is (3/8π) at every φ, so the ratio is 1.000 — there is no peak for it to fall away from.
AnswerBoth m = ±1 densities are φ-independent rings; (Y₁(−1) − Y₁(+1))/√2 ∝ x/r is 2pₓ and gains the yz-plane as a node. At φ = 30° the pₓ density is 0.750 of its peak, while the m = +1 density is unchanged.