core34 min studyLesson 376 in Physics

Hydrogen wavefunctions and orbital nodes

Separate radial and angular zeros and distinguish probability per volume from probability per radial shell.

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What you’ll understand

  • Count positive-radius Coulomb radial nodes from n and ℓ.
  • Locate angular zeros in a specified real orbital.
  • Explain why the most probable shell radius differs from the maximum spatial density.

An orbital picture represents a wavefunction or a probability density, not a tiny classical path. Its lobes, phase signs, radial shells, and angular nodal surfaces encode different mathematical features. Hydrogen makes these distinctions explicit because its wavefunctions separate into known radial and angular factors. The main practical skill is to identify what function is being plotted and what integration measure turns it into a probability.

The radial polynomial controls spherical nodes

For the point-Coulomb Hamiltonian write ψnℓm=Rnℓ(r)Yℓm(θ,ϕ)\psi_{n\ell m}=R_{n\ell}(r)Y_{\ell m}(\theta,\phi). Let a=aμ/Za=a_\mu/Z and ρ=2r/(na)\rho=2r/(na). The radial factor is proportional to ρℓe−ρ/2Ln−ℓ−12ℓ+1(ρ)\rho^\ell e^{-\rho/2}L_{n-\ell-1}^{2\ell+1}(\rho), with an associated Laguerre polynomial. Its positive roots produce spherical nodal surfaces. The origin factor for nonzero ℓ and asymptotic decay at infinity are not counted as positive-radius radial nodes.

The radial-node number is n−ℓ−1n-\ell-1. Increasing n at fixed ℓ adds a radial node, whereas increasing ℓ within one Coulomb shell removes one. This is a property of the separated Coulomb eigenfunctions, not a general rule that every three-dimensional state of a given energy has one fixed nodal geometry. Superpositions within a degenerate shell can change the shape and phase structure, and arbitrary superpositions need not retain one radial and one angular factor.

Rnℓ(r)∝ρℓe−ρ/2Ln−ℓ−12ℓ+1(ρ),ρ=2rna,Nradial=n−ℓ−1R_{n\ell}(r)\propto\rho^\ell e^{-\rho/2}L_{n-\ell-1}^{2\ell+1}(\rho),\qquad \rho=\frac{2r}{na},\qquad N_{\rm radial}=n-\ell-1
The length a includes reduced mass and nuclear charge. ρ is dimensionless. The proportionality omits the normalization constant, which has units length⁻³ᐟ². The node count concerns positive finite radii of these separated bound eigenfunctions, excluding r = 0 and infinity.↔ Scroll wide equations horizontally.

Angular nodes depend on the chosen orbital basis

An s orbital has constant angular factor and no angular nodal surface. A real pzp_z orbital has angular factor cos⁡θ\cos\theta, so the plane z = 0 is a node. Real pxp_x and pyp_y orbitals are rotated combinations of the complex m eigenfunctions. Their differently oriented planes do not create different Coulomb energies. A lobe colour normally indicates wavefunction sign or phase; it does not indicate positive versus negative probability or opposite electric charges.

The familiar count of ℓ angular nodes is used for the conventional real separated orbital basis, with planes or cones counted in the usual way. It should not be interpreted as a basis-independent claim about every complex spherical harmonic's nodal surfaces. For example, Y11∝sin⁡θeiϕY_{11}\propto\sin\theta e^{i\phi} has azimuthal phase winding rather than the nodal plane of a real p orbital. Specifying the basis is therefore essential when translating a quantum number into a particular three-dimensional picture.

A density maximum is not a shell maximum

For the normalized 1s orbital, spatial density decreases monotonically from the origin. Yet a thin shell at radius r has area proportional to r2r^2, and its probability per unit radius initially grows from zero. Combining the geometric factor with exponential decay produces a shell-density maximum at a nonzero radius. There is no contradiction: probability in a fixed tiny volume near the nucleus and probability in a whole spherical shell are different measurements.

For any normalizable continuous spatial density, the probability of one exact point is zero even if its density value is finite. An s orbital's finite density at the nucleus is therefore not a finite probability assigned to a point. Likewise, a radial node is a sphere where amplitude vanishes, while the sign change across it can affect interference with another state. Plot captions should distinguish amplitude, density per volume, and density per radius before a maximum or zero is interpreted.

ψ1s(r)=e−r/aπa3,P1s(r)=4πr2∣ψ1s∣2=4r2a3e−2r/a,∫0∞P1s(r) dr=1\psi_{1s}(r)=\frac{e^{-r/a}}{\sqrt{\pi a^3}},\qquad P_{1s}(r)=4\pi r^2|\psi_{1s}|^2=\frac{4r^2}{a^3}e^{-2r/a},\qquad \int_0^\infty P_{1s}(r)\,dr=1
The spatial wavefunction is length⁻³ᐟ², spatial density is length⁻³, and radial shell density P is length⁻¹. P(r)dr is a probability after integrating all angles. This exact formula is the ideal hydrogenic 1s state with a = a_μ/Z.↔ Scroll wide equations horizontally.

Where is a 1s electron most likely?

For the normalized hydrogenic 1s wavefunction given above, locate the maximum of spatial density and the maximum of radial shell density. Calculate the probability that r is smaller than a. Explain why the two maximum locations answer different questions.

  1. Spatial density is ∣ψ1s∣2=e−2r/a/(πa3)|\psi_{1s}|^2=e^{-2r/a}/(\pi a^3). Its derivative is negative for r > 0, so its maximum value occurs at the origin. This is a density per volume, not a probability assigned to an exact point.

  2. For positive r, differentiate the logarithm of shell density: dlog⁡P/dr=2/r−2/ad\log P/dr=2/r-2/a. It vanishes at r=ar=a, where the growing shell area and decreasing spatial density balance. The shell density itself vanishes at r = 0.

  3. Using s=2r/as=2r/a, the probability below a is 12∫02s2e−sds=1−5e−2=0.3233\tfrac12\int_0^2s^2e^{-s}ds=1-5e^{-2}=0.3233. The antiderivative follows by two integrations by parts; both the lower limit and the spherical measure matter.

The result

Spatial density is greatest at r = 0, while probability per radial interval is greatest at r = a. Only about 32.3% lies inside radius a, so a density maximum should not be confused with an enclosed-probability threshold.

Explore the model

Interactive lab

Interactive lab · function

Isolate the radial geometric factor

Reset and set both controls to a = 1 and x = 0.5; predict y = 0.25. Reset and set both controls to a = 1 and x = 1; predict y = 1. Interpret x as normalized radius and this quadratic only as the shell-area factor. The lab omits the exponential radial density, angular harmonics, and nodes; it does not locate the actual 1s shell maximum.

Isolate the radial geometric factorThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forGeometry contributes an r² weight to radial probability, but the complete shell distribution also needs the physical wavefunction.

Check your understanding

For a normalized 1s orbital with finite nonzero spatial density at the origin, which statement is correct?

Try it yourself

A hydrogenic 2s radial function is proportional to (2−r/a)e−r/(2a)(2-r/a)e^{-r/(2a)}. Locate its positive-radius radial node and count its angular nodes. Compare with a real 2p_z orbital, whose radial factor is proportional to re−r/(2a)r e^{-r/(2a)} and angular factor to cos⁡θ\cos\theta. Identify its nodal surface.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Separate zeros of radial factors at positive radius from zeros of the angular factor; do not count the origin as a radial node.

Compare with the solution

For 2s the polynomial vanishes at r=2ar=2a, giving one spherical radial node, in agreement with n−ℓ−1=2−0−1=1n-\ell-1=2-0-1=1. Its constant s angular factor gives no angular node. The 2p radial factor has no zero for r > 0, matching 2−1−1=02-1-1=0. Its real p_z angular factor vanishes at θ=π/2\theta=\pi/2, giving the xy nodal plane. The radial factor's origin zero is not an additional spherical node. These statements concern the specified real orbitals; a complex m = 1 combination has different angular phase and zero geometry.

Take it with you

  • Coulomb radial nodes occur at positive roots of the radial polynomial.
  • Angular nodal pictures require a specified real or complex orbital basis.
  • Spatial density and shell probability use different measures and can peak at different radii.

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