Radial Wavefunctions

Introduction

When the Schrödinger equation for the hydrogen atom is separated in spherical coordinates, the wavefunction splits into two independent pieces: an angular part that fixes the shape and orientation of an orbital, and a radial part that tells us how the wavefunction changes as we move away from the nucleus. The radial part controls the size of an orbital, the number of spherical nodes it has and how quickly the electron density fades into empty space. This page examines the structure of those radial functions term by term.

Core explanation

The product form. For a hydrogen-like atom with nuclear charge Z, every orbital can be written as

ψ(n,l,mₗ)(r, θ, φ) = R(n,l)(r) × Y(l,mₗ)(θ, φ)

The spherical harmonic Y depends on l and mₗ; the radial function R depends on n and l but not on mₗ. That is why all three 2p orbitals share one radial function.

Three factors in every radial function. Using the dimensionless distance ρ = Zr/a₀, each radial function has the general form

R(n,l)(r) = N × ρ^l × L(ρ) × e^(−ρ/n)

- N is a normalisation constant containing (Z/a₀)^(3/2). - ρ^l is a power factor . It makes R go to zero at the nucleus whenever l > 0. - L(ρ) is an associated Laguerre polynomial of degree n − l − 1. Its roots are the radial nodes. - e^(−ρ/n) is the exponential decay that guarantees the wavefunction is square-integrable.

The first few functions (Z = 1 gives hydrogen):

Orbital R(n,l)(r) --- --- 1s 2(Z/a₀)^(3/2) e^(−ρ) 2s (1/2√2)(Z/a₀)^(3/2) (2 − ρ) e^(−ρ/2) 2p (1/2√6)(Z/a₀)^(3/2) ρ e^(−ρ/2) 3s (2/81√3)(Z/a₀)^(3/2) (27 − 18ρ + 2ρ²) e^(−ρ/3) 3p (4/81√6)(Z/a₀)^(3/2) (6ρ − ρ²) e^(−ρ/3) 3d (4/81√30)(Z/a₀)^(3/2) ρ² e^(−ρ/3)

Near the nucleus. As r → 0 the exponential tends to 1 and the lowest power of ρ dominates, so R ∝ r^l. An s function (l = 0) is finite and non-zero at the nucleus; p, d and f functions vanish there. Physically, the centrifugal term l(l + 1)ħ²/2mr² in the radial equation pushes electrons with angular momentum away from r = 0.

Far from the nucleus. At large r the exponential wins over any polynomial, so every R decays to zero. The decay constant Z/na₀ becomes smaller as n increases, so higher shells spread further out. The decay length na₀/Z is linked directly to the energy: a less tightly bound electron has a more slowly decaying tail.

Units. Because ∫R²r²dr = 1, R has units of (length)^(−3/2), which is why the prefactor contains (Z/a₀)^(3/2).

Formulae

R(n,l)(r) = N ρ^l L(ρ) e^(−ρ/n), with ρ = Zr/a₀ and a₀ = 4πε₀ħ²/(mₑe²) ≈ 52.9 pm. Normalisation: ∫₀^∞ R(n,l)² r² dr = 1. Number of radial nodes = degree of L(ρ) = n − l − 1.

Step-by-step reasoning

To interpret any radial function quickly:

1. Read off ρ^l: this gives the behaviour at the nucleus (zero unless l = 0). 2. Find the degree of the polynomial: n − l − 1 roots means that many radial nodes. 3. Read the exponent: decay constant Z/na₀ sets the overall size. 4. Combine: the function starts at the value set by ρ^l, oscillates through its nodes and finally decays exponentially.

Visual explanation

Plot R against r for 1s, 2s and 3s on the same axes. The 1s curve falls smoothly from a maximum at r = 0. The 2s curve starts positive, crosses zero once at 2a₀ and then shows a small negative lobe that decays away. The 3s curve crosses zero twice. The 2p and 3d curves start at zero, rise to a single hump and fade, reaching further out as n grows.

Real-world analogy

A radial function is like the sound of a bell heard at increasing distance. Close to the bell the pattern of loud and quiet zones (nodes) depends on how it is struck, but far away every sound fades steadily. The polynomial sets the pattern of zones; the exponential sets the fading.

Real-world example

Computational chemistry programs build molecular orbitals from atom-centred functions. Slater-type orbitals copy the hydrogen form r^(n−1)e^(−ζr) directly, and their exponent ζ is tuned for each element. The exponential tail of the radial function is what makes orbital overlap, and therefore bond strength, fall off rapidly as atoms are pulled apart.

Why?

Why must every bound radial function contain a decaying exponential? The Schrödinger equation at large r reduces to d²u/dr² ≈ (2mₑ E /ħ²)u, whose solutions are e^(+κr) and e^(−κr). Only the decaying solution can be normalised, and demanding that the series solution terminates as a polynomial is exactly what quantises the energy.

Common misconception

"Only s orbitals have radial functions." Every orbital has a radial part. What is special about s orbitals is that their radial function is non-zero at the nucleus, whereas p, d and f radial functions start at zero because of the ρ^l factor.

Worked example

Question: Locate the radial node of the hydrogen 2s orbital.

Reasoning: The radial node occurs where the polynomial factor (2 − ρ) is zero. With Z = 1, ρ = r/a₀, so 2 − r/a₀ = 0 gives r = 2a₀. The exponential never vanishes at finite r, so this is the only node.

Answer: r = 2a₀ ≈ 106 pm.

Quick check

1. Why is the 2p radial function zero at the nucleus while the 2s function is not? Answer: The 2p function contains the factor ρ¹ (since l = 1), which vanishes at r = 0; the 2s function has l = 0, so no such factor.

Exam focus

Be able to write R = N ρ^l L(ρ) e^(−ρ/n) and say what each factor does. Examiners often ask you to find radial nodes by setting the polynomial equal to zero, or to state that R ∝ r^l near the nucleus. Remember that R does not depend on mₗ.

Advanced insight

The associated Laguerre polynomials form an orthogonal set, which is why radial functions with the same l but different n are orthogonal. Functions with different l need not be orthogonal radially; their orthogonality comes from the spherical harmonics instead. The terminating-series condition in the radial equation gives n = nᵣ + l + 1, where nᵣ is the number of radial nodes, which explains why n must exceed l.

Summary

The hydrogen wavefunction factorises into R(n,l)(r)Y(l,mₗ)(θ,φ). Each radial function combines a power factor ρ^l, a Laguerre polynomial of degree n − l − 1 whose roots are radial nodes, and an exponential e^(−Zr/na₀) that ensures decay. R ∝ r^l near the nucleus, so only s functions are non-zero there, and orbitals grow larger as n increases.

Practice questions

1. State the degree of the polynomial factor in R(4,1) and hence the number of radial nodes. Answer: Degree n − l − 1 = 4 − 1 − 1 = 2, so the 4p radial function has two radial nodes. 2. Show that the 3p radial function has a node at r = 6a₀ for hydrogen. Answer: The polynomial is 6ρ − ρ² = ρ(6 − ρ). Apart from the origin, which is not counted as a node, it vanishes at ρ = 6, that is r = 6a₀. 3. How does the decay constant of the exponential change from 1s to 3s, and what does this mean for orbital size? Answer: It falls from Z/a₀ to Z/3a₀, so the 3s function decays three times more slowly and the orbital is much larger. 4. What are the units of R(n,l), and why? Answer: Length to the power −3/2, because R²r²dr must be dimensionless when integrated to give a probability of 1. 5. How does the 1s radial function of He⁺ compare with that of hydrogen? Answer: With Z = 2 the exponential decays twice as fast, so the function is more compact, and the prefactor is larger by 2^(3/2) to keep it normalised.