Radial Distribution Functions

Introduction

The probability density ψ² tells us how likely the electron is to be found in a tiny volume at one particular point. Chemists, however, often want a different answer: how likely is the electron to be found at a given distance from the nucleus, in any direction? That question is answered by the radial distribution function , P(r). It turns out to look very different from ψ², and it is the function that reveals the shell structure of atoms.

Core explanation

From density to distance. Imagine a thin spherical shell of radius r and thickness dr centred on the nucleus. Its volume is its surface area times its thickness:

dV = 4πr² dr

For an s orbital, ψ² has the same value everywhere on the shell because the function is spherically symmetric. The probability of finding the electron in the shell is therefore

P(r) dr = ψ² × 4πr² dr, so P(r) = 4πr²ψ²

For orbitals with angular dependence, integrating over all angles gives the general form P(r) = r²R(n,l)². For an s orbital the two expressions agree because Y(0,0)² = 1/4π.

Two competing factors. P(r) is a product of a factor that grows with distance, r², and a factor that shrinks with distance, ψ². At the nucleus r² = 0, so P(0) = 0 even for a 1s electron whose ψ² is largest there. Far away ψ² decays exponentially and wins. In between there is a maximum.

The hydrogen 1s case. With ψ₁ₛ² = (1/πa₀³)e^(−2r/a₀),

P(r) = (4r²/a₀³) e^(−2r/a₀)

This rises from zero, peaks at r = a₀ = 52.9 pm and then decays.

Shells and nodes. Each radial node of ψ becomes a zero of P(r), so an orbital with n − l − 1 radial nodes shows n − l peaks. For hydrogen:

Orbital Number of peaks Main maximum --- --- --- 1s 1 a₀ 2s 2 about 5.2a₀ (small inner peak near 0.76a₀) 2p 1 4a₀ 3s 3 about 13a₀ 3d 1 9a₀

The main maxima cluster by n: the n = 2 functions peak around 4–5a₀, the n = 3 functions around 9–13a₀. This clustering is the quantum-mechanical origin of the idea of electron shells .

Penetration. The small inner peaks of 2s and 3s place some electron density very close to the nucleus. A 2s electron therefore penetrates the 1s core more than a 2p electron does. In hydrogen this makes no difference to the energy, but in many-electron atoms it lowers the energy of s orbitals below that of p orbitals of the same shell.

Formulae

P(r) = r²R(n,l)(r)², which for s orbitals equals 4πr²ψ². Normalisation: ∫₀^∞ P(r) dr = 1. Probability of lying between r₁ and r₂: ∫ P(r) dr from r₁ to r₂.

Step-by-step reasoning

To sketch P(r) for any orbital:

1. Start at zero: P(0) = 0 for every orbital because of the r² factor. 2. Count radial nodes (n − l − 1); each is a zero of P(r), giving n − l peaks. 3. Place the outermost, largest peak at roughly n²a₀/Z for the highest-l case, a little further for lower l. 4. Let the curve decay exponentially at large r.

Visual explanation

Picture an onion cut in half. The layers are the spherical shells; P(r) measures how much electron "stuff" sits in each layer. Plotting ψ² against r gives a curve that is highest at the centre, but P(r) starts at zero, because the innermost layers are tiny, and rises to a peak where both the density and the layer size are substantial.

Real-world analogy

Think of people in a city whose population density is highest in the centre. The central square is crowded, but it is small. A ring road at 5 km has lower density but a much greater area, so more people live at that distance. P(r) counts the people on each ring rather than the crowding at a point.

Real-world example

When the total radial electron density of an argon atom is calculated, it shows three distinct maxima corresponding to the n = 1, 2 and 3 shells. Electron-scattering measurements on atomic gases probe this same density, and they support the shell picture used throughout the periodic table.

Why?

Why does the r² factor appear at all? Because probability is density times volume. There are many more points at a large distance than at a small one: the number of points in a thin shell grows as its area 4πr². Ignoring this factor would give the probability at a point, not at a distance.

Common misconception

"The 1s electron is most likely to be found at the nucleus." The density is greatest at the nucleus, but the most probable distance is a₀. The volume near r = 0 is so small that the probability of being within a tiny radius of the nucleus is almost zero.

Worked example

Question: Using the hydrogen 1s radial distribution function, compare P(r) at r = a₀ and r = 2a₀.

Reasoning: P(r) = (4r²/a₀³)e^(−2r/a₀). At r = a₀: P = (4/a₀)e^(−2) = 0.541/a₀. At r = 2a₀: P = (16/a₀)e^(−4) = 0.293/a₀. The ratio is 0.293/0.541 ≈ 0.54.

Answer: P(2a₀) is about 54% of P(a₀), so being at twice the Bohr radius is roughly half as probable per unit radius.

Quick check

1. How many maxima does the radial distribution function of a hydrogen 4s orbital have? Answer: Four, because a 4s orbital has n − l − 1 = 3 radial nodes, which separate four peaks.

Exam focus

Know the derivation P(r) = 4πr²ψ² from the shell volume, and explain clearly the difference between ψ² (per unit volume) and P(r) (per unit radius). Be ready to sketch 1s, 2s and 2p curves with the correct number of peaks and zeros.

Advanced insight

Summing P(r) over all occupied orbitals of an atom gives its total radial electron density. For heavier atoms the shell maxima move inwards as Z increases, roughly as n²a₀/Z eff, and the peaks become less distinct for outer shells. Hartree–Fock calculations reproduce these shells, and they are the basis of Slater's shielding rules.

Summary

The radial distribution function P(r) = r²R² (4πr²ψ² for s orbitals) gives the probability per unit radius of finding the electron at distance r. It is zero at the nucleus, has n − l peaks separated by radial nodes and decays at large r. Its main maxima group by n, revealing shells, and its inner peaks show how s electrons penetrate the core.

Practice questions

1. Explain why P(r) is zero at r = 0 for a 1s orbital even though ψ² is largest there. Answer: P(r) contains the factor r², the relative volume of a shell at radius r, which is zero at the nucleus. 2. Write P(r) for the hydrogen 2p orbital given R(2,1) ∝ r e^(−r/2a₀), and find where it is largest. Answer: P ∝ r⁴e^(−r/a₀). Setting the derivative to zero gives 4r³ − r⁴/a₀ = 0, so r = 4a₀. 3. Which hydrogen orbital, 2s or 2p, has more electron density very close to the nucleus, and why does this matter in lithium? Answer: 2s, because of its small inner peak. In lithium this penetration makes the 2s electron feel more of the nuclear charge, so 2s lies below 2p. 4. State the normalisation condition for P(r). Answer: The integral of P(r) from zero to infinity equals 1, since the electron must be found at some distance.