Mean Radius and Expectation Values in Hydrogen

Introduction

The most probable radius picks out the single peak of the radial distribution, but many measurable properties depend on averages taken over the whole electron cloud. The average distance ⟨r⟩ relates to atomic size, ⟨r²⟩ to how atoms respond to magnetic fields, and ⟨1/r⟩ directly to the potential energy. This page shows how to calculate these expectation values for hydrogen and what they reveal about the balance of kinetic and potential energy in an atom.

Core explanation

Expectation values with radial functions. For any operator that depends only on r, the angular integration gives 1 (the spherical harmonics are normalised), leaving

⟨f(r)⟩ = ∫₀^∞ f(r) P(r) dr = ∫₀^∞ f(r) R(n,l)² r² dr

The standard integral. Hydrogen integrals reduce to

∫₀^∞ rᵏ e^(−br) dr = k!/b^(k+1)

⟨r⟩ for 1s. With P(r) = (4/a₀³)r²e^(−2r/a₀) and b = 2/a₀:

⟨r⟩ = (4/a₀³) ∫ r³ e^(−2r/a₀) dr = (4/a₀³) × 3!/(2/a₀)⁴ = (4/a₀³) × 6a₀⁴/16 = 1.5a₀ ≈ 79 pm

This is 50% larger than the most probable radius, a₀, because P(r) has a long tail at large r.

⟨1/r⟩ for 1s.

⟨1/r⟩ = (4/a₀³) ∫ r e^(−2r/a₀) dr = (4/a₀³) × 1!/(2/a₀)² = 1/a₀

Note that ⟨1/r⟩ ≠ 1/⟨r⟩: 1/a₀ is larger than 1/(1.5a₀), because regions close to the nucleus weigh heavily in 1/r.

General formulae for a hydrogen-like atom:

⟨r⟩ = (a₀/2Z)[3n² − l(l + 1)]

⟨1/r⟩ = Z/(n²a₀)

Orbital ⟨r⟩/a₀ (H) ⟨1/r⟩ × a₀ (H) --- --- --- 1s 1.5 1 2s 6 0.25 2p 5 0.25 3s 13.5 0.111 3d 10.5 0.111

⟨r⟩ grows roughly as n², and for a given n orbitals with higher l are slightly more compact on average. ⟨1/r⟩ depends only on n.

Average potential energy. The potential energy is V = −Ze²/(4πε₀r), so

⟨V⟩ = −(Ze²/4πε₀)⟨1/r⟩ = −Z²e²/(4πε₀n²a₀)

For hydrogen 1s, ⟨V⟩ = −e²/(4πε₀a₀) = −27.2 eV (one hartree). Since the total energy is E₁ = −13.6 eV, the average kinetic energy is ⟨T⟩ = E − ⟨V⟩ = +13.6 eV. In general ⟨V⟩ = 2E and ⟨T⟩ = −E, a result of the virial theorem for a Coulomb potential.

Formulae

⟨r⟩ = (a₀/2Z)[3n² − l(l + 1)]; ⟨1/r⟩ = Z/(n²a₀); ⟨V⟩ = 2E; ⟨T⟩ = −E; E = −13.6 Z²/n² eV. For 1s hydrogen, ⟨r²⟩ = 3a₀².

Step-by-step reasoning

To evaluate any radial expectation value:

1. Write P(r) = r²R² for the orbital. 2. Multiply by the function f(r) to be averaged. 3. Collect powers of r and identify k and b in rᵏe^(−br). 4. Apply k!/b^(k+1) and simplify. 5. Check the dimensions: ⟨r⟩ in units of a₀, ⟨1/r⟩ in units of 1/a₀.

Visual explanation

On a graph of P(r) for 1s, mark three vertical lines: the peak at a₀ (most probable), the balance point of the area at 1.5a₀ (mean) and the median at about 1.34a₀, where half the area lies on each side. The long tail drags the mean to the right of the other two.

Real-world analogy

Household incomes are distributed in the same lopsided way. The most common income is lower than the median, which is lower than the mean, because a few very large incomes stretch the tail. Likewise ⟨r⟩ exceeds r mp because of the electron's occasional excursions far from the nucleus.

Real-world example

The diamagnetic susceptibility of atoms is proportional to the sum of ⟨r²⟩ over their electrons. Large, diffuse atoms such as xenon are therefore much more diamagnetic than helium. Similarly, the nuclear magnetic shielding of a proton depends on ⟨1/r⟩ of the surrounding electron density, which is why NMR chemical shifts are sensitive to electron-withdrawing groups.

Why?

Why is ⟨1/r⟩ the same for 2s and 2p even though their ⟨r⟩ values differ? ⟨V⟩ is fixed by the energy through the virial theorem, ⟨V⟩ = 2E, and in hydrogen E depends only on n. Therefore ⟨1/r⟩ must depend only on n as well, even though the shapes of the distributions differ.

Common misconception

"The average potential energy equals the potential energy at the average radius." Since V ∝ 1/r and ⟨1/r⟩ ≠ 1/⟨r⟩, the potential at 1.5a₀ is −18.1 eV, not the correct average of −27.2 eV. Averages of non-linear functions must be calculated directly.

Worked example

Question: Calculate ⟨r²⟩ for the hydrogen 1s electron.

Reasoning: ⟨r²⟩ = (4/a₀³) ∫ r⁴ e^(−2r/a₀) dr. Here k = 4 and b = 2/a₀, so the integral is 4!/(2/a₀)⁵ = 24a₀⁵/32 = 0.75a₀⁵. Multiplying by 4/a₀³ gives 3a₀².

Answer: ⟨r²⟩ = 3a₀², so the root-mean-square radius is √3 a₀ ≈ 1.73a₀.

Quick check

1. What is the mean radius of the 2p orbital in He⁺, given ⟨r⟩ = (a₀/2Z)[3n² − l(l + 1)]? Answer: (a₀/4)(12 − 2) = 2.5a₀, since Z = 2, n = 2 and l = 1.

Exam focus

Be fluent with the integral ∫rᵏe^(−br)dr = k!/b^(k+1); nearly every hydrogen expectation value uses it. Show that ⟨r⟩₁ₛ = 1.5a₀ and ⟨1/r⟩₁ₛ = 1/a₀, and use ⟨V⟩ = 2E and ⟨T⟩ = −E to connect them to the energy.

Advanced insight

The spread of the radial distribution can be measured by Δr = (⟨r²⟩ − ⟨r⟩²)^(1/2). For 1s this is (3 − 2.25)^(1/2)a₀ ≈ 0.87a₀, comparable with ⟨r⟩ itself, showing how broad the electron cloud is. Expectation values of r⁻³ appear in spin–orbit coupling and hyperfine interactions, and ⟨δ(r)⟩, the density at the nucleus, controls the Fermi contact interaction seen in ESR and NMR.

Summary

Expectation values of functions of r are integrals of f(r)P(r). For hydrogen 1s, ⟨r⟩ = 1.5a₀, ⟨1/r⟩ = 1/a₀ and ⟨r²⟩ = 3a₀². In general ⟨r⟩ = (a₀/2Z)[3n² − l(l + 1)] and ⟨1/r⟩ = Z/n²a₀. The average potential energy is twice the total energy and the average kinetic energy is its negative, so the 1s electron has ⟨V⟩ = −27.2 eV and ⟨T⟩ = +13.6 eV.

Practice questions

1. Calculate ⟨r⟩ for the hydrogen 3p orbital. Answer: (a₀/2)(27 − 2) = 12.5a₀. 2. Using ⟨1/r⟩ = Z/n²a₀, find ⟨V⟩ for hydrogen n = 2 in electronvolts. Answer: ⟨V⟩ = −27.2 eV × (1/4) = −6.8 eV, which is twice E₂ = −3.4 eV. 3. What is the average kinetic energy of the electron in He⁺ in its ground state? Answer: E = −13.6 × 4 = −54.4 eV, so ⟨T⟩ = −E = +54.4 eV. 4. Explain why ⟨1/r⟩ is greater than 1/⟨r⟩ for the 1s orbital. Answer: 1/r is very large at small r, so the portions of the cloud close to the nucleus raise the average of 1/r more than they lower ⟨r⟩.