Radial probability and the virial theorem

|ψ|² is a probability per unit volume, in m⁻³: multiply it by a small volume and you get a probability. Ask only how far out the electron is, though, and direction drops out: the probability you want is the one in the thin shell between r and r + dr, and that shell has volume 4πr²dr. Writing ψ = Rₙₗ(r)Yₗₘ(θ, φ) and integrating the angles out, with the spherical harmonic normalised so its angular integral is 1, leaves P(r) = r²|Rₙₗ(r)|², the radial probability density, in m⁻¹ — probability per unit radius, integrating to 1 over all r. The r² is geometry, the growing area of the shell; it is not part of the wavefunction. That one factor is why the 1s state has |ψ|² largest at the nucleus while P(r) is zero there: the density is highest exactly where there is almost no shell to put it in. The substitution u = rR has already swallowed the factor, so P(r) = |u(r)|² and the norm is the integral of |u|² dr with nothing further attached.

Put R₁₀ = 2a₀^(−3/2)e^(−r/a₀) into P = r²|R|² to get P₁ₛ(r) = (4r²/a₀³)e^(−2r/a₀). Differentiating, dP/dr goes as 2r − 2r²/a₀, which vanishes at r = a₀: the most probable shell sits exactly at the Bohr radius, 0.0529 nm — the radius of Bohr's orbit, now meaning something different. The mean is a different question with a different answer. ⟨r⟩ = ∫rP dr = 1.5a₀ = 0.0794 nm, half as far out again, because P is skewed: it climbs from zero over a width of order a₀ and then decays over a comparable width, but only the decaying side has a tail. Integrating P confirms the shape — 32.3% of the probability lies inside a₀, and 57.7% inside ⟨r⟩ itself, so more than half the cloud is closer in than the mean. Neither number is 'the radius of hydrogen'. Both are honest summaries of one distribution that has no edge at all.

Every 1s expectation value is the same integral, ∫₀^∞ rᵏe^(−br) dr = k!/b^(k+1), taken with b = 2/a₀. Normalisation is the k = 2 case: (4/a₀³)(2!/(2/a₀)³) = (4/a₀³)(a₀³/4) = 1, as it must be. Then ⟨r⟩ is k = 3: (4/a₀³)(6a₀⁴/16) = 1.5a₀. ⟨r²⟩ is k = 4: (4/a₀³)(24a₀⁵/32) = 3a₀². And ⟨1/r⟩ is k = 1: (4/a₀³)(a₀²/4) = 1/a₀. Those three numbers already say something the peak alone cannot. The spread is Δr = √(⟨r²⟩ − ⟨r⟩²) = √(3 − 2.25) a₀ = 0.866a₀, so Δr/⟨r⟩ = 1/√3 = 0.577. The 1s cloud is 58% as wide as it is big. An atomic radius quoted to three figures without saying which measure it is has quoted noise in the last two.

For a general hydrogenic state, ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)]: it grows as n², shrinks as 1/Z, and at fixed n a larger l is slightly more compact — 2s gives 6a₀ against 2p's 5a₀, because although the 2s state reaches in close to the nucleus through its inner lobe, its outer lobe, which carries most of the weight, reaches further out. ⟨1/r⟩ = Z/(n²a₀) is stranger: it takes the same value for every l allowed at that n. You can get it without an integral by differentiating Eₙ with respect to the coupling e² (the Hellmann-Feynman theorem), and it is exactly the l-degeneracy of the Coulomb spectrum written as an average. ⟨1/r²⟩ = Z²/[n³(l + ½)a₀²] does remember l. That alone splits nothing in hydrogen, because the centrifugal energy it measures is traded against radial kinetic energy inside the same Eₙ; but the fine-structure corrections, weighted the same way towards small r, do split what the Coulomb energy could not, and they combine so the result depends on n and j alone. And these are averages of different functions: for 3d, ⟨1/r⟩ = 1/(9a₀) while 1/⟨r⟩ = 1/(10.5a₀).

For any stationary state, 2⟨T⟩ = ⟨r ⋅ ∇V⟩. If V goes as r^k this collapses to 2⟨T⟩ = k⟨V⟩, and the Coulomb potential has k = −1, so 2⟨T⟩ = −⟨V⟩. Combine that with E = ⟨T⟩ + ⟨V⟩ and the whole split comes free: ⟨T⟩ = −Eₙ and ⟨V⟩ = 2Eₙ. For 1s that reads ⟨T⟩ = +13.606 eV and ⟨V⟩ = −27.211 eV against E₁ = −13.606 eV, and ⟨1/r⟩ checks it independently — ⟨V⟩ = −(e²/4πε₀)⟨1/r⟩ = −27.211/n² eV, which is 2Eₙ for every n and every l. Two things are worth keeping. A bound state is not a state of small kinetic energy: the electron pays 13.6 eV of kinetic energy to sit in a 27.2 eV well, and the binding energy is only what is left over. And the result is a test — compute ⟨T⟩ and ⟨V⟩ any other way, and if they miss −E and 2E you have not got an eigenstate.

⟨T⟩ = 13.606 eV for 1s converts to an rms speed of √(2⟨T⟩/mₑ) = 2.19 × 10⁶ m s⁻¹, which is c/137 — the fine-structure constant showing up as a velocity. That is also the size of what you neglected: relativistic corrections enter at (v/c)² ≈ 5 × 10⁻⁵ of the energy, which is precisely the scale of hydrogen fine structure. The averages then scale predictably, since ⟨r⟩ goes as n²/Z. He⁺ in 1s has ⟨r⟩ = 0.75a₀ = 0.040 nm with ⟨T⟩ = 54.4 eV, while a Rydberg s state at n = 50 has ⟨r⟩ = 3750a₀ ≈ 0.20 μm — an atom whose size is optical. The cost is identical in every case: these are moments of a distribution that never terminates. About a quarter of the 1s probability, 23.8%, lies beyond r = 2a₀, where a classical electron of energy −13.6 eV could not go at all.