University Physics IV · The Hydrogen Atom · 10.7
Radial Probability & the Virial Theorem
The radial function is only half an answer. Turn it into probability per unit radius, learn which average of r a question is really asking for, and let the virial theorem hand you the kinetic and potential shares of the energy without another integral.
Build the model
Connect the measurement to the mechanism.
The radial equation handed you Rₙₗ(r); this topic turns it into numbers an experiment can check, and the first move is the one people get wrong. |ψ|² is a probability per unit volume, but nothing measures a volume at a point: what a measurement answers is how much probability lies in the shell between r and r + dr, and that shell has area 4πr². Integrating the angles out therefore leaves P(r) = r²|Rₙₗ(r)|² = |u(r)|², a probability per unit radius — and every radius an atom has is a moment of that one function. The moments disagree, and they are meant to: for 1s the most probable shell is at a₀, the mean is 1.5a₀, and ⟨1/r⟩ is 1/a₀ rather than 1/(1.5a₀).
What ties the moments back to physics is the virial theorem, which for an inverse-square force splits the energy with no further integration: ⟨T⟩ = −Eₙ and ⟨V⟩ = 2Eₙ, so hydrogen's ground state pays 13.6 eV of kinetic energy to sit in a 27.2 eV well. That is the payoff, and the bill comes with it — every one of these numbers is an average over a distribution with no edge, so "the size of the hydrogen atom" names a chosen moment, not a property of the atom.
- Simple definition
- The radial probability density P(r) = r²|Rₙₗ(r)|² is the probability per unit radius of finding the electron in the shell between r and r + dr, and every radius an atom is said to have — most probable, mean, or 1/⟨1/r⟩ — is one moment of it.
- Example
- For hydrogen 1s, P(r) = (4r²/a₀³)e(−2r/a₀) peaks at r = a₀ = 0.0529 nm, yet ⟨r⟩ = 1.5a₀ = 0.0794 nm and only 32.3% of the probability lies inside a₀.
The r² is the shell's area, not part of the wavefunction — it is what makes P vanish at the nucleus while |ψ|² is largest there.
P in m⁻¹, probability per unit radius; |ψ|² is in m⁻³, per unit volume; u = rR
Two honest answers to 'how big is hydrogen'. The tail is one-sided, so the mean lands 50% beyond the most probable shell.
a₀ = 0.0529 nm, so the peak sits at 0.0529 nm and the mean at 0.0794 nm
Size grows as n² and shrinks as 1/Z. At fixed n a larger l is more compact: 2s gives 6a₀ where 2p gives 5a₀.
a₀ in metres, built from the reduced mass; Z the nuclear charge; l at most n − 1
It fixes ⟨V⟩ = −27.21Z²/n² eV, so the Coulomb energy cannot depend on l — the l-degeneracy written as an average.
units m⁻¹; identical for every l allowed at that n, and never equal to 1/⟨r⟩
Averages weighted towards small r do separate states of one n, which is where fine structure and the quantum defect come from.
units m⁻²; the centrifugal energy is ħ²l(l+1)/2μ times this
For 1s: ⟨T⟩ = +13.6 eV inside a well of ⟨V⟩ = −27.2 eV. Both halves of E, with no second integral.
general form 2⟨T⟩ = k⟨V⟩ for V ∝ rk; here k = −1; Eₙ = −13.606Z²/n² eV
Two densities, and the r² between them
|ψ|² is a probability per unit volume, in m⁻³: multiply it by a small volume and you get a probability. Nothing measures a volume at a point, though — the question an experiment answers is how much probability lies 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.
The 1s shell: peak at a₀, mean at 1.5a₀
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 one number Bohr's orbit got right, 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.
One standard integral does most of the work
Every 1s expectation value is the same integral, ∫₀^∞ rⁿe(−br) dr = n!/b(n+1), taken with b = 2/a₀. Normalisation is the n = 2 case: (4/a₀³)(2!/(2/a₀)³) = (4/a₀³)(a₀³/4) = 1, as it must be. Then ⟨r⟩ is n = 3: (4/a₀³)(6a₀⁴/16) = 1.5a₀. ⟨r²⟩ is n = 4: (4/a₀³)(24a₀⁵/32) = 3a₀². And ⟨1/r⟩ is n = 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 moment it is has quoted noise in the last two.
Which averages remember l, and which forget
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 the 2s state dips into the core through its inner lobe but 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, which is why the centrifugal term and the fine-structure corrections, both weighted towards small r, split what the Coulomb energy could not. And these are averages of different functions: for 3d, ⟨1/r⟩ = 1/(9a₀) while 1/⟨r⟩ = 1/(10.5a₀).
The virial theorem splits E without another integral
For any stationary state, 2⟨T⟩ = ⟨r · ∇V⟩. If V goes as rk 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.
What the numbers buy, and what they cost
⟨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.
Change one variable at a time
Make the relationship visible.
Leave n at 1 and push the r-power from 0 to 2: at 0 the marker sits on the axis, because |R|² for 1s is largest at the nucleus; at 2 it lands on the dashed line at r = a₀. Then hold the power at 2 and raise n — the faint mean line closes on the bold mode line as the shell goes classical.
PEAK OF DRAWN CURVE0.00 a₀
MEAN RADIUS1.5 a₀
MEAN KINETIC ⟨T⟩13.61 eV
MEAN POTENTIAL ⟨V⟩-27.21 eV
Live interpretationPEAK OF DRAWN CURVE: 0.00 a₀. MEAN RADIUS: 1.5 a₀. MEAN KINETIC ⟨T⟩: 13.61 eV. MEAN POTENTIAL ⟨V⟩: −27.21 eV
Catch the common trap
Explain before calculating.
A hydrogen atom sits in the 3d state (n = 3, l = 2), for which ⟨r⟩ = 10.5a₀. What is ⟨V⟩, the expectation value of its Coulomb potential energy? Take e²/4πε₀a₀ = 27.21 eV.
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe hydrogen 1s radial function is R₁₀(r) = 2a₀(−3/2)e(−r/a₀). Find the radius of the most probable shell and the mean radius ⟨r⟩, and say why the two differ. Take a₀ = 0.0529 nm.
- Build the radial probability density: P(r) = r²|R₁₀|² = (4r²/a₀³)e(−2r/a₀). This is probability per unit radius, and the r² shell factor is already in it.
- Most probable shell: dP/dr = (4/a₀³)e(−2r/a₀)(2r − 2r²/a₀) = 0. Discard r = 0, where P is a minimum, and the bracket gives r = a₀ = 0.0529 nm.
- Mean radius: ⟨r⟩ = ∫₀^∞ rP dr = (4/a₀³)∫₀^∞ r³e(−2r/a₀) dr = (4/a₀³) × 3!/(2/a₀)⁴ = (4/a₀³)(6a₀⁴/16) = 1.5a₀ = 0.0794 nm.
- They differ because P is one-sided: it is pinned to zero at the nucleus but decays with an exponential tail, and the tail drags the mean 50% past the peak. Integrating P confirms it — 57.7% of the probability lies inside ⟨r⟩.
AnswerMost probable radius a₀ = 0.0529 nm; mean radius 1.5a₀ = 0.0794 nm. The r² pins P to zero at the nucleus, and the exponential tail pulls the mean out past the peak.
MediumFor the hydrogen 3d state (n = 3, l = 2), find ⟨V⟩, ⟨T⟩ and E from ⟨1/r⟩ = 1/(n²a₀), then compare the mean radius with the most probable radius. Use e²/4πε₀a₀ = 27.211 eV and ⟨r⟩ = (a₀/2)[3n² − l(l+1)].
- ⟨1/r⟩ = 1/(9a₀), so ⟨V⟩ = −(e²/4πε₀)⟨1/r⟩ = −27.211/9 = −3.023 eV.
- The virial theorem for an inverse-square force gives 2⟨T⟩ = −⟨V⟩, so ⟨T⟩ = +1.512 eV, and E = ⟨T⟩ + ⟨V⟩ = −1.512 eV — which is −13.606/9 eV, the level energy, as it has to be.
- Mean radius: ⟨r⟩ = (a₀/2)[3(9) − 2(3)] = (a₀/2)(27 − 6) = 10.5a₀.
- Most probable radius: 3d is a circular state (l = n − 1), so R₃₂ goes as r²e(−r/3a₀) and P goes as r⁶e(−2r/3a₀). Setting 6/r = 2/3a₀ gives r = 9a₀, that is n²a₀.
- So the peak sits at 9a₀ and the mean at 10.5a₀. Taking −27.211/10.5 = −2.592 eV instead would be using 1/⟨r⟩ in place of ⟨1/r⟩ — an error of 14%, and the energy would then fail to come back as −1.512 eV.
Answer⟨V⟩ = −3.023 eV, ⟨T⟩ = +1.512 eV, E = −1.512 eV; most probable radius 9a₀ against a mean of 10.5a₀.
HardFor hydrogen 1s, find the fraction of the probability lying beyond the classical turning point, and the width Δr of the radial distribution. Take E₁ = −13.606 eV, e²/4πε₀a₀ = 27.211 eV, a₀ = 0.0529 nm and P(r) = (4r²/a₀³)e(−2r/a₀).
- Classical turning point: a classical electron of energy E₁ stops where V(r) = E₁, that is −27.211(a₀/r) = −13.606 eV, giving r = 2a₀. Beyond it the classical kinetic energy would be negative.
- Change variable to X = 2r/a₀, under which P dr becomes ½X²e(−X) dX, and the tail integral from X to infinity is ½e(−X)(X² + 2X + 2).
- The turning point is X = 4, so the forbidden fraction is ½e(−4)(16 + 8 + 2) = 13e(−4) = 13 × 0.018316 = 0.2381.
- Width: ⟨r²⟩ = (4/a₀³) × 4!/(2/a₀)⁵ = (4/a₀³)(24a₀⁵/32) = 3a₀², and ⟨r⟩ = 1.5a₀, so Δr = √(3 − 2.25) a₀ = 0.866a₀ = 0.0458 nm.
- Relative width Δr/⟨r⟩ = 0.866/1.5 = 1/√3 = 0.577. The ground state is 58% as wide as it is big, and almost a quarter of it sits where classical mechanics forbids the electron to be at all.
Answer23.8% of the 1s probability lies beyond r = 2a₀, in the classically forbidden region; Δr = 0.866a₀ = 0.0458 nm, a relative width of 1/√3 = 0.577.