University Physics IV · The Hydrogen Atom · 10.3
The Radial Equation & Effective Potential
Half the hydrogen atom's difficulty is furniture. Substitute u = rR and the furniture vanishes: the first-derivative term goes, the Jacobian folds into the norm, and what is left is a bead-on-a-wire problem in a well you can read by eye — barrier, floor, and turning points — before you solve anything.
Build the model
Connect the measurement to the mechanism.
The three-dimensional atom does not need three-dimensional intuition. Once the angular factor is fixed as a spherical harmonic, everything left depends on one coordinate, and the substitution u = rR tidies it completely: the awkward (2/r) dR/dr term cancels, the r² Jacobian disappears into the norm, and u obeys a strictly one-dimensional Schrödinger equation on the half-line r ≥ 0. The price of the tidiness is paid in two places.
The potential picks up ħ²l(l+1)/2μr² — the angular kinetic energy, prepaid as a repulsive centrifugal barrier, exact because l(l+1)ħ² is the exact eigenvalue of L̂². And the origin becomes an infinite wall, u(0) = 0, because keeping u finite there would plant a δ function in the equation that nothing balances. What the tidiness buys is the entire 1D toolkit: sketch Veff = −e²/4πε₀r + ħ²l(l+1)/2μr², and its floor at −13.6/l(l+1) eV already forbids a 1p state, its turning points at n² ± n√(n² − l(l+1)) Bohr radii bracket each orbital's shell, and the exponential tail beyond them is ordinary 1D tunnelling.
The only genuine approximation on the page is the point-charge −1/r itself.
- Simple definition
- The radial equation is the one-dimensional Schrödinger equation obeyed by u(r) = rR(r) on the half-line r ≥ 0, in an effective potential that adds the centrifugal barrier ħ²l(l+1)/2μr² to the actual central potential.
- Example
- For hydrogen's l = 1 states the effective well bottoms out at r = l(l+1)a₀ = 2a₀ ≈ 0.106 nm, at a depth of −13.6/2 = −6.8 eV — so a 1p state, which would need E₁ = −13.6 eV, has nowhere to sit.
One substitution buys the whole 1D toolkit — node counting, turning points, exponential tails.
μ = reduced mass (kg); E in J; identical in form to the 1D Schrödinger equation on r ≥ 0
The angular kinetic energy, prepaid as repulsion — exact at every r, since l(l+1)ħ² is the eigenvalue of L̂².
l = 0, 1, 2, …; the second term is the centrifugal barrier, in J for r in m
|u|² is the probability per unit radius directly; plot u, not R, to see where the electron lives.
the shell factor r² is already inside u = rR — never insert it twice
u(0) = 0 for every l: the origin is an infinite wall, and each unit of l flattens u by one more power of r.
bound states have E < 0, so κ is real (m⁻¹); the r(−l) solution is discarded
Bound states must sit above the floor, so n² > l(l+1): the restriction l ≤ n − 1, read straight off the well.
a₀ = 4πε₀ħ²/(μe²) = 0.0529 nm; for l = 1 the floor is −6.8 eV at 2a₀
Brackets the shell where u oscillates; beyond r₊ the orbital dies exponentially — its soft edge.
from Veff(r±) = Eₙ = −13.6 eV/n²; real only when l ≤ n − 1
One substitution kills the first derivative
After separation, the radial factor R(r) obeys −(ħ²/2μ)[R″ + (2/r)R′] + [V(r) + ħ²l(l+1)/2μr²]R = ER, and that middle term (2/r)R′ is what stops it being an ordinary one-dimensional problem. The fix is the identity (1/r²)(r²R′)′ = (1/r)(rR)″. Define u = rR, multiply through by r, and the first derivative vanishes: −(ħ²/2μ)u″ + Veff u = Eu, with nothing left to distinguish it in form from a particle in a one-dimensional well. The same substitution settles the bookkeeping. Probability in three dimensions integrates over shells of volume 4πr² dr, so the norm of R is ∫|R|²r²dr; written in u this is ∫|u|²dr — the Jacobian has been absorbed, once, and must not be inserted again. Nothing here used the Coulomb form: any central potential — the 3D oscillator, a screened nucleus, a molecular ion — reduces the same way, which is why this one page of algebra is reused for the rest of the course.
The centrifugal term is a bill, not a model
The centrifugal term is not modelling; it is the angular motion, already solved, sent to the radial equation as a bill. Kinetic energy in a central field splits into a radial piece plus L̂²/2μr², and acting on the factored state R(r)Yₗₘ the operator L̂² returns the exact number l(l+1)ħ². So an electron with angular momentum pays ħ²l(l+1)/2μr² of kinetic energy just to be at radius r, and the radial problem sees that cost exactly as a classical orbit sees L²/2μr² — the sole quantum edit is l(l+1)ħ² in place of L². Two things follow. Because 1/r² beats 1/r as r → 0, any l ≥ 1 walls the origin off behind a barrier that diverges faster than the Coulomb well dives. And the coefficient is l(l+1), not l²: even the smallest nonzero angular momentum costs 2ħ², which is why p states are pushed off the nucleus so decisively.
The wall at the origin
The half-line needs a condition at r = 0, and it is u(0) = 0 for every l. The quick reason: R = u/r must stay finite. The honest reason is sharper. Suppose u(0) = c ≠ 0 in an s state; then ψ ≈ c/r near the origin, and ∇²(1/r) = −4πδ³(r), so the Schrödinger equation acquires a δ-function term that neither the Coulomb potential nor the energy can balance — the candidate fails exactly at the point being argued about. Near the origin the equation itself sorts the options: with u″ ≈ l(l+1)u/r² dominating, u behaves as r(l+1) or r(−l), and the second is discarded — for l ≥ 1 it is not normalisable, for l = 0 it is the δ-function impostor. So the physical solution rises off the origin as r(l+1): each unit of l flattens the wavefunction near the nucleus by another power of r. And because u(0) = 0 is exactly the boundary condition of an infinite wall, 1D intuition must be borrowed in its half-line form: of the bound states of the full-line well −e²/4πε₀|x|, hydrogen's s spectrum keeps only the odd ones.
Read the well before solving it
Before solving anything, nondimensionalise and stare. Measuring r in Bohr radii (ρ = r/a₀) and energy in units of 13.6 eV, hydrogen's effective potential becomes v(ρ) = −2/ρ + l(l+1)/ρ² — one curve per l, no other parameters. For l = 0 it is the bare Coulomb dive. For l ≥ 1 the barrier wins at small ρ, the well bottoms out at ρ = l(l+1), and the floor sits at −13.6/l(l+1) eV: −6.8 eV at 2a₀ for l = 1, −2.27 eV at 6a₀ for l = 2, −1.13 eV at 12a₀ for l = 3. That floor is already a theorem. A bound state must lie above the minimum of its well, so Eₙ = −13.6/n² eV with angular momentum l requires n² > l(l+1), i.e. n ≥ l + 1. The famous restriction l ≤ n − 1, which the series solution derives properly in the next topic, is visible here as bare energy bookkeeping: there is no 1p state because the p-well's floor at −6.8 eV sits above the −13.6 eV a 1-state would need.
Turning points bracket the shell
Setting Veff(r) = Eₙ gives the classical turning points — the radii where a classical electron of that energy and angular momentum would stall and swing back. In dimensionless form the condition is a quadratic, ρ² − 2n²ρ + n²l(l+1) = 0, so ρ± = n² ± n√(n² − l(l+1)). For the 2p state that is 4 ± 2√2: an allowed shell from 1.17a₀ to 6.83a₀ (0.062 nm to 0.361 nm), the quantum cousin of the perihelion and aphelion of a Kepler ellipse. The turning points organise the whole shape of u. Between them E > Veff and u curves toward the axis, oscillating — its n − l − 1 radial nodes all live there. Outside them E < Veff and u curves away, decaying as e(−κr) at large r with κ = √(−2μE)/ħ: the orbital's soft exponential edge is one-dimensional tunnelling into the classically forbidden region, nothing more exotic.
The one approximation on the page
Every term above is exact except one. The centrifugal barrier carries no approximation — l(l+1)ħ² is an eigenvalue, not an estimate. What is idealised is −e²/4πε₀r itself, which assumes a point proton. The real proton has a charge radius near 0.84 fm ≈ 1.6 × 10⁻⁵ a₀; inside it the potential flattens instead of diving. Which states notice follows from the small-r analysis: u ~ r(l+1) means only l = 0 states put appreciable probability inside the nucleus, so finite nuclear size nudges s levels and barely touches the rest. The nudge is tiny in ordinary hydrogen but measurable — it separates isotopes, and in muonic hydrogen, where the roughly 200-times-smaller orbit magnifies the overlap, it is the effect behind the modern proton-radius measurements. Knowing which term to trust and which to correct is what it means to own the model.
Change one variable at a time
Make the relationship visible.
Raise l at fixed n: the centrifugal barrier walls off the origin and the allowed shell narrows. Then set n = 2, l = 2: the width readout hits zero because E₂ = −3.40 eV lies below the l = 2 well floor of −2.27 eV — the figure's way of saying l stops at n − 1.
LEVEL Eₙ-3.40 eV
INNER TURNING r−1.17 a₀
OUTER TURNING r+6.83 a₀
ALLOWED SHELL WIDTH5.66 a₀
Live interpretationLEVEL Eₙ: −3.40 eV. INNER TURNING r−: 1.17 a₀. OUTER TURNING r+: 6.83 a₀. ALLOWED SHELL WIDTH: 5.66 a₀
Catch the common trap
Explain before calculating.
In the u = rR formulation of hydrogen's radial problem, which statement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFor a hydrogen electron with l = 2, find the radius at which the effective potential is least and the depth of the well floor there, in eV. Take a₀ = 0.0529 nm, e²/(4πε₀a₀) = 27.2 eV, and ħ²/(2μa₀²) = 13.6 eV.
- Write the well: Veff(r) = −e²/(4πε₀r) + ħ² l(l+1)/(2μr²), with l(l+1) = 6.
- Set dVeff/dr = 0: e²/(4πε₀r²) = ħ²⋅6/(μr³), so rₘᵢₙ = 4πε₀ħ²⋅6/(μe²) = 6a₀.
- Numerically rₘᵢₙ = 6 × 0.0529 nm = 0.317 nm.
- Substitute back: Veff(6a₀) = −27.2/6 + 13.6⋅6/36 = −4.53 + 2.27 = −2.27 eV, matching the general floor −13.6 eV/l(l+1).
- Read a consequence off the floor: E₃ = −1.51 eV lies above −2.27 eV, so 3d fits; E₂ = −3.40 eV lies below it, so no 2d state can exist.
Answerrₘᵢₙ = 6a₀ ≈ 0.317 nm, floor Veff = −2.27 eV. The 3d level (−1.51 eV) sits above the floor and exists; a 2d level (−3.40 eV) would sit below it, so there is none.
MediumFind the classical turning points of hydrogen's 2p state (n = 2, l = 1), in Bohr radii and in nm, and state what u(r) does on each side of them.
- Nondimensionalise with ρ = r/a₀ and energies in units of 13.6 eV: Veff/13.6 eV = −2/ρ + l(l+1)/ρ², and E₂/13.6 eV = −1/4.
- Set them equal with l(l+1) = 2: −1/4 = −2/ρ + 2/ρ². Multiply through by −4ρ²: ρ² − 8ρ + 8 = 0.
- Solve the quadratic: ρ = [8 ± √(64 − 32)]/2 = 4 ± 2√2, so ρ− = 1.17 and ρ+ = 6.83.
- Convert: r− = 1.17 × 0.0529 nm = 0.062 nm; r+ = 6.83 × 0.0529 nm = 0.361 nm.
- Between them E > Veff, so u curves toward the axis and oscillates — with n − l − 1 = 0 radial nodes, a single arch. Outside them E < Veff, so u curves away and decays, at large r as e(−κr) with κ = √(−2μE₂)/ħ.
Answerr− = (4 − 2√2)a₀ ≈ 1.17a₀ ≈ 0.062 nm and r+ = (4 + 2√2)a₀ ≈ 6.83a₀ ≈ 0.361 nm; u oscillates between them and decays exponentially outside.
HardShow that u(r) = C r e(−r/b) solves hydrogen's l = 0 radial equation only for one value of b, and find that b and the energy it forces. Use ħ = 1.055×10⁻³⁴ J s, μ = 9.109×10⁻³¹ kg, and e²/4πε₀ = 2.307×10⁻²⁸ J m.
- For l = 0 the equation is −(ħ²/2μ)u″ − [e²/(4πε₀)] u/r = Eu. The trial u = C r e(−r/b) already meets both boundary conditions: u(0) = 0 and decay at infinity.
- Differentiate twice: u′ = C(1 − r/b)e(−r/b), and u″ = C(r/b² − 2/b)e(−r/b).
- Insert and divide by C e(−r/b), using u/r = C e(−r/b): −(ħ²/2μ)(r/b²) + ħ²/(μb) − e²/(4πε₀) = E r.
- This must hold at every r, so match powers of r separately. The r⁰ terms give ħ²/(μb) = e²/(4πε₀), i.e. b = 4πε₀ħ²/(μe²) — the Bohr radius, forced rather than assumed. The r¹ terms give E = −ħ²/(2μb²).
- Numbers: b = (1.055×10⁻³⁴)² / (9.109×10⁻³¹ × 2.307×10⁻²⁸) = 1.113×10⁻⁶⁸ / 2.102×10⁻⁵⁸ = 5.29×10⁻¹¹ m.
- E = −ħ²/(2μb²) = −e²/(4πε₀ · 2b) = −2.307×10⁻²⁸ / (1.058×10⁻¹⁰) = −2.18×10⁻¹⁸ J = −13.6 eV.
Answerb is forced to the Bohr radius a₀ = 5.29×10⁻¹¹ m and E to −13.6 eV: the 1s ground state. Any other b leaves an unbalanced 1/r term — the equation, not a postulate, fixes the atom's size.