University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.5
Truncation: n = nᵣ + l + 1 and the Bohr energies derived
Quantum Mechanics I quoted −13.6 eV/n²; this is where you earn it. One vanishing numerator in the radial recursion names the integer n, caps l at n − 1, and turns into the Bohr energy and radius with the reduced mass and Z in place. Then the result is tested against spectroscopy, and the corrections it leaves out are named and sized.
Build the model
Connect the measurement to the mechanism.
With the asymptotic factors removed, the hydrogen radial function is u(ρ) = ρ(l+1) e(−ρ) v(ρ), where ρ = κr and κ = √(−2μE)/ħ, and topic 02.04 gave v = Σ cⱼρʲ with cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)]. The series converges at every negative energy, but unless a numerator vanishes its coefficients approach those of e(2ρ), so u grows like e(+ρ): the rising exponential rejected at infinity returns through the tail, and ∫u² dr diverges. The ρ(−l) branch was already rejected at the origin, so there is one chain with one seed, and quantisation is a single condition: 2(nᵣ + l + 1) = ρ₀ for some nᵣ = 0, 1, 2, ….
Naming n = nᵣ + l + 1 gives ρ₀ = 2n, so n is a positive integer, and l ≤ n − 1 because nᵣ cannot be negative. The energy was hidden in ρ₀ = 2/(κa), so κ = 1/(na) and Eₙ = −ħ²/(2μa²n²), which is −½μc²α²/n², Z² times deeper with radii a/Z once e² becomes Ze². The condition sees only the sum nᵣ + l, so each level holds n² states, a degeneracy the derivation produces but does not explain.
The model stops at its assumptions: it is non-relativistic and spinless, so it misses corrections of relative order (Zα)², and its nucleus is a point. Hydrogen's measured 1s–2s interval lies 9.3 parts per million above it; hydrogen-like iron's Lyman-α line lies 1% above.
- Simple definition
- Truncation is the requirement that the radial series v(ρ) = Σ cⱼρʲ end at a finite power ρ(nᵣ), which happens only when ρ₀ = 2(nᵣ + l + 1) = 2n, and so only at the hydrogen energies Eₙ = −13.6 eV/n².
- Example
- For l = 1 at ρ₀ = 6, c₁ = [2(0 + 2) − 6]c₀/(1 × 4) = −c₀/2 and c₂ = [2(1 + 2) − 6]c₁/(2 × 5) = 0, so v = c₀(1 − ρ/2): the 3p state, nᵣ = 1, n = 3, E₃ = −13.598 eV/9 = −1.511 eV, one radial node at r = 6a = 0.318 nm.
The numerator of cⱼ₊₁/cⱼ vanishes at j = nᵣ, and every later coefficient with it. Any other ρ₀ leaves u growing like e(+ρ).
nᵣ, l and ρ₀ = 2/(κa) are dimensionless; nᵣ is the index of the last non-zero cⱼ, so v has degree nᵣ
l ≤ n − 1 is not a separate rule: it is nᵣ ≥ 0 rewritten. A 2d state would need nᵣ = −1.
all dimensionless integers; nᵣ counts the radial nodes, l labels the angular factor Yₗᵐ
From ρ₀ = 2Z/(κa) = 2n, hydrogen's ρ₀ with e² → Ze². The tail e(−ρ) = e(−Zr/na) stretches with n and shrinks with Z: excited states are larger as well as more weakly bound.
κ in m⁻¹, a in m; μ in kg, e in C, ħ in J s, ε₀ in F m⁻¹; hydrogen (Z = 1): a = 5.2947 × 10⁻¹¹ m
From E = −ħ²κ²/(2μ) with κ = Z/(na), so derived from normalisability rather than assumed from Bohr's orbits.
μ = mₑM/(mₑ + M) in kg for nuclear mass M; Eₙ in J. Hydrogen: E₁ = −13.598 eV, E₂ = −3.400 eV
Binding is (Zα)²/2n² of the rest energy and the 1s speed is v/c ≈ Zα, so the neglected relativistic terms are smaller by a further (Zα)².
α = e²/(4πε₀ħc) = 1/137.036, dimensionless; μc² in J; ½mₑc²α² = 13.606 eV
The energy depends on nᵣ and l only through their sum. The m-degeneracy is rotational symmetry; the l-degeneracy needs an extra symmetry of the 1/r potential and waits for topic 02.09.
dimensionless count: 1, 4, 9, 16 for n = 1 to 4; spin doubles it to 2n²
An unstopped series brings back the growing exponential
Topic 02.04 left u(ρ) = ρ(l+1) e(−ρ) v(ρ), with ρ = κr and κ = √(−2μE)/ħ, and v = Σ cⱼρʲ obeying cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)]; the energy sits only in ρ₀. For large j the ratio cⱼ₊₁/cⱼ tends to 2/j, the ratio for e(2ρ) = Σ(2ρ)ʲ/j!. So the series converges for every ρ at every negative energy, yet its sum grows like e(2ρ), and u like e(+ρ): the rising solution rejected at infinity in 02.03 re-enters through the tail. Take l = 0 and ρ₀ = 3, which is E = −6.04 eV: c₁ = −0.500c₀, c₂ = −0.0833c₀, c₃ = −0.0208c₀, and every later coefficient keeps the minus sign. At ρ = 10 the sum makes u = −43c₀, where the 1s function ρe(−ρ) has fallen to 4.5 × 10⁻⁴. Unlike the oscillator there is only one chain: the ρ(−l) branch was discarded at the origin, so c₀ is the only seed and no second series has to be switched off. Normalisability imposes exactly one condition, and the recursion itself must meet it.
One vanishing numerator fixes ρ₀ = 2n and caps l
The chain stops only if a numerator is zero: 2(j + l + 1) − ρ₀ = 0 at some j = nᵣ. The next coefficient and every one after it then vanish, and v is a polynomial of degree nᵣ, which 02.06 identifies as an associated Laguerre polynomial whose nᵣ roots are real and positive, the radial nodes. Define n = nᵣ + l + 1. The condition becomes ρ₀ = 2n, and because nᵣ and l are non-negative integers, n = 1, 2, 3, … and l ≤ n − 1. The bound on l is nᵣ ≥ 0 rewritten, not an extra postulate. For l = 2 at ρ₀ = 4 the root of 2(j + 3) = 4 is j = −1, an index the series does not have, and every numerator 2j + 2 is positive. For n = 3 the allowed pairs are (nᵣ, l) = (2, 0), (1, 1) and (0, 2): the 3s, 3p and 3d functions carry 2, 1 and 0 radial nodes. A ρ₀ that is not an even integer, or an even integer below 2(l + 1), leaves u growing, so that l has no state at that energy.
From ρ₀ back to the energy and the Bohr radius
The energy entered through ρ₀ = μe²/(2πε₀ħ²κ), which is 2/(κa) with a = 4πε₀ħ²/(μe²). Setting it equal to 2n gives κ = 1/(na), and E = −ħ²κ²/(2μ) then gives Eₙ = −ħ²/(2μa²n²) = −(μ/2ħ²)(e²/4πε₀)²/n². With the hydrogen reduced mass μ = 0.999456 mₑ this is E₁ = −13.598 eV, E₂ = −3.400 eV and E₃ = −1.511 eV, and a = 5.2947 × 10⁻¹¹ m, 0.054% longer than a₀ = 5.2918 × 10⁻¹¹ m, which uses mₑ. Dimensions: ħ²/(μa²) is (J s)²/(kg m²) = kg m² s⁻² = J. The length scale grows with n, since u decays as e(−r/na). The limit n → ∞ takes Eₙ to 0 from below, the ionisation threshold, and the gap to the level beneath, about 2|E₁|/n³, closes. A classical circular orbit of energy Eₙ has radius n²a and speed αc/n, so its orbital frequency is exactly 2|E₁|/(hn³). At n = 100 that is 6.576 GHz, and the 100 → 99 photon is 6.676 GHz: 1.5% apart, a gap that falls as 3/(2n), as correspondence requires.
Charge Z, reduced mass μ, and the order in α
Nothing in the derivation used the size of the nuclear charge, so a hydrogen-like ion follows from e² → Ze²: a becomes a/Z and Eₙ = −Z²(μ/mₑ)(13.606 eV)/n², where 13.606 eV = ½mₑc²α² is the Rydberg energy for an infinitely heavy nucleus. With α = e²/(4πε₀ħc) = 1/137.036 the energies read Eₙ = −½μc²(Zα)²/n²: binding is a fraction (Zα)²/2n² of the rest energy, and the 1s electron moves at about v/c = Zα. Relativistic corrections are therefore smaller than Eₙ by a further (Zα)², and that ratio grows as Z². For C⁵⁺ (Z = 6, μ = 0.999954 mₑ, a/Z = 8.82 pm) the formula gives an ionisation energy of 489.78 eV; the tabulated 489.99 eV is higher by 4.3 × 10⁻⁴, close to the (Zα)²/4 = 4.8 × 10⁻⁴ that the Dirac equation, quoted in Unit 8, predicts for a 1s level. For hydrogen the measured excess is 1.1 × 10⁻⁵ against α²/4 = 1.3 × 10⁻⁵, so the residual grows by a factor of about 40 between Z = 1 and Z = 6, close to the Z² = 36 the α-counting predicts; the Lamb shift, which Unit 8 adds, accounts for the rest.
n² states per level, and a degeneracy left unexplained
The energy depends on nᵣ and l only through n = nᵣ + l + 1. For each l the angular solutions of Unit 1 supply 2l + 1 values of m, so level n holds the sum of 2l + 1 from l = 0 to n − 1, which is n² states: 1, 4, 9 and 16 for n = 1 to 4. The m-degeneracy is expected, because any central potential is rotationally symmetric and turning a state cannot change its energy. The l-degeneracy is not. Change V(r) away from −e²/(4πε₀r) and it goes: sodium's valence electron sees a nuclear charge screened differently at different radii, and its 3s level lies 2.10 eV below 3p, the energy of the 589 nm yellow D line. Nothing in the recursion says why 1/r is special; at fixed n it simply trades nᵣ for l. Topic 02.09 supplies the reason, a conserved quantum Runge–Lenz vector. Spin, which this Hamiltonian ignores, doubles the count to 2n², and the fine structure of Unit 8 splits the l-degenerate levels only partly, leaving 2s and 2p₁/₂ together until the Lamb shift separates them.
What the Bohr energies leave out, measured
Three assumptions were built in, and each has a size. Non-relativistic kinematics and the missing spin–orbit coupling give corrections of relative order α² = 5.3 × 10⁻⁵, the fine structure treated in Unit 8, where the relativistic input is quoted rather than derived. Quantum electrodynamics adds the Lamb shift, smaller again by about α times a logarithm, also quoted in Unit 8. A nucleus of radius R rather than a point shifts s levels by a fraction of order (R/a)² ≈ 2.5 × 10⁻¹⁰ for the proton, which Unit 6 estimates by perturbation theory. The measured numbers confirm this ranking. The formula gives the 1s–2s interval as ¾ × 13.598287 eV = 10.198715 eV, a frequency of 2.466038 × 10¹⁵ Hz; the measured value is 2.466061 × 10¹⁵ Hz, 9.3 parts per million higher, which is order α². The ionisation energy comes out at 13.598287 eV against a measured 13.598434 eV, a residual of 1.1 × 10⁻⁵. Everything the derivation assumed survives to one part in 10⁵ in hydrogen; what is left over is the physics of Units 6, 8 and 9.
Change one variable at a time
Make the relationship visible.
At ν = 2.60 with l = 1 the tail dives off the bottom; slide to 3.00 and it settles onto the axis as 3p, one node at ρ = 2. Just past 3 the tail flips to the top. Then set l = 3: nothing settles at ν = 1, 2 or 3, because 2(j + 4) = 2ν has no root j ≥ 0 until ν = 4.
TRIAL ENERGY −13.598 eV/ν²-2.012 eV
ρ₀ = 2ν5.20
NUMERATOR ZERO, j = ν − l − 10.60
NEAREST ALLOWED LEVEL-1.511 eV
Live interpretationTRIAL ENERGY −13.598 eV/ν²: −2.012 eV. ρ₀ = 2ν: 5.20. NUMERATOR ZERO, j = ν − l − 1: 0.60. NEAREST ALLOWED LEVEL: −1.511 eV
Catch the common trap
Explain before calculating.
In hydrogen the radial recursion cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)] is run for l = 2 at the energy E₂ = −3.40 eV, where ρ₀ = 4. What does the series do, and what follows?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn hydrogen the radial recursion cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)] is run for l = 2 at an energy where ρ₀ = 8. Show that the series stops, identify nᵣ and n, give the energy in eV and the position of every radial node in nm, and list the other states that share this energy.
- Termination needs a zero numerator: 2(j + 2 + 1) − 8 = 0 gives j = 1. So nᵣ = 1 and n = nᵣ + l + 1 = 1 + 2 + 1 = 4, consistent with ρ₀ = 2n = 8.
- Run the chain: c₁ = [2(0 + 3) − 8]c₀/[(1)(0 + 6)] = −2c₀/6 = −c₀/3, and c₂ = [2(1 + 3) − 8]c₁/[(2)(1 + 6)] = 0, so every later coefficient is zero and v = c₀(1 − ρ/3), a polynomial of degree nᵣ = 1.
- Energy: E₄ = −13.598 eV/4² = −0.8499 eV, using the reduced-mass value E₁ = −13.598 eV. The decay constant is κ = 1/(4a), so ρ = r/(4a).
- Node: v = 0 at ρ = 3, so r = 3 × 4a = 12a = 12 × 5.2947 × 10⁻¹¹ m = 6.354 × 10⁻¹⁰ m = 0.635 nm. The factor ρ³ in u = ρ³e(−ρ)v vanishes only at the origin, so this is the only radial node.
- The same n = 4 is reached by l = 0, 1 and 3 with nᵣ = 3, 2 and 0: the 4s, 4p and 4f states. Counting m values, 1 + 3 + 5 + 7 = 16 spinless states share E₄.
AnswerThe series stops after c₁: nᵣ = 1, n = 4 (the 4d state), E₄ = −0.850 eV, v = c₀(1 − ρ/3) with one radial node at r = 12a = 0.635 nm; 16 spinless states share the level.
MediumA negative muon captured by a helium-4 nucleus forms the hydrogen-like ion (μ⁻ ⁴He)⁺. Taking rest energies 105.658 MeV for the muon, 3727.379 MeV for the α particle and 0.510999 MeV for the electron, find the reduced mass, the length scale a/Z, E₁, E₂ and the 2p → 1s photon energy. Then use the α particle's charge radius, 1.68 fm, to judge the point-nucleus assumption against hydrogen's.
- Reduced mass: μc² = (105.658 × 3727.379)/(105.658 + 3727.379) MeV = 102.746 MeV, so μ/mₑ = 102.746/0.510999 = 201.07. The nucleus is only 35 times heavier than the muon, so μ sits 2.8% below the muon mass, against 0.054% for hydrogen.
- Length scale: a/Z = a₀(mₑ/μ)/Z = 5.2918 × 10⁻¹¹ m/(201.07 × 2) = 1.316 × 10⁻¹³ m = 131.6 fm.
- Energies: Eₙ = −Z²(μ/mₑ)(13.606 eV)/n² gives E₁ = −4 × 201.07 × 13.606 eV = −10.94 keV and E₂ = E₁/4 = −2.736 keV. Check with the α form: ½μc²(Zα)² = 0.5 × 102.746 MeV × (2/137.036)² = 10.94 keV.
- Photon: 2p → 1s carries E₂ − E₁ = ¾ × 10.943 keV = 8.207 keV, so λ = 1239.84 eV nm/8207 eV = 0.1511 nm, an X-ray.
- Point nucleus: R/(a/Z) = 1.68 fm/131.6 fm = 0.0128, and the finite-size correction is of the order of its square, 1.6 × 10⁻⁴. The relativistic fraction (Zα)² = 2.1 × 10⁻⁴ does not depend on the mass, so here the two are comparable. In hydrogen (0.84 fm/52 947 fm)² = 2.5 × 10⁻¹⁰, some 10⁵ times below α² = 5.3 × 10⁻⁵.
Answerμc² = 102.75 MeV (201.07 mₑ), a/Z = 131.6 fm, E₁ = −10.94 keV, E₂ = −2.736 keV, and the 2p → 1s photon is 8.21 keV (0.151 nm). The nucleus spans 1.3% of the orbit scale, so the point-nucleus error, about 10⁻⁴, rivals the relativistic one.
HardHot plasma in galaxy clusters emits the Lyman-α doublet of hydrogen-like iron, Fe²⁵⁺ (Z = 26, nuclear mass 55.9207 u), whose rest-frame energies, before the cluster's redshift lowers them, are 6951.9 eV and 6973.1 eV. Predict the line from the Bohr energies with the reduced mass, compare it with the statistically weighted mean of the doublet, decide whether the gap has the size the neglected physics predicts, and then use the fine-structure formula of Unit 8 to see how much of the gap it closes.
- Reduced mass: mₑ/M = 5.4858 × 10⁻⁴ u/55.9207 u = 9.81 × 10⁻⁶, so μ/mₑ = 0.99999. Then E₁ = −26² × 0.99999 × 13.6057 eV = −9197.4 eV, E₂ = E₁/4 = −2299.3 eV, and the Bohr line is ¾ × 9197.4 eV = 6898.0 eV.
- The doublet comes from 2p₃/₂ (four states) and 2p₁/₂ (two states), so the weighted mean is (2 × 6973.1 + 6951.9)/3 = 6966.0 eV. The Bohr line is 68.0 eV low, a fraction 68.0/6966.0 = 0.98%.
- Size check: the 1s speed is v/c ≈ Zα = 26/137.036 = 0.190, so relativistic corrections are of relative order (Zα)² = 0.0360, with a coefficient of order ¼ for the ground state: about 0.9%. The gap has that size. In hydrogen the same fraction is 26² = 676 times smaller, which is why the formula holds there to 10⁻⁵.
- Quote Unit 8's result Eₙⱼ = Eₙ[1 + ((Zα)²/n²)(n/(j + ½) − ¾)]. For 1s₁/₂ the bracket is 1 + (Zα)²/4 = 1.00900, so E = −9280.1 eV; for 2p₁/₂ it is 1 + (5/16)(Zα)² = 1.01125, giving −2325.2 eV; for 2p₃/₂ it is 1 + (Zα)²/16 = 1.00225, giving −2304.5 eV.
- Predicted lines: 9280.1 − 2304.5 = 6975.6 eV and 9280.1 − 2325.2 = 6954.9 eV, against the rest-frame 6973.1 and 6951.9 eV. The 68 eV gap shrinks to 2.5 to 3.0 eV, 4 × 10⁻⁴ of the line, the size of the next relativistic terms, of order (Zα)⁴, and of the QED Lamb shift of the 1s level, both beyond this lesson.
AnswerThe Bohr line is 6898.0 eV, 68 eV (0.98%) below the doublet's weighted mean of 6966.0 eV: the (Zα)²/4 ≈ 0.9% size the non-relativistic assumption predicts. The quoted fine-structure formula gives 6954.9 and 6975.6 eV, within 3 eV of the doublet.