The Hydrogen Atom
The hydrogen atom is the only atom quantum mechanics solves exactly, and every orbital picture in chemistry descends from it. This part does the full solution — the separation, the radial equation, the three quantum numbers falling out as conditions rather than assumptions — and then spends as much time on what the answers mean, because that is what the exam tests: nodes, radial distribution functions, the difference between ψ2 and 4πr2ψ2, and why the 2s and 2p levels are degenerate here and nowhere else. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.
The 10 sections in Part 5
- 1The two-body problem and the reduced mass Free below
- 2The Coulomb potential and separation in spherical polars
- 3The radial equation, the Laguerre polynomials, and where n comes from
- 4The three quantum numbers and the constraints among them
- 5Radial wavefunctions and radial nodes
- 6Angular functions, real orbitals and the shapes you draw
- 7The radial distribution function, and why its maximum is not where ψ² peaks
- 8Expectation values, the size of the atom, and the virial theorem
- 9The spectrum, selection rules, hydrogen-like ions and fine structure
- 10Where this leads — a pointer to Part 6
The two-body problem and the reduced mass
Free extractSection E.1 of Part 5, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.
Two particles orbiting each other look like one particle orbiting a fixed point, provided you give it the right mass. That mass is μ.
A hydrogen atom is two particles: a proton and an electron, both free to move, attracting each other. Writing the Schrödinger equation for that honestly gives an equation in six coordinates — three for each particle — which nobody can solve directly.
The escape is a change of variables you already know from classical mechanics. Instead of tracking where each particle is, track two other things: where the centre of mass is, and where the electron is relative to the proton. When you do this, the Hamiltonian splits cleanly into two independent pieces:
The first piece describes the atom as a whole drifting through space. It is a free particle; its energy is ordinary kinetic energy and it has nothing to do with chemistry, so we discard it. The second piece is a one-particle problem: a single fictitious particle of mass μ, at position r relative to the origin, moving in the potential V(r). That is the problem we solve for the rest of the part.
For hydrogen, M/me = 1836.15, so μ = 0.9994557 me. The correction is about one part in 1836 — a fraction of a tenth of a percent. You may reasonably ask why anybody bothers. The answer is that spectroscopy measures wavelengths to eight or nine significant figures, so a correction in the fourth figure is enormous by that standard, and the difference between hydrogen and deuterium spectra is entirely this effect.
The coordinate change in full, and what it does and does not assume.
Start with the exact two-particle Hamiltonian, particle 1 the nucleus and particle 2 the electron:
Define the centre-of-mass and relative vectors
The chain rule, applied twice and with some patience, converts the sum of the two Laplacians into
with Mtot = m1 + m2 and 1/μ = 1/m1 + 1/m2. The crucial structural point is that the cross terms cancel exactly. There is no residual ∇R·∇r coupling. That is why the separation is exact rather than approximate, and it is why the reduced-mass step is not an approximation at all — unlike, say, the Born–Oppenheimer separation of Part 7, which looks superficially similar and is an approximation.
Since V depends on r alone, the total wavefunction factorises as Ψ(R,r) = Χ(R)ψ(r) with Etotal = Etrans + E. The translational factor is a plane wave with a continuous energy; we set it aside and never mention it again. Everything called ‘the energy of the hydrogen atom’ from here on is the internal energy E.
Three consequences worth carrying forward:
- Every me in the standard results should be a μ. The Bohr radius, the Rydberg constant and the energies all carry it. Most textbooks quietly write me and mean ‘in the infinite-nuclear-mass limit’.
- Isotope effects in atomic spectra are pure reduced-mass effects. Deuterium's lines are shifted from hydrogen's by the ratio μD/μH, and that shift is how deuterium was discovered by Urey in 1931–32.
- Positronium (e⁺e⁻) has μ = me/2 exactly, so every energy level is half the hydrogen value and every wavelength is doubled. It is a favourite exam variant precisely because it makes you show that you know where μ enters.
| Quantity | Value | Where it comes from |
|---|---|---|
| R∞ | 10973731.568 m⁻¹ = 109737.32 cm⁻¹ | infinite nuclear mass |
| RH | 10967758.34 m⁻¹ = 109677.58 cm⁻¹ | R∞ × 0.99945568 |
| RD | 10970742.66 m⁻¹ = 109707.43 cm⁻¹ | R∞ × 0.99972763 |
| μH/me | 0.99945568 | Mp/me = 1836.153 |
| μD/me | 0.99972763 | Md/me = 3670.483 |
| μPs/me | 0.5 | positronium, both masses equal |
Predict the isotope shift of the Hα line (deuterium vs hydrogen) Medium
⚠ Common mistakes & exam traps
- Forgetting μ entirely and then being surprised that RH ≠ R∞. If a question quotes 109677.6 cm⁻¹ it is talking about hydrogen; 109737.3 cm⁻¹ is the infinite-mass constant. They are not the same number and a question that supplies one and expects the other is testing exactly this.
- Using μ = m1m2/(m1−m2). The denominator is a sum. A quick check: μ must always be smaller than the smaller of the two masses.
- Thinking the reduced-mass step is an approximation. It is exact. What is approximate is neglecting the centre-of-mass motion — and even that is not an approximation, it is a separation of a degree of freedom we do not care about.
- Applying the μ correction to Z but not to a0. The Bohr radius itself contains μ: a0 = 4πε0ℏ²/μe². For a consistent finite-mass calculation, correct everything or nothing.
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Read the rest of Part 5
The remaining 9 sections of this part — The Coulomb potential and separation in spherical polars, The radial equation, the Laguerre polynomials, and where n comes from, The three quantum numbers and the constraints among them… — and all nine parts of Quantum Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.
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