The Bohr model and its failures

Classical physics alone allows every circular orbit. For an electron circling a nucleus of charge Ze, Newton's second law with Coulomb's force gives mₑ v²/r = Zke²/r², with k = 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻². That is one equation in two unknowns, so r can take any value, the energy −Zke²/(2r) varies continuously, and the emitted spectrum should be continuous. The orbit is also unstable: an accelerating charge radiates, and the Larmor rate makes a classical orbit of radius 52.9 pm spiral into the proton in about 1.6 × 10⁻¹¹ s. Bohr's 1913 answer was three postulates that classical physics cannot supply. Certain orbits are stationary and do not radiate. They are the ones with angular momentum L = nħ. Light is emitted or absorbed only in a jump between two of them, with hf = Eᵢ − E_f. Nothing justifies the postulates except whether the numbers they produce match the spectrum.

Put v = nħ/(mₑ r) from the rule into the force balance: mₑ n²ħ²/(mₑ² r³) = Zke²/r², so rₙ = n²ħ²/(Z mₑ ke²) = n²a₀/Z. Check the dimensions: (J s)² divided by kg × J m is J s² kg⁻¹ m⁻¹, and with J = kg m² s⁻² that is m. Put rₙ back into the rule and vₙ = Zke²/(nħ) = Zαc/n, which is 2.19 × 10⁶ m s⁻¹, or c/137, in hydrogen's ground state. For the energy, the force balance says K = ½mₑ v² = Zke²/(2r) while U = −Zke²/r, so E = K + U = −K = U/2, the virial theorem for an inverse-square force. Substituting rₙ gives Eₙ = −(mₑ k²e⁴/(2ħ²)) Z²/n² = −13.6 eV × Z²/n². Each power of Z has a source: r ∝ 1/Z because the pull is Z times stronger, v ∝ Z, and E ∝ Z² because the larger charge and the smaller radius both deepen the well.

Bohr stated L = nħ; de Broglie, in 1924, gave it a reason. An electron of momentum p carries a wavelength λ = h/p. Confine that wave to a circular track and it meets itself after one lap. If the circumference holds a whole number of wavelengths the wave returns in phase and reinforces itself; if not, each lap arrives shifted and the sum over many laps cancels. The condition 2πr = nλ = nh/(mₑ v) is exactly mₑ v r = nħ. Combined with the Coulomb force balance, a circular orbit of radius r holds N = √(Zr/a₀) wavelengths, so in hydrogen the orbits at a₀, 4a₀ and 9a₀ close with N = 1, 2 and 3, while one at 3a₀ holds 1.73 wavelengths and is forbidden. The argument explains the rule without proving it. It puts the electron wave on a one-dimensional line, but the real wave fills three dimensions: in the Schrödinger solution n − 1 counts radial and angular nodes together, and the ground state has no circulating wave at all.

Three checks decide whether a Bohr-model number can be trusted. First, the nucleus is not infinitely heavy: electron and nucleus both orbit their centre of mass, and the fix is to replace mₑ by the reduced mass μ = mₑ M/(mₑ + M). For hydrogen μ = 0.99946 mₑ, which moves every line by 0.054%; deuterium's μ is larger by 2.7 × 10⁻⁴, enough to shift Hα by 0.18 nm, which is how deuterium was found in 1932. Second, the speed ratio v/c = Zα/n must be small for a nonrelativistic model: 0.0073 in hydrogen, 0.60 for n = 1 in lead. Third, correspondence. The photon for n → n − 1 has f = (13.6 eV/h) Z²[1/(n − 1)² − 1/n²], and the classical orbital frequency is f_orb = v/(2πr) = Z²αc/(2πa₀n³). Their ratio n(2n − 1)/(2(n − 1)²) is 3 at n = 2 but 1.01 at n = 150, so jumps between neighbouring high orbits radiate at the classical frequency, as they must.

Bohr's ground state is a circular orbit with L = ħ. Measurement says the hydrogen ground state has zero orbital angular momentum: it carries no orbital magnetic moment, and its magnetism comes from electron spin, which the model does not contain. The orbit also lies in a plane, so a Bohr atom would be a flat disc with a preferred axis, while ground-state hydrogen is spherically symmetric. The uncertainty principle rules the orbit out. On the n = 1 orbit p = mₑ v = ħ/a₀; pinning the radius to within 10%, Δr = 0.1a₀, needs Δp ≥ ħ/(2Δr) = 5ħ/a₀, five times the momentum itself, so no definite trajectory of that size can exist. The energies survive because the Coulomb potential is special: its energy depends only on n = nᵣ + l + 1, one more than the total count of radial and angular nodes, so a model that counts n correctly gets Eₙ right whatever it says about l. For n = 1 the only possibility is l = 0.

Bohr's rule quantises one electron on one orbit. With two electrons it has no principle for arranging them, and the arrangements that can be computed miss: putting helium's two electrons on opposite sides of one ring overshoots the first ionisation energy by about 18%, as the Hard example shows, while He⁺, with one electron, comes out right. The model is also silent on everything but line positions. It gives frequencies but no intensities, no lifetimes and no selection rules. In hydrogen the 2p state decays to 1s in 1.6 ns, while 2s, the same Bohr orbit with the same energy, has no single-photon dipole route to 1s and lives about 0.12 s, nearly 10⁸ times longer. The model has one n = 2 state; nature has several, and a high-resolution spectrometer shows them: Hα resolves into fine-structure components spread over about 0.02 nm. Sommerfeld's elliptical orbits patched the fine structure. Nothing patched the rest.