Limitations of Bohr — fails for multi-electron atoms, fine structure: Question bank

Before we start, some symbols and words you must own (each anchored here in plain language):

Definition The vocabulary these traps rely on

One-electron species ::: an atom or ion with exactly one electron orbiting the nucleus (H, He⁺, Li²⁺, Be³⁺, …). The number of protons $Z$ can be anything; what matters is that only one electron is present.
Fine structure ::: the splitting of a single Bohr spectral line into several very closely spaced lines, visible only at high resolution, caused by spin–orbit coupling and relativity — see Electron Spin and Spin-Orbit Coupling.
The $r_{12}$ term ::: the electron–electron repulsion energy $+ k e^{2} / r_{12}$ , where $r_{12}$ is the distance between two electrons. It exists only when there are ≥2 electrons, and it is the single term that wrecks Bohr for real atoms.
Reduced mass $μ$ ::: the effective mass $μ = \frac{m _{e} M}{m _{e} + M}$ (with $M$ = nuclear mass) that replaces the bare electron mass once you account for the nucleus also orbiting the common centre of mass — see Reduced Mass Correction.
Total angular momentum $j$ ::: the quantum number you get by combining the orbital angular momentum $l$ with the spin $s$ ; the fine-structure energy depends on $j$ , a number Bohr never had (see Electron Spin and Spin-Orbit Coupling and Quantum Numbers n, l, m, s).

True or false — justify

Bohr's energy formula gives numerically wrong answers for hydrogen.

False. With the reduced-mass correction it matches hydrogen's spectrum to many decimal places; Bohr fails elsewhere by missing physics, not by fumbling arithmetic on H.

He⁺ is a one-electron species, so Bohr describes it correctly.

True. He⁺ has

Z = 2

but only one electron, so the

r_{12}

term never appears and the clean

E_{n} = - 13.6 Z^{2} / n^{2}

formula applies exactly (up to reduced mass).

Neutral helium and He⁺ should both obey Bohr since both have

Z = 2

False.

Z

is not the deciding factor; the number of electrons is. Neutral He has two electrons, so the

k e^{2} / r_{12}

repulsion appears and Bohr collapses.

Fine structure is caused by electron–electron repulsion.

False. Repulsion is the multi-electron problem. Fine structure shows up even in hydrogen (one electron), so its cause is spin–orbit coupling + relativity, not repulsion.

Bohr predicts that a magnetic field splits spectral lines.

False. Bohr energy depends only on

n

, so it predicts no splitting in a field; the observed Zeeman splitting is one of the effects it cannot explain.

The reduced-mass correction is why hydrogen and deuterium lines differ.

True. Deuterium's nucleus is roughly twice as heavy, so its reduced mass

μ

is slightly larger, shifting its lines by about 1 part in a few thousand relative to hydrogen.

Because Bohr assumes a fixed orbit, it is consistent with Heisenberg's uncertainty principle.

False. A fixed circular orbit specifies position and momentum at once, which violates

Δ x Δ p \geq ℏ/2

— see Heisenberg Uncertainty Principle.

Fine-structure splitting is large enough to see with any spectrometer.

False. It scales as

α^{2} \approx (1/137)^{2} \approx 5 \times 1 0^{- 5}

of the main energy, so you need a high-resolution instrument — see Fine-Structure Constant.

Bohr can tell you which spectral lines appear but not how bright they are.

True. Bohr gives transition energies (line positions) but has no mechanism for transition probabilities, so it cannot predict line intensities.

Since Bohr is "outdated," it fails for every atom including hydrogen.

False. It is exact for one-electron species. It fails only when there is more than one electron, or when you demand fine detail (fine structure, fields, intensities).

Why questions

Why does the

r_{12}

term make the problem unsolvable analytically, when a fixed nuclear charge did not?

r_{12}

is the distance between two moving electrons — it constantly changes with no fixed value, so there is no single clean orbit radius to plug in; it is a quantum three-body problem with no closed-form solution.

Why does Bohr succeed for He⁺ but not for neutral He?

He⁺ has one electron so the repulsion term vanishes and the hydrogen-like formula applies; neutral He has two electrons, adds

k e^{2} / r_{12}

, and Bohr has no variable for it.

Why can Bohr not explain the periodic table or subshells (s, p, d)?

Its energy depends only on

n

, so it has no notion of orbital shape

l

or of shielding/penetration — the very features that build shells, subshells, and periodicity — see Quantum Numbers n, l, m, s.

Why is the fine-structure constant

α

the natural scale for the splitting?

α = k e^{2} / (ℏ c)

(with

k

the Coulomb constant,

ℏ

the reduced Planck constant,

c

the speed of light) measures how "relativistic" the orbiting electron is; both spin–orbit coupling and the relativistic mass shift scale as

α^{2}

, so

α^{2} E_{n}

sets the size of the splitting.

Why does treating the electron as a point on a flat circle already conflict with modern physics?

A point on a definite circular path fixes position and momentum simultaneously (violating uncertainty) and ignores the electron's de Broglie wave nature and the true 3D probability-cloud shape of orbitals.

Edge cases

What happens to Bohr's accuracy in the limit of an infinitely heavy nucleus (

M \to \infty

Then the reduced mass

μ = m_{e} M / (m_{e} + M) \to m_{e}

and the reduced-mass correction vanishes, so the "bare-mass"

- 13.6

eV becomes exact — the correction is largest for the lightest nucleus (hydrogen).

Is fine structure present in the hydrogen ground state (

n = 1, l = 0

Yes — even though spin–orbit coupling vanishes for

l = 0

, the relativistic correction (the electron's mass changing as it speeds up, shifting the energy by a factor

\propto α^{2}

) still shifts the level, so the

n

-only Bohr picture is incomplete even here.

For a one-electron ion with very large

Z

(e.g. U⁹¹⁺), is Bohr's simple formula still trustworthy?

Not fully — with one electron the

r_{12}

term is still absent, but the electron moves so fast that relativistic and fine-structure corrections (

\propto Z^{2} α^{2}

) become large, so the naive

- 13.6 Z^{2} / n^{2}

deviates noticeably.

In the limit of zero applied field, do the Zeeman and Stark splittings vanish?

Yes — both splittings are caused by the external field, so at zero field they collapse to the field-free (still fine-structure-split) levels; the fine structure itself remains because it needs no external field.

Does Bohr's model predict hyperfine splitting (the coupling of the electron to the nuclear spin)?

No — Bohr has no electron spin, let alone nuclear spin, so it cannot produce the even tinier hyperfine splitting (the coupling between the electron's magnetic moment and the nucleus's own magnetic moment) that shows up as, e.g., hydrogen's famous 21 cm line.

Can Bohr account for the Lamb shift seen in ultra-high-resolution hydrogen spectra?

No — the Lamb shift is a tiny splitting between levels Bohr treats as identical, caused by the electron interacting with fluctuations of the electromagnetic field (a quantum-electrodynamics effect); Bohr has no field-fluctuation physics at all, so this correction lies entirely beyond it.

Recall One-line summary of every trap

Bohr is exact for one-electron species with the reduced-mass tweak; it breaks whenever there are ≥2 electrons ( $r_{12}$ ), whenever you resolve fine detail (spin–orbit + relativity, scale $α^{2}$ , and even finer: hyperfine and Lamb shift), or whenever you apply fields (Zeeman/Stark) — because its energy carries only $n$ and $Z$ , and nothing else.

Question bank — Limitations of Bohr — fails for multi-electron atoms, fine structure

True or false — justify

Spot the error

Why questions

Edge cases