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

Before we start, three constants you will reuse. Each is a plain-English quantity, not a magic symbol:

Level 1 — Recognition

L1.1 For which of these species is Bohr's formula $E_{n} = - 13.6 Z^{2} / n^{2}$ eV exactly valid? Options: (a) H, (b) He, (c) He⁺, (d) Li²⁺, (e) Li.

Recall Solution L1.1

What decides it: Bohr's clean derivation used one electron feeling one Coulomb field. The instant a second electron appears, the repulsion term $+ k e^{2} / r_{12}$ enters and there is no fixed $r_{12}$ — the formula collapses.

(a) H: 1 electron ✔
(b) He: 2 electrons ✘
(c) He⁺: helium that lost one electron → 1 electron ✔
(d) Li²⁺: lithium that lost two electrons → 1 electron ✔
(e) Li: 3 electrons ✘ Answer: (a), (c), (d) — the one-electron species.

L1.2 True/False: "Bohr's model correctly predicts how bright each spectral line is."

Recall Solution L1.2

False. Bohr tells you which jumps ( $n_{i} \to n_{f}$ ) are allowed and their energies, but it says nothing about the intensity (probability) of each transition. Intensities need quantum-mechanical transition rules — see Schrödinger Equation for Hydrogen.

Level 2 — Application

L2.1 Compute Bohr's ground-state ( $n = 1$ ) energy for He⁺ ( $Z = 2$ ).

Recall Solution L2.1

Why Bohr applies: He⁺ has one electron → hydrogen-like → formula valid. $E_{1} = - 13.6 \frac{Z ^{2}}{n ^{2}} = - 13.6 \frac{2 ^{2}}{1 ^{2}} = - 13.6 \times 4 = - 54.4 eV .$ Answer: $- 54.4$ eV. (This does match experiment for He⁺, because there is no repulsion term to spoil it.)

L2.2 Neutral helium's measured first ionization energy is $24.6$ eV. If you naively used Bohr's one-electron $Z = 2$ formula for a helium electron, what ionization energy would you predict, and by roughly what factor is Bohr off?

Recall Solution L2.2

Naive Bohr: treat one He electron as hydrogen-like with $Z = 2$ : $∣ E_{1} ∣ = 13.6 \times 4 = 54.4 eV (predicted ionization energy) .$ Compare to reality: measured $= 24.6$ eV. $ratio = \frac{54.4}{24.6} \approx 2.21.$ Bohr overshoots by a factor $\approx 2.2$ . Why: the naive formula ignores that the other electron partly shields the nucleus (each electron does not feel the full $Z = 2$ ) and adds repulsion energy. Bohr has no term for either. This is exactly Failure 1.

Level 3 — Analysis

L3.1 Estimate the fractional size of the fine-structure splitting relative to the main Bohr energy, using $α \approx 1/137$ . Then say why you need a high-resolution spectrometer.

Recall Solution L3.1

Why $α^{2}$ : the fine-structure shift scales as $α^{2} E_{n}$ (from spin–orbit coupling + relativity — see Electron Spin and Spin-Orbit Coupling). $\frac{Δ E _{fine}}{E _{n}} \sim α^{2} = (\frac{1}{137})^{2} = \frac{1}{18769} \approx 5.3 \times 1 0^{- 5} .$ Meaning: the split lines sit about 1 part in ~19000 apart. An ordinary spectrometer sees them as one fat line; you need resolving power better than $\sim 1 0^{5}$ to separate them. That is why Bohr — and low-resolution instruments — miss it entirely.

L3.2 The Sommerfeld/Dirac fine-structure energy is $E_{n, j} \approx - 13.6 \frac{Z ^{2}}{n ^{2}} [1 + \frac{Z ^{2} α ^{2}}{n} (\frac{1}{j + \frac{1}{2}} - \frac{3}{4 n})] .$ For hydrogen ( $Z = 1$ ), level $n = 2$ , compute the correction bracket for the two allowed $j$ values $j = \frac{1}{2}$ and $j = \frac{3}{2}$ , and show they differ (this is the splitting).

Recall Solution L3.2

Fixed pieces: $Z = 1$ , $n = 2$ , so $\frac{Z ^{2} α ^{2}}{n} = \frac{α ^{2}}{2}$ and $\frac{3}{4 n} = \frac{3}{8}$ .

For $j = \frac{1}{2}$ : $\frac{1}{j + \frac{1}{2}} = \frac{1}{1} = 1$ , so inner term $= 1 - \frac{3}{8} = \frac{5}{8}$ . $bracket_{1/2} = 1 + \frac{α ^{2}}{2} \cdot \frac{5}{8} = 1 + \frac{5 α ^{2}}{16} .$ For $j = \frac{3}{2}$ : $\frac{1}{j + \frac{1}{2}} = \frac{1}{2}$ , so inner term $= \frac{1}{2} - \frac{3}{8} = \frac{1}{8}$ . $bracket_{3/2} = 1 + \frac{α ^{2}}{2} \cdot \frac{1}{8} = 1 + \frac{α ^{2}}{16} .$ They differ by $\frac{5 α ^{2}}{16} - \frac{α ^{2}}{16} = \frac{4 α ^{2}}{16} = \frac{α ^{2}}{4}$ . Since the two $j$ -values give two different energies at the same $n$ , Bohr's " $n$ only" line splits into two. That gap is fine structure. The $j = 1/2$ state is pulled lower (more negative energy) because its bracket is larger.

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

Level 4 — Synthesis

L4.1 The hydrogen ( $m_{e}$ , proton mass $M_{H}$ ) and deuterium ( $m_{e}$ , nucleus mass $M_{D} \approx 2 M_{H}$ ) share the same $Z$ and same $n$ -jumps, yet their spectral lines differ slightly. Bohr's bare formula says they should be identical. Explain the missing ingredient and estimate the fractional shift for hydrogen. (Use $m_{e} / M_{H} \approx 1/1836$ .)

Recall Solution L4.1

Why Bohr fails here: his derivation assumed the nucleus is infinitely heavy and fixed. Really, electron and nucleus both orbit their common centre of mass, so the mass in the energy formula is the reduced mass $μ = \frac{m _{e} M}{m _{e} + M}$ (see Reduced Mass Correction). The Rydberg energy scales as $μ$ , so $\frac{μ}{m _{e}} = \frac{M}{m _{e} + M} = \frac{1}{1 + m _{e} / M} \approx 1 - \frac{m _{e}}{M} .$ For hydrogen: shift $\approx m_{e} / M_{H} \approx \frac{1}{1836} \approx 5.45 \times 1 0^{- 4}$ — about 0.05%. For deuterium: $M_{D} \approx 2 M_{H}$ , so its shift is about half as large; the lines land at slightly different wavelengths. That measurable difference is how deuterium was discovered. Bohr's bare $m_{e}$ formula can't produce it — it needs the $μ$ refinement. Relates to Hydrogen Spectrum & Rydberg Formula.

L4.2 Classify each observation by the specific Bohr limitation it exposes: (i) a hydrogen line splits into two in a magnetic field; (ii) neutral lithium's spectrum can't be predicted from $n$ alone; (iii) a hydrogen line splits in a strong electric field; (iv) the electron cannot have a known position and momentum at once.

Recall Solution L4.2

(i) Magnetic-field splitting → Zeeman effect — Bohr predicts no field response (Zeeman and Stark Effects).
(ii) Multi-electron ( $Z = 3$ , 3 electrons) → repulsion + $l$ -dependence missing (Failure 1).
(iii) Electric-field splitting → Stark effect — unexplained by Bohr (Zeeman and Stark Effects).
(iv) Fixed orbit fixes both $x$ and $p$ → violates the Heisenberg Uncertainty Principle, $Δ x Δ p \geq ℏ/2$ .

Level 5 — Mastery

L5.1 A student claims: "Since Bohr gives He⁺ perfectly and He⁺ has $Z = 2$ , I can get neutral He's ground energy by doubling — two electrons each at $- 54.4$ eV, so $- 108.8$ eV total, and its ionization energy is $54.4$ eV." Identify every flaw, and state what physics must be added to get near the true value ( $\approx - 79.0$ eV total, ionization $24.6$ eV).

Recall Solution L5.1

Flaw 1 — no repulsion term. Adding two independent $- 54.4$ eV electrons ignores $+ k e^{2} / r_{12}$ , the mutual repulsion. Repulsion is positive energy, so it makes the true total energy less negative than $- 108.8$ eV. Flaw 2 — no shielding. Each electron does not feel the full $Z = 2$ ; the other electron partly screens the nucleus, so the effective nuclear charge is $< 2$ . That also raises (makes less negative) the energy. Flaw 3 — ionization ≠ one orbit energy. Ionization energy is (energy of He⁺) − (energy of He), not a single electron's $∣ E_{1} ∣$ . Check the direction: naive $- 108.8$ eV vs true $- 79.0$ eV — the repulsion + shielding push it up by $- 79.0 - (- 108.8) = 29.8 eV,$ exactly the kind of large correction Bohr has no term for. Getting near the true value needs the full two-electron Hamiltonian with the $r_{12}$ coupling — the Schrödinger equation generalised, solved approximately (variational / perturbation methods).

L5.2 Build the full argument: why does the appearance of the label $j$ (total angular momentum) in the Dirac energy formula prove Bohr was structurally incomplete, not just numerically imprecise? Use one sentence per logical step.

Recall Solution L5.2

Bohr's energy depends on one label: $n$ . Two states with the same $n$ are, to Bohr, identical in energy.
Experiment shows two states with the same $n$ can have different energies (the split lines).
Therefore reality needs another label to distinguish them — the Dirac formula supplies $j$ .
$j$ combines orbital angular momentum ( $l$ ) and spin ( $s = \frac{1}{2}$ ) — see Quantum Numbers n, l, m, s — neither of which exists in Bohr's picture.
So Bohr's failure is not "the number $- 13.6$ is slightly wrong"; it is "the model lacks the variables ( $l, s, j$ ) that nature uses." Conclusion: the fix is structural (add spin, orbital shape, relativity), which is precisely what quantum mechanics did — not a numerical tweak.

Recall Self-test: one line each

Which species does Bohr solve exactly? ::: One-electron species only: H, He⁺, Li²⁺, Be³⁺ … What term breaks multi-electron atoms? ::: The electron–electron repulsion $+ k e^{2} / r_{12}$ (no fixed $r_{12}$ ). What causes fine structure in hydrogen? ::: Spin–orbit coupling + relativity, scale $\sim α^{2}$ — not repulsion. Why do H and D lines differ? ::: Reduced-mass correction $μ$ ; the nucleus is not infinitely heavy. Why is $j$ appearing fatal for Bohr? ::: Bohr has only $n$ ; nature needs the extra label $j$ (from $l$ and spin), which Bohr lacks.