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

Definition The anchors we build on

Coulomb's constant $k$ . When two charges attract or repel, the force is $F = \frac{k q _{1} q _{2}}{r ^{2}}$ . The number $k$ just sets the strength of that electric push/pull: $k \approx 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2}$ (equivalently $k = \frac{1}{4 π ε _{0}}$ , where $ε_{0}$ is how easily empty space "permits" an electric field). Physically $k$ is what turns "two charges a distance $r$ apart" into an actual force in newtons. It appears everywhere below — in the nucleus-pull term $k Z e^{2} / r^{2}$ , in the electron–electron repulsion $k e^{2} / r_{12}$ , and inside $α$ .
Bohr ground energy of hydrogen: $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}} eV$ . Here $Z$ = number of protons in the nucleus (the "pull"), and $n$ = which orbit (the shell number, $n = 1, 2, 3, \dots$ ). The minus sign means the electron is bound — you must spend energy to rip it out.
Fine-structure constant: $α = \frac{k e ^{2}}{ℏ c} \approx \frac{1}{137} \approx 0.00730$ . Every symbol is now earned: $k$ (just defined), $e$ = the electron's charge, $ℏ$ = Planck's constant $h$ divided by $2 π$ (nature's unit of angular momentum), $c$ = speed of light. Multiply and divide these and, astonishingly, all the units cancel — $α$ is a pure number that measures "how strong is electromagnetism". We use it because every small relativistic/spin correction turns out to be a power of $α$ — see Fine-Structure Constant.
The photon shortcut $h c = 1240 eV\cdotnm$ . A photon's energy is $E = \frac{h c}{λ}$ ( $λ$ = wavelength). Rather than carry $h = 4.136 \times 1 0^{- 15} eV\cdots$ and $c = 2.998 \times 1 0^{17} nm/s$ separately, multiply them once: $h c = (4.136 \times 1 0^{- 15}) (2.998 \times 1 0^{17}) \approx 1240 eV\cdotnm$ . So for any wavelength in nm, $E (eV) = 1240/ λ (nm)$ . This is a packaged constant, not magic — we derive it here so Ex 9 can lean on it.

The scenario matrix

Every problem in this topic lands in one of these cells. Read the figure below first — it is the map of the whole page; each branch is a worked example.

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

The same map as a table (tag = which example hits that cell):

#	Case class	What decides it	Bohr's fate	Example
C1	One electron, $Z = 1$ (hydrogen)	electron count $= 1$	Exact	Ex 1
C2	One electron, $Z > 1$ (He⁺, Li²⁺, …)	electron count $= 1$ , big $Z$	Exact ( $Z^{2}$ scaling)	Ex 2
C3	Two-or-more electrons (He, Li…)	$e$ – $e$ repulsion term $k e^{2} / r_{12}$ appears	Fails	Ex 3
C4	Reduced-mass patch (H vs D)	finite nuclear mass $M$	Fixable (small shift)	Ex 4
C5	Fine-structure size ( $α^{2}$ )	spin-orbit + relativity	Missing, but tiny	Ex 5
C6	Fine-structure splitting ( $j$ levels)	total ang. momentum $j$	No variable for it	Ex 6
C7	External field (Zeeman/Stark)	field on/off	Predicts no split	Ex 7
C8	Degenerate / limiting cases ( $n \to \infty$ , $Z \to 0$ )	boundary values	Consistent limit	Ex 8
C9	Real-world word problem	which telescope line?	mixed	Ex 9
C10	Exam twist (spot the false claim)	reasoning trap	—	Ex 10
C11	Ultra-high-resolution (hyperfine, Lamb)	nuclear spin, QED	Utterly absent	Ex 11

Ex 1 — Hydrogen, the case Bohr nails (Cell C1)

Worked example Ground-state ionization energy of hydrogen

Q: Using Bohr, find the energy to remove the electron from ground-state hydrogen ( $Z = 1$ , $n = 1$ ).

Forecast: Guess before reading — $10$ , $14$ , or $50$ eV? Why these three? $10$ eV is the visible-light photon scale (too small to unbind), $50$ eV is the He⁺ scale you'll meet in Ex 2 (too big for $Z = 1$ ), and $14$ eV brackets the truth — the benchmarks are chosen to force you to feel the $Z^{2}$ dependence before you compute it.

Write $E_{n} = - 13.6 \frac{Z ^{2}}{n ^{2}}$ . Why this step? This is the only formula Bohr gives, and for one electron it is exact — nothing is missing.
Put $Z = 1, n = 1$ : $E_{1} = - 13.6 \frac{1}{1} = - 13.6$ eV. Why? Ground state means the lowest orbit, $n = 1$ .
Ionization = raise the electron to $n = \infty$ (free, $E_{\infty} = 0$ ). Energy needed $= E_{\infty} - E_{1} = 0 - (- 13.6) = + 13.6$ eV. Why? "Removing" the electron means climbing all the way out of the well.

Verify: Measured hydrogen ionization energy is $13.6$ eV. ✔ Units: eV, an energy. ✔ This is why Bohr was celebrated.

Ex 2 — He⁺: one electron, big charge (Cell C2)

Worked example Why He⁺ still obeys Bohr

Q: He⁺ has $Z = 2$ and one electron. Find its ground-state energy and ionization energy.

Forecast: Bigger nucleus pulls harder — will the energy be $2 \times$ or $4 \times$ hydrogen's? Why these two? If binding scaled with the charge $Z$ you'd say $2 \times$ ; if it scaled with $Z^{2}$ (as the Bohr algebra actually gives) you'd say $4 \times$ . The choice is a direct test of whether you remember the squared $Z$ .

Electron count $= 1 \Rightarrow$ no $r_{12}$ term $\Rightarrow$ Bohr is exact. Why this step? The repulsion term $k e^{2} / r_{12}$ only exists with two electrons; with one, it is absent. See Hydrogen Spectrum & Rydberg Formula.
$E_{1} = - 13.6 \frac{Z ^{2}}{n ^{2}} = - 13.6 \cdot \frac{2 ^{2}}{1 ^{2}} = - 13.6 \cdot 4 = - 54.4$ eV. Why? Energy scales as $Z^{2}$ , not $Z$ — the $Z$ appears squared in the Bohr derivation (once in the force, once in the radius).
Ionization energy $= + 54.4$ eV. Why? Same climb-out logic as Ex 1.

Verify: Experimental He⁺ ionization energy $\approx 54.4$ eV. ✔ Exactly $4 \times$ hydrogen — the $Z^{2}$ law confirmed. ✔

Ex 3 — Neutral helium: where Bohr breaks (Cell C3)

Worked example The two-electron catastrophe

Q: Try to predict neutral He's ( $Z = 2$ , two electrons) ionization energy the naive Bohr way, and compare with reality (24.6 eV).

Forecast: Will naive Bohr over- or under-estimate? Why it's a real fork: repulsion pushes electrons out (less bound $\to$ Bohr over-estimates), but a naive person might expect the strong $Z = 2$ pull to dominate and under-estimate. Pick before step 2.

Naive Bohr: treat each electron as feeling the full $Z = 2$ nucleus, ignore the other. Each sits at $E_{1} = - 54.4$ eV, so removing one "should" cost $54.4$ eV. Why this step? We deliberately push Bohr past its limits to see it fail.
Reality: $24.6$ eV — less than half. Why the gap? The second electron repels the first ( $+ k e^{2} / r_{12}$ ), pushing its energy up (less bound), so less energy is needed to remove it. Bohr has no term for this.
Conclusion: Bohr's error here is $\sim 30$ eV — not a rounding slip, a missing physics error. Why? The full Hamiltonian is a non-separable three-body problem; no clean orbit exists.

Verify: Bohr naive $54.4$ eV vs measured $24.6$ eV; ratio $\approx 2.2$ , i.e. Bohr is off by $121%$ . ✔ Confirms Bohr fails the moment electron count $> 1$ .

Ex 4 — The reduced-mass patch: H vs D (Cell C4)

Worked example Why deuterium's lines shift

Q: Hydrogen's nucleus is 1 proton ( $M \approx 1836 m_{e}$ ); deuterium's is a proton+neutron ( $M \approx 3672 m_{e}$ ). By what fraction does the Rydberg energy of H differ from the infinite-mass value, and how does D compare? See Reduced Mass Correction.

Forecast: Will the shift be near $0.05%$ or $5%$ ? Why these two? $5%$ would be a visible, coarse shift; $0.05%$ is "needs a good spectrometer". Since the electron is $\sim 1836 \times$ lighter than the proton, the correction is $\sim 1/1836$ — the benchmarks pin down whether you feel that ratio.

Replace $m_{e}$ by reduced mass $μ = \frac{m _{e} M}{m _{e} + M}$ . Why this step? Both nucleus and electron orbit their common centre of mass; the "effective" orbiting mass is $μ$ , slightly less than $m_{e}$ .
The correction factor is $\frac{μ}{m _{e}} = \frac{M}{m _{e} + M} = \frac{1}{1 + m _{e} / M}$ . For H: $\frac{1}{1 + 1/1836} \approx 0.999456$ . Why? Energy $\propto μ$ , so energies drop by this factor from the ideal $13.6$ eV.
Shift from infinite-mass value $\approx 1 - 0.999456 = 5.44 \times 1 0^{- 4} = 0.0544%$ . Why? That $\sim$ 1 part in 1836 is the classic reduced-mass magnitude.
For D: factor $= \frac{1}{1 + 1/3672} \approx 0.999728$ , shift $\approx 0.0272%$ — half of hydrogen's. Why? Heavier nucleus $\Rightarrow$ $μ$ closer to $m_{e}$ $\Rightarrow$ smaller shift. Hence H and D lines sit at slightly different wavelengths — this is how deuterium was discovered.

Verify: H shift $\approx 5.44 \times 1 0^{- 4}$ , D shift $\approx 2.72 \times 1 0^{- 4}$ , ratio $\approx 2.0$ . ✔ Matches the mass ratio $M_{D} / M_{H} \approx 2$ .

Ex 5 — Fine structure: how big? (Cell C5)

Worked example Order-of-magnitude of the splitting

Q: Estimate the fine-structure energy shift for the hydrogen $n = 2$ level, as a fraction of $E_{2}$ , using $\sim α^{2} E_{n}$ . See Electron Spin and Spin-Orbit Coupling.

Forecast: $1%$ ? $0.005%$ ? Smaller? Why these two? $1%$ is a shift you'd catch with a cheap grating; $0.005%$ is $α^{2}$ -sized (recall $α^{2} \approx 5 \times 1 0^{- 5}$ ). Guessing tests whether you remember that fine structure is an $α^{2}$ effect.

Shift $\sim α^{2} ∣ E_{n} ∣$ with $α \approx 1/137$ . Why this step? Spin-orbit and relativistic corrections both scale as $α^{2}$ — that's why we introduced $α$ : it is the natural yardstick for these tiny terms.
$α^{2} = (1/137)^{2} \approx 5.33 \times 1 0^{- 5}$ . Why? This pure number is the fractional size of the correction.
$∣ E_{2} ∣ = 13.6/4 = 3.4$ eV, so shift $\sim 5.33 \times 1 0^{- 5} \times 3.4 \approx 1.8 \times 1 0^{- 4}$ eV. Why? Multiply fraction by the level energy to get an actual energy.

Verify: Fractional shift $\approx 5.3 \times 1 0^{- 5}$ ( $\approx 0.005%$ ). ✔ That's why you need a high-resolution spectrometer — the split is 5 parts in 100,000.

Ex 6 — Fine structure needs $j$ , which Bohr lacks (Cell C6)

Worked example Splitting the

$n = 2$ level with the Dirac formula Q: Use $E_{n, j} \approx - 13.6 \frac{Z ^{2}}{n ^{2}} [1 + \frac{Z ^{2} α ^{2}}{n} (\frac{1}{j + \frac{1}{2}} - \frac{3}{4 n})]$ for hydrogen $n = 2$ . The allowed total angular momenta are $j = \frac{1}{2}$ and $j = \frac{3}{2}$ . Find the energy gap between them. See Quantum Numbers n, l, m, s.

Forecast: Will the two $j$ -values give the same energy (like Bohr) or different? Why this fork: if energy truly depended on $n$ alone (Bohr), the gap is exactly zero; any nonzero answer proves Bohr is missing the variable $j$ .

Compute the bracket for $j = \frac{1}{2}$ : $\frac{1}{j + 1/2} = \frac{1}{1} = 1$ ; for $j = \frac{3}{2}$ : $\frac{1}{2}$ . Why this step? The whole point: energy now depends on $j$ — a variable Bohr's " $n$ -only" formula does not have.
Prefactor $- 13.6 \cdot \frac{1}{4} \cdot \frac{α ^{2}}{2} = - 13.6 \cdot 0.25 \cdot \frac{5.33 \times 1 0 ^{- 5}}{2} = - 9.06 \times 1 0^{- 5}$ eV. Why? This is the coefficient multiplying the $j$ -dependent piece $(\frac{1}{j + 1/2} - \frac{3}{4 n})$ ; $Z = 1$ , $n = 2$ .
Difference: gap $= ∣ prefactor ∣ \times (1 - \frac{1}{2}) = 9.06 \times 1 0^{- 5} \times 0.5 = 4.53 \times 1 0^{- 5}$ eV. Why? Only the $\frac{1}{j + 1/2}$ term differs between the two $j$ ; the $- \frac{3}{4 n}$ cancels in the subtraction.

Verify: Computed gap $\approx 4.5 \times 1 0^{- 5}$ eV, matching the known H $n = 2$ fine-structure splitting ( $\approx 4.5 \times 1 0^{- 5}$ eV $\approx$ 10.9 GHz). ✔ Bohr predicts a gap of zero — that is the failure.

Ex 7 — External field: Bohr predicts no split (Cell C7)

Worked example Zeeman splitting Bohr can't see

Q: A hydrogen line is observed in a magnetic field $B = 1$ T. The Zeeman shift is $Δ E = μ_{B} B$ per unit of magnetic quantum number, with the Bohr magneton $μ_{B} = 5.79 \times 1 0^{- 5}$ eV/T. Find the shift and contrast with Bohr's prediction. See Zeeman and Stark Effects.

Forecast: Bohr says the line does what in a field — split, shift, or nothing? Why these three: "split/shift" require an orientation variable ( $m$ ); "nothing" is what a formula with only $n$ must predict. The fork exposes exactly the missing variable.

$Δ E = μ_{B} B = 5.79 \times 1 0^{- 5} \times 1 = 5.79 \times 1 0^{- 5}$ eV. Why this step? The field couples to the electron's magnetic moment; different $m$ -states (orientations) gain different energy — a splitting.
Bohr's energy $= - 13.6 Z^{2} / n^{2}$ contains no $B$ and no $m$ . Why? Bohr has no orbital-orientation variable, so his line cannot respond to a field: predicted shift $= 0$ .
So observed lines split by $\sim 5.79 \times 1 0^{- 5}$ eV while Bohr insists they don't move at all. Why it matters: any nonzero split refutes Bohr — the Stark (electric-field) effect fails for the identical reason.

Verify: Zeeman shift $\approx 5.79 \times 1 0^{- 5}$ eV at 1 T. ✔ Bohr prediction $= 0$ . ✔ Contradiction confirms limitation #3–4.

Ex 8 — Degenerate & limiting cases (Cell C8)

Worked example The boundaries:

$n \to \infty$ , small $Z$ , and large $Z$ Q: (a) What is $E_{n}$ as $n \to \infty$ ? (b) What is the energy gap between $n = 1000$ and $n = 1001$ in hydrogen — does Bohr stay sensible? (c) Small- $Z$ limit: as $Z \to 0$ (an imaginary "chargeless nucleus"), what happens to $E_{1}$ , and is that physically sensible? (d) Large- $Z$ check: C⁵⁺ ( $Z = 6$ , one electron).

Forecast: Does the model blow up, or fade smoothly, at every edge?

(a) $E_{\infty} = - 13.6/ \infty^{2} = 0$ . Why this step? As orbits grow huge, binding vanishes — the electron is essentially free. Bohr behaves correctly at the boundary.
(b) $E_{1000} = - 13.6/100 0^{2} = - 1.36 \times 1 0^{- 5}$ eV; $E_{1001} = - 13.6/100 1^{2} = - 1.357 \times 1 0^{- 5}$ eV; gap $\approx 2.72 \times 1 0^{- 8}$ eV. Why? High- $n$ levels crowd together (energies $\to 0$ ); the model stays finite and smooth — no divergence.
(c) Small $Z$ : $E_{1} = - 13.6 Z^{2}$ . As $Z \to 0$ , $E_{1} \to 0^{-}$ smoothly and quadratically (e.g. $Z = 0.1 \Rightarrow E_{1} = - 0.136$ eV; $Z = 0.01 \Rightarrow E_{1} = - 1.36 \times 1 0^{- 3}$ eV). Why? No pull $\Rightarrow$ no binding — the electron drifts free. The $Z^{2}$ law degrades gracefully to zero; nothing diverges or goes positive, so the boundary is physically sane. This is the true $Z \to 0$ boundary the title promised.
(d) C⁵⁺: one electron, $Z = 6$ , $E_{1} = - 13.6 \cdot 36 = - 489.6$ eV. Why? One-electron $\Rightarrow$ Bohr exact, and $Z^{2} = 36$ gives a large but finite, physically real binding.

Verify: $E_{\infty} = 0$ ✔; gap $(1000 \to 1001) \approx 2.7 \times 1 0^{- 8}$ eV ✔; $E_{1} (Z = 0.1) = - 0.136$ eV ✔ and $\to 0$ as $Z \to 0$ ✔; C⁵⁺ ground $= - 489.6$ eV ✔ (all limits well-behaved).

Ex 9 — Real-world word problem (Cell C9)

Worked example Which line is the astronomer seeing?

Q: A telescope records a hydrogen emission line at $656.3$ nm. Which Bohr transition is it, and does Bohr correctly predict this line's position (ignore fine structure)?

Forecast: Balmer or Lyman? High or low $n$ ? Why the fork: Lyman lines ( $n \to 1$ ) are ultraviolet (invisible to an optical telescope); Balmer lines ( $n \to 2$ ) are visible red/blue. Seeing a visible line already tells you the series before any arithmetic.

Photon energy $E = \frac{h c}{λ} = \frac{1240 eV\cdotnm}{656.3 nm} = 1.889$ eV. Why this step? $E = h c / λ$ , and we packaged $h c = 1240$ eV·nm in the anchors box above — no faith required, it's just $h \times c$ .
Test the $3 \to 2$ transition (Balmer H-α): $E_{3} - E_{2} = - 13.6 (1/9) - (- 13.6/4) = - 1.511 + 3.4 = 1.889$ eV. Why? Visible hydrogen lines are the Balmer series ( $n \to 2$ ); try the smallest jump first.
Match! It is the $n = 3 \to 2$ H-α line. Why Bohr succeeds here: hydrogen is one-electron, and we only asked for the coarse position — exactly Bohr's domain.

Verify: $E (3 \to 2) = 1.889$ eV $\Rightarrow λ = 1240/1.889 = 656.3$ nm. ✔ Matches the observed H-α wavelength. (High-res would reveal the fine-structure doublet of Ex 6 — that part Bohr misses.)

Ex 10 — Exam twist: spot the false claim (Cell C10)

Worked example Which statement is WRONG?

Q: Exactly one statement is false. Which? (A) Bohr gives Li²⁺'s spectrum exactly. (B) Fine structure in hydrogen is caused by electron–electron repulsion. (C) Bohr obeys the uncertainty principle poorly because it fixes both orbit radius and speed. (D) Neutral helium's ionization energy is not reproduced by naive Bohr.

Forecast: Pick before reading the analysis.

(A) true — Li²⁺ has one electron ( $Z = 3$ ), so Bohr is exact. Why? No $r_{12}$ term.
(B) false — hydrogen has only one electron, so there is no repulsion; fine structure comes from spin-orbit + relativity. This is the trap. See Heisenberg Uncertainty Principle for the next one.
(C) true — a fixed $r$ and $v$ means known $x$ and $p$ together, violating $Δ x Δ p \geq ℏ/2$ .
(D) true — see Ex 3, off by $\sim 30$ eV.

Verify: For Li²⁺, $E_{1} = - 13.6 \cdot 9 = - 122.4$ eV (one-electron, so exact — confirms A). The false statement is (B). ✔

Ex 11 — Ultra-high resolution: hyperfine & the Lamb shift (Cell C11)

Worked example The last two cracks Bohr never sees

Q: Beyond fine structure, two even tinier splittings exist in hydrogen's ground state. (a) The 21-cm hyperfine line comes from the electron spin flipping relative to the proton's spin — its photon has wavelength $λ = 21.1$ cm. What is its energy in eV, and compare to the fine-structure scale? (b) Name the Lamb shift and why Bohr (and even Dirac) miss it.

Forecast: Will the hyperfine energy be bigger or smaller than fine structure ( $\sim 5 \times 1 0^{- 5}$ eV)? Why: fine structure came from $α^{2}$ ; hyperfine involves the proton, which is $\sim 1836 \times$ heavier, so its magnetic moment is $\sim 1836 \times$ weaker — expect much smaller.

(a) Convert the wavelength: $λ = 21.1 cm = 2.11 \times 1 0^{8} nm$ , so $E = \frac{1240}{2.11 \times 1 0 ^{8}} \approx 5.9 \times 1 0^{- 6} eV$ . Why this step? Same $E = h c / λ$ packaged constant — this is the famous 21-cm radio line radio astronomers map the galaxy with.
Compare: $5.9 \times 1 0^{- 6}$ eV vs fine structure $\sim 4.5 \times 1 0^{- 5}$ eV — hyperfine is roughly $10 \times$ smaller. Why? It depends on the proton's magnetic moment (nuclear spin), which is weak because the proton is heavy. Bohr has no spin at all, electron or nuclear, so it predicts no such line. See Electron Spin and Spin-Orbit Coupling.
(b) The Lamb shift slightly separates two levels ( $2 s_{1/2}$ and $2 p_{1/2}$ ) that fine structure predicts to be identical. Its cause is the electron interacting with the flickering vacuum electromagnetic field (quantum electrodynamics). Why Bohr fails: Bohr has no spin, no fields, and certainly no quantum vacuum — it is blind to hyperfine and Lamb effects entirely. These are the finest failures of all.

Verify: 21-cm line energy $\approx 5.9 \times 1 0^{- 6}$ eV ✔; ratio to fine structure $\approx 0.13$ (about one-tenth) ✔. Bohr prediction for both hyperfine and Lamb $= 0$ . ✔

Recall Quick self-test

Bohr is exact for which species? ::: Any one-electron species: H, He⁺, Li²⁺, Be³⁺, C⁵⁺ … Cause of fine structure in hydrogen? ::: Spin–orbit coupling + relativistic correction (NOT repulsion). Fractional size of fine-structure shift? ::: About $α^{2} \approx 5 \times 1 0^{- 5}$ . He⁺ ground energy vs H? ::: $Z^{2} = 4 \times$ deeper: $- 54.4$ eV. Why does D's line differ from H's? ::: Heavier nucleus $\Rightarrow$ larger reduced mass $\Rightarrow$ smaller shift. As $Z \to 0$ , what happens to $E_{1}$ ? ::: $E_{1} = - 13.6 Z^{2} \to 0^{-}$ smoothly — no binding, no divergence. What causes the 21-cm hyperfine line? ::: Electron spin flipping relative to the proton spin. What is $k$ in $k e^{2} / r_{12}$ ? ::: Coulomb's constant $\approx 8.99 \times 1 0^{9} N\cdotm^{2} / C^{2} = 1/4 π ε_{0}$ .