Bohr's model works remarkably well for hydrogen — it predicts line spectra, quantised energy levels, and the Rydberg formula — but it hits a hard wall the moment you move beyond one-electron systems. NEET expects you to know exactly where and why it fails.
Limitation 1 — Multi-electron atoms. Bohr's model cannot explain the spectra of atoms with two or more electrons. It treats the electron–nucleus interaction in isolation and has no way to account for electron–electron repulsion. Helium's spectrum, for instance, is beyond its reach (NCERT Class 11 Chemistry Chapter 2, page 50).
Limitation 2 — Fine structure. Even hydrogen's spectral lines, under high-resolution spectroscopy, split into closely spaced components (fine structure). Bohr's model predicts single lines and cannot account for these splittings, which arise from relativistic effects and spin–orbit coupling.
Limitation 3 — Zeeman and Stark effects. When atoms are placed in external magnetic (Zeeman) or electric (Stark) fields, spectral lines split further. Bohr's framework has no mechanism to predict these field-induced splittings.
Limitation 4 — Flat circular orbits. Bohr assumed electrons travel in fixed circular orbits with definite radii and velocities. This contradicts the Heisenberg uncertainty principle (Δx·Δp ≥ h/4π), which forbids simultaneous precise knowledge of position and momentum. The modern quantum mechanical model replaces orbits with probability-based orbitals.
Limitation 5 — No chemical bonding explanation. Bohr's model cannot explain how atoms bond to form molecules — it lacks the framework of orbital overlap and electron sharing.
Watch-out for NEET: A common confusion is applying Bohr's energy formula E_n = −13.6 Z²/n² eV to multi-electron atoms. This formula is valid only for hydrogen-like (one-electron) species: H, He⁺, Li²⁺, Be³⁺. The moment a second electron is present, the model breaks down — that is precisely the limitation NCERT emphasises.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following is NOT a limitation of Bohr's model of the atom?
Answer: A. Bohr's model successfully explains the line spectrum of hydrogen — that is its central achievement, not a limitation. The Rydberg formula derived from Bohr's postulates accurately predicts all hydrogen spectral series (NCERT Class 11 Chemistry Chapter 2, page 50).
Why B is wrong: B describes a genuine limitation — Bohr's model has no mechanism to predict splitting of spectral lines in an external magnetic field.
Why C is wrong: C describes a genuine limitation — Bohr's model fails for any atom with more than one electron, including helium.
Why D is wrong: D describes a genuine limitation — high-resolution spectroscopy reveals fine structure that Bohr's single-line prediction cannot account for.
Bohr's model fails for multi-electron atoms primarily because it does not account for:
Answer: C. The Bohr model treats the system as a single electron interacting with the nucleus. In multi-electron atoms, electron–electron repulsions significantly alter energy levels, and the model has no way to incorporate these interactions (NCERT Class 11 Chemistry Chapter 2, page 50).
Why A is wrong: A is incorrect — Bohr's model does include nuclear charge (Z appears in the energy and radius formulas). The failure is not about ignoring Z but about ignoring other electrons.
Why B is wrong: B is incorrect — quantisation of angular momentum (mvr = nh/2π) is one of Bohr's own postulates; it is built into the model, not missing from it.
Why D is wrong: D is incorrect — Coulombic attraction between electron and nucleus is the central force in Bohr's model. The model handles this well for one-electron systems.
The splitting of spectral lines when an atom is placed in an external electric field is called:
Answer: D. The Stark effect is the splitting of spectral lines under an external electric field. Bohr's model cannot explain this splitting — it is cited as one of its limitations (NCERT Class 11 Chemistry Chapter 2, page 50).
Why A is wrong: A is incorrect — the Zeeman effect is splitting in a magnetic field, not an electric field. Both are limitations of Bohr's model, but they are distinct phenomena.
Why B is wrong: B is incorrect — the Compton effect describes the increase in wavelength of X-rays scattered by electrons, not spectral line splitting in an electric field.
Why C is wrong: C is incorrect — the photoelectric effect describes ejection of electrons from a metal surface by light, unrelated to spectral line splitting.
Bohr's energy formula E_n = −13.6 Z²/n² eV is applied to calculate the ground-state energy of a species. For which of the following is this application valid?
Answer: B. Be³⁺ has lost three electrons and retains only one — it is a hydrogen-like ion. The Bohr energy formula is valid only for one-electron (hydrogen-like) species. E₁ = −13.6 × 4²/1² = −217.6 eV for Be³⁺ (NCERT Class 11 Chemistry Chapter 2, page 50).
Why A is wrong: A is incorrect — neutral helium has 2 electrons. Bohr's model cannot handle electron–electron repulsion, so the formula does not apply. The hydrogen-like helium species would be He⁺, not He.
Why C is wrong: C is incorrect — Li⁺ retains 2 electrons after losing one from neutral Li (Z = 3). With two electrons, electron–electron repulsion invalidates Bohr's formula. The hydrogen-like lithium species is Li²⁺.
Why D is wrong: D is incorrect — neutral lithium has 3 electrons. Bohr's model breaks down for any multi-electron atom. Only Li²⁺ (one electron) would be hydrogen-like.
A student uses Bohr's model to predict that the 3p → 2s transition in sodium should produce a single spectral line at a specific wavelength. In reality, the observed spectrum shows:
Answer: C. Sodium's famous D-line is actually a doublet (589.0 nm and 589.6 nm) arising from spin–orbit coupling — a fine-structure effect. Bohr's model predicts only single lines and cannot account for this splitting (NCERT Class 11 Chemistry Chapter 2, page 50).
Why A is wrong: A is incorrect — the observation of fine structure (doublets, triplets) in multi-electron atoms is precisely the limitation Bohr's model cannot explain. A single line would mean the model works, which it does not for sodium.
Why B is wrong: B is incorrect — sodium absolutely emits light. The characteristic yellow sodium D-line is one of the most well-known emission features in spectroscopy.
Why D is wrong: D is incorrect — while Bohr's model may not predict the correct wavelength for sodium (a multi-electron atom), the key limitation demonstrated here is the doublet splitting, not merely a wavelength error.
Which of Bohr's postulates is directly contradicted by the Heisenberg uncertainty principle?
Answer: A. Bohr's postulate of definite circular orbits assigns simultaneous precise values to both position (radius r) and velocity (momentum mv). The Heisenberg uncertainty principle (Δx·Δp ≥ h/4π) forbids this — if position is precisely known, momentum is fundamentally uncertain, and vice versa (NCERT Class 11 Chemistry Chapter 2, page 50).
Why B is wrong: B is incorrect — angular momentum quantisation is actually supported by quantum mechanics (though the quantum mechanical expression differs from Bohr's: L = √(l(l+1))ℏ rather than L = nℏ). The uncertainty principle does not directly contradict quantisation of angular momentum.
Why C is wrong: C is incorrect — the postulate about radiation emission during transitions is consistent with quantum theory. Atoms do emit photons during energy-level transitions; quantum mechanics refines the mechanism but does not contradict the core idea.
Why D is wrong: D is incorrect — discrete energy quanta (E = hν) are fundamental to quantum mechanics. Planck's quantum hypothesis and the Bohr frequency condition are upheld, not contradicted, by modern theory.
A researcher argues: "Since Bohr's model gives correct energy levels for He⁺ (Z = 2), it should also work for neutral He (Z = 2)." This reasoning is flawed because:
Answer: D. He⁺ has only one electron, making it hydrogen-like — the Bohr formula E_n = −13.6 Z²/n² applies correctly. Neutral He has two electrons, introducing electron–electron repulsion that Bohr's model cannot account for. Same Z, fundamentally different physics (NCERT Class 11 Chemistry Chapter 2, page 50).
Why A is wrong: A is incorrect — both He⁺ and neutral He have the same nuclear charge Z = 2. The difference is the number of electrons (1 vs 2), not the nuclear charge.
Why B is wrong: B is incorrect — the Bohr model works for any hydrogen-like species regardless of Z: H (Z = 1), He⁺ (Z = 2), Li²⁺ (Z = 3), etc. There is no minimum Z requirement.
Why C is wrong: C is incorrect — the Bohr model works well for He⁺ because it is a one-electron species. The energy levels predicted by E_n = −13.6 × 4/n² eV match experimental values.
A student calculates the ground-state energy of He⁺ using the Bohr formula as E₁ = −13.6 eV (using Z = 1 by mistake). The correct value differs from this by a factor of:
Answer: B. For He⁺ (Z = 2), E₁ = −13.6 × Z²/n² = −13.6 × 4/1 = −54.4 eV. The student used Z = 1, getting −13.6 eV. The ratio is 54.4/13.6 = 4. The student's error is forgetting the Z² factor — for He⁺, Z² = 4 (trap: dropping Z² when applying Bohr formulas to hydrogen-like ions).
Why A is wrong: A is incorrect — the scaling factor is Z² = 2² = 4, not Z = 2. The energy formula has Z squared in the numerator, so forgetting Z gives an error of Z² = 4, not just Z = 2.
Why C is wrong: C is incorrect — 8 = 2³, but no cubic power of Z appears in the Bohr energy formula. The Z dependence is strictly Z², giving a factor of 4.
Why D is wrong: D is incorrect — 16 = 2⁴. This would apply if both Z² and n² were wrong, but the student only forgot Z², not n. With n = 1 correct, the error factor is Z² = 4.
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Given
• Species: He⁺ (Z = 2, one electron — hydrogen-like)• Electron transitions from n = 3 to n = 1
Required
Energy of the photon emitted during this transition.
Concept
For hydrogen-like species, the energy of an electron in orbit n is E_n = −13.6 × Z²/n² eV. The photon energy equals the difference |E_final − E_initial|.
Formula
ΔE = 13.6 × Z² × (1/n₁² − 1/n₂²) eV, where n₁ < n₂.
Here n₁ = 1 (final), n₂ = 3 (initial).
Substitution
ΔE = 13.6 × (2)² × (1/1² − 1/3²) eV
ΔE = 13.6 × 4 × (1 − 1/9) eV
Calculation
1 − 1/9 = 8/9
ΔE = 13.6 × 4 × 8/9
ΔE = 13.6 × 32/9
ΔE = 435.2/9
ΔE = 48.36 eV
Note on exact values: Z = 2 is an exact integer (nuclear charge is a counting number), and n₁ = 1, n₂ = 3 are exact quantum numbers. The constant 13.6 eV limits the significant figures.
Final answer
The photon emitted has energy 48.36 eV (3 significant figures, following the precision of 13.6).
Compare: for hydrogen (Z = 1), the same 3 → 1 transition gives 13.6 × 1 × 8/9 = 12.09 eV. He⁺ gives exactly 4× more — this is the Z² factor at work.
Common trap
Forgetting Z² for He⁺. A student who uses Z = 1 (or drops Z entirely) would get 12.09 eV — the hydrogen answer, not the He⁺ answer. On NEET, this wrong value is a standard distractor. Always check: is the species hydrogen-like with Z ≠ 1? If yes, include Z².
Similar NEET-style question
Calculate the energy required to remove the electron from the ground state of Li²⁺ (Z = 3). [Answer: 13.6 × 9/1 = 122.4 eV — the Z² = 9 factor makes lithium(2+) nine times more tightly bound than hydrogen.]
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Could not explain: spectra of multi-electron atoms; finer details (Zeeman, Stark effects); chemical bonding. Failed because it treated electron as a particle in defined orbit.
-- NCERT Class 11 Chemistry, Ch. 2, p. 49More in Structure of Atom: 4 exam traps and mistakes · 5 formulas · 3 question patterns from its other lessons.
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