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 22).
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?
Bohr's model fails for multi-electron atoms primarily because it does not account for:
The splitting of spectral lines when an atom is placed in an external electric field is called:
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?
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:
Which of Bohr's postulates is directly contradicted by the Heisenberg uncertainty principle?
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:
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:
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.] ---
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. 22Energy of nth orbit. Negative (bound). Ground state H: -13.6 eV.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E_n | orbit energy | eV |
| Z | nuclear charge | - |
| n | principal | - |
Radius of nth Bohr orbit for hydrogen-like atom of nuclear charge Z.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | principal quantum number | - |
| Z | nuclear charge | - |
| r_n | orbit radius | Å |
Wavelength associated with moving particle of momentum mv.
| Symbol | Quantity | SI Unit |
|---|---|---|
| h | Planck 6.626e-34 | J*s |
| m | mass | kg |
| v | velocity | m/s |
Position and momentum cannot both be known with arbitrary precision.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Δx | position uncertainty | m |
| Δp | momentum uncertainty | kg*m/s |
Spectral wavelengths of hydrogen-like atoms. Lyman (n1=1, UV), Balmer (n1=2, visible), Paschen (n1=3, IR).
| Symbol | Quantity | SI Unit |
|---|---|---|
| lambda | wavelength | m |
| R_H | Rydberg 1.097e7 | 1/m |
| Z | nuclear charge | - |
| n1, n2 | integers, n2>n1 | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student writes Cr as [Ar]3d⁴4s² (expected) instead of actual [Ar]3d⁵4s¹. Same for Cu: actual [Ar]3d¹⁰4s¹ (one e⁻ promoted from 4s to 3d).
Question asks for ground-state electronic configuration of Cr (Z=24) or Cu (Z=29).
Half-filled (d⁵) and fully filled (d¹⁰) configurations have extra stability from exchange energy and symmetry. Cr and Cu adopt these configurations by promoting one 4s electron.
Category: Similar Terms
Student forgets Z² scaling when applying Bohr formulas to He⁺ (Z=2) or Li²⁺ (Z=3).
Question involves hydrogen-like ion (He+, Li2+, etc.).
E_n = -13.6 × Z²/n² eV. r_n = (0.529/Z) × n² Å. He+: 4× more bound than H. Li²⁺: 9× more bound. Always include Z².
Root cause: concept gap
Half-filled (d⁵) and full-filled (d¹⁰) configurations have extra exchange-energy stability. Cr and Cu adopt these by promoting one 4s electron.
Root cause: formula misuse
Always include Z². E_n = -13.6 × Z²/n². For He+: 4× more energetic than H. For Li²⁺: 9×.
9 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
From the following pairs of ions which one is not an iso-electronic pair? Fe2+, Mn2+
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
misses cr cu anomaly
Writes 3d⁴4s² instead of 3d⁵4s¹
forgets z squared
Drops Z² for hydrogen-like
uses h instead of h over 4pi
Drops 4π factor
uses wrong l range
Uses l ≤ n instead of l < n
Test yourself on this topic with real past-paper questions:
Practice this topic →Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.