(1) Electrons orbit nucleus in stable circular orbits without radiating. (2) Angular momentum quantised: L = mvr = nℏ (n = 1, 2, 3, ...). (3) Energy emitted/absorbed when electron jumps between orbits: hν = E_2 - E_1.
-- NCERT Class 12 Physics, Ch. 12, p. 293Bohr Model
8 MCQs2 revision cards9-step worked example
Source: NCERT Atoms and NucleiPYQ coverage: NEET 2023, 2025, 2026Official key: NTA-verifiedLast updated: 25 Sep 2026
Bohr Model, explained for NEET
The scaling in Bohr's model is where marks are lost. Radius goes as n², energy as 1/n² with a minus sign. A student who remembers "n²" for one and applies "n" to the other gets a plausible wrong number, and the paper always supplies that number as an option. Write both exponents down before substituting.
Bohr's three postulates (NCERT Class 12 Physics, Chapter 12, page 293) repair Rutherford's fatal defect: an orbiting electron should radiate continuously and spiral in. Bohr's answer was that only certain orbits are allowed, and in them the electron does not radiate.
- Stationary orbits. The electron moves in circular orbits without emitting energy. The Coulomb attraction supplies the centripetal force.
- Angular momentum quantisation. Only orbits with L = mvr = nh/2π are permitted, n = 1, 2, 3, …
- Frequency condition. Radiation is emitted or absorbed only when the electron jumps between two stationary orbits, with hν equal to the energy difference.
Postulate 2 is the whole model in one line. Feed it into the force balance and the allowed radii and energies drop out (Chapter 12, page 299):
r_n = n²a₀, with a₀ = 0.529 Å, and E_n = −13.6/n² eV.
Read the signs. Energy is negative because the electron is bound; the zero is set at infinite separation. So E₁ = −13.6 eV is the lowest energy, not the highest, and removing the electron from n = 1 costs 13.6 eV. As n grows, levels crowd towards zero, so the gap between consecutive levels shrinks — a fact NEET tests directly.
Watch out for scope. The model is exact only for one-electron systems. Extending it to helium, or to any atom with two or more electrons, is outside what it can do.
Can you answer these Bohr Model MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In Bohr's second postulate, the quantity that is restricted to integer multiples of h/2π is the
Show answer and why every option is right or wrong
Answer: C. The quantisation condition is L = mvr = nh/2π — angular momentum, stated as the second postulate in NCERT Class 12 Physics, Chapter 12, page 293.
Why A is wrong: A is wrong because the radius is not itself postulated to be quantised; r_n = n²a₀ is a consequence derived from the angular-momentum condition plus the force balance, and its steps are n², not integer multiples of a fixed unit.
Why B is wrong: B is wrong because kinetic energy follows from the postulate rather than being assumed; it scales as 1/n², not as an integer multiple of anything.
Why D is wrong: D is wrong because the revolution frequency is a derived quantity that decreases as 1/n³; no postulate constrains it to integer steps.
MCQ 2Easy RecallPractice
The defect of Rutherford's atom that Bohr's first postulate was introduced to remove is that a classical orbiting electron would
Show answer and why every option is right or wrong
Answer: A. An accelerating charge radiates classically, so the electron would lose energy continuously and collapse into the nucleus; Bohr's first postulate asserts that in certain orbits the electron does not radiate (NCERT Class 12 Physics, Chapter 12, page 293).
Why B is wrong: B is wrong because the electron is negative and the nucleus positive, so the Coulomb force is attractive — that attraction is what supplies the centripetal force in the model.
Why C is wrong: C is wrong because classical mechanics places no restriction on orbital size at all; the difficulty is energy loss by radiation, not a radius that is too large.
Why D is wrong: D is wrong because the ground-state angular momentum in Bohr's model is h/2π, not zero, and in any case a zero value would be a feature of the new model rather than a defect of the old one.
MCQ 3Easy RecallPractice
According to Bohr's third postulate, a photon is emitted by a hydrogen atom when the electron
Show answer and why every option is right or wrong
Answer: B. The frequency condition states that radiation is emitted only during a transition between two stationary orbits, with hν equal to the energy difference (NCERT Class 12 Physics, Chapter 12, page 293).
Why A is wrong: A is wrong because the first postulate says precisely the opposite: while the electron remains in a stationary orbit it does not radiate, however fast it moves.
Why C is wrong: C is wrong because complete removal is ionisation, which requires energy to be supplied to the atom, not released by it.
Why D is wrong: D is wrong because going from a₀ to 4a₀ is a move from n = 1 to n = 2, an upward jump; that transition absorbs a photon rather than emitting one.
MCQ 4Direct ApplicationPractice
For the hydrogen atom, the radius of the n = 3 orbit is
Show answer and why every option is right or wrong
Answer: B. r_n = n²a₀, so r₃ = 3² a₀ = 9a₀ ≈ 4.76 Å (NCERT Class 12 Physics, Chapter 12, page 300).
Why A is wrong: A is wrong because it applies a linear scaling in n; the radius goes as n², which is the most frequently penalised slip in this topic.
Why C is wrong: C is wrong because it inverts the dependence, borrowing the 1/n² behaviour that belongs to the energy and applying it to the radius. Radius grows with n; energy magnitude falls.
Why D is wrong: D is wrong because it uses n³. No Bohr quantity scales as n³ except the period of revolution, which is not what was asked.
MCQ 5Direct ApplicationPractice
The energy of the n = 2 level of the hydrogen atom is
Show answer and why every option is right or wrong
Answer: C. E_n = −13.6/n² eV, so E₂ = −13.6/4 = −3.40 eV (NCERT Class 12 Physics, Chapter 12, page 300).
Why A is wrong: A is wrong because it divides by n rather than n², giving −13.6/2. The exponent on n is 2 in the energy formula as well as in the radius formula — only the sign of the exponent differs.
Why B is wrong: B is wrong because it multiplies by n instead of dividing by n², giving −13.6 × 2, so it makes the n = 2 electron more tightly bound than the ground state. Higher n must always mean energy closer to zero.
Why D is wrong: D is wrong because the sign is positive. Bound-state energies in this convention are negative, with zero taken at infinite separation; a positive value would describe a free electron.
MCQ 6Direct ApplicationPractice
The energy needed to remove the electron completely from a hydrogen atom in its ground state is
Show answer and why every option is right or wrong
Answer: C. Ionisation from n = 1 means going from E₁ = −13.6 eV to E = 0 at n → ∞, so the energy required is 13.6 eV (NCERT Class 12 Physics, Chapter 12, page 300).
Why A is wrong: A is wrong because 3.40 eV is |E₂|, the energy to ionise from the first excited state, not from the ground state named in the stem.
Why B is wrong: B is wrong because 10.2 eV is the n = 1 to n = 2 excitation energy, −3.40 − (−13.6). Excitation lifts the electron to a higher bound level; ionisation takes it to zero energy.
Why D is wrong: D is wrong because 1.51 eV is |E₃|, the ionisation energy from the second excited state.
MCQ 7CalculationPractice
In a hydrogen atom the electron moves from the n = 1 orbit to the n = 4 orbit. The radius increases by a factor X and the magnitude of the energy decreases by a factor Y. The values of X and Y are
Show answer and why every option is right or wrong
Answer: D. r ∝ n² gives r₄/r₁ = 16, and |E| ∝ 1/n² gives |E₁|/|E₄| = 16, so both factors are 16 (NCERT Class 12 Physics, Chapter 12, page 300).
Why A is wrong: A is wrong because it uses a linear n for both quantities; both carry an exponent of 2, so the factor is 4² = 16, not 4.
Why B is wrong: B is wrong because it scales the radius correctly as n² but treats the energy as linear in n. The energy magnitude falls by 4² = 16.
Why C is wrong: C is wrong because the two errors are interchanged: the radius is scaled linearly while the energy is scaled by n². The radius factor is 16.
MCQ 8Concept TrapPractice
A student applies E_n = −13.6/n² eV to a singly ionised helium atom and then to a neutral helium atom. The result is
Show answer and why every option is right or wrong
Answer: D. The Bohr treatment is built on one electron orbiting a nucleus, so it extends to hydrogen-like ions such as He⁺ once the nuclear charge is accounted for, but it cannot handle the electron–electron interaction in neutral helium (NCERT Class 12 Physics, Chapter 12, page 293).
Why A is wrong: A is wrong because the postulates were framed for a single electron in a central Coulomb field; a second electron introduces a repulsion the model has no way to include.
Why B is wrong: B is wrong because the formula is not confined to the ground state — it gives every level n = 1, 2, 3, … of hydrogen. The real limitation is the electron count, not the value of n.
Why C is wrong: C is wrong because the nuclear charge alone is not the obstacle. A one-electron system with Z = 2 is exactly the case the model still handles.
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Bohr Model: quick recall before you leave
How do you solve a Bohr Model question? A worked example
- 1
Given.
Hydrogen atom, electron in the orbit with n = 3 (exact, a counting integer).
Bohr radius a₀ = 5.29 × 10⁻¹¹ m.
Ground-state energy constant = 13.6 eV. - 2
Required.
The radius of the orbit, and the energy required to ionise the atom from this orbit.
- 3
Concept.
Bohr's angular-momentum quantisation fixes the allowed orbits. The radius rises as n² and the binding energy magnitude falls as 1/n². Ionisation from level n means raising the electron from E_n to zero energy, so the energy required is |E_n|.
- 4
Formula.
r_n = n² a₀
E_n = −13.6/n² eV
Ionisation energy from level n = 0 − E_n = |E_n|. - 5
Substitution.
r₃ = 3² × (5.29 × 10⁻¹¹ m)
E₃ = −13.6/3² eV - 6
Calculation.
r₃ = 9 × 5.29 × 10⁻¹¹ m = 4.76 × 10⁻¹⁰ m
E₃ = −13.6/9 eV = −1.51 eV
Ionisation energy = 1.51 eV
The quantum number n = 3 and the factor 9 = 3² are exact counting integers, so they impose no limit on significant figures. The answers carry three significant figures, inherited from a₀ = 5.29 × 10⁻¹¹ m and from 13.6 eV. - 7
Final answer.
r₃ = 4.76 × 10⁻¹⁰ m, and 1.51 eV is required to ionise the atom from n = 3.
- 8
Common trap.
Mixing the two exponents. The frequent wrong pair is r₃ = 3a₀ = 1.59 × 10⁻¹⁰ m (linear in n) together with E₃ = −13.6/9 (correct), or the reverse — r₃ = 9a₀ paired with E₃ = −13.6/3 = −4.53 eV. Both wrong values appear as options. Write "r ∝ n², E ∝ 1/n²" in the margin before substituting anything.
A second slip at step 7: quoting the ionisation energy as −1.51 eV. The level energy is negative; the energy you must supply is positive. - 9
Similar NEET-style question.
In a hydrogen atom, the electron is in the orbit of radius 2.12 × 10⁻¹⁰ m. Taking a₀ = 5.29 × 10⁻¹¹ m, the energy of the electron in this orbit is
(a) −13.6 eV (b) −6.80 eV (c) −3.40 eV (d) −1.51 eV
Route: r/a₀ = 4, so n² = 4 and n = 2; then E₂ = −13.6/4 = −3.40 eV. Answer (c). The step that catches people is taking n = 4 from the ratio instead of n = 2.
What to remember before solving Bohr Model questions
Formula
Bohr orbit radius and energy
r_n = n² a_0 (a_0 = 0.53 Å, Bohr radius). E_n = -13.6/n² eV. n = 1 (ground): E = -13.6 eV. n = ∞: E = 0 (ionised).
-- NCERT Class 12 Physics, Ch. 12, p. 295Which Bohr Model formulas do you need for NEET?
2 formulas — click to collapse
Bohr energy levels
Energy of n-th Bohr orbit in hydrogen. n=1 ground state -13.6 eV; ionisation energy = 13.6 eV.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E_n | energy of n-th level | eV |
| n | principal quantum number | - |
Valid when
- Hydrogen atom
- Non-relativistic
Bohr orbit radius
Allowed radii in Bohr model of hydrogen atom. Quantised by integer n.
| Symbol | Quantity | SI Unit |
|---|---|---|
| r_n | n-th orbit radius | m |
| a0 | Bohr radius 5.29e-11 | m |
| n | quantum number | - |
Valid when
- Hydrogen-like atom (single electron)
- Non-relativistic
More in Atoms and Nuclei: 3 exam traps and mistakes · 4 formulas · 4 question patterns from its other lessons.
Bohr Model questions from past NEET papers
3 questions from NEET 2023, 2025, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
14
23
32
41
NTA Answer: Option 4(final)
NEET 2025
1r ∝ n4/3; ν ∝ n–1/3
2r ∝ n1/3; ν ∝ n1/3
3r ∝ n1/3; ν ∝ n2/3
4r ∝ n2/3; ν ∝ n1/3
NTA Answer: Option 4(final)
NEET 2023
11.06 Å
21.59 Å
34.77 Å
40.53 Å
NTA Answer: Option 3(final)
How does NEET ask about Bohr Model?
1 recurring pattern from past papers — click to collapse
Bohr orbit: r_n = n² a_0; E_n = -13.6/n² eV. Compute for given n.
Direct ApplicationMedium
Common distractors
uses 1 n not 1 n squared
Misuses scaling
Sources
NCERT refs: Class 12 Physics Chapter 12, p.293
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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