1) Electron orbits nucleus in stationary states without radiating. 2) Angular momentum mvr = nh/(2π), n = 1, 2, 3,... 3) Electron transitions: ΔE = hν.
-- NCERT Class 11 Chemistry, Ch. 2, p. 46Bohr Model Hydrogen
8 MCQs2 revision cards9-step worked example
Source: NCERT Structure of AtomPYQ coverage: NEET 2022, 2024, 2025Official key: NTA-verifiedLast updated: 26 Sep 2026
Bohr Model Hydrogen, explained for NEET
The Z² trap that costs you marks on hydrogen-like ion questions.
Bohr's model applies to any one-electron (hydrogen-like) system: H, He⁺, Li²⁺, Be³⁺. The two core formulas you need are:
- Energy: E_n = −13.6 × Z²/n² eV
- Radius: r_n = 0.529 × n²/Z Å
Both come from NCERT Class 11 Chemistry Chapter 2 (pages 46–48). The model treats the electron as circling the nucleus in fixed orbits where angular momentum is quantised. The key outputs are orbit energy and orbit radius — both depend on the principal quantum number n and the nuclear charge Z.
Where aspirants lose marks: The high-frequency trap is forgetting Z² when the question shifts from hydrogen to He⁺ or Li²⁺. For hydrogen (Z = 1), Z² = 1 and the factor is invisible. The moment Z ≠ 1, dropping it gives the wrong answer by a factor of Z².
Concrete check: He⁺ ground-state energy = −13.6 × 4/1 = −54.4 eV, not −13.6 eV. If your answer for He⁺ equals the hydrogen value, you forgot Z².
Radius scales oppositely. Energy goes as Z² (more bound), but radius goes as 1/Z (orbits shrink). He⁺ ground-state radius = 0.529/2 = 0.2645 Å — half the hydrogen value, not the same.
Watch-out: When a question says "the second orbit of Li²⁺," substitute n = 2 and Z = 3 into both formulas. Don't default to hydrogen values and then try to "correct" afterward — plug in Z from the start.
Can you answer these Bohr Model Hydrogen MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
The ground-state energy of the hydrogen atom in the Bohr model is −13.6 eV. What is the energy of the electron in the second orbit (n = 2) of hydrogen?
Show answer and why every option is right or wrong
Answer: C. E₂ = −13.6 × 1²/2² = −13.6/4 = −3.4 eV. Direct substitution into the Bohr energy formula (NCERT Class 11 Chemistry Chapter 2, page 47).
Why A is wrong: A is wrong because −13.6 eV is the ground-state (n = 1) energy, not n = 2. This confuses the orbit number.
Why B is wrong: B is wrong because −6.8 eV results from dividing by 2 instead of 2² = 4 — a common arithmetic slip with the n² denominator.
Why D is wrong: D is wrong because −1.51 eV corresponds to n = 3 (−13.6/9). This substitutes the wrong principal quantum number.
MCQ 2Direct ApplicationPractice
The ground-state energy of He⁺ (Z = 2) in the Bohr model is:
Show answer and why every option is right or wrong
Answer: B. E₁ = −13.6 × Z²/n² = −13.6 × 4/1 = −54.4 eV. The Z² factor is essential for hydrogen-like ions (NCERT Class 11 Chemistry Chapter 2, page 48).
Why A is wrong: A is wrong because −13.6 eV is the hydrogen ground-state energy — this answer drops the Z² factor entirely (trap: forgetting Z² for hydrogen-like ions).
Why C is wrong: C is wrong because −27.2 eV results from using Z instead of Z² (multiplying by 2 rather than 4). The formula requires Z squared.
Why D is wrong: D is wrong because −6.8 eV corresponds to hydrogen's n = 2 energy. This confuses the orbit level and ignores the Z² scaling.
MCQ 3Easy RecallPractice
Which of the following is a postulate of Bohr's model of the atom?
Show answer and why every option is right or wrong
Answer: A. Bohr's quantisation condition states mvr = nh/2π, where n = 1, 2, 3, … This is a core postulate (NCERT Class 11 Chemistry Chapter 2, page 46).
Why B is wrong: B is wrong because Bohr's model specifically postulates that electrons do NOT radiate energy in stationary orbits — this was the key departure from classical electrodynamics.
Why C is wrong: C is wrong because Bohr's model restricts electrons to specific quantised orbits with fixed energies, not arbitrary ones.
Why D is wrong: D is wrong because energy becomes more negative (more bound) as n decreases. The electron is more tightly bound at lower n, meaning lower (more negative) energy, not higher.
MCQ 4Direct ApplicationPractice
The radius of the first Bohr orbit of hydrogen is 0.529 Å. The radius of the first orbit of Li²⁺ (Z = 3) is:
Show answer and why every option is right or wrong
Answer: D. r₁ = 0.529 × 1²/3 = 0.529/3 = 0.1763 Å. The radius scales as n²/Z, so higher Z shrinks the orbit (NCERT Class 11 Chemistry Chapter 2, page 48).
Why A is wrong: A is wrong because 0.529 Å is the hydrogen value (Z = 1). Using this for Li²⁺ ignores the 1/Z scaling — the classic trap of forgetting Z for hydrogen-like ions.
Why B is wrong: B is wrong because 1.587 Å = 0.529 × 3, which multiplies by Z instead of dividing. The radius formula has Z in the denominator, not the numerator.
Why C is wrong: C is wrong because 4.761 Å = 0.529 × 9, which uses Z² in the numerator. The radius depends on n²/Z, not n² × Z².
MCQ 5Easy RecallPractice
In the Bohr model, the energy of the electron in the nth orbit of a hydrogen-like atom is proportional to:
Show answer and why every option is right or wrong
Answer: A. E_n = −13.6 × Z²/n² eV, so the energy is proportional to Z²/n² (NCERT Class 11 Chemistry Chapter 2, page 48).
Why B is wrong: B is wrong because the energy depends on Z squared, not Z to the first power. Using Z instead of Z² underestimates the nuclear-charge effect.
Why C is wrong: C is wrong because n²/Z² is the inverse of the correct proportionality. This would imply energy increases with n, contradicting the bound-state nature of the model.
Why D is wrong: D is wrong because the denominator must be n², not n. The energy levels scale as 1/n², producing the characteristic convergence of energy levels at high n.
MCQ 6CalculationPractice
An electron in He⁺ transitions from n = 2 to n = 1. The energy released is:
Show answer and why every option is right or wrong
Answer: D. E₁ = −13.6 × 4/1 = −54.4 eV; E₂ = −13.6 × 4/4 = −13.6 eV. ΔE = E₁ − E₂ = −54.4 − (−13.6) = −40.8 eV. Energy released = 40.8 eV. Two steps: compute each level with Z², then take the difference.
Why A is wrong: A is wrong because 10.2 eV is the hydrogen (Z = 1) transition energy for n = 2 → 1 (i.e., 13.6 × (1 − 1/4) = 10.2 eV). This drops the Z² = 4 factor entirely (trap: forgetting Z² for hydrogen-like ions).
Why B is wrong: B is wrong because 54.4 eV is the ground-state energy of He⁺, not the n = 2 → 1 transition energy. The transition releases |E₁ − E₂|, not |E₁|.
Why C is wrong: C is wrong because 13.6 eV is the ground-state energy of hydrogen, not the transition energy. This confuses a single energy level with an energy difference.
MCQ 7CalculationPractice
The ratio of the radii of the second orbit of He⁺ to the third orbit of hydrogen is:
Show answer and why every option is right or wrong
Answer: A. A is correct. r = a₀n²/Z. r(He⁺, n=2) = 0.529 × 4/2 = 1.058 Å. r(H, n=3) = 0.529 × 9/1 = 4.761 Å. Ratio = 1.058/4.761 = 0.222 = 2:9. Two steps: compute each radius with n²/Z, then divide — and note that He⁺ carries Z = 2, which is what halves 4:9 to 2:9.
Why B is wrong: B is wrong because 2:3 comes from simply taking the ratio of the quantum numbers (n₁/n₂ = 2/3) without applying the full n²/Z formula for each species.
Why C is wrong: C is wrong because 4:9 is the ratio of n² alone (4 for He⁺, 9 for hydrogen), which leaves the nuclear charge out. He⁺ has Z = 2, so its radius is 4/2 = 2 in units of a₀, not 4 — and 4:9 becomes 2:9.
Why D is wrong: D is wrong because 4:3 results from computing n² for He⁺ correctly (4) but using n instead of n² for hydrogen (3 instead of 9).
MCQ 8Easy RecallPractice
According to Bohr's model, an electron in a stationary orbit:
Show answer and why every option is right or wrong
Answer: B. Bohr's second postulate states that an electron in a stationary orbit has a definite energy and does not radiate. Radiation occurs only during transitions between orbits (NCERT Class 11 Chemistry Chapter 2, page 46).
Why A is wrong: A is wrong because continuous emission was the classical prediction that Bohr's model was designed to overcome. Stationary orbits are defined by the absence of radiation.
Why C is wrong: C is wrong because Bohr's quantisation condition restricts angular momentum to integral multiples of h/2π. Arbitrary values are not allowed.
Why D is wrong: D is wrong because spiraling inward is the classical expectation for an accelerating charged particle. Bohr's model avoids this by postulating stable, non-radiating orbits.
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Bohr Model Hydrogen: quick recall before you leave
How do you solve a Bohr Model Hydrogen question? A worked example
Pattern: Bohr energy transition for hydrogen-like ion (P.CHE.U02.BOHR_ENERGY_TRANSITION, observed 2021/2023/2025).
- 1
Given
A Li²⁺ ion (Z = 3) has its electron in the n = 3 orbit. Calculate the energy released when the electron transitions to the ground state (n = 1).
- 2
Required
Energy released during the n = 3 → n = 1 transition of Li²⁺.
- 3
Concept
Bohr model: the energy of each orbit in a hydrogen-like atom depends on both the principal quantum number n and the nuclear charge Z. The photon energy emitted equals the difference between the initial and final orbit energies.
- 4
Formula
E_n = −13.6 × Z²/n² eV
ΔE = E_final − E_initial (the magnitude gives the energy released) - 5
Substitution
E₃ = −13.6 × 3²/3² = −13.6 × 9/9 = −13.6 eV
E₁ = −13.6 × 3²/1² = −13.6 × 9/1 = −122.4 eV - 6
Calculation
ΔE = E₁ − E₃ = −122.4 − (−13.6) = −108.8 eV
Note on exact values: Z = 3, n = 1, and n = 3 are exact integers (quantum numbers). The constant 13.6 eV is a defined numerical value of the Bohr energy for hydrogen. These do not introduce rounding uncertainty. - 7
Final answer
Energy released = 108.8 eV.
The answer carries 4 significant figures, matching the precision of the 13.6 eV constant used. - 8
Common trap
If you forget Z² and use Z = 1 (hydrogen values): E₃ = −1.51 eV, E₁ = −13.6 eV, ΔE = 12.09 eV. That is 9× too small. The factor of Z² = 9 for Li²⁺ scales every energy level. Check: if your Li²⁺ answer equals a hydrogen answer, you dropped Z².
- 9
Similar NEET-style question
"Calculate the energy required to remove the electron from the second orbit of He⁺ (Z = 2) to infinity." (Answer: E₂ = −13.6 × 4/4 = −13.6 eV; ionisation from n = 2 requires 13.6 eV.)
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What to remember before solving Bohr Model Hydrogen questions
Formula
Bohr radius and energy
Hydrogen-like atoms: r_n = (0.529 × n²)/Z Å; E_n = -13.6 × Z²/n² eV. For H: r_1 = 0.529 Å, E_1 = -13.6 eV.
-- NCERT Class 11 Chemistry, Ch. 2, p. 48Which Bohr Model Hydrogen formulas do you need for NEET?
Bohr energy (hydrogen-like)
Energy 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 | - |
Valid when
- Hydrogen-like atom
- Non-relativistic
Bohr radius (hydrogen-like)
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 |
Valid when
- Hydrogen-like (one-electron) atom
- Non-relativistic
Where do students lose marks on Bohr Model Hydrogen?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student forgets Z² scaling when applying Bohr formulas to He⁺ (Z=2) or Li²⁺ (Z=3).
When it triggers
Question involves hydrogen-like ion (He+, Li2+, etc.).
How to avoid
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: formula misuse
Correction
Always include Z². E_n = -13.6 × Z²/n². For He+: 4× more energetic than H. For Li²⁺: 9×.
More in Structure of Atom: 2 exam traps and mistakes · 3 formulas · 2 question patterns from its other lessons.
Bohr Model Hydrogen questions from past NEET papers
3 questions from NEET 2022, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
Energy and radius of first Bohr orbit of He+ and Li2+ are [Given RH = 2.18 × 10⁻¹⁸ J, a0 = 52.9 pm]
1En(Li 2+) = –8.72 × 10⁻¹⁶ J; rn(Li2+) = 17.6 pm En(He+ ) = –19.62 × 10⁻¹⁶ J; rn(He+ ) = 17.6 pm
2En(Li 2+) = –19.62 × 10⁻¹⁸ J; rn(Li2+) = 17.6 pm En(He+ ) = –8.72 × 10⁻¹⁸ J; rn(He+ ) = 26.4 pm
3En(Li 2+) = –8.72 × 10⁻¹⁸ J; rn(Li2+) = 26.4 pm En(He+ ) = –19.62 × 10⁻¹⁸ J; rn(He+ ) = 17.6 pm
4En(Li 2+) = –19.62 × 10⁻¹⁶ J; rn(Li2+) = 17.6 pm En(He+ ) = –8.72 × 10⁻¹⁶ J; rn(He+ ) = 26.4 pm
NTA Answer: Option 2(final)
NEET 2024Revised key
1–x
2−x/9
3–4x
4−(4/9)x
NTA Answer: Option 1(revised_final)
NEET 2022
1158.7 Å
2158.7 pm
315.87 pm
41.587 pm
NTA Answer: Option 2(final)
How does NEET ask about Bohr Model Hydrogen?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Bohr energy difference between two levels; compute photon energy/wavelength using Rydberg or E_n formulas.
Multi StepMedium
Common distractors
forgets z squared
Drops Z² for hydrogen-like
Sources
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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