Bohr's Model and Hydrogen Spectrum – Spectrum of the hydrogen atom: NEET MCQs with Answers

15 free questions on Bohr's Model and Hydrogen Spectrum – Spectrum of the hydrogen atom for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Spectrum of the hydrogen atom. Bohr model of a hydrogen atom - its postulates, derivation of the relations for the energy of the electron and radii of the different orbits, limitations of Bohr's model;

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Question 1

Previous-year question
Two statements are given below. In the light of these statements, choose the correct answer from the options given. [NEET (Sawai Madhopur) 2024]
Statement I: The energy of the He⁺ ion in the n = 2 state is the same as the energy of the H atom in the n = 1 state.
Statement II: It is possible to determine simultaneously the exact position and exact momentum of an electron in the H atom.
  1. Both Statement I and Statement II are false.
  2. Statement I is true but Statement II is false.
  3. Statement I is false but Statement II is true.
  4. Both Statement I and Statement II are true.
Show answer
Answer: (B) Statement I is true but Statement II is false.
Explanation: Statement I is true: Eₙ = −13.6 Z²/n² eV; for He⁺ (Z = 2, n = 2) E = −13.6 × 4/4 = −13.6 eV, the same as for H in n = 1. Statement II is false: by Heisenberg's uncertainty principle the exact position and exact momentum of an electron cannot be determined simultaneously.

Question 2

Previous-year question
Which of the following series of transitions in the spectrum of the hydrogen atom falls in the visible region? [NEET 2019]
  1. Lyman series
  2. Balmer series
  3. Paschen series
  4. Brackett series
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Answer: (B) Balmer series
Explanation: The Lyman series (n₁ = 1) lies in the ultraviolet, while the Paschen (n₁ = 3) and Brackett (n₁ = 4) series lie in the infrared. Only the Balmer series (n₁ = 2) has lines in the 400–700 nm visible range.

Question 3

Previous-year question
The angular speed of the electron in the nth orbit of the Bohr hydrogen atom is [NEET 2018]
  1. directly proportional to n
  2. inversely proportional to √n
  3. inversely proportional to n²
  4. inversely proportional to n³
Show answer
Answer: (D) inversely proportional to n³
Explanation: Angular speed ω = v/r. Since v ∝ 1/n and r ∝ n², we get ω ∝ (1/n)/n² = 1/n³, i.e. inversely proportional to n³.

Question 4

Previous-year question
If r is the radius of the first orbit, the radius of the nth orbit of the H-atom is given by [AIPMT 1988]
  1. rn²
  2. rn
  3. r/n
  4. r²n²
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Answer: (A) rn²
Explanation: The Bohr radius formula is rₙ = (n²h²)/(4π²me²Z) = 0.529 × (n²/Z) Å. For the H-atom, Z = 1, so rₙ ∝ n². If the first orbit (n = 1) has radius r, then r₁ = r × 1² = r. Therefore rₙ = r × n² = rn². Example: the n = 3 orbit is 9 times the radius of the ground-state orbit.

Question 5

Previous-year question
Which of the following statements does not form a part of Bohr’s model of the hydrogen atom? [AIPMT 1989]
  1. Energy of the electrons in the orbits are quantised
  2. The electron in the orbit nearest the nucleus has the lowest energy
  3. Electrons revolve in different orbits around the nucleus
  4. The position and velocity of electrons in the orbit cannot be determined simultaneously
Show answer
Answer: (D) The position and velocity of electrons in the orbit cannot be determined simultaneously
Explanation: Quantised orbit energies, the lowest energy for the innermost orbit, and electrons revolving in fixed orbits are all Bohr postulates. The impossibility of determining position and velocity simultaneously is Heisenberg's uncertainty principle (1927), which actually contradicts Bohr's fixed-orbit picture.

Question 6

Previous-year question
The energy of an electron in the nth Bohr orbit of the hydrogen atom is [AIPMT 1992]
  1. −13.6/n⁴ eV
  2. −13.6/n³ eV
  3. −13.6/n² eV
  4. −13.6/n eV
Show answer
Answer: (C) −13.6/n² eV
Explanation: From Bohr’s derivation, Eₙ = −(2π²me⁴Z²)/(n²h²), which for hydrogen (Z = 1) reduces to the standard result Eₙ = −13.6/n² eV atom⁻¹. Verify with known values: n = 1 gives −13.6 eV (the ground state), n = 2 gives −3.4 eV, n = 3 gives −1.51 eV — all standard. The inverse-SQUARE dependence on n is the signature of the Bohr model.

Question 7

Previous-year question
If the ionisation potential for the hydrogen atom is 13.6 eV, then the ionisation potential for He⁺ will be [AIPMT 1993]
  1. 54.4 eV
  2. 6.8 eV
  3. 13.6 eV
  4. 24.5 eV
Show answer
Answer: (A) 54.4 eV
Explanation: For a one-electron species in the ground state, IE = 13.6 Z² eV. For He⁺, Z = 2, so IE = 13.6 × 4 = 54.4 eV; 24.6 eV is the first IE of neutral He, to which this formula does not apply.
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Question 8

Previous-year question
When an electron of charge e and mass m moves with a velocity v about a nuclear charge Ze in a circular orbit of radius r, the potential energy of the electron is given by [AIPMT 1994]
  1. Z²e²/r
  2. −Ze²/r
  3. Ze²/r
  4. mv²/r
Show answer
Answer: (B) −Ze²/r
Explanation: The electrostatic potential energy of an electron (−e) at distance r from a nucleus of charge +Ze is PE = −Ze²/r. It is negative because the two attract, with zero taken at infinite separation; mv²/r is the centripetal force, not an energy.

Question 9

Previous-year question
The radius of the hydrogen atom in the ground state is 0.53 Å. The radius of the Li²⁺ ion (atomic number = 3) in a similar state is [AIPMT 1995]
  1. 0.17 Å
  2. 0.53 Å
  3. 0.265 Å
  4. 1.06 Å
Show answer
Answer: (A) 0.17 Å
Explanation: For hydrogen-like species, rₙ = 0.53 × (n²/Z) Å. ‘A similar state’ means the ground state, n = 1. For Li²⁺: Z = 3, n = 1 ⇒ r = 0.53 × (1²/3) = 0.177 ≈ 0.17 Å. Physically, the greater nuclear charge of Li²⁺ pulls its single electron three times closer than in hydrogen.

Question 10

Previous-year question
In Bohr’s model of an atom, when an electron jumps from n = 1 to n = 3, how much energy will be emitted or absorbed? [AIPMT 1996]
  1. 2.389 × 10⁻¹² ergs
  2. 0.239 × 10⁻¹⁰ ergs
  3. 2.15 × 10⁻¹¹ ergs
  4. 0.1936 × 10⁻¹⁰ ergs
Show answer
Answer: (D) 0.1936 × 10⁻¹⁰ ergs
Explanation: Using Eₙ = −1312/n² kJ mol⁻¹ for hydrogen: E₁ = −1312/1² = −1312 kJ mol⁻¹ E₃ = −1312/3² = −145.8 kJ mol⁻¹ ΔE = E₃ − E₁ = −145.8 − (−1312) = +1166.2 kJ mol⁻¹ (energy is ABSORBED, since the electron moves outward). Per atom: ΔE = (1166.2 × 10³)/(6.023 × 10²³) = 1.936 × 10⁻¹⁸ J. Converting to ergs (1 J = 10⁷ erg): 1.936 × 10⁻¹⁸ × 10⁷ = 1.936 × 10⁻¹¹ = 0.1936 × 10⁻¹⁰ ergs.

Question 11

Previous-year question
What will be the longest wavelength line in the Balmer series of the hydrogen spectrum? [AIPMT 1996]
  1. 546 nm
  2. 656 nm
  3. 566 nm
  4. 556 nm
Show answer
Answer: (B) 656 nm
Explanation: The longest wavelength corresponds to the SMALLEST energy transition. In the Balmer series (n₁ = 2), that is the n = 3 → n = 2 transition (the Hα line). Using the Rydberg equation, 1/λ = Rₕ (1/n₁² − 1/n₂²) with Rₕ = 109677 cm⁻¹: 1/λ = 109677 [(1/2²) − (1/3²)] = 109677 × (5/36) = 15233 cm⁻¹. λ = 1/15233 = 6.56 × 10⁻⁵ cm = 6.56 × 10⁻⁷ m = 656 nm. This is the familiar red line of the hydrogen spectrum.

Question 12

Previous-year question
The spectrum of He is expected to be similar to that of [AIPMT 1998]
  1. H
  2. Li⁺
  3. Na
  4. He⁺
Show answer
Answer: (B) Li⁺
Explanation: Atomic spectra depend on the number of electrons and the electronic configuration. Neutral He has 2 electrons (1s²). Li⁺ also has 2 electrons (1s²) — the two species are ISOELECTRONIC, so their spectra have the same general pattern. H (1 electron) and He⁺ (1 electron) are one-electron species with a completely different, simpler spectral pattern; Na has 11 electrons. Hence the spectrum of He resembles that of Li⁺.

Question 13

Previous-year question
The Bohr radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) is (in Å) [AIPMT 1998]
  1. 0.13
  2. 1.06
  3. 4.77
  4. 2.12
Show answer
Answer: (D) 2.12
Explanation: For hydrogen r ∝ n². The first excited state is n = 2, so r = 0.530 × 2² = 2.12 Å.

Question 14

Previous-year question
According to Bohr’s theory, the energy required for an electron in the Li²⁺ ion to be emitted from the n = 2 state is (given that the ground state ionisation energy of the hydrogen atom is 13.6 eV) [AIPMT 1999]
  1. 61.2 eV
  2. 13.6 eV
  3. 30.6 eV
  4. 10.2 eV
Show answer
Answer: (C) 30.6 eV
Explanation: For a hydrogen-like ion, Eₙ = −13.6 × (Z²/n²) eV. For Li²⁺, Z = 3 and n = 2: E₂ = −13.6 × (3²/2²) = −13.6 × (9/4) = −30.6 eV. To emit (remove) the electron, it must be raised to n = ∞ where E = 0. Energy required = 0 − (−30.6) = 30.6 eV.

Question 15

Previous-year question
Who modified Bohr’s theory by introducing elliptical orbits for the electron path? [AIPMT 1999]
  1. Hund
  2. Thomson
  3. Rutherford
  4. Sommerfeld
Show answer
Answer: (D) Sommerfeld
Explanation: Arnold Sommerfeld extended Bohr’s model in 1916 by allowing the electron to move in ELLIPTICAL orbits in addition to circular ones. This introduced a second quantum number (the azimuthal quantum number) to describe the shape of the orbit, and successfully explained the fine structure of spectral lines that Bohr’s purely circular model could not. Hund gave the rule of maximum multiplicity; Thomson proposed the plum-pudding model; Rutherford proposed the nuclear model.

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