NEETAceToday's NEET Daily

Bohr model, energy levels and hydrogen spectrum: NEET MCQs with answers

54 questions on this topic of Atoms and Nuclei (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Atoms and Nuclei freeAll Atoms and Nuclei questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. If the de Broglie wavelength of an electron in the nth Bohr orbit of a hydrogen atom is λ, then the circumference of that orbit is equal to:

  1. n λ
  2. λ / n
  3. n² λ
  4. 2π n λ
Show answer and explanation

Answer: (A) n λ

According to de Broglie's hypothesis explaining Bohr's postulate, an integral number of electron de Broglie wavelengths must fit into the circumference of the orbit: 2π rₙ = n λ. Therefore, the circumference is nλ.

medium

2. Given that the Rydberg constant is R, what is the maximum wavelength of radiation emitted in the Paschen series of the hydrogen spectrum?

  1. 9 / R
  2. 144 / (7 R)
  3. 16 / (7 R)
  4. 7 / (144 R)
Show answer and explanation

Answer: (B) 144 / (7 R)

Maximum wavelength corresponds to the minimum energy transition in the Paschen series, which occurs from n = 4 to n = 3. Using 1/λ = R(1/3² - 1/4²) = R(1/9 - 1/16) = 7R / 144, we get λ = 144 / (7R).

hard

3. An electron in a hydrogen atom undergoes a transition from an initial state with principal quantum number n to the ground state (n = 1). If the angular momentum of the electron decreases by 2h/π during the transition, what is the value of n?

  1. 3
  2. 5
  3. 4
  4. 6
Show answer and explanation

Answer: (B) 5

According to Bohr's quantization condition, angular momentum is Lₙ = n(h/(2π)). The change in angular momentum is ΔL = Lₙ - L₁ = (n - 1)h/(2π). Given ΔL = 2h/π = 4h/(2π), we have n - 1 = 4, which yields n = 5.

medium

4. The energy of the photon emitted when an electron in a hydrogen atom falls from n = 6 to n = 3 is:

  1. 0.378 eV
  2. 1.13 eV
  3. 2.27 eV
  4. 1.51 eV
Show answer and explanation

Answer: (B) 1.13 eV

Bohr energy levels are Eₙ = −13.6/n² eV, so ΔE = 13.6(1/3² − 1/6²) eV = 1.13 eV.

easy

5. If the wavelength of the first line of the Lyman series of the hydrogen spectrum is λ, what is the wavelength of the first line of the Balmer series?

  1. 27/5 λ
  2. 5/27 λ
  3. 32/5 λ
  4. 5/32 λ
Show answer and explanation

Answer: (A) 27/5 λ

For the first line of the Lyman series (n = 2 to n = 1), 1/λ = R(1/1² - 1/2²) = 3R/4. For the first line of the Balmer series (n = 3 to n = 2), 1/λ' = R(1/2² - 1/3²) = 5R/36. Dividing the two equations gives λ'/λ = (3/4) / (5/36) = 27/5, so λ' = (27/5)λ.

hard

6. When a hydrogen atom in its ground state absorbs a photon of energy 12.75 eV, the orbital angular momentum of the electron increases by an amount equal to:

  1. 4h/(2π)
  2. 2h/(2π)
  3. h/(2π)
  4. 3h/(2π)
Show answer and explanation

Answer: (D) 3h/(2π)

The energy absorbed is ΔE = 12.75 eV. The energy of the excited state is Eₙ = E₁ + 12.75 = -13.6 + 12.75 = -0.85 eV. Since Eₙ = -13.6/n², n² = 13.6/0.85 = 16, so n = 4. The change in orbital angular momentum is ΔL = (n - 1)h/(2π) = (4 - 1)h/(2π) = 3h/(2π).

easy

7. Which of the following transitions in a hydrogen atom results in the absorption of radiation with the longest wavelength?

  1. n = 1 → n = 2
  2. n = 2 → n = 3
  3. n = 3 → n = 4
  4. n = 4 → n = 5
Show answer and explanation

Answer: (D) n = 4 → n = 5

Absorption requires a transition from a lower state to a higher state. The energy of the absorbed photon is ΔE = hc / λ. For longest wavelength, the transition must have the minimum energy difference ΔE. Since energy levels get closer together as n increases, the difference ΔE is smallest for n = 4 → n = 5 (ΔE = -0.54 - (-0.85) = 0.31 eV).

easy

8. In a hydrogen atom, the ratio of the change in orbital angular momentum when an electron drops from the n = 4 state to the n = 2 state, to the angular momentum of the electron in its ground state, is:

  1. 1 : 1
  2. 2 : 1
  3. 3 : 1
  4. 4 : 1
Show answer and explanation

Answer: (B) 2 : 1

According to Bohr's quantization postulate, L_n = n(h / 2π). For the transition from n = 4 to n = 2, ΔL = L₄ - L₂ = (4 - 2)(h / 2π) = 2(h / 2π). The ground state angular momentum is L₁ = 1(h / 2π). Therefore, the ratio ΔL / L₁ = 2 / 1 = 2 : 1.

medium

9. If the radius of the ground state orbit of hydrogen atom is a₀, what is the radius of the first excited state of a singly ionized helium atom (He⁺)?

  1. a₀
  2. 2 a₀
  3. 4 a₀
  4. 0.5 a₀
Show answer and explanation

Answer: (B) 2 a₀

The radius in Bohr's model is given by r = a₀(n²/Z). For the first excited state of He⁺, n = 2 and Z = 2. Therefore, r = a₀(2² / 2) = 2 a₀.

medium

10. Consider the following two statements regarding the Bohr model of the hydrogen atom: Statement I: The total energy of an electron in any Bohr orbit is negative, which signifies that the electron is bound to the nucleus. Statement II: As the principal quantum number n increases, the potential energy of the electron decreases. In the light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Total energy E_n = -13.6/n² eV is negative, signifying a bound system, so Statement I is correct. Potential energy U_n = 2E_n = -27.2/n² eV. As n increases, -27.2/n² becomes less negative, meaning the potential energy increases (approaching 0 as n → ∞). Thus, Statement II is incorrect.

medium

11. The energy of the photon emitted when an electron in a hydrogen atom falls from n = 4 to n = 2 is:

  1. 3.4 eV
  2. 5.1 eV
  3. 2.55 eV
  4. 0.85 eV
Show answer and explanation

Answer: (C) 2.55 eV

Bohr energy levels are Eₙ = −13.6/n² eV, so ΔE = 13.6(1/2² − 1/4²) eV = 2.55 eV.

medium

12. An electron in a hydrogen-like atom transitions from state n to state (n - 1). Which of the following correctly describes the change in kinetic energy (ΔK) and potential energy (ΔU) of the atom during this transition?

  1. ΔK > 0 and ΔU < 0
  2. ΔK < 0 and ΔU > 0
  3. ΔK > 0 and ΔU > 0
  4. ΔK < 0 and ΔU < 0
Show answer and explanation

Answer: (A) ΔK > 0 and ΔU < 0

As n decreases (dropping to a lower orbit), the electron moves closer to the nucleus. Its speed increases, so its kinetic energy increases (ΔK > 0). The potential energy becomes more negative, meaning it decreases (ΔU < 0). In fact, ΔU = -2ΔK.

Keep going with Atoms and Nuclei. All 254 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Atoms and Nuclei