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Electromagnetic radiation, photoelectric effect and hydrogen spectrum: NEET MCQs with answers

90 questions on this topic of Atomic Structure (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. When a hydrogen atom in an excited state transitions to the ground state, it emits a single photon with wavenumber equal to 0.75 R_H. The principal quantum number of this initial excited state is:

  1. 5
  2. 3
  3. 4
  4. 2
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Answer: (D) 2

For transition to the ground state, n₁ = 1. The wavenumber is ν̅ = R_H [1/1² - 1/n₂²]. Given ν̅ = 0.75 R_H = (3/4) R_H. Equating gives 1 - 1/n₂² = 3/4, so 1/n₂² = 1/4, which yields n₂ = 2.

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2. What is the ratio of the difference between the 1st and 2nd lines of the Lyman series (in terms of wavenumber) to the Rydberg constant R_H for a hydrogen atom?

  1. 5 / 36
  2. 7 / 144
  3. 1 / 12
  4. 3 / 16
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Answer: (A) 5 / 36

For Lyman series, ν̄ = R_H (1/1² - 1/n²). For line 1 (n=2): ν̄₁ = R_H (1 - 1/4) = 3/4 R_H. For line 2 (n=3): ν̄₂ = R_H (1 - 1/9) = 8/9 R_H. The difference is ν̄₂ - ν̄₁ = R_H (8/9 - 3/4) = R_H (32 - 27)/36 = 5/36 R_H. Thus, the ratio is 5/36.

hard

3. What is the ratio of the shortest wavelength line in the Balmer series to the longest wavelength line in the Lyman series for atomic hydrogen?

  1. 64 / 27
  2. 27 / 128
  3. 128 / 27
  4. 27 / 64
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Answer: (C) 128 / 27

The wavenumber for n = 4 → 2 is ν̅₁ = 1/λ₁ = R_H(1/4 - 1/16) = 3R_H / 16, so λ₁ = 16 / (3R_H). For n = 3 → 1, ν̅₂ = 1/λ₂ = R_H(1/1 - 1/9) = 8R_H / 9, so λ₂ = 9 / (8R_H). The ratio λ₁ / λ₂ = [16 / (3R_H)] / [9 / (8R_H)] = (16 × 8) / (3 × 9) = 128 / 27.

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4. What is the ratio of the energy of a photon emitted during the transition n = 2 → n = 1 to that emitted during n = 3 → n = 2 in a hydrogen atom?

  1. 4 : 1
  2. 9 : 4
  3. 27 : 5
  4. 5 : 27
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Answer: (C) 27 : 5

The energy of an emitted photon is given by ΔE ∝ (1/n₁² - 1/n₂²). For transition 2 → 1: ΔE₁ ∝ (1/1² - 1/2²) = 3/4. For transition 3 → 2: ΔE₂ ∝ (1/2² - 1/3²) = 5/36. The ratio ΔE₁ / ΔE₂ = (3/4) / (5/36) = (3/4) × (36/5) = 27/5.

easy

5. For a hydrogen atom, which electronic transition results in the emission of radiation with the longest wavelength in the Brackett series?

  1. n = 6 → n = 4
  2. n = 5 → n = 4
  3. n = ∞ → n = 4
  4. n = 4 → n = 3
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Answer: (B) n = 5 → n = 4

For the Brackett series, n₁ = 4. The longest wavelength corresponds to the minimum energy difference, which occurs for the transition from the immediate next level, i.e., n = 5 to n = 4.

hard

6. A beam of light contains two wavelengths, 400 nm and 500 nm, with equal total intensity, and falls perpendicularly on a clean metal surface with a work function of 2.60 eV. What will be the maximum kinetic energy of the emitted photoelectrons? (Take hc = 1240 eV·nm)

  1. No photoelectrons are emitted
  2. 0.12 eV
  3. 2.60 eV
  4. 0.50 eV
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Answer: (D) 0.50 eV

Energy of 400 nm photons is E₁ = 1240 / 400 = 3.10 eV. Energy of 500 nm photons is E₂ = 1240 / 500 = 2.48 eV. Since E₂ (2.48 eV) is less than the work function W₀ (2.60 eV), the 500 nm photons cannot cause photoelectric emission. Only the 400 nm photons eject photoelectrons, with K_max = E₁ - W₀ = 3.10 eV - 2.60 eV = 0.50 eV.

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7. What is the ratio of the frequency of the second line of the Lyman series to that of the first line of the Lyman series in a hydrogen atom?

  1. 32 : 27
  2. 27 : 32
  3. 4 : 3
  4. 9 : 8
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Answer: (A) 32 : 27

For Lyman series, n₁ = 1. The first line corresponds to n₂ = 2, so ν₁ ∝ (1 - 1/4) = 3/4. The second line corresponds to n₂ = 3, so ν₂ ∝ (1 - 1/9) = 8/9. The ratio ν₂/ν₁ = (8/9) / (3/4) = 32/27.

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8. A monochromatic source emits yellow light of wavelength 589 nm with an output power of 66.3 W. How many photons does this source emit in 20 seconds? (h = 6.63 × 10⁻³⁴ J·s, c = 3.0 × 10⁸ m/s)

  1. 3.93 × 10²⁰
  2. 1.96 × 10²²
  3. 1.96 × 10²¹
  4. 3.93 × 10²¹
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Answer: (D) 3.93 × 10²¹

Energy of one photon is E = hc / λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) / (589 × 10⁻⁹) = 3.377 × 10⁻¹⁹ J. Total energy emitted in 20 seconds is Total Energy = Power × time = 66.3 W × 20 s = 1326 J. Number of photons N = 1326 / (3.377 × 10⁻¹⁹) ≈ 3.93 × 10²¹ photons.

easy

9. The energy of a photon of light is 3.0 eV. What is its approximate wavelength in angstroms? (Take hc ≈ 12400 eV·Å)

  1. 4133 Å
  2. 3100 Å
  3. 2480 Å
  4. 6200 Å
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Answer: (A) 4133 Å

Using the relation λ (in Å) = hc / E (in eV) = 12400 / 3.0 ≈ 4133.3 Å. Hence, 4133 Å is the correct wavelength.

easy

10. In which of the following transitions in a hydrogen atom will the emitted photon have the highest frequency?

  1. n = 2 → n = 1
  2. n = 3 → n = 2
  3. n = 4 → n = 3
  4. n = 5 → n = 2
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Answer: (A) n = 2 → n = 1

Energy of an emitted photon is ΔE = 13.6 × (1/n₁² - 1/n₂²) eV. For n = 2 → 1, ΔE = 13.6 × (1 - 1/4) = 10.2 eV. For n = 3 → 2, ΔE = 13.6 × (1/4 - 1/9) = 1.89 eV. For n = 5 → 2, ΔE = 13.6 × (1/4 - 1/25) = 2.856 eV. For n = 4 → 3, ΔE = 13.6 × (1/9 - 1/16) = 0.66 eV. Since frequency ν is directly proportional to ΔE, transition n = 2 → 1 produces the highest frequency.

easy

11. Consider the following statements regarding the photoelectric effect: Statement I: The saturation photoelectric current depends directly on the intensity of the incident radiation. Statement II: The stopping potential for photoelectrons depends on both the frequency of the incident radiation and the nature of the metal surface. In 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
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Answer: (A) Both Statement I and Statement II are correct

Photoelectric current is proportional to the number of photoelectrons emitted per second, which depends directly on the intensity of incident light. Stopping potential is given by eV₀ = hν - W₀, which depends on both the frequency of incident radiation (ν) and the work function (W₀, characteristic of the metal). Hence, both statements are correct.

easy

12. The limiting line (series limit) of the Paschen series of the hydrogen spectrum corresponds to an electronic transition from:

  1. n = ∞ to n = 3
  2. n = ∞ to n = 2
  3. n = 3 to n = 1
  4. n = 4 to n = 3
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Answer: (A) n = ∞ to n = 3

For the Paschen series, the lower energy level is n₁ = 3. The series limit (shortest wavelength, highest frequency) corresponds to a transition from n₂ = ∞ to n₁ = 3.

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