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Matter waves and de Broglie relation: NEET MCQs with answers

69 questions on this topic of Dual Nature of Matter and Radiation (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. If the momentum of a moving particle is increased by 100%, the percentage decrease in its de Broglie wavelength is:

  1. 50%
  2. 25%
  3. 75%
  4. 100%
Show answer and explanation

Answer: (A) 50%

The de Broglie wavelength is λ = h/p. When momentum is increased by 100%, the new momentum becomes p' = 2p. The new wavelength is λ' = h/(2p) = λ/2, which represents a decrease of [(λ - λ/2)/λ] × 100% = 50%.

medium

2. Starting from rest, an electron is accelerated through 200 V. Using λ = 1.227/√V nm, its de Broglie wavelength is:

  1. 0.868 nm
  2. 0.00614 nm
  3. 0.174 nm
  4. 0.0868 nm
Show answer and explanation

Answer: (D) 0.0868 nm

λ = h/√(2meV) = 1.227/√V nm = 1.227/√200 = 0.0868 nm.

Quick tip: For electrons, λ = 1.227/√V nm (or 12.27/√V Å). Watch the unit.

NCERT Class 12, Chapter 11

hard

3. A photon and an electron each possess an energy of 100 eV. If λph is the wavelength of the photon and λe is the de Broglie wavelength of the electron, then the ratio λph / λe is of the order of:

  1. 10⁻²
  2. 1
  3. 10⁴
  4. 10²
Show answer and explanation

Answer: (D) 10²

For the photon, λph = hc/E. For the electron, λe = h/√(2mE). Hence, λph/λe = c √(2m/E). Substituting m = 9.1 × 10⁻³¹ kg, E = 100 × 1.6 × 10⁻¹⁹ J = 1.6 × 10⁻¹⁷ J, and c = 3 × 10⁸ m/s gives λph/λe ≈ 3 × 10⁸ × √(1.82 × 10⁻³⁰ / 1.6 × 10⁻¹⁷) ≈ 3 × 10⁸ × 3.37 × 10⁻⁷ ≈ 100.

medium

4. A proton and an electron have the same de Broglie wavelength. If E_p and E_e are their respective kinetic energies, and c is the speed of light, which of the following is correct regarding their non-relativistic kinetic energies?

  1. E_e / E_p = m_p / m_e
  2. E_p > E_e
  3. E_p = E_e
  4. E_e / E_p = √(m_p / m_e)
Show answer and explanation

Answer: (A) E_e / E_p = m_p / m_e

Kinetic energy is given by E = p² / (2m). Since both particles have identical de Broglie wavelength λ, their momenta p = h/λ are identical. Hence, E_e / E_p = (p² / 2m_e) / (p² / 2m_p) = m_p / m_e.

easy

5. A proton and a deuteron (mass ≈ 2m, charge = e) have the same de Broglie wavelength. Which of the following statements is correct regarding their physical properties?

  1. Both have equal kinetic energy.
  2. The deuteron has half the kinetic energy of the proton.
  3. The deuteron has greater linear momentum than the proton.
  4. The velocity of the deuteron is twice that of the proton.
Show answer and explanation

Answer: (B) The deuteron has half the kinetic energy of the proton.

Since de Broglie wavelength λ = h/p, equal wavelength implies equal linear momentum (p_p = p_d). Kinetic energy is given by K = p²/(2m). Therefore, K_d/K_p = m_p/m_d = 1/2, meaning the deuteron has half the kinetic energy of the proton.

hard

6. A particle of mass m is confined to move in a one-dimensional potential box of length L. If the matter wave forms a standing wave with 3 antinodes, the momentum of the particle is:

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

Answer: (A) 3h / (2L)

For a standing matter wave with 3 antinodes (third harmonic), L = 3(λ / 2), which gives λ = 2L / 3. Since momentum p = h / λ, we have p = h / (2L / 3) = 3h / (2L).

medium

7. An electron at rest is accelerated through a potential difference of 200 V. Its de Broglie wavelength is approximately:

  1. 0.0868 nm
  2. 0.174 nm
  3. 0.868 nm
  4. 0.00614 nm
Show answer and explanation

Answer: (A) 0.0868 nm

λ = h/√(2meV) = 1.227/√V nm = 1.227/√200 = 0.0868 nm.

Quick tip: For electrons, λ = 1.227/√V nm (or 12.27/√V Å). Watch the unit.

NCERT Class 12, Chapter 11

medium

8. An electron at rest is accelerated through a potential difference of 700 V. Its de Broglie wavelength is approximately:

  1. 0.464 nm
  2. 0.325 nm
  3. 0.00175 nm
  4. 0.0464 nm
Show answer and explanation

Answer: (D) 0.0464 nm

λ = h/√(2meV) = 1.227/√V nm = 1.227/√700 = 0.0464 nm.

Quick tip: For electrons, λ = 1.227/√V nm (or 12.27/√V Å). Watch the unit.

NCERT Class 12, Chapter 11

medium

9. Starting from rest, an electron is accelerated through 400 V. Using λ = 1.227/√V nm, its de Broglie wavelength is:

  1. 0.00307 nm
  2. 0.613 nm
  3. 0.245 nm
  4. 0.0614 nm
Show answer and explanation

Answer: (D) 0.0614 nm

λ = h/√(2meV) = 1.227/√V nm = 1.227/√400 = 0.0614 nm.

Quick tip: For electrons, λ = 1.227/√V nm (or 12.27/√V Å). Watch the unit.

NCERT Class 12, Chapter 11

medium

10. An electron and a photon each have a de Broglie wavelength of 1.00 nm. What is the ratio of the momentum of the electron to the momentum of the photon?

  1. 1 : 1
  2. 1 : 2
  3. 2 : 1
  4. 1836 : 1
Show answer and explanation

Answer: (A) 1 : 1

For any entity having wave nature, momentum is given by p = h / λ. Since both the electron and the photon have the same wavelength λ = 1.00 nm, their momenta must be identical, yielding a ratio of 1 : 1.

medium

11. Starting from rest, an electron is accelerated through 800 V. Using λ = 1.227/√V nm, its de Broglie wavelength is:

  1. 0.0434 nm
  2. 0.00153 nm
  3. 0.434 nm
  4. 0.347 nm
Show answer and explanation

Answer: (A) 0.0434 nm

λ = h/√(2meV) = 1.227/√V nm = 1.227/√800 = 0.0434 nm.

Quick tip: For electrons, λ = 1.227/√V nm (or 12.27/√V Å). Watch the unit.

NCERT Class 12, Chapter 11

medium

12. A particle of charge q and mass m is accelerated from rest through a potential difference V. If another particle has charge 2q and mass 8m, what potential difference must it be accelerated through to have the same de Broglie wavelength as the first particle?

  1. 16V
  2. V / 4
  3. 4V
  4. V / 16
Show answer and explanation

Answer: (D) V / 16

The de Broglie wavelength is λ = h / √(2mqV). For the two particles to have the same wavelength, m₁q₁V₁ = m₂q₂V₂. Here, m(q)(V) = (8m)(2q)(V₂), which simplifies to mqV = 16mq V₂, hence V₂ = V / 16.

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