A Level Physics 22.3 Wave-Particle Duality MCQs – 30 Practice Questions with Answers (Cambridge 9702)

This page covers Section 22.3 of the Cambridge International AS & A Level Physics (9702) syllabus, Wave-Particle Duality. It is an A Level (A2) topic within Quantum Physics, and these 30 original practice questions cover the de Broglie wavelength, electron diffraction and comparisons between matter waves and electromagnetic radiation, each with a full worked answer.

Formula recap: de Broglie wavelength λ = h / p = h / (m v). For an electron accelerated from rest through a potential difference V, its kinetic energy is eV, so p = √(2 me e V) and λ = h / √(2 me e V). Constants: Planck constant h = 6.63 × 10-34 J s; electron mass me = 9.11 × 10-31 kg; elementary charge e = 1.6 × 10-19 C; proton mass = 1.67 × 10-27 kg. See the full A Level Physics Formula Sheet.

De Broglie Wavelength

Q1. De Broglie’s hypothesis states that every moving particle has an associated wavelength λ given by

  • (A) λ = p / h
  • (B) λ = h / p
  • (C) λ = h p
  • (D) λ = h / (m c2)
Show Answer

Answer: B – The de Broglie wavelength is λ = h / p = h / (m v), where p is the momentum of the particle and h is the Planck constant.

Q2. An electron travels at a speed of 2.0 × 106 m/s. Its de Broglie wavelength is

  • (A) 7.3 × 10-10 m
  • (B) 1.8 × 10-10 m
  • (C) 3.6 × 10-10 m
  • (D) 1.5 × 10-9 m
Show Answer

Answer: C – p = m v = 9.11 × 10-31 × 2.0 × 106 = 1.82 × 10-24 kg m/s. λ = h / p = 6.63 × 10-34 / 1.82 × 10-24 = 3.6 × 10-10 m.

Q3. The speed of a particle is doubled, its mass remaining constant. Its de Broglie wavelength

  • (A) is halved
  • (B) is doubled
  • (C) is quartered
  • (D) is unchanged
Show Answer

Answer: A – λ = h / (m v), so λ is inversely proportional to v. Doubling v doubles the momentum and halves the wavelength.

Q4. A ball of mass 0.15 kg is thrown at 30 m/s. Its de Broglie wavelength is approximately

  • (A) 1.5 × 10-30 m
  • (B) 1.5 × 10-32 m
  • (C) 4.4 × 10-35 m
  • (D) 1.5 × 10-34 m
Show Answer

Answer: D – p = m v = 0.15 × 30 = 4.5 kg m/s. λ = 6.63 × 10-34 / 4.5 = 1.5 × 10-34 m, far too small for any wave behaviour to be detected.

Q5. An electron is accelerated from rest through a potential difference of 100 V. Its de Broglie wavelength is

  • (A) 1.2 × 10-9 m
  • (B) 1.2 × 10-10 m
  • (C) 3.9 × 10-11 m
  • (D) 1.2 × 10-12 m
Show Answer

Answer: B – Kinetic energy = e V = 1.6 × 10-19 × 100 = 1.6 × 10-17 J. p = √(2 me E) = √(2 × 9.11 × 10-31 × 1.6 × 10-17) = 5.4 × 10-24 kg m/s. λ = 6.63 × 10-34 / 5.4 × 10-24 = 1.2 × 10-10 m.

Q6. The accelerating potential difference for an electron beam is increased from 100 V to 400 V. The de Broglie wavelength of the electrons

  • (A) doubles, to 2.5 × 10-10 m
  • (B) stays the same
  • (C) halves, to 6.1 × 10-11 m
  • (D) quarters, to 3.1 × 10-11 m
Show Answer

Answer: C – Since p = √(2 m e V), the wavelength varies as 1 / √V. Quadrupling V doubles p, so λ falls from 1.2 × 10-10 m to half that value, 6.1 × 10-11 m.

Q7. Everyday objects such as balls and cars are never observed to diffract because

  • (A) their de Broglie wavelength is far smaller than any gap or spacing they could meet
  • (B) their momentum is too small
  • (C) their de Broglie wavelength is too large to pass through gaps
  • (D) only charged particles have a de Broglie wavelength
Show Answer

Answer: A – Diffraction is only noticeable when the wavelength is comparable to the size of the gap or obstacle. A macroscopic object has λ of order 10-34 m or less, which is vastly smaller than any real gap, so no diffraction is observed.

Q8. A neutron of mass 1.67 × 10-27 kg travels at 1.0 × 103 m/s. Its de Broglie wavelength is

  • (A) 7.3 × 10-10 m
  • (B) 4.0 × 10-13 m
  • (C) 2.0 × 10-10 m
  • (D) 4.0 × 10-10 m
Show Answer

Answer: D – p = m v = 1.67 × 10-27 × 1.0 × 103 = 1.67 × 10-24 kg m/s. λ = 6.63 × 10-34 / 1.67 × 10-24 = 4.0 × 10-10 m.

Q9. An electron has a de Broglie wavelength of 1.0 × 10-10 m. Its momentum is

  • (A) 6.6 × 10-44 kg m/s
  • (B) 6.6 × 10-24 kg m/s
  • (C) 6.6 × 10-25 kg m/s
  • (D) 7.3 × 10-24 kg m/s
Show Answer

Answer: B – p = h / λ = 6.63 × 10-34 / 1.0 × 10-10 = 6.6 × 10-24 kg m/s.

Q10. A proton and an electron travel at the same speed. Compared with the electron, the proton has

  • (A) the same de Broglie wavelength
  • (B) a larger de Broglie wavelength
  • (C) a smaller de Broglie wavelength, because its momentum is larger
  • (D) no associated wavelength, because it is heavier
Show Answer

Answer: C – At the same speed, p = m v is much larger for the proton (its mass is about 1836 times the electron mass), so λ = h / p is much smaller.

Electron Diffraction

Q11. A beam of electrons passed through a thin graphite film produces a pattern of concentric rings on a fluorescent screen. This is evidence that

  • (A) electrons can behave as waves
  • (B) electrons are negatively charged
  • (C) the atoms in graphite are arranged in circles
  • (D) electrons always travel in straight lines
Show Answer

Answer: A – Rings are a diffraction pattern, and diffraction is a wave property. The electrons are being diffracted by the regularly spaced atoms of the graphite crystal.

Q12. For electrons to be diffracted noticeably by a crystal, their de Broglie wavelength must be

  • (A) much larger than the spacing between the atoms
  • (B) exactly equal to the diameter of an electron
  • (C) much smaller than the spacing between the atoms
  • (D) comparable to the spacing between the atoms in the crystal
Show Answer

Answer: D – Diffraction effects are greatest when the wavelength is about the same size as the gap or spacing involved. Atomic spacings are about 10-10 m, so the electrons need λ of about 10-10 m.

Q13. In an electron diffraction experiment, the accelerating voltage is increased. The ring pattern

  • (A) moves further apart (larger ring diameters)
  • (B) moves closer together (smaller ring diameters)
  • (C) disappears completely
  • (D) stays in exactly the same position
Show Answer

Answer: B – A higher voltage gives the electrons more momentum and so a smaller de Broglie wavelength. A smaller wavelength is diffracted through smaller angles, so the rings shrink towards the centre.

Q14. The rings in electron diffraction move closer together when the accelerating voltage is raised because

  • (A) the electrons gain extra charge
  • (B) the de Broglie wavelength of the electrons increases
  • (C) the electrons gain momentum, their wavelength decreases, and so the diffraction angles decrease
  • (D) the spacing of the graphite atoms decreases
Show Answer

Answer: C – p increases as √V, so λ = h / p decreases. Smaller wavelengths diffract through smaller angles, giving smaller rings.

Q15. When electrons are sent through a graphite film one at a time, the screen shows

  • (A) no pattern, because single electrons cannot diffract
  • (B) a complete ring produced by each individual electron
  • (C) many spots produced simultaneously by each electron
  • (D) one localised spot for each electron, with the spots gradually building up the diffraction pattern
Show Answer

Answer: D – Each electron is detected as a single localised particle (one spot), yet after many electrons the accumulated spots form the wave diffraction pattern. This is wave-particle duality in action.

Q16. Neutron diffraction is used to investigate the structure of crystals because

  • (A) slow (thermal) neutrons have de Broglie wavelengths comparable to atomic spacings
  • (B) neutrons are attracted to nuclei by the electrostatic force
  • (C) neutrons have no momentum
  • (D) neutrons travel at the speed of light
Show Answer

Answer: A – A neutron moving at about 103 m/s has λ of about 4 × 10-10 m, similar to the spacing of atoms in a crystal, so it is diffracted by the crystal just as electrons are.

Q17. The accelerating voltage in an electron diffraction tube is reduced from 100 V to 25 V. The de Broglie wavelength of the electrons, and the ring pattern, change as follows:

  • (A) wavelength halves; rings move closer together
  • (B) wavelength is unchanged; rings are unchanged
  • (C) wavelength doubles; rings appear at larger angles
  • (D) wavelength increases four times; rings appear at larger angles
Show Answer

Answer: C – λ varies as 1 / √V. Reducing V by a factor of 4 increases λ by a factor of √4 = 2, and a larger wavelength diffracts through larger angles.

Q18. Which observation demonstrates the particle behaviour of electrons, rather than wave behaviour?

  • (A) Diffraction rings produced by a graphite film
  • (B) Discrete scintillations (flashes) when individual electrons strike a fluorescent screen
  • (C) Interference fringes produced by an electron beam
  • (D) The spreading of an electron beam after passing through a narrow slit
Show Answer

Answer: B – Each scintillation is a single, localised transfer of energy at one point on the screen, which is particle-like behaviour. Diffraction, interference and spreading are all wave properties.

Q19. Electrons must have a de Broglie wavelength of 1.0 × 10-10 m to be diffracted by a crystal. The potential difference through which they should be accelerated from rest is about

  • (A) 1.5 V
  • (B) 15 V
  • (C) 1500 V
  • (D) 150 V
Show Answer

Answer: D – p = h / λ = 6.63 × 10-34 / 1.0 × 10-10 = 6.63 × 10-24 kg m/s. Kinetic energy = p2 / (2 m) = (6.63 × 10-24)2 / (2 × 9.11 × 10-31) = 2.4 × 10-17 J. V = E / e = 2.4 × 10-17 / 1.6 × 10-19 = 150 V.

Q20. For protons to be diffracted by the same crystal as electrons, so that their wavelength is also about 1.0 × 10-10 m, the protons must travel

  • (A) much more slowly than the electrons
  • (B) much faster than the electrons
  • (C) at exactly the same speed as the electrons
  • (D) at the speed of light
Show Answer

Answer: A – The same wavelength needs the same momentum, p = 6.6 × 10-24 kg m/s. For a proton, v = p / m = 6.6 × 10-24 / 1.67 × 10-27 ≈ 4.0 × 103 m/s, far slower than the electron value of about 7.3 × 106 m/s.


Duality Calculations & Comparisons

Q21. An electron travelling at 1.0 × 106 m/s has a de Broglie wavelength of 7.3 × 10-10 m. A proton travelling at the same speed has a de Broglie wavelength of about

  • (A) 7.3 × 10-10 m
  • (B) 4.0 × 10-13 m
  • (C) 1.3 × 10-6 m
  • (D) 4.0 × 10-16 m
Show Answer

Answer: B – At the same speed, λ is inversely proportional to mass. λ(proton) = 7.3 × 10-10 / 1836 = 4.0 × 10-13 m.

Q22. An electron and a proton have the same kinetic energy. The ratio (de Broglie wavelength of electron) / (de Broglie wavelength of proton) is about

  • (A) 1
  • (B) 1836
  • (C) 43
  • (D) 1 / 1836
Show Answer

Answer: C – For the same kinetic energy, λ = h / √(2 m E), so λ varies as 1 / √m. The ratio is √(mp / me) = √1836 ≈ 43. The lighter particle has the larger wavelength.

Q23. An electron microscope can resolve much finer detail than a light microscope because

  • (A) the de Broglie wavelength of the electrons can be made far smaller than the wavelength of visible light
  • (B) electrons are easier to see than photons
  • (C) electrons travel faster than light
  • (D) the de Broglie wavelength of the electrons is far larger than the wavelength of visible light
Show Answer

Answer: A – Resolution is limited by diffraction to details roughly the size of the wavelength used. Electrons accelerated through kilovolts have λ of order 10-11 m, compared with about 5 × 10-7 m for visible light, so far finer detail can be resolved.

Q24. An electron is accelerated from rest through a potential difference of 10 kV. Its de Broglie wavelength is

  • (A) 1.2 × 10-10 m
  • (B) 1.2 × 10-12 m
  • (C) 3.9 × 10-11 m
  • (D) 1.2 × 10-11 m
Show Answer

Answer: D – Kinetic energy = 1.6 × 10-19 × 1.0 × 104 = 1.6 × 10-15 J. p = √(2 × 9.11 × 10-31 × 1.6 × 10-15) = 5.4 × 10-23 kg m/s. λ = 6.63 × 10-34 / 5.4 × 10-23 = 1.2 × 10-11 m. (Equivalently, λ scales from the 100 V value by 1 / √100 = 1/10.)

Q25. A photon and an electron have the same wavelength. They must also have the same

  • (A) kinetic energy
  • (B) momentum
  • (C) speed
  • (D) mass
Show Answer

Answer: B – For both photons and particles, p = h / λ. Equal wavelengths therefore mean equal momenta. Their energies differ: for the photon E = p c, while for the electron the kinetic energy is p2 / (2 m).

Q26. X-rays of wavelength 1.2 × 10-10 m and electrons of wavelength 1.2 × 10-10 m both produce diffraction patterns when passed through the same crystal. This is because

  • (A) diffraction depends on the wavelength, and both the X-rays and the electrons exhibit wave behaviour
  • (B) X-rays are made up of electrons
  • (C) electrons are a form of electromagnetic radiation
  • (D) a crystal can only diffract particles
Show Answer

Answer: A – The diffraction pattern from a given crystal is determined by the wavelength. Since the wavelengths are the same, the patterns are similar, showing that the electrons behave as waves of that wavelength.

Q27. An alpha particle of mass 6.6 × 10-27 kg travels at 1.0 × 107 m/s. Its de Broglie wavelength is

  • (A) 1.0 × 10-11 m
  • (B) 4.0 × 10-13 m
  • (C) 1.0 × 10-14 m
  • (D) 1.0 × 10-17 m
Show Answer

Answer: C – p = m v = 6.6 × 10-27 × 1.0 × 107 = 6.6 × 10-20 kg m/s. λ = 6.63 × 10-34 / 6.6 × 10-20 = 1.0 × 10-14 m.

Q28. The speed at which an electron has a de Broglie wavelength of 5.0 × 10-11 m is

  • (A) 7.3 × 106 m/s
  • (B) 1.5 × 107 m/s
  • (C) 1.5 × 106 m/s
  • (D) 3.6 × 107 m/s
Show Answer

Answer: B – p = h / λ = 6.63 × 10-34 / 5.0 × 10-11 = 1.33 × 10-23 kg m/s. v = p / m = 1.33 × 10-23 / 9.11 × 10-31 = 1.5 × 107 m/s.

Q29. Which statement about wave-particle duality is correct?

  • (A) Electrons diffract as waves but are always detected as localised particles
  • (B) Electrons behave as waves when moving and as particles only when at rest
  • (C) Duality applies to light but never to matter
  • (D) The wavelength of a moving particle can be measured directly with a metre rule
Show Answer

Answer: A – Which aspect of behaviour is observed depends on the experiment: propagation and diffraction reveal wave behaviour, while detection and collisions (as in the photoelectric effect for light) reveal localised, particle-like behaviour.

Q30. A molecule of mass 1.0 × 10-26 kg moves at 5.0 × 102 m/s. Its de Broglie wavelength is

  • (A) 1.3 × 10-13 m
  • (B) 1.3 × 10-7 m
  • (C) 6.6 × 10-10 m
  • (D) 1.3 × 10-10 m
Show Answer

Answer: D – p = m v = 1.0 × 10-26 × 5.0 × 102 = 5.0 × 10-24 kg m/s. λ = 6.63 × 10-34 / 5.0 × 10-24 = 1.3 × 10-10 m, comparable to atomic spacings, so even large molecules can show diffraction effects.

These are original practice questions written for QuaidianPoint. They are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.