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A2 · Practice questions · Wave-particle duality

Every moving particle has a wavelength.

Six original Cambridge-style questions on the de Broglie wavelength λ = h/p, electrons accelerated through a potential difference, electron diffraction, and which observations show the wave and which the particle nature.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Momentum first, then wavelength.

01
Analysis
[2 marks]

State what is meant by the de Broglie wavelength of a particle. Explain why the ring pattern produced when a beam of electrons passes through a thin film of graphite is evidence for the wave nature of electrons.

  • The de Broglie wavelength is the wavelength associated with a moving particle, given by λ = h / p where p is the momentum of the particle. ✓
  • The rings are a diffraction pattern, and diffraction is a property only of waves; the electrons are diffracted by the regularly spaced atoms of the crystal, which act like a grating. ✓
02
Calculation
[3 marks]

A neutron of mass 1.67 × 10⁻²⁷ kg moves at 2.2 × 10³ m s⁻¹. Calculate its momentum and its de Broglie wavelength, and explain why a beam of such neutrons is diffracted by a crystal.

  • p = mv = 1.67 × 10⁻²⁷ × 2.2 × 10³ = 3.7 × 10⁻²⁴ N s. ✓
  • λ = h / p = 6.63 × 10⁻³⁴ / 3.67 × 10⁻²⁴ = 1.8 × 10⁻¹⁰ m. ✓
  • This wavelength is comparable to the spacing of atoms in a crystal (about 10⁻¹⁰ m), and diffraction is only significant when the gap or spacing is comparable to the wavelength. ✓
03
Calculation
[3 marks]

Electrons are accelerated from rest through a potential difference of 150 V. Calculate the kinetic energy of an electron in joules, its momentum, and its de Broglie wavelength. (Electron mass = 9.11 × 10⁻³¹ kg.)

  • Ek = eV = 1.60 × 10⁻¹⁹ × 150 = 2.4 × 10⁻¹⁷ J. ✓
  • p = √(2meEk) = √(2 × 9.11 × 10⁻³¹ × 2.4 × 10⁻¹⁷) = 6.6 × 10⁻²⁴ N s. ✓
  • λ = h / p = 6.63 × 10⁻³⁴ / 6.61 × 10⁻²⁴ = 1.0 × 10⁻¹⁰ m. ✓
04
Analysis
[3 marks]

Light can be described both as a wave and as a stream of particles. State one observation that requires the wave model and one that requires the particle model, and explain why physicists keep both models rather than choosing one.

  • Wave model: interference (for example two-slit fringes) or diffraction through a grating, which need waves that superpose. ✓
  • Particle model: the photoelectric effect, in which energy is delivered in packets of hf to single electrons, with a threshold frequency and instantaneous emission. ✓
  • Neither model alone explains every observation: a wave cannot give a threshold frequency, and a particle cannot produce interference, so each model is used where it applies (wave-particle duality). ✓
05
Calculation
[3 marks]

An electron (mass 9.11 × 10⁻³¹ kg) and a proton (mass 1.67 × 10⁻²⁷ kg) each have a de Broglie wavelength of 0.10 nm. Compare their momenta, and calculate the speed of each particle.

  • Since λ = h / p and the wavelengths are equal, the momenta are equal: p = 6.63 × 10⁻³⁴ / (0.10 × 10⁻⁹) = 6.6 × 10⁻²⁴ N s for both. ✓
  • Electron: v = p / m = 6.63 × 10⁻²⁴ / 9.11 × 10⁻³¹ = 7.3 × 10⁶ m s⁻¹. ✓
  • Proton: v = 6.63 × 10⁻²⁴ / 1.67 × 10⁻²⁷ = 4.0 × 10³ m s⁻¹, about 1800 times slower because it is about 1800 times more massive. ✓
06
Calculation
[4 marks]

In an electron diffraction tube, electrons are accelerated from rest through 5.0 kV and strike a thin graphite film, producing rings on a fluorescent screen. (Electron mass = 9.11 × 10⁻³¹ kg.)

(a) Show that the de Broglie wavelength of the electrons is about 1.7 × 10⁻¹¹ m. (b) The accelerating voltage is reduced to 1.25 kV. Calculate the new wavelength. (c) State and explain what happens to the radius of the rings.

  • (a) Ek = eV = 1.60 × 10⁻¹⁹ × 5000 = 8.0 × 10⁻¹⁶ J, so p = √(2 × 9.11 × 10⁻³¹ × 8.0 × 10⁻¹⁶) = 3.8 × 10⁻²³ N s. ✓
  • λ = h / p = 6.63 × 10⁻³⁴ / 3.82 × 10⁻²³ = 1.7 × 10⁻¹¹ m. ✓
  • (b) λ is proportional to 1 / √V; V falls to one quarter, so λ doubles to 3.5 × 10⁻¹¹ m. ✓
  • (c) The rings grow larger: with a longer wavelength the electrons are diffracted through larger angles by the same atomic spacing. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.