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9702 · 22.3

Wave-particle duality — practice questions

Practice and worked examples for 9702 Wave-particle duality. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Light of frequency 7.5 × 10¹⁴ Hz is incident on a metal surface with a work function of 2.1 eV. Calculate the maximum kinetic energy of the emitted photoelectrons in Joules. (Planck's constant, h = 6.63 × 10⁻³⁴ Js; 1 eV = 1.6 × 10⁻¹⁹ J)

Show solution outline
  1. Convert the work function to Joules: Φ = 2.1 eV × (1.6 × 10⁻¹⁹ J/eV) = 3.36 × 10⁻¹⁹ J.
  2. Calculate the incident photon energy: E = hf = (6.63 × 10⁻³⁴ Js) × (7.5 × 10¹⁴ Hz) = 4.9725 × 10⁻¹⁹ J.
  3. Apply the photoelectric equation: hf = Φ + KE_max.
  4. Rearrange to find KE_max: KE_max = hf - Φ.
  5. Substitute the calculated values: KE_max = (4.9725 × 10⁻¹⁹ J) - (3.36 × 10⁻¹⁹ J) = 1.6125 × 10⁻¹⁹ J.
  6. The maximum kinetic energy of the photoelectrons is approximately 1.6 × 10⁻¹⁹ J (to 2 significant figures).

Worked example 2

An electron is accelerated from rest through a potential difference of 500 V. Calculate its de Broglie wavelength. (Planck's constant, h = 6.63 × 10⁻³⁴ Js; mass of electron, mₑ = 9.11 × 10⁻³¹ kg; elementary charge, e = 1.60 × 10⁻¹⁹ C)

Show solution outline
  1. Calculate the kinetic energy (KE) gained by the electron. The energy gained from an electric field is KE = qV. KE = (1.60 × 10⁻¹⁹ C) × (500 V) = 8.00 × 10⁻¹⁷ J.
  2. Calculate the momentum (p) of the electron. We can relate kinetic energy to momentum using KE = p²/2m. Rearranging for p: p = √(2m × KE).
  3. Substitute the values for mass and kinetic energy: p = √(2 × 9.11 × 10⁻³¹ kg × 8.00 × 10⁻¹⁷ J) p = √(1.4576 × 10⁻⁴⁶ kg²m²s⁻²) = 1.2073 × 10⁻²³ kgms⁻¹.
  4. Use the de Broglie equation to find the wavelength (λ): λ = h/p.
  5. Substitute the values for h and p: λ = (6.63 × 10⁻³⁴ Js) / (1.2073 × 10⁻²³ kgms⁻¹) = 5.49 × 10⁻¹¹ m.
  6. The de Broglie wavelength of the electron is 5.49 × 10⁻¹¹ m (or 54.9 pm). This is comparable to the spacing of atoms in a crystal, which is why electron diffraction is observable.
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