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9702 · 22.2

Photoelectric effect — practice questions

Practice and worked examples for 9702 Photoelectric effect. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Monochromatic light of frequency 7.5 × 10¹⁴ Hz is incident on a clean sodium surface. The work function of sodium is 2.28 eV. Calculate: (a) The energy of an incident photon in Joules. (b) The maximum kinetic energy of an emitted electron in Joules.

Show solution outline
  1. Convert the work function to Joules: Φ = 2.28 eV × (1.60 × 10⁻¹⁹ J/eV) = 3.648 × 10⁻¹⁹ J
  2. Calculate the energy of an incident photon (hf): E = hf = (6.63 × 10⁻³⁴ Js) × (7.5 × 10¹⁴ Hz) E = 4.9725 × 10⁻¹⁹ J
  3. Apply Einstein's Photoelectric Equation to find the maximum kinetic energy: KE_max = hf - Φ KE_max = (4.9725 × 10⁻¹⁹ J) - (3.648 × 10⁻¹⁹ J) KE_max = 1.3245 × 10⁻¹⁹ J
  4. Round to an appropriate number of significant figures (3 s.f.): KE_max ≈ 1.32 × 10⁻¹⁹ J

Worked example 2

Light of wavelength 420 nm is incident on a metal surface. The maximum kinetic energy of the emitted photoelectrons is found to be 0.95 eV. Calculate: (a) The work function of the metal in Joules. (b) The threshold frequency of the metal.

Show solution outline
  1. Calculate the energy of the incident photon (E = hc/λ): E = (6.63 × 10⁻³⁴ Js × 3.00 × 10⁸ m/s) / (420 × 10⁻⁹ m) E = 4.7357 × 10⁻¹⁹ J
  2. Convert the maximum kinetic energy to Joules: KE_max = 0.95 eV × (1.60 × 10⁻¹⁹ J/eV) KE_max = 1.52 × 10⁻¹⁹ J
  3. Find the work function using Einstein's equation (Φ = hf - KE_max): Φ = E - KE_max Φ = (4.7357 × 10⁻¹⁹ J) - (1.52 × 10⁻¹⁹ J) Φ = 3.2157 × 10⁻¹⁹ J Φ ≈ 3.22 × 10⁻¹⁹ J (to 3 s.f.)
  4. Calculate the threshold frequency (f₀ = Φ/h): f₀ = (3.2157 × 10⁻¹⁹ J) / (6.63 × 10⁻³⁴ Js) f₀ = 4.8502... × 10¹⁴ Hz f₀ ≈ 4.85 × 10¹⁴ Hz (to 3 s.f.)
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