ZIMSEC A Level · Physics · Form 6
The photoelectric effect
Light consists of photons with energy E = hf = hc/λ (h = 6.63 × 10⁻³⁴ J s). Electrons are emitted only if hf exceeds the work function φ: hf = φ + ½mv²max. Intensity changes the number of electrons, not their maximum energy – evidence for the particle nature of light.
Worked example: Light of wavelength 400 nm falls on a metal with work function 3.0 × 10⁻¹⁹ J. Find the maximum kinetic energy of the photoelectrons.
E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) ÷ 400 × 10⁻⁹ = 4.97 × 10⁻¹⁹ J
KEmax = 4.97 × 10⁻¹⁹ − 3.0 × 10⁻¹⁹ = 1.97 × 10⁻¹⁹ J
Wave–particle duality and energy levels
- de Broglie wavelength λ = h/p; electron diffraction shows particles behave as waves.
- Atoms emit or absorb photons when electrons move between discrete energy levels: hf = E₁ − E₂, giving line spectra.
The nucleus
- Nucleon number A, proton number Z; isotopes differ in neutron number.
- Mass defect and binding energy: E = Δmc² (1 u = 931.5 MeV). Fission of heavy nuclei and fusion of light nuclei release energy because binding energy per nucleon increases.
Radioactive decay
- α (helium nucleus), β⁻ (electron and antineutrino), γ (photon). Balance A and Z in equations: ²³⁸U → ²³⁴Th + ⁴He.
- Activity A = λN; N = N₀e^(−λt); half-life t½ = ln 2/λ.
Worked example: A sample has a half-life of 8.0 days. What fraction remains after 24 days?
24 ÷ 8 = 3 half-lives → (½)³ = 1/8
Check yourself
- Find the energy of a photon of frequency 5.0 × 10¹⁴ Hz.
- Why does increasing the intensity of light not increase the maximum kinetic energy of photoelectrons?
- Find the decay constant for a half-life of 10 s.
Show answers
- 3.3 × 10⁻¹⁹ J
- Each electron absorbs one photon; intensity increases the number of photons, not the energy of each photon.
- λ = 0.693/10 = 0.0693 s⁻¹
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